ML20132D727
ML20132D727 | |
Person / Time | |
---|---|
Site: | Millstone |
Issue date: | 12/13/1996 |
From: | Champagne M, Ewing R, Hundley D NORTHEAST NUCLEAR ENERGY CO. |
To: | |
Shared Package | |
ML20132D715 | List: |
References | |
96-ENG-01528E2, 96-ENG-01528E2-R00, 96-ENG-1528E2, 96-ENG-1528E2-R, NUDOCS 9612200086 | |
Download: ML20132D727 (301) | |
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AMPACITY DERATING OF CABLES DUE TO THERMO-LAG TITLE 96-ENG-01528E2 00 CALCULATION # REV# Vendor Calc #
System MSC/2391 Structure N/A Component CABLES _
l Executive Summary l This calculation addresses the cables in cable trays and conduit encased in Thermo-Lag fire wrap at MP2. For cable tray, j a model was developed using standard heat transfer equations. The validity of this model was verified by comparing the 2
calculated surface temperature of the wrap and its associated ACF (ampacity correction factor) produced by the model to actual test results on wrapped cable tray tested by Texas Utilities. The model was then used to determine an ACF for the various cable tray installations at MP2, the worse case value determined was then utilized. For the conduit installations a model using the same heat transfer type equations was employed and its associated ACF determined.
The ampacity allowed of cables in tray were determined by the simultaneous solution of the heat transfer equations (Stolpe equations). For cables in conduit, the IPCEA values were taken directly from Standard P54-426 and derated based upon directions provided in the standard to adjust for temperature and the number of conductors contained within the conduit.
l The resultant ampacities were then multiplied by thc ACF factor and an Imax ampacity determined. This value was finally l compared to the load current on the cable (appropriate multiplication factor used to ensure conservatism).and evaluated against the acceptance criteria of the calculation .
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Does this calculation:
- 1. Suppon a DCR, MMOD, an independent review method for a DCR ,or confirm test results for Yes ONo @ )
an installed DCR7 If yes, indicate the DCR, MMOD number and/or Test Procedure number.
- 2. Support independent analysis? If yes, indicate the procedure, work control or other reference Yes ONo E it supports.
- 3. Revise, supersede, or void existing calculations? If yes, indicate the calculation number and Yes ONo @
revisions.
. 4. Involve QA or QA-related systems, components or structures? Yes ENo O
- 5. Impact the Unit licensing basis, including technical specifications, FSAR, procedures or Yes ONo S licensing commitments? If yes, identify appropriate change documents.
Approvals (Print / Signature)
Preparer Dan Hundley //1##d Date: 12/13/96 Independent Reviewer 4fdAut.iove'+'A91-/hhw Date: n.-# 9 t,
) Supervisor M ELdA C'F Date: /,2-15 -f 6 1 a., se.m +
(-rk., e~t~ to uc;,,waly r&)W afisp,%.\ erl A;>a N'o. 96 -GH NUC DCM FORM 5-1A 9612200086 961213 Rev'03 PDR ADOCK 05000336 P PDR
Calc. No. 96-ENG -01528E2 Rev. 00 Page 2 of 44 A
M., .
Y mpy +9%
/
CTP DATA BASE INPUTS [CT Unit Only]
Calculation Number: 96 ENG 01528E2 Revision 00 Date 12/13/96 (prefix) (sequence (suffix) number)
Vendor Calculation Number /Other:
CCN# Superseded By: QA @ Yes C No Supercedes Cale:
Unit EWR Number Component Id Computer Code Rev. l
- / Level 2 N/A N/A N/A 00 PMMS CODES Structure System Component Reference Calculation Reference Drawings Sheet N/A MSC/2391 N/A N/A N/A l
Comments:
NUC DCM FORM 5-1B Rev 03
Calc. No. 96-ENG -01528E2 Rev. 00 Page 3 of 44 TABIT OF CONTENTS i ,
4 I
ITEM # DESCRIPTION PAGE# i Cover Sheet 1 i
CTP Data Base Inputs 2 Table of Contents 3 i
j 1.0 Purpose 4 2.0 Summary of Results 4 !
1 4
3.0 References / Design Inputs 6
- 4.0 Assumptions 13 5.0 Method of Calculation 16 !
i 6.0 Body of Calculation 19 7.0 Reviewers Comments and Resolutions 41 8.0 Attachments 43
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l Calc. No. 96-ENG -01528E2 Rev. 00 i Page 4 of 44 I I
1.0 Purpose '
The objective of this calculation is to determine if the cables covered with Thermo-Lag fire barrier material at Millstone Nuclear Power Station Unit 2 are within their ampacity design limitations. This will be accomplished by determining the load on each affected conductor and comparing this value to the maximum allowed derated current due to the presence of Thermo-Lag. i
- Acceotance Criteria: The maximum derated ampacity (Imax) must be greater than the actual (load) current in each conductor.
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. 2.0 Summary of Results 4160 Volts Tha 4160 volt cables used to backfeed from Unit I between the Z5 bus and the Z1 and Z2 diesel buses were found to be ofinsufficient size to carry 3MVA as allowed by OP - 2343 (Ref. 3.1.11).
CABLE DESCRIPTION Z5A501A/A Z5 SAFETY BUS TIE Z5A505A/B 25 SAFTEY BUS TIE Cable tray Z52FA10 was wrapped with conduit 5T540 for approximately an eight foot (8) section. This was beyond the model's limits. From the 120/125v section that follows, it is obvious that no appreciable heat is contributed by the cables in the conduit. Cable Z5A505A/B after reducing its ampacity by the derate specified by the model still has a 75 amp spare capacity and when powering the 25 (24E) bus would by inspection be acceptable. This cable, if used as a backfeed from Unit I as identified above, is not capable of picking up the load of a diesel and has been identified as failing.
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Calc. No. 96-ENG -01528E2 Rev. 00 Page 5 of 44 2.0 Summary of Results Continued:
480 Volt Continued:
l l 480 Volts l
The cables tabulated and identified below could not meet ampacity requiremente CABLE DESCRIPTION l Z1B5103/A Charging Pump MP18A (No.1)
ZlB5105/A Charging Pump MP18B (No. 2) l Z2B6102/H Charging Pump MP18C (No. 3)
Z2B6105/F Charging Pump MP18B (No.2) 120/125 Volt i
All the 120/125 volt cables passed with an acceptable margin.
Conduit 5T540 which contains a 3/C #6 cable (5A602/C) has been boxed in with cable tray Z52FA10 and is considered beyond the model's limits. The continuous load on this cable is less than 1 amp. After using the projected model derate, the cable still has a capacity of 47 amps. By inspection this installation is acceptable.
An ACR has been generated (ACR M2-96-0757)(Ref. 3.1.84) to identify the cables that did not pass the calculations acceptance criteria.
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Calc. No. 96-ENG -01528E2 Rev. 00 Page 6 of 44 3.0 References / Design Inputs:
3.1-
References:
3.1.1 Standard Handbook for Electrical Engineers, Tenth Edition.
3.1.2 Drawing 25203-32020, Sheet 49, Revision 5.
3.1.3 Drawing 25203-34077, Sheet 15, Revision 3.
. 3.1.4 The National Electric Code Handbook,1993 edition.
3.1.5 IPCEA P46-426, Power Cable Ampacities, AIEE S-135-1-62 Volume 1 Copper Conductors and S-135-2-62 Volume 11 Aluminum conductors.
3.1.6 Millstone Nuclear Power Site - Unit 2, Final Safety Analysis Report (MNPS-2 FSAR), Section 8.0.
3.1.7 Drawing 25203-34077, Sheet 13, Revision 3 3.1.8 Tray Schedule Millstone Unit 2 -Volume 1 Pages 1-1045 dated March 27, 1996 and Volume II pages 1046-2011 dated March 27,1996 (note both volumes contain work in progress to date).
3.1.9 Drawing 25203-30022 Sheet 3HA, Revision 1.
3.1.10 3M Letter, dated June 17,1986.
3.1.11 Operating Procedure 2343, Revision 17.
3.1.12 TU Electric Ampacity Derating Test Report No. 12340-94583 95166, 95246, dated March 19,1993.
1 3.1.13 Thermo-Lag PDCR 2-57-86.
3.1.14 OPAL Load Management System Program, SP-EE-344.
3.1.15 Drawing 25203-30011 Sheet 41, Revision 18.
3.1.16 Okonite Cable Technical Data Catalog,1984.
i 3.1.17 Calculation #95-ENG-01327-E2, Revision 0, Millstone Unit 2 -600-1000 volt cable Ampacity review tray Section Z15NA40.
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1 Calc. No. 96-ENG -01528E2 Rev. 00 Page 7 of e
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- 3.0 References /Denien Innuts continued: l i
Referenced continued: ,
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- 3.1.18 Bechtel Calculation #E-204-3, dated February 21,1972. l 1
1 3.1.19 Drawing 25203-32009 Sheet 43A, Revision 3.
3.1.20 Drawing 25203-30001, Revision 11.
i 3.1.21 Bechtel Calculation E-209-1, dated April 5,1976.
1 3.1.22 Drawing 25203-32021 Sheet 15, Revision 5.
3.1.23 Foxboro Catalog cut Mi-2A0-130, dated May 1978, t
4 3.1.24 Drawing 25203-30011 Sheet 21, Revision 15.
1 3.1.25 Drawing 25203-32009 Sheet 42A, Revision 1.
I j 3.1.26 MP2 TSO2 Cable Raceway Schedule (Computerized) 11/23/96 l 3.1.27 ICEA P-54-440, Ampacities Cables in Open-top Cable Trays, Revision 2, ,
i 1979 and Revision 3,1986.
l 3.1.28 Drawing 25203-30011 Sheet 34, Revision 12.
3.1.29 Drawing 25203-30022 Sheet 28, Revision 3.
l 3.1.30 Drawing 25203-32020 Sheet 49A, Revision 2. !
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j 3.1.31 Drawing 25203-39360 Sheet 1, Revision 3.
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3.1.32 Drawing 25203-32009 Sheet 40, Revision 15.
- 3.1.33 Drawing 25203-32009 Sheet 42, Revision 9.
3.1.34 Drawing 252003-34001 Sheet 11, Revision 5.
i 3.1.35 Drawing 25203-30022 Sheet 3 Revision 10.
3.1.36 Drawing 25203-32009 Sheet 43, Revision 14.
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. . - - - - . - - .- .. . ._- ._ - . - ~ - - _ _ . _ - . - . . . . - _ .
Calc. No. 96-ENG -01528E2 Rev. 00
- Page 8 of 44 1
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i 3.0 References /Denien Innuts continued: -
i References continued:
l ,
r 3.1.37 Drawing 25203-30022 Sheet 10, Revision 2.
l 3.1.38 Drawing 25203-30011 Sheet 40, Revisien 21.
3.1.39 Drawing 25203-30011 Sheet 39, Revision 16. .
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3.1.40 Drawing 25203-30022 Sheet 12, Revision 23.
3.1.41 Drawing 25203-30012 Sheet 12, Revision 4. l t
3.1.42 Kerite Engineering Memorandum No.178A, May 1,1979.
j 3.1.43 Cable Description Report, Dated April 9,1996 (computer generated) 3.1.44 EPRI Volume 4. !
l l 3.1.45 EWR M2-96-024 '
l 3.1.46 Drawing 25203-32021 Sheet 12, Revision 5.
. 3.1.47 Drawing 25203-32009 Sheet 46, Revision 5. i l
I l 3.1.48 Drawing 25203-32031 Sheet 41, Revision 1.
j 3.1.49 Drawing 25203-32009 Sheet 38, Revision 8.
l 3.1.50 Drawing 25203-32009 Sheet 35, Revision 4. l 3.1.51 Drawing 25203-32009 Sheet 37, Revision 5.
l 3.1.52 Drawing 25203-32031 Sheet 40, Revision 3.
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! 3.1.53 Drawing 25203-32012 Sheet 22, Revision 10.
3.1.54 Drawing 25203-32009 Sheet 54, Revision 6.
3.1.55 Drawing 25203-30011 Sheet 42, Revisioa 21. !
! 3.1.56 Drawing 25203-32033 Sheet 9, Revicion 3. I l
! Cale. No. 96-ENG -01528E2 Rev. 00 Page 9 of 4 4-i ==
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3.0 References /Denian Innuts continued: i References continued:
. l l 3.1.57 Drawing 25203-32033 Sheet 13, Revision 7.
3.1.58 Drawing 25203-32009 Sheet 6, Revision 11.
3.1.59 Drawing 25203 32009 Sheet 53, Revision 6.
3.1.60 Drawing 25203-32009 Sheet 39, Revision 6.
3.1.61 Drawing Specification 7604-E-15 "S&8 Kv Power Cable".
3.1.62 Conduit Schedule Millstone Unit 2- pages 1-600 dated March 27,1996.
( note volume contains work in progress to date).
3.1.63 Drawing 25203-32007 Sheet 56, Revision 1.
3.1.64 ASCO Catalog No. NP-1.
l l 3.1.65 Drawing 25203-32020 Sheet 42, Revision 10.
l 3.1.66 Drawing 25203 30011 Sheet 43, Revision 17 I I
3.1.67 Drawing 25203- 32009 Sheet 44, Revision 4. I 3.1.68 Drawing 25203-28500 Sheet 604B, Revision 4.
3.1.69 Drawing 25203-32025 Sheet 33, Revision 6.
3.1.70 Drawing 25203-32025 Sheet 32, Revision 6. l
! l 3.1.71 Drawing 25203-32009 Sheet 7, Revision 9. !
3.1.72 Drawing 25203-32009 Sheet 60, Revision 3. l l
3.1.73 Drawing 25203-28500 Sheet 1100, Revision 2.
1 3.1.74 Drawing 25203-39282, Revision 2.
i 3.1.75 Drawing 25203-39284, Revision 2.
Calc. No. 96-ENG -01528E2 Rev. 00 Page 10 of 44-3.0 References / Design Inputs continued:
References continued:
3.1.76 Drawing 25203-34077, Sheet 2, Revision 3.
3.1.77 Drawing 25203-34077, Sheet 25, Revision 3.
3.1.78 Drawing 25203-34077, Sheet 24, Revision 3.
3.1.79 Drawing 25203-34077, Sheet 20, Revision 3.
3.1.80 Drawing 25203-34077, Sheet 16, Revision 4.
3.1.81 Drawing 25203-34077, Sheet 14, Revision 3.
3.1.82 J.H. Neher, M. H. McGrath " Power Apparatus Systems", October 1967.
3.1.83 Drawing 25203-30022, Sheet 1WH, Revision 1.
3.1.84 ACR M2-96-0757.
3.1.85 Drawing 25203-30011, Sheet 29, Revision 17.
3.1.86 Drawing 25203-30011, Sheet 24, Revision 35.
3.1.87 Specification 7604-E-15A,"5Kv Power Cable".
3.1.88 Specification 7604-E-17, " Low Voltage Power Cable".
3.1.89 Specification 7604-E-18, "Multiconductor Control Cable".
3.1.90 Drawing 25203-30022 Sheet 12DC, Revision 1.'
3.1.91 Drawing 25203-32021 Sheet 14, Revision 5.
3.1.92 Drawing 25203-32021 Sheet 13, Revision 5.
3.1.93 Bechtel Calculation E-200-2 dated February 15,1971.
3.1.94 Drawing 25203-30044 Sheet 10, Revision 7.
Calc. No. 96-ENG -01528E2 Rev. 00 Page 11 of 44 '
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i Reis.rences/nesign Inouts continued:
References continued:
3.1.95 G.E. Heat Transfer and Fluid Flow Data Book,1971.
I 3.1.96 NUSCO Calculation 96-ENG-1559E1, Rev. 00.
l 3.1.97 EEQ Walk down Attachemnt B page B1, dated 8/29/90 ;
3.2 Design Inputs:
3.2.1 Allowable ampacities for cables in conduit and cables in air were taken l from IPCEA Power Cable Ampacities P-46-426 copy right 1962. '
Maintained spacing ampacities were taken from Bechtel Calculation E-204-3 (Ref. 3.1.18) unless otherwise noted.
3.2.2 Allowable ampacities for cable in cable tray (random fill) were developed in ,
Attachment 2 or taken from NEC Table 310-68 (Ref. 3.1.4) unless otherwise noted.
3.2.3 Texas Utilities Thermo-Lag Test ampacity derate values for various sizes of cables covered with Thermo-Lag (Ref. 3.1.12).
3.2.4 Ambient Temperature is provide by the FSAR as 50' C (Ref. 3.1.6) l 3.2.5 The power cable installed at MP2 is rated 90*C (Ref.3.1.6).
The control cable installed at MP2 is rated 90*C. (Ref.3.1.42). l l
3.2.6 The copper and aluminum wire properties are taken from Okonite Cable (Ref. 3.1.16) 3.2.7 The Unit 2 cables covered by Thermo-Lag are identified in Attachment C. The cables were identified from field walk down . The input was cross verified by using the PDCR that installed the Thermo-Lag (Ref. 3.1.13) and the cable raceway schedule (Ref. 3.1.8) [see Attachment 7].
l 3.2.8 The loading on cables listed in tables in Section 6.0 were obtained from the one lines and motor data obtaired from References 3,1.20,3.1.15,3.1.38, 3.1.39 3.1.25,3.1.28,3.1.29,3.1.30,3.1.31,3.1.32 and 3.1.33 or as specified in various sections of this calculation. In addition the trays and conduits which are fire wrapped are shown on the 25203-34077 series drawings. l (Ref. 3.1.76, 3.1.7, 3.1.81, 3.1.3, 3.1.80, 3.1.79, 3.1.78, 3.1.77)
Cc1c. No. 96-ENG -01528E2 Rev. 00 -
Page 12 of 44 l 3.0 References / Design Innuts continued:
Design Inputs continued:
3.2.9 Thermo I my Thickness Conversion <
l To convert from the Thermo-Lag thickness values used by TU to the 1" Thermo-Lag thickness used by NU see Attachment A.
3.2.10 Ambient Temnerature Conversion The ampacity of cables taken from various sources are not all based upon an ambient of 50*C. To adjust to 50*C the following formula can be used:
I2 =
1TC - TA2 - ATD I,
TC - TA3 - ATD where: i I is the current at the existing ambient ( A3 )
3 l
I is 2 the current at the desired new ambient (A2 ) I TC is the conductor temperature in*C l ATD is the dielectric loss temperature rise.(see Assumption 4.5) ;
3.2.11 Load Conversions G nhase) i Knowing either horse power (HP), Kilowatts (KW) or Kilovolt-amps (KVA) the following formula can be used to determine load ampacity for 3 phase l systems.
l Where HP is known: '
l RP x746 I = 45 x E x %eff x pf Where KW is known:
KW X1000 I = 45 x E x pf Where KVA is known:
I = K7A x 1000 5xE l
Calc. No. 96-ENG -01528E2 Rev. 00 Page 13 of 44-
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3.0 References / Design Innuts continued:
Design Inputs continued:
3.2.12 Denth of fill (heat Intensity) to Amnacity
, Knowing the depth of fill, the heat intensity can be determined, with the diameter of the conductor, nurnber of conductors and resistance per foot the allowed ampacity for a cable can be determined utilizing the formula below !
(Attachment D):
3=D2 M nR See Attachment B for solutions for Q for various cable trays and for specific cables.
3.1.13 All unincorporated design changes against the referenced drawings have been reviewed to ensure that the latest design was reviewed for impact. 1 1
4.0 Assumptions
I 4.1 All loads are considered balanced, each phase carrying the same current.
Justification - Loading of distribution panels, power panels, lighting panels, !
etc, employ standard engineering practices that optimize phase loading. 1 4.2 Feeder cables load will be determined based upon 1.25 times the largest load l with the remaining loads added directly to this value. Justification - -l l National Electric Code Article 430-24 allows development of feeder cable l loads employing this method (Ref. 3.1.4) l 4.3 Motor operated valves do not contribute any appreciable heat to the j raceways. Justification -cables must be energized continuously to reach their equilibrium temperature. Motor operated valve cables only carry loads for minutes and are off the majority of the time. The cables are not energized long enough to contribute appreciable heat to a raceway. ,
4.4. Spare cables act as heat sinks and will reduce the overall temperature of a i raceway. This will be ignored, unless specifically identified within the calculation, and the spare cables included in the depth calculation and ampacity ratings of the raceway (Ref. 3.1.44).
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Calc. No. 96-ENG -01528E2 Rev. 00 Page 14 of 4 t
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4.0 Assumptions continued:
4.5 The dielectric temperature loss on a cable is very small and will have a negligible impact upon ampacity calculations since it is subtracted fiom both the numerator and denominator of the Temperature adjustment equation.
Justification - as an example using a typical value for ATD of 0.19 and a temperature difference of 40
- C to 50* C ambient the differences are:
90 0.19 90 - 50 vs or 90 0.19 90 - 40 0.894000616 vs 0.894427191 or less than .048%
This assumption is valid for lower voltage cables and only becomes of interest at voltages above 15KV (Ref. 3.1.44).
4.6 Cables that are wTrpped using conduit wrap (not in conduit) in cable tray will have the cabl', s ampacity determined based upon cable fill. The derate ACF for the cable will use the worse case (tray or conduit ACF) to ensure that the cables are evaluated conservatively. The conduit wrap around cables without the conduit by inspection is bounded by the conduit model (one conduit barrier is eliminated) thus the conduit model ACF will be utilized as it is more conservative than the free air wrapped cable (s) .
4.7 The 4160 volt cables installed in cable tray were installed using uniform spacing. Justification - Standard installation practices employed at MP2 required higher voltage cables to be installed using maintained spacing.
(Ref. 3.1.34, Note 17). Random spacing will be used unless cables are specifically identified as being installed with uniform spacing.
4.8 Junction Boxes will use the same ampacity as used for conduit.
Justification, the box is fully enclosed as is a conduit but has a larger I
surface area to dissipate heat.
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Calc. No. 96-ENG -01528E2 Rev. 00 Page 15 of 4+
4.0 Assumntions continued:
4.9 For 460 volt motors the power factor and efficiency are assumed to be 0.85 and 0.88 respectively. Justification- values are taken from OPAL which is conservative (Ref. 3.1.14).
4.10 Cable diameters were taken from Reference 3.1.8 and 3.1.43. For ampacity I calculations, a -10% value will be used. Justification - the smaller diameter ;
will ensure conservatism. j i
4.11 Load Multipliers will be used to adjust for potential voltage fluctuations. l For motors 1.25 will be used,1.2 for transformers and for all other loads a :
1.1 multiplier will be used. Justification -Ref. 3.1.44. l i
4.12 All conduits, unless field verified will be assumed to be in a 6 x 6 groupmg and a multiplier of 0.68 will be used against the ampacity. Justification - -
this is a worse case conservative approach based upon IPCEA Table IX ;
(Ref. 3.1.5). i t
4.13 Electronic cards mounted components utilizing transistors and intergrated circuits are limited by heat (5 Watts) by design limitations. For transistor and integrated circuits this value will be used. Justification - very conservative assumption. ;
4.14 MOV's by their nature are intermittent loads and do not contribute to the steady state temperature of the tray other than by providing diversity. !
MOV cables will be evaluated under transient conditions. For example, .
from Referenace 3.1.97, MOV for 2-CH-501 has a rise of 7.5
- C for a 15 !
minute duty cycle. I Tc(f) = 50 + 7.5 = 57.5
- C. !
For other similar valves, the final temperature, the temperature at any time (t) can be determined using the following equation:
1 T,( t) = T,( 0 ) + (T,( f ) - T,( 0 )](1 - e * ) (Ref. 3.1.44)
This confirms that the MOV's do not contribute appreciable heat and actually contribute diversity.
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Calc. No. 96-ENG -01528E2 Rev. 00 Page 16 of 44-
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5.0 Method of Calculation:
This calculation will identify all cables in MP2 where Thermo-Lag has been installed. For ease of organization the cables will be grouped by voltage class (4160,480,120-125 and instrumentation).
The physical installation of the cable will be determined, i.e. whether it is in tray, conduit, free air, or wire-way and the allowable ampacity determined (Ref. 3.1.76 through 3.1.81).
Conduit:
For cables in conduit , the ampacity will be derated based upon the number of conductors in the conduit (Ref. 3.1.44), the grouping factor (assumption 4.12) and then multiplied by the Thermo-Lag derating factor from Attachment 1. The cable load will be determined and sdjusted to reflect voltage fluctuations. These values then will be compared and the cable's installation will be considered acceptable if the load current is below the derated maximum value (Imax).
Trau For cables in tray the process is slightly more complicated than conduit ,
installations. The allowed ampacity will differ based upon the possible field l installation variations.
Uniform Spacing For cables that hase beu identified as installed utilizing uniform spacing the allowable ampacity will be determined directly from the ampacity tables in Calculation E204-3 (Ref. 3.1.18). The Thermo-Lag derating from Attachment 1 l will be applied to this ampacity and the maximum allowable current (Imax) obtained. This value is then compared to the cable's load current to determine if the installation is acceptable. ,
Random Fill Determination of the maximum ampacity a cable can withstand and not exceed its jacket rating when installed in cable tray is dependent upon the cable tray i dimensions and depth of fill.
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The ampacities of the cables will be determined based upon the ampacities developed using the Stolpe equations and constants provided in the ICEA publication 54-440 (Ref. 3.1.27),
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Calc. No. 96-ENG -01528E2 Rev. 00 Page 17 of 44 5.0 Method of Calculation:
Random Fill continued:
After the maximum permissible ampacity is identified (conductor temperature cannot exceed 90' C), the load ampacity on the cable is determined. Load values that are below the maximum allowable (Imax) will be deemed acceptable.
Free Air Cable The maximum allowable ampacity for free air cables as required will be obtained from IPCEA (Ref. 3.1.5) or the National Electrical Code (Ref. 3.1.4) for smaller cables not included in IPECA. This value will be multiplied by the appropriate !
Thermo-Lag derating value from Attachment 1. The cable load current will be compared to the maximum allowable ampacity (Imax) and ifit is more than the load current the cable will be considered acceptable as installed.
Load Ampacities:
The loads used for cables will be determined utilizing a conservative sequence approach. This will ensure the calculation is conservative and will reduce the need for excessive load research. The search may be halted once an ampacity is determined not to exceed the maximum allowably ampacity. Should the cable still have an ampacity too high after using all five approaches, depicted below, additional more detailed evaluations will be conducted using other means of evaluation as specified later in this methodology. The order of determining the appropriate value ofload is :
- 1. 80% of breaker rating 2.100% fuse rating
- 3. calculation E-209-1
- 4. OPAL
- 5. Drawing / vendor documentation review To adjust for voltage fluctuations, efficiencies and other conditions which may affect ampacity, a multiplication factor (EPRI recommended, Ref. 3.1.44 ) will be used to ensure conservatism. Motors -1.25, Transformers- 1.2, other loads - 1.1.
For feeder cables, the load will be determined by multiplying the largest motor load by a 1.25 multiplier. The remaining loads will then be summed with this value (Ref. 3.1.4, Section 430-24).
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l Calc. No. 96-ENG -01528E2 Rev. 00 Page 18 of 44
==- m 5.0 Method of Calculation: l Additional Evaluations :
Load ampacities for trays that are not overfilled will be evaluated at its existing fill and the increased ampacity used to evaluate the loads.
l It is important to note that a great deal of conservatism is also included in this method since ;
every cable is assumed to be fully loaded and at 90*C. Should the maximum value as l
determined by the above method still be insufficient to allow acceptance of the installation, !
specific cable ampacities will be utilized and diversity will be taken into consideration.
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Calc. No. 96-ENG -01528E2 Rev. 00 Page 19 of M 6.0 Body of Calculation 6.1 4160 Volt Cables A total of four (4) 4160 volt cable trays were identified in the plant that are covered with Thermo-Lag. The cables associated with these trays are tabulated in Table 6.1.
From the tray schedule (Ref. 3.1.8) the cables are eitna aluminuun triplexed 750 MCM cable or triplexed 350 MCM aluminum. (Note that mamtained spacing ampacity values were not used for these cables).
Tray Z52EA10 ,
This tray is 38% full ( Ref. 3.1.8) which equates to a depth of Depth = % fill x Denth of Trav = .38 x 4 x 4 + n = 1.935".
100n/4 From Attachment B, Appendix B1, the value (I3 ) for 750 MCM at an ambient of 40' C and a depth of 1.935" is 609.38 amps.
To adjust to 50' C, the plant ambient (Ref. 3.2.10), the formula provided in Section 310-15 of Reference 3.1.4. [shown below] is used (Note that ATD is very small and negligible (Assumption 4.5).
TC - TA, - ATD TC - TA1 - ATD 750MCM - 609. 3 8 = 6 09. 3 8 x 0. 8 9 = 54 2. 3 5 amps 90 - 40 The cable is limited to 80% of the free air value or 418.66 amps.
Derating for the Thermo-Lag installed on the cable tray ( Attachment A) provides the maximum ampacity the cable can carry and not go above its 90*C rating.
750MCM - Imax = 41S.66 x 0.60 = 251.2 amps 1
Cele. No. 96-ENG -01528E2 Rev. 60 Page 20 of 44
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6.0 Body of Calculation continued:
[
6.1 4160 Volt Cables continued:
Tray 52BA10 <
This tray is also at 38% fill. The allowed ampacity for 750MCM will be the same as for the tray above,Imax = 251.2 amps.
I Tray Z52FA10 l
This tray is also at 38% fill. The allowed ampacity for 750 MCM will be the same !
as for the tray above, Imax = 251.2 amps.
Trm Z22EA10 Thu tray is at 23% fill. From Attachment B, Appendix B2,350 MCM at an ambient ofO' C at a 1.1714" depth is 508.46 amps.
i To adjust to 50* C , the plant ambient (Ref. 3.2.10), the formula provided in i Section 310-15 of Reference 3.1.4. (shown below] is used (Note that ATD is very j small and negligible Assumption 4.5). <
(Tc - TA, - bTD k TC - TA, - ATD 90 - 50
= 508. 46 = 5 0 8. 4 6 x 0. 89 = 4 52. 53 amps 90 - 40 The cable is limited to 80% of the free air value or 262.02 amps.
Derating for the Thermo-Lag installed on the cable tray ( Attachment 1) provides the maximum ampacity the cable can carry and not go above its 90*C rating.
350MCM - Imax = 262.02 x 0.60 = 157.21 amps The four trays and their associated cables are tabulated on the following page along 1 with the Imax determined using the above method. In addition the table includes j the cable's load values determined as follows:
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Calc. No. 96-ENG -01528E2 Rev. 00 Page 21 of 44
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i 6.1 4160 Volt Cables continued:
TABLE 61 Tray Cable Code Type Isoe Imax Load P/F 252EA10 Z5A501A/A A14 1-3/C750 418.66 251.2* 174.64 P 38% Backfeed 418.66 251.2* 416.36 F 52BA101 5A505/A A14 1-3/C750 418.66 251.2* 87.32 P 38% 5A505/P A14 1-3/C750 418.66 251.2* 87.32 P Backfeed 418.66 251.2* 208.18 P Z52FA10 Z5A505A/B A14 1-3/C750 418.66 251.2* 174.64 P 38% Backfeed 418.66 251.2* 416.36 F Z22EA10 Z2A407/A A16 1-3/C350 262.02 157.21 76.5 P 23 %
- Unit I calculation (Ref. 3.1.96) must also be reviewed for worst case Imax value.
Load Determination Cable Z5A501 A/A is the Z5 (24E) swing bus feed from Bus 24C. The load on the Z5 consists of the following ( Ref. 3.1.20,3.1.62,3.1.93 and 3.1.94):
DEVICE RATING FLA VOLT MULT LOAD P5B 450HP 61.2 4000 1.25 76.5 PilB 350HP 45 4000 1.00 45 P41B 400HP 52.5 4000 1.00 52.5 XFMS KVA/O 4160 1.00 0.64 TOTAL 174.64 (Note that the Unit 1 inter-tie is considered open for this analysis.)
The HP (horse power), E (voltage, motor nameplate), FLA values for the motors identified (PSB, P11B and P41B) are available in the Bechtel Calculation (Ref. 3.1.93) .
Ampacity for the transformers was calculated from a 5 ohm load at 120v or 600va related to 4160v = 0.144 amps + .5 amp PT = 0.64 amps (Ref. 3.1.94). In accordance with Reference 3.1.4 the largest load on a feeder cable uses a 1.25 load multiplier, all other loads are summed directly to the total.
Calc. No. 96-ENG -01528E2 Rev. 00 Page 22 of 44-L r
! 1 6.0 P-av or Calculation continued: ,
i t.t 4160 Volt Cables continued: l
! Cable Z5A505A/B is a feed to swing bus 25 from bus 24D and will see the same j load identified for cable Z5A501 A/A (174.64 amps) (Ref. 3.1.20).
Cables 5A505/A and 5A505/P (parallel feeds from Unit 1) for Unit 2 would carry the Z5 (24E) bus load identified above,174.64/2 or 87.32 amps) (Ref. 3.1.20).
, ]
] OP-2343 (Ref. 3.1.11) specifies a load of 3MVA be provided via Unit 1.
I = 3,000,000 = 416.36 amps 4160x V3 Resolution of the 3MVA backfeed capability will be addressed via ACR M2 , 0757 (Ref. 3.1.84)
Cable Z2 A407/A is the power feed to Service Water Pump PSC (Ref.3.1.93). Load current is obtained by multiplying the full load amps (Ref. 3.1.14) by 1.25.
61.2 x 1.25 = 76.5 amps.
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4 Calc. No. 96-ENG -01528E2 Rev. 00 Page 23 ofd l
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6.0 Body of Calculation continued:
! 6.2 480 Volt Cables l A total of ten (10) cable trays were identified in the plant that were covered with j Thermo-Lag and six (6) conduits. The cables associated with these trays are j
tabulated on the following pages. From the cable schedule (Ref. 3.1.26) the cables l can be identified as power cables. The cables in the tray were installed in i accordance with Reference 3.1.18, meeting the requirements of random fill. Note, conduits listed identify all voltage class cables to allow determination of the total number of conductors and any appropriate derating factor.
2 Imax Evaluations:
The following is an example of how the loads were determined for the cables listed in Table 6.2: For tray Z23HA10 at 10% full, the allowed ampacity for 500MCM j triplexed at 50' C ambient is calculated in Attachment B, I1 = 412.96 amps. Imax
! is obtained by using the Thermo-Lag derating,412.96 x 0.6 = 247.78 amps.
i TABLE 6.2 key follows -
. Raceway cable Code Type Isoe Made Imax Load P/F
. Z23HA10 Z2B0610/A B14 3-1/C500 412.96 TPLX 247.78 198.69 P i Z2B0610/B 10% B14 3-1/C500 412.96 TPLX 247.78 198.69 P i
j Z23HB103 Z280610/A B14 3-1/C500 412.96 TPLX 247.78 198.69 P l 19 % Z2B0610/B B14 3-1/C500 412.96 TPLX 247.78 198.69 P Z23GE10: Z2B0610/A B14 3-1/C500 412.96 TPLX 247.78 198.69 P j 22 % Z2B0610/B B14 3-1/C500 412.96 TPLX 247.78 198.69 P Z2B0607/A B10 3-1/C250 230.55 TPLX 138.33 102.3 P NOTE 1 Z2B0606/A B14 3-1/C500 412.96 TPLX 247.78 225.4 P Z2B0606/B B14 3-1/C500 412.96 TPLX 247.78 225.4 P
, Z2B0607/A B10 3-1/C250 213.06 TPLX 127.84 102.3 P 25 %
Z23FA25p Z2B0607/A B10 3-1/C250 202.54 TPLX 12L52 102.3 P "MW . i NOTE 1 ;
27 %
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! Calc. No. 96-ENG -01528E2 Rev. 00 i Page 24 of 44 i
Raceway cable Code Type Isoe Made Imax Load P/F j Z14FM20 ZlB5103/A B09 3-1/C4/0 238.52 TPLX 143.1 147.5 F 9% ZlB5105/A B09 3-1/C4/0 238.52 TPLX 143.1 147.5 F l
ZlB5145/A B01 3/C#12 22.78 13.67 MOV P IB5158/A B01 3/C#12 22.78 13.67 MOV P l
) 224FL20, Z2B6105/A B09 3-1/C4/0 TPLX SPARE P l 8% Z2B6102/H B29 1/C4/0 166.34 99.80 147.5 F Z2B6105/F B29 1/C4/0 166.34 99.80 147.5 F
{
, Z25BG20' Z286105/A B09 3-1/C4/0 TPLX SPARE P j 10 % 2B41 A16/A B03 3/C#8 43.85 26.31 23.5 P j Z2HT007/H B03 3/C#8 SPARE P
! Z2B6102/A B09 3-1/C4/0 TPLX SPARE P Z2HT032/H B03 3/C#8 SPARE P
- 2B41A55/A B01 3/C#12 SPARE P l
l Z14FM10 ZlB5103/A B09 3-1/C4/0 238.52 TPLX 143.11 147.5 F i
12 % ZlB5105/A B09 3-1/C4/0 238.52 TPLX 143.11 147.5 F ZlB5145/A B01 3/C#12 19.34 11.6 MOV P IB5158/A B01 3/C#12 19.34 11.6 MOV P 1B3238/c B01 3/C#12 19.34 11.6 1.57 P IB31 A08/A B02 3/C#10 SPARE P IB31 A08/G B03 3/C#8 46.54 27.92 23.6 P ZIA636 ZlB5103/A B09 3-1/C4/0 247 TPLX 90.53 147.5 F 35 % Z1McV02/B C02 2/C #14 20.05 1KV 7.51 0.55 P ZlVA1010/A C62 2/C#10 32.8 IKV 12.02 1.82 P Z2A209 Z2B6102/H (3)B29 1/C4/0 227 118.86 147.5 P 21 %
Z2A201 Z2B6105/F (3)B29 1/C4/0 227 83.2 147.5 F 30 % Z286105/G C86 7/C #14 20.5 7.51 3.52 P Z2B6105/K C86 7/C #14 20.5 7.51 3.52 P ,
Z2A1074 Z2B6105/F (3)B29 1/C4/0 227 83.2 147.5 F 53 % Z2B6102/H (3)B29 1/C4/0 227 83.2 147.5 F Z2MCV04/J C85 3/C #14 20.5 7.51 0.55 P Z2B6105/G C86 7/C #14 20.5 7.51 3.52 P
Calc. No. 96-ENG -01528E2 Rev. 00 Page 25 of 44
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Raceway Cat.le Code Type I, Make Imax Load P/F l so Z2A1074 Z2B6105/K C86 7/C #14 20.5 7.51 3.52 P 4
Z24FL10 Z2B6102/H B29 1/C4/0 241.90
) 145.14 147.5 F
! 4% Z2B6105/F B29 1/C4/0 241.90 145.14 147.5 F l
Il
! Z2A1070 2B6168/A B01 3/C#12 24.6 9.47* 7.84 P
! 33 % 2B6169/A B01 3/C#12 24.6 6.44 2.35 P i
- Z2HV5279ff C81 2/C #16 14.76 3.86 0.044 P
! Z2B6102/K C86 7/C #14 20.5 5.37 3.52 P Z2B6105/J C85 3/C #14 20.5 5.37 3.52 P j Z2B6117/F C86 7/C #14 20.5 5.37 4.95 P l 2B6168/C C07 7/C #14 20.5 5.37 0.66 P j 2B6169/F C07 7/C #14 20.5 5.37 0.66 P l d
Z2CH517/F C86 7/C #14 20.5 5.37 0.309 P l Z2CH519/F C86 7/C #14 20.5 5.37 0.153 P
! Z2HV2525/F C87 9/C #14 20.5 5.37 0.153 P
- 2K01160/B C85 3/C #14 20.5 5.37 0.04 P 2K04028/B C85 3/C #14 20.5 5.37 0.04 P
{ Z2NF09/G C86 7/C #14 20.5 5.37 1.76 P 2QR006B/H C85 3/C #14 20.5 5.37 0.093 P i
l Z2A1073 Z2B6102/H B29 1/C4/0 227 71.3 147.5 F I 38% Z2B6105/F B29 1/C4/0 227 71.3 147.5 F Z2B6102/J C87 9/C #14 20.5 6.44 3.52 P l
j Z286105/G C86 7/C #14 20.5 6.44 3.52 P j Z2B6105/H C87 9/C #14 20.5 6.44 3.52 P j Z2B6117/D C86 7/C #14 20.5 6.44 4.95 P
! Z2B6117/E C85 3/C #14 20.5 6.44 4.95 P i l l Key for Table 6.2 i NOTE I . Load is taken from OPAL, 81.5 KVA or 102.3 amps.
} I for tray is obtained through calculation (solving the simultaneous Stolpe equations in Attachment D). This j is the method employed by ICEA in the development of the ampacity tables in Publication P-54-440.
! For conduits include all voltage classes of cable to ensure of proper derate for the number of conductors.
j IPCEA P46-426 is used if possible, NEC is used for wire smaller than #8 AWG,
- group factor of 1.0 from field inspection h
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- -. -. - .- _ -. . _ - . _ - . .=
Cele. No. 96-ENG -01528E2 Rev. 00
, Page 26 of 44-6.0 Body of Calculation continued:
6.2 480 Volts continued:
Immu Evaluations :
Trav TMHB10 the 500MCM cable's allowed ampacity at 19% fill (depth = 0.968 in)is obtained from Attachment B,412.96 amps. Derating for Thermo-Lag,412.96 x 0.6 =
247.78 amps.
Trav Z23GE10 the 500MCM cable's allowed ampacity at 22% fill is (depth = 1.12 in) l obtained from Attachment B,412.96 amps. Derating for Thermo-Lag,412.96 x 0.6 =
- 247.78 amps.
The 250MCM cable's allowed ampacity at 22% fill is also obtained from Attachment B, 230.55 amps. Derating for Thermo-Lag,230.55 x 0.6 = 138.33 amps.
Trav ZI4FM20_ at 9% fill Depth = 0.458 inches i
For 3/C 4/0 Triplex, free air at 40' C is 335 amps , adjusting to 50' C gives 335 x .89 =
298.15 amps and 80% is equal to 238.52 amps. Derating for Thermo-Lag gives 238.52 x 0.6 = 143.1 amps.
I i The 3/C#12 cable's allowed ampacity at 9% fill is also obtained from Attachment B, 22.78 3 amps. Derating for Thermo-Lag,22.78 x 0.6 = 13.67 amps. j l i j Trav Z24FL20 is at 8% fill and a depth of 0.407 in.
j The 4/0 Triplexed cable is limited to 80% ofits free air value at 50' C. See Tray Z14FM20 {
j for allowed ampacity. For non-triplexed 4/C see Attachment B,166.34 amps. Derating for ,
Thermo-Lag,166.34 x 0.6 = 99.80 amps.
Trav Z25BG20 is at 10% fill and a depth = 0.509 inches. The 3/C #8 allowed ampacity J
from Attachment B is 43.85 amps. Derating for Thermo-Lag, 43.85 x 0.6 = 26.31 amps.
IInv Z14FM10 is at 12% fill and a depth = 0.611 inches.
The 80% of free air values calculated for the 4/0 triplex cables in the tray above will also be applicable for this tray,238.52 amps. Derating for thermo-lag 238.52 x 0.6 = 143.11 amps.
, For the 3/C #12, from Attachment B the ampacity is 19.34 amps. Derating for Thermo-Lag,19.34 x 0.6 = 11.6 amps.
Calc. No. 96-ENG -01528E2 Rev. 00 i
Page 27 of 4+
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6.0 Body of Calculation continued:
6.2 480 Volts continued:
Z1 A636 Conduit This conduit is 35% full, the 3-1/C4/0 TPLX from IPCEA (REF. 3.1.5) is 278 amps at 40' C. Adjusting to 50' C ambient gives 278 x .89 = 247 amps. Derating for the total number of conductors in the conduit (7), requires a 0.7 derate (Ref. 3.1.44). Or 247 x .7 = 172.9 amps, derating for grouping factor,172.9 x .68 =l17.57 amps, derating for Thermo-Lag 117.57 x 0.77 = 90.53 amps.
From Reference 3.1.4:
For #14 AWG the allowed ampacity is 25 amps at 30' C, adjusted to 50* C gives 25 x 0.82
= 20.5 amps. Adjusting for the number of conductors in the conduit,20.5 x .7 = 14.35 amps derating for grouping factor ,14.35 x .68 = 9.76 amps and derating for Thermo-Lag 9.76 x 0.77 = 7.51 amps.
For #10 AWU the allowed ampacity is 40 amps at 30' C, adjusted to 50' C gives 40 x 0.82 = 32.8 amps. Adjusting for the number of conductors in the conduit (7),32.8 x .7 =
22.96 amps, derating for grouping factor,22.96 x 0.68 = 15.61 amps and derating for Thermo-Lag 15.61 x 0.77 = 12.02 amps.
1 Z2 A209 Conduit This conduit is 21% full, the 3-1C 4/0 cable from IPCEA (Ref. 3.1.5) is i 255 amps at 40* C. Adjusting to an ambient of 50' C, 255 x 0.89 = 227 amps. ( The total number of current carrying conductors in the conduit is 3). Derating for grouping factor, 227 x .68 = 154.36 amps and derating for Thermo-Lag gives 154.36 x 0.77 = 118.86 amps.
i Z2 A201 Conduit This conduit is 30% full, the 3-1C 4/0 cable from IPCEA (Ref. 3.1.5) is l 255 amps at 40' C. Adjusting to an ambient of 50' C, 255 x 0.89 = 227 amps. Adjusting for the total number of conductors in the conduit (17), requires a derate of 0.7 (Ref.
3.1.44). 227 x .7 =158.9 amps, derating for grouping factor,158.9 x .68 = 108.05 and j derating for Thermo-Lag 108.05 x 0.77 = 83.2 ampt l
For #14 AWG the allowed ampacity is 25 amps at 30' C, adjusted to 50' C gives 25 x 0.82 = 20.5 amps. Adjusting for the number of conductors in the conduit,20.5 x .7 = i 14.35 amps, derating for grouping factor,14.35 x .68 = 9.76 amps and derating for Thermo-Lag 9.76 x 0.77 = 7.51 amps.
Z2A1073 Conduit This conduit is 38% full, the 3-1/C 4/0 cable from IPCEA (Ref. 3.1.5) is 255 amps at 40' C. Adjusting to an ambient of 50' C, 255 x 0.89 = 227 amps. ( The total number of conductors in the conduit is 41, requiring a derate of.6). Adjusting for the number of conductors in the conduit,227 x .6 = 136.2 amps , adjusting for group factor, 136.2 x .68 = 92.6 amps and derating for Thermo-Lag 92.6 x 0.77 = 71.3 amps.
l Calc. No. 96-ENG -01528E2 Rev. 00
- Page 28 of +4
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I 6.0 Body of Calculation continued:
4 6.2 480 Volts continued:
e Z2A1073 Conduit
. The #14 AWG cable from above is rated at 20.5 amperes at 50' C , adjusting for 41 conductors,20.5 x 0.6 = 12.3 amps adjusting for grouping factor,12.3 x .68 = 8.36 amps.
- Deratng for Thermo-Lag 8.36 x 0.77 = 6.44 amps.
Trav Z24FL10 is at a 4% fill Depth = 0.2037 inches.
I' From Attachment B the 3-1/C 4/0 ampacity is 241.90 amps. Derating for Thermo-Lag j gives 241.90 x 0.6 = 145.14 amps.
Z2A1074 Conduit This conduit is 53% full, the conduit length is 3 feet. This short distance allows the greater than 40 % fill without doing damage to the cables. The ampacity of the 3-1/C 4/0 from IPCEA (Ref. 3.1.5)is 255 amps at 40' C. Adjusting to an ambient of I 50' C, 255 x 0.89 = 227 amps. The total number of conductors in the conduit is 23, derating for the conductors,227 x 0.7 = 158.9 amps. Derating for grouping factor,158.9 x {
.68 = 108.05 aamps and derating for Thermo-Lag,108.05 x 0.77 = 83.2 amps. i The #14 AWG cable from above is rated at 20.5 amperes at 50' C , adjusting for 23 conductors,20.5 x 0.7 = 14.35 amps. Derating for grouping factor,14.35 x 0.68 = 9.758 .
i amps and deleting for Thermo-Lag 9.758 x 0.77 = 7.51 amps.
Z2A1070 Conduit This conduit is 33% full, the #12 AWG cable is 30 amps at 30' C ,
adjusting to 50* C,30 x 0.82 = 24.6 amps. Adjusting for 78 conductors,24.6 x 0.5 = 12.3 amps. Derating for group factor 12.3 x 0.68 = 8.36 amps and derating for Thermo-Lag 8.36 x 0.77 = 6.44 amps.
l The #14 cable is rated at 20.5 amperes at 50' C , adjusting for 78 conductors,20.5 x 0.5 =
10.25 amps, adjusting for group factor,10.25 x 0.68 = 6.97 amps and deleting for Thermo-Lag,6.97 x 0.77 = 5.37 amps.
The #16 cable is rated at 18 amperes at 30' C , adjusting to 50* C 18 x .82 = 14.76 amps.
The total number of conductor in the conduit is 78,14.76 x 0.5 = 7.38 amps. Derating for grouping factor,7.38 x .68 = 5.02 amps and derating for Thermo-Lag,5.02 x 0.77 = 3.86 amps.
i Calc. No. 96-ENG -01528E2 Rev. 00 Page 29 of 44-l l
1 6.0 Body of Calculation continued:
6.2 480 Volts continued:
Load Determination:
Cables Z2B0610/A and Z2B0610/B are parallel feeder cables to MCC B61. Continuous load on MCC B61 is shown on Attachment 5 [ 397.38 amps].
397.38 + 2 = 198.69 amps Cables Z2B0606/A and Z2B0606/B are parallel feeder cables to MCC B62. Continuous load on MCC B62 is shown on Attachment 6, [450.8 amps] i 450.8 + 2 = 225.4 amps i
Cables Z1B5103/A, Z1B5105/A ,Z2B6102/H and Z2B6105/F for Charging Pumps P18A and P18B and P18C are all 100HP motors per (Ref. 3.1.28,3.1.39).
The nameplate full load amps is 118 amps. Adjusting for voltage fluctuations 118 x 1.25 gives 147.5 amps. l I
Cables Z2B6102/A,1 B31 A08/A, Z2B6105/A,2B41 A55/A, Z2HT032/H and Z2HT007/H i are spare (Ref. 3.1.26)
Cables ZlB5145/A and 1B515S/A are MOV feeds (see Assumption 4.14, Ref. 3.1.58 and 3.1.71).
Cable 2B41 A16/A is the degasifier Pump #2 motor is 15 HP (Ref. 3.1.85)
HP x 6 I = 4 x E x eff x pf = 18.8 x 1.25 = 23.5 amps Cable 1.B3238/C is RM8997, a 1 hp motor (Ref. 3.1.86)
= 1. 2 5 x 1. 2 5 = 1. 57 AMPS I = 4 x 460x. 88x. 89 Cable 2B6168A is for a Shp motor MP72B (Ref. 3.1.38)
I = 4 x 460x. 88x. 89 = 6. 27 x 1. 25 = 7.84 amps
Calc. No. 96-ENG -01528E2 Rev. 00 Page 30 of 44 i ,
i 6.0 Body of Calculation continued:
6.2 480 Volts continued:
Load Determination continued:
Cable 2B6169A is for a 1.5 hp motor MP125 (Ref. 3.1.38) t I 1. 5 x 746 I= = 1. 88 amps x 1.25 = 2.35 amps Cable IB31 A08/G is for a 15 hp motor, NP16A1. (Ref. 3.1.24) i i 15 x 746 ,
I= = 18. 88 ampsx 1.25 = 23.6 amps
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. Cable Z2B0607/A, Battery Charger 201B, per OPAL /FSAR Table 8.3-2 maximum load i on diesel during LOCA from charger is 81.5KVA (Ref. 3.1.31).
4 I = 81. 5 x 1000 = 102. 3 amps !
. V3 x 480 j Cable Z1MCV02/B, load is 0.5 amps ( Ref. 3.1.67) x 1.1 = 0.55 amps Cable Z1VA1010/A, load is 1.65 amps (Ref. 3.1.37) x 1.1 = 1.82 amps Cable Z2MCV04/J, load is 0.5 amps (Ref. 3.1.47) x 1.1 =0.55 amps ,
Cable Z2B6105/G, load is 3.2 amps (Ref. 3.1.25) x 1.1 = 3.52 amps i
Cable Z2B6105/K, load is 3.2 amps (Ref. 3.1.25). x 1.1 = 3.52 amps
Calc. No. 96-ENG -01528E2 Rev. 00 Page 31 of 44
. 6.0 Body of Calculation continued:
6.2 480 Volts continued:
Load Determination continued:
4 Cable Z2HV5279ff, V =i10vde, R = 250fl, load is 0.04 amps (Ref. 3.1.68 and 3.1.23 ) x 1.1 = 0.044 amps Cable Z2B6102/K, load is 3.2 amps (Ref. 3.1.36) x 1.1 = 3.52 amps 4
Cable Z2B6105/J, load is 3.2 amps (Ref. 3.1.25) x 1.1 = 3.52 amps Cable Z2B6117/F, load is 4.5 amps (Ref. 3.1.49) x 1.1 = 4.95 amps Cable 2B6168/C, load is 0.6 amps (Ref. 3.1.56) x 1.1 = 0.66 amps Cable 2B6169/F, load is 0.6 emps (Ref. 3.1.57) x 1.1 = 0.66 amps Cable Z2CH517/F, load is 0.281 amps (Ref. 3.1.64 and 3.1.50)X 1.1 = 0.309 amps Cable Z2CH519/F, load is 0.139 amps (Ref. 3.1.64 and 3.1.51) x 1.1 = 0.153 amps Cable Z2HV2525/F, load is 0.139 amps (Ref. 3.1.64) x 1.1 = 0.153 amps Cable 2K01160/B, load is 0.04 amps (Ref. 3.1.52) x 1.1 = 0.04 amps Cable 2K04028/B, load is 0.04 amps (Ref. 3.1.48) x 1.1 = 0.04 amps Cable 2QR006B/H, load is 0.0842 amps (Ref. 3.1.63) x 1.1 = 0.0926 amps Cable Z2B6102/J load is based on a 3.2 amp fuse (Ref. 3.1.19) x 1.1 = 3.52 amps.
Cable Z2B6105/H load is based on a 3.2 amp fuse (Ref. 3.1.25) x 1.1 = 3.52 amps.
Cable Z2B6117/D load is based on a 4.5 amp fuse (Ref. 3.1.49) x 1.1 = 4.95 amps.
1 Cable 22B6117/E load is based on a 4.5 amp fuse (Ref. 3.1.49) x 1.1 = 4.95 amps.
Cable Z2NF09/G load is based upon a 200 ohm load at 125VDC (Ref. 3.1.65) 1.6 amps x 1.1 = 1.76 amps.
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Calc. No. 96-ENG -01528E2 Rev. 00 1 Page 32 of 44-i 6.0 Body of Calculation continued:
j 6.3 120ac/125dc Volts Power
)
i A total of five cable trays, a wire-way, five conduit and one box were identified in j the plant as covered with Thermo-Lag.
l l TABLE 6.3 l Tray / Cable Code Type I,so Imax Load P/F j Conduit 5T540 5A602/C B04 3/C#6 61.5 32.2 1 P l 10 %
' Z2A1078 Z2DV2008/A B46 2/C#10 32.8 13.74 0.91 P Z2VA2004/A B46 2/C#10 32.8 13.74 1.76 P l J603 see Z25XA10 l Z25XA10 - Z2DV2008/A B46 2/C#10 32.8 12.63 0.91 P
} Z2VA2004/A B46 2/C#10 32.8 12.63 1.76 P f Z25V4188/K C90 7/C#12 24.6 9.47 0.91 P l Z2SV4188/M C94 2/C#10 32.8 12.63 0.91 P l Z2SV4188/L C90 7/C#12 24.6 9.47 0.185 P j Z2SV4188/J C94 2/C#10 32.8 12.63 0.185 P l Z2HV5279/P C89 3/C#12 24.6 9.47 0.39 P
! Z2HV5279/R C89 3/C#12 24.6 9.47 0.434 P
! Z2HV5279ff C81 2/C#16 14.76 5.68 0.92* P
! Z2HV5279/U C81 2/C#16 14.76 5.68 0.55 P l Z2B6102/J C87 9/C#14 20.5 7.89 3.52* P Z2B6102/K C86 7/C#14 20.5 7.89 3.52* P Z286105/H C87 9/C#14 20.5 7.89 3.52' P Z2B6105/J C85 3/C#14 20.5 7.89 3.52 P Z236105/K C86 7/C#14 20.5 7.89 0.13 P Z2B6117/D C86 7/C#14 20.5 7.89 4.95* P Z2B6117/E C85 3/C#14 20.5 7.89 4.95* P Z2B6117/F C86 7/C#14 20.5 7.89 4.95 P Z2CH517/E C86 7/C#14 20.5 7.89 0.168 P Z2CH517/F C86 7/C#14 20.5 7.89 0.168 P Z2CH519/E C87 9/C#14 20.5 7.89 0.168 P Z2CH519/F C86 7/C#14 20.5 7.89 0.168 P
- LOAD IDENTIFIED FOR 480 V SECTION
7 __. _ . _ _ _ . .. _
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l Calc. No. 96-ENG -01528E2 Rev. 00 i Page 33 of 44-l 6.0 Body of Calculation continued:
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l 6.3 120ac/125dc Volts Power
} Tray / Cable Code Type Iso Imax Load P/F j Conduit Z25XA10 Z2HV2525/F C87 9/C#14 20.5 7.89 0.213 P Z2HV2525/G C86 7/C#14 20.5 7.89 0.213 P Z2HV2525/H C85 3/C#14 20.5 7.89 0.085 P l
Z2NF09/F C86 7/C#14 20.5 7.89 1.73 P f Z2NF09/G C86 7/C#14 20.5 7.89 1.73 P 2K01160/B C85 3/C#14 20.5 7.89 's.055 P l
Z2MCV04/J C85 3/C#14 20.5 7.89 0.55* P i Z2HV5279/S C76 9/C#12 24.6 9.47 0.37 P i
Z2T871 Z2SV4188/C C91 9/C#12 24.6 9.02 0.91 P l Z2SV4188/D C96 4/C#10 32.8 12.02 0.91 P .
l Z2NF09/F C86 7/C #14 20.5 7.51 1.73 P l Z25BG20 - Z2HT007/C C62 2/C#10 SP P j 10 % Z2HT028/C C62 2/C#10 SP P Z2HT032/C C62 2/C#10 SP P 2B6102/F B01 3/C#12 9.984 5.99 2.3 P 2HT007/F CO2 2/C #14 SP P 2HT028/F CO2 2/C #14 SP P 2HT032/F CO2 2/C #14 SP P 2K02049/B CO2 2/C #14 7.68 4.61 0.04 P 2K02051/B C02 2/C #14 7.68 4.61 0.04 P 2K02053/B CO2 2/C #14 7.68 4.61 0.04 P 2K02055/B C02 2/C #14 7.68 4.61 0.04 P 2CH910/B C03 3/C #14 6.153 3.68 1.65 P 2CH910/C C03 3/C #14 6.153 3.68 1.65 P 2FT9860/A C03 3/C #14 6.153 3.68 0.04 P 2HC2152/B C03 3/C #14 SP P Z2HT028/H C62 2/C#10 SP P i
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F Calc. No. 96-ENG -01528E2 Rev. 00 l Page 34 of +4 j = = -
6.0 Body of Calculation continued:
1 6.3 120ac/125dc Volts Power J / Cable Code Type I% Imax Load P/F l 224FL20., 2B41A16/F C07 7/C#14 4.96 2.97 0.334 P l Z2B6105/G C86 7/C#14 4.96 2.97 2.5 P j Z286105/K C86 7/C#14 4.96 2.97 0.146 P Z2HV8133/C CO2 2/C#14 7.68 5.10 0.153 P l
1 l
j Z24FL20 Z2HV8133/D C07 7/C#14 4.96 2.97 0.301 P l' Z2HV8248/C C02 2/C#14 7.68 4.61 0.301 P Z2HV8248/D C07 7/C#14 4.96 2.97 0.301 P l Z2MCV04/B CO2 2/C#14 7.68 4.61 0.55 P
} Z2VA2010/A C02 2/C#14 7.68 4.61 1.82 P
- Z2MCV04/J C85 3/C#14 6.153 3.69 0.55* P
! Z14FM10 IB5103/F B01 3/C#12 9.984 6.3 1.01 P
! 12 % IB5105/C B01 3/C#12 9.984 6.3 1.01 P ZlVA1010/A C62 2/C#10 13.358 8.87 1.82* P
, IB31A08/F C07 7/C#14 4.96 2.97 0.34 P ZlB5145/C C07 7/C#14 4.96 2.97 1.29 P IB5158/B C07 7/C#14 4.96 2.97 0.66 P ZlCH192/B C07 7/C#14 4.96 2.97 0.853 P ,
) ZlHV8247/C C07 7/C#14 4.96 2.97 0.153 P
! ZlHV8247/D C07 7/C#14 4.96 2.97 0.153 P i ZlHV8249/C C02 2/C#14 4.96 2.97 0.153 P !
{ ZlHV8249/D C07 7/C#14 4.96 2.97 0.153 P 1K96001/B CO2 2/C#14 7.68 4.61 0.04 P !
j IK96002/B CO2 2/C#14 7.68 4.61 0.04 P ZIMCV02/B C02 2/C#14 7.68 4.61 0.123 P l
1RM8997/D C02 2/C#14 7.68 4.61 0.83 P l IVAAS/E CO2 2/C#14 7.68 4.61 SP P IVAAS/EE C02 2/C#14 7.68 4.61 SP P ZlCH196/B C07 7/C#14 4.96 2.97 0.254 P
)
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! Calc. No. 96-ENG -01528E2 Rev. 00 Page 35 of 4 +
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i 6.0 Body of Calculation continued:
l l 63 120ac/125dc Volts Power continued:
TABLE 63 Tray / Cable Code Type I3oe Imax Load P/F l Conduit j Z14FM20 ZlVA1010/A C62 2/C#10 13.358 8.02 1.82* P l 9% IB5103/F B01 3/C#12 9.984 5.99 3.2 P l 1B5105/C B01 3/C#12 9.984 5.99 2.31 P l ZlB5145/C C07 7/C #14 4.96 2.97 1.29 P 1B5158/B C07 7/C #14 4.96 2.97 0.66 P
, ZlCH192/B C07 7/C #14 E6 2.97 0.854 P I ZlHV8247/C C02 2/C#14 7.68 4.61 0.153 P I
ZlHV8247/D C07 7/C #14 4.96 2.97 0.153 P
$ ZlHV8249/C C02 2/C#14 7.68 4.61 0.153 P j ZlHV8249/D C07 7/C #14 4.96 2.97 0.153 P ZlMCV02/B CO2 2/C#14 7.68 4.61 0.124* P ZlCH196/B C07 7/C #14 4.96 2.97 0.28 P i
i Z2A201 Z2B6105/G C86 7/C #14 20.5 7.51 2.52* P Z2B6105/K C86 7/C #14 20.5 7.51 0.134* P i
l Z24FL10 Z2MCV04/J C85 3/C #14 6.153 3.69 0.55* P i 4% Z2B6105/G C86 7/C #14 17.57 10.54 3.52* P Z2B6105/K C86 7/C #14 17.57 10.54 3.52* P i
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ZlA636 ZlVA1010/A C62 2/C#10 32.8 12.02 0.55* P i Z1MCV02/B CO2 2/C #14 20.5 7.51 1.82* P
- load identified for 480 volt section.
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Cdc. No. 96-ENG -01528E2 Rev. 00 Page 36 of M
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6.0 Body of Calculation continued:
6.3 120mc/125dc Volts Power continued:
Imax Evalnation 5T540 Conduit, the ampacity of SA602/C .: 3/C #6 cable in conduit is 75 amps (Ref. 3.1.4) at 30' C. Adjusting to 50' C,75 x 0.82 = 61.5 amps. This is the only cable in the conduit. Derating due to grouping factor,61.5 x .68 = 41.82 amps and derating for Thermo-Lag,41.82 x 0.77 = 32.20 amps.
Z2A1078 Conduit. the ampacity of #10 AWG cable in conduit is 40 amps at 30' C.(Ref.3.1.4) Adjusting for 50* C,40 x .82 = 32.8. Derating for 4 conductors in a conduit (Ref.3.1.44),32.8 x 0.8 = 26.24 amps. Derating for group factor,26.24 x
.68 = 17.84 amps and derating for Thermo-Lag,17.84 x 0.77 = 13.74 amps.
Z25XA10 Wire-Way , will be treated as a conduit. For #12 AWG (Ref. 3.1.4) cable is rated 30 amps at 30* C. Adjusting for 50' C,30.x 0.82 = 24.6. Adjusting for the number of conductors (REF. 3.1.44),146 gives 24.6 x 0.5 = 12.3 amps.
Derating for the Thermo-Lag,12.3 x 0.77 = 9.47 amps.
For # 10 AWG (Ref. 3.1.4) cable is rated at 40 amps at 30* C, adjusting for 50' C.,
40 x 0.82 = 32.8. Adjusting for the number of conductors 146,32.8 x 0.5 = 16.4 amps. Derating for Thermo-Lag,16.4 x 0.77 = 12.63 amps.
For # 14 (Ref. 3.1.4) is good for 25 amps at 30' C.. Adjusting to a 50*C ambient,25 x 0.82 = 20.5 amps. Adjusting for 146 conductors,20.5 x 0.5 = 10.25 amps.
Derating for Thermo-Lag,10.25 x 0.77 = 7.89 amps.
For #16 (Ref. 3.1.4) cable is rated at 18 amps in a 30*C ambient, adjusting for a 50*C ambient 18 x 0.82 = 14.76, adjusting for 146 conductors ,14.76 x 0.5 = 7.38 amps. Deraing for Thermo-Lag,7.38 x 0.77 = 5.68 amps.
Z25BG20 TRAY the ampacities for the control cables will use the 35% fill values for conservatism (actual 10% fill) . From Attachment 2, Appendix AD,2/C
- 10 in tray can carry 13.358 amps, at 50' C. Derating for Thermo-Lag,13.358 x 0.6 = 8.02 amps.
The 3/C #12 can be obtained from Attachment 2, Appendix AC, using the 35% fill value, provides 9.984 amps at 50' C. Derating for Thermo-Lag,9.984 x 0.6 = 5.99 amps.
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Calc. No. 96-ENG -01528E2 Rev. 00 Page 37 of 4 6.3 120ac/125dc Volts Power continued:
, Imax Evaluation continued:
Z25BG20 TRAY continued:
The 3/C #14 at a fill of 35 % (1.78") per Attachment 2, Appendix AA provides an allowed ampacity of 6.153 amps at 50' C. Derating for Thermo-Lag,6.153 x 0.6 =
3.69 amps.
For 2/C #14, Attachment 2, Appendix AB allowed ampacity is 7.68 amps at 50* C i
. Derating for Thermo-Lag 7.68 x 0.6 = 4.61 amps.
For Trav Z14FM20, Z24FL20. and Trav Z14FM10. the fill and the values i
developed for cables in Z25BG20 above are also applicable.
For 7/C #14 (C86) adjusting for the diameter difTerence:
I2 = It = 6.163 = 7. 57 amps adjusting for number of conductor difference,I = 7.57 f = 4.96 amps Derating for Thermo-Lag,4.96 0.6 = 2.97 amps.
Z2T871 Conduit. From Reference 3.1.4,3/C #12 in conduit can carry 30 amps. at 30' C, adjusting to 50* C,30 x .82 = 24.6. Adjusting for the number of:onductors (20), 24.6x 0.7 = 17.22 amps. Derating for group factor,17.22 x .68 = 11.71 am ps and derating for Thermo-Lag,11.71x 0.77 = 9.02 amps.
For 4/C #10 the allowed ampacity is 40 amps (Ref. 3.1.4) at 30' C, for a 50* C ambient,40 x .82 = 32.8. Adjusting for the number of conductors (20), 32.8 x 0.7
= 22.96 amps. Derating for grouping factor,22.96 x 0.68 = 15.61 and derating for Thermo-Lag,15.61 x 0.77 = 12.02 amps.
The #14 cable is rated at 20.5 amperes at 50' C (Ref. 3.1.4), adjusting for 17 conductors,20.5 x 0.7 = 14.35 amps. Derating for grouping factor,14.35 x .68 =
9.76 amps and deleting for Thermo-Lag,9.76 x 0.77 = 7.51 amps.
Z2 A201 Conduit. The #14 cable is rated at 20.5 amperes at 50' C (Ref. 3.1.4),
adjusting for 17 conductors,20.5 x 0.7 = 14.35 amps. Derating for grouping factor, 14.35 x .68 = 9.76 amps and deleting for Thermo-Lag,9.76 x 0.77 = 7.51 amps.
Z24FL10 Trav From Attachment 2, Appendix X, the cable is rated at 17.57 amps at 50* C. Derating for Thermo-Lag 17.57 x .6 = 10.54 amps.
Calc. No. 96-ENG -01528E2 Rev. 00 Page 38 of 44
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4 6.3 120ac/125de Volts Power continued:
Load Determination :
- Cable 5A602/C provides 125 vde to the 4160v switch gear from MP14160v switchgear bus 14H cubicle A610 Unit 105A. The breaker trip requirement is approximately 12 amps but is only momentary. The circuit breaker closing is less than 7 amps and is also only momentary. The indicating lights constitute the only continuous load and are less than 1 amp (Ref. 3.1.3 and 3.1.83).
Cable Z2VA2004/A powers C09 control room panel. The cable is the power supply to a Foxboro power supply (N-2ARPS 05) which transforms the 120 to 24 volts, this breaks down into two circuits one requiring 5 amps in a de loop and one 3 amps ac. These ampacities are combined as a conservative measure and the total load is taken as 8 amps at 24 ac. The VA rating of the transformer is thus 24 x 8 =
192 va. The cable would see 192 /120 = 1.6 amps (Ref. 3.1.90).
Cable Z2DV2008/A powers C10 control room panel. The cable is fed from DV20 circuit 8 which is a 30 amp 250 volt de breaker. From DM2-5-1213-95 Sheet 6 (Ref. 3.1.35) the starting load on the panel is 17.5 amps. The normal load seen by the cable is 0.8268 amps and will be used as the load (Ref. 3.1.60,3.1.9,3.1.53) ;
Cables Z2SV4188/K. Z2SV4188/M, Z2SV4188/C and Z2SV4188/D are all associated with the controls of C10 and see part or all of the 0.8268 amps identified above. For conservatism all will use a load of 0.8268 amps (Ref. 3.1.9,3.1.2).
Cable 2B6102/F per Drawing 25230-32009 Sheet 43 (Ref. 3.1.36) the motor heater is 110 watts at 120, the current is 110/120 = 0.9167 amps (rounded up to 0.92).
For Tray Z14FM20 the 1/C#12 cables (1B5103/F and 1B5105/C) feed 110 watt heaters (Ref. 3.1.32 and 3.1.33). The load is 110 watts,110/120 = 0.9167 amps (rounded up to 0.92). Cable Z1VA1010/A is on breaker 10 of Panel VA10 (Ref.
3.1.37) and is loaded to 1.6472 amps (rounded up to 1.65)
For Tray Z14FM10 see the above tray for common cables.
Cable Z2MCV04/B (Ref. 3.1.47) uses a 0.5 amp fuse in the circuit.
Cable Z2VA2010/A (Ref. 3.1.37) has a total load of 1.648 amps on this cable.
Calc. No. 96-ENG -01528E2 Rev. 00 Page 39 of 44-
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6.3 120ac/125dc Volts Power continued:
Load Determination continued:
Cable Z2B6105/H has a 3.2 amp fuse in the circuit, current value used will be the fuse value (Ref. 3.1.25 and 3.1.30).
Cable Z2HV5279/U (Ref. 3.1.68) has a 250 ohm resister in series with the load.
Using the highest DC voltage (125 v) as its source, I =V/R or 125/250 = 0.5 amps.
Cable 1B31 A08/F (Ref. 3.1.70) has a 6 amp fuse, the worse case conductor for this cable has 4 relays (0.l'2 amps each). This worse case value will be used as the load (0.52 amps).
Cable 2CH910B and 2CH910C (Ref.3.1.72) has a 1.5 amp load.
Cable 2FT9860/A (Ref. 3.1.73) has a 0.04 amp load.
Cables starting with K after the facility indicator are connected to alarm circuitry cards which directly interface with transistors and other electronic components. A conservative 5 Watts will be used (Assumption 4.13) Sw + 120v = 0.04 amps.
Control circuits where multi-conductors are used have been evaluated on a conductor by conductor basis. The largest load on any conductor is used as the load for all remaining conductors. See Attachment 8 for the appropriate ampacities.
Cable Z2SV4188/L (Ref. 3.1.2) identifies a 0.1684 amp load. l Cable Z2SV4188/J (Ref. 3.1.2) identifies a 0.1684 amp load.
Cable Z2B6105/J (Ref. 3.1.25) identifies a 3.2 amp fuse in the circuit.
Cable Z2B6117/F (Ref. 3.1.49) identifies a 4.5 amp fuse in the circuit.
Cable 2HC2152/B (Cable is not installed) 6.4 Instrumentation Cable From Attachment 3 a total of 3 cable trays 4 conduits one wire-way and one box were identified in the plant that were covered with Thenno-Lag. Although the instrumentation cable will not be impacted by Thermo-Lag since it only carries minimal current this section is included to document a!! cables covered by Thermo-Lag in MP2. See Table 6.4.
i Calc. No. 96-ENG -01528E2 Rev. 00 l Page 40 of 44 j 6.0 Body of Calculation continued:
) '
- TABLE 6.4 -
l 2
fray /j ; ,, Cable Code Type l l, Conduit :1, l
3
'rire way; p 42- -
- D6TA104 2P9934/B 1 54? 2/C#16
- D4 ' 2/C#16 2QR011/C 2QR0ll/D 134M 2/C#16
- 2PT102B/D B4> 2/C#16 4
2PT102B/E 2/C#16 j
T34~
)
Z2QR032/K 134 - 2/C#16 4 .
Z2OR032/L 134.. 2/C#16 j - Z2QR033/S D4. 2/C#16 Z2QR033/R B4 s 2/C#16 Z2LT5282/D 134: 2/C#16 Z2LT5282/E B4.. 2/C#16
, 2QR035B/D B4 2/C#16 j e Z20R003B/BB B4. 2/C#16 I i Z2OR003B/AA D6 - 4/C#16 j Z2QR003B/R 136 - 4/C#16 Z2QR003B/Z D4 2/C#16 Z2QR003B/CC D4 2/C#16 l 2QR035B/D 134- 2/C#16 ,
j w Z2PT4224/H 134 V 2/C#16 l Z2PT4224/K D4" 2/C#16 r
- Z2PT4224/G 1341- 2/C#16
+- Z2FT5278B/D 434 " 2/C#16 Z2FT5278B/E D4) 2/C#16 j 2LT208/B 134 ~~ 2/C#16
- 2LT208/C B4" 2/C#16
! 2LT206/C 134 2/C#16 i 2LT206/B B4: . 2/C#16 4
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l Calc. No. 96-ENG -01528E2 Rev. 00 Page 41 of 44
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l 7.0 Reviewers Comments and Resolutions I
7.1 Additional references were identified as needed to document loads.
Resolution: References were added.
7.2 Revision levels of several drawings were updated.
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Resolution: Latest revisions were reviewed and confirmed that no impact resulted due to the changes implemented on the new revision.
7.3 Several 120 volt loads were revised.
Resolution: All 120 volt loads revised except 1 (one) were revised downward, one load was erroneous. All changes were evaluated and agreed to.
7.4 Two cables were identified as entered in a table under an incorrect tray.
I Resolution: Cables were entered in their proper location. l 7.5 Additional loads were identified on the MCC's (Attaciunents E and F). ]
l Resolution: The loads were checked and entered.
7.6 Spare cables were identified that were not shown as spare in the calculation.
Resolution: Cables were confirmed as spare and identified as such in the calculation.
7.7 Tray width dimensions were identified as incorrect.
Resolution: All trays were rechecked and the appropriate dimensions on three (3) trays corrected.
7.8 Discrepancy with raceway fill was identified.
Resolution: Comment was rejected, fills were correct, wording was revised for clarity.
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CrJc. No. 96-ENG -01528E2 Rev. 00 i
Page 42 of 44 t
l l 7.0 Reviewers Comments and Resolutions continued:
5
!; - 7.9 Uniformity of Ampacity (I) values in the tables for each voltage section were
! identified.
Resolution: Tables were made uniform to eliminate confusion.
! 7.10 Review of TS02 computerized raceway schedule identified two discrepancies in the i schedule pertaining to fire wrap.
1
- l. Resolution: The schedule was determined to be incorrect. DCN DM2-00-1439-96 l_ was initiated to correct.
a The independent reviews were completed by 6 personnel, (Bob Blodgett, Mike
- Champagne, Jose Gomez, Khwaja Haque, Jack Padden and Mike Relyea).
1 B. Blodgett: was responsible for reviewing the reference design drawings / documents and j . for reviewing the electrical loads on these documents. In addition, Bob also reviewed the ,
design documents to verify that the cables that are wrapped with thermo-lag were {
- identified, this was accomplished by review ofdrawings, field walk downs and review of l photographs of the installations.
J. Gomez: was responsible for reviewing the cable in conduit deratings and for verifying that the applicable information was transferred into the main body of the calculation.
]
K. Haque: was responsible for reviewing Attachments A(excluding thermal models) and B (excluding thermal models) for data inputs, the cable in tray deratings and for verifying that the applicable information was transferred into the main body of the calculation. l M. Champagne: was responsible for reviewing the engineering references, design inputs , j assumptions , Attachments E and F , the method of calculation, the thermal model was i reviewed (Attachment's A and B) and the independent overall review of the calculation.
J.Padden: was responsible for reviewing the reference design drawings / documents and for reviewing the electrical loads on these documents.
Mike Relyea: was responsible for reviewing the reference design drawings / documents and for reviewing the electrical loads on these documents.
- . . - . . ,- - - - - - . - - . - . . _ ~ _ . - - - _ = - . . . - . - . -
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3 Calc. No. %-ENG -01528E2 Rev. 00 l
- Page 43 of 44 j i
l 8.0 Attachments INDEX Attachment A ........Thermo-Lag Derating Model.
j Attachment B......... Ampacities for specific cable tray cables i
1 Z23HA10.................. Appendix A 3
Z23HB10.................. Appendix B 1 i Z23GE10................... Appendix C !
Appendix D j Z14FM20................... Appendix F l i Appendix H l j Z24FL20.. ................. Appendix I Appendix i 3
Z25BG20................... Appendix K
- Appendix AA j Appendix AB Appendix AC Appendix AD l 4
Z14FM10................... Appendix E Appendix M >
i Appendix N i Z24FL10 Appendix G 3
Z23FA30 Appendix V Z23FA25 Appendix W Z52EA10 Appendix B1 Z22EA10 Appendix B2 TESTING VERIFICATION Test #1 3/C #8 1.5" depth.......... Appendix P and PP Test #2 3/C #8 2.0" depth......... Appendix Q and QQ Test #3 500MCM 1.5" depth......... Appendix T and TT ,
Test #4 500MCM 3.0" depth.... .... Appendix U and UU l' Test #5 4/0............. 3.0" depth Appendix R and RR Test #6 4/0........... . 2.5" depth......... Appendix S and SS
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Calc. No. 96-ENG -01528E2 Rev. 00 Page 44 of 44 e 1
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T l 8.0 Attachments INDEX continued: ,
Attachment C.......... Selected 120/125v Control Circuit Ampacities .
l Attachment D.......... Stolpe IEEE Paper I 1
Attachment E........... MCCB61 Coatinuous Load Tabulation j i
Attachment F........... MCCB62 Continuous Load Tabulation t
Attachment G........... Independent Reviewer Evaluation )
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l Calc. No. 96-ENG-01528E2 Rev. 00 Page Al of MNo ATTACHMENT A This attachment provides the thermal models for the TSI covered cable tray and conduits at MP2 and the development of the ACF (ampacity correction factor) for each.
The model was developed using standard heat transfer equations.
~
The model developed predicts a given surface temperature for a specific material and
- installation. The model was then compared to existing test results performed on TSI 4
wrap at Texas Utilities. The comparison shows that the model conservatively predicts the surface temperature [0.675 vs 0.68].
The only difference between the TU and MP2 model is the thickness of the TSI material. I With only one variable changing, i.e. thickness, the model is re-run to address the ,
maximum and minimum possibilities of TSI installation at MP2 for all wrapped tray and l the worse case determined (see the following page). Note that the change in thickness of the TSI wrap has a minimal affect upon the resultant temperature. This is to be expected as the TSI has a relatively low resistance to heat transfer.
The tray model was also run increasing the dimensions of the cable tray to much larger installation [48" width]. As expected the surface are becomes the governing factor and the temperature of the surface reduces as expected [62.3* C down to 60.7
- C]. This indicates that the model behaves as expected when the size of the installation becomes large.
The conduit model, uses the same methodology as the cable tray model and is based upon the heat transfer through a cylinder. The Ampacity Correction Factor (ACF) for the two different models is tabulated below.
ACF Cable Tray 0.60*
Conduit 0.77
- This conservative value is utilized for ease of calculation unless noted.
. - . - - . . ....-. - _ . - . . . . - . . - . . - . . _ - -- . - - =.- ___._. - - . - -
Calc. No. 96-ENG-01528E2 Rev.00 Page NL of AAo ATTAC HMENT A
SUMMARY
l The values in the table are a summary of the values obtained from the following pages of Attachment A and from Attachment B.
- Appendix Tray depth Q Min ACF Max ACF f A Z23HA10 0.509 13.137 0.693 0.609 l l B Z23HB10 0.968 10.157 0.735 0.623 I C Z23GE10 1.12 5.161 0.685 0.614 D Z23GE10 1.12 5.161 0.685 0.614 E Zl4FM10 0.611 10.661 0.691 0.61 F Zl4FM20 0.458 14.747 0.694 0.609 G Z24FL10 0.2037 35.701 0.698 0.606 H Z14FM20 0.458 14.797 0.694 0.609 I Z14FL20 0.407 16.88 0.694 0.608 i K Z25BG20 0.509 13.137 0.693 0.609 M Z14FM10 0.611 10.661 0.d1 0.61 N Z14FM10 0.611 10.661 l0591 0.61 B1 Z52EA10 1.935 5.07 0.792 0.689 B2 Z22EA10 1.1741 11.313 0.864 0.689 V Z23FA30 0.2037 4.407 0.617 0.608 W Z23FA25 0.2037 3.983 0.615 0.607 X Z24FL10 0.2037 35.701 0.698 0.606 For conservatism 0.6 will be used as the ACF for cable tray.
Calc. No. 96-ENG41528E2 Rev. 00 Z23HA10 min ATTACHMENT A.
ap M of M AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data input .
at pneraW for a lahr tray D3HA10 (Anadment 2)
Q~ 13.137 Ta := 50 . Ampient temperature MP2)
Conductor maximum allowed temperature Tc ~ 90 Surface area of tray, sqft/ft 24" wide x 11ong x top and bottom)
^ ._
'- 4
( Ref. 3.1.8)
~9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) a := 5.3- 10 c := 0.1 Emissivity of galvanized steel tray cover (Re!. 3.1.1)
" OO 9 "# "Y ACFcov ::0.59 (Ref. 3.1.10) ,
etsi := 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.0 Thickness of TSI (minimum) -
el := 4 Tray innerheight, inches e2 := 24 Tray inner width, inches
Calc. No. 96-ENG-01528E2 Rev. 00 A1TACHMENT A Page di of #8" This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray Qt := Q Allowable heat generation for a ladder type tray Qt = 13.137 d := 0.509 d is adjusted by pi over4 ,
Quntray ::Qt 24 d Allowed heat generation of a 24" wide by d i nches deep uncoveredi tray, Watts /ft (adjusted to circular area of cable)
Quntray = 160.482 Qcover : ACFcov 2-Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 55.864 Watts / ft Calculate the termal resistance at the tray surface Ts : 50 Initial guess -
Given 1
Qcover=0.402-( Ts - Ta )# + o A E-( ( Ts + 273.16 )# - ( Ta + 273.16 )# )
Ts := Find ( Ts )
Ts = 90.997 cable tray cover surface temperatre Tc - Ts Qmu R = -0.018 degC-ft/ watts
- - - _ _ _ ~ _ - - - _ _ - - - _ = _ - - _ _ - _ _ _ _ _ _ - _ - _ _ _ _ _ _ _ _ _ _ _ _ - _ - _ - _ _ _ - - _ - _ _ _ _ _ _ _ - _ _ _ _ - _ . _ - - _ - _ . _ - . _ - _ _ _ _ .- - - _ - -
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page AS of *'e Modelfor the TSI Encased Cable Tray bl ::Ttsi Top thickness of TSI, inches b2 ::Ttsi Side thickness of TSI, inches
(
b3 ::Ttsi Bottom thickness of TSI, inches b4 := Ttsi Side thickness of TSI, inches .
Calculate ThermalResistance of TSI I i Rk := i I I f I fel f
!fel + 0 54l + 1 e2 + 0 54 l + l - + 0 54 l + 1 e2 + 0 54 l
( bl .j (b2 ./ (b3 ./ (b4 .)
8 Rk = 0.326 degC-ft/ watts Calculate the Thermal resistance between the conductor and outer Surface of TSI Rws :: R + Rk Totalthermal resistance form i conductor to outer surface of TSI i
Rws = 0.308 i
__ _ _ _ _ _ _ _ _ _ _ . . __ __ __ _ _ _ _ _ - . . _ _ _ _ . _ _ _ _ . . _ _ _ _ _ _ _ _ _ . . _ _ . _ _ _ _ _ _ . . _ _ . _ _ _ . _ . _ _ _ . _ _ _ _ _ _ _ _ _ _ _ - _ _ . _ . _ _ _ _ _ _ _ . _ _ _ . . - - . _ _ _._______.__.____m_..__________. - _ _ _ _ _ _ _ _
Calc. No. 96-ENG-01528E2 Rev. 00 -
i UI ATTACHMENT A Sofve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96)
.t i '
f 3,8 1 0.25' fe2 + 2-bli 4
Al :: 0.27- l 12- l -i l 0.527 t( c2 + 2 blj ( 12 j Convection coefficient for the top surface of the Al = 0.294 tray I 1.8 3 0.25' d+2bl i A2:= 0.12 l 12- I 0.527 t ( e2 + 2 blj 12 .
A2 = 0.131 Convection coefficient for the bottom surface of the wrapped tray.
I 1.8 3 0.25 el + 2 bl A3 := 0.29 I 12- l
-2 0.527 )
( el + 2 blj (12)
A3 = 0.211 Cuwechon coefficsent for the side surfaces of the wrapped tray A1 + A2 + A3 = 0.636 Given i
5 Te - T, ~ Iel + e2 + bl + b21 !
Al + A2 + A3 )-( Ts - Ta)# + 5.3 10 ' mit I f -( ( Ts + 273.16 )# - (Ta + 273.16 )# )
Rws i 12 j Ts := Find ( Ts )
i Ts = 66.236 Wrapped tray surface temperature degC.
I i
[
4
h Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page A7 of MW Calculate Oc'and Or' j
-1 5
Heat transmission from surface of TSI to ambient Qac ( AI + A2 + A3 )-( Ts - Ta )4 air due to corvection, Watts /ft i Qcc = 20.716
-g Qir ::-0.53 10 -c 2 el + c2 + bl + b2 (( Ts + 273.16)4 - ( Ta + 273.16 )4 ) -
12 Qrc = -6.261 ,
f QTSI := 90 - Ts !
Rws l
QTSI = 77.061 i i
Calculate Derating Factor of tray covered with TSI fire wrap l t <
QTSI !
ACFtsi ::
l 1 Quntray ;
. t ACFtsi = 0.693 i
L I
! [
] !
i t
i P
______..--_.__m.__ _ . _ . _ _ _ . . _ _ _ . . _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ . _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ - _ _ _ _ . - _ _ _ _ _ _ . _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ . _ . _ _ _ _ _ _ _ _ _- _ _ _ _ _ _ __
COlc. No. 96-ENG-01528E2 Rev. 00 3HA10 MAX ATTACHMENT A ap O of M AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data input:
A a a peaW for a la@r tray N40 (Machnt 2)
Q :: 13.137 Ta ::50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc " W Surface area of tray, sqft/ft 24" wide x 11ong x top and bottom)
^ ~. ~_ 4
( Ref. 3.1.8) o :: 5.3 10~9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1) m n r for @ me mhy ACFcov ::0.59 (Ref. 3.1.10) ctsi :: 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.5 Thickness of TSI(maxernum) e1:=4 Tray inner height, inches e2 := 24 Tray inner width, inches
Calc. No. 96-ENG-01528E2 Rev. 00 89* U' ATTACHMENT A This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray Qi ::Q Allowable heat generation for a ladder type tray Qt = 13.137 d ::0.509 d is adjusted by piover 4.
Quntray ::Qt 24 d Allowed heat generation oof 24* wide and 3" deep uncovered fill tray, Watts /ft (adjusted to circular area of cable)
Quntray = 160.482 Qcover : ACFcov 2-Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 55.864 Watts / ft Calculate the termal resistance at the tray surface Ts : 50 Initial guess Given 5,,
Qcover=0.402-(Ts-Ta)# + o A c-(( Ts + 273.16)# - ( Ta + 273.16 )# )
Ts := Find ( Ts ) ;
Ts = 90.997 cable tray cover surface temperatre ,
R ._ Tc - Ts Qcover R = -0.018 W aus o
_ _ _ . . _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ - _ _ __..______.__.-_m_m.m._____m- ___,._m__.- __.___ m_.. . _ - - _ _ _ . _ ._ ___m__.__.__-__________ ________m_. _ _ _ _ _ . _ _ _ _ _ - - ___ ____
i Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page' AM of 49 i
Modelfor the TSI Encased Cable Tray bl ::Ttsi Top thickness of TSI, inches ,
b2 ::Ttsi Side thickness of TSI, inches ;
b3 ::Ttsi Bottom thickness of TSI. inches b4 ::Ttsi Side thickness of TSI, inches !
Calculate Thermal Resistance of TSI i
k 1.8.2928 (Ref. 3.1.95 )
Rk ': f f i f I I 1 Ic2 l el + 0.54 l + 1 e2 + 0.54 l + i el + 0.541 + 1 - + 0.54 !
(bl j (b2 / (b3 j (b4 / !
Rk = 0.48 degC-ft/ watts r Calculate the Thermal resistance between the conductor and outer Surface of TSI j l
Rws ::R + Rk Totalthemmiresstance form '
conductor to outer surface of TSI Rws = 0.463 -
i I
t 6
b i
a i
i
I Calc. No. 96-ENG-01528E2 Rev. 00 Page A81 of 4**
ATTACHMENT A Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) ;
f 1.8 3 0.25' i A1 ::0.27- i 12-e2 + 2 blj l - {fe212 + 2-bl
! 0.527
( ( )
n coehnt foN top sudace of t%
A1 = 0.303 wrapped tray l.8 3 0 25' e2 + 2 bl A2 := 0.527 0.12-(12- e2 + 2 bij l
12 A2 = 0.135 Convection coefficient for the bottom surface of the wrapped tray.
I 1.8 3 0.25 el + 2 bl I A3 : 0.291 12- l 2 0.527
( el + 2 blj ( 12)
A3 = 0.236 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.674 .
Given 2 fel + e2 + bl + b21 Tc - Ts d
( A + A2 + A3 )-(Ts - Ta)d + 5.3 10'9 usi-2-( j l -( ( Ts + 273.16 )4 -(Te me )
Rws a 12 Ts :: Find ( Ts )
Ts = 62.44 Wrapped tray surface temperature degC
m _. .._.
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page 41L of M #
Calculate Oc' and Or' 5
Heat transmissen from surface of TSI to amtwent Qx ::( Al + A2 + A3 )-( Ts- Ta )4 air due to convection,WattsMt Qoc = 15.739
-g i Qre : -0.53-10 c 2 el + e2 + bl + b2-( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
12 f Qrc = -4.871 l QTSI :: 90- Rw:,
Ts ;
QTSI = $9.578 i
1 Calculate Derating Factor of tray covered with TSI fire wrap QTSI '
ACFtsi :=
1 Quntray t
ACFtsi = 0.U/ ,
i t
i b
L t
i I
I 5
P u_____-._---_____.______--_-_____-__-____-______ _ _ _ _ ____- _ _ _____ _- - - . . . _ __ - ___ =
Calc. No. 96-ENG-01528E2 Rev. 00 Z23HB10 min
"" U' ATTACHMENT A AMPACITY MODELING OF APPENDIX R WRAPPED TRAY ;
r Data input:
A a at gemW for a lahray NB10 (Anach 2)
Q := 10.157 Ta:= 50 Ampient temperature MP2) !
r Conductor maximum allowed temperature !
Tc ': 90 Surface area of tray, sqft/ft (12" wide x 1'long x top and bottom) i A ._2
( Ref. 3.1.8) [
o :: 5.3 10~9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) t c := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
ACFcoy := 0.59 ^ " "" ' '9 # Y (Ref. 3.1.10) ,
Etsi::0.9 Emissivity of TSI surface (Attachment 1)
> k : 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.0 Thickness of TSI(minimum) ,
el := 4 Tray innerheight, inches c2 :: 12 Tray inner width, inches I
I i
+
i
t Calc. No. 96-ENG-01528E2 Rev. 00 ,
- ATTACHMENT A Page Ai+ of Aivo This sechon of the modelis for covered tray Calculation for the allowable heat generation in a covred tray
}
Qt : Q Allowable heat generation for a ladder type tray i
i t
Qt = 10.157 ,
d := 0.%8 d is adjusted by piover 4 i Quntray := Qt-12 d Allowed heat generation of a 24" wide by d i nches deep uncoveredi tray, Watts #t (adjusted to circular area of cable) :
Quntray = 117.984 :
2 Qcover := ACFcov -Quntray Allowable heat generation for a covered, unwrapped tray, watts #t l
Qcover = 41.07 Watts / ft i
Calculate the termal resistance at the tray surface I 2
Ts ::50 Initial guess I
Given i i
N.
Qcover=0.24-( Ts - Ta )4 + o A c-( ( Ts + 273.16 )# - ( Ta + 273.16 )# ) [
Ts :: Find ( Ts) i Ts = 100.164 cable tray cover surface temperatre R ._ Tc - Ts Qcover i i
R = -0.247 t degC-fthvatts ,
1 i
C11c. No. 96-ENG-01528E2 Rev. 00 Attachment A Page AtIof AWo 5
Modelfor the TSI Encased Cable Tray bl ::Ttsi Top thickness of TSI, inches b2 := Ttsi Side thickness of TSI, inches !
b3 := Ttsi Bottom thickness of TSI, inches i
b4 := Ttsi Side thickness of TSIinches !
Calculate Thermal Resistance of TSI 1
._ k 1.8 .2928 (Ref. 3.1.95 )
'~ '
\ fe2 y fel \ fc2 8 l fel + 0.54! + 1 - + 0.54 l +l - + 0.54 l + 1 - + 0.541 <
( bl j (b2 j (b3 ) (b4 ) :
Rk = 0.555 degC-ft/ watts '
Calculate the Thermal resistance between the conductor and outer Surface of TSI t F
l Rws ::R + Rk Totalthermal resistance form !
conductor to outer surface of TSI Rws = 0.308 f
I i
k
)
t
-- - - _ _ - . _. - _ - - - = _ _ _ _ _ _ _ _ _ _-- - __- __-- _ _-- _ _-- - - ___ _ ______-__ - _ - - - - _ . _ _ _ _ _- _ .- __ - -
_. _ . _ . _ . . _ . . _ _. _ _ ..._ _ .__ _ . . . _ _ . . . _ . . . . . . . ..m_._ _ __. _ . . _
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page Alk of Al" Solve Heat Balance Equation, to determine outer Surface Temperature of TSI e Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 1.8 3 0.25' fe2 + 2 bl i Al : 0.27- l 12- I ! J 0.527
( e2 + 2 blj ( 12 j I
Convection coefficient for the top surface of the A1 = 0.185 wrappesItray ;
f 1,8 1 0.25' c2 + 2 bi A2 := 0.12 i 12- 1 0.527
( e2 + 2 blj 12 A2 = 0.082 Convection coefficient for the bottom surface of the wrapped tray. >
f 1.8 3 0.25 el + 2 bl A3*:0.29-l 12- l 2 0.577 i el + 2 blj ( 12 )
A3 = 0.21' Convection coefficient for the side surfaces of the wrapped tray .
Al + A2 + A3 = 0.479 Given 5
Tc - Ts -9 fel + c2 + bl + b21 .
- ( Al + A2 + A3 )-( Ts - Tis)7 4- 5.3 10 ttsi 2-! l-( ( Ts + 273.16)4 - ( Ta + 273.16 )# ) i Rws ( 12 )
Ts := Find (Ts) ,
Ts = 70.346 Wrapped tray surface temperature degC.
_____2_.._________.____m__ _ _ _ _ _ _ . _ _ _ _ . . _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ , _ _ , . _ _ _ _ _ _ _ . . , _ _ _ _ . _
Calc. No. 96-ENG-01528E2 Pev. 00 ,
Attachment A Page Al7 of Al*
Calculate Oc' and Or' 5
at tensen fm suh MSI to aN Qec := ( Al + A2 + A3 )-( Ts - Ta ) air due to convection, Watts /ft Qcc = 20.645 l
-8 el + e2 + bl + b2 Qre := -0.53 10 c 2- -( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 ) "
12 Qic = -4.797 QTSI := 90- Ts l Rws QTSI = 63.82 ;
Calculate Derating Factor of tray covered with TSI fira wrap !
4 7
QTSI ACFtsi := [
1 Quntray ,
ACFtsi = 0.735 {
.?
t I
i t
I t
I e
I
_ _ _ _ _ - - _ _ _ _ . = _ _ . _ _ _ _ - _ _ - _ _ _ _ _ _ - _ _ - - - - - - -- - - -_ _ _ _ _ _ _ -- __ _ ___ __
Calc. No. 96-ENG-01528E2 Rev. 00 Z23HB10 MAX p,F p g, g% :
ATTACHMENT A AMPACITY MODELING OF APPENDIX R WRAPPED TRAY t
Data input aW MaWraWw a mdmy H2O (AMment 2) l Q: 10.157 Ta:=50 Ampient temperature MP2) ;
Conductor maximum allowed temperature Tc 'W ,
Surface area of tray, sqtt/ft 12% x 11ong x top and bottom)
A _2-( Ref. 3.1.8) o :: 5.3- 10' Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1) ,
A w fw caw tray j ACFcoy := 0.59 (Ref. 3.1.10) etsi := 0.9 Emissivity of TSI surface (Attavinnent 1) !
k := 0.1 Thermal conductnnty of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.5 Tikki-ss of TSI (mannum) cl := 4 Tray innerheight, inches e2 := 12 Trayinner width, inches i
i
CCic. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A age M of M This section of the modelis forcovered tray Calculation for the allowable heat generation in a covred tray Qt ::Q Allowable heat generation for a ladder type tray Qt = 10.157 d : 0.%8 d is adjusted by pi over 4.
Quntray := Qt 12 d Allowed heat generation oof 24* wide and 3* deep uncovered fill tray, Watts /ft (adjusted to circular area of cable)
Quntray = 117.984 2
Qcover ::ACFcov -Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 41.07 Watts / ft Calculate the termal resistance at the tray surface Ts:=50 Initial guess Given 1
Qcover=0.24-( Ts - Ta )# + o A c-( ( Ts + 273.16)4 - ( Ta + 273.16 )4 )
Ts := Find ( Ts )
Ts = 100.164 cable tray cover surface temperatre
._ Tc - Ts Qcover degC-It/ watts
Calc. No. 96-ENG-01528E2 Rev. 00 t Attachment A ap M of M i
' Model for the TSI Encased Cable Tray bl *:Ttsi Top thickness of TSI, inches b2 ::Tisi Side thickness of TSI, inches b3 := Ttsi Bottom thickness of TSI, inches b4 ::Ttsi Side thickness of TSI, inches Calculate Thermal Resistance of TSI 1
k 1.8.2928 (Ref. 3.1.95 )
Rk *: f f i I f I fe2 I 4 el + 0.54 l + l e2 + 0.541 + l el + 0.54 } +
l i - + 0.54 l (bl ) (b2 ) (b3 / (b4 /
I Rk = 0.808 degC-ft/ watts Calculate the Thermal resistance between the conductor and outer Surface of TSI t t
Rws := R + Rk Totalthermal resistance form !
conductor to outer surface of TSI f
Rws = 0.56 ,
k
?
t
{
r
. _ _ _ . _ _ _ _ _ _ _ . . _ _ _ _ _ _ _ _ _ - - - - - _ - - _ _ _ - - - - - _ _ _ _ _ ._ _ ..__ __ _.._._.._-__..L
Calc. No. 96-ENG-01528E2 Rev. 00 .
Page A24 of O a L ATTACHMENT 1 Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 1.8 3 0.25' fe2 + 2 bli A1 := 0.27- 1 12- l I l 0.527
( c2 + 2 blj ( 12 j Convection coefficient for the top surface of the A1 = 0.195 - wrapped tray I 1.8 1,0.25' c2 + 2 bl A2 := 0.12-l 12- I 0.527
( e2 + 2 bij 12 A2 = 0.087 Convection coefficient for the bottom surface of the wrapped tray. i f 1.8 3 0.25 el + 2 bl A3 ::0.29 i 12- j 2-0.527 i
, ( el + 2 blj ( 12 ) j A3 = 0.236 Convection coefficient for the side surfaces of the wrapped tray :
Al + A2 + A3 = 0.518 Given 5,
Tc - Ts 1 Al A2 + A3 )-( Ts- Ta)4 + 5.3 10'9 Rws Etsi 2- [el + e2 + bl + b2[( ( Ts + 273.16)4 - ( Ta + 27 1 12 Ts : Find (Ts)
Ts = 64.362 Wrapped tray surface temperature degC.
I e
. _ _ _ _ . . _ _ . . _.___.__._.1.__.___.m. _ _ _ _ _ _ _ _ . _ _ _ . _ . _ _ _ _ _ _ _ _ _ . _ _ _ _ _ ___m__ . _ __. - _ _ _ _ _ _ _ _ _ _ _ - _ . _ . - -
= , , _w. . .,-.v.~e , ... .n-_ ,-. .wm. , , , . , ..*..-,
- ~ . _ . _ . - . - - . _
i CCic. No. 96-ENG-01528E2 Rev. 00 j Attachment A Page AIS of **
Calculate Oc' and Or' 5
Heat transmission from surface of TSI to ambient air due to convechon, Watts /ft j Qcc != ( Al + A2 + A3 )-( Ts- Ta)
- Qcc = 14.475
-8 el + e2 + bl + b2 :
Qic :=-0.5310 c 2- -( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 ) !
12 Qre = -3.477 QISI := 90 - Ts ,
Rws i QTSI = 45.771 i
Calculate Deratng Factor of tray covered with TSI fire wrap I
ACFtsi ::
1 Quntray ACFtsi = 0.623 i
t l
t
Calc. No. 96-ENG-01528E2 Rev. 00 Z23GE10 min ATTACHMENT A age M of AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data Input A!! wa at genemWor a Mr kay Ma0 Wachnt 2)
Q ::5.161
~
Ta::50 An1pient temperature MP2)
Conductor maximum allowed temperature Tc ' %
Surface area of tray, sqft/ft (24" wide x 11ong x top and bottom)
^ ._
~~ 4
( Ref. 3.1.8) o :: 5.3- 10' Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
A coMm fador for @t er mMmy ACFcov := 0.59 (Ref. 3.1.10)
Etsi := 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.0 Thickness of TSI(minimum) e1:=4 Tray innerheight, inches e2 := 24 Tray inner width, inches
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A ap M of M This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray ;
i Qt := Q Allowable heat generation for a ladder typ tray Qt = 5.161 d := 1.12 d is adjusted by piover 4 Quntray ::Qt 24 d Allowed heat generation of a 24" wide by d i nches deep uncoveredi tray, Watts #t (adjusted to circular area of cable)
Quntray = 138.728 Qcover : ACFcov 2-Quntray Allowable heat generation for a covered, unwrapped tray, wattsMt ,
i
.Qcover = 48.291 Watts / ft !
Calculate the termal resistance at the tray surface Ts:=50 Initial guess ,
Given ,
j 5 j Qcover=0.402-( Ts - Ta )# + o A c-( ( Ts + 273.16 )# - ( Ta + 273.16 )# )
i
~
Ts := Find ( Ts )
L Ts = 86.422 cable tray cover surface temperatre i R .-
._ Tc - Ts
+
Qcover ;
R = 0.074 h
f
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page Mof #f*
Modelfor the TSI Encased Cable Tray i
bl : Tisi Top thickness of TSI, inches b2 ::Ttsi Side thickness of TSI, inches .
b3 ::Ttsi Bottom thickness of TSl, inches b4 ::Ttsi Side thickness of TSI, inches Calculate Thermal Resistance of TSI I !
k 1.8 .2928 (Ref. 3.1.95 )
Rk ~:
fel i f I f I fe2 i r
! - + 0.54 l + L e2 + 0.54l + l el + 0.54 j + l - + 0.54 l (b1 / (b2 ) (b3 ) (b4 /
Rk = 0.326 degC-fthratts ;
i Calculate the Thermal resistance between the conductor and outer Surface of TSI i
Rws :: R + Rk Totalthermal resistance form conductor to outer surface of TSI i
Rws = 0.4 i
b i
5
_ . _ . _ _ _ _ _ _ _ _ _ _ _ . . . . _ _ _ . _ _ . . _ _ . . . . . . . _ _ _ _ . _ _ _ . _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _. - - , _ _ . , . . _ _ _ _ _ _ _ , . _ _ . _ . _ . , . . _ ___.__,.___._.,._._..,__.l
Calc. No. 96-ENG-01528E2 Rev. 00 -
ATTACHMENT 1 A Page Mb of A
Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convecten and radiation coefficents for the TSI surface (Ref. 3.1.96) -
f 1.8 3 0.25' fd+2bd Al :: 0.27- l 12- l i
- t c2 + 2 blj ( 12 )l 0.527 n cwht for h top sWau of tM A1 = 0.294 wmppuu tray 3 0.25' '
r 3,g d + 2 bl A2 :: 0.12-l 12- ! 0.527
( e2 + 2 blj 12 A2 = 0.131 Convection coun&-4 for the bottom surface of the wrapped tray. ,
r g,g 3 0.25 el + 2 bl A3 ::0.29 I 12- -
2 0.527 >
( cl + 2 blj} (12) :
A3 = 0.211 Convecton coefficient for the side surfaces of the wrapped tray ,
Al + A2 + A3 = 0.636 Given 5
Tc - Ts ~9 el + e2 + hl + b21 Al A2 + A3)-(Ts - Ta) + 5.3 10 mid -(( Ts + 273.16)4 - (Ta + 273.16)4 )
Rws i 12
'I Ts :: Fmd(Ts) l Ts = 63.954 Wrapped tray surface temperature degC b
I
.. - -. . .. . . - . . - = _ _ . . - - ._~. . - . - . . . - - . - - - - . . - . _- ~. . . . - ._ .- .
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page AI7 of M Calculate Oc' and Of 5
at tensehm sudace d TSI to a@N Qec := ( Al + A2 + A3 )-( Ts - Ta ) air due to convection, WattsMt Qec = 17.143
~8 e Qre :=-0.53 10 -c 2 l + e2 + bl + b2-( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 ) -i 12 ;
i Qec = ~5324 t
QTSI := 90 - Ts Rws Q131 = 65.%3 ,
i Calculate Derating Factor of tray covered with TSI fire wrap QTSI ACFtsi :: :
Quntray ACFtsi = 0.685 !
t
-t f
i t
t i
_ _ _ _ _ . . . _ _ _ . _ _ _ . . . _ _ _ _ . . _ _ _ . _ . .__._.__._______._.__.__._.__.____._.________.m._m._.________.._.._____.._=___._m__._________.._m__________.____m. _
Calc. No. 96-ENG-01528E2 Rev. 00 E10 W ap M of M ATTACHMENT A AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data input at mmW for a mdmy N80 Wachnt 2) 4 Q ::5.161 Ta:=50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc ' %
Surface area of tray, sqft/ft (24" wide x 11ong x top and bottom)
^ ._
'- 4
( Ref. 3.1.8) o :: 5.3 10-9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
A coM% fadodo@ m mMmy ACFcov := 0.59 (Ref. 3.1.10) etsi := 0.9 Emissivity of TSI surface (AttaGment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.5 Thickness of TSI(maximum) el::4 Tray inner height, inches e2 := 24 Tray inner width, inches
Calc. No. 96-ENG-01528E2 Rev. 00 i
ATTACHMENT A Page Alf of A*
This section of the model is for covered tray.
Calculation for the allowable heat generation in a covred tray Qt := Q Allowable heat generation for a ladder type tray Qt = 5.161 d := 1 12 d is adjusted by piover4.
Quntray := Qt 24 d Allowed heat generation oof 24* wide and 3* deep uncovered fill tray, Watts #t (adjusted to circular area of cable)
Quntray = 138.728 :
2 Qcover ::ACFcov -Quntray Allowable heat generation for a covered, unwrapped tray, watts #t Qcover = 48.291 Watts / ft t
Calculate the terrnal resistance at the tray surface Ts := 50 Initial guess Given 5
I Qcover=0.402-( Ts - Ta ) + a- A c-( ( Ts + 273.16 )# - ( Ta + 273.16 )# )
Ts := Find (Ts)
Ts = 86.422 cable tray cover surface ternperatre
. R .-._ Tc - Ts Qcover
= 0.074 degC-ft/ watts I
~ _ - _ _ - . _ _ _ _ _ _ _ _ _ - _ _ - _ _ _ _ . . _ _ _ - - _ _ - _ _ _ _ . - - - . __ _ - _ - _ _ _ . - - - _ - _ _ _ - - _ _ - - _ - - - . _ - _ - _ _ _ - _ _ - _ _ - _ _ _ _ _ - - _ _ _ - . _ _ - - _ _ - - - _ - - . - - - - _ - _ _ _ - . _ _ _ - - _ . - _ _ - _ .
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page A9 of #*
Model for the TSI Encased Cable Tray l bl ::Ttsi Top thickness of TSI, inches ,
b2 ::Ttsi Side thickness of TSI, inches b3 ::Ttsi Bottom thickness of TSI, inches b4 ::Ttsi Side thickness of TSI, inches Calculate Thermal Resistance of TSI !
l
._ k 1.8.2928 (Ref.3.1.95 ) t i f I f I fe2 I l fel + 0.541 + ie2 + 0.541 + ! el + 0.541 +
- l - + 0.54 l i (bl j '(b2 / (b3 / (b4 /
Rk = 0.48 degC-ft/ watts :
Calculate the Therrnal resistance between the conductor and outer Surface of TSI +
1 Rws :: R + Rk Totalthermalresistance fonn ;
conductor to outer surface of TSI ,
Rws = 0.555 l
l
?
l I
I
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A ap $ of M Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref.3.1.96) f 1.8 y 0.25' .fe2 + 2 bli Al := 0.27- 1 12- I I l 0.527 i e2 + 2 blj ( 12 j o on sda of Al = 0.303 the wrapped tray -
f 1.8 3 0.25' e2 + 2 bl A2:: 0.12-{ 12- 1 0.527
( e2 + 2 blj 12 A2 = 0.135 Convection coefficient for the bottom surface of the wrapped tray.
f 1.8 1 025 el + 2 bl A3 := 0.29- l 12- l 2 0.527
( el + 2 blj ( 12 )
A3 = 0.236 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.674 Given 5
Tc - Ts ~9
( A + A2 + A3 )-( Ts- Ta)4 + 5.3- 10 u si 2- fel + e2 + hl + b21 -(( Ts + 273.16 - ( Ta +)4273.16 )4 )
Rws i 12 /
Ts := Find (Ts)
Ts = 61.042 Wrapped tray surface temperature degC.
i f
= _ __ . _ . _ _. . . _ _ . _ _ . _ _ _ _ _ _ . _ . .
_ ________.___.____________,___________m_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ . . _ _ _ _ _ _ . _ . . _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ , _ _ _ _ _ _ _ _ _ _
CCic. No. 96-ENG-01528E2 Rev. 00 Attachment A Page 4Rof 4f*
- t Calculate Oc' and Or' E Heat transmission from surface of TSI to ambient Qr := ( Al + A2 + A3 )-( Ts - Ta )4 air due to convection, WattsMt i Or = 13.56 I
Oc :=-0.53 10'8 c 2 el + e2 + bl + b2 (( Ts + 273.16)4 - ( Ta + 273.16 )4 )
12 i
Qc = -4.2%
I
] QT3I := 90- Rws Ts Q131 = 52.221 Calculate Derating Factor of tray covered with TSI fire wrap I
QTSI ACFtsi :: ,
1 Quntray ;
ACFtsi = 0.614 I i F
f 1
i i
i
Calc. No. 96-ENG-01528E2 Rev. 00 -
Z14FM10 min ATTACHMENT A age W of h AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data input A a pm r a lah tmy ZM10 (Anachment 2)
Q := 10.o61 Ta ::50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc 3 90 Surface area of tray, sqft/ft (24%ide x 1'long x top and bottom)
^ ._
~~ 4
( Ref. 3.1.8) o :: 5.3-10-9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c := 0.1 Emessmty of galvanized steel tray cover (Ref. 3.1.1)
ACFcov := 0.59
^ 9 " "Y (Ref. 3.1.10) ctsi :: 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.0 Thickness of TSI(rrunemum) el::4 Tray innerheight, inches e2 ::24 Trayinnerwidth, inches I
I
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A age & of M This section of the model is for co, 3 red tray Calculation for the allowable heat generation in a covred tray Qt ::Q Allowable heat generation W a ladder type tray Qt = 10.661 d := 0.611 d is adjusted by piover 4 Quntray ::Qt-24 d Allowed heat generation of a 24" wide by d i nches deep uncoveredl tray, Watts /ft (adjusted to circular area of cable)
Quntray = 156.333 Qcover : ACFcov 2-Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 54.419 Watts / ft
~
Calculate the termal resistance at the tray surface ;
Ts := 50 Initial guess Given 5
Qcover=0.402-( Ts - Ta )4 + & A c-( ( Ts + 273.16)4 - ( Ta + 273.16 )4 )
Ts : Find (Ts)
I Ts = 90.136 cable tray cover surface temperatre Tc - Ts Qcowr -
R = 1002 degC-ft/ watts .
I
_ _ _ _ . _ . _ _ _ _ _ _ _ . . _ _ . , _______.m__._ _ _ . . . _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ . _ _ _ _ _ _ _ _ _ . _ . _ _ _ _ . _ . _ _ _ _ _ _ _ _ _ . -__ m _ _ _ _ _ . - _ _ _ _ _ _ _ _
r
- Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page Mof AIV*
Modelfor the TSI Encased Cable Tray bl : Tisi Top thickness of TSI, inches b2 ::Tisi Side thickness of TSI, inches b3 ::Ttsi Bottom thickness of TSI, inches b4 ::Ttsi Side thickness of TSI, inches Calculate Thermal Resistance of TSI 1
Rk := i f i f i fe I fel + 0 54l + l e2 + 0 54 i + 1 el + 0 54 l + l 2 + 0 54 l l - . .
(bl j (b2 .) Ab3 .) (b4 /
Rk = 9.326 degC-ft/ watts Calculate the Thermal resistance between the conductor and outer Surface of TSI Rws :: R + Rk Totalthermal resistance form conductor to outer surface of TSI Rws = 0.324
______....____.._-___._._.._._.___m.._. ._.._
CElc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page M of Al*
Solve Heat Balance Equation, to determine outer Surface Ternperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 3,g 3 0.25' rd + 2 bl i AI ::0.27- l 12- 1 -l l-0.527
( c2 + 2 blj ( 12 /
Convection coefficient for the top surface of the Al = 0.294 wrapped tray I 1.8 3 0.25' e2 + 2 b1 A2 :: 0.12-l 12- 1 0.527
( e2 + 2 blj 12 A2 = 0.131 Convection coefficient for the bottom surface of the wrapped tray.
I I.8 3 0.25 el + 2 bl A3 := 0.29- l 12- 1 -2 0.527
( el + 2 blj (12) t A3 = 0.211 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.636 Given 5
~9 ' *
=( Al + A2 + A3 )-( Ts - Ta ) + 5.3- 10 -Etsi 2- l-(( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
Rws ( 12 j ,
Ts : Fiml(Ts)
I Ts = 65.8 Wrapped tray surface temperature degC
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment 1 A Page M7 of elfo i Calculate Oc' and Of 5
at tensdssbn fm suh OUSl to a@d Qir := { Al + A2 + A3 )-( Ts - Ta ) air due to convection, Watts /ft +
Qec = 20.025
~8 el + e2 + bl + b2 Qre :=-0.53 10 c 2- -( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 ) .
12 l Qre = -6.081 QTSI := 90- Ts Rws Q13I = 74.75 Calculate Derating Factor of tray covered with TSI fire wrap f
ACFtsi := '
1 Quntray ACFtsi = 0.691 i
i i
l i
L h
Calc. No. 96-ENG-01528E2 Rev. 00 1 10 MAX Page 436 of /N ATTACHMENT A AMPACITY MODEUNG OF APPENDIX R WRAPPED TRAY Data Input-Allowable heat generated for a ladder tray Z14FM10 (Attachment 2)
Q := 10.661 Ta := 50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc ~ W
- Surface area of tray, sqft/ft (24" wide x 11ong x top and bottom)
A ._4 ~~
( Ref. 3.1.8) o := 5.3 10' Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c := 0.1 Emissmty of galvanized steel tray cover (Ref. 3.1.1)
ACFcov := 0.59
"" Y (Ref. 3.1.10) etsi := 0.9 Ermssmty of TSI surface (Attachment 1) k := 0.1 Thermal conductmty of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.5 TI-A,;u=ss of TSI (maximum) el:=4- Tray inner height, inches e2 := 24 Tray inner width, inches 4
I
C alc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A age M of %
This section of the model is for covered tray Caledation for the allowable heat generation in a covred tray Qt := Q Allowable heat generation for a ladder type tray Qt = 10.661 d := 0.61I d is adjusted by pi over 4.
Quntray : Qt-24 d Allowed heat generation oof 24" wide and 3" deep uncovered fill tray, Watts /ft (adjusted to circular area of cable)
Quntray = 156.333 2
Qcover ::ACFcov -Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 54.419 Watts / ft Calculate the termal resistance at the tray surface Ts := 50 Initial guess Given 1
Qcover=0.402-( Ts - Ta)4 + o A c-(( Ts + 273.16)4 - ( Ta + 273.16 )4 )
Ts := Find ( Ts)
Ts = 90.136 cable tray cover surface temperatre
~
- R ::
Qcover R = 1.002 degC-ft/ watts
Calc. No. 96-ENG-01528E2 Rev. 00 "98 I Attachment A Model for the TSI Encased Cable Tray bl :: Ttsi Top thickness of TSI, inches b2 ::Ttsi Side thickness of TSI, inches b3 :=Ttsi Bottom thickness of TSI, inches b4 ::Ttsi Side thickness of TSI, inches Calculate Thermal Resistance of TSI I
Rk := i i f f f I I fel + 0 54 + l e2 + 0.54 l + l el + 0.54 l + l e2 + 0.541 (bl ib2 / (b3 ; (b4 j Rk = 0.48 degC-ft/ watts Calculate the Thermal resistance between the conductor and outer Surface of TSI Rws :: R + Rk - Total thermal resistance form conductor to outer surface of TSI Rws = 0.478
.m_ . . _ _- . . . - _ . . - . _ _ _ _ _ . - - - _ _ . _ . . _ . .-.____.s __.-_.__-.m._____.__.---m m._m_.._._m__--___________.__m-_ .-.-_m____--.a.______... -_ _ _ . - - . _ _ _ _ - _ _ . _ _ . - __ _ _ _ _ _ _ _ _ _ _ _ .
- _ _ _ _ . _ _ _ _um___...r._-m_m
Calc. No. 96-ENG-01528E2 Rev. 00 Page A4 of Affe ATTACHMENT A Solve Heat Balance Equation, to determine outer Surface Ternperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96)
I I.8 I.0.25' fe2 + 2 bli Al :: 0.27- ! 12- l -l l 0.527
( e2 + 2 blj ( 12 )
m n coehnt for tM toop surface of Al = 0.303 the wrapped tray f
1.8 3 0.25' e2 + 2 bl A2 := 0.12 1 12- ) 0.527
( c2 + 2 blj 12 A2 = 0.135 Convection coefficient for the bottom surface of the wrapped tray.
A3 := 0.29 J 12- I 2 0.527
( el + 2 blj ( 12 )
A3 = 0.236 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.674 Given 5
Tc - Ts -9 fel + e2 + bl + b21
=( AI + A2 + A3 )-( Ts - Ta)4 + 5.3 10 - etsi-2 1 j -( ( Ts + 273.16 )# - ( Ta + 273.16 )4 )
Rws ( 12 j Ts := Find ( Ts )
Ts = 62.181 Wrapped tray surface temperature degC.
_ _ _ _ - - _ _ _ _ - - - _ _ _ _ - - _ _ _ - - _ _ _ _ _ _ _ _ _ - - _ _ _ - _ _ - - - _ _ - _ - _ _ _ _ _ - - _ . - _ _ - _ - _ _ _ _ - - _ _ _ _ _ _ _ _ _ _ = _ - _ . .
_-. . .... . .. .- - . . - _ - - ~ . - . . . .-_ . - ...- . . -- . . . --
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page M of 4f4p:
Calculate Oc' and Or' 1 Heat transmisson from surface of TSI to amtnent Qx := ( Al + A2 + A3 )-( Ts - Ta)4 air due to convection, Watts /ft Qx = 15.331
-8 .
Qre :=-0.5310 c 2 ei + e2 + bl + b2-( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
12 Qic = -4.764 QTSI := 90- Ts Rws QTSI = 58.206 Calculate Derating Factor of tray covered with TSI fire wrap ACFtsi :=
1 Quntray ACFtsi = 0.61 1
_ _ _ . _ _ _ _ . _ _ m. _. -
_ _ _ _ . . _ _ _ _.._...__.m- _ _ _ _ _ _______..m .._._.--_________m_m_--_m. . - _ _ .____.--m-_._._ _ _ _ __._ ___ ____m___..._ . - . _ _ _ ____ _____.-_a_2_______ __ .______.______m_ __.______m
- - . . . . . . .-. . - .~ .-- . . - . - ~ . - . . . - . .. . - , _ . ~ . - - - . . - . . . . - _ .
Calc. No. 96-ENG-01528E2 Rev. 00 Z14FM20 min i ATTACHMENT A age M of M i AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data input: l
._ g g97 Allowable heat generated for a ladder tray Z14FM20 (Attachment 2)
Ta := 50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc' 90 Surface area of tray, sqft/ft (24" wide x 1'long x top and bottom)
A ._4
( Ref. 3.1.8) i
-9 o := 5.3- 10 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) i e := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
A m n fador fo@t me mW tmy ACFcov := 0.59 (Ref. 3.1.10) ctsi := 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft4egF (Attachment 1)
Tisi := 1.0 Thickness of TSI(minimum) el := 4 Tray innerheight, inches c2 := 24 Tray inner width, inches b
4
._ -__ .__ -_ _-_ ____- - - - _ _ - - - . ~ .. .. . . . . . . _-
Cele. No. 96-ENG-01528E2 Rev. 00 ,
ATTACHMENT 1 A ap N of M This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray Qt := Q Allowable heat generation for a ladder type tray i
Qt = 14.797 d := 0.458 d is adjusted by piover 4 Quntray := Qt 24 d Allowed heat generation of a 24* wide by d i nches deep uncoveredi tray, Watts /ft (adjusted to circular area of cable) ,
Quntray = 162.649 2
Qcover := ACFcov -Quntray Allowable heat generation for a covered, urmiapped tray, watts /ft Qcover = 56.618 Watts / ft Calculate the termal resistance at the tray surface Ts ::50 Initial guess Given N.
I Qcover=0.402-(Ts- Te)# + o A e(( Ts + 273.16)# - ( Ta + 273.16 )4 )
Ts : Find (Ts) s Ts = 91.445 cable tray cover surface temperatre R .- ._ T: - Ts -
Qcover ,
R = -0.026 h ans
.-._.,_._____.___--__-m_. _ . _ _ . . _ ___.-_.u_.m .____.-_..._.________.________m_ - . _ . - _ _ . . - - - -
CCdc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page Nof 48*
Modelfor the TSI Encased Cable Tray bl ::Ttsi Top thickness of TSI, inches b2 ::Ttsi Side thickness of TSI, inches ,
, b3 ::Ttsi Bottom thickness of TSI, inches b4 ::Ttsi Side thickness of TSI, inches Calculate Thermal Resistance of TSI I l
- 1. @ef.3.1.95 ) t Rk :: i i f i i fel fe2 fe2 l - + 0.54 l + l - + 0.541 + 1 el + 0.54 l + ! - + 0.54 j (bl j (b2 j (b3 / (b4 j ,
Rk = 0.326 degC-ft/ watts I
l Calculate the Thermal resistance between the conductor and outer Surface of TSI Rws ::R + Rk Totalthermal resistance form :
conductor to outer surface of TSI [
Rws = 0.301 f 8
t i
l
]
- - - ~ . _ - - _ - - _ _ _ . _ _ _ . - - . _ _ _ - - - _- _ _ - _ - _ - - - - _ - _ _ - - _ - - - _ _ _ - _ - _ _ - - - - ~ - _ -
---_-----_n-_
Calc. No. 96-ENG-01528E2 Rev. 00 Page A4 of AfW ATTACHMENT A i
Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) -
f 0.25' fe2 + 2-blI i 1.8 Al : 0.27- l 12- I i 0.527 i e2 + 2 bl/ 1 12 j Convection coefficient for the top surface of the Al = 0.294 wrapped tray
- f 1.8 3 0.25' d + 2 bl A2 := 0.12- l 12- 1 -0.527
( c2 + 2 blj 12
- A2 = 0.131 Convection coefficient for the bottom surface of the wrapped tray.
2 0.527 A3 := 0.29 ! 12-el+2blj)
( ( 12 )
A3 = 0.211 Convection coefhcient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.636 Givm 5
~9
-( A1 + A2 + A3 )-( Ts - Ta ) + 5.3 10 ctsi i -( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
Ts := Find (Ts)
Ts = 66.463 Wrapped tray surface temperature degC.
i
. - . - . - . _ . - . . - - _ . . - - - . - - _ - _ - - - - _ _ . _ _ . _ - - _ . _ - . . _ . _ _ . _ - - - - _ _ _ _ _ _ _ . . _ _ _ . _ _ _ _ - _ - _ . _ . _ _ _ _ _ _ - _ _ _ _ - _ _ _ - - _ - _ - _ _ . - - _ . ~ - - -
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page M7 of AW Calculate Oc' and Or' -
5 a mnse f.m suda of TSI to aN Qec :=( Al + A2 + A3 )-( Ts- Ta ) air due to convection,WattsMt Qx = 21.079
-8 Qic :=-0.53 10 -c-2 el + e2 + bl + b2 (( Ts + 273.16 )4- ( Ta + 273.16 )4 )
12 Qre = -6.355 90 - Ts QTSI ::
Rws QISI = 78.273 Calculate Derating Factor of tray covered with TSI fire wrap ACFtsi :=
1 Quntray ACFisi = 0.694
Calc. No. 96-ENG-01528E2 Rev. 00 14 20 MAX ap M of M ATTACHMENT A AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data input flowable heat generated for a ladder tray Z14FM20-1 (Attachment 9 ._ ,4g Ta := 50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc " 90 Surface area of tray, sqft/ft (24" wide x 1'long x top and bottom)
^ ._
'- 4
( Ref. 3.1.8) o := 5.3 10-9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
AmcWWbn MododgM cow @ tmy ACFcov := 0.59 (Ref. 3.1.10) etsi := 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.5 Thickness of TSI(maximum) .
el := 4 Tray inner height, inches c2 := 24 Tray inner width, inches
! Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page Mi of 4'@ [
I This section of the model is for covered tray Calculation for the allowable heat vermistion in a covred tray [
i 4
Qt: Q Allowable heat generation for a ladder type tray Qt = 14.797 ,
l d := 0.458 d is adjusted by piover4. ;
i h
Quntray := Qt 24 d Allowed heat generation oof 24" wide and 3* deep uncovered fill tray, Watts /ft (adjusted to circular area of cable) l Quntray = 162.649 f l
2 Qcover ::ACFcov -Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft [
t i'
Qcover = 56.618 Watts / ft f
Calculate the termal resistance at the tray surface j Ts ::50 Initial guess ;
I Given ,
a l 5 l Qcover=0.4024 Ts - Ta ) + o A c-((Ts + 273.16)4 - ( Ta + 273.16 )4 ) !
Ts := Find (Ts)
Ts = 91.445 cable tray cover surface temperatre E
R .-
._ Tc - Ts r i
Qcover R = -0.026 degC-ft/ watts l
)
w --i
.mm_________. _ --%.- __ -- _--,
Calc. No. 96-ENG-01528E2 Rev. 00 ;
Attachment A Page 360 of d'* - ;
Modelfor the TSI Encased Cable Tray bl ::Ttsi Top thickness of TSI, inches b2 :=Ttsi Side thickness of TSI, inches b3 :=Ttsi Bottom thickness of TSI, inches M :: R si Side thickness of TSI, inches Calculate Thermal Resistance of TSI I .
t k- 1.8 .2928 (Ref.3.1.95 ) ;
f i f i f I f i L l el + 0.54 l + l e2 + 0.541 + l el + 0.54 j + l c2 + 0.541 !
(bl j (b2 / (b3 ) (M J- !
Rk = 0.48 degC-ft/ watts ;
r i
Calculate the Thermal resistance between the conductor and outer Surface of TSI !
t Rws ::R + Rk Totalthermal resistance form :
conductor to outer surface of TSI }
Rws = 0.455 h
i l i i
4
_ _ . . ._. __.._--m.___.-...m.-m-_ -
.__~____-____.m..__.._~__mm_.__-_.______m.________m__._ ___.___m____.__ m.m ____ --- _.__________m.- -
_-- _ . . _ . - ....m... . . _ - . _ _ - - . . _ _ . . _ - _ . . . _ - - . . _ . _ - . . - _ . . _ = - . - ._ _ . . - .- _ . . . . _.._ _ . _ - - ..- . . . - _ . .
CCic. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page M of At* -
Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref 3.1.96) f 1.8 3 0.25' fd + 2 bli AI : 0.27- l 12- 1 -l l 0.527
( e2 + 2 blj ( 12 j n a coe t to p sudace of h Al = 0 303 wrapped tray f 1.8 3 0.25' d + 2 bl A2 := 0.12 I 12- l, -0.527
( e2 + 2 bl/ 12 A2 = 0.135 Convection coefficient for the bottom surface of the wrapped tray. ,
I 1.8 3 025 el + 2-b1 A3 :: 0.29- l 12- I 2 0.527
( el + 2 blj ( 12 )
A3 = 0.236 Convection coefficsent for the side surfaces of the wrapped tray Al + A2 + A3 = 0.674 !
Given 3
Tc - Ts 1 Al + A2 + A3 )-( Ts - Ta)4 + 5.3- 10~9 Etsi 2-(Iel + e2 + bl +l-((Ts b2 + 273.16)4 - ( Ta + 273.16)# ) !
Rws i 12 j 1 I'
Ts := Find ( Ts)
, Ts = 62.574 Wrapped tray surface temperature degC 1 L I
t i
t t
t l
. _ _ _ _ . . _ _ _ _ _ - . . _ _ _ . _ . . . . ._ _ . . _ . - _ - . _ _ . _ _ . . _ . _ . _ , _ . . _ . _ _ _ . _ _ _ - ~ . _ - - - . - - _ _
._..._ . . . . . . - . . _ . _ _ .. .._ . _ . . _ _ . _ . _ . _ _ . __.m._ . ..- ... . ._ . - .
Calc. No. 96-ENG-01528E2 Rev. 00 ;
Attachment A Page M of ANo Calculate Oc' and Ot' s 5
Heat transmission from surface of TSI to ambient ,
Qcc ::( Al + A2 + A3 )-(Ts- Ta)4 air due to convection,Wattsift [
Qtr = 15.951 ;
~3 '
Qre ::-0.5310 c 2 el + e2 + bl + b2 (( Ts + 273.16)4 - ( Ta + 273.16 )4 )
{
12 .
Qre = -4.926 l t
QTSI := 90 - Ts Rws l QTSI = 60.289 ;
\
Calculate Derating Factor of tray covered with TSI fire wrap ,
i' QTSI ACFtsi := i 1 Quntray l I
ACFtsi = 0.609 i
i I
4
[
t i
i i
i-
1 Calc. No. 96-ENG-01528E2 Rev. 00 Z24FL10 min ATTACHMENT A ap M of M AMPACITY MODELING OF APPENDIX R WRAPPED TRAY ;
Data input ;
A at genemtWor a W tmy mMO (AMant 2)
Q := 35.701 Ta ::50 Ampient temperature MP2)
Conductor maximum allowed temperature j Tc":90 Surface area of tray, sqft/ft (24W x 11ong x top and bottom)
^ ._
'- 4
( Ref. 3.1.8) a : 5.3 10~9 Stefan- Boltzmann constant, Wats /sqft K^4 (Ref. 3.1.27) ,
e := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
- "#" "Y ACFcov := 0.59 (Ref. 3.1.10) ctsi :: 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.0 Thickness of TSI(min'msn) et:=4 Tray innerheight, inches e2 := 24 Tray inner width, inches ,
i
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page A-Sf_ og At*
This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray Qt ::Q Allowable heat generation for a ladder type tray s
Qt = 35.701 d := 0.2037 d is adjusted by piover 4 Quntray ::Qt 24 d Allowed heat generation of a 24* wide by d i nches deep uncoveredi tray, Watts /ft (adjusted to circular area of cable) ,
Quntray = 174.535 2
Qcover ::ACFcov -Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 60.756 Watts / ft i
Calculate the termal resistance at the tray surface Ts : 50 Initialguess Given 1
Qcowr=0.402-( Ts - Ta )4 + o A c-(( Ts + 273.16)4 - ( Ta + 273.16 )4 )
Ts := Find ( Ts) ,
Ts = 93.88 cable tray cover surface temperatre Tc - Ts QCow R = 1N degC-ft/ watts
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page A55 of A*
Modelfor the TSI Encased Cable Tray bl ::Ttsi Top thickness of TSI, inches b2 : Ttsi Side thickness of TSI, inches i
b3 ::Ttsi Bottom thickness of TSI, inches b4 :=Ttsi Side thickness of TSI, inches ,
Calculate Thermal Resistance of TSI I
' ' ~
Rk: i i f I i fel fe2 l - + 0.54 l + l fe2 + 0.54 l + 1 el + 0.54 l + i - + 0.541 (bl ) ( b2 ) (b3 ) (b4 )
r Rk = 0.326 degC-ft/ Watts t
Calculate the Thermal resistance between the conductor and outer Surface of TSI Rws :: R + Rk Totalthermal resistance form conductor to outer surface of TSI Rws = 0.262 l
i b
I P
- - _ - - . _ _ . . . - _ _ _- _ _ . . - - . - --..-____-.w.-----._____ _ _ _ _ _ . _ - - _ _ - _ _ . - _ _ _ _ _ . _ _ _ _ _ _ _ - -.--_-_-__---_--__-x----.--_-__-_-__ _ - _ _ _ . -- - _ _--.____.--_- - _ - _ _ _ . _ _ - _ . _ ____ - _ - -_- _ - _ _ . - _ - ._ _ -
CCic. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page _O__ of A84*
Solve Heat Balance Equaticn, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96)
I 0 25' 1.8 3 re2 + 2 bli Al :: 0.27- i 12- ! i 1 0.527
( e2 + 2 blj ( 12 j Convection coefficknt for the top suiface of the Al = 0.294 wrapped tray f 1.8 y 0.25' e2 + 2 bl A2 := 0.12- l 12- 0.527 ,
t
( e2 + 2 blj 12 A2 = 0.131 Convection coefficient for the bottom surface of the wrapped tray. ,
I 1.8 3 0.25 el + 2 bl A3 : 0.29- l 12- l 2-0.527
( el + 2 blj ( 12 )
A3 = 0.211 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.636 Given i i
1 Iel + e2 + bl + b21 Tc - Ts AI + A2 + A3 )-( Ts - Ta )4 + 5.3- 10' etsi 2- ( -( ( Ts + 273.16 )4 - ( Ta + 273.I6 )4 )
Rws i 12 Ts := Find ( Ts ) l Ts = 67.708 Wrapped tray surface temperature degC.
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page Nof 48# !
I Calculate Oc'and Of 5
at mnsmisse fmm suh ORSI to a@nt Qx := ( Al + A2 + A3 )-( Ts - Ta ) air due to convection, Watts /ft -
Qcc = 23.091 t
Qre :=-0.53- 10'8 c-2 el +12- e2 + bl + b2 (( Ts + 273.16)4 - ( Ta + 273.16 )4 )
i Qrc = -6.875 ;
QTSI :: 90- Ts Rws .[
QTSI = 84.963 j
t Calculate Derating Factor of tray covered with TSI fire wrap ACFtsi :=
1 Quntray ACFtsi = 0.698 l
j t I
1
Calc. No. 96-ENG41528E2 Rev. 00 Z24FL10 MAX ~
ATTACHMENT A age M of F AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data Input Q := 35.701 mm f r a la@r tray MMO (Ament 2)
Ta::50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc *:90 udace am of tray, @ (24% x 1bng x top aM Mom)
A -
4
( Ref. 3.1.8) a :: 5.3 10-9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) e := 0.I Emissivity of galvanized steel tmy cover (Ref. 3.1.1)
A co or fo@ cow caw imy ACFcov := 0.59 (Ref. 3.1.10) ctsi :: 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductmty of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.5 Thickness of TSI (minimum) el::4 Tray innerheight, inches e2 := 24 Trayinner width, inches
. . . . _ _ _ _ - _ _ - - _ _ _ - _ _ _ - _ _ _ _ = _
_ _ _ - - ~ - . . . . _ _ _ _ _ _ - - _ - - - - - - - - _ - __ - -_ _ _ - _ . . - - - _ - _ . _ _
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page AST of 4/*
This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray Qt := Q Allowab'e heat generation for a ladder type tray Qt = 35.701 d := 0.2037 d is adjusted by pi over4 Quntray := Qt 24 'd Allowed heat generation of a 24" wide by d i nches deep uncoveredi tray, WattsMt (adjusted to circular area of cable)
Quntray = 174.535 Qcover ::ACFcov 2-Quntray Allowable heat generation for a covered, unwrapped tray, wattsat Qcover = 60.756 Watts / ft Calculate the terma! resistance at the tray surface Ts:: 50 Initial guess Given 5
Qcover=0.402-( Ts - Ta )# + o A c-( ( Ts 4- 273.16 )# - ( Ta + 273.16 )# )
Ts :: Find ( Ts)
Ts = 93.88 cable tray cover surface temperatre Tc - Ts Qcover R = 1N degC-ft/ watts
CCic. No. 96-ENG-01528E2 Rev. 00 Attachment A Page M ' of A@
Model for the TSI Encased Cable Tray bl := Ttsi Top thickness of TSI, inches b2 ::Ttsi Side thickness of TSI, inches b3 ::Ttsi Bottom thickness of TSI, inches b4 := Ttsi Side thickness of TSI, inches Calculate Thermal Resistance of TSI 1
Rk:=
f i f I f I f i l el + 0.54 l + i e2 + 0.54l + j( b3el + 0.541e2++l 0.54l (bl j '(b2 / ) '(b4 )
. Rk = 0.48 degC-ft/ watts Calculate the Thermal resis*ance between the conductor and outer Surface of TSI Rws :: R + Rk Totalthermal resistance form conductor to outer surface of TSI Rws = 0.417 r
I
m___ _. - _ _ _ _ . _ _ . _ . .
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page A4.of M '
Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96)
I f 1.8 1.0.25' fc2 + 2 bli A1 ::0.27- l 12- I l 1-0.527 i e2 + 2 blj ( 12 /
for t p sudace of tM !
Al = 0.303 wrapped tray ;
l T g,g 3 0.25' e2 + 2 bl A2 := 0.12-I 12- 0.527
- i e2 + 2 blj 12 A2 = 0.135 Convection coefficient for the bottom surface of the wrapped tray.
I 1.8 3 0.25 el + 2 bl A3 :: 0.29- l 12- I
-2 0.527 i el + 2 blj (12)
A3 = 0.236 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.674 Given 5
- ~
'r( Al + A2 + A3 )-( Ts - Ta ) + 5.3- 10'-ctsi-2 ! 12 j
-( ( Ts + 273.16 )# - ( Ta + 273.16 )# )
Rws (
Ts := Find ( Ts )
Ts = 63.291 Wrapped tray surface temperature degC.
5
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page Mof M Calculate Oc' and Or' 5
Heat transmission from surface of TSI to amtnent Qec := ( Al + A2 + A3 )-( Ts - Ta )4 air due to convechon, Watts /ft Qec = 17.096 g e t Qre :=-0.53 10 c 2 l + e2 + bl + b2 (( Ts + 273.16)4 - ( Ta + 273.16 )4 ) -
12 Qre = -5.225 !
QTSI := 90- Ts Rws QTSI = 64.117 Calculate Derating Factor of tray covered with TSI fire wrap ACFtsi :=
1 Quntray ACFtsi = 0.606 i
i t
I 4
1
____-._m.__m___._______ .
Calc. No. 96-ENG41528E2 Re/. 00 14 0 min ap M of M ATTACHMENT A AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data Input:
gg Allowable heat generated for a ladder tray Z14FL20 (Attachment 2)
Ta := 50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc' 90 Surface area of tray, sqft/ft (24% x 11ong x top and bottom)
A ._4
( Ref. 3.1.8)
~9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) a := 5.3- 10 c := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
ACFcoy ::0.59
"" Y (Ref. 3.1.10) ctsi : 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.0 Thickness of TSI(maxrnum) et := 4 Tray inner height, inches e2 := 24 Tray inner width, inches
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page N of 4'
- This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray Qt ::Q Allowable heat generation for a ladder type tray Qt = 16.88 d := 0.407 d is adjusted by pi over 4.
Quntray := Qt 24-d Allowed heat generation oof 24" wide and 3* deep uncovered fill tray, WattsMt (adjusted to circular area of cable)
Quntray = 164.884 2
Qcover := ACFcov -Quntray Allowable heat generation for a covered, unwrapped tray, watts #t Qcover = 57.3% Watts / ft Calculate the termal resistance at the tray surface
' Ts := 50 Initial guess Given 1
Qcover=0.402-( Ts - Ta )# + o A c-( ( Ts + 273.16 )# - ( Ta + 273.16 )# )
Ts :: Find (Ts)
Ts = 91.906 cable tray cover surface ternperatre R .-
._ Tc - Ts Qcover R = -0.033 ~
C- W aMs
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page MI of A'"
Modelfor the TSI Encased Cable Tray ;
I bl ::Ttsi Top thickness of TSI, inches b2 ::Tesi Side thickness of TSI inches ,
b3 ::Ttsi Bottom thickness of TSI, inches b4 ::Ttsi Side thickness of TSI, inches ;
Calculate Thermal Resistance of TSI i 1
k 1.8.2928 (Ref. 3.1.95 )
Rk *:
f I fe2 i f I fe2 i !
l el + 0.54 l + l - + 0.54 l + l el + 0.54 l + l - + 0.541 (bl j (b2 ) (b3 ) (b4 )
Rk = 0.326 degC-ft/ watts ;
Calculate the Thermal resistance between the conductor and outer Surface of TSI l i
Rws :: R + Rk Totalthermal resistance form conductor to outer surface of TSI Rws = 0.293 i
i i
i
)
. - . . ._------;.----..-_---_-_---.---.__...---__.-,_.=-__.---- - - - - - - --e >---,,.---,--n,. ,,,-----,-,,--.,>r - - - ~ . --,,. -- ,v, -- ,,-. . . , -- ,--a . -- -v-. w-+
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A F ' .
Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96)
I f 1.8 3 0.25e fe2 + 2 bli Al ::0.27- l 12- -l j -0.527
( e2 + 2 bljl ( 12 )
on n coehnt for h toop sudam of Al = 0.294 the wrapped tray t
t I 1.8 3 o.25' e2 + 2 bl .
A2 :: 0.12 i 12- l 0.527 !
( c2 + 2 blj 12 A2 = 0.131 Convection coefficient for the bottom surface of the wrapped tray. ,
I 1.8 I.0.25 el + 2 bl =
A3 :: 0.29- l 12- -
2 0.527
( el + 2 bljl ( 12 )
A3 = 0.211 Convection coefficient for the side surfaces of the wrapped tray i
Al + A2 + A3 = 0.636 k
Given 5
Tc - Ts -9 f el + e2 + bl + b21
"( Al + A2 + A3 )-( Ts - Ta )7 + 5.3 10 etsi 2-1 l-(( Ts + 273.16)4 - ( Ta + 273.16 )4 ) r Rws ( 12 )
Ts := Find (Ts)
Ts = 66.697 Wrapped tray surface temperature degC. .
I 5
. - - ___m_m. _ _ _ _ _ _ _ _ _ . - _ . _ _ _ - _ _ . , . . , , , . __ _ _ _ ,%,. ,_,. ,.
Calc. No. 96-ENG-01528E2 Rev 00 Attachment A Page A67 of A*
Calculate Oc' and Or' 1 Heat transmission from surface of TSI to amtwent Qec := ( Al + A2 + A3 )-( Ts- Ta )4 air due to convection, Watts /ft Qcc = 21.455
-g el + e2 + bl + b2 Qre :=-0.5310 c -( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
12 Qre = -6.452 QTSI := 90- Ts Rws QTSI = 79.525 Calculate Derating Factor of tray covered with TSI fire wrap ACFtsi :=
i Quntray ACFtsi = 0.694
1 Calc. No. 96-ENG-01528E2 Rev. 00 Z14FL20 MAX Pap M 4 of & '
ATTACHMENT A l AMPACITY MODELING OF APPENDIX R WRAPPED TRAY ;
Data input: 5 1 A at pmWm a laWe My ZM20 (Anahnt 2)
Q := 16.88 Ta ::50 Ampient temperature MP2) [
Conductor maximum allowed temperature -
Tc' 90 Surface area of tray, sqft/ft (24* wide x 1'long x top and bottom)
^ ~. _ 4
~
( Ref. 3.1.8) a :: 5.3 10-9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c :: 0.1 Emissivity of valvanized steel tray cover (Ref. 3.1.1) ,
ACFcoy ::0.59
^ "" #" Y (Ref. 3.1.10) ctsi :: 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductmty of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.5 Thickness of TSI(maxrnum) el::4 Tray innerheight, inches e2 : 24 Tray innerwidth, inches j I
t 5
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page MT of AM' This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray Qt := Q Allowable heat generation for a ladder type tray Qt = 16.88 d := 0.407 d is adjusted by piover4.
- Quntray := Qt 24 d Allowed heat generation oof 24* wide and 3* deep uncovered fill tray, Watts /ft (adjusted to circular area of cable)
Quntray = 164.884 Qcover ::ACFcov 2-Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 57.3% Watts / ft Calculate the termal resistance at the tray surface Ts ::50 Initialguess Given 5.
Qc6 er=0.402-( Ts - Ta )# + o- A e( ( Ts + 273.16)4 - ( Ta + 273.16 )4 ) ,
Ts := Find (Ts) '
Ts = 91.906 cable tray cover surface temperette i
Tc - Ts Qcover i R = 1.033 degC-ft/ watts i
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A . Page 478 of AT*
I Model for the TSI Encased Cable Tray ;
bl := Ttsi Top thickness of TSI, inches t b2 := Ttsi Side thickness of TSI,incher :
b3 ::Ttsi Bottom thickness of TSI, inches ,
b4 ::Tisi Side thickness of TSI, inches j Calculate Thermal Resistance of TSI I f I
k- 1.8 .2928 (Ref. 3.1.95 )
Rk ': f i 1 fe2 1 .f fe2 1 l el + 0.54 l + 1 - + 0.54 l + i el + 0.54 i + 1 - + 0.541 ,
ibt j (b2 j lb3 ) (h4 ) j I
Rk = o.48 degC-ft/ watts Calculate the Thermal resistance between the conductor and outer Surface of TSI j i
i Rws := R + Rk Totalthermal resistance form conductor to outer surface of TSI !
Rws = o.447 l l
I i
i t
f i
_ _ _ _ _ - _ _ _ _ _ _ _ - - - _ _ _ _ _ _ _ _ ~ . - . _ _ _ . . , _ _ . . - + , - ~
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A PageA7/ of M Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the corivection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 1,g 3 0.25' re2 + 2 bli Al :: 0.27- i 12- 1 j 0.527 i e2 + 2 bli ( 12 )
nv d on a for h toop surface of Al = 0 303 the wrapped tray f 1.8 3 0.25' e2 + 2 b1 A2 :: 0.12 l 12- J 0.527
( e2 + 2 bli 12 A2 = 0.135 Convection coefficient for the bottom surface of the wrapped tray.
'+
-2 0.527 A3 := 0.29- I 12- l
( el + 2 blj ( 12 )
A3 = 0.236 Convection coefficient for the side surfaces of the wrapped tray A1 + A2 + A3 = 0.674 Given 5
-( Al + A2 + A3 )-( Ts - Ta ) + 5.3- 10'9 ttsi 2- -( ( Ts + 273.16 )# - ( Ta + 273.16 )# )
Ts :: Find ( Ts )
Ts = 62.71i Wrapped tray surface temperature degC
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page 47Aof Affo 1
Calculate Oc' and Or' [
1 Heat transmission from surface of TSI to amtxent Qx := ( Al + A2 + A3 )-( Ts - Ta)4 air due to convection,WattsMt Qx = 16.169
-8 e Qre ::-0.5310 -c 2 l + e2 + bl + b2-(( Ts + 273.16)4 - ( Ta + 273.16 )4 ) -
12 Qre = -4.983 Q131 :: 90 - Ts Rws QTSI = 61.019 Calculate Deratog Factor of tray covered with TSI fire wrap ACFtsi :: '
1 Quntray ACFtsi = 0.608 t i
I
?
i
+
f
- t
.m . . _ _ _ _ _. . . . . . . - _ _ _ _ . _ . . . ..._. _ . . .
Calc. No. 96-ENG-01528E2 Rev. 00 Z25BG20 min Page A73 of 4/*
ATTACHMENT A AMPACITY MODELING OF APPENDIX R WRAPPED TRAY -
Data input:
A aW Mat genemW for a W tmy H5BG20 (Anachnt 2)
Q :: 13.137 Ta ::50 Ampient temperature MP2) ,
Conductor maximum allowed temperature Tc* 90 Surface area of tray, sqft/ft (24* wide x 11ong x top and bottom) ;
^ ". -_ 4
( Ref. 3.1.8) o :: 5.3 10-9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) e : 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1) cohn Mor for W cow @ tmy ACFcov := 0.59 (Ref. 3.1.10) etsi := 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.0 Thickness of TSI(minimum) el:=4 Tray innerheight, inches ,
l e2 := 24 Tray inner width, inches 1
- . _ _ . .-- ._.-.-......._._.----_-.______-.__._-__._._-__-aa---_ .-- --, _- - , -a - , -r- - - - --- - ,
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page/7f of AM This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray Qt ::Q Allowable heat generation for a ladder type tray Qt = 13.137 d := 0.509 d is adjusted by pi over 4 Quntray := Qt 24 d Allowed heat generation of a 24* wide by d i nches deep uncoveredt tray, Watts /ft (adjusted to circular area of cable)
Quntray = 160.482 Qcover ::ACFcov 2-Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 55.864 Watts / ft -
Calculate the termal resistance at the tray surface Ts::50 Initial guess Given ,
5 Qcover=0.402-( Ts - Ta )# + o- A c-( ( Ts + 273.16 )# - ( Ta + 273.16 )# )
Ts := Find ( Ts )
Ts = 90.997 cable tray cover surface ternperatre Tc - Ts Qcover R = -0.018 Chans
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page 476of 8/#
Model forthe TSI Encased Cable Tray bl ::Ttsi Top thickness of TSI, inches b2 ::Ttsi Side thickness of TSl, inches b3 ::Ttsi Bottom thickness of TSI, inches b4 ::Ttsi Side thickness of TSI, inches Calculate Thermal Resistance of TSI I
*~~' }
Rk := i I i i f fel fe2 l fel + 0 541 .
l + 0 54 l + i - + 0 54 l + I - + 0 541
+ e2 . .
( bl / (b2 / (b3 ) (b4 Rk = 0.326 degC-ft/ watts Calculate the Thermal resistance between the conductor and outer Surface of TSI Rws :: R + Rk Total thermal resistance form conductor to outer surface of TSI Rws = 0.308 e
._. _ ._ _ _. _ _ _ _ . _ . _ . __..m..__ . _ . _ . . ____._-___-_____.m._______.. -
_ . _ _ ..___-_m-__ -
_ _ . _ _________.____.-___m_ _____._.____ __ _ _ _ _ _ _ . _ ___.-_____
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENTA Page Ano f AN*
Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 1.8 3 0.25' fe2 + 2 bli Al ::0.27- 1 12- j -1 H 0.527
( e2 + 2 blj ( 12 /
nv na ent for tM top sudace of t%
A1 = 0.294 wrapped tray I 1.8 3 0.25' e2 + 2 bl A2 : 0.12 l 12- 1 0.527
( c2 + 2 blj 12 ,
A2 = 0.131 Convection coefficient for the bottom surface of the wrapped tray. i f 1.8 1 0.25 el + 2 bl A3 := 0.29 l 12- l -2 0.527
( el + 2 blj ( 12 ) i A3 = 0.211 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.636 Given 5
Tc - Ts -9 1 -
=( Al + A2 + A3 )-( Ts - Ta)7 + 5.3 10 Etsi 2-lfel + e2 + bl +l-((b2Ts + 273.16)4 - ( Ta + 273.16 )# ) t Rws ( 12 )
i Ts *: Find ( Ts ) ;
~
Ts = 66.236 Wrapped tray surface temperature degC. i i
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page 4W of Ar#
Calculate Oc' and Or' 5
at tmnsdssn from mdace CSIto ah Qx := ( Al + A2 + A3 )-( Ts - Ta ) air due to convection, WattsMt Qcc = 20.716
-g Qrc::-0.5310 c 2 el + e2 + bl + b2 (( Ts + 273.16)4 - ( Ta + 273.16 )4 ) -
12 Qre = -6.261 QTSI := 90 - Ts Rws t QTSI = 77.061 i
Calculate Derating Factor of tray covered with TSI fire wrap
., t ACFtsi := --QTSI 1 Quntray -;
ACFtsi = 0.693 t
i 1
4 i
. . _ . . _ . . _ . _ . _ _ _ _ _ _ _ . . . . _ _ _ _ _ _ . . _ . _ . _ _ . _ _ _ . _ . . . _ . _ _ _ _ _ . . _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ . _ _ . _ _ _ _ _ . _ _ _ _ _ _ _ - _ _ _ _ _ _ _ ______.__________m..___._ _ _.______._.________m
Calc. No. 96-ENG41528E2 Rev. 00 Z25BG20 MAX ATTACHMENT A Page kN of M#
AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data input:
A a aWmWor a W tmy D5GB20 (Anachment 2)
Q := 13.137 Ta ::50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc ' W Surface area of tray, sqft/ft (24" wide x 1'iong x top and bottom)
A ._4
( Ref. 3.1.8) -
-9 o := 5.3-10 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) t := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
A n fador for t@t e caw hy ACFcov := 0.59 (Ref. 3.1.10) ttsi :: 0.9 Emissmty of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.5 Thickness of TSI(maxwnum) el := 4 Tray innerheight, inches e2 := 24 Tray inner width, inches
. - _ - . _- ._ __ _ _ _ _ - - - _ _ - - _ _ _ _ _ _ _ - _ _ _ _ _ - _ _ - - _ _ _ _ = _ _ _ _ _ - _ - _ _ _ _ _ _ _ _ _ - _ - - - _ _ - _ _ _ _ _ _ - _ _ _ - _ _ _ _ _ - - __ _ _ _ _ - _ -__ .
.- . . . .- . . _ . . . . - _ - - - - . - . _ . - ~- . - . . . . - - ~ - - . _ . . . . - - - . . - . - . . . _ - .
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page 4D of M i
This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray i
k Qt := Q Allowable heat generation for a ladder type tray Qt = 13.137 l t
d := 0.509 d is adjusted by pi over 4. ,
r Quntray := Qt 24 d Allowed heat generation oof 24' wide and 3* deep uncovered fill '
tray, Watts /ft (adjusted to circular area of cable)
Quntray = 160.482 ,
Qcover ::ACFcov 2-Quntray Allowable heat generation for a covered, unwrapped tray, wattsat j Qcover = 55.864 Watts / ft Calculate the termal resistance at the tray surface .
Ts := 50 Initial guess Given 5 i Qcover=0.402-(Ts- Ta)4 + o- A c-((Ts + 273.16)# - ( Ta + 273.16 )4 )
?
Ts := Find ( Ts )
i Ts = 90.997 cable tray cover surface temperatre t
i R .- ,_ Tc - Ts Qcover R = 1.018 gg, i
Calc. No. 96-ENG-01528E2 Rev. 00 -
Attachment A Page 4# of M ,
-1 Modelfor the TSI Encased Cable Tray bl :Ttsi Top thickness of TSI, inches b2 ::Ttsi Side thickness of TSI, inches b3 ::Ttsi Bottom thickness of TSI, inches ,
Side thickness of TSI, inches !
b4 :=Ttsi Calculate ThermalResistance of TSI -
1
' I Rk ::
f i fe2 i f I fe2 I l el + 0.54 l + 1 - + 0.54 l + 1 el + 0.54 l + l - + 0.541 (hl j (b2 ) (b3 ) (b4 j .
Rk = 0.48 degC-ft/ watts ;
Calculate the Thermal resistance between the conductor and outer Surface of TSI i
Rws :: R + Rk Totalthermal resistance form conductor to outer surface of TSI t Rws = 0.463 V
b
___________________m_._ _ _ _____________________m_.______________________.___m_mm __m____ _A___.____ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ - - - , . _ -.,--m.-_ - . . . ,,- v --..-- .. -- -- v---.
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page M/ of h*
Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 1.8 3 0.25' fe2 + 2 bl i Al := 0.27- l 12- ! i 0.527 i e2 + 2 blj ( 12 j n on cwhnt fo% toop suda of A1 = 0.303 the wrapped tray f 1.8 3 0.25' e2 + 2 bl A2:= 0.12-l 12- 1 0.527
( e2+ 2 blj ,
12 A2 = 0.135 Convection coefficient for the bottom surface of the wrapped tray.
I 1.8 I.0.25 el + 2 bl A3 :: 0.29- l 12- I
-2 0.527
( el + 2-blj (12)
A3 = 0.236 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.674 Given 2
Tc- Ts -9 eI + c2+ bi + b21 Al + A2 + A3 )-( Ts - Ta )4 + 5.3 10 etsi-2 l -( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
Rws k 12 j Ts :: Fiml( Ts )
Ts = 62 44 Wrapped tray surface temperature degC.
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page 48kof AN*
Calculate Oc' and Or' 1 Heat transmission from surface of TSI to ambient Qcc := ( AI + A2 + A3 )-( Ts - Ta )4 air due to convection, Watis/ft Qcc = 15.739 8
Qre : -0.5310 - c el + e2 + hl + b2 ( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
12 Qre = -4.871 Q'I31 :: 90 - Ts Rws QTSI = 59.578 Calculate Derating Factor of tray covered with TSI fire wrap ACFtsi ::
1 Quntray ACFtsi = 0.609
Calc. No. 96-ENG-01528E2 Rev. 00 2EA10 nn age M of M ATTACHMENT A AMPACITY MODELING OF APPENDlX R WRAPPED TRAY Data input Al pm w a W tmy E2EA10 (AhWnt 2) ,
- Q
- = 5.07 Ta : 50 Ampient temperature MP2)
Conductor maxernum allowed temperature j Tc* 90 Surface area of tray, sqtt/ft (6" wide x 1'long x top and bottom)
A ._ ~~ I '
f
( Ref. 3.1.8)
-9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) a := 5.3 10 e : 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
ACFcoy .= 0.59 Ampacity correction factor for tight cover cable tray (Ref. 3.1.10)
Etsi := 0.9 Emissivity of TSI surface (Attsdiiiio-it 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.0 Thickness of TSI(maximum) ;
el::4 Tray inner height, inches i
e2 :: 6 Tray inner width, inches !
1 2
t i
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A a p M of M This section of the modelis for covered tray Calculation for the allowable heat generation in a covred tray Qt ::Q Allowable heat generation for a ladder type tray Qt = 5.07
-d:=1.935 d is adjusted by piover4.
Quntray ::Qt 6 d Allowed heat generation oof 6" wide and 3" deep uncovered fill tray, Wattsat (adjusted to circular area of cable)
Quntray = 58.863 2
Qcover ::ACFcov -Quntray Allowable heat generation for a covered, unwrapped tray, watts #t !
Qcover = 20.49 Watts / ft Calculate the termal resistance at the tray surface '
Ts ::50 Initialguess Given ,
5 Qcover=0.143-( Ts - Ta ) + o- A c-( ( Ts + 273.16)4- ( Ta + 273.16 )# )
Ts := Find ( Ts)
Ts = 94.765 cable tray cover su face temperatre R .-._ Tc - Ts Qcover i R = -0.233
-Watts
-l Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page M of 4'@
i i
Model for the TSI Encased Cable Tray t
bl ::Ttsi Top thickness of TSI, inches t
b2 :=Ttsi Side thickness of TSI, inches .
b3 ::Ttsi Bottom thickness of TSI, inches b4 ::Ttsi Side thickness of TSI, inches ,
Calculate Thermal Resistance of TSI !
I l k 1.8.2928 (Ref. 3.1.95 ) i Rk .-
._ f fel 1 / I I fe2 I .
1 - + 0.54 14 i c2 + 0.54 l + 1 el + 0.54 l + 1 - + 0.54 j !
(bl j \b2 / (b3 j (b4 ) .
Rk = 0.856 degC-fthvatts I
Calculate the Thermal resistance between the conductor and outer Surface of TSI i i
Rws :: R + Rk Totalthermal resistance form conductor to outer surface of TSI f f
Rws = 0.624 i b
b I
i 1
I t
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A' Page 4M of #/*
Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 1.8 3 0.25' fe2 + 2 bli AI := 0.27- l 12- 1 I l 0327
( e2 + 2 blj ( 12 j nv ne nfr op suda of i Al = 0'122 the wrapped tray f 1.8 e2 + 2 bl A2 := 0.12- l 12-( c2 + 2 blj 3l 0.25 '12 0.527 A2 = 0.054 Convection coefficient for the bottom surface of the wrapped tray.
I 1.8 3 0.25 el + 2 bl A3 :: 0.29- l 12- j -
2 0.527
( el + 2 bij ( 12 )
A3 = 0.211 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.386 i
Given 5 t Tc - Ts -9 f el + e2 + bl + b21
=( Al + A2 + A3 )-( Ts - Ta )4 + 5.3 10 ctsi 2 ! l-( ( Ts + 273.16)4 - ( Ta + 273.16 )4 ) t Rws ( 12 /
Ts := Find (Ts)
Ts = 66.%8 Wrapped tray surface temperature degC. . l r
i I
_ _ .- _ _ . . - . _ _ _ . . . _ _ . . _ . . _ - . . . _._._-.-..m.. . . - . _ . . . - _ - . . . _ - . . _ _ _ . . . . _ . . . _ . . .. ._ . .
Calc. No. 96-ENG-01528E2 Rev. 00 ,
Attachment A Page 261 of A/*
Calculate Oc' and Or' 5 Heat transmission from surface of TSI to ambient Qec := ( Al + A2 + A3 )-( Ts - Ta )4 air due to convection, Watts /ft Qcc = 13.298 Qre :=-0.53 10'8 c 2 el + e2 + bl + b2 ( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
12 Qre = -2.626 ,
QTSI := 90 - Ts }
Rws -
i QTSI = 36.931 i Calculate Derating Factor of tray covered with TSI fire wrap ,
i QTSI ACFtsi :=
1 Quntray ACFtsi = 0.792 4
l b
I
Calc. No. 96-ENG-01528E2 Rev. 00 10 MAX age M of M ATTACHMENT A AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data input:
^ #* '"'* "Y
- Q ::5.07 Ta ::50 Ampent temperature MP2)
Conductor maximum allowed temperature Tc ': W Surface area of tray, sqft/ft (6" wide x 11ong x top and bottom)
^ "._ I -
( Ref. 3.1.8) o : 5.3 10 ' Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c := 0.1 Eiriios;vity of galvani.% teel tray cover (Ref. 3.1.1)
ACFcov ::0.59
" " ' 9 "# Y (Ref. 3.1.10)
Etsi::0.9 Emissivity of TSI surface (Attachment 1) k :: 0.1 Thermal conductmty of TSI, BTU /hr-ft-degF (Attachment 1)
'nsi :: 1.5 Tri'u-ss of TSI(mammum) i e1::4 Tray inner height, inches e2 ::6 Trayinner width, inches
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A age M of M This section of the modelis for covered tray Calculation for the allowable heat generation in a covred tray Qt ::Q Allowable heat generation for a ladder type tray Qt = 5.07 d::I.935 d is adjusted by pi over 4.
i Quntray := Qt 6 d Allowed heat generation oof 6" wide and 3" deep uncovered fill tray, WattsMt (adjusted to circular area of cable)
Quntray = 58.863 j 2
Qcover ::ACFcov -Quntray Allowable heat generation for a covered, unwrapped tray, wattsMt Qcover = 20.49 Watts / ft Calculate the termal rnistance at the tray surface ;
i Ts : 50 Initial guess ;
Given i i
k 2.
Qcover=0.143-( Ts - Ta)# + o- A c-(( Ts + 273.16)# - ( Ta + 273.16 )# ) i Ts := Find (Ts)
Ts = 94.765 cable tray cover surface temperatre [
R ,_ .- Tc - Ts l
t Qcover i
R = -0.233 h atts
. __,___s-....-m-_ m_us._ _.._.__m._.--__._.w_____.__m.-_---.___.__u - . - _ _ _ - ___ . ,~s. - - + e-,. ,w wei n..e- -
_e--% w .- -- - -. -
Calc. No. 96-ENG-01528E2 Rev. 00 I
t Attachment A Page AfD of A '"
Modelfor the TSI Encased Cable Tray bl ::Ttsi Top thekness of TSI, inches [
b2 ::Ttsi Side thickness of TSI, inches f b3 ::Ttsi Bottom thickness of TSI, inches j b4 ::Ttsi Side thickness of TSI, inches t
Calculate Thermal Resistance of TSI i
1 i k 1.8.2928 (Ref. 3.1.95 )
Rk ~
fel i f i fel I f i l - + 0.541 + l e2 + 0.541 + l - + 0.54 l + 1 e2 + 0.541 (bl J (b2 f (b3 ) (b4 j ,
- Rk = 1.225 degC-fthvatts ,
Calculate the Thermal resistance between the conductor and outer Surface of TSI
, Rws := R + Rk Totalthermal resistance form i conductor to outer surface of TSI j Rws = 0.992 ,
I r
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page A9/ of 4/@
Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 1,8 3 0.25' fc2 + 2 bl i Al :: 0.27- l 12- l l l 0.527
( c2 + 2 blj ( 12 /
n na n f r tM tg suda of A1 = 0.133 the wrag, ped tray I 1.8 3 0.25' e2 + 2 bl,
( e2 + 2 blj 12 A2 = 0.059 Convection coefficient for the bottom surface of the wrapped tray, A3 :: 0.29- 1 12- I ' 2 0.527
( el + 2 bl/ ( 12 )
A3 = 0.236 Convechon coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.428 Given 1
Tc - Ts el + e2 + bl + b21 AI + A2 + A3 )-( Ts - Ta )4 + 5.3 10~9 Etsi 2, -( ( Ts + 273.16 )4- ( Ta + 273.16 )4 )
Rws i 12 j Ts := Find ( Ts)
Ts = 62.27 Wrapped tray surface temperature degC.
_ . ~._ _ . _.. _ - ._ . . - _ . _ .
_ . . ._ . _ . . . . . ~ . . . . . .
Calc. No. 96-ENG-01528E2 Rev. 00 >
Attachment A Page A92- of A'* ,
Calculate Oc' and Or' 5 -
Heat transmission from surface of TSI to ambient t Qcc := ( Al + A2 + A3 )-( Ts - Ta )4 air due to convection, Watts /ft !
Qcc = 9.833
-8 Qrc := -0.53 10 c 2 el + e2 + bl + b2 (( Ts + 273.16)4 - ( Ta + 273.16 )4 )
12 Qic = -2.013 1 QTSI := 90- Rws Ts QTSI = 27.951 i
i Calculate Derating Factor of tray covered with TSI fire wrap i
ACFtsi :=
1 Quntray ACFisi = 0.689 t
i I
t
Calc. No. 96-ENG-01528E2 Rev. 00 0M age N of M ATTACHMENT A -
i AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data input >
Q :: 11.313 Allowable heat generated for a ladder tray Z22EA10 (Attachment 2)
Ta::50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc 90 Surface area of tray, sqft/ft (6" wide x 1*long x top and bottom)
^ ._
'- I
( Ref. 3.1.8) a :: 5.3- 10~9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) r.:= 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
?.Scoy := 0.59 A m n r f r tigM com @ tmy (Ref. 3.1.10) ,
ttsi := 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1) l Ttsi := 1.0 Thickness of TSI(maximum) et:=4 Tray innerheight, inches e2 := 6 Trayinner width, inches 1
. _ . . . . .- _ _ _ _ _ _ .___ -.-. _ . . _. . ._..__. _ _.___.._...___..___.m.--
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A ap N of M n
This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray t
Qt ::Q Allowable heat generation for a ladder type tray Qt = 11.313 ;
d : 1.1714 d is adjusted by pi over4.
Quntray := Qt 6 d Allowed heat generation of 6" wide and 3* deep uncovered fill tray, Watts /It (adjusted to circular area of cable)
Quntray = 79.512 Qcover ::ACFcov 2-Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft i Qcover = 27.678 Watts / ft 1 l
Calculate the termal resistance at the tray surface .
I Ts:=50 Initialguess !
Given t t
5 !
Qcover=0.143-( Ts - Ta ) + o A t-( ( Ts + 273.16 )# - ( Ta + 273.16 )# ) j Ts := Find ( Ts) cable tray cover surface temperatre Ts = 106.987 l
f;
, R ._ Tc - Ts Qcover R = -0.614 !
degC-ft/ watts ;
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page A6of M ,
Model for the TSI Encased Cable Tray bl := Ttsi Top thickness of TSI, inches b2 := Ttsi Side thickness of TSI, inches b3 :=Ttsi Bottom thickness of TSI, inches b4 : Ttsi Side thickness of TSI, inches l Calculate Thermal Resistance of TSI I ,
k- 1.8 .2928 (Ref. 3.1.95 ) i Rk M i i f fe2 f i I I l el + 0.54 l + 1 - + 0.54 l + i el + 0.54 l + 1 c2 + 0.541 (b1 j (b2 ) (b3 ) (b4 ) l Rk = 0.856 degC-ft/ watts e
Calculate the Thermal resistance between the conductor and outer Surface of TSI Rws ::R F Rk Totalthermal resistance form conductor to outer surface of TSI [
Rws = 0.242 l i
_ _-_ _ .____ _ - . _ . . . _ . _ . . _ . _ _ . _ . _ _ _ . . . . ._ - _ . .. ._ m_ _ _ .
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page M of AI*
Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface ({Ref. 3.1.96)
'f 1.g 3 0.25' fc2 + 2 bl i A1 ::0.27- l 12- 1 -l 1 0.527
~( e2 + 2 blj ( 12 j Convection coefficient for the top surface of the AI = 0122 wrapped tray f 1.8 3 0.25' e2 + 2 bl I A2 :: 0.12 I 12- I 0.527 i
( c2 + 2 bl/ 12 A2 = 0.054 Convection coefficient for the bottom surface of the wrapped tray.
f 1.8 3 0.25 el + 2-bl A3 :: 0.29- l 12- 1 -
2 0.527
( el + 2 blj ( 12 ) ,
A3 = 0.211 Convection coefficient for the side surfaces of the wrapped tray [
i i
Al + A2 + A3 = 0.386 Given 5
Tc - Ts -9 fel + e2 + bl + b21
=( Al + A2 + A3 )-( Ts - Ta )7 + 5.3-etsi 102 l 1 -( ( Ts + 273.16 )4 -- ( Ta + 273.16 )4 )
Rws ( 12 /
Ts := Find ( Ts )
Ts = 75.607 Wrapped tray surface temperature degC.
P l
-- _ - - _ = _ _. _ _ _ - - _ _ - - _ _ - -- _ -___ __ - - - . .
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page A47 of Br*-
Calculate Oc' and Or' S Hest transmission from surface of TSI to ambient Qoc := ( A1 + A2 + A3 )-( Ts - Ta)4 air due to convection, Watts /ft Qtr = 22.244 Qre :=-0.53 10"8 c 2 el +12c2 + bl + b2 (( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
Qre = -4.123 QTSI := 90- Ts Rws QTSI = 59.354 Calculate Derating Factor of tray covered with TSI fire wrap ACFtsi :=
1 Quntray i
t ACFtsi = 0.864 l
i i
i o
I t
L
_.. .- _ . . _ _ _ . . . . _ . _ _ _ __ __..__..m____-m._._____m__ _ . _ _ _ _, . - . - _,- . . . . . . , . . ~ , . . , . . _ . . . . . . , - . _ . . . ~ . - , _ . , _ _ _ . , ~ . , . . _ , . , , . . . _ , . , , . - , _ _ _
Calc. No. 96-ENG-01528E2 Rev. 00 1
2EA10 MAX ATTACHMENT A "98 U AMPACITY MODELING OF APPENDIX R WRAPPED TRAY I
Data input i A a at ymmW for a Mer t ay U2EA10 (Anachnt 2)
Q := 11.313 Ta ::50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc' 90
^ ._ Surface area of tray, sqft/ft (6" wide x 11ong x top and bottom)
'- y
( Ref. 3.1.8) a :: 5.3 10-9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
ACFcoy ::0.59 A coWM fador fodght me mW tmy (Ref. 3.1.10) etsi := 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.5 Thekness of TSI(maximum) el :: 4 Tray inner height, inches c2 := 6 Tray inner width, inches
- . , ._ . -._ .. . . . . . . . ~ . .- _ . = . - . . . . _ . . _ - . -
Calc. No. 96-ENG-01528E2 Rev. 00 :
ATTACHMENT A ""
This section of the rnodelis for covered tray Calculation for the allowable heat generation in a covred tray Qt ::Q Allowable heat generation for a ladder type tray Qt = 11.313 i i
d := 1.1714 d is adjusted by pl over 4.
Quntray ::Qt 6-d Allowed heat generation of 6* wide and 3" deep uncovered fill tray, Watts /ft (adjested to circular area of cable)
Quntray = 79.512 Qcover ::ACFcov 2-Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 27.678 Watts / ft ,
Calculate the termal resistance at the tray surface Ts:: 50 initial guess Given L
1 Qcover=0.143-( Ts - Ta)4 4- o A-c-( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
Ts := Find ( Ts )
Ts = 106.987 cable tray cover surface temperatre Tc - Ts Qcover R = 1.614 degC-ft/ watts
Galc. No. 96-ENG-01528E2 Rev.00 Attachment A Page 480of A'/*
Modelfor the TSI Encased Cable Tray bl : Ttsi Top thickness of TSI, inches i b2 := Ttsi Side thickness of TSI, inches b3 :=Ttsi Bottom thickness of TSI, inches b4 :=Ttsi Side thickness of TSI, inches Calculate Thermal Resistance of TSI I i Rk :=
~ ' ' }
i fe2 i f i f i ,
fel + 0 541 + l- + 0 54 l + l el + 0 54 j + 1 e2 + 0 541 1 i -
(bl .j (b2 j (b3
) (b4 .)
Rk = 1.225 degC-ft/ watts Calculate the Thermal resistance between the conductor and outer Surface of TSI Rws ::R + Rk Totalthermal resistance form conductor to outer surface of TSI L Rws = 0.611 i
t i
L
__.,__...__...,..m m %. ~ - - . _ . _ _ - _ .._.-_.-...-___.--m__.a_. _..__-..__._-___.m_ a __ _ . _ -
.._.___.__m_.._-_._________ ___ _ _ _ _ _ _ . -- _ _e__ _ _ -
-r-' _ _. -t.P- --
_.- . _ . - . _ ~ _ . . . . . _ _ . . . ._ . - _ _ . . . .~. _ - _._.._._. . . _ . _ - . . _ . - .
. - _ _-. ~_. _ . . - . _ - . .
i Calc. No. 96-ENG-01528E2 Rev. 00 1
4 ATTACHMENT A Page Alo/ of AfW Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 0 25' 1,8 3 fe2 + 2 bli j A1 :: 0.27- l 12- l i 0.527
( e2 + 2 blj ( 12 j Convection coefficient for the top surface of the Al = 0.133 wrapped tray f 1,8 3 0.25' e2 + 2 bl I A2 :: 0.12 1 12- l 0.527
( e2 + 2 blj 12 A2 = 0.059 Convection coefficient for the bottom surface of the wrapped tray. ,.
I 1.8 3 0.25 el + 2 bl A3 : 0.29- l 12- I 2 0.527
( el + 2 bl/ ( 12 )
A3 = 0.236 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.428 Given 5
Tc - Ts 9 fel + e2 + bl + b2 1 Rws
=( Al + A2 + A3 )-( Ts - Ta )7 + 5.3ctsi 102- l 12 l-((Ts + 273.16)4 - ( Ta + 273.16 )4 )
( ) [
l Ts : Find (Ts) ;
i Ts = 66.358 Wrapped tray surface temperature degC. I I
L f
r L
h
. _ _ _ _ _ _ - - - _ _ _ - . _ _ _ - _ . - . _ _ _ _ . ._-______-____._______________-_.-.__-__-----___=.----____.-____-____-.__.__-__A
I k
Calc. No. 96-ENG-01528E2 Rev. 00 i
Attachment A Page 4t** of AIP Calculate Oc' and Or' .
1 Heat transmission from surface of TSI to amtnent Qcc := ( Al + A2 4- A3 )-( Ts - Ta )4 air due to convection, Watts /ft Qx = 14.086
-8 e
, Qre :=-0.53 10 c 2 l + e2 + bl + b2 (( Ts + 273.16)4 - ( Ta + 273.16 )4 ) -
12 ,
Qic = -2.735 QTSI :: 90- Ts Rws Q131 = 38.7 ;
Calculate Derating Factor of tray covered with TSI fire wrap i
ACFtsi ::
1 Quntray ACFtsi = 0.698 i
i P
t
)
Calc. No. 96-ENG-01528E2 Rev. 00 Z23FA30 min ATTACHMENT A ageM of M AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data input:
AHow ee genem r a lamer tray M4MO (Anachnt 2)
Q ::4.407 Ta:=50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc *: 90 Surface area of tray, sqftJft (24* wide x 1'long x top and bottom) _
^ ._
'- 4
( Ref. 3.1.8) a :: 5.3 10~9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c := 0.1 Emissivity of gahranized steel tray cover (Ref. 3.1.1)
Ampacity correction factor for tight cover cable tray ACFcov := 0.59 (Ref. 3.1.10) ctsi := 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.0 Thickness of TSI (minimum) et:=4 Tray inner height, inches e2 := 24 Tray inner width, inches
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page Alof of A&
This section of the model is for covered tray Calculation for the a!!owable heat generation in a covred tray Qt ::Q Allowable heat generation for a ladder type tray Qt = 4.407 d := 0.2037 d is adjusted by piover 4 Quntray ::Qt 24 d Allowed heat generation of a 24" wide by d i nches deep uncoveredi tray, Watts /ft (adjusted to circular area of cable)
Quntray = 21.545 Qcover : ACFcov 2-Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 7.5 Watts / ft Calculate the termal resistance at the tray surface Ts ::50 Initial guess ,
Given 1
Qcover=0.402-( Ts - Ta )4 + o- A c-( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
Ts := Find ( Ts )
Ts = 57.752 cable tray cover surface temperatre Tc - Ts Qcover degC-ft/ watts
i.
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page Afefof 4'40 Modelfor the TSI Encased Cable Tray >
bl :: Ttsi Top thickness of TSI, inches b2 ::Ttsi Side thickness of TSI, inches b3 ::Ttsi Bottom thickness of TSI, inches b4 ::Ttsi Side thickness of TSI, inches Calculate Thennal Resistance of TSI I
' *~
Rk := i
) f I fc2 I l fel + 0 54l + 1fe2 + -
0.54 l + i el + 0.541 + l - + 0.54 l
( bl ./ (b2 j (b3 ) (b4 )
i Rk = 0.326 degC-ft/ watts Calculate the Thermal resistance between the conductor and outer Surface of TSI Rws :: R + Rk Totalthermal resistance form conductor to outer surface of TSI Rws = 4.626 i
?
i h
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page *dof **
t Solve Heat Balance Equation, to determine outer Surface Temperature of TSI j Calculate the convection and radiation coefficents for the TSI sudace (Ref. 3.1.96) f 1.8 3 0.25. I e2 + 2-bli !
L A1 := 0.27- l 12- l -l 1 0.527
( e2 + 2 blj ( 12 j' f Convection coefficient for the top surface of the . i Al = 0.294 wrapped tray I 1.8 3 0.25' e2 + 2 bl A2 := 0.12 1 12- l 0.527 r
( e2 + 2 blj 12 5
t A2 = 0.131 Convection coefficient for the bottom surface of the wrapped tray.
I 1.8 3 0.25
' A3 :: 0.29- l 12- f -el + 2 bl -2 0.527 '
( el + 2 bl/ ( 12 )
, A3 = 0.211 Convection coefficient for the side surfaces of the wrapped tray j i
Al + A2 + A3 = 0.636 Given l '
5 Tc - Ts Iel + e2 + bl + b21
( A + A + A3 )-( Ts - Ta)# + 5.3- 10~9 -ctsi 2 l -
( Ta + 273.16 )# )
Rws ( 12 f -(( Ts + 273.16 )4 f i
Ts := Find ( Ts)
Ts = 52.046 Wrapped tray surface temperature degC. i t
4 i
t
...m_... _. ..m_ . _ _ .-_ . . _ . _ _ _ . . -
_ . . _ _ . . _ _ _ _ _ _ . . . __..__.m. ._.,_____._________m____.-._ m- ____m__________ ____._m ._ - _ - _
Cric. No. 96-ENG-01528E2 fiev. 00 Attachment A Page #7 of A*
Calculate Oc' and Or*
5 Qec := ( Al + A2 + A3 )-( Ts - Ta ) air due to convection, WattsMt Qtr = 1.555
~8 el + e2 + bl + b2 Qre :=-0.53-10 -c -( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
12 Qm = -0.739 90 - Ts QTSI ::
Rws QTSI = 8.204 Calculate Derating Factor of tray covered with TSI fire wrap ACFtsi ::
1 Quntray ACFtsi = 0.617
Calc. No. 96-ENG-01528E2 Rev. 00 I 223FA30 MAX ATTACHMENT A o
'age Af** of d'f'
- t AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data input:
A aW W WWor a lah tmy MNO (AMet 2)
Q := 4.407 Ta ::50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc~ 90 Surface area of tray, sqft/ft (24" wide x 11ong x top and bottom)
A '._ 4
( Ref. 3.1.8) !
o := 5.3 10~9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) c :: 0.1 Errussmty of galvanized steel tray cover (Ref. 3.1.1)
A coWWMor fodgM com caw tmy ACFcov := 0.59 (Ref. 3.1.10) etsi := 0.9 Errussivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.5 Ti-A,ki-s= of TSI (mwwmum) el:=4 Tray innerheight, inches e2 := 24 Trayinner width, inches
._ ._.__m__.__.._m_ _______.___._.____.- _ ___ __ _ _ _ ._______ __._.m____.m__.___m____
. . . _ . _ _ _ _ _ . . _ . . _ . _ . . _. _ _ _ _ . . _ _ . . _ . _ . _ _ _ . .m_. _ .m. . . . . .. m. _ . _ .
C2ic. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page N81 of BI*
This section of the modelis for covered tray Calculation for the allowable heat generation in a covred tray Qt ::Q Allowable heat generation for a ladder type tray Qt = 4.407 d := 0.2037 d is adjusted by piover4 Quntray : Qt-24 d Allowed heat generation of a 24* wide by d i nches deep uncoveredi tray, Watts /ft (adjusted to circular area of cable)
Quntray = 21.545 2
Qcover ::ACFcov -Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 7.5 Watts / ft Calculate the termal resistance at the tray surface Ts := 50 Initial guess ,
Given 5
@ aver =0.402-( Ts - Ta ) + o A c-(( Ts + 273.16 )# - ( Ta + 273.16 )# )
Ts := Find (Ts)
Ts = 57.752 cable tray cover surface temperatre R ._
.- Tc - Ts Qcover R = 4.3 !
degC-ft/ watts
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page Alto of Arvo Model for the TSI Encased Cable Tray bi ::Ttsi Top thickness of TSI, inches b2 :=Tisi Side thickness of TSi, inches b3 ::Ttsi Bottom thickness of TSI, inches b4 ::Ttsi Side thickness of TSI, inches Calculate Thermal Resistance of TSI 1
Rk :=
i f i f i fe2 i '
l fel + 0.541 j + e2 1 + 0 541 + 1el + 0 54 l +
. l - + 0 54 l .
(bl (b2 ; (b3 .) (b4 j ;
Rk = 0.48 degC-ft/ watts Calculate the Thermal resistance between the conductor and outer Surface of TSI Rws :: R + Rk Totalthermal resistance form conductor to outer surface of TSI Rws = 4.78 t
h i
C&lc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page A_f of 84*
1 Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 1,g 3 0.25' fe2 + 2 bl i Al := 0.27- 1 12- 1 l l-0.527
( e2 + 2 blj' ( 12 j ,
Convection coefficient for the top surface of the 4
A1 = 0.303 wrapped tray I 1.8 3 0.25' c2 + 2 bl A2 :: 0.12- l 12- ! 0.527 i e2 + 2 blj 12 ;
A2 = 0.135 Convection coefficient for the bottom surface of the wrapped tray.
i i f 1.8 3 0.25 el + 2'bl A3 := 0.29- l 12- j -
-2 0.527 ;
( el + 2 blj ( 12 ) ;
A3 = 0.236 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.674 Given 5
Tc - Ts I I Rws
-( Al + A2 + A3 )-( Ts - Ta )7 + 5.3 10 9 Etsi 2-lfel 12
+ c2l-(( +
Tsbl + b2 - ( Ta + 273.16 )4 )
+ 273.16)4 i
( j Ts := Find (Ts) i Ts = 51.92 Wrapped tray surface temperature degC.
i r
l t
f L
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page An of AV6 4
Calculate Oc' and Or' l "e tensdssbn fmm sudace of TSIto amt .
Qec := ( AI + A2 + A3 )-( Ts - Ta ) air due to convection, Watts /ft i l
Qtr = I.522 .
i
-8 Qre := -0.53 10 t 2 el + c2 + bl + b2 (( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
i 12 ,
Qre = -0.716 i
90 - Ts .
QTSI :=
Rws QTSI = 7.966 i
Calculate Derating Factor of tray covered with TSI fire wrap ;
ACFtsi :=
1 Quntray ACFtsi = 0.608 ,
l t
t 5
l l
i
- .-. . . . ~ . . . _ . - . - . .
Celc. No. 96-ENG-01528E2 Rev. 00 l Z23FA25 min ATTACHMENT A Pa9e Ad of 4@ t AMPACITY MODELING OF APPENDIX R WRAPPED TRAY ;
Data Input Allowable heat generated for a ladder tray Z24FL10 (Attachment 2)
Q ._3.983 L
Ta := 50 _ Ampient temperature MP2) .
.,, . 90 Conductor maximum allowed temperature Surface area of tray, sqft/ft (24* wide x 1'long x top and bottom) [
A ._4 -
( Ref. 3.1.8) [
o :: 5.3 10 ' Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27)
P c : 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1) l ACFcoy ::0.59 A rf r cow @ hy t (Ref. 3.1.10) casi :: 0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Themmi conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi := 1.0 TiaAumss of TSI(minimum) .
I el::4 Tray innerheight, inches e2 := 24 Tray inner width, inches t
i i
i v
_ _ - = _ _ _ - _ - - _ _ _ _ _ _ _ _ _ - _ _ _ _ _ - - _ _ _ - _ _ . _ _ _ _ . . . .. -. . . . - - - . . . .
b Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page8/// of #IA This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray Qt:= Q Allowable heat generation for a ladder type tray ;
Qt = 3.983 d := 0.2037 d is adjusted by piover4 Quntray ::Qt 24 d Allowed heat generation of a 24" wide by d i nches deep uncoveredt tray, Watts /ft (adjusted to circular area of cable)
Quntray = 19.472 !
t 2
Qcover ::ACFcov -Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 6.778 Watts / ft Calculate the termal resistance at the tray surface ,
Ts := 50 initial guess l
Given 5
Qcover=0.402-( Ts - Ta )4 + & A c-( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
Ts := Find ( Ts )
I Ts = 57.117 cable tray cover surface temperatre i
Tc - Ts -
Qcover R = 4.851 degC-ft/ watts
. ~_ .. . .__._ _ _ . .- __ . _ . . _ . .. . . . _ . _ . _ -
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page#ttI of Al*
Model for the TSI Encased Cable Tray bl ::Ttsi Top thickness of TSI, inches b2 ::Ttsi Side thickness of TSI, inches b3 ::Ttsi Bottom thickness of TSI, inches b4 ::Ttsi Side thickness of TSI, inches .
Calculate Thermal Resistance of TSI 1
Rk := f i f i f i f i l el + 0.541 + l e2 + 0.54 l + l el + 0.54 l + l e2 + 0.541 (bl j (b2 / (b3 j (b4 j' Rk = 0.326 degC-ft/ watts i Calculate the Thermal resistance between the conductor and outer Surface of TSI k
Rws :: R + Rk Totalthermal resistance form conductor to outer surface of TSI Rws = 5.177 r
4
~- ,. . . . . . - ~ - . - . , . . . - . . . . = . .- . .-.
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page@8of 4/*
Solve Heat Balance Equation, to determine outer Surface Temperature of TSI Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 3.g 3 0.25' f e2 + 2 bl i Al :: 0.27- l 12- 1 -l l 0.527
( c2 + 2 blj 1 12 /
nc nf tpsWamd h Al = 0.294 wrapped tray I 1.8 3 0.25' c2 + 2-bl A2 := 0.12- l 12- l 0.527
( e2 + 2 blj 12 A2 = 0.131 Convection coefficient for the bottom surface of the wrapped tray.
- ' 2 0.527 A3 := 0.29- 12- !
( el + 2 bij ( 12 )
A3 = 0.211 Convection coefficient for the side surfaces of the wrapped tray Al + A2 + A3 = 0.636 Given 2
Tc - Ts Iel + e2 + bl + b21 ,
AI + A + A3 )-( Ts - Ta)# + 5.3- 10' etsi 2 J Rws i 12 [-((Ts + 273.16)# -- ( Ta + 273.16)4 )
Ts := Fi:n1{ Ts) .
Ts = 51.848 Wrapped tray surface temperature degC. ,
Calc. No. 96-ENG5 01528E2 Rev. 00 Attachment A Page diff of #t*
Calculate Ckt and Of 5
Heat transtrusson from surface of TSI to amtnent Qtr ::( Al + A2 + A3 )-( Ts *la )4 air due to convection, Watts /ft Qec = 1.369
-g Qre ::-0.53- 10 c-2 el + c2 + bl + b2-( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 )
12 ;
Qre = -0.667 i
QTSI : 90- Rws Ts Q131 = 7369 t
Calculate Derating Factor of tray covered with TSI fire wrap
- ACFtsi :: Jy Quntray i
ACFisi = 0.615 j i
i F
r e
Calc. No. 96-ENG-01528E2 Rev. 00 Z23FA25 MAX Page AI/S of A/*'
ATTACHMENT A AMPACITY MODELING OF APPENDIX R WRAPPED TRAY Data Input: ,
at mmWor a lahmy NUO (Ament 2) t Q := 3.983 Ta::50 Ampient temperature MP2)
Conductor maximum allowed temperature Tc ~= 90 Surface area of tray, sqft/ft (24* wide x 11ong x top and bottom) ;
^ '. _ 4
( Ref. 3.1.8)
~9 Stefan- Boltzmann constant, Wats /sqft-K^4 (Ref. 3.1.27) a :: 5.3- 10 c := 0.1 Emissivity of galvanized steel tray cover (Ref. 3.1.1)
A coWe Mor for @ m caw tmy ACFcov := 0.59 ;
(Ref. 3.1.10)
Etsi:0.9 Emissivity of TSI surface (Attachment 1) k := 0.1 Thermal conductivity of TSI, BTU /hr-ft-degF (Attachment 1)
Ttsi : ".5 Thickness of TSI(minimum) e1::4 Tray inner height, inches t
c2 ::24 Tray inner width, inches i
Y I
I
.__.m - .. . . . . . _ . . . . _ _ _ . . _ . . . . . . _ _ _ _ . _ . . . . _ _ . ____ . _ __ ...__. __ . . .. .. .
Calc. No. 96-ENG-01528E2 Rev. 00 ATTACHMENT A Page AI'f of B'*
This section of the model is for covered tray Calculation for the allowable heat generation in a covred tray Qt ::Q Allowable heat generation for a ladder type tray Qt = 3.983 d : 0.2037 d is adjusted by piover 4 Quntray ::Qt 24 d Allowed heat generation of a 24* wide by d i nches deep uncoveredt tray, Watts /ft (adjusted to circular area of cable)
Quntray = 19.472 Qcover := ACFcov2-Quntray Allowable heat generation for a covered, unwrapped tray, watts /ft Qcover = 6.778 Watts / ft Calculate the termal resistance at the tray surface Ts := 50 Initial guess Given 4 5
Qcover=0.402-( Ts - Ta )# + o A-c-( ( Ts + 273.16)4 - ( Ta + 273.16 )4 )
Ts := Find ( Ts)
Ts = 57.117 cable tray cover surface temperatre i
R .-
_ Tc - Ts -
Qcover R = 4.851 degC-ft/ watts
__ ._. ._ _ ..__.-..___________.-.__..__.________.._m_._=_._______m_._-..______ _ ___ _____ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ . _ _ _ .m __ _____ _ _ __
Calc. No. 96-ENG-01528E2 Rev. 00 Attachment A Page M8"of d'*
Model for the TSI Encased Cable Tray bl 2:Ttsi Top thickness of TSI, inches ,
b2 ::Ttsi Side thickness of TSIinches t b3 ::Ttsi Bottom thickness of TSI, inches i b4 ::Ttsi Side thickness of TSI, inches i
Calculate Thermal Resistance of TSI I
Rk:= f i i fel i fe2 i i l el + 0.54 l + 1 fe2 + 0.54 l + 1 - + 0.54 l + 1 - + 0.541 Abl j ( b2 ) (b3 / (b4 ) .
i l
Rk = 0.48 degC-ft/ watts ?
Calculate the Thermal resistance between the conductor and outer Surface of TSI
- t Rws : R + Rk Totalthermal resistance form conductor to outer surface of TSI Rws = 5.332 ;
l t
I l
i h
___m.___.____ _ _ - _ . _ ______ _ _ - __m -- ..c ......---,,.---,---=,...-~,-r--+ .,, . .- - . , , , - - . - , , - - , . - .y. ,.m3, ,., ,m..,9 % .
~. - .. . - . .. ._. .. _ ,... _ _ _. . _ . _ . . . _ . . . _ . . . _ _ - . _ . . . . _ _ . __ ~ . _ _ _ _ ~ _ . _ . . - _ . - - - . . . - _ . _ .._ .-_. . .. _
, F Calc. No. 96-ENG-01528E2 Rev. 00 ;
i ATTACHMENT A Page&__s/og new Solve Heat Balance Equation, to determine outer Surface Temperature of TSI ;
Calculate the convection and radiation coefficents for the TSI surface (Ref. 3.1.96) f 1.8 3 0.25' fe2 + 2 bli !
Al := 0.27- l 12- 1 I l 0.527 ;
( e2 + 2 blj ( 12 ) l 6
Convection coefficient for the top surface of the '
Al = 0.303 wrapped tray i
I 1.8 3 0.25' c2 + 2 bl l A2 := 0.12- l 12- 1 -0.527
( e2 + 2 bij 12 s
A2 = 0.135 Convection coefficient for the bottom surface of the wrapped tray.
f 1.8 1.0.25 el + 2 bl -
A3 := 0.29- l 12- 1 2 0.527
( el + 2 blj ( 12 )
A3 = 0.236 Convection coefficient for the side surfaces of the wrapped tray i
i Al + A2 + A3 = 0.674 Given f i
5 '
Tc - T, el + e2 + bl + b21 A2 + A3 )-( Ts- Ta)# + 5.3 10 etsi 2. j ((Ts + 273.16)d - ( Ta + 273.16)4 )
Al -1 Rws L 12 i Ts := Find ( Ts) i Ts = 51.739 Wrapped tray surface temperature degC. ,
1 [
i i
a I
I i
_---,,.___-___~_~--___a.---_____.__u_----_._-_a - - __ . . - ---------.-w_____-__--__-_ - . _ _ = . _ _ _ . - - - - _ _- _ , - - - , - _ _ .
Calc. No. 96-ENG-01528E2 Rev. 00 .
Attachment A Page #iS' of Ar*
i Calculate Oc' and Or' 5
g := ( AI + A2 + A3 )-( Ts - Ta )
a mm m Ito a h air due to convection, WattsMt i
Qcc = I.345 ,
8 e Qre := -0.53- 10 c 2 l i e2 + bl + h2-( ( Ts + 273.16 )4 - ( Ta + 273.16 )4 ) >
12 i
Qre = -0.648 QTSI := 90 - Ts '
Rws QTSI = 7.176 Calculate Derating Factor of tray covered with TSI fire wrap ;
ACFtsi :=
1 Quntray ACFtsi = 0.607 t
- . _ _ . _ _ _ _ _ _ _ _ _- __..____._.______._.____.m3-____m_-._______.m.__m_._.___..__ _ _ _ _ __________m__m_ m-___ m:_.____. _ _ _ _ _ . _ _ _ m ._
Calc.No. 96-ENG-01528E2 Rev.00 Page&) of BM ATTACHMENT A 1.25 in thick TSI CONDUlT MODEL The following is a model of Thermo-Lag on Conduit using the TU Test Results .
The thermal resistance model of a cable in conduit consists of the following thermal resistance components.
Ri thermal resistance of the insulationfjacket Rsd thermal resistance between the outer cable jacket to the inside conduit wall Rd thermal resistance of the conduit Rag thermal resistance between the outer surface of the conduit and the inside surface of thermal Lag shellproduct.
Rthi thermal resistance of the thermal tag shell Re thermal resistance between the outer surface of the thermal Lag and ambient air Rtot is the sum of the above thermal resistnace terms / thermal resistance between the conduutor and ambient air.
Methodology to develop an ampacity Model of a power calbe in Thermal Lagged conduit using TU test:
- 1. Calculate the heat generated by one conductor during both the base line condition (O) and thermal lagged conduit conditions (Op).
- 2. Using a known value of Op, calculate Rtot, thermal resistance between conductor and ambient air, during test of thermal Lag conduit.
- 3. Calculate thermal resistance of thermal Lag (Rhtl) using accepted equation for calculating
, the thermal resistance of a cylinder. Equation 38 from Neher-McGrath technical paper is used to calculate Rthl.
- 4. Using equation for Re, thermal resistance between thermal Lag and ambient air, calculate Ts, surface temperature of thermal lag. Ts is solved using MathCad givene/ find function.
Once Ts is solved, Re is then calculated.
I
- 5. With Re, Riot, and Rthi known, the thermal resistance between conductor and inside surface of thermal Lag (Ri_thl) can be calculated. Rl_thiis considered to be constant regardless of the Thermal Lag.
- 6. A new Rthi is calculated for a new theraml Lag thickness, for MS2 the thickness is 1.25 inch.
- 7. Two equations are defined with respect to Opp, heat generated for new thickness of thermal Lag. These equations are set equal to each and solved for TS. Opp is then solved using the value of Ts.
B. The ampacity correction factor (ACF) for the thermal tag conduit is then solved.
Calc.No.96-ENG-01528E2 Rev.00 Page#Mof A*
ATTACHMENT A
- 1. Calculate Parameters Associated with Test Conditions INPUT:
j ETSI := 0.9 emissivityof thermalLag TSI :: 0.5 thickness of thermal Lag in test, inches pTSI := 577.7 thermal resistivity of thermal lag, C-cm/ watt (calculated from a value of thermal conductivity of 1.0 BTU /hr-ft-F.
Dend : 5.5 outer diameter of conduit, inch.
-3 Rac := 0.023 10 ac resistance of 750 kemil Cu Ib :: 571 base current in TU test for the thermo-lagged 5 in conduit, amps Ip :: 510 base current in Tu test, amps np:= 4 number of conductors in conduit Tc ::90 conductor termperature, C Ta ::40 ambient temperature, C O, heat generated in Test Base Line Conditions 2
Q :: Ib Rae heat generated by single conduitro, in base condition (watts /ft).
Q = 7.499 watts /ft Op, heat generated in test conditions of 1/2 inch thermal tag conduit Qp :: 1p2Rae heat generated by single conductor, in derated conditions of test.
Qp = 5.982 watts /ft Rtot, Total Thermal resistance betwesen conductor and ambier't during Test Rtot := thermal resistance between conductor and ambient air QP Rtot = 8.358 C-ft/ watt
Calc.No.96-ENG41528E2 Rev.00 Paged /(of h#
ATTACHMENT A t
' Re, thermal resistance between Thermal Lag surface and ambient during test 1
! Dsp :: Dervi + 2 TSI Ts :: 50 Initial guess surface temperature Re = "E 0.25 f fa T
-+- Tsi '
Dsp- Dsp l(Tsj - Ts3 l + 1.6 eTSI-1 + 0.0167 1 (2
l
- 2) ,
Calculate Ts, Ri-thi I
Dsp *: Dend + 2 TSI Outer diameter of thermal lagged conduit, where the outer diameter of the conduit is 5.5 inch.
Dsp = 6.5 inch Op= (Ts-Ta)/Re Equation relating Re and Op Ts = 50 Initial guess Given Ts - Ta QP-15.6 np l f Ts - Ta3 0.25 f <
Dsp 1 + 1.6 cTSI- 1 + 0.0167- I Ta + Tsi ' f !
( Dsp jl (2 2j,,
1 i
Ts *: Find ( Ts )
l l
Ts = 70.306 degree C ;
l I
i l
- _ _ - . . -. . . . .. . - = - - - . - - , - - . . . ~ ..
Calc.No.96-ENG-01528E2 Rev.00 Pagedatof 8No
, ATTACHMENT A Confirmation of Given/ Find calculation:
, Ts - Ta 2 = 5.982 -
15.6 np i watts /ft-2 i
fTs- Ta3025 f i' /in^2 Dsp 1 + 1.6 ETSI- 1 + 0.0167 I Ta + Ts l
( Dsp jl (2 2j,,
l i
e i 5.982 is equal to Op thus conferming the value of Ts.
J Calculate Re, thermal resistance between Thermal Lag surface and ambient air
- 'np I Re ::
.ITs - Ta3 0.25
-+-Tsi T
/a '
i
! Dsp 1 l + 1.6 cTSI- 1 + 0.0167 l 1 i
( Dsp / (2 2j '
Re = 5.066 C-ft/ watt l Calculate Rthl, thermal resistance of thermal lag, based on thickness used in TU Test I Dspi thermal resistance of thermal lag , equation Rth! ::0.012 np pTSI log I adapted from Equation 38 of Neher /McGrath
! Rth! = 2.012 C-ft/ watt Calculate RLthl, thermal resistance between conductor and thermal Lag i
Ri_thi :: Rtot - Re - Rth) thermal resistance between conductor and inside wallof thermaltag Ri_th! = 1.28 C-ft/ watt J
4 9
Calc.No.96-ENG41528E2 Rev.00 Page4/27 of 4/#
ATTACHMENT A
- 2. Calculate New Conditions TSI:=1.25 new thickness of TSI, inch Dsp *: Dend + 2 TSI outer diameter of thermallagged conduit, inch Dsp = 8 inch f Dspi thermal resistance of thermal lag, equation Rth! *:0.012 np pTSI log l
( Dend) adapted from Equation 36 of Neher-McGrath (Ref. 3.1.82).
Rthi = 4.512 The new Opp can be defined as follows:
Tc - Ts Ri_thi + R:hl Ts - Ta 15.6 np qpp .,
f Ts - Ta3 0.25 f Dsp- l + 1.6 ETSI- 1 + 0.0167 1 Ta + Tsi l '
( Dsp jl (2 2/ ;
Combining these two equations we can solve for Ts Ts 50 Initial guess of surface temperature ,
Given Tc - Ts _ Ts - Ta Ri_th! + Rth! 15.6 np
~"
Dsp- ( 'Dsp j + 1.6 ETSI- 1 + 0.0167 I + l 1 (2 2f,, j i
Ts := Find ( Ts )
Ts = 62.613 new surface temperature of thermal lag for new thermal lag thickness
_ . . . _ . _ . . _. _ .__. . _ _ . _ . . _ _ _ _ . . . _ - _ . _ _ _ . ~ _ . . _ _ _ _ . _ . . _ . _ . _ . . . _ . . . _ _ . _ _ - . . . . . = . _ _ . _
1 7
J Calc.No.96-ENG41528E2 Rev.00 Page##6dA#
1 i
! ATTACHMENT A 1
' Using the Ts and Opp values determined above:
Tc - Ts Qpp := Ri_thi + Rth!
Qpp = 4.728 i
E ACFtsi :=
1 Q Ampacity correction factor for 1.25 in TSI conduit wrap J ACFtsi = 0.794 ACFTU := $
1 1 Q Ampacity correction factor for 1/2 in TS! determined by TU ACFTU = 0.893 l
1 4
4 4 .
1 1
I i
a 4
j u
i d
Calc.No.96-ENG-01528E2 Rev.00 P ge A2'/of &fLo ATTACHMENT A 1.0 in thickTSI CONDUlT MODEL The following is a model of Thermo-Lag on Conduit using the TU Test Results .
The thermal resistance model of a cable in conduit consists of the following thermal resistance components.
Ri thermal resistance of the insulationfjacket Rsd thermal resistance between the outer cable Jacket to the inside condui wall Rd thermal resistance of the conduit Rag thermal resistance between the outer surface of the conduit and the inside surface of thermal Lag shellproduct.
Rthi thermal resistance of the termal tag shell Re thermal resistance between the outer surface of the thermal Lag and ambient air Rtot is the sum of the above thermal resistance terms / thermal resistance between the Ri_thl conductor and ambient air.
Methodology to develop an ampacity Model of a power cable in Thermal Lagged conduit using TU test:
- 1. Calculate the heat generated by one conductor during both the base line condition (O) and thermal lagged conduit conditions (Op).
- 2. Using a known value of Op, calculate Rtot, thermal resistance between conductor and ambient air, during test of thermal Lag conduit.
- 3. Calculate thermal resistance of thermal Lag (Rhtl) using accepted equation for calculating the thermal resistance of a cylinder. Equation 38 from Neher-McGrath technical paper is used to calculate Rthl.
- 4. Using equation for Re, thermal resistance between thermal Lag and ambient air, calculate Ts, surface temperature of thermallag. Ts is solved using MathCad given/ find function.
l
' l Once Ts is solved, Re is then calculated. '
- 5. With Re, Rtot, and Rthi known, the thermal resistance between conductor and inside surface of thermal Lagt (Ri_thl) can be calculated. Ri_thiis considered to be constant regardless of I the Thermal Lag. l l
- 6. A new Rthi is calculated for a new theraml Lag thickness, for MS2 the thickness is 1.25 inch.
- 7. Two equations are defined with respect to O", heat generated foi new thickness of thermal Lag. These equations are set equal to each and solved for TS. O" is then solved using the value of Ts.
- 8. The ampacity correction factor (ACF) for the thermallag conduit is then solved.
a
.m. . . . _ . _ _ _ _ _ . _ _ . . . _ _ _ . _ . _ _ . - . _ . _ . . . . _ - _ _ _ _ . . _ .
M Calc.No.96-ENG41528E2 Rev.00 Page Oof 4/#o ATTACHMENT A
- 1. Calculate Parameters Associated with Test Conditions i INPUT:
eTSI := 0.9 emissivityof thermalLag TSI := 0.5 thickness of thermal Lag in test, inches pTSI :: 577.7 thermal resistivity of thermallag, C-cm/ watt (calculated from a value of thermal conductivity of 1.0 BTU /hr-ft-F.
Dend : 5.5 outer diameter of conduit, inch.
f Rac : 0.02310~3 ac resistance of 750 kemil Cu, page 13 of ETP104.1-0 lb := 571 base current in Tu test, amps Ip := 510 base current in Tu test, amps
[ np:=4 number of conductors in conduit Tc :: 90 conducctor termperature, C
, Ta:=40 ambient temperature, C j
l 0, heat generated in Test Base Line Conditions 2
Q :: Ib Rac heat generated by single conduitro, in base condition (l^2R).
Q = 7.499 watts Mt
]
Op, heat generated in test conditions of 1/2 inch thermal lag conduit Qp :: Ip2Rac heat generated by single conductor, in derated conditions of test.
3 Qp = 5.982 watts /ft Rtot, Total Thermal resistance betwesen conductor and ambient during Test Rtot': thermal resistance between conductor and ambient air QP 2
Rtot = 8.358 C-ft/ watt i
l l
i 4
a i
I
Calc.No.96-ENG-01528E2 Rev.00 PIge4/# of M ATTACHMENT A Re, thermal resistance between Thermal Lag surface and ambient during test Dsp :: Dend + 2 TSI Ts::50 Initial guess surface temperature
., !5.6 np
~
Dsp- ( 'Dsp j + 1.6 ETSI- 1 + 0.0167- + l
, (2 2) , ,
i Calculate Ts, Ri-thi j i
Dsp :: Derx! + 2 TSI Outer diameter of thermallagged conduit, where the outer diameter of the conduit is 5.5 inch.
Dsp = 6.5 inch Op= (Ts-Ta)/Re Equation relating Re and Op Ts = 50 Initial guess l Given Ts - Ta QP-15.6 np
'~ "
Dsp-l + 1.6 ETSI- 1 + 0.0167 I + l
( Dsp jl (2 2),,
l Ts : Find (Ts)
Ts = 70.306
. .. -. -.-=~. ~. - ~ . ~ .
Calc. No. 96-ENG 01528E2 Rev.00 ' Page&of A#
ATTACHMENT A Confirmation of Given/ Find calculation: ,
Ts - Ta
= 5.982 15.6 np f Ts - Ta3 0.25 f. .
Dsp I l + 1.6 eTSI- 1 + 0.0167 1 j,, + g.,3 -
4 Dsp j (2 2 jl , ,
5.982 is equal to Op thus conferming the value of Ts.
Calculate Re, thermal resistance between Thermal Lag surface and ambient air Re := "E f Ts - Ta3 0.25 I Dsp 1 i + 1.6 ETSI- 1 + 0.01671 Ta + Tsi '
( Dsp j (2 2 jl ,
Re = 5.066 C-ft/ watt Calculate Rthl, thermal resistance of thermallag, based on thickness used in TU Test fDspi thermal resistance of thermal lag , equation Rth! := 0.012 np pTSI logI adapted from Equation 38 of Neher /McGrath Rth! = 2.012 C-ft/ watt -
Calculate Rl_thl, thermal resistance between conductor and thermal Lag Ri_th! :: Rtot - Re - Rthi thermal resistance between conductor and inside wallof thermallag
, Ri_th! = 1.28 C-ft/ watt l
i i
l l
Calc.No.96-ENG-01528E2 Rev.00 Page43 Eof 4/W ATTACHMENT A
- 2. Calculate New Conditions TSI := 1.0 new thickness of TSI, inch Minimum Dsp :: Dend + 2 TSI outer diameter of thermal lagged conduit, inch Dsp = 7.5 inch IDspi thermal resistance of thermallag, equation I Rth! := 0.012 np pTSI log (Dend; ad:.:pted from Equation 38 of Neher-McGrath Rth! = 3.735 i l
The new Opp can be defined as follows:
Tc - Ts Ri_thi + Rthi >
Ts - Ta Qpp ::
- ~ "
Dsp- + 1.6 eTSI- 1 + 0.0167- l +
( Dsp jl (2 2)j ,
Combining these two equations we can solve for Ts Ts::50 Initial guess of surface temperature Given Tc - Ts _ Ts - Ta Ri_thl + Rth! 15.6 np Dsp 1 fTs - Ta3 0.25 l + 1.6 eTSI- 1 + 0.0167 l -+-Tsi '
T fa l
( Dsp j (2 2j, Ts := Find ( Ts )
Ts = 64.648 new surface temperature of thermal tag for new thermal tag thickness i
- . . . . -_. ~ - . . . . _ - - - _ . . . ~ - . . . . . - - . . . . . . - . . - . - . _ ~ . . . _ . . . . _ .. . . _ _ ,
1 1 Calc.No.96-ENG41528E2 Rev.00 Page A/Wof 4/#
i
! ATTACHMENT A 4
) Using the Ts and Opp values determined above:
i 4
l . Tc - Ts 4
i QPP : Ri_th! + Rthi d
- Qpp = 5.055 i OPP ACFtsi
- :
1 Q I . Ampacity correction factor for 1.0 in TSI conduit wrap 1 ACFtsi = 0.821 1
i I
i j
4
}
2 i
i k
i a
d 4
4 1
i 4
! 1 I
E i
k
- 4
Calc.No.96-ENG41528E2 Rev.00 Pge#/$fBM ATTACHMENT 1 A 1.5 in thick TSI CONDUIT MODEL The following is a model of Thermo-Lag on Conduit using the TU Test Results .
The thermal resistance model of a cable in conduit consists of the following thermal resistance components.
Ri thermal resistance of the insulation /acket J
Rsd thermal resistance between the outer cable jacket to the inside conduit wall Rd thermalresistanceof theconduit Rag thermal resistance between the outer surface of the conduit and the inside surface of thermal Lag shell product.
Rthi thermal resistance of 'he termal lag shell Re thermal resistance between the outer surface of the themial Lag and ambient air Rtot is the sum of the above thermal resistance terms / thermal resistance between the Ri_th! conductor and ambient air.
Methodology to develop an ampacity Model of a power cable in Thermal Lagged conduit using TU test:
- 1. Calculate the heat generated by one conductor during both the base line condition (O) and thermal lagged conduit conditions (Op).
- 2. Using a known value of Op, calculate Rtot, thermal resistance between conductor and ambient air, during test of thermal Lag conduit.
- 3. Calculate thermal resistance of thermal Lag (Rhtl) using accepted equation for calculating the thermal resistance of a cylinder. Equation 38 from Neher-McGrath technical paper is used to calculate Rthl.
- 4. Using equation for Re, thermal resistance between thermal Lag and ambient air, calculate Ts, surface temperature of thermal lag. Ts is solved using MathCad given/ find function.
Once Ts is solved, Re is then calculated.
. 5. With Re, Rtot, and Rthl known, the thermal resistance between conductor and inside surface of thermal Lagf (Ri_thi) can be calculated. Ri_thiis considered to be constant regardless of the Thermal Lag.
- 6. A new Rthiis calculated for a new theraml Lag thickness, for MS2 the thickness is 1.25 inch.
- 7. Two equations are defined with respect to O", heat generated for new thickness of thermal Lag. These equations are set equal to each and solved for TS. O' is than solved using the value of Ts.
- 8. The ampacity correction factor (ACF) for the thermallag conduit is then solved.
Calc.No.96-ENG41528E2 Rev.00 Page4# of #*
ATTACHMENT A
- 1. Calculate Parameters Associated with Test Conditions INPUT:
ETSI :: 0.9 emissivityof thermalLag TSI := 0.5 thickness of thermal Lag in test, inches pTSI :: 577.7 thermal resistivity of thermal lag, C-cm/ watt (calculated from a value of thermal conductivity of 1.0 BTU /hr-ft-F.
Dend : 5.5 outer diameter of conduit, inch.
~3 Rac := 0.02310 ac resistance of 750 kemil Cu, page 13 of ETP104.1-0 lb :: 571 base current in Tu test, amps Ip :: 510 base current in Tu test, amps np ::4 number of conductors in conduit Tc :: 90 conductor termperature, C
'la *: 40 ambient temperature, C O, heat generated in Test Base Line Conditions 2
Q :: Ib Rac heat generated by single conduitro, in base condition (l^2R).
Q = 7.499 watts /ft Op, heat generated in test conditions of 1/2 inch thermal lag conduit l 2
Qp :: Ip Rac heat generated by single conductor, in derated conditions of test.
Qp = 5.982 watts /ft Rtot, Total Thermal resistance betweeen conductor and ambient during Test Rtot :: thermal resistance between conductor and ambient air Qp Rtot = 8.358 C-ft/ watt
Calc.No.96-ENG41528E2 Rev.00 Page#57 of #N3 ATTACHMENT A Re, thermal resistance between Thermal Lag surface and ambient during test Dsp *:Dend + 2 TSI Ts:=50 Initial guess surface temperature nP
. Re :=
0 i Dsp l fTs- Ts3 25 l + 1.6 cTSI- 1 + 0.0167 1 fa T
-+- Tsi '
1
( Dsp j (2 2),
Calculate Ts, Ri-thi i
Dsp :: Dend + 2 TSI Outer diameter of thermallagged conduit, where the outer diameter nf the conduit is 5.5 inch.
Dsp = 6.5 inch
.I l Op= (Ts-Ta)/Re Equation relating Re and Op l
Ts = 50 initial guess Given Ts - Ta i g-15.6 np f Ts - Ta3 0.25 j Dsp l
( . Dsp j I + 1.6 ETSI- 1 + 0.01671 -+-Tsi '
fTa l
(2 2),,
i l Ts := Find ( Ts )
i Ts = 70.306 3
i l
3 1
i j
4 j
4 1
4 I
i
Calc.No.96-ENG41528E2 Rev.00 Page M 8 of#<#
]
1 ATTACHMENT A J
Confirmation of Given/ Find calculation:
j Ts - Ta a
= 5.982
{ 15.6 np i f f Ts - Ta3 0.25 Dsp I l + 1.6 eTSI- 1 + 0.0167 ! Ta + Tsi i '
j.
( Dsp j , (2 2j,,
a 5.982 is equal to Op thus conferming the value of Ts.
Calculate Re, thermal resistance between Thermal Lag surface and ambient air i.
15.6 np 4' Re := -
fTs - Ta3025 f l Dsp-I l + 1.6 eTSI- 1 + 0.0167-l Ta + Tsi '
l ( Dsp j (2 2 jl , !
i 1 4 1 1
I Re = 5.066 C-ft/ watt i Calculate Rthi, thermal resistance of thermal lag, based on thickness used in TU Test f DspI thermal resistance of thermallag , equation Rth! ':0.012 np pTSI logl(M;l adapted from Equation 38 of Neher /McGrath i Rth! = 2.012 C-ft/ watt i
. Calculate Rl_thl, thermal resistance between conductor and thermal Lag 9
- 3. Ri_th! :: Riot - Re - Rthl thermal resistance between conductor and inside l wallof thermallag Ri_th! = 1.28 C-ft/ watt 1
i 1
.._______m.__.. . . - _ . _ _ . _ _ . _ . . _ - . . . _ . . . - . _ _ . . _ . _ _ . _ _ . . _ . . . . _ . . - _ _ ~ . _ _
Calc.No.96-ENG41528E2 Rev.00 Page AM of AtW )
i ATTACHMENT A 2 Calculate New Conditions l l
TSI : 1.5 new thickness of TSI, inch Minimum Dsp :: Dend + 2 TSI outer diameter of thermal lagged conduit, inch Dsp = 8.5 inch IDspi thermal resistance of thermallag, equation Rth! := 0.012 np pTSI logl
( D nd; adapted from Equation 38 of Neher-McGrath Rth! = 5.242 The new Opp can be defined as follows:
Tc - Ts Qpp ::
Ri_thl + Rth!
Ts - Ta Qpp ::
f Ts - Ta3 0.25 f Dspl I + 1.6 cTSI- 1 + 0.0167 1 Ta + Tsi
( Dsp j , (2 2 jl ,
Combining these two equations we can solve for Ts l
Ts:: 50 initial guess of surface temperature Given Tc - Ts _ Ts - Ta Ri_thi + Rthi 15.6 np f Ts - Ta3 0.25 f Dsp1 I + 1.6 ETSI- 1 + 0.01671 Ta + Tsi l '
( Dsp j (2 2),,
Ts := Find ( Ts ) :
Ts = 60.918 new surface temperature of thermal tag for new thermal lag thickness l l
1 i !
l l
4 l
Calc.No.96-ENG41528E2 Rev.00 Page A@of M@
p ATTACHMENT A ,
Using the Ts and Opp values determined above:
Tc - Ts QPP _ Ri_thl+ Rth!
2 Qpp = 4.458 P
ACFtsi :=
1 Q Ampacity correction factor for 1.5 in TSI conduit wrap ACFtsi = 0.771 i l
1 l
I 1
i l
i CALC. NO. 95-ENG-01528E2 REV.00 PAGE / OF 4'e r- l ATTACHMENT B TABLE OF CONTENTS APPENDIX TRAY CABLE l A Z23HA10 500 MCM CU TPLX l B Z23HB10 500 MCM CU TPLX l C Z23GE10 500 MCM CU TPLX I D Z23GE10 250 MCM CU TPLX i
E Z14FM10 3/C #8 AWG CU F Z14FM20 3/C #12 AWG CU G Z24FL10 4/0 AWG CU H Z14FM20 4/0 AWG CU TPLX l 1 Z24FL20 4/0 AWG CU i K Z25BG20 3/C #8 AWG CU M Z14FM10 3/C #12 AWG CU N Z14FM10 4/0 AWG CU TPLX l V Z23FA30 250MCM CU TPLX W Z23FA25 250MCM CU TPLX X Z24FL10 7/C #14 AWG CU AA Z25BG20 3/C #14 AWG CU AB Z25BG20 2/C #14 AWG CU AC Z25BG20 3/C #12 AWG CU AD Z25BG20 2/C #10 AWG CU B1 Z52EA10 750MCM ALTPLX ,
B2 Z22EA10 350M CM AL TPLX TEST RUNS
SUMMARY
SHEET TEST APPENDIX (ICEA-1986) APPENDIX (ICEA-197!!)
- 1 P PP 2 0 QQ 3 T TT 4 U UU
. 5 R RR 6 S SS This attachment contains the calculations for the allowed ampacity in each of the above listed trays and the type of cable installed in the tray. Each appendix is a separate calculation performed in MathCad on a Macintosh Centris 650 computer. Verification that the calculations are correct are based upon the results of the 6 test runs included herein. The test runs were ran at a fill and wire diameter taken from the ICEA P-54-440 and the values compared against the values presented in the two years identified above of the standard. The results indicate that the calculations are being performed correctly.
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE81 OF M APPENDIX A PAGE 1 THERMAL MODEL Z23HA10 Values 10% fill on 4" 1.0 Tray Thermal Data 0.509 in/B14 -
w := 24 in Wkith.inTray R=0.03135?'5/1000/ft cad =2.178 d *:0.509 in Depth of cables in Tray 500MCMTPLX Ta ::( 273.16 + 40)- K Ampient Temperature Tc := unknown K Sunace Temperature Tm := ( 273.16 + 90) K Max CableTemperature f 4 I watt Overall Convective Heat transfer Coeffic ient
, .= g, , g , . Tc _ Tai LK Kj 2 ft - K h := ( 0.101 )-
2 ft K
~8 **"
o : 0.530 Stephan-Boltzman Constant !
2 4 rt.x nnal MsW of 6 mass a@my c := 0.8 surface.
p := 400- K- effective thermal resistivity of cable mass watt
.a ft2 Surface area of cable mass per unit length As ._4 T sides excluded Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) 1 I 4 4 T
fc Tai fTc 'I'd l - l h-( As p d) l -
l h-( As p d) o As-E p d,. 4 (K K) (K K) _ o- As e p d, 4 Tc : m e 8w 8- w 8w 8w Tc = 357.577 *K
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B - PAGE U OF M APPENDIX A PAGE 2 j 2.2 Calculated value of total heat (W) per unit length generated in cable tray 3
1 I Tc Tai d 4 4 W := h 1 -
I - As-( Tc - Ta ) + o As-c-( Tc - Ta ) ,
(K Kj ;
W = 160.481
- ft 2.3 Recalculated Max Cable Temperature (Tm) ;
i 1 ;
I Tc # 4 Tai fTc Tai 4
hi -
I -( As p d) hl -
l - As p dTa 4 4 :
(K (K K) + o As-t r d Tc - o As E p dTa KJ Tm :: +1 -Tc-8w 8w 8w 8w Tm = 363.16 ,
2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise !
ATc : Tm - Tc .
ATc = 5.583 K i Wpd l ATc := !
8- w ATc = 5.583 *K 2.4.2 Air Temperature Drop i
ATa = Tc - Ta :
i ATa = 44.417 'K 2.4.2 Total temperature Drop [
AT := ATc + ATa i
AT = 50 *K ;
2.5 Heat generated per unit area W
- watt Q := d w !
Q = 13.137 ft2 in L
_ - _ _ _ - - - - _ _ - - _ - _ - _ ____-_=__ _-__ -_ __ _ -. - __ - __ _ _ _ _ _-_- _-_ J
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE0+ OF M APPENDIX A PAGE 3 Ampaaty Calculation n:=3 number of conductors cad :: 2.1781 in Cable outer diameter R ::0.0313575- Resistance taken from Okonite book 1000- ft I40..- cad Qx 2 1nR i
140 = 721.338
- amp Tray Z23HA10 Ampacity at40C 150 :: 140-0.89 Correction factor to get to 50C 150 = 641.991
- amp Arnpacity at 50C From IPCEA 80% of the arnpacity of 500MCM at 40C is 464 amps (580 x .8) Adjusting to 50C gives:
lair 80 := 464 0.89 amp ;
lair 80 = 412.%
- amp nuximum ampacity allowed at 50C i
i i
l h
k t
h
CALC. NO. 96-ENG-01528E2 REV.0 ATTACHMENT B PAGE 6 OF M A AGE 1 THERMAL MODEL Z23HB10 Values 19% fill on 4" 1.0 Tray Thermal Data 0.968 irdB14 w :: 12 in WKith in Tray H=0.0313575/1000ft cad =2.178 d :: 0.%8-in Depth of cables in Tray 500MCMTPLX Ta : ( 273.16 + 40) K Ampient Temperature Tc :: unknown K Surface Terrgerature Tm ::( 273.16 + 90) K Max CableTemperature 1
un mR o w Heat tmnsfehhnt h ::0.101- 1 -
(K Kj 2 ft K h := ( 0.101 }-
2 ft K
-8 **"
o :: 0.530 10 - --
Stephan-Boltzman Constant ,
2 4 ft , g i
. Effective thermal emrnesivity of cable mass and tray i e _ 0.8 surface. t
- i p := 400- K- effective thermal resistivity of cable mass watt t 2
._ ft Surface area of cable mass per unit length
^5 ~~ 4 T sides excluded i
l Tc := Tm + Ta ,
2 Tc = 328.16 *K Initial assumed value for Surface Temp. (Tc) r 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) i I I ITc Tai # fTc Tai 4 I l - l h-( As p d) ' -
l h-( As p d) .
Tc := m As c p d, 4 (K Kj _
(K Kj _ o As E p d, 4
- 8w 8w 8w 8- w w
Tc = 347.548 *K i
CALC. NO. 96-ENG-01528E2 REV.0 ATTACHMENT B PAGE N OF M APPENDIX B PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray ,
I f Tc Tai # 4 4 W := h- l -
l - As-( Tc - Ta ) + o As c-( Tc - Ta )
LK K/ i W = 117.978
- ft 2.3 Recalculated Max Cable Temperature (Tm)
- I I fTc Tai # fTc Tai #
h-i -
l -( As-p d) h-l -
l - As p-dTa 4 4 ;
Tm : (K Kj
+1 Tc -
(K Kj + c- As t p-d Tc - o As t p dTa i 8w 8w 8w 8w j Tm = 363.16 t
2.4 Temperature Rises f 2.4.1 Cab le Mass temperature Rise f
ATc := Tm - Tc l ATc = 15.612 *K '
W p-d ATc *:
8w ATc w 15.612 *K l 2.4.2 Air Temperature Drop ATa := Tc - Ta 7 ATa = 34.388 *K h 2.4.2 Total temperature Drop ,
AT :: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area !
W Q::dw watt ft .
Q = 10.157
- 2 in 1
_ _ _ _ _ _ _ _ . _ _ . _ _ .._______m______ . . _ . _ . _ _ _ _ _ _ . _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ . _ . _ . _ _ _ _ _ _ . _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ ._ _ _ _ _ . _ _ - . _ _ _ _ _ __
CALC. NO. 96-ENG41528E2 REV.0 ATTACHMENT B PAGE 67 OF A" APPENDIX B ~ PAGE 3 Ampacdy Calculation n := 3 number of conductors cad :: 2.178 in - Cable outer diameter R ::0.0313575- Resistance taken from Okonite book 1000- ft I40._ cad Qx 2 inR 140 = 634.226
- amp Tray Z23HB10 Ampaoty at40C 0.89 Correction factor to get to 50C 150 = 564.462
- amp Ampacdy at 50C From IPCEA 80% of the ampacdy of 500MCM at 40C is 464 amps. Adjusting to 50C gives:
lair 80 ::464 0.89 amp lair 80 = 412.96
- amp maximum ampacdy allowed at SOC
. _ _ - _ - . _ - _ ~ . . -
l 7 CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGEO* OFM A AGE 1 THERMAL MODEL Z23GE10 Values 22% fill 4* tray
- 1.0 Tray M Ma 1.12 in/ B14 w := 24 in Width in Tray - R=0.0313575/1000ft cad =2.178 d :: 1.12 in Depth of cables in Tray 500MCMTPLX Ta := ( 273.16 + 40) K Ampient Temperature Tc := unknown K Surface Temperature Tm ::( 273.16 + 90) K Max CableTemperature 1
wan mR aMmnsfehk knt h ::0.101- - -
AK Kj n 2 ,g h:=(0.101)-
ft2- K a : 0.530- 10~8-
~
Stephan-Boltzman Constant 2 #
ft K c := 0.8 Effectrve thermal emmisivity of cable mass and tray surface.
p := 400- K- effective thermal resistiwty of cable mass watt 2
Surface area of cable mass per unit length As ._4 ft N ed W Tc :: Tm + Ta Tc = 338.16 'K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) 1 I Ic T Tal # T fc Tal l -
I h-( As p d) 1 -
l -h-( As p-d)
Tc := m As c p d 4 (K Kj iK Kj , _ o As- c p d, ,,, ,,
4 .rc 8w 8w 8w 8w Tc = 352.541 *K
CALC.NO.96-ENG4152SE2 REV.0 ATTACHMENT B PAGEOf OF M APPENDIX C PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray J.
I Tc #
Tai 4 W := h- l -
1 - As-( Tc - Ta ) + o- As-t-( Tc - Ta# )
iK Kj W = 138.718
- ft 2.3 Recalculated Max Cable Temperature (Tm)
I I I 4 I Tc 4 Tai hl Tc -Tai- I -( As p-d) hl -
l ' As p dTa (K Kj (K K/ o As c p-dTc4 _ o- As c p-d Ta4 Tm ': , _
8w 8w 8- w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc ::Tm - Tc ATc = 10.619 *K W p-d ATc *:
8w ATc = 10.619 *K 2.4.2 Air Temperature Drop ATa != Tc - Ta ATa = 39.381 *K 2.4.2 Total temperature Drop AT := ATc + ATa AT = 50 *K 2.5 Heat generated per unit area W
watt Q::dw Q = 5.161
- 2ft in .
,,____m ... . _ . . . _ . . . . - . . . _ _ _ _ _ . _ - . _ . . _ _ . _ _ . ._ .- . - _ . _ . ..
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE hOF 8#" !
~
APPENDIX C PAGE 3 ,
4 Ampacity Calculation n::3 number of conductors ,
cad := 2.1781 in Cable outer diameter ;
R ::0.0313575- Resistance taken from Okonite book 1000- ft 140.-_ cad Qx 2 1 nR 140 = 452.109
- amp Ampacity at40C .
t i
150 := 140-0.89 A aN 150 = 402.377
- amp ,
f
(
rom IPCEA 80% of the ampacity of 500MCM at 40C is 464 amps. '
Adjusting to 50C gives:
i 0
lair 80 :=464 0.89 amp lair 80 = 412.96
- amp Maximum ampacity allowed at 50C. ,
t 1
I i
l
_ .m . _ . _ .. ._- . ~.__. ..._____ . ._. . . _ . . _ _ . . . , _ _ . _ _ _ - . . _
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE &' OF 8##
APPENDIX D PAGE1 THERMAL MODEL Z23GE10 Values 22% fill on 4*
1.0 Tray Thermal Data 1.12 in/ B10 w := 24 in Wdth in Tray R=0.0594925/1000ft <
1- 9 d :: 1.12 in Depth of cables in Tray p Ta :=( 273.16 + 40) K Ampient Temperature Tc *: unknown K Surface Temperature i Tm ::( 273.16 + 90) K Max CableTemperature I
fc T Tai watt Overall Convective Heat transfer Coeffic ient h ::0.101 i -
l (K K) 2 nK watt l h := ( 0.101 )-
2 ft , g
~8 **"
o :: 0.530 10 - Stephan-Boltzman Constant +
n2.g4 Ie mass aM hy E ::0.8 surface.
p ::400 K- effective thermal resistivity of cable mass watt e
2
._ ft Surface area of cable mass per unit length
^5 '- 4y sides excluded Tm + Ta 2 '
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) i 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) ,
1 I i I 4 I Tc Tal Tal #
l Tc - j -h-( As p d) l - j -h-( As p d) .
As c p-d 4 (K K) , _ _
(K Kj _ & As t p d 4 l Tc : mi , ,
8w 8w 8w 8w Tc = 352.541 *K
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE M OF8 f t
APPENDIX D PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray
.I.
Ic T Tal 4 4 4 W:=hl - 1 As-( Tc- Ta) + o As c-( Tc - Ta )
(K Kj W = 138.718
- ft 2.3 Recalculated Max Cable Temperature (Tm) 1 I Ic T Tai fc T Tai #
h1 - l -( As p d) h-l -
1 - As p-d Ta 4 4 ,
(K KJ (K Ki + a As t p d Tc o As E p-dTa Tm :: +1 Tc -
8w 8w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc ::Tm - Tc ATc = 10.619 *K W- p d gw ATc = 10.619 *K t
2.4.2 Air Temperature Drop ATa ::Tc - Ta ATa = 39.381 *K 2.4.2 Total temperature Drop AT := ATc + ATa AT = 50 *K 2.5 Heat generated per unit area W watt Q: dw ft Q = 5.161
- in2
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGEBISOF APPENDIX D PAGE 3 i
Arnpacity Calculatkxi n:=3 number of conductors ;
cad := 1.719 in Cable outer diameter .
4 dm R := 0.0594925- Resistance taken from Okonite book 1000 ft ,
QR I40 .-._ C2O-2 1 nR ,
140 = 259.049
- amp TrayZ23GE10 Ampacity at40C 150 := I40 0.89
" 'W Ampacity at 50C
- 150 = 230.553
- amp ;
From IPCEA the 80% of the free air value 299,180 > 140 180 .-._374 .8-amp thus the 150 value is the correct value to use.
180 = 299.2
- amp 6
L 4
i I >
CALC. NO. 96-ENG-01528E2 ' REV.0 ATTACHMENT B PAGE M OF M APPENDIX E PAGE 1 THERMAL MODEL Z14FM10 VALUE 1.0 Tray Thermal Data 12% FILL in 4" Tray 0.611 in/ B03 w := 24 in Wdth in Tray R=0.84875/1000ft d := 0.611-in Depth of cables in Tray Ca =E C#8 Ta ::( 273.16 + 40) K Ampient Temperature Tc :: unknown K Surface Temperature Tm := ( 273.16 + 90)-K Max CableTemperature I
wa m a mnsh Mb M h :: 0.101- l 1-( K- K) g 2 -K h ::( 0.101 )-
ft2- K
~8 watt o :: 0.530 10 - Stephan-Boltzman Cc.1stant a2.g4 c := 0.8 Effective thermal emmesivity of cable mass and tray surface.
- p :: 400 K- effechve thermal resistivity of cable mass watt
._ ft2 Surface area of cable mass per unit length
^5 '- 4 T sides excluded Tm + Ta Tc ::
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculaten 2.1 calculated Value for Surface Temperature (Tc) 1 I f Tc I #
Tai # Tal I -
l h-( As p d) lTc -
l -h-( As p d)
^ Ta#, Tc Tc := root Tc# + +1 -Tc - Tm - Ta -
l 8w 8- w 8w 8w Tc = 356.63i *K
. - . . . - _ _ _ - - - - - - _ _ _ .__ _ _ - - _ _ _ _ . _ _ - _ _ - _ - _ _ _ - _ _ - _ - _ _ _ - - - _ _ - _ __ _ - _ _ - - - _ _ _ - _ . _ _ _ _ _ _ - _ _ _ _ = _ _ _ _ . _ _ _ _
CALC. NO.695-ENG-01528E2 REV.0 ATTACHMENT B - PAGE 26._OF AD "
APPENDIX E PAGE 2 22 Calculated value of total heat (W) per unit length generated in cable tray
.I I 4 Tai W := h l Tc - - 1 4
As-( Tc - Ta ) + o As c-( Tc - Ta# )
(K Kj W = 156.331
- 2.3 Recalculated Max Cable Temperature (Tm) It 1 i f Tai # T Ic Tai h-lTc -
l -( As p d) h-l - j As p dTa 4 4 (K K) (K K) + o As e p dTc - o As-e p-d Ta Tm *: +1 Tc -
8w 8w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc 3 Tm - Tc ATc = 6.529 *K Wpd O"
- g. w ATc = 6.529 *K 2.4.2 Air Temperature Drop ATa ::Tc- Ta ATa = 43.471 *K 2.4.2 Total temperature Drop AT : ATc + ATa .
AT = 50 *K 2.5 Heat generated per unit area W watt q:= dw Q = 10.661
- 2ft in
CALC.NO.695-ENG-0 528E2 REV.0 ATTACHMENT B PAGEGNOFikov ,
APPENDIX E PAGE 3 Ampacdy Calculation n: 3 number of conductors cad :2 0.774 in Cable outer diameter r
R : 0.84875- Resistance taken from Okonite book ,
1000-ft 140._ cad Qx >
t 2 1 nR i 140 = 44.385
- amp Tray Z14FM10 Ampacdy at40C i
O*
- O 150 := I40-0.89 i
A at E 150 = 39.502
- amp Since 150 is less than 80% of the free air value at 50C 150 is the correct value.
i t
I L
F l
r i
r r
8 h l
i
i 1
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE681 OF APPENDIX F PAGE1 THERMAL MODEL Z14FM20 i Values l 9% fill on 4" 1.0 Tray ThermalData 0.458 in w := 24 in Wdthin Tray R=2.15/1000ft Depth of cabbs in Tray cad =0.603 d := 0.458 in 3/C#12AWG :
Ta ::( 273.16 + 40) K Ampient Ternperatere Tc :: unknown-K Surface Temperature Tm : ( 273.16 + 90) K Max CableTemperature
.I. ;
wan mH e mnsfer Cm knt h ::0.101- -
(K K) 3 2 .g h:=(0.101)- 2 ft , g i
~8 watt o :: 0.530-10 - Stephan-Boltzman Constant !
R 2.g4 m sW of 6 mass a@ ray ;
e := 0.8 surface. ,
p ::492.496 K '" effective thermal resistivity of cable mass j watt -
2 Surface area of cable mass per unit length As:: 4.R sides excluded i ft Tm + Ta 2 ,
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) a 2.0 Thermal Calculation 1 calculated Value for Surface Temperature (Tc) !
I 1 i
Ic T Tai 4 I'I'c "l'ai # --
i - 1 h-( As p d) l -
I h-( As p d)
Tc ::m - As t p d, 4 (K Kj (K K) ,
_ c As e p d, 4 8w 8w 8w ; 8- w -
Tc = 357.062 *K ,
.. -. = .- .. -- .- .- . . - . . . . . . . .
CALC.NO.9&ENG4)1528E2 REV.0 ATTACHMENT B PAGE 689 OFMV APPENDIX F PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray i
_1 f Tc Tai # 4 d W := h I -
l - As-( Tc - Ta ) + o As c-( Tc - Ta )
(K Kj W = 158.217
- ft 2.3 Recalculated Max Cable Ternperature(Tm)
I 1 I #
fc T Tai #
Tai hI -
I -( As p d) h-lTc -
l As p dTa 4 4 Tm :: +1 Tc - + # ^*' # - # "#
- 8- w 8w 8- w 8w ,
Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ,
ATc ::Tm - Tc ATc = 6.098 *K W p-d ATc :
8w ATc = 6.098 *K 2.4.2 AirTemperature Drop ATa :: Tc - Ta ATa = 43.902 *K 2.4.2 Total temperature Drop i AT :: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area W watt Q := d w ft 1 Q = 14.394
- 2 in
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE BeiOF 88V APPENDIX F PAGti 3 c Ampacity Calculation n ::3 number of conductors .
cad ::0.603 in Cable outer diameter R := 2.15- Resistance taken from Okonite book 1000 ft Q-x I40 _ cad 2 1nR 140 = 25.245
- amp TrayZ14FM20 Ampacity at40C stion fadorM pt to E _
150 :: 140 0.89 A aN 150 = 22.468
- amp i
i Since 80% of the free air value is 26.24 amps 150 above is less than this value,150 is the correct value to use.
i t
i I
t
- - . _2 m _.__.m__._._.___.__.2-___._._._a_. -
- - - - . --..___.____m.m._ m._-w.._ ___--.m__.._.. _ _ _ _ _ . , . . _ _ _ , _. . _ . . _ _ _ _ ___--m_. .__._.-- _ - - - _ _ _ ma__ __-_: __A- ._.
CALC. NO. 96-ENG-01528E2 REV.0 ATTACHMENT B PAGE Sao op &*
APPENDIX G Page 1 THERMAL MODEL Z24FL10 t 1.0 Tray Thennal Data VALUE' w := 24 in Widthin Tray 4% FILL in 4* Tray d := 0.2037 in Depth of cables in Tray 0.2037 in/B29 R=0.06890625/1000ft Ampient Temperature cad =0.738' Ta :=( 273.16 + 40) K 4W CM Tc :: unknown-K Surface Temperature Tm := ( 273.16 + 90) K Max CableTemperature 3
wau H n e Hea mnsfer NR knt h :: 0.101- -
1-(K Kf RK2 h:=(0.101)- 2 ft K >
c :: 0.530 8, wan Stephan-Boltzman Constant R 2.g4 e := 0.8 Effective thermal emmisivity of cable mass and tray surface.
p ::400 K- effective thermal resistivity of cable mass watt l
2 As : 4 g Surface area of cable mass per unit length ft sides excluded Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc)
I I fTc Tai j # T fc Tai !
I -
h-( As p d) l -
1 h-( As p d) 4 -Ta#, Tc Tc :: root -Tc + +1 Tc - Tm - - Ta -
8w 8w 8w 8w Tc = 360.729 *K
+
f
CALC. NO. 96-ENG-01528E2 REV.0 ATTACHMENT B PAGE E OF V APPENDIXG PAGE2 2.2 Calculated value of total heat (W) per unit length generated in cable tray i
I Tc tai 4 4 4
! W :: h- l -
1 - As-( Tc - Ta ) + o As c-( Tc - Ta )
iK Kj W = 174.535
- ft 2.3 Recalculated Max Cable Temperature (Tm) 1 I I ,fTc #
Tai 4 Tal h-lTc -
l -( As p-d) hi -
l - As p-d Ta 4 4
'E # #
- Tm :: +1 Tc - + -
8w 8w 8w 8w Tm = 363.159 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc ::Tm - Tc ATc = 2.43 *K ATc :: #
8w ATc = 2.43 *K 2.4.2 AirTemperature Drop ATa ::Tc - Ta ATa = 47.569 'K 2.4.2 Total temperature Drop AT : ATc + ATa AT = 49.999 *K 2.5 Heat generated per unit area W
Q:=dw watt It Q = 35.701
- in 2
. _ _ _, __ _ _ . . . .. ..._._m. _... , ____.. . .
CALC. NO. 96-ENG-01528E2 REV.0 ATTACHMENT B PAGEM OF #f0
- APPENDIX G Page 3
-i Ampacity Calculation n := 3 number of conductors y cad ::0.738 in Cable outer diameter R ::0.06890625- Resistance taken from Okonite book i 1000 ft Q-x I40._ cad -
2 1nR I40 = 271.802
- amp Tray Z24FL10 Ampacity at400 fetor to get to E 150 :: 140 0.89 t
150 = 241.903
- amp
^
- 180 (80% of the free air value) is 320 amps (400 amp x 0.8) . This is greater than the 271.802 amps thus the 150 value is the correct value to use. l i
. . . - - ._ . _ . _ _ _ . . . _ _ _ _ . _ . _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ , _._,._,._.______,___________,_,,._____.,.___,__,m_____,,
CALC. NO. 96-ENG-01528E2 REV.0 ATTACHMENT B PAGE6ts o r brF PAGE1 THERMAL MODEL Z14FM20 VALUE 9% FILL in 4* Tray 1.0 Tray Thermal Data 0.458 in/B09 R=0.06890625/1000ft w := 24 in Width in Tray cad =1.485*
4/0MCMTPLX d := 0.458 in Depth of cablesin Tray Ta :=( 273.16 + 40) K Ampient Temperature Tc : unknown K Surface Temperature Tm ::( 273.16 + 90) K Max CableTemperature I
wan mH n e eat tensfeWoek W h :: 0.101- - -
(K Kj R 2 ,g h := ( 0.101 )-
2 ft-K
-8 watt o :: 0.530 10 - Ste han-Boltzman Constant h 2.g4 e := 0.8 Effective thermal emmisivity of cable mass and tray surface.
p :: 400- K- effective thermal resistivity of cable mass watt 2
Surface area of cable mass per unit length As ._4 ft sides excluded Tc _ Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc)
I I f Tal 4 fc T Tai 4 lTc - I -h-( As p d) l - l h-( As p-d)
Tc : root As c p-d, e4 (K K) , _ _
(K Kj , c As c-p d, ,4 , c 8w 8w 8w 8w Tc = 358.068 *K
- _ __ _-. -_ _ _ - - . _ _ _ - - _ _ _ - . - _ _ _ . - _ = _ - - - _ _ _ _ _ - - _ _ - _ - - _ - _ _ _ - _ _ _ - _ _ _ _ - _ _ _ - _ _ - - _
CALC. NO.95-ENG41528E2 REV.0- ATTACHMENT B PAGE_eaf0F OlW APPENDIX H PAGE2 l 2.2 Calculated value of total heat (W) per unit length generated in cable tray
_I ITc Tai 4 4 W::hl -
l - As-( Tc - Ta ) + c As e-( Tc - Ta )
(K Kj watt W = 162.65
- i ft !
2.3 Recalculated Max Cable Temperature (Tm) {
i I I t I 4 Tai # fTc Tal h-lTc - 1 -( As p-d) hl -
l - As p dTa 4 4 (K K4 (K K/ + c As-E-p d Tc - c As c p dTa ,
Tm := +1 Tc - l 8w 8w 8w 8w Tm = 363.16 2.4 Temperature Rises .
2.4.1 Cab le Mass temperature Rise j ATc ::Tm - Tc ATc = 5.092 *K ,
Wpd 8w ATc = 5.092 *K I 2.4.2 AirTemperature Drop ATa ::Tc- Ta ATa = 44.908 *K 2.4.2 Total temperature Drop AT := ATc + ATa I AT = 50 *K l 2.5 Heat generated per unit area -
W watt Q:=dw_ ft '
Q = 14.797 * !
2 in t
CALC. NO. 96-ENG41528E2 REV.0 ATTACHMENT B PAGE_6__0F & l APPENDIX H PAGE 3 Ampacity Calculation i
n := 3 number of conductors [
cad :: 1.485 in Cable outer diameter R := 0.06890625- Resistance taken from Okonite book
, 1000- ft f
I40 ._ cad -
QK n- R . I 2 1 i
i 140 = 352.104
- amp TrayZ14FM20 Ampacity at 40C l I
U* ""
150 := 140 0.89
' at E 150 = 313.372
- amp 180 is 268 amps (335 x 0.8) at 40C. Since 180<l40 the 180 value adjusted to SOC is used fair 80 := 268 0.89-amp i Iair80 = 238.52
- amp This is the maximum allowable arnpacity. ,
4 i
i 4
e - -- -- , ,,ev , , -,%e, .- - . , ,
l t CALC. NO. 96-ENG-01528E2 REV.O ATTACHWENT B PAGE826OF 0105" - i APPENDIX l PAGE 1 THERMAL MODEL Z24FL20 VALUE' i 8% FILL in 4* Tray i 1.0 Tray ThermalData 0.407 in/B29 t w := 24 in Wxfthin Tray R=0.06890625/1000ft i cad =0.738' d : 0.407 in Depth of cables in Tray 4/0MCM ;
Ta := ( 273.16 + 40)-K Ampient Temperature ;
Tc :: unknown-K Surface Temperature j Tm :=( 273.16 + 90) K Max CableTemperature ,
I f Tc Tai 4 watt Overall Convective Heat transfer Coeffe ient h ::0.101 j -
l- 3 (K Kj 2 ft K ,
+
h := ( 0.101 )-
ft2- K J: 0.530 8, wan Stephan-Boltzman Constant ft , g 2 4 Ie f mass aM tmy c :: 0.8 surface.
p := 400- K "" effective thermal resistivity of cable mass watt i
.= Surface area of cable mass per unit length ,
i gg sides excluded ,
t Tc :: Tm + Ta 2-L Tc - 338'16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) 1 3 f Tc 4 Tai # ITc T2 l -
l h-( As p d) l -
! -h-( As p d) i As t p-d, 4 (K K) (K Kj _ o As e p-d 4 Tc *: m , _ _ ,
8w ,
8w 8w 8w t
Tc = 358.573 *K i
CALC. NO. 96-ENG41528E2 REV.0 ATTACHMENT B PAGE orJ OF M1 APPENDIXl PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray 3
i f Tc Tal # 4 W := h- l - l As-( Tc- Ta) + o As c-( Tc# - Ta )
iK K/ i W = 164.887
- ft 2.3 Recalculated Max Cable Temperature (Tm) 1 I f Tal # f Tc Tai #
h-lTc -
l -( As-p d) h1 -
j As p-d Ta 4 4 Tm .: .
(K K) +1 Tc - (K K) + c As e p d Tc - c As c-p dTa 8w 8w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc ::Tm - Tc ATc = 4.587 *K W p-d ATc :: ATc = 4.587 *K 8w 2.4.2 AirTemperature Drop ATa :: Tc - Ta ATa = 45.413 *K ;
2.4.2 Total temperature Drop AT :: ATc + ATa ,
AT = 50 *K 2.5 Heat generated per unit area j i
W watt Q := d w ft Q = 16.88
- 2 in l
l
CALC. NO. 96-ENG-01528E2 REV.O ATTACHMENT B PAGE6* OF8181 .
APPENDIX l PAGE 3 i
Ampacity Calculation n :3 number of conductors ,
i cad := 0.738 in Cable outer diameter R := 0.06890625- Resistance taken from Okonite book 1000 ft -
cad Q-x I40 := -
2 1 nR f
I40 = l86.897
- amp TrayZ24FL20 Ampacity at40C t
150 := 140- 0.89 or to Wo E ;
{
t 150 = 166.338
- amp Ampacity at 50C l 180 is 320 amps (400 x .8). Since this is > than 140, the 150 value is the value to use.
l r
k i
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_ _ __.m._.__. ._ __ _.m._.. _ . - _ . _ . _ _ . . _ . ________m__.___.__..._.__________.______m_ _ _ _ . . _ _ . _ _ _ . _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __________________-_e,- - - _-_ e - e- _ _ -__
CALC. NO. 96-ENGK)1528E2 REV.O ATTACHMENT B PAGE 6'81OF M APPENDIX K PAGE 1 THERMAL MODEL Z25BG20 VALUE ,
1.0 Tray Them1al Data 10% FILL in 4" Tray '
0.509 in/ B03 '
w := 24 in Wxfth in Tray R=0.84875/1000ft Depth of cables in Tray cad =0.774*
d := 0.509-in 3/C#8 i Ta := ( 273.16 + 40) K Ampient Temperature Tc :: unknown K Surface Temperature .
l Tm ::( 273.16 + 90) K Max CableTemperature ,
1 .
wu mH e at tensfehk knt h := 0.101- - j -
(K K4 2 '
ft K h = ( 0.101 )-
2 ft K watt Stephan-Boltzman Constant o :: 0.530-10'8-ft K 2 d EMm hlemmisW of caw mass aM tray c *: 0.8 surface.
p :: 400- K '" effective thermal resistivity of cable mass watt .
._ ft2 Surface area of cable mass per unit length AS *- 4 T sides excluded ,
Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) 1 I 4
f Tc Tai f Tc Tai #
l - l h-( As p d) l -
! h-( As p-d) 4 -Ta# Tc Tc :: root Tc + +1 Tc - Tm - Ta -
8w 8w 8w 8w t Tc = 357.577 K I h
CALC. NO. 96-ENG-01528E2 REV.O ATTACHMENT B PAGE6A OF8" "
APPENDIX K PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray ITc tai 4 4 W :: h- i -
1 As-( Tc - Ta) + o As c-( Tc - Ta )
(K Kj W = 160.481
- ft 2.3 Recalculated Max Cable Temperature (Tm) 1 I f Tai # I Tc Tai 4 h-lTc -
I -( As p d) h-l -
j - As p d Ta LK K) (K KJ c As c p dTc4 _ o As c p d ,4
.g. ,.= , , _
8w 8- w 8w 8w Tm = 363.16 2.4 Temperature Rises l 2.4.1 Cab le Mass temperature Rise ATc := Tm - Tc ATc = 5.583 *K +
W p- d ATc *:
8w ATc = 5.583 *K 2.4.2 AirTemperature Drop ATa := Tc - Ta !
ATa = 44.417 *K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area ,
W ,
watt Q::dw ft Q = 13.137
- 2 in
CALC. NO. 96-ENG41528E2 REV.O ATTACHMENT B PAGElh_OFM APPENDIX K PAGE 3 Ampacity Calculation n := 3 number of conductors cad ::0.774 in Cable outer diameter R := 0.84875- Resistance taken from Okonite book 1000 ft Q-n 140.-._ cad 2 1nR I40 = 49.27
- amp
^
- Since 150 is less than 80% of the free air value at 50C [52.48 amps] 150 is the correct value to use.
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE 808 OF de*
APPENDIX M PAGE1 THERMAL MODEL- Z14FM10 Values 12% fill on 4*
1.0 TrayThermalData 0.611 in/ B01 w := 24 in Widthin Tray R=2.15/1000ft d := 0.611 in Depth of cables in Tray
- f2AWG Ta ::( 273.16 + 40) K Ampient Temperature Tc := unknown K Surface Temperature Tm :=( 273.16 + 90) K Max CableTemperature i
.I.
h := 0.101- -
AK Kj gK 2 !
t h ::( 0.101 )-
2 ft K watt I o :: 0.530 10-8, Stephan-Boltzman Constant a2.g4 e := 0.8 i ,- mass ad W surface.
p := 400 K " effective thermal resistivity of cable mass '
watt -
2 ft Surface area of cable mass per unit length As _4 sides excIluded
- T .
.g Tm + Ta ,
2 Tc = 338.16 *K Initial assumed value for Surface Temp. (I c) +
2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) ;
i 1 1 I Tai # f Tai. #
lTc - I -h-( As p-d) lTc -
l h-( As p-d) '
o As t p d, 4 (K K) (K K/ , _ a As t p dTa#, Tc Tc := w , _ _
8w 8w 8w 8w Tc = 356.631 *K
CALC.NO.96-ENG-01528E2 RcV.0 ATTACHMENT B PAGE6MOF ##
APPENDIX M PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray I
W *: h-(K
-h Kj As-( Tc - Ta ) + o As c-( Tc# - Ta#)
W = 156.331
- ft '
2.3 Recalculated Max Cable Temperature (Tm)
I 1 4 4 f Tc Tai fTc Tai hI - l -( As p-d) h-l - ! As-p dTa 4 4 Tm :: +1 Tc - +"' -
8w 8w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc := Tm - Tc ATc = 6.529 *K Wpd ATc :: ATc = 6.529 *K 8w 2.4.2 AirTemperature Drop ATa := Tc - Ta ATa = 43.471 *K 2.4.2 Total temperature Drop AT *: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area
- watt q := .
dw Q = 10.661
- 2n in
CALC.NO.96-ENG41528E2 REV.0 ATTACHMENT B PAGE offOFN#k APPENDIX M PAGE 3 Ampacity Calculation n :: 3 number of conductors cad := 0.603 in Cable outer diameter R := 2.15- Resistance taken from Okonite book 1000 ft .
Ox !
140 _ cad 2 1nR 140 = 21.726 amp TrayZ14FM10 Ampacdy at40C on orto ptto E 15 0 : : I40-0.89 A at E 150 = 19.336
- amp Since 80% of the free air value is greater than 150 above is less than this value,150 is the '
i correct value to use.
t i
t l-i
._ . _ _ _ _ _ _ _ . _ _ _ _ _ . _ _ _ . _ _ . _ . _ ~ _ _ . _ _ _ . _ _ _ _ _ _ . _ _ . . ~ . _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ -
_________m.._ ______.______m___m________ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ . _ _ __.__ _ _ _ _ _ _ _ _
CALC. NO. 96-ENG-01528E2 REV.O ATTACHMENT B PAGE# OF##
APPENDIX N PAGE 1 THERMAL MODEL Z14FM10 VALUE 12% FILL in 4* Tray 1.0 TrayThermal Data 0.611 in/ B09 w := 24 in Width in Tray R=0.06890625/1000ft Depth of cables in Tray cad =1.485*
d ::0.6t t in 4/0MCMTPLX Ta ::( 273.16 + 40) K Ampient Temperature Tc ::unknow K Surface Temperature Tm := ( 273.16 + 90 ) K Max CableTemperature J.
wan au Qnde Heat hnsfer Mb snt h ::0.101- - -
(K Kj g2 -K h := ( 0.101 )-
2 ft K
~8 **"
o :: 0.530 10 - Stephan-Boltzman Constant 2 4 ft K c := 0.8 Effective thermal emmisivity of cable mass and tray surface.
p := 400- K- effective thermal resistivity of cable mass watt ft2 Surface area of cable mass per unit length A8 ._
- - 4' sides excluded ft Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) 1 I f Tc # #
Tai T fc Tal l - I h-( As p d) l - j h-( As-p d)
Tc := root
'^P Te# + iK K) , , _ _
(K K) . _ c As c p d-Ta# 'I'c 8w 8w 8w 8w Tc = 356.631 *K
CALC. NO. 96-ENG41528E2 REV. O ATTACHMENT B PAGE831'OF M PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray
_I 4
W := h- l l As-( Tc - Ta ) + o As c-( Tc - Ta# )
(K - Kj W = 156.331
- ft 2.3 Recalculated Max Cable Temperature (Tm)
I i I Tal 4 I Tc Tai #
h-lTc -
l -( As p-d) hj - l As p d Ta 4 Tm :: (K Kj
+1 Tc -
(K K) + o As c p d Tc - o- As c p-d 8w 8w 8- w 8w Tm = 363.16 2.4 Temperature Rises 2-.4.1 Cab le Mass temperature Rise ATc ::Tm - Tc ATc = 6.529 *K W p-d 8w ATc = 6.529 *K 2.4.2 AirTemperature Drop ATa : Tc- Ta ATa = 43.471 *K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area W
Q := d w watt ft Q = 10.661
- 2 in
_=.. . . .. .
CALC. NO. 96-ENG-01528E2 REV.0_ ATTACHMENTB PAGE 6FOF&
APPENDIX N PAGE 3 ,
f 1
Ampacdy Calculation I n ::3 number of conductors t
cad := 1.485 in Cable outer diameter L
i R :=0.06890625 ohm Resistance taken from Okonite book '
1000 ft 140 := -
2 1nR I40 = 298.867
- amp TrayZ14FM10 Ampacdy at40C 150 := 140 0.89 orto W to E A at E 150 = 265.991
- amp i
The ampacdy in air at 40C is 335,80% is equal to 268 amps. Smce 268 amps is < 140 the free
- air value (80%) adjusted to 500 is the correct value to use.
180 := 268 amp 0.89 .
180 = 238.52
- amp a@M to E i
r r
[
I
__ _.-__-.__._-_______.m _ _ _ _ _ _ _ _ _ - . _ _ _ . . - .-_.._..____.m___-______--_.__.___-___--_._.m.-____.,_______.-m_-_.-_____:.__.___..__.___________--_.-_--_ m
CALC.NO.96-EPG01528E2 REV.0 ATTACHMENT B PAGES9*OF F APPENDIX V PAGE1 THERMAL MODEL Z23FA30 Values 25% fill on 4*
1.0 TrayThermal Data 1.27 in/ B10 w ::24 in Wdth in Tray R=0.0594925/1000ft d :: 1.27 in Depth of cables in Tray cad =1.719 250MCMTPLX Ta ::( 273.16 + 40) K Ampient Temperature Tc : unknown K Surface Temperature Tm := ( 273.16 + 90 ) K Max CableTemperature I
wa mH aMmnsfe@HN.t h :: 0.10l* ! - -
(K Kl 2 ft K h := ( 0.101 )-
2 ft K
~8 watt o := 0.530 Ste han-Boltzman Constant ,
2 4 ft K Iem sW of caw mass ahy E: 0.8 surface. ,
p:=400 K '" effective thermal resistivity of cable mass watt
. ft2 Surface area of cable mass per unit length '
^8*-4'T sides excluded Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) ;
I I l
- I '
ITc Tai Tai 4 l -
I
- h-( As p d) lTc -
l -h-( As p d) o As E p d, 4 (K K/ , _ _
(K K) , _ o- As c* p d, 4 Tc ::rs 8w 8- w 8w 8w ;
I Tc = 351.5 *K P
_ . _ _ _ _ _ _ . . _ _ _ _ _ _ . _ _ _ _ _ _m_____m _____ _ _ _ _ _________._____-_._.____-___.__m__2 ___ __i._._-.___ _ _ _ _ e
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGEA97OFOF APPENDIXV PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray
.L I Tc Tal # 4 W := h I - j As-( Tc - Ta ) + o As-c-( Tc - Ta# )
iK Kj W = 134.325 -
ft 2.3 Recalculated Max Cable Temperature (Trn)
I 1 I Tc # I Tc #
Tai Tal hi -
! -( As p d) hl - l - As p dTa 4 4 Tm :: +1 Tc - + ^*' ' P # - ' ^ ' 'E
- 8w 8w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc : Tm - Tc ATc = 11.66 *K Wpd 8w ATc = 11.66 *K t
2.4.2 AirTemperature Drop ATa ::Tc- Ta ATa = 38.34 *K 2A.2 Total temperature Drop AT :: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area W watt Q := d w Q = 4.407
- ft 2 in
i CALC,NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE de OF O'8" APPENDIX V PAGE 3 ,
1 Ampacity Calculation n ::3 number of conductors .
cad :: 1.719 in Cable outer diameter R : 0.0594925- Resistance taken from Okonite book 1000 ft I40._
.- cad Qx 2 1nR 140 = 239.387
- amp TrayZ23FA30 Ampacity at40C l
f rto ptto E 150 := 140 0.89 150 = 213.055
- amp aN Since 80% of the free air value is 299 amps,150 is the correct 180 := 374 .8 amp ,
value to use.
180 = 299.2
- amp i
[
I t
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGESW OF M APPENDIX W PAGE1 THERMAL MODEL Z23FA25 Values 27% fill on 4*
1.0 TrayThennal Data 1.375 in/ B10 w := 24 in Width in Tray R=0.0594925/1000ft d : 1.375 in Depth of cables in Tray C 5 Pu Ta *:( 273.16 + 40)-K Ampient Temperature Tc := unknown K Surface Temperature Tm ::( 273.16 + 90) K Max CableTemperature
.I
- " **" * * "'*"* U" "
h : 0.101 1 - -
iK Kj 2 ft , g 1 := ( 0.101 )-
2 ft K
-8 **"
o :: 0.530-10 - Stephan-Boltzman Constant 2 4 It K nnale o caw mass aM tmy c := 0.8 surface.
p ::400 K effective thermal resistivity of cable mass watt g .= 4[ Surface area of cable mass per unit length sides excluded ft Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) 1 I f Tc Tal IcT Tai #
l -
) h-( As p d) l -
J h-( As p-d)
As c p d,. c4 (K Kj
, (K K) _ o- As c p d, 4 Tc ::m , _ _ ,
8w 8w 8w 8- w Tc = 350.808 *K
CALC.NO.96-ENG01528E2 REV.0 ATTACHMENTB PAGE & OF M "
APPENDIXW PAGE 2 22 Calculated value of total heat (W) per unit length generated in cable tray L
I Tc Tai 4 4 4 .
W: h- l -
l As-( Tc- Ta) + c As c-( Tc - Ta ) -
(K Kj watt W = 131.427
- ft !
2.3 Recalculated Max Cable Temperature (Tm) i I 4
I Tai fTc Tal . #
h-lTc -
I -( As-p d) h-l -
l As p dTa 4 4 ,
Tm : (K K) +1 Tc -
(K K) + o As t-p d Tc - o- As t p-dTa 8w 8w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass twinpwreiure Rise ATc *:Tm - Tc '
ATc = 12.352 *K W p-d 8w ATc = 12.352 *K 2.4.2 Air Temperature Drop ATa ::Tc- Ta ATa = 37.648 *K ,
2.4.2 Total temperature Drop AT *: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area W watt Q := _d w a Q = 3.983
- 2 in
- _ . _ _ _ _ _ _ _ . . _ _ _ _ _ _ _ _ _ _ _ _ __ _ . . _ _ _ _ _ _ . _ . _ _ _ _ _ . _ _ _ _ _ _ _ _ _ . _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ . . _ _ . _ _ _ . _ _ _ _ _ _~ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ . _ _ . .
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENTB PAGE M OF APPENDlX W PAGE 3 Ampacity Calculation n::3 number of conductors cad :: 1.719 in Cable outer diameter R ~:0.0594925 ohm Resistance taken from Okonite book 1000- ft 140.-._ cad Qn 2 1 n- R I40 = 227.569
- amp TrayZ23FA2S Ampacity at 40C omteim to get to E 150 :: I40-0.89 A at E 150 = 202.537
- amp Since the 80% of free air value is > 140 use 150 as correct 180 ::374.8 amp value.
180 = 299.2
- amp
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGE M OF **"
APPENDIX X PAGE 1 THERMAL MODEL Z24FL10 Values 4% fir on 4*
1.0 Tray Thennal Data 0.2037 in/C86 w : 24 in Widthin Tray R=3.41/1000ft d :: 0.2037 in Depth of cablesin Tray CDO 7 A G Ta : ( 273.16 + 40) K Ampient Temperature Tc :: unknown K Surface Temperature Tm ::( 273.16 + 90 )- K Max CableTemperature ,
I au mH on e a mnsfehk M h := 0.101- l -
LK K) hg 2 h = ( 0.101 )-
2 ft - K
~8 **"
o :: 0.530-10 - Stephan-Boltzman Constant n2.g4 nnal MsW of caw mass aM tray e := 0.8 surface.
p :: 400- K- effective thermal resistivity of cable mass watt 2
._ Surface area of cable mass per unit length
^5 '- 4' ft sides excluded g
Tm + Ta 2
I Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc)
I 1 ITc Tal Ic T Tai 4 l - l h-( As p d) l -
l h-( As p-d) o- As e p d, 4 (K K) (K Kj _ o As c p-d, 4 Tc := w 1 Tc - h - ,
8w 8w 8w 8w ,
}
Tc = 360.729 *K
CALC.NO.96-ENG41528E2 REV.0 ATTACHMENT 2 PAGE M OFM" APPENDIX X PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray 1
f Tc Tai 4 4 4 W :: h- l -
1 As-( Tc- Ta) + o As c-( Tc - Ta )
(K Kj W = 174.535
- ft 2.3 Recalculated Max Cable Temperature (Tm)
I I f Tc # #
Tai fTc Tal h-l -
l -( As p dp hI -
l - As p-dTa 4 4 Tm : +1 -Tc - + -
8w 8w 8w 8w Tm = 363.159 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc := Tm - Tc ATc = 2.43 *K Wpd ATc 3
- g. w Tc = 2.43 *K 2.4.2 AirTemperature Drop ATa ::Tc- Ta ATa = 47.569 'K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 49.999 *K 2.5 Heat generated per unit area W watt Q := d w_ ft Q = 35.701
- 2 in
CALC.NO.96-ENG41528E2 REV.0 ATTACHMENT 2 PAGEM6OF F APPENDlX X PAGE 3 Ampacity Calculation n := 7 number of conductors cad := 0.576 in Cable outer diameter R := 3.41- Resistance taken from Okonite book 1000- ft I40._ cad Qx 2 1nR 140 = 19.742
- amp TrayZ14FM20 Ampacity at40C 150 := I40-0.89
"' ' O 150 = 17.57
- amp
^ "
80% of free air value is 22.96 amps at 50 C.
150 is correct value to use.
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGE #fOF Ar*V ,
APPENDIX AA PAGE 1 -THERMALMODEL Z25BG20 Values 35% fill on 4" >
1.0 TrayThermal Data 1.78 in w := 24 in Wdth in Tray . R=3.41/1000ft Depth of cables in Tray cad =0.468 d := 1.78 in 3/C#14AWG ,
I Ta *:( 273.16 + 40) K Ampent Temperature Tc := unknown K Surface Temperature Tm ::( 273.16 + 90) K Max CableTemperature f Tc #
Tai watt Overall Convective Heat transfer Coeffic ient h ::0.101 1 - l (K K) 2 ft K l h:=(0.10I)- 2 ft K t
-8 **"
o :: 0.530 10 - Stephan-Boltzman Constant i 2 4 ft , g m mass aM W c := 0.8 surface. i p := 400- K- effective themial resistmty of cable mass watt i
I 2 Surface area of cable mass per unit length As::4 3 sides excluded i ft i Tc .-
._ Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation i
2.1 calculated Value for Surface Temperature (Tc) 1 1 t f Tc # I Tc Tai Tai # r l - l h-( As p d) 1 -
l h-( As p-d) t
. 0- As t p-d, 4 (K K) , , _
(K K) ,
_ o- As t p d 4 8w 8w 8w 8w Tc = 348.389 *K l
CALC. NO.96-EPG01528E2 REV.0 ATTACHMENT 2 PAGE M OF#r8*
APPENDIX AA PAGE 2 ,
2.2 Calculated value of total heat (W) per unit length generated in cable tray -;
_1 fcT Tal # 4 4 W := h 1 - 1 As-( Tc - Ta ) + o As e ( Tc - Ta ) <
LK K/
W = 121.411
- ft 2.3 Recalculated Max Cable Temperature (Tm) 1 I i I f Tai #
Tc Ta - # -( As p-d) I i
h-l l h-!Tc - l As p-d Ta 4 4 :
(K Kj (K Kg + o As c p d Tc - o As c p dTa Tm :: +1 Tc -
8w 8w 8- w 8w j Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ,
s t
ATc := Tm - Tc ATc = 14.771 *K Wpd
- g. w ATc = I4.771 *K ,
2.4.2 AirTemperature Drop I
t ATa *: Tc - Ta ATa = 35.229 *K 2.4.2 Total temperature Drop l AT : ATc + ATa AT = 50 *K 7 2.5 Heat generated per unit area f W ,,ig g=dw Q = 2.842
- ft 2 in ,.
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGE#_ff_OF F .
APPENDIX AA PAGE 3 Ampacity Calculation n ': 3 number of conductors i
cad ::9.468 in Cable outer diameter 7 I
ohm i R ::3.41- Resistance taken from Okonite book 1000 ft ,
140._ .- cad Qx 2 1nR 140 = 6.913
- amp TrayZ14FM20 Ampacity at40C o Etor to pt to E 150 :: I40 0.89 i A at E t 150 = 6.153
- amp r
l i
h J
i
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGE 8P OFM*"
APPENDIX AB PAGE 1 THERM ML ME20 Valuea 35% fill on 4* -
1.0 Tray Thermal Data 1.78 in/CO2 w := 24 in Width in Tray R=3.41/1000ft i
- d := 1.78 in Dep'h of cables in Tray cad =0.45 2/C#14AWG t Ta := ( 273.16 + 40) K Ampient Temperature .l Tc :: unknown K Surface Temperature Tm := ( 273.16 + 90)-K Max CableTemperature t I !
fc T Tai 4 watt Overall Convective Heat transfer Coethe ient h :: 0.101 1 -
l !
(K K) h2 ,g h := ( 0.101 )-
2 ft , g
~8 "'"
o := 0.530 Stephan-Boltzman Constant g2-K#
I " "" , mass aM M e := 0.8 surface, p :: 400 K '" effechve thermal resistmty of cable mass watt
.= 4, A Surface area of cable mass per unit length a . sides excluded T" + T*
Tc :
2 Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation ,
- 2.1 calculated Value for Surface Tanywaure(Tc) 1 1 fc T Tai 4 fc T Tai 4
-h-( As p d) -h-( As p-d) l -
l 1 -
l As t-p d 4 (K K) , (K Kj c As t p d, 4 Tc ::m , _ ,
8w- 8w 8w 8w Tc = 348.389'K
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGE #9 OF#*k i APPENDIX AB PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray 3
i fc T Tai 4 4 4 W := h l -
l - As-( Tc - Ta) + o As c-( Tc - Ta )
iK Kj W = 121.411
- ft 2.3 Recalculated Max Cable Temperature (Tm)
I I f Tc Tal 4 ITc Tai 4' h-l - j -( As p d) hl -
l As p dTa 4 (K Kj (K K) + c As-c p d Tc o- As t p d Ta4 Tm :: +1 Tc - -
8w 8w 8w 8w ;
Tm = 363.16 i 2.4 Temperature Rises ,
2.4.1 Cab le Mass temperature Rise ATc := Tm - Tc ATc = 14.771 *K W p- d 8w ATc = 14.771 *K .
2.4.2 Air Temperature Drop ATa : Tc- Ta ATa = 35.229 *K 2.4.2 Total temperature Drop AT := ATc + ATa AT = 50 *K 2.5 Heat generated per unit area ,
W watt Q := d w_
Q = 2.842 * ,ft in'
___ _ _ _ _ . . . . _ _ _ . _ . _ _ _ . _ _ . _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ . . _ _ _ . _ _ _ _ . _ _ _ _ . _ _ . _ . _ _ _ _ _ __.________________________________._._.____.___________________________..___._________._____.___.__.__.____.______.m__..
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGE WPOFM APPENDIX AB PAGE 3 Ampacity Calculaticn n::2 number of conductors cad :: 0.45-in Cable outer diameter R := 3.41- Resistance taken from Okonite book 1000 ft 14 0 : : -
2 1nR 140 = 8.141
- amp TrayZ14FM20 Ampacity at40C m or to pt to E 150 :: 140-0.89 A at E 150 = 7.246
- amp
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGEMIOF **"
APPENDIX AC PAGE 1 THERMAL MODEL Z25BG20 Values 35% fill on 4*
1.0 Tray Thermal Data 1.78 in/B01 w := 24 in Width in Tray R=2.15/1000ft ,
d :: 1.78 in Depth of cables in Tray Ca C G Ta :=( 273.16 + 40) K Ampient Temperature Tc := unknown K Surface Temperature Tm := ( 273.16 + 90 ) K Max CableTemperature I r wau waH &nde Nat tmnsfehk knt h :: 0.101- - -
iK Kl 2 ft K h ::( 0.101 )-
2 ft K a :: 0.53010~8- Stephan-Boltzman Constant
- f,2. g4
. Effective thermal emmisivity of cable mass and tray c _ 0.8 surface.
p := 400- K '" effective thermal resistivity of cable mass watt
.= 4,[ Surface area of cable mass per unit length ft sides excluded Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Cak:ulation 2.1 calculated Value for Surface Temperature (Tc) 1 1 I Tc Tai 4 .fTc Tai # ,
I -
l -h-( As p d) l - '
-h-( As p-d)
'^' 4 -Ta#, Tc Tc := root Tc + +1 Tc - Tm - - Ta - -
8w 8w 8w 8w Tc = 348.389 *K
_ .m . . _ . . _ . . . .__ _ . . . _ . . ..__. _.. . . . . _
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGE8kOF M APPENDIX AC PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray
.I T
Ic Tai 4 4 W := h- i -
1 - As-( Tc - Ta) + o As c-( Tc - Ta )
(K Kj W = 121.411
- ft i 2.3 Recalculated Max Cable Temperature (Tm) 1 I I # I h-lTc -
Tai I -( As p-d) h-l Tc Tal. - i #- As p-d Ta 4 4 (K KJ (K K) + o As t p-d Tc - o As t p dTa Tm :: +1 -Tc -
8w 8w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc ::Tm - Tc ATc = 14.771 *K Wpd 8w ATc = 14.771 *K 2.4.2 AirTemperature Drop ATa ::Tc - Ta ATa = 35.229 *K 2.4.2 Total temperature Drop AT := ATc + ATa AT = 50 *K 2.5 Heat generated per unit area W watt Q::dw Q = 2.842
- 2rt in i
i CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGE M OF/_p "
i APPENDlX AC PAGE 3 :
Ampacity Cabulation n ::3 number of conductors cad := 0.603 in Cable outer diameter R := 2.15- Resistance taken from Okonite book 1000- ft .
I40 := cad- -Qx 2 1, n- R ,
140 = 11.217
- amp TrayZ14FM20 Ampacity at40C f
fE r ptto E 150 :: I40-0.89 A at E 150 = 9.984
- amp
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGE G OFhoV APPENDIX AD PAGE 1 THERMAL MODEL Z25BG20 Values 35% fill on 4*
1.0 Tray Thermal Data 1.78 in/C62 w := 24 in WxithinTray R=1.35/1000f t d := 1.78 in Depth of cables in Tray cad =0.522 2/C#10AWG Ta :=( 273.16 + 40)-K Ampient Temperature Tc := unknown K Surface Temperature Tm ::( 273.16 + 90 ) K Max CableTemperature I !
wan mH nv at tensfe@k ent h := 0.101- l - -
(K Kj ft ,
2 g h := ( 0.101 )-
2 ft K
~8 **"
o :: 0.530 10 - Stephan-Boltzman Constant 2 4 ft , g nnahsW of caw mass ahy e :: 0.8 surface.
p := 400- K '" effective thermal resistivity of cable mass watt 2
._ ft Surface area of cable mass per unit length
^8 ' '
ft sides excluded Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc)
I I 4 ITc 4
.fTc Tai Tai i -
l -h-( As p d) j - j h-( As p d) o As t-p d, 4 (K Kj
, (K K/ _ o As e p-d 4 Tc : root , _ _
8w 8w 8w 8w Tc = 348.389 *K
CALC.NO.96-ENG41528E2 REV.0 ATTACHMENT 2 PAGEAE_OF Mf6" t
APPENDtX AD PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray -
_1 Ic T Tai # 4 4 W := h l -
1 - As-( Tc- Ta ) + o- As-c-( Tc - Ta )
(K Kj ,
W = 121.411
- 2.3 Recalculated Max Cable Ternperature(Tm) ~
1 !
i' f Tai 4 fTc Tai 4 h-lTc - l -( As p d) h-l -
I - As p-dTa 4 Tm :
(K K) +1 Tc -
(K K/ + o As t p-d Tc - o As E p-d a i 8w 8w 8w ,
8w Tm = 363.16 .
2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise !
ATc ::Tm - Tc t ATc = 14.771 *K -
Wpd ATc :: ATc = 14.771 *K 8w L
2.4.2 Air Temperature Drop ATa ::Tc- Ta ATa = 35.229 'K 2.4.2 Total temperature Drop AT : ATc + ATa AT = 50 *K 2.5 Heat generated per unit area W watt Q: dw rg Q = 2.842 * ,
in'
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGE#9'OF A
- APPENDIX AD PAGE 3 Ampacity Calculation n ::2 number of conductors cad := 0.522 in Cable outer diameter .
i R := 135- Resistance taken from Okonite book 1000- ft 140.-._ cad _ Qn 2 1 nR I40 = 15.009
- amp TrayZ14FM20 Ampacdy at40C ISO := 140 0.89 -
U' E A at E 150 = 13358
- amp L
i i
l
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGE#ff OF M APPENDIX B1 PAGE 1 THERMAL MODEL Z52EA10ygg,
- 38% fill on 4' '
1.0 Tray Thermal Data 1.935 in/B14 w : 6-in Widthin Tray R=0.033251456/1ktt cad =3.05 d := 1.935 in Depth of cables in Tray 750MCMTPLX AL Ta ::( 273.16 + 40)-K Ampient Temperature Tc := unknown-K Surface Temperature Tm : ( 273.16 + 90) K Max CableTemperature 1
wau emH at tensfer Mc knt h := 0.101- - -
(K Kj gg2 , K h := ( 0.101 )-
2 ft K
~8 **" Stephan-Boltzman Constant a := 0.530 2 4 ft , g e := 0.8 Effective thermal emmisivity of cable mass and tray surface.
p := 400- K- effective thermal resistivity of cable mass watt
._ ft2 Surface area of cable mass per unit length
^5 '- 4 T sides excluded Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) 1 I Tal, 4 #
ITc Ic T Tal i -
I h-( As p d) l -
l h-( As p d)
As c p-d-Tc# + (K K) , (K K) _ o As t p d, 4 Tc ': m , _ _ ,
8w 8w 8w 8w Tc = 332.019 *K
CALC.NO.96-ENG41528E2 REV.0 ATTACHMENT 2 PAGE$& OF#fF APPENDIX B1 PAGE 2 2.2 Caiculated value of total heat (W) per unit length generated in cable tray
-1 I Tc Tai 4 4 4 W :: h- l - 1 As-( Tc - Ta ) + o As c-( Tc - Ta )
kK Kj W = 58.864
- ft 2.3 Recalculated Max Cable Temperature (Tm)
I I fTc 4 ITc Tai # Tai hl - j -( As p d) h-l - - l - As p d Ta 4 Tm :: (K -Kj
+1 Tc -
(K KJ + a As c p-d Tc -
o As c p-d a 8w 8w 8w 8w Tm = 363.16 ,
2.4 Temperature Rises 2.4.1 Cab le Mass imiperature Rise ATc := Tm - Tc ATc = 31.141 *K Wpd ATc :
8w ATc = 31.141 *K 2.4.2 Air Temperature Drop ATa ::Tc - Ta f ATa = 18.859 *K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 50 *K i 2.5 Heat generated per unit area W
Q::dw watt ft Q = 5.07
- in2
.- .__ _ _ _ _ _ _ . _ _ _ . _ . -_._____.;_ .._._._._.__.-m_. _..____.-____.________._________.____.-m_:.m ______.___._______.____.__...__._____.____.____._____-__.____._._..________.___._____-<-e*w+ ,
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT 2 PAGEh/ OF #fe
APPENDIX B1 PAGE 3 Ampacity Calculation n ::3 number of conductors cad := 3.05-in Cable outer diameter '
R ::0.033251456- Resistance taken from Okonite book !
1000 ft ;
I40._ .- cad
'Qx 2 1nR I
140 = 609.377
- amp Tray Ampacdy at40C i i
150 :: 140-0.89 Correction factor to get to SOC 150 = 542.346
- amp Ampacdy at 50C i
I air - 588 amps at 40 C 180 :: 588 .8 .89 ;
180 = 418.656 e
l s
l i
CALC.NO.96-ENG41528E2 REV.0 ATTACHMENT 2 PAGE NIOF ,
APPENDIX B2PAGE 1 THERMAL MODEL Z22EA10 y,g, 23% fill on 4" 1.0 Tray hl Ma 1.1714 in/B14 '
i
, := 6 in Widthin Tray R=0.06543487/1000ft cad =2.39 ,
d := 1.1714 in Depth of cables in Tray 350MCMTPLX '
AL Ta :=( 273.16 + 40) K Ampient Temperature Tc := unknown- K Surface Temperature Tm ::( 273.16 + 90) K Max CableTemperature i 1 l wan H on e aMramfehk knt h := 0.101 1 -
l -
(K Kj g 2 -K l h := ( 0.101 )-
ft , g 2
a : 0.530- 10~8- Stephan-Boltzman Constant 2 4 ft - K c := 0.8 Effectrve thermal emmisivity of cable mass and tray surface.
- I p::400 K- effective thermal resistivity of cable mass watt 4
2 ft Surface area of cable mass per unit length j As ._4 T sides excluded 4 4
Tm + Ta 2 !
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) [
t 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) ;
i 1 f Tc 4 I Tc 4 Tal T2 {
l -
l h-( As p d) l -
l h-( As p d)
As t- p- d, ,, 4 , (K Kj (K K) _ o As c p d 4 ,
Tc ::e y _ _
8w 8w 8w 8w ,
t i
Tc = 337.6% *K I
CALC. NO. 96-ENG41528E2 REV.0 ATTACHMENT 2 PAGE M OFM APPENDIX B2 PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray 4
' W :: h- -
l As-( Tc - Ta) + o As t-( Tc - Ta# )
(K Kj W = 79.509
- ft 2.3 Recalculated Max Cable Temperature (Tm)
I I f Tc 4 Tai fcT Tai 4 hj -
l -( As p d) hi -
l - As p-d Ta 4 Tm ._ .-
(K K/ +1 Tc -
(K K) + o As c p d Tc - o- As-e p-d 8w 8- w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc ::Tm - Tc ATc = 25.464 *K Wpd 3.rc ':
8w ATc = 25.464 *K 2.4.2 Air Temperature Drop ATa ::Tc - Ta ATa = 24.536 K 2.4.2 Total temperature Drop AT := ATc + ATa
, AT = 50 *K 2.5 Heat generated per unit area W
Q watt aw ft Q = 11.313
- 2 in
CALC.NO.96-ENG41528E2 REV.0 ATTACHMENT 2 PAGE NPOF 8 "
APPENDIX B2 PAGE 3 Ampacity Calculation n := 3 number of conductors cad := 2.39 in Cable outer diameter R := 0.06543487- Resistance taken from Okonite book 1000 ft I40._.- cad Qx 2 1 nR I40 = 508.461
- amp Tray Ampacity at400 150 :: I40-0.89 Correction factor to get to 500 150 = 452.53I
- amp Ampacity at 50C
! air = 368 amps at 40C.
180 := 368 .8 .89 180 = 262.016
1 4
CALC. NO. 96-ENG-01528E2 REV. 00 Page 445of Bem- l j ATTACHMENT B )
i TESTS FOR MATHCAD VERIFICATION i
l To prove the validity of the MathCad ampacity several tests were run using the depth and ,
cable diameters provided in the ICEA tables. The ampacities presented in the ICEA table l for each test made was then compared to the calculated value to ensure the calculation 1 was providing conservative values. The results are tabulated below:
l j TEST Appendix Cable Depth ICEA MATHCAD Vanance 1 PP 3/C #8 1.5 38 36.766 3.249 %
l 3 2 QQ 3/C #8 2.0 31 30.137 2.784 % ;
$ 3 TT 500MCM 1.5 438 397.554 9.234 %
4 UU 500MCM 3.0 272 242.456 10.863 %
l 5 RR 4/0 3.0 135 124.212 7.991 %
6 SS 4/0 2.5 153 142.199 7.059 %
4 From the above it is apparent that the values calculated in all cases are conservative.
i i
I 4
l i
CALC. NO. 96-ENG-01528E2 REV. 00 Page*' of mz !
ATTACHMENT B TESTS FOR MATHCAD VERIFICATION i
l To prove the validity of the MathCad ampacity several tests were run using the depth and i cable diameters provided in the ICEA tables. The ampacities presented in the ICEA table for each test made was then compared to the calculated value to ensure the calculation was properly functioning. The results are tabulated below:
TEST Appendix Cable Depth ICEA MATHCAD Variance 1 P 3/C #8 1.5 38 33.799 11.056 %
2 0 3/C #8 2.0 31 27.924 9.922 %
3 T 500MCM 1.5 438 365.473 16.559 %
4 U 500MCM 3.0 272 227.309 16.431 %
5 R 4/0 3.0 135 116.425 13.739 %
6 S 4/0 2.5 153 132.606 13.329 %
From the above it is apparent that the values calculated in all cases are conservative. ,
I
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE d47 OFAre" Appendix P PAGE1 '
THERMAL MODEL TEST 1 1.0 Tray Thermal Data 1.5 in/ B03 - >
w := 24 in Width in Tray R=0.84875/1000ft -!
d := 1.5 in Depth of cables in Tray cad =1.02" 3/C#8 AWG Ta := ( 273.16 + 40) K . Ampient Temperature TABLE 3-16 (ICEA)
~
Tc := unknown K . Surface Temperature P-54-440,1986 Tm :=( 273.16 + 90) K Max CableTemperature I !
wan H n Hea ransfer Coek knt h := 0.101- -
(K K) gK 2 watt h := ( 0.101 )- ;
g2 .g o := 0.530- 10-8, watt Stephan-Boltzman Constant n 2.g4 e := 0.8 Effective thermal emmisivity of cable mass and tray surface.
p := 400- K- effective thermal resistivity of cable mass watt j t
g2 As := 4 - Surface area of cable mass per unit length h neglecting tray sides '
Tc := Tin + Ta ,
2 ,
i Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) j 2.0 Thermal Calculation ;
2.1 calculated Value for Surface Temperature (Tc)
E { l f 4 Tal 4 Tai i lTc -
l -h-( As p d) ' I'I'cl l - h-( As p d) -t As c p d, ,e4 (K Kj
, (K Kj _ o As c p d, 4 Tc := w , _ _
8w 8w 8w 8w ,
Tc = 350.021 *K i
. _ - - - . _. ..=-- . -. -. . . - . - . . . . . .
- . . . . ~ . . . _ _ .
CALC.NO.695-ENG-01528E2 REV.0 ATTACHMENT B PAGE M OF ""
t APPENDIX P PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray
_I_
fc T Tai # 4 W:=h! -
1 As-(Tc - Ta ) + o As c-( Tc - Ta4) '
(K K/
W = 128.148
- 2.3 Recalculated Max Cable Temperature (Tm) ft i 1 I 4 4 Tal ITc Tal .
h-lTc -
1 -( As p-d) h-l -
l As p dTa 4 4 Tm :: (K Kj
+1 Tc -
(K Kj + 0 As c p d Tc - o As c p dTa 8w 8w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc ::Tm - Tc ATc = 13.139 *K W p-d 8w ATc = 13.139 K 2.4.2 Air Temperature Drop ATa :: Tc - Ta ATa = 36.861 *K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area W
Q::dw watt ft Q = 3.56 -
in 2
CALC.NO.695-ENG-01528E2 REV.0 ATTACHMENT B PAGE NT O FM" APPENDIX P PAGE 3 Ampacity Calculation n::3 number of conductors cad := 1.02-in Cable outer diameter R := 0.84875- Resistance taken from Okonite book 1000- ft 140 := -
2 1nR I40 = 33.799
- amp Ampacdy at40C ICEA TABLE 16 VALUE IS 38 AMPS. RESULTS ARE CONSERVATIVE.
~f 14 0 i V := '
!-1 - 100 (38 amp)
V = 11.056 % VARIATION t
I
?
CALC. NO. 96-ENG-01528E2 REV.0 KTACHMENT B PAGE M8 OF M Appendix PP PAGE1 THERMAL MODEL TEST 1 1.0 Tray Thermal Data 1.5 in/ B03
- w := 24 in Wdthin Tray R=0.84875/1000ft d := 1.5 in Depth of cables in Tray cad =1.02" l 3/C#8 AWG Ta := ( 273.16 + 40)- K Ampient Temperature TABLE 16 (ICEA)
Tc := unknown K Surface Temperature P-54-440,1979 ,
Tm := ( 273.16 + 90 ) K Max CableTemperature i
, .= g, , g , , f c _ Ta'
- T watt Overan Convective Heat transfer Coeffic ient (K Kj gK 2 !
h := ( 0.201 )-
2 g ft
~8 watt Stephan-Boltzman Constant a := 0.530 2 4 ft , g e := 0.8 . Effective thermal emmisivity of cable mass and tray ,
surface.
p := 400 K "" effective thermal resistivity of cable mass watt ;
t 2
ft As := 4 - Surface area of cable mass per unit length R neglecting tray sides Tm + Ta 2
! Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) .
[ I I I 4 Tc -- Tai # T fc Tal I l l -h-( As p d) 1 -
l h-( As p-d) c As-c p d,. ,c4 + (K K) , , , (K K) ,
_ o As E p d, 4 8w 8w 8w J 8w Tc = 347.614 *K
. - . __-___..___m_ _ _ _ . _ _ - ____.___&_ . _ ._m.-_mm..__ ._.______.m_ _..____m_m____m___._____ _ . _ _ _ . m .c,..4
. . . . _ . _ _ . - . . ._ . _ . . . .-. - . _ . . _ . . - . - __-..._. .. . _m -. . . - - - .. --. ~ _ . _ . _ . _ __ _ . . . .
CALC.NO.695-ENG-01528E2 REV.0 ATTACHMENT B PAGE gil_OF 88"
- APPENDIX PPPAGE 2GE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray r
.I ITc tai 4 4 -!
W := h l -
l As-( Tc - Ta ) + o As c-( Tc - Ta ) '
(K Kj watt I W = 151.633
- I 2.3 Recalculated Max Cable Temperature (Tm) I' i
I 1 f 4 Tai 4 f'I'c ~I'ai h-lTc - -( As p-d) -
l As p-d Ta Tm ::
(K Kj
+1 -Tc -
hIK ( K) + o As e p dTc 4
- o As e p-dTa 4
8- w 8w 8w 8w Tm = 363.16 '
2.4 Temperature Rises ,
2.4.1 Cab le Mass temperature Rise !
ATc ::Tm - Tc ATc = 15.546 *K Wpd ATc :: ATc = 15.546 *K
- g. w 2.4.2 AirTemperature Drop ATa :: Tc - Ta
! ATa = 34.454 *K j 2.4.2 Total temperature Drop AT :: ATc + ATa ;
AT = 50 *K 2.5 Heat generated per unit area W !
n := _d w
- watt ,
t ft t Q = 4.212 * !
2 in l
_ _ _ _ _ . . _ _ . _ _ _ _ ._ _. ._ _. _....__..____.__...____ _ __._____ _ _ _..______. _ ___ _ ______ _._._ ____ _ _=_ __.__._____ _ ___ _ _ _ __ _ _
4 _ _ _ _ _ _ _ - . _ _ _ _ _ _ _ < ,. - - -,4
. _. .. _ . . . . . . _ _ _ . . .. _ . _ _ _ _ _ . . . . . . ~ . _ _ _ . _ . . _ _ _ . _ . . __ . _ . _ . _ _ . . _ . . _. . . . _ _ . . _ _ . _ . .
CALC.'NO.695-ENG41528E2 REV.0 ATTACHMENT B PAGE815OFM APPENDIX PPPAGE 3 Ampacity Calculation n::3 number of conductors cad := 1.02 in Cable outer diameter R ::0.84875- Resistance taken from Okonite book 1000- ft t
cad Qx I4e := -
2 1nR I40 = 36.766
- amp Ampacity at40C ICEA TABLE 16 VALUE IS 38 AMPS. RESULTS ARE CONSERVATIVE.
V :: !(38 ampfI-1 -100 V = 3.249 % VARIATION l
i f
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE 61) OF 60 " '
APPENDIX Q PAGE 1 THERMAL MODEL TEST 2 1.0 Tray Thermal Data 2.0 in/ B03 w ::24 in Width in Tray R=0.84875/1000ft d := 2.0 in Depth of cables in Tray Ca =
,C#8 Ta ::( 273.16 + 40) K Ampient Temperature
^ '
Tc := unknown-K Surface Temperature 40,1986 Tm :=( 273.16 + 90) K Max CableTemperature
.l.
h : 0.101 1 -
wau eraH n e amnsfer Coek snt (K Kj gK 2 h := ( 0.101 )-
R2 -K
~8 **"
o := 0.530 10 - Stephan-Boltzman Constant n 2.g4 e := 0.8 Effective thermal emmisivity of cable mass and tray surface.
p := 400 K- effective thermal resistivity of cable mass watt 2
Surface area of cable mass per unit length
^8 '- # n T sides of tray not included.
Tc ::
2 Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) 1 I I # f Tc
' Tal Tai 4 lTc - I h-( As-p d) l "- j h-( As p d)
Tc :: root
- As e-p d 4
-Tc +
(K K) , _ _
(K KJ ,
_ o As e p d 4 8w 8w 8w 8w Tc = 347.217 *K
. . - - . ~. . .. . . ,- . . _ _ ~ . . . -- . . . . . ~ _ . - _ . . - . _ . - - _ - .. , , -
CALC NO.695-ENG-01528E2 REV.0 ATTACHMENT B PAGE87f OF08" APPENDlX Q PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray
.I.
4 I Tc tai 4 4 W :: h-l - l As-( Tc - Ta ) + o As c-( Tc - Ta )
(K Kj W = 116.63
- 2.3 Recalculated Max Cable Temperature (Tm) ft r
1 I fTc f 4 i Tai 4 TM h-l - l -( As p d) IrlTc - l As p dTa 4 4 (K Kj
+1 (K K) + a As e p-d Tc - o As t p dTa Tm ..- Tc -
8w 8w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc := Tm - Tc '
ATc = 15.943 *K W- p d t I
8w ATc = 15.943 *K 2.4.2 Air Temperature Drop ATa := Tc - Ta ATa = 34.057 *K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area ,
W Q: dw watt Q = 2.43
- 2It in
i CALC.NO.695-ENG01528E2 REV.0 ATTACHMENT B PAGE M OF M APPENDlX Q PAGE 3 Ampacity Calculation t
n ::3 number of conductors l
cad :: 1.02 in Cable outer diameter t
R := 0 84875- Resistance taken from Okonite book 1000- ft I
t Q-n ,
140._ .- cad-2 1nR 140 = 27.924
- amp Ampacity at40C I L
ICEA TABLE 16 VALUE IS 31 AMPS. RESULTS ARE CONSERVATIVE. I f 14 0 I '
V :: l 1 100 !
(31 amp) ,
V = 9.922 % VARIATION !
L i
i 5
i I
t f
i
._ _ . _ . . ~ . _ . . _ . . ~ . . _ m_ _ _ _ . , _ _ _- _ _ _ . _ . _ _ . _ _ _ . . _ _ _ . _ _ _ _ _ _ _ _ . . _ _ _ . . _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _____ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ . _ ____ ___m______.______ -_ _ _ ____ - ___
CALC.NO.96-ENGK)1528E2 REV.0 ATTACHMENT B PAGE 4# OFM APPENDIX QQ PAGE1' THERMAL MODEL TEST 2 1.0 Tray Thennal Data 2.0 in/ B03 w := 24 in Widthin Tray R=0.84875/1000ft d := 2.0-in Depth of cablesin Tray Ca =
,C#8 Ta ::( 273.16 + 40) K Ampient Temperature TABLE 16 (ICEA)
Tc := unknown K Surface Ternperature P-54-440,1979 Tm := ( 273.16 + 90 ) K Max CableTemperature I
- " mH e at transfehk ent h ::0.101 1 - - -
(K Kj ft .
2 g h : ( 0.201 )-
2 ft-K
~8 watt Stephan-Boltzman Constant o := 0.530 2 #
ft K e : 0.8 Effective thennal emmesivity of cable mass and tray surface.
p:=400 K- effective thermal resistivity of cable mass watt
.._ ft2 Surface area of cable mass per unit length
^5~-4'T sides of tray not included.
i Tm + Ta 2
Tc = 338.16 *K - Initial assumed value for Surface Temp. (Tc) 2.0 Themwil Calculation 2.1 calculated Value for Surface Temperature (Tc) 1 I 4
Ic T Tai 4 fc T *I'd 1 - j h-( As-p d) l -h-( As p d)
(K Kj (K KjI _ o As e p d_ 4 Tc ::m As c p d,. .c # + , _ _ ,
8w . 8w 8w Sw Tc = 344.589 *K
CALC.NO.695-ENG-01528E2 REV.0 ATTACHMENT B PAGE #U OF8'*
APPENDIX OO PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray L
I Tc Tal 4 4 4 W := h I - I As-( Tc - Ta) + o As c-( Tc - Ta )
(K Kj W = 135.847
- 2.3 Recalculated Max Cable Temperature (Tm) ft i I I Tc Tal 4 I Ta , 4 i hl -
! -( As p-d) h-lTc - f As p-dTa 4 4 (K K) (K K) a- As-c p dTc _ o As c p d Ta Tm *: , _
8w 8- w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc ::Tm - Tc ATc = 18.571 *K Wpd 8w ATc = 18.571 *K 2.4.2 Air Temperature Drop ATa ::Tc- Ta ATa = 31.429 *K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area w
Q:=dw watt ft Q = 2.83
- in2
CALC.NO.695-ENG-01528E2 REV.0 ATTACHMENT B PAGE8M OF#
APPENDIXOO PAGE3 Arnpacity Calculation n ::3 number of conductors cad := 1.02 in Cable outer diameter R ::0.84875- Resistance taken from Okonite book 1000-It 140 :=
2 1 nR I40 = 30.137 amp Ampacdy at40C ICEA TABLE 16 VALUE IS 31 AMPS. RESULTS ARE CONSERVATIVE.
V :: I -1 -100 (31 ampj V = 2.784 % VARIATION
CALC.NO.96-ENG41528E2 REV.0 ATTACHMENT B PAGE84 OFM APPENDIXT PAGE1 THERMAL MODEL TEST 3 1.5 in 1.0 Tray Thermal Data R=0.0313575/1000ft w :: 24-in Widthin Tray cad =2.12 500MCMTPLX d : 1.5-in Depth of cables in Tray TABLE 3-8 Ta ::( 273.16 + 40) K Ampient Temperature (ICEA,P54-440,1986) l'c :: unknown K Surface Temperature Tm := ( 273.16 + 90) K Max CableTemperature ,
.I.
wa mmu on e ea mnsfer Coek ient h : 0.101- -
i -
(K Kj RK2 h::(0.10I)- 2 ft K
~8 watt o := 0.53010 - Stephan-Boltzman Constant 2 4 ft K m tMnnal Mskh of caw mass aM tmy c := 0.8 surface.
p :: 400- K '" effective thermal resistivity of cable mass watt ,
ft2 Surface area of cable mass per unit length As ._4-sides not included g
Tc ,_ Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) ,
r -
g i 4 4 ITc Tai fTc TE .
l -
l h-( As p d) 1 -
l h-( As p d)
As c p d, 4 (K K) , (K Kj _ o- As- e p- d 4 Tc ::m _ _ ,
8w 8w 8w 8w Tc = 350.021 *K
. . - - . _ .. _ . . _ . . - _ . _ . . . _ . _ - _ _ . . _ . . . . _ _ _ . . . . _ -.___m ._. . . . . . - - . . _ _
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGENOF M ~
APPENDIX T PAGE 2 l 2.2 Calculated value of total heat (W) per unit length generated in cable tray ,
. .I.
ITc Tal 4 4 4 ,
W : h l 1 As-( Tc - Ta ) + c As c-( Tc - Ta )
(K Kj W = 128.148
- ft 2.3 Recalculated Max Cable Temperature (Tm) 1 I ITc Tal, 4 I Tai 4 t h-l (K
KJ I -( As p d) h-lTc(K K)
- l - As p-d Ta 4
- o As c p dTa 4
Tm .:. +1 Tc - + 0- As t p-d Tc 8w 8w 8w 8w ,
Tm = 363.16 '
2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc :: Tm - T, N- 13.139 *K ATc ::
8w ATc = 13.139 *K ,
2.4.2 Air Temperature Drop '
t ATa ::Tc - Ta ATa = 36.861 *K 2.4.2 Total temperature Drop AT :: ATc + ATa ,
AT = 50 *K l
2.5 Heat generated per unit area W
watt i Q: dw t ft Q = 3.56 * ,
2
- in s \
9 CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE0# OF&b&
APPENDIXT PAGE 3 Ampacdy Calculation n::3 number of conductors cad ::2.C in Cable outer diameter R := 0.0313575- Resistance taken from Okonite book 1000- ft ,
cad Q-x .
14 0.. :
2 1nR I40 = 365.473
- amp Ampacdy at40C t I
ICEA VALUE AT 40 C IS 438 AMPS. THIS VALUE IS CONSERVATIVE.
f 14 0 I ;
V := l -1 1-100 (438 amp j- ;
V = 16.559 % VARIATION i
f r
i i
6 s
r
_ _ _ - _ . . _ . _ _ . _ _ _ _ _ _ _ _ _ .______.____m_..._ _____._ _ _ _ _ _ _ __._____. _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _. . _ _ _ _ _ _ _ _ . _ . _ _ . _ _ _ _ _ _ _ _
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGEM OF M '
APPENDIX TT PAGE1 THERMAL MODEL TEST 3 1.5 in 1.0 Tray Thermal Data R=0.0313575/1000ft w : 24 in Width in Tray cad =2.12 500MCMTPLX d :: 1.5 in Depth of cables in Tray.
TABLE 8 (ICEA)
Ta :=( 273.16 + 40) K Amp.ient Temperature (P-54-4401979)
Tc :: unknown K Surface Temperature Tm := { 273.16 + 90 ) K Max CableTemperature J.
rah Conde Heat tensfehR knt h : 0.101- b\
K/
a (K ft2K -
h ::( 0.201 )- 2 ft . g
~8 **"
a :: 0.530- 10 - Stephan-Boltzman Constant 2 4 rt . g nnalemdsW of caw mass aM tray e := 0.8 surface.
p := 400- K effective thermal resistivity of cable mass watt 2
._ ft Surface area of cable mass per unit length
^8 '- 4 T sides not included Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc)
I 1 4 #
fTc Tal fc T Tal
) h-( As p d) 1 -
) h-( As p-d)
As t-p d 4 (K K) , (K K) , _ o- As E p d, 4 Tc ::m _ _
8w 8w 8w 8w Tc = 347.614 *K n._.. _ - - _ - . _ _ - . . _ . . _ _ . _ _ _ - - - - - . _ _ _ _ . - . _ - - . _ _ . . - _ - _ - _ . . . - - - _ _ _ -- -
e
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE89)OFA* '
APPENDIXTT PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray
_I Ic T Tai # 4 4 W :: h- i -
I As-( Tc - Ta) + o- As c-( Tc - Ta )
(K Kj W = 151.633
- fl 2.3 Recalculated Max Cable Temperature (Tm) 1 I !
I Tc 4 Tai 4 .fTc Tai hI -
l -( As p d) h! - I As p dTa 4 4 '
(K Kj
+1 (K K) + o As t p d Tc - o As e p-dTa Tm :: -Tc - .
Sw 8w 8w 8w l l
Tm = 363.16 2.4 Temperature Rises i 2.4.1 Cab le Mass temperature Rise l i
ATc := Tm - Tc ATc = 15.546 *K [
W p-d ATc := i 8w ATc = 15.546 *K 2.4.2 Air Temperature Drop :
i ATa :: Tc - Ta ATa = 34.454 *K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area ;
QI W dw watt Q = 4.212
- ft in
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE6*fOF8886 ,
APPENDIX TT PAGE 3 Ampacity Calculation n := 3 number of conductors ,
cad := 2.12 in Cable outer diameter ,
R := 0.0313575- Resistance taken from Okonite book 1000 ft !
Q-x I40 .-
._ cad- i 2 1nR I40 = 397.554
- amp Ampacity at40C l ICEA VALUE AT 40 C IS 438 AMPS. THIS VALUE IS CONSERVATIVE.
f 140 i V := l -1 l-100 (438 amp j V = 9.234 % VARIATION i
i
. __ . _ _ _ _ _ _ . _ _ _ . _ . . . - . . . _ _ _ . . . _ _ . . _ . . _ . - _ _ _ _ _ - _ -- . _ _ _ _ - . .m._ . _ . . _ . - _. . .
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGEMOF8f4 APPENDIX U PAGE 1 THERMAL MODEL TEST 4 3.0 in 1.0 TrayM Data R=0.0313575/1000ft '
w := 24 in Widthin Tray cad =2.12 500MCMTPLX d := 3.0 in Depth of cables in Tray V
TABLE 3- 8 (ICEA)
Ta :=( 273.16 + 40) K Ampient Temperature (P54-440,1986)
Tc :: unknown- K Surface Temperature Tm := ( 273.16 + 90) K Max CableTemperature watt OmH Convh Heat transfer Mb knt h := 0.101- - -
iK Kl 2 ft K watt !
h := ( 0.101 )-
ft2- K a := 0.530 8, watt Stephan-Boltzman Constant 2 4 f1-K '
Effective thermal emmesivity of cable mass and tray g . 0.8 surface.
p : 400- K effective thermal resistivity of cable mass ,
watt ft2 Surface area of cable mass per unit length As.. sides mt imlM i
Tm + Ta Tc = 338.16 K Initial assumed value for Surface Temp. (Tc) t 2.0 Thermal Calculation
' 2.1 calculated Value for Surface Temperature (Tc) i i.
f 4 Tai T Ic Tai 4 l Tc - h-( As p d) h-( As p d) 1 1 l Tc ::m a- As c p d Tc4 + (K Kj (K Kj , _ o- As c p d 4 8- w 8w 8w 8w Tc = 342.83 *K :
CALC.NO.96-ENGK)1528E2 REV.0 ATTACHMENT B PAGE64 OFbf0F APPENDIX U PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray 8
4 W:=h! -
As-( Tc - Ta ) + o As e-( Tc - Ta# )
(K K/.
W = 99.144
- ft 2.3 Recalculated Max Cable Temperature (Tm) 1 I f Tc Tai 4 fTc Tal #
hl - i
-( As p-d) h-l -
I - As p dTa 4 4 Tm ::
(K K) +1 - Tc - (K K) + a- As c p d Tc - o As c p d Ta 8w 8w 8w 8- w Tm = 363.159 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc := Tm - Tc ATc = 2033 *K ATc :=
8w ATc = 20.33 *K 2.4.2 AirTemperature Drop ATa ::Tc - Ta ATa = 29.67 *K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 49.999 K 2.5 Heat generated per unit area w
watt Q :: d w ft Q = 1377
- in2
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENTB .PAGE6M OFStW APPENDIX U PAGE 3 Ampacity Calculation n := 3 number of conductors cad := 2.12 in Cable outer diameter R := 0.0313575- Resistance taken from Okonite book i 1000- ft I40.--_ cad l Q x 2 ( n- R >
P I40 = 227.309
- amp Ampacity at40C ICEA VALUE AT 40 C IS 272 AMPS. THIS VALUE IS CONSERVATIVE. i f 14 0 i V := l -I j -100 (272 amp j V = 16.431 % VARIATION i
i i
=
L
_ _ _ _ . . _ - _ _ _ _ _ _ . _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ . _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ m_____________- __ _____ _ _ _ . __ ___ m ___
CALC.NO.96-ENG41528E2 REV.0 ATTACHMENT B PAGE @ W APPENDIX UUPAGE 1 THERMAL MODEL TEST 4 3.0 in 1.0 Tray M M R=0.0313575/1000ft w := 24 in Widthin Tray cad =2.12 500?ACMTPLX 4 d := 3.0-in Depth of cables in Tray TABLE 8 (ICEA)
Ta ::(273.16 + 40) K Ampient Temperature Surface Temperature P-54-440' 1979 Tc := unknown K Tm := ( 273.16 + 90)-K Max CableTemperature I
wan mH on e t transfehk M h :: 0.101-(K - K) 3-K2 !
h:=(0.201)-
a2 ,g
~8 **"
o :: 0.530 10 - Stephan-Boltzman Constant ,
a 2,g4 c := 0.8 Effechve thermal emmisivity of cable mass and tray surface.
p : 400- K- effective thermal resistivity of cable mass watt 2
uda area d @ mass p W @ t A5 *-
- - 4 T sides not included
+
i Tc _ Tm + Ta 2
Tc = 338.16'K initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation '
2.1 calculated Value for Surface Temperature (Tc) 1 1 ,
I f Tc 4 f 4 I' Tal Tal l -
I -h-( As p d) lTc - l h-( As p d)
Tc : root -Tc +
4 (K K/ , _ _
(K K) ,
_ o As c p d 4 8w 8w 8- w 8w Tc = 340.031 *K
__.___..,_.___.__..__._.._.._._.-_.__..___._..._.._____.._...__.__I
CALC.NC.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE847OF6pS-APPENDIX UU PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray
.I '
ITc Tai 4 4 4 W := h-l -
1 - As-( Tc - Ta ) + o As c-( Tc - Ta )
(K Kj j i
watt !
W = 112.797
- l ft -
2.3 Recalculated Max Cable Temperature (Tm) 1 I >
.fTc Taij 4 I Tai 4 l h-l -
-( As p d) h-lTc - l As p-d Ta 4 4 i Tm :: AK Kj
+1 Tc -
(K Kj + 0- As t-p-d Tc _ o- As c p d Ta 8w 8w 8w 8w l Tm = 363.16 t 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc ::Tm - Tc I ATc = 23.129 *K Wpd ATc :: ,
8w ATc = 23.129 *K 2.4.2 AirTemperature Drop .
ATa ::Tc - Ta I
ATa = 26.871 *K 2.4.2 Total temperature Drop j AT :: ATc + ATa I AT = 50 *K 2.5 Heat generated per unit area W <
watt I Q::dw t Q = 1.567
- 2ft in [
[
_ _ - . .-__-.___-..i
~ . . _ . . ... - - --.-_ - -.-. . - - , _ . - . ~ ~ .. - . .. - . - - _ - - - . - . . .. . . . .. . . . _ . - . . ~
CALC.NO.96-ENG-01528E2 REV.0 ATTACHMENT B PAGE h OFD F APPENDIX UU PAGE 3 i
Ampacity Calculation n ::3 number of conductors ;
cad :: 2.12-in Cable outer diameter ohm R ::0.0313575- Resistance taken from Okonite book :
1000- ft .
140 _ cad -
QR 2 1nR i
140 = 242.456
- amp Ampacdy at400 l
lCEA VALUE AT 40 C IS 272 AMPS. THIS VALUE IS CONSERVATIVE. -i f 14 0 1 l V :: l -1 l-100 '
( 272 amp j r
i V = 10.862 % VARIATION i l
t I
t-
?
i
CALC. NO. 96-ENG-01528E2 REV.O ATTACHMENT B PAGE P' OFD_fe._2.
APPENDIX R Page1 THERMAL MODEL TEST 5 3.0 in 1.0 Tray Thermal Data R=0.06890625/1000f t w := 24 in WK!!hin Tray cad =1.61" 4/0MCM d ::3.0 in Depth of cables in Tray TABLE 3- 9 ICEA S M O,1986 Ta : ( 273.16 + 40) K Ampient Temperature Tc :: unknown K Surface Temperature Tm ::( 273.16 + 90) K Max CableTemperature 1
wan rH m we Heat '.mnsfer Cd knt h ::0.101 1 -
LK K/ 2 ft K h := ( 0.101 )-
2 It K
~8 **" Stephan-Boltzman Constant a != 0.530 2 4 ft , g ml s ass aM tmy c :: 0.8 surface.
p := 400- K '" effective thermal resistivity of cable mass i watt
._ fl 2 Surface area of cable mass per unit length
^8 *- 4' sides nnt in cluded.
ft Tm + Ta 2
Tc = 338.16 *K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) 1 1 ,
- I ITc Tai #
l -
h-( As p d) lTc -
l h-( As p-d)
Tc :: m - As c p d,.,,c 4 + (K K , _ _
(K Kj ,
_ o As c p d 4 8- w 8w 8w 8- w Tc = 342.83 *K
CALC. NO. 96-ENG-01528E2 REV.O ATTACHMENT B PAGE@ OFOSS-APPENDIX R PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray L
4 I Tc Tai 4 4 W ::h j -
l As-( Tc - Ta) + o- As c-( Tc - Ta )
(K Kj W = 99.144
- ft 2.3 Recalculated Max Cable Temperature (Tm) 1 1 4
ITc Tal T Ic Tai 4 h-i -
l -( As p d) hI -
j - As p d Ta (K K) (K Kj c- As-c p d c4 _ o As e p-dTa#
Tm *: , _
8- w 8w 8w 8w Tm = 363.159 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise -
ATc := Tm - Tc ATc = 20.33 *K W- p d 8w ATc = 20.33 *K 2.4.2 Air Temperature Drop ATa :: Tc - Ta ATa = 29.67 *K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 49.999 *K 2.5 Heat generated per unit area ,
wau Q := d w Q = 1.377
- 2a in _
p 6
- . _ . _ _ _ _ - _ . _ _ _ _ . . - _ - . - . _ _ _ _ _ _ _ - . _ . _ _ . - -_..--__.----_-___x-- - - . _ _ . _ _ _ - - . _ . - - - - . _ - - _ - - - - - - _ - - - _ - - - . - - - . _ _ _ _ _ _ - -- - - _ _ . - - -
. . - . _ . ___m.__ .. _._ _.-... . _ - . . . . _ _ _ --. _ _ _ _ . - . . - . - . . . __--_.m - - - . . . .
CALC. NO.96-ENG41528E2 REV.O ATTACHMENT B PAGE O OFGf'1 APPENDIX R PAGE 3 t
Ampacity Calculation n: 3 number of conductors cad := 1.61 in Cable outer diameter R := 0.06890625- Resistance taken from Okonite book ,
1000- ft 140._
.- cad Qx ;
2 1nR i
140 = 116.452 amp Tray Z24FL20 Ampacity at 40C ICEA TABLE 9 VALUE IS 135 AMPS. THIS VALUE IS CONSERVATIVE.
f 140 t
1 L
v := 1 -I P-loo tl35 amp j l r
V = 13.739 % VARIANCE i
[
'l
_ _ . _ . . _ _ _ _ _ _ _ _ . _ _ _ _ _ - . , _ _ _ _ . . _ _ _ . _ _ _ _ _ _ _ _ _ . _ . . _ _ _ . _ _ _ _ . _ _ _ _ _ _ . . . _ . _, .._..____mm _=___m_ _ _ _ . . ___ _ _ _ __ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ . _ _ __ _ _ _ _ _ . _ . _ _ _ , _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ --
CALC. NO. 96-ENG-01528E2 REV.O ATTACHMENT B PAGE M OF 8"1 APPENDIX RR Page9 THERMAL MODEL TEST 5 3.0 in 1.0 Tray ThermalData R=0.06890625/1000ft w := 24 in Wdthin Tray cad =1.61" 4/0MCM d := 3.0 in Depth of cables in Tray TABLE 9 ICEA 440,1 W9 Ta ::( 273.16 + 40) K Ampient Temperature Tc :: unknown K Surface Temperature Tm := ( 273.16 + 90) K Max CableTemperature 1
w mH n e aH nsfeW@ MM h := 0.101 1 -
l-(K Kj 2 3 -K h := ( 0.201 )-
2 ft K
-8 **"
o :: 0.530 Stephan-Boltzman Constant 2 4 ft K i sW of caw mass aM tray c := 0.8 surface.
p :: 400- K '" effective thermal resistivity of cable mass watt ft2 Surface area of cable mass per unit length As _4 T sides not in cluded.
Tm + Ta 2
e= .16*K Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc)
I 1 ITc Tai # f Tc Tai 4 1 -
l 'h-( As p-d) l - 1 -h-( As p d)
Tc ::m As c p-d,.,.c4 iK Kj y , _ _
(K Kj , . _ o As c p d 4 8- w 8w 8w 8w Tc = 340.031 *K
CALC. NO. 96-ENG41528E2 REV.O ATTACHMENT B PAGE86OFOC1 APPEND 0(RR PAGE2 2.2 Calculated value of total heat (W) per unit length generated in cable tray 3
4 W: h 1 -
As-( Tc - Ta ) + o As c-( Tc - Ta# )
iK Kj W = 112.797
- ft ,
2.3 Recalculated Max Cable Temperature (Tm) 1 1 I 4 I #
Tai Td h-lTc - j -( As p-d) hiTc -
l As p-dTa 4 4 (K K/ (K Kj 0- As e p dTc _ o As e p-d Ta
.g. ,.= , , _
8w 8w 3w 8w Tm = 363.16
- 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc := Tm - Tc ATc = 23.129 'K Wpd 8w ATc = 23.129 *K 2.4.2 AirTemperature Drop ATa : Tc- Ta ATa = 26.871 *K 2.4.2 Total temperature Drop AT : ATc+ ATa AT = 50 *K 2.5 Heat generated per unit area wau Q := d w ft Q = 1.567
- in 2
CALC. NO. 96-ENG-01528E2 REV.O ATTACHMENTB PAGED 4 OF#D2-1 APPENDIX RRPAGE 3
-t Ampacity Calculation n := 3 number of conductors cad := 1.61 in Cable outer diameter !
l r
R :=0.06890625 ohm Resistance taken from Okonite book !
1000- ft [
i 140._
.- cad Qx 2 1 nR l i
140 = 124.212
- amp Tray Z24FL20 Arapacity at 40C i
ICEA TABLE 9 VALUE IS 135 AMPS. THIS VALUE IS CONSERVATIVE. j f 14 0 I i V := l -1 1-100 !
(135-amp j i V = 7.991 % VARIANCE f i
i I
L i
I i
i i
. . . _ . . . _ _ . m .____.. _ _ - __,.m____-m_m.______,_.___m. i__m__.____.__t_.__
__.___ =_. ___-_-__--a .-e-wwe-----we-meem.4 ,_-c - e e-+--*
- .- --. . - - . ~. -.- . - ..--c -. .- ~ . . _ ,
CALC. NO. 96-ENG-01528E2 REV.O ATTACHMENT B PAGE## OF 6tt3 APPENDIX S PAGE1 THERMAL MODEL TEST 6 2.5 in 1.0 Tray Thennal Data R=0.0658875/1000ft w ::24 in Width in Tray cad =1.61" 4/0MCM d := 2.5 in Depth of cables in Tray 3- 9 lC Ta : ( 273.16+ 40) K Ampient Temperature p 49, j 98 Tc :: unknown K Surface Temperaturo Tm ::( 273.16 + 90) K Max CableTempe _re wan mH nde Heat tmnsfehk snt h := 0.101- - j -
(K Kj 2 ft K h:=(0.201)- 2 ft - K o := 0.530- 10"8 -
Stephan-Boltzman Constant 2 4 ft . g nnaWsW of caw mass aM tray e := 0.8 surface.
p : 400- K #"
effective thermal resistivity of cable mass watt 2
Surface area of cable mass per unit length A5 ._'- 4' ft sides not included ft Tc .- _ Tm + Ta 2
c= .16 %
Initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation 2.1 calculated Value for Surface Temperature (Tc) 1 1 I Tc #
Tal # ITc Tai l -
l -h-( As p d) 1 - I -h-( As p-d)
Tc := m As c p d, 4 (K K/ , _ _
(K K) c- As c p d,. ,,,
4 ,,,e 8- w 8w 8w 8w Tc = 342.109 *K
CALC. NO. 96-ENG-01528E2 REV.O ATTACHMENT B PAGEAf' OF001 APPENDIX S PAGE 2 2.2 Calculated value of total heat (W) per unit length generated in cable tray fTc Tal 4 4 W := h-l -
1 As-( Tc - Ta ) + o As c-( Tc - Ta# )
(K Kj W = 123.192
- ft 2.3 Recalculated Max Cable Temperature (Tm)
I 1 fc T Tai # fTc 'I'd hl -( As p d) h-i l As p-d Ta o As t p-dTa 4
! 4 Tm :: (K K;
+1 Tc -
(K K) + o As t p- d Tc -
8- w ,
8w 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc ::Tm - Tc ATc = 21.051 *K W p-d 8w ATc = 21.051 *K 2.4.2 AirTemperature Drop ATa ::Tc - Ta
- ATa = 28.949 *K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 50 *K 2.5 Heat generated per unit area wau Q :: d w a Q = 2.053
- 2 in
CALC. NO. 96-ENG-01528E2 REV.0 ATTACHMENT B PAGE6f70FOf't1 i
APPENDIX S PAGE 3 Ampacity Calculation n:=3 number of conductors .
I cad :: 1.61 in Cable outer diameter R := 0.06890625- Resistance taken from Okonite book 1000- ft 140 :=
2 1nR L I40 = 142.199
- amp Tray Z24FL20 Ampacity at40C I
ICEA TABLE 9 VALUE IS 153 AMPS. THIS VALUE IS CONSERVATIVE.
f 140 T i V := l -I l-100 (153 amp /
V = 7.059 % VARIATION .
I i
CALC. NO. 96-ENG41528E2 REV.O ATTACHMENT B PAGED /*OF 002.
APPENDIX SS PAGE 1 THERMAL MODEL TEST 6 2.5 in 1.0 Tray Thermal Data R=0.06890625/1000f1 w := 24 in Wdth in Tray cad =1.61" 4/0MCM d := 2.5-in Depth of cables in Tray TABLE 9 ICEA Ta ::( 273.16 + 40) K Ampient Temperature P 54-4401986 Tc :: unknown K Surface Temperature Tm := ( 273.16 + 90 ) K Max CableTemperature I
wan mH n e Heat tensfehk ient h ::0.101- -
(K K) 2 ft - K h := ( 0.201 )-
ft2- K
~8 **
o := 0.530- 10 - Stephan-Boltzman Constant n2.g4 sW om mass aM tray c := 0.8 surface.
p :: 400- K effective thermal resistivity of cable mass watt F
2 Surface area of cable mass per unit length l' As ._4 ft sides not included ft l
Tc :: Tm + Ta ;
2 Tc = 338.16 %
initial assumed value for Surface Temp. (Tc) 2.0 Thermal Calculation ,
2.1 calculated Value for Surface Temperature (Tc) 1 I I Tai # f '
Tai Tc :: root As c p d,.,,c4 lTc (K
K)
I h-( As p d) lTc (K
-)K,
-h-( As p d)
_ o As-c p d, 4 1
8w 8w 8w 8w Tc = 342.109 *K -a r
r
__..._.__.___m_____.____m.______ - _______.._.____.______.._______....______m _ _ _ . _ _ _ _ _ _ _ _ . . ~ _ _ . _ . _ _ . _ _ _ _ - _ _ _ _ . . _ _ . _ _ . _ . _ . . _ _ _ . _ _ _ _ _ _ _ _ _
CALC. NO.96-ENGK)1528E2 REV.0 ATTACHMENT B 'PAGE6#!OFCl*3 APPENDIX SS PAGE 2 2.2 Calculated value of total neat (W) per unit length generated in cable tray 4 d W::h1 -
1 As-( Tc - Ta ) + o As c-( Tc - Ta )
iK Kj W = 123.192
- Il 2.3 Recalculated Max Cable Temperature (Tm) 1 1 I 4 f Tal Tai 4 h-lTc -
l -( As p d) hITc -
l As p dTa o As c p dTc4 _ o As c p dTa 4
Tm *: (K Kj (K K, 8w Sw 8w 8w Tm = 363.16 2.4 Temperature Rises 2.4.1 Cab le Mass temperature Rise ATc *:Tm - Tc ATc = 21.051 *K W-p d 8w ATc = 21.051 *K 2.4.2 Air Temperature Drop ATa ::Tc - Ta ATa = 28.949 *K 2.4.2 Total temperature Drop AT :: ATc + ATa AT = 30 *K 2.5 Heat generated per unit area wau Q::dw a Q = 2.053
- 2 in
_. . . . - _ . . . . _ . ...s . . _ . . . . . _ _ . . _ . . . . . _ . . . _ .- . _ .....~ . . .__.._. ._. ..._ _.- .-. m - . _. _ _.. .. _ _ , ..... .
CALC. NO. 96-ENG-01528E2 REV.0 ATTACHMENT B PAGEO#10F M I APPENDIXSS PAGE3
< i Ampacity Calculation n ::3 number of conductors ;
cad :: 1.61 in Cable outer diameter R ::0.06890625 ohm Resistance taken from Okonite book 1000- ft Qx I40._
.- cad
- 2 1nR 140 = 142.199 amp Tray Z24FL20 Ampacity at40C ICEA TABLE 9 VALUE IS 153 AMPS. THIS VALUE IS CONSERVATIVE.
V :: 1 f I40 I
-I l -100 (153 amp j ;
V = 7.059 % VARIATION u
P L
1 L
t L
i
_.m ____________.m.m... _ _ _ ._._.__ _._______m._____ _ . _ _ _ . _ _ . _ _ _ . _ _ _ . ____.-____m_ __ _ _ ____
ATTACHMENT C.1 CALC. N'i 96-!ENG-01528E2 R00 PAGE C I OF C{_
CONTROL CABLE LOADING l
TRAY Z14FM10
@ [f5M 3 UST LOAD M58/SM UST LOAD l X1 33X+BGR+ X1 42O+42C+RG 0.6 i 420+42C 1.17 11 42 0.5 21 42C 0.5 12 42 0.5 22 420 0.5 3R R 0.045 3R R 0.045 3G G 0.045 3G Gi 0.045 51!B,33X 0.0775 32009 6 R11 i 32009 7 R9 I
Z,1CH192/B;j UST.
LOAD 31HV82475j LIST LOAD P11FG 0.09 2 HY8247 0.1392 1lFY192 0.776 N1 HY8247 0.1392 l N1 lFY192 0.776
]
3RiR 0.045 l 3GlG 0.045 32009 53 R6 ! 32021 12 R5 Z1HV8247/D J LIST LOAD 22HV8249/C5 UST LOAD l 12lHY8247 0.1392 2 HY8249 0.1392 22;HY8247 0.1392 N1 HY8249 0.1392 l 3RiR 0.045 3G'G 0.045 N1!RE 0.135 3 bib 0.045 32021 12 R5L 32021 14 R5 l
f5N '8N9/d2UST LOAD R1MCV02[BM LIST LOAD 12lHY8249 0.1392 1 42X +R 0.112 22iHY8249 0.1392 11 42X +R -0.112 3RlR 0.045 3GlG 0.045 N1lRE 0.135 i
3B B 0.045 32021 14 R5 32509 44 R4 l
! I l
l l
Page 1 l
l
i ATTACHMENT C.1 4 :
CALC. NO. 96-iENG-01528E2 R00 PAGE G1 OF_Gf_
l
, CONTROLCABLE LOADING J
$[]{196/_B73 UST LOAD
, P1 RGB+74 0.1535 1 FY196 0.0776
- N1 RGB+74+ 0.2311 i FY196 0.0776 3R R 0.045 3G,G 0.045 3B B 0.045 I 32009 54 R6 a
1 4
f l 5
' I i
a
- 1. '
.i ,
I 1
Page 1
1 ATTACHMENT C.2 CALC. NO. 96-;ENG 01528E2 R00 PAGE C5 OF A f._
l CONTROL CABLE LOADING Z25XA10 M 5{7 5 3 UST LOAD ~~~57534 UST LOAD HYS17 11 0.0776 HYS17 11 0.0776 R 33R 0.045 R 33R 0.045 G 33G 0.045 G 33G 0.045 RTN N11 0.1676 RIN N11 0.1676 l 32009 35 R4. 32009 35 R4.
M@5/5][] UST LOAD %)19tESh UST LOAD HY519 11 0.0776 HYS19 11 0.0776 !
R 33R 0.045 RTN N1 0.1676 G l 33G 0.045 R 3R 0.045 RTN l N11 0.1676 G ~
3G 0.045 3200937R5] 32009 37 R5 t
filIV5555/C UST LOAD RR8V2525IG3 UST LOAD RTN i P1 0.2126 HY2525 41 0.0776 HY2525 i 21 0.0776 HY2525 51 0.0776 HY2525 '
11 0.0776 RTN P11 0.2126 HY2525 e 1 0.0776 G 33G 0.045 HY2525 I N1 0.0776 R 33R 0.045 R ! 3R 0.045 B 33B 0.045 G 3G 0.045 l B l 3B 0.045 32009 39 R6 4 32009 39 R6 22N 2525/H.; LIST LOAD MEQ9/Ej$ UST LOAD HY2525 l 61 0.0776 H21 ARM 11 1.57 HY2525 i N11 0.0776 H21 FLD 12 0.807 32009 39 R6 ARM H21 13 1.57 FLD H21 N11 0.852 32020 42 R10 I l f.2M~09/GC) LIST LOAD @,MhM LIST LOAD H21FLD 1 1 1.57 HY5279 21 0.281 H21FLD i 2 0.807 HY5279,74-1,B N 0.3445 H21 ARM i 3 1.57 H21 FLD,W t N1 0.852 32020 42 R10 32012 22 R10 i
Page 1
- . .- ._= -. .- . .. . . . . _ - _ _ . - . - . - . - _ . - . - -
ATTACHMENT C.2 CALC. NO. 96-l ENG-01528E2 ROO PAGE C4 OF _GjE i
CONTROL CABLE LOADING l l
EdY52Z9/82 LIST LOAD . __'_3MI} LIST LOAD 741.8,G,R.HY527o P 0.4343 74-1,B N11 0.281 R 33R 0.045 74-1,B N21 0.0635 G 33G 0.045 HYS279 51 0.281 I 74 1 P 0.0185 74-1,B 31 0.0635 74-1,B 41 0.0635 RTN P11 0.3335 3LT 3R 0.135 l 3LT 3G 0.135 i 32012 22 R10 32012 22 R10 l l
l 1
i l
I Page 1
ATTACHMENT C.3 CALC. NO. 96-! ENG-01528E2 R00 PAGEG5 OF _{ _[
ATTACHMENT 8 CONTROL CABLE LOADING l
TRAY Z24FL20 RSSpj05[55!UST LOAD M@g[QUST LOAD X31 l42X+42X+42X 11 42X 6 0.0667 l42 2.2831 12 42X-6+42X-7 0.1334 4l42 2.083 13 42X-8 0.0667 191lNONE 1 42X-6 0.0667 U ! 42X+42X+42X 0.2001 2 42X-7 0.0667 3 42X-8 0.0667 32009 42A R1 32009 42A R1 i I ZiiNV5155ICluST LOAD IEihV81.SS/Dj UST LOAD 2 lHY8133 0.1392 12 HY8133 0.1392 N1 'HY8133 0.1392 22 HY8133 0.1392 l 3R R 0.045 1 3G G 0.045 I N1 HY8133+RGB 0.2742 l 3B B 0.045 32021 13 R5 i 32021 13 RS 2NV8248/Ci UST LOAD (Z5NV8N8/.D] LIST LOAD l
_ 2 lHY8248 0.1392 12 HY8248 0.1392_ l N1 'HY8248 0.1392 22 HY8248 0.1392 !
i 3K R 0.045 3G G 0.045 N1 RGB,74-1 0.154 3B B 0.045 32021 15 R5' 32021 15 R5 l
58415i6C LIST LOAD 5sID[964/T3 LIST LOAD 21 lFX+F+CR+63 0.304 1 42X 0.0667 31 l FX+F4CR 0.174 11 42X 0.0667 41l 63(HGA) 0.0325 X1;m 0.09 3RIR 0.045 3G!G 0.045 32025 33 R6 I I 32009 46 R5 Page 1
Caic. No. 95-ENG-01528E2 RR Page DL of D4 ATTACHMENT D 1 in wen tLt t$e arne,ar) ei a c6te is directl) eroporuunal to its two low for the Lrge cunductor uns. Le that for the thin wall c*erall Jt. meter iD; Thies. ineresung the insubnun thietness on a XU inzutated ables. the amrsatin are ewn lower than for the phen cien.svet e m.:ce Hs Jismewr and thus ineresws its ampaaty tJuci mall rubber cbles. and the soiety of the persent 'ampaesto shen installed in a .:itar tny. (or a greert percent irsy idt and the would be enn more n,t,iessionsble, wme te:nrrrature utmts . . .
Tliis point is made ,to surplement one of the favorable g Here it rmns de pomsed out that the ampacties ei the bulky crtres of the small Jimmeter XU celes. SpectScily, more .\.
ruthr msubica cables is trays are not at all the nme as arapaanes cshks can be installed m a cab &c tray than otner kinds of insubted for the wrull erassiiried raah ethylene insutsted esbles with very thin czbies, and thus there is cronomy in using fewer nble trays Along insutations. For esamrie, a numter 12 AWC rubber insutsted able with being able to install more cables in a tr:y at is essential that the with a dumeter os .24 inches may have an allowable best intensity thin wall cabic city leu c1:rrent than the honer insubted abiet if Ifrom Fipre 4.% pive an ampseity oi:4 amre;the sarr.tonductor this is not done. there will be overnesung of the XU cbles and the insubted with crossimLed pnhethylene would have a daarneter ef screiented loss of csble life resulung m premature chie failures, j only ah'ut .le mdes and therrinn. frorn envation t9 t.an ampactry of 16 .mrt it t.*.us tv.mmes necessary to distinguish between thin The best otrscrr: tion to be made from the theory is reisted to nil and 1.ien; mailicsubted cablest throughout this paper.rciercoer heat generation ist chie trays being in proportion to the crow.
Io poi)etr.)lene e.b.e isnpues thin wallinsulation and rubber irnpises secuorial arcs of each cble. A somewhat eddent justiGotion for (b.
requirement en be seen from the io!!o*1mg reasoning.
truck = asi mwianon. ]
. The he Jiiference m .rnraaty coaes from the fact that for a The most elementsry e:usuon dese:sbing convection hest 11ow geen rer.cnt tray G!L more wroutinked putyethylene ithin wsill is intabted .md .s rs na be radeJ meu the iny than rubber ethid m:llisr.sulaird (*.<sduerus Since the lutal amount of hest which truy g a gi de perisratte t.1 "' trs) must remum const:nt. the heat per cun-Juctor muu te seu (Jr tr.e small JLmeier cables than for the large shere b is the convectiun heat trannier cucinaent. A,is the autisce ones area convecting hest lo the alt :nd iT is temperature differense i between the cble surher and the ambent air. The haue ettuauon !
With ett .r )* ,ble heat tsienstres from Figure 4 and uung - for conducaon hest ir:nsier is t herr: In twaatwn e d s. tr.c ampaatws of sevent abic uns and ,7 pereer.: trJ) tau .sfr he octemed. The results are up=n in Ferre 5. q = lay #
=hkn h a pr#ml .intaary t bic fur typici unple condu. or rusher : wiatcJ arter cer luctars insia: led in Mnch by *a-mch where k is the thermal conducuvuy of the hest conducties rnetjium. l cutiin tras s. F *r ceme.insen. the rrnendy publided amoae:ues tor A ga the cross-secuonal r.rr: throup wnich heat flows, and AT is the the wme ts re htte .re at=0 riottect: they are (or the suumed cv of tempenture drop over 2 distar.ce it in the direction of hest Oow,.
maurium Je tune
- hich n for JJ or more condue: ors m the tny. >*oie that convection heat flow is preocrtional to surisce area w'
- nd thus .re Fr ' ci ths amtsaty ci a three conduetor c:bie :: air. conducion hcs llow is pruroruunal to cross sectinnel arc. Sin. I.
a concurtion is the reverning method of hest flow wuhan a uphily racied chic mass. we suculd be concerned with cross secuonal arens of abies rather than reripheral or wrisce are31.
IOCC- 20
--9*.-J41 *p*
TEST PROCIDCRE
, s Dt*<t A
- ' :4 I i i IllU //V/ Tive different able tray arnntements have been thorouchly f
SCC '
j' l l lll [f f --
tested in order to determ ne the hest tnasier properties of ench arrangement. Two of the tests ir.vohed :nJomly ar anged cables or g ,gv einous stres iri 2Mneh wide trays snJ three tests were performed on 1 inch wide trzys alth only one able sUr in the tr:y. Tabic I
%, -y ,
E 7/4 ' . i 84
- 46 i- summarues the enrious tests shich were performed and Tigdre 6 b 3 ! h/, l llil tl l llll .
shows the overall test setup.
w 2C' TARE I - Suirmary of Tests Ccmoucted to Support i.is Analytical tesults g , . . .
i.e .i i i , l o ne i e si .
TW PER;DfT C.ML! 312E5 INSULATION
- _ritt mt
- , !l :!..!UI l lldll$ 10 0 l llIll. !!.
4000 5lz' its m IC 8
CONOU: Test s!M (hi:W) 3*a2P 20 d12 to 4/0 Rubber dg 3 124* 15 plZ to 4/0 Ruboer "'II
- 3*a12* 40 3/C-al2 Rubber .
3*a12* AD 3/C *12 29 thin Ft. .% Amrauws edespiraituhher utsuLrtrJ rnppre rubles m J"x 29 mall 3*a12* 50 1/C_-500
.J" rrsrs .ts Jewens:.rrJ Fr this stuJe and numpared wuh JPCC.4 and .\TC :.thws s.ar trees twowsmns owe than U c..aawr rs. W C..ne earmt semeentrure in a J# C ambwnt. Some details of the testirit which were common to alt tests enn be seen from Figure 6. 600 volt rated copper conductor esbies were This g=rhical comr.iruon. along eith test resulu presented hid in a 24-foot long clait tny and termpentures were rnessured later, inskes it quite etcar tr.JI the prewnr ampac2 ties for tnys with three different tray coss secuenst one wu in the mid4engt,h of II /
tray and two others at the quarter lengths. In nsny cases cables bis.n percent nils are tro bp. ice small conductor sizes.while bemt 96$
. . . .~. - . - - . - - . - . .
- i 5
i Calc. No. 95-ENG-01528E2 R00 Page QL of D4 )
l ATTACHMENT D ;
{
a
.e .
test results to be piracnted later show thes value to be validTor either The systens terripentureirott'is the sum of the drop through, the packed esbie mass (AT,) and the drop through the air (AT,1 , rubber or polyethylene insulased embles which are t yhtfy p
^*
- asound the cable tray. /
At this point we must define able tray Ivrernt Gil as the suni g of the crosNectional ares,s of J!l ables in the any tincludag
, Therefore * * * . conductor. insulation and iackett dieided by the total availabie i .
crosHecuonalarea an the esbit tray Iwidth times height 1. It can be AT = AT, + ATs . H)
- wen that a cabir uzy which is pacted as ushi as possible and level
. acrcss she top is filled to dout,7! *. Neauw about 2$*; of the in) a The drof* through the able snass (AT,1 en be ottained from " 4 8 for a rectanguist si:b with inniform )'
- arra is%d aren b'et4rn the etreular cab #es. Frem the shrive percent
$ .the equation smn by Hohnaa I inn . a . ,* *p i .v "" , @, g7ji nil dennit$ it is a* ppa'r'ent th'.it'a Nash deep any with 20i fill .
- i . internal best pencrat ion.* ~~.-.y ~ ~
Dis the sarne der'th of"racte'd calde' ai a' 3-irEh deep Ine with 407
,,ljr,JA.inss%-er:W U.*.' b8f $.y . A~h" ,* #*
- d " " ~ '
~~'~ .* . ~
..:.,.:.v.1x A*s:aca AT,8"- s*.:d' .j t . .
- E, ,,
r= ^ ~ '.!' c%
- si="~fm.* '"Y. e, *
~
-r s~ " s
.* W64 s.psti. ass ea ; M C 8. WrO* * * *
-."#'. ..it . . . . . .. . .
s <
"* i c , In appleing equ:tions f li and e *:io pet the ampactry of specific - .
where A a effective :hcrr.alresistivity of cable mass "
- *' [ ennductor mz's e in s ymn able tray an mteressmy oberresuun c:n 8 1 d 'a depth of dble mass "**
~~~
{
- w = width of cable mass and tray ? .,
,be made.The chie ,mpsesy in n sma by
- .s . W =' the total heat renerzted in the tray per unit length., , , , g ign l= _
. . "" 4
~ huntion(Jiis speeme:!)y for one dirnensional hest now out the top ,
and 'subeututing for the circubr croisectional s.re. c(e:chhMe ( A e "and baWom of the tray and is ignores'any heat flow out the sides of
- the tray. Dis is 'a reshstic srmplific:uon which is accurate for -Mneh=~ "'. unrw.get,m ... .
+
.and wider esble in>s . *. ' " [, .- :. .
6* O W- aiu l . .
, ' l=_
j .
The tempernture drep through the air (AT,lia obtamed frem a
- heat bal:nce between converuon and rddistion hrst der. Using b: sic equaticas from bleAdam/ we find -
ru w = hA,.tT,
- e Ag e!T,4. T,#1 10 25
= the hest less from the t.ay due to 20 NI I il 1Ill where AA,?.7, conveeuen 3 g3 \
" sA3s[Tgd . T,#1 = the best loss from the tray due to ,
radistion h- A . .
s
$ and h = ovenit conweenon best tr:nsfer eorflicient .,
g for rny
= surface area of cubie man per umt 515'O tx . s. ,-
9C*c j A, '/e 75*c I trzy length a s i \t \
g' I N l g , s/ j' f 60*c !
j e = siefan-netta.msnn ennstant
- e
= cffeetny thermal emis>mty of y g g\; g g l l cWe mass and any surfs.e w 6 T, = sverage esbir mass surfaw temperature E \ l h \l l * ! 'l 4
The three equations IJa. f5 L and f(.) have three unknowns sad the> b4
\
can be 5c}ved to get the total alkwable hrst wlaish can be geneinsed in a catie'tr=y (W1. Since cadauen 166 is eune non bnear. the w ' ' \
so6ution'to the three equations must be ebtained by itersucc!thus. N3 4
for perIeral appliention the soluuon for T is done most easily on a $g \i computet.
7 .. < 2 \
^
\*
Maring the total hest pencrated in the cable in), the heas *\
rener: tion per unit ares rs simply
. . \ \
.} th g
'.... w * *
\ N
- (h \
- O .=I d It w I t
\
4 The ampseity of each cable in the iny is fmally deterrnined I 10 iS 20 DE M to 60 N sichg{y*. equations PER0EWT TRAT FILL
, .tf1)[A CL +
" l- ,
b - 'T!!EORETICAL Hr.5ULTS *
. ty; !.
g '.
D solution to etruations 141. ($1. and tM for W and several deprecs of cable trvy fill mill result in curves surular to thr*se shn*n m Fvure 4,11 is seen that as the cable tray percent fill merciws.the sunm%le best intensary de:resses due in prester temperature drer in Fig. af. Allr8avhle f.rve ser=wre tvl sa srwms.r4 rubbr'l* '#
1 I f=drethrt. n. r,Nn .er sh. reqrns iemtwvsr,ac en .t.mehrs the tightly packed eshic rnass. Ficure J da made sur an eficetm drer be JJ.,*rkn u.k 2,ur ,.r.erstm.c en . .rm( emba nt o
thermal resiirity of the cable ma>> hems .idwt.em wat:. a the ,
Calc. No. 95-ENG-01528E2 R00 Page D3 of D4 ATTACHMENT D by in. Inp thl an - wtmnalarcaelexh.ompe=sreaNr. Dius o*vr2O heat transfer mechansam is russble. A. ,sunple analyaxal the (feektil now semesnt to e4,pleda the alkemable heas mis 9Neyint solution to the beat tr2nsfer frora the general. bypothetinj oble tray m Figure J has been stade,and some rather subine rmdings i,,,
varmanahte ers) essenrurau. .
the anahsas wdl be gointed cut, The re:nnemit crewnied thus far is vs.mtacantly different from
- that used for cble tray ratines we now use.,To show this.conwder a 1.arye able tray ranJomly fdled mich,say 300 tightly pacted 600 voit cbles of assorted waet Accurdag to the ratings published so far,
- enry esble in this Irry must N Jerated to 30e of tfie amraegy for a J conduelor esble in air.U.oJ Figure 2 shows that neven sanew conductor fl2.cbles can verury about the same arta sin the tray Ja one t!O chte. Cumparmg the hest whieti is gener:ted wiihin the g#
4 equal stess of c bles it c:n be seen that three to (tar tunes more heJe
- , AT g is pruduced in the hndle of seven all cables.:.s in a ungle =J 0 ,
- c2bler, even though ISJce conferurnuoris occur.r the same 3res in
"* . \
the lilled tr21[hss etiret is etJetly what we want so cummate m a' CA81.E MASS
- MDie trzy snstallation Ncuse at is pusuble to get bundles of smag WifH UNtFCAM
- H EAT CENERATICN f c:bles whicts phiduer luestly mienw heat snurres and result m hot spots within the cble tray ercsheetsurt.
w 1T3 . J. 5.mpluicJ anetrtsral muJet for hear transfer Imne e sighair
.~ . _
turLrst echle and,v eunwmore wil1.nwer suble.
Beinre proceeding 37th the analysis, two additional conditions must he socctfied. De first condition is enbacs in any tny must be mstalled Jt a cunstant. or uniform. depth. This is to Trennt embles trem king hesprd on one side of 3 iny with a resulting vacant space en the other side.Tlne scepnd rundiuon il to 3hsun e. at litst. tr.at all the 6abkt m the trzy are power ethles which mill unsformly generste hast taruughout the tray. Thew conditivata allow thc. Jandom msuun of able tu be tressed as a homritnevus teet:ngular at:s1 mith umicrm tw:t generzeiurt.
The tasi now is in simrly Gnd the ItowaNe hest intenney 301 let trays contaminir eartable amounts of oble. Once we tinJ the hest fee. l. (e.o.scrrna thre fenen s midwak arguent. .hewli tv.t..!
mierrut), the hett which a:"Jn be gener3 ed by each mdr6 rvAlv trJr e:nduetor sq s esa le cieu!sted (rom ,
r s GPCuP y
/ O F 2 .g . ci)
' \ n
- U SLES
' \
/- g *sincttCABLE 'hete a
- nuinner u(conJueiurs m ecle i A = erou-wtiarul area of the n.eunJuctor chie.
/ Q8 allowable best per unit arcs gene sted in the tra l -
\
\ /
/
\ ' anJ.cicestse. ,
j, l
~ ~ ) .,,e '
g a l- R 62) l Dweci..e - .
= nere i
l.' matsmum allowable current for 3 Jondueter !
R
- 3.c. resistance of conductor
- ti the masm[um oyrstirig temreraturr of the insulation msttrtslin the c:ble any.
lh . fit.:sitwl st:r mmtversen ur is pn.at rubber uswLa.J able. l lleat genersted in sny tightly packed emble tray mus,t rass Vii Nmparamus can be fr:sJe oser and owf with the prewnt ' through two mede: 11 the cshle mass. and 21 the air immcJutel) !
.rr'und the iny. Siinv heat 11ams throush the media there is a ampaettin for cables in trays The result is that wuall eunJueter sier resultms temperature dror in' rach as shEOn in Fipre .t. 3T, cbks are allowed to "wrick" h rJer stan the th;r we eJbles 3 hen
- through the esbits and AT, thruugh the sir,
, they are all pla.,cJ an J enmmon raridum tr:y. Atrusity. all cables should N wncked umiermly by coming in the same crerstmg ,
To Jetermine the total amount ot* hest s%'s which .an be j
temrersture in the tray. * .h.arated by a enble tray in an ambient temrersture a sf t and rnamtein its hr; hest temrvrsrure at or below the atentini 8#*F'*
ANALYTICAL MODEL ature iT,1 ef the esble insutstion in the trzy. we rnwit limi
./
system temperature dror itT)in
%enewt esble amrJeinws c n N estabhshed with eJtevlations. d' t.T = Tm-Ta instead of an empirieJi Jrrreach. J better understandany et tr.c
?O
Calc. No. 95-ENG-01528E2 R00 PageEM of 3 3.
ATTACHMENT D .;
I ..* k. * . , . .;* AMPACITIES FOR CABLL3 IN RA.NDOh!LY FILLED TRAYS
' L 5toire *
. Southern Calisurius Edrsen Company j l.os Angrles.Californi . .n tasTRACT- ,. ,
PitoutDI DETNTION
- The allowable currens which as) be carried by a given condue The first of many variables to be ex:mineJ is the extent to tar see cable has hern thoroughly investipsmi in almod'cvery ., s hi.:h the ables in 'any tray are packcJ. lt a aprirent that cables in enneerrible.tyfe of bbk instausuon. One ares which has not had a irry" loose rrsngement are earniinity nnadrsed m air"which can snuch,stinzion 'up to nem itl the allouable currenjhich cd bc ;,, freely flow through the veant stber m a* trig At {he sipace,between errwd by rabies in emble trays or troustis. ~
This paper presents a : . cables is reduced. by rocking ables c!oser toertrer, free flow'of air '
compirted ;eneral method for niculatmg the ar5ps'citics of chles in % through she, pad is graduacy retrietcJ. Tsung this to tlw poirat cshle trays:lt has been derrved froen ric;nentary heat transfer theory ". wherr adiacent c:birs are touching each other en au sides the and smriy wrsTwJ with man) full +tsic tes'as.The method shoes th:. continuous free spect betweEn. cables becomes practicsny non.
currently published sinpacuies for small cables in highly fDied in>1 c.ustent and only small air podsts, rem:in betw een the c=bles.
roust be reduced but the large czble ampsetties cn be safely incrrased. N- . ,
Appt.ing this reasoning to br:t Ocw frum cbles in a c'b'id tray
- me ser that a loose packing as desir:Plc since air can naturally now around esch oble. The hest will ihen rime out ce the pack and be 7.,,, . ..t . INT RODL'CTION ,
i 4. . v.; . .. .. n.." . ....rn replaced with cuales air frorn alw hnliom. Men cNes he:u ne in the studies which base been m Je on the current err >ug tightly par 6'rd. there is no mar flow through the bursil6 and thus best abjliry of rie:trie pomvr cabks. the.niust sample cw of eme chw natural air now. in (s.t. the operating in air" h s heEn rap's'nd'ed to mult'tple cables in a conduit.I canot unty . 2)be forcarried out of heat to 11nw tir ofbundk ou's t h)'he tight bundle is >> hest multip'.t chin or rundu ts in siseked hanks.* and several rabin conducuen thruugh the conglomeratwn c( cable cenJuetors.ansula.
' pulkJ intu'stert ruerein.3 The results of thew studies arr incorpo- tion, anJ air pudets.
a r:trd te v:rious cairnts in both the A!D.irGA Fower Cabic Ampacith1 and the Nauonal Elrrtrte Cnde. Cabi ampacuies in randomly fukJ in>s must he based on the assurnpiirin that c: bits arr tight!) packed at.d that we canot depend The ampseities, or der: ting ,fs: tors, which in:n been orier. on heat bemg c:rned out of the bundle by air Dowing through it.
mined so far art fer c:bles whkh art m some form of an orderly W(thout quntson this ti;htly packed rendiumn does not tzist is arrangement: a further sirnplif.ction which h:s ben c:sily justified ever> esble trzy, but it dras nndumly ocevt eften enough that.for in the past is all the conductors ceasidered we r the - sairt). ca:h csbic tray mu>t be designed as thous.t. at ras going to bc 15crtunately, this samph% treatment cannot be justiGed when J ::chtly packed. It is not even necessary that the er.ure cable tr:y be
[t.onsiderm; g
amnaettirs of randomly arrangeJ chin in travs.I taphaly racked. since a packed width es only about thsre inches is suificrens to produes a het spot in an otheroe cr# in). !
A tvriest cable tny instal!suon wnkh ta found in the electne poerr generation and distributmn mdustry ran be naualired as a Warh the criterion af light eshte packm; rstabirshed n a then '
Nnch deer 24. inch wiJe metal truugh euntAmny anywirre from :0 required la determine hne the heat grneranon i. Jmribuerd m the !
tra) croe.=ecuon The naar>> czble sun posubh. both sangle and I in .500 randomly aresnped single ne mutusonductor power and control czeles ranging in site from a 1:Ary .to 730 Mril. Thn arrey mulu.ecaducuw. and r:ch carrpng a difterent current apparerttly of cables is uwaDy seevreJ along YsIcTo~il tray with sonw lin to makn il quier difficult to place alloople currents on such a prewn the enbles airesdy in the tray frem iMftmg if adJitional heterogeneous masture. Ilowever looking at the probirm from the cables should be pulled into the tray. Dunng constru: tion a. cables standrums that we do nne want anc hougos, se tne esbie tray.the are scruved in the ersy, group by greur, the) can become packrd problem enn he sorved.
together tight enough that air is unable to carrulate thenush the mass of esbies. With the physic =l ties and the normal eibrstmn which is liet spots in a thermal system are produerJ by locally intense present in mnsi plant's. rven tr.any of the iniusil) icev cable arrzvs heat sourers; thua. in ewry arts of the rable in) u t must flWnIABaBa j can he e: Fried to settle and thus become mere or less restneting to such conj;nons. In other words, the hrst renerated m every ares of a air flow, rable int reou,resion must % uniinrm. Tha v. the key to the entire frobypn.of ampacures for_ randoml) arrangris,3atars ut cmby, trays.
Sewal other enriables tend to complicite the determmation of ,and,jhr concrpt of umform,he'at, generanon ,cnnot.be..ovete ampaeines of obles in trays. Some of the more arparent ones are the emphasiard._
~
fullness of a tray. diversity of loading of cables in a tray, determirung the, locssion of the hottest spot over the tny crospercuott. and the Consider Figure i sienn; a hypothene.1 sher of ana frorn a amount of power cable twhich genentr> hestIin proportion to the 131 i:21. tightly packed cabk tray.The heat intensny withm ,cacb uni,t ,
amount"of control able !which pencretes neghgible heal)in a tray. .
area. expenard in watts /ft per square anch of'crou arctional'arrA All the' $bkridsb'lb Nn be, and are' accounted for in the m'rthod - maist be constant all the uay dnwn to the smallest' unit area inside desc-ibed b'ertirC* , , .
the tny. wnich is the smaDnt cable in the tray. We therefore place amparitin of cables, such as shnwn in Figure 1.in prepartion to the overall crosewetional area of the andmJual cabin, sneluding the Paner 70 TP M7.PWR recommended and approved by the lasula- conductor arid insulation, ted Cenductors Committee of the IUL Power Group for prnent,- -
tien at the IEET Summer Power Merung and E!!Vronierena.1.cn 9reles C.lif July 1217.1970, blanusenri submitted September If oc know the' att..eable hest inienwe) for a tiern cable *
. lwr:made available for pr2 nuns Aptd 28.1970. tra), we can unmedaately place ampusun on rvery cable in the tray M
i 4
Calc. No. 96-ENG-01528E2 Rev. 00 Page El of Et i ATTACHMENT E B61 MCC Bus Loading
Reference:
l Drawing 25203 30011 Sheet 39 Rev.16 (Ref. 3.1.39)
Drawing 25203 30011 Sheet 40 Rev. 21 (Ref. 3.1.38)
Drawing 25203 30011 Sheet 41 Rev.18 (Ref. 3.1.15)
EWR M296024 (Ref. 3.1.45) I Only the continuous loads specified on these drawings are considered. MOV loads are not continuous and do not contribute to heat loading of cable in tray. (Assumption 4.3) (Note the feeder cable load is determined in accordance with Assumption 4.2)
BREAKER RATING AMPS MULT. LOAD
. AF4 100HP 125.18 1.25 156.47 BF1 .25HP .31 1.00 .31 CFI 25HP 31.29 1.00 31.29
. HF1 45KVA 17.90* 1.0 17.90* I
. HF3 0.75HP 0.94 1.00 0.94 l
BR3 6.75KW 0 1.00 0 BR4 6.75KW 0 1.00 0 CR3 15KVA 18.04 1.00 18.04 CR4 15KVA 18.04 1.00 18.04 CR5 1SKVA 18.04 1.00 18.04 DR2 5HP 6.26 1.00 6.26 i DR4 15KVA 3.15' l.00 3.15*
ER2 4.5KW 6.37 1.00 6.37 ER3 0.75HP 0.94 1.00 0.94 l FR1 1.93KW 2.73 1.00 2.73 i GR1 5.8KW 8.21 1.00 8.21 0 GR4 15HP 18.78 1.00 18.78 HR4 40HP 50.07 1.00 50.07 l
- ER4B 30KVA 36.08 1.00 36.08 ER5 3HP 3.76 1.00 3.76 !
TOTAL 397.38
- See following page for details.
HP x 746 l FLA = [EFF = 0.88, PF = 0.85]
KW x 1000 AMPS =
E x PF x 5 KVA x 1000 EX4
._ . __ ~ . _ _ . - - . ._..__ _ _ _ . . - . _ _ _ _ _ . _ . _ . _ _ _ . _ _ _ . . . _ . _ _ _ . _ _ . _ . _ .
4 4
4-l Calc. No. 96-ENG-01528E2 Rev. 00 t Page E2. ofE2_-
] ATTACHMENT E i
i
- l. Note 1 - per Drawing 25203-30012 Sheet 12 (Ref. 3.1.41) the connected load is 2220
} watts.
- I= '2??0 = 3.15 amps [DR4]
J-480 x 4 x 0.85
$ Note 2 - The Boric Acid Tank Heaters are no longer required since the concentration of j boric acid has been reduced. The load for these heaters is thus 0 amps. [BR3, BR4]
{' Note 3 - UAC2 per Drawing 25203-30022 Sheet 12 Revision 23 (Ref. 3.1.40) has a
- connected load of12648 watts.
I= 12648 = 17.90 amps [HFl]
- 480 x 4 x 0.85 1 i
1 i
i i
1 4
- I I l 4
l l ;
l t
1 j
)
t o .. - _ , . _,
Cake. No. 96-ENG-01528E2 Rev.00 :
Page F( of FI ATTACHMENT F B62 MCC Bus Loading
Reference:
Drawing 25203 30011 Sheet 42 Rev. 21(Ref. 3.1.55)
Drawing 25203 30011 Sheet 43 Ref.17 (Ref. 3.1.66)
Only the continuous loads identified on the drawing (s) are considered. MOV loads are not continuous and do not contribute to heat loading of cable in tray. (Assumption 4.3)
(Note the feeder cable load is determined in accordance with Assumption 4.2)
BREAKER RATING AMPS MULT. LOAD DF4 30KVA 32* 1.00 32* )
EF2 UPS 81** 1.00 81 "
BR2 3HP 3.76 1.00 3.76 CR2 5HP 6.26 1.00 6.26 CR3 15KVA 18.04 1.00 18.04 l CR4 15KVA 18.04 1.00 18.04 CR5 15KVA 18.04 1.00 18.04 DR1 25HP 31.29 1.00 31.29 ,
DR2 7.5HP 9.39 1.00 9.39 j DR3 15KVA 18.04 1.00 18.04 l FR1 20HP 25.04 1.00 25.04 FR4 59KW 83.49 1.00 83.49 CR1 15HP 18.78 1.00 18.78 ER2B 40HP 50.07 1.25 62.59 FR2 20HP 25.04 1.00 25.04 TOTAL 450.8
- = 80% of breaker value used.
"= References 3.1.74 and 3.1.75 identify the battery charger and inverter loads as 31 and 50 amps respecitively.
FLA = 5 x E x EFF x PF (EFF = .88, PF = .85)
AMPS =
E x PF x 5 AMPS =
Ex4 r
I
Calc. No. 96-ENG-01528E2 REV. 00 ATTACHMENT G i l
Page G1 OF G4 INDEPENDENT REVIEWER EVALUATION (Enter "X" or "NA" to indicate applicability to review)
- 1. Are the commitments provided in the Safety Analysis Report (SAR) and the Design Inputs X documents correctly incorporated into the design documents?
- 2. Does the proposed design affect or modify a Unit Safety Technical Specification in any way? N/A If yes, will the initiation of a Technical Specification Change Request (TSCR) be required? (Re:
NGP 4.02/ NARC)
- 3. Will the implementation of the proposed design require the initiation of a Final Safety Analysis N/A Report (FSAR) change? (Re: NGP 4.03/ NARC)
- 4. Are assumptions necessary to perform the design activity adequately described and reasonable? X Are the assumptions identified for subsequent reverification when the detailed design actisities are complete?
- 5. Does the design meet the requirements of applicable codes, standards, and regulatory X requirements?
- 6. Has applicable construction and operating experience been considered? X
- 7. Have the design interface requirements been satisfied? X
- 8. Was an appropriate design method used? X
- 9. Are the specified parts, equipment and processes suitable for the required application and have X all conditions been considered?
- 10. Are accessibility and other design provisions adequate for performance of needed in-senice N/A
, inspections, maintenance, and repair? Are adequate maintenance features included and requirements satisfied? Is adequate accessibility provided for performance of the expected insenice inspection required during plant life?
- 11. Has the design properly considered radiation exposure to the public and Unit personnel? N/A (ALARA, Reg Guide 8.8, NGP 5.16)
- 12. Have adequate preoperational and subsequent periodic test requirements been appropriately N/A incorporated into the design?
- 13. Are appropriate Quality Assurance and ANS Safety Classification requirements specified such N/A as inspections to be performed, acceptance criteria, specialized training / skills needed, personnel qualifications, material procurement, and material handling requirements?
- 14. Are the applicable codes, standards, and regulatory requirements, including issue and addenda X properly identified?
- 15. Are the inputs correctly stated and incorporated into the design package and the output X reasonable compared to inputs?
NUC DCM FORM 4-1A Rev.03 Page1of4
CL!c. N3. 96-ENG-01528E2 REV. 00 ATTACHMENT G Page G2 OF G4 INDEPENDENT REVIEWER EVALUATION
- 16. Are the specified materials compatible with each other and with the design environmental N/A conditions to which the material will be exposed? (Nonmetallic materials require an evaluation of their suitability relative to temperature and radiation environments - Normal and Post-Accident.) Are non-metallic components or parts relied upon structurally, where there integrity under varied conditions is not assured by engineering analysis?
- 17. Have adequate maintenance features and requirements been specified? N/A
- 18. Are the acceptance criteria mcorporated in the design documcats sufficient to allow verification X that the design requirements have been satisfactorily accomplished? Are adequate preoperational and subsequent periodic test requirements appropriately specified?
- 19. Are adequate handling, storage, cleaning, and shipping requirements specified? N/A
- 20. Are adequate identification requirements specified? X
- 21. Are requirements for record preparation, review, approval, retention, etc., adequately specified? X
- 22. Is the design such that potentid crud traps are not built into radioactive fluid lines (long-radius N/A elbows, quantity of valves muumized, ball valves used to extent possible)?
- 23. Have new equipment tag numbers been identified? N/A
- 24. Are equipment failure effects on existing critical components addressed? N/A
- 25. Has the affect on seismic structures been addressed? N/A
- 26. Are all welds identified (allow for use of piping being removed)? N/A
- 27. Are installatica fire safeguards identified? N/A
- 28. Will the authorized inspector be contacted and a repair package submitted? N/A
- 29. Are necessary spare parts identified? N/A
- 30. Are any barriers (i.e., fire, CO2, halon, ventilation, water, flood, tornado, high energy line break, N/A and radiation) being altered or penetrated and are the barrier design basis requirements being met?
- 31. Have human factors requirements been considered (workplace design, accessibility, lighting and N/A noise, human computer interaction, man machine interface, labeling, layout, etc.)?
- 32. Is the design consistent with Station Spill Prevention Control and Countermeasure Plan (SPCC)? N/A
- 33. Would the design change increase the potential for flooding, reduce the capability to isolatt or N/A cope with local compartment ficgiina, or locate essential equipment where it would be suceptible to flooding?
NUC DCM FORM 4-I A Rev.03 Page 2 of 4
l Cole. No. 96-ENG-01528E2 REV. 00 A'ITACHMENT G Page G3 OF G4 INDEPENDENT REVIEWER EVALUATION
- 34. Have the failure modes and affects with adjacent high energy piping systems been considered for N/A pipe whip, jet impingement, and/or emironmental effects (Stress Ana'ysis Engineering should be contacted if a reviewis required).
- 35. Has consideration been given in the design process for power supply surge withstand capability N/A and muumizing the probability, affect a id/or generation of electromagnetic noise interference on the design modification or adjacent systems?
- 36. Has NPRDS component failure history been considered? N/A
- 37. Has the design change properly considered all operational aspects; i.e., has the design change X taken challenges to the operator into consideration?
- 38. Arc appropriate measures specified to prevent debris / foreign material from entering Unit N/A systems, and is the potential impact of debris / foreign material on relevant Unit systems assessed? Has the design considered the non-use of materials which could plug containment sumps (Re: NRC Bulletin 93-02)? l
- 39. Have all Quality Software requirements been considered? X
- 40. Have applicable vendor manuals been checked, to ensure designs incorporate appropriate N/A installation guidelines?
- 41. Has the use oflow Cobalt material been specified for systems which communicate with the N/A reactor coolant system (e.g., valve seats)?
- 42. Has the Design incorporated ALARA principles to reduce occupational exposure?(Re: NGP N/A 5.27/ Engineering Design Standard 37120)
- 43. Is Structural Integrity assured (including Fatigue and Corrosion)? N/A l
- 44. Have weep holes for drainage been installed in non-safety related components (lighting panels, N/A termination boxes, and motor connection boxes) which are located in harsh emironments? (Re:
ISEG Report E91 -006)?
- 45. Have all SBOQA(Station Blackout) requirements been considered? X
- 46. Has Relay Selection (electrical contact rating and life cycles) been analyzed, and documented? N/A
- 47. Have pressure locking and thermal binding of gate valves been considered? N/A
- 48. Have additions / deletions to the contamment inventories of Zinc, Aluminum and Steel been N/A recorded and the Contamment Inventory Tracking Program updated?
- 49. Are conclusions drawn in the 10 CFR 50.59 Evaluation fully supported by adequate discussion X in the text or the 10 CFR 50.59 Evaluation itself?
NUC DCM FORM 4-1 A Rev.03 Page 3 of 4
Calc. N2. 96-ENG-01528E2 REV. 00 t ATTACHMENT G Page G4 OF G4 INDEPENDENT REVIEWER EVALUATION I
- 50. Has the integrated design package considered appropriate supplemental resiews by other X engineering disciplines (seismic, electrical, etc.) and affected departments (Operations, Maintenance, etc.)?
- 51. Are drawings, sketches, calculations, r:ferences, etc. included in the design package X appropriately?
- 52. Are calculations included or referenced in the design package that requires revision, resiewed X and approved with appropriate changes?
- 53. Has the design considered the elimination of possible obstructions to ladders (OSHA 1910.27)? N/A
- 54. Has the design ensured that adjacent unit (s) interfaces were addressed? X 1 have completed my Independent Review of this package by utilizing the " Design Review" method. All concerns were addressed satisfactorily WLDh)( "
/2 ~l 3- 9G PreTarer(s) Signature (s) Date W
NUC DCM FORM 4-1 A Rev.03 Page 4 of 4