ML18107A332
| ML18107A332 | |
| Person / Time | |
|---|---|
| Site: | Salem  |
| Issue date: | 08/29/1996 |
| From: | Ed Miller MPR ASSOCIATES, INC. |
| To: | |
| Shared Package | |
| ML18107A331 | List: |
| References | |
| 108-173-EJM-01, 108-173-EJM-01-R00, 108-173-EJM-1, 108-173-EJM-1-R, NUDOCS 9906020087 | |
| Download: ML18107A332 (28) | |
Text
{{#Wiki_filter:ATTACHMENT I NLR-N990254 SUPPLEMENTAL INFORMATION GENERIC LETTER 95 "PRESSURE LOCKING AND THERMAL BINDING OF SAFETY RELATED POWER OPERA TED GATE VALVES" SALEM GENERATING STATION UNITS 1AND2 9906020087-990524 PDR ADOCK 05000272 p PDR
MPR Associates, Inc. 320 King Street Alexandria, VA 22314 CALCULATION TITLE PAGE Client Public Service Electric & Gas Company Page 1 of 5 plus attachments Project Task No. Salem MOV Evaluations 108-173 Title Opening Force Required to Overcome Pressure I .ockipg of Calculation No. Pressurizer Block Valves 1PR6, IPR7, 2PR6, and 2PR7 108-173-EJM-Ol Preparer I Date Checker I Date Reviewer I Date Rev. No. 1(/........ /(!// .
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Eric J. Miller
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320 King Street Alexandria, VA 22314 Calculation No. P~a"By /;/ / Check~d By Page of 108-173-EJM-01
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MPR Associates, Inc. 320 King Street Alexandria, VA 22314 Calculation No. Checked By ,. Page 108-173-EJM-01 3 of 5 V//,...,-n--...__; ( PURPOSE This calculation determines the stem force required to open the pressurizer block valves at Salem Units 1 and 2 under postulated pressure locking conditions. A bounding approach is used to calculate an opening force that will be capable of overcoming pressure locking and packing forces in any of the four valves considered. RESULTS Maximum required opening force is 12.484 lbf. The table below summarizes results with and without thermal binding loads, and with two different assumptions for bonnet pressure during closing. No bonnet cooling, Bonnet cooling, Bonnet Pressure =2235 psig Bonnet Pressure = 1457 psig Without Thermal Binding 9092 lbf 9139 lbf Forces With Thermal Binding 12437 lbf 12484 lbf Forces PRESSURE LOCKING SCENARIO Closing Conditions Nonnal operating conditions at the pressurizer block valves are 2235 psig and 653F (Ref 4a). The valves are assumed to be closed under these conditions. For calculating the wedging interference, these conditions are applied upstream, downstream, and in the valve bonnets. It is assumed that the valves are closed with the thrust recommended in Reference 3. Consistent with the bounding approach of this calculation, the largest thrust recommended in Reference 3 for the four v_alves is used in this calculation.
- Opening Conditions Should a steam generator tube leak occur during loss of offsite power conditions, the operator may need to open the block valves to reduce primary side pressure. Reference 4b provides a plot of primary side pressure under such a scenario. The minimum RCS pressure before the PORV's are assumed to be opened is shown as 1457 psig. It is assumed that the block valves are opened when the RCS pressure has reached this minimum value.
Two scenarios were evaluated for bonnet pressure during opening. One scenario assumed the bonnet pressure remained at full normal operating pressure as the RCS was depressurized. The second scenario assumed that the bonnet cooled during the transient, reducing the pressure in the bonnet.
MPR Associates, Inc. 320 King Street Alexandria, VA 22314 Calculation No. 108-173-EJM-Ol CALCULATION Description of Method - The force due to pressure locking of the valve disk is calculated using.a model developed by MPR Associates (Ref 6) and implemented using Mathcad 5.0. The model is based on the basic geometry of a flexible valve disk. It uses fundamental relations between the disk and body forces and deflections during valve closure to calculate the amount of wedging interference that occurs when the valve is wedged shut. The calculated wedging interference is used as a starting point for applying the force and deflection relationships necessary to determing the unwedging force. The model considers differences in pressure and temperature when the valve is closed and when it is to be opened. The basis for the inputs to the Mathcad model are documented in an Excel spreadsheet. A printout from this spreadsheet is attached to this calculation. A printout of the Mathcad calculation of the disk unwedging force is also also attached to this calculation~ *Note that four cases were run. These cases calculate required stem force for the two different bonnet pressure assumptions and with and without thermal binding loads. Thermal binding loads are calculated using the equations in Reference 10. The total opening force is calculated using the unwedging force calculated from the Mathcad model, the packing force, and the stem rejection force. Results of these calculations are at the end of each Mathcad printout.
MPR Associates, Inc. mMPR 320 King Street Alexandria, VA 22314 Calculation No. / / " ./hecked By Page of 108-173-EJM-01 5 5
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REFERENCES
- 1. Velan Calculation DC-055, "Required Thrust/Weak Link Calculations," for VeJan PO P9-77592-K, faxed to MPR on 8/2611996.
- 2. MPR Calculation 108-085-CSK-Ol, "Calculation of Required Opening Thrust for Selected Gate Valves after Bonnet Pressurization," Rev. 1.
- 3. MPR Report 1693, "Evaluation of Salem Valves for Pressure Locking and Thermal Binding,"
Rev. 1, J~nuary 1996.
- 4. PSE&G Fax from Bob Lewis to Paul Damerell, August 23, 1996, containing:
a) PSE&G Memo from R. Lewis to C. Gassell, on "NRC GL 95-07, RAI dated July l, 1996," memo is dated July 29, 1996. b) Background Information for Westinghouse Owners Group Emergency Response Guideline E~ 3, ..Steam Generator Tube Rupture," HP-Rev 1, September l, 1983.
- 5. ASME Boiler and Pressure Vessel Code, Section ill, Appendix I, 1983 Edition.
- 6. MPR Calculation 002-156-DHH-816, "Method to Predict Opening Thrust," for "Gate Valve Opening Thrust," Rev. 0.
- 7. ASME Steam Tables
- 8. MPR Report 1409, "Algorithms for Estimating Friction Coefficients at Sliding Contacts in a Gate Valve, Rev. 2, September 1994.
- 9. PSE&G Document NC.DE-PS.ZZ-0033(Q)-A8, Rev 4 and 5, Section 3.1.8, "Assumption Verification and Setpoint Determination." Excerpts attached to this calculation.
- 10. MPR Calculation 108-085-TW 1, "Screening of Salem Valves for Thermal Binding Susceptiblity," Rev. 0.
MPR Associates, Inc.
~MPR 320 King Street Alexandria, VA 22314 Calculation No. P~'9d By hecked By I . AJ
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Inputs for Prsssure Lock Calcu/aUon v Sheet 1 of 2 Problem ID PR7 ~ Valves Covered 1PAS, 1PR7, 2PR6, 2PR7 1. References are listed in calculation text. Manufacturer Velan 2. Shaded blocks contain input numbers. Size 3in Un-shaded numbers are calculated from inouts. Pressure Class 1500 lb Mathcad Variable Value Unfts Name Reference Plmenslons Mean Disk Thickness 0.761 in t Ref 1 "Plate Thickness" Seat OD 2.750 In Ref t Dimension D Seat ID 2.375 In Ref t Dimension d Mean Disk Seating Radius 1.281 in a =(seaUD+seat_OD)1212 Disk Hub Radius I *o.630jin Ref t Dimension b Disk Total Width at Center t.959 in dw =2x(h+t) Body Wall Thickness 0.761 in tb Assumed same as t Half Disk Wedge Angle I
- sldeg theta Ref 1 phi Reference Temperature 70 F Tr Assumed
~JQ1Ing 121:1~ E1:trc1t Closing Stem Force 10966 lbf F.stem Recommended closing force in MPR* 1693, Rev 1 (Rel 3).
.. Votes data from Ref 9. The maximum baseline static SBLJMIST force recorded for the four valves being Packing Load, Max 38Z7 lbf F.pack considered. PSS force used.
Votes data from Ref 9. The minimum baseline static SBLJMIST force recorded for the four valves being Packing Load, Min 803 lbf considered. l:::lul2 L~c!llll ~al!::ulill112a:1 I *~:959,in Total bottom width on 4" Velan Valve Ref 2, Appendix A, Note 4 for SJt 2113 Bottom width lor Disk thickness, 3" Valve 1.522 In =2xt Hub Length at bottom, 3" Valve 0.437 in 2h =bottom_width_4"valve
- bottom_thickness_3"valve Hub Length from Center, 3" Valve 0.219 In h =hub_length_at_bottom 12 Materials Disk A*351*CF8 or A182*F316 Ref 3, Appendix A Seat Ring A*351-CF8 Ref 3, Appendix A Body A182-F316 Ref 3, Appendix A Seat Ring Face Stelllte 6 Rel 2, Appendix A, Assumed same as valves SJ12/13 Disk Face Stellite 6 - Ref 2, Appendix A, Assumed same as valves SJ12113
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MPR Associates, Inc. mMPR 320 King Streat Alexandria, VA 22314 Calculation No. By
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(/ lnpw for Pressure Lock Cslculstlon Sheet2 of 2 Mathcad Variable Value Units Name Reference Closing Condjt!ona Upstream Pressure I 2235lpsig Puc Pbc Normal operaUng conditions, Ref 4a Assumed-same-as Puc Bonnet Pressure 2235 pslg Downstream Pressure 2235 psig Pde Assum~ same as Puc Disk Temperature Body Temperature I 6s3IF 653 F Tdc Tbc Nonna! operating conditions, Ref 4a Assumed same as Tdc Disk Elastic Modulus 2.50E+07 psi Ede Interpolated from data in Ref 5 forTdc Disk Poisson's RaUo 0.3 udc Assumed Disk Coeff. of Expansion 9.61E-06 in/in/F adc Ref 5forTdc Body Elastic Modulus 2.50E+07 psi Ebe Interpolated from data In Rel 5 for Tbc Body Poisson's Ratio 0.3 \lbc Assumed Body Coeff. of Expansion 9.61E*06 in/in/F abc Ref 5forTbc Opening Conditions Upstream Pressure I 1457!psig Puo Scaled from Rel 6, lowest pressure before PORV is opened Bonnet Pressure 1457 pslg Pbo Assumed equal to upstream pressure Downstream Pressure o psig Pdo Assumed Disk Temperature 593 F Tdo Saturation temperature at Puo, Ref 7 Body Temperature 593 F TbO Saturation temperature at Puo, Ref 7 Disk Elastic Modulus 2.53E+07 psi Edo Interpolated from data in Ref 5 forTdc Disk Poisson's Ratio 0.3 udo Assumed Disk Coeff. of Expansion 9.53E*06 in/in/F ado Ref 5forTdo Body Elastic Modulus 2.53E+07 psi Ebo Interpolated from data in Ref 5 for Tbo Body Poisson's Ratio 0.3 ubo Assumed Body Coeff. of Expansion 9.53E-06 in/in/F abo Ref 5 for Tho Area ca1culatlons Seat OD 2.750jin Ref 1 Dimension D Seat ID 2.375_in Ref 1 Dimension d Seat Mean Diameter 2.563 In =(Seat_OD+ Seat_ID)/2 Seat Area 5.157 ln"2 :::SeaLmean_diam x pi /4 Seat Ring Face Area 1.509 in112 =(Seat_OD"2 - Seat_1D112rpi/4 Clos)ng Condjtlons and Erl'1lon Factor Closing DP ! 2235lpsig Set equal to bonnet pressure Closing Contac1 Stress 8 ksl =DP*Seat_area I Seat_ring_face_area Closing Temperature I 653lF Normal operating condiUons, Rel 4a Closing Mu 0.400 µc Ref 8, Stellite-on-Stellite, Flat-on-Flat In Water, 10ksl contact stress, max mu at 650F Opening CondHlons and Frl~tlon Factor Opening DP l 1457]pslg Set equal to bonnet pressure Opening Contact Stress 5 ksl =DP"SeaLarea I Seat_rlng_face_area Opening Temperature I 593IF Assur:ned equal to disk temperature at openinQ, Tdo Opening Mu 0.44 µc contact stress, interpolated using max mu's at 550F and 650F I
MPR Associates, Inc., e Sheet1 Checked by:- -'-/ -'"~-~_:. . :1,~ ;.:. -:.: . . - -.=- *=- - EVALUATION OF GATE VALVE OPENING THRUST PURPOSE: This calculation evaluates the equations for disk opening force derived in calculation 002-156-DHH-816. VALVES EVALUATED: Salem Pressurizer Block Valves 1 PR7, 1 PR6, 2PR7, 2PR6 pjmenslon:t Mni~cl"I ~[QR~i:ll~:i Closing Opening Disk Thickness: t :=0.76Hn 6 6 Disk Elastic Modulus.: Ede :=25.0*10 *psi Edo := 25.3* 10 *psi Disk Seating Radius: a:= l.281-in Disk Poisson's Ratio: vdc :=0.3 vdo :=0.3 Disk Hub Radius: b := 0.630-in -6 -I -6 *1 Disk Coet of Expan.: adc:=9.61*10 *deg ado :=9.53* 10 *deg Hub Length from Center: h := 0.219*in 6 6 Body Elastic Modulus: Ebe :=25.0*10 *psi Ebo := 25.3* 10 *psi Disk Total Width at Center. dw := l.959*in Body Poisson's Ratio: vbc :=0.3 vbo :=0.3 Body Wall Thickness: tb := 0.761*in -6 -I -6 -I Body Coef. of Expan.: abc := 9.61-10 *deg abo := 9.53* 10 *deg Half Disk Wedge Angle: e :=5*deg Stem Diameter: D := 1.125*in Stroke L :=2.688*in Reference Temperature: Tr:=70*deg Ambient Temperature: T amb := 120*deg mdttlons Closing Opening
- upstream Pressure: Puc := 2235* psi Puo := 1457*psi Bonnet Pressure: Pbc :=2235*psi Pbo := 1457*psi Downstream Pressure: Pde :=2235*psi Pdo :=O*psi Disk Temperature: Tdc :=653*deg Tdo :=593*deg Body Temperature: The := 653*deg Tho :=593 *deg Friction Coefficient µc := 0.40 µo :=0.44 Closing Forces Closing Stem Force: F stem:= 10966*1bf Packing Force: F pack:= 3827*lbf Fth.stem :=3*~*L*(Tbc-T amb)
Thermal Binding, Stem Expansion after Closing: m*dcg F th.stem =4298 *!bf lbf Thermal Binding, bonnet F lh b
- onn
- = 2.25*--*L*(Thc- Tho) in*deg F th.bonn =363 *lbf cooling:
* :=Pbc*D -~
2 Stem Rejection Force: F reJ.c 4 F rcj.c =2222*lbf
".:losing Disk Force: Fe := F stem+ F th.stem+ F th.bonn - F pack- F rej.c Fe =9578*1bf Mathcatt 5.0 - EJM01.MCD 5:01:27 PM on 8129196
MPR Associates, Inc., &20 King S~reet,Al xandria, VA 22314 e I; I Sheet 2. Calculation No. 108-173-EJM-01 Prepared by:
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uc.IERMINATION OF CLOStNG FORCES 2 2 Pbc*1t*b
-n*a -(Puc - Pbc) - - - -
cos(9)
-2.797*103 0-lbf 0
2 Pbc*x*b *tan(6) 243.815 Pc:= O*lbf Pc= a *!bf 2 Pbc*x*b *tan(0) . 243.815 2 -2.797*103 l 2 Pbc*x*b
- n*a *(Pde- Pbc) - - - - 3 eos(0) 9.578*10 Fe
-1 0 0 0 0 0
-µe*sin(9) cos(0) 0 0 0 0 -I
µc*cos(9) sin(9) -1 0 0 0 0 Mc:= 0 0 0 µc*sin(9) -cos(9) 0 Solving for the force vector, Sc, 0 0 0 µc*cos(9) sin{9) -1 0 1 Sc:=Mc- *Pc 0 0 0 1 -1 0 0 0 0 1 0 0 0 The forees are:
3 9.862*10 Sue :=Se0 Sue = 9862 *lbf 4 1.266-10 Hue :=Sc 1 Hue = 12659 *lbf 4.789-1<>3 Puc :=Sc2 Fue =4789*lbf 3 *lbf Sc= 9.862*10 Sdc :=Sc3 Sdc = 9862 *lbf 4 1.266-10 Hdc :=Sc4 Hdc = 12659*lbf 3 4.789* 10 Fdc :=Sc5 Fdc =4789*Jbf 4 l.227*10 He :=Sc6 He= 12267 *!bf 5:01:44 PM on 8129196 Malhcad 5.0 - EJM01.MCD
MPR Associates, Inc., lexandria, VA 22314 e /} I . Sheet 3 Checked by: d1~c.1/---
.*:"TERMINATION OE WEDGING INTERFERENCE Flexibilities for Closing 6 6 4 2 2 4 a *(7*m+ 3) + b -(m- I) - a *b *(m+ 7)- a *b *(7*m- 5) ...
1)].1n(~)- 1 2 m:=- 2 2 2 2 4 2 2 ) +-4*a *b *[a *(5*m- 1) + b *(m+ 16-a -b *(m+ 1 )*ln(.!) vdc 3* ( m - I b b
~:- . ~-~~~~~-------'--'-------~-..!......!-
2 2 16*m2 a -Cm-t-l)+b -(m-l) 3*(m 2
- t) .
a4-(3*m+ 1 )- b4*(m- 1)- 2*a2*b2*(m+ 1)- 8*m*a2*b2*In (a)b - 4*a2*b2*(m+ 1)-ln (a)b 2
'I':=
2 2 2 4*m *7t a *(m+l)+b *(m- I) h II>> 'P Ghc:=--- Gpc:=-- Gsc:=-- Hub
.,..b2*Edc Edc*t3 Edc*t3 Disk, Pressure Gpc =2.446-10-s *in* psf 1 dw 2 Ksc:=------ dw Gsc =1.373*10-s *in*lb(
1 4*n*(a -r tb)*tb*Ebc Kpc:=--- Disk, Seat Force 4*tb*Ebc 1 Body, Seat Forces Ksc = 4.013* 10-9 *in*lbf 1 Body, Bonnet Pressure: Kpc = 5.043* 10-8 *in*psC Wedging lnterterence at Reference Conditions 1r :=2*Ghc*Hc+ (Gsc + Ksc)*(Suc + Sdc) + Gpc*(Puc+ Pde- 2*Pbc)+ 2*Kpc*Pbc ...
-4
+-dw*adc*(Tdc- Tr)+ dw*abc*(Tbc-Tr) wr =7.47758*10 *in DETERMINATION OF OPENING FORCES Flexibilities fur Opening 6 6 4 2 2 4 a *(7*m+ 3) + b *(m- 1 )- a *b *(m+ 7)- a *b *(7*m- 5) ...
m:=- I 2 vdo 2[ 2 J*(m2 _ 1) +-4*a2*b *a *(5*m- l)+b '(m+ 1) *In 2 J (a) b -16*a4*b2-(m+ l)*ln (a) b
~:= . ~---~-~----~---'--'---~-----~-
2 2 16*m2 a *(m-r l) + b *(m- 1) 4 4 2 2 2 2 (a) 2 2 (a) 2.
'1':= *(m2 - I). a *(3*m+ 1)-b -(m-1)-2*a *b *(m+ 1)- B*m*a *b *ln b -4*~ *b *(m+ l)*ln b 3
4*m 2
*1t
[ 2 2 a *(m+ 1) t- b -{m- 1) h qi 'I' 1 Gho:=--- Gpo:=-- Gso:=-- Hub Oho= 6.942* 10-9 *in*lbf n:*b2*Edo Edo*t3 Edo*t3 Disk, Pressure Gpo =2.417*10--a *in*psf 1 dw dw 2 Disk, Seat Force Gso = 1.357* 10-g *in* lbf 1 Kso : = - - - - - - Kpo:=-- 4*11:*(a-r tb)*tb*Ebo 4*tb*Ebo -I Body, Seat Forces Kso =3.965" 10-9 *in* !bf Body, Bonnet Pressure: Kpo =4.983*10
-a *in*psi-1 Wedging lntederence at Openjng Conditions wo := wr + dw*ado*(Tdo- Tr) - dw*abo*(Tho- Tr) WO =7.47758*10-4 *in Define: W := wo- Gpo-(Puo+ Pdo- 2*Pbo) - 2*Kpo*Pbo W =6.37759*10-4 *in Mathcad 5.0
- EJMD1.MCO 5:02:55 PM on 8129196
MPR Associates, Inc., e 320 King S~ret, :le andria, VA 22314 e /Jh I *d
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rr-~ Calculation No.108-173-EJM-01 2 2 Pbo*1t*b
-Jt*a *(Puo- Pbo)-
cos(9)
-1.824* 103 O*lbf 0
2 Pbo*Jt*b *tan(B) 158.943 O*lbf 0 Po:=: 2 Pbo*it*b *tan(B) Po= *lbf 158.943 2 2 Pbo*n*b 5.688*103
-Jt*a *(Pdo- Pbo) -
cos(B) 0 0-lbf 3.637* 104 w Gso+ Kso
-1 0 0 0 0 0 0
µo*sin(9) cos(B) 0 0 0 0 -I 0
-µo*cos(0) sin( 9) 0 0 0 0 0 0 0 0 -µo*sin(9) -cos(9) 0 0 Mo:= 0 0 0 - µo-cos( B) sin( 8) 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 -1 2*Gho*cos(B) 0 0 0 0 0 Gso+ Kso The forces are:
Solving for the opening forces, So: 9.839-103 1 So :=Mo- *Po 1.16&10 4 Suo := So 0 Suo =9839 *!bf 3.455* la3 Huo :=So 1 Huo =11663 *lbf Fuo :=So2 Fuo =3455 *!bf 1.707* 104 So= *!bf Sdo :=So 3 Sdo =17072*Ibf 4 1.138*10 Hdo:=So4 Hdo = 11384 *lbf 6.65*103 Fdo :=So5 Fdo = 6650 *!bf 4 1.2* 10 Ho :=So 6 Ho= 11996*lbf 4 1.01*10 Fo := So7 Fo =10105 *!bf Opening Forces Calculated Disk Opening Force Fo =10105*lbf Pbo = 1457 *psi F th.stem =4298 *!bf 3 Packing load F pack= 3.827* 10 *lbf F th.bonn c 363 *!bf
*4:n:
Stem rejection force 2 F rej.o :=Pbo-D F rej.o =1448*1bf Stem opening Force F open:= Fo + F pack- F rej.o F open= 12484*lbf Mathcad 5.0 - EJM01.MCD 5:03: 18 PM on 8129198
MPR Associates, Inc., Calculation No.108-173-EJM-01
. 0 King Prepared S~,.,t xandrla, VA 22314 by:_---<~_..c..--0....._____
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, Shee\1 EVALUATION OF GATE VALVE OPENING THRUST PURPOSE: This calculation evaluates the equations for disk opening force derived in calculation 002-156-DHH-816.
VALVES EVALUATED: Salem Pressurizer Block Valves 1 PR7, 1 PR6, 2PR7, 2PR6 Djmenslons Mab:rlDI Ero12etll~!i Closing Opening 6 6 Disk Thickness: t :=0.76l*in Disk Elastic Modulus.: Ede :=25.0* 10 -psi Edo := 25.3* 10 *psi Disk Seating Radius: a:= 1.281-in Disk Poisson's Ratio: vdc := 0.3 vdo :=0.3
-6 -1 -6 -I Disk Hub Radius: b :=0.63CHn Disk Coef. of Expan.: adc := 9.61*10 **deg ado:=9.53*10 *deg 6 6 Hub Length from Center: h :=0.219*in Body Elastic Modulus: Ebe:= 25.0* 10 -psi Ebo :=25.3*10 -psi Disk Total Width at Center: dw := 1.959*in Body Poisson's Ratio: vbc :=0.3 vbo :==0.3
-6 -I -6 -I Body Wall Thickness: tb :=0.761*in Body Coef. of Expan.: abe := 9.61*10 *deg . abo := 9.53* 10 *deg Half Disk Wedge Angle: e :=5*deg Stem Diameter: D := 1.125*in Stroke L := 2.688*in Reference Temperature: Tr :=70*deg Ambient Temperature: T amb := 120*dcg 1dltlons Closing Opening Upstream Pressure: Puc := 2235*psi Puo := 1457*psi Bonnet Pressure: Pbc := 2235*psi Pbo :=2235*psi Downstream Pressure: Pde := 2235*psi Pdo :=O*psi Disk Temperature: Tdc :=653*deg Tdo :=593*deg Body Temperature: The:= 653*deg Tho := 593-deg Friction Coefficient: µc := 0.40 µa :=0.44 qoslryg Forces Closing Stem Force: F stem : =10966* lbf Packing Force: F pack:= 3827*lbf Thermal Binding, Stem Expansion after Closing:
Fth.stem*:=3*~*L*(Tbc-T m*dcg amb) F th.stem =4298*lbf Thermal Binding, bonnet Fth b lbf
- =2.25*--*L-(Tbc-Tbo) F th.bona =363 *lbf
- onn in*deg cooling:
2 Stem Rejection Force: * := Pbc*D F reJ.C *!: F rej.c =2222*lbf 4 losing Disk Force: Fe:= F stem+ F th.stem+ F th.bonn - F pack- F rej.c Fe = 9578 *lbf 5:03:52 PM on 8129198 Mathcad S.D - EJM01.MCD
MPR Associates, Inc., .320Kin~ndria,VA 22314 * / Sheet2
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Calculation No. 108-173-EJM-01 Prepared by: Checked by:/.'-** / .\*tV/v~ I ...£fERMINATION OF CLOSING FORCES 2 2 Pbc*7t*b
-it*a *(Puc - Pbc)- - - -
cos( B) 3
-2.797*10 O*lbf 0
2 Pbc*it*b *tan( e) 243.815 Pc := O*lbf Pc= 0 *!bf 2 Pbc*7t*b *tan(9) 243.815 2 Pbc*7t*h 2 -2.797* 103
-7t*a *(Pde- Pbc)- - - - 3 cos( 0) 9.578* 10 Fe
-1 0 0 0 0 0
- µc*sin(B) cos(9) 0 0 0 0 -1
µc*cos( 9) sin(B) - 1 0 0 0 0 Mc:= 0 0 0 µc*sin(S) -cos(B) 0 1 Solvil)g for the force vector, Sc, 0 0 0 µc.-cos(9) sin(B) -1 0 1 Sc :=Mc- *Pc 0 0 0 l -I 0 0 0 0 0 0 0 The forces are:
9.862* Ht Sue :=Sc0 Sue =9862 *lbf 1.266-104 Hue :=Sc1 Hue = 12659 *lbf 4.789-Hr Fuc :=Sc2 Fuc =4789 *lbf 3 *Ibf Sc= 9.862*10 Sdc :=Sc3 Sdc =9862*lbf 4 1.266-10 Hdc :=Sc4 Hdc = 12659 *lbf 3 4.789" 10 Fdc :=Sc5 Fdc =4789*Ibf 4 1.227* 10 He :=Se6 He= 12267 *lbf 5:04:11 PM on M!9/96 Malhcad 5.0
- EJM01.MCD
MPA Associates, Inc., e 320 King Ale ndria, VA 22314 a Str~et, Sheet 3 Calculatlon No.108-173-EJM-01 Prepared by: ~
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Checked by: If& (J./v';,,,_,____
.:TERMINATION Of WEDGING INTERFERENCE flexibilities for Closing 6 6 4 2 2 4 a *(7*m+ 3) + b *(m- 1)- a *b *{m+ 7)- a *b *(7*m- 5) ...
2 l)+b 2 *(m+l)J.1n(~)-16*a4 *b2-(m+l)*ln(~) 1 2 2 m:=- ) +-4*a2*b *[a *(5*m-2 vdc ell := 3* ( m - I . _ _ _ _ _ _ _ _ _ _ _ _ ___,_h_,__ _ _ _ _ _ _..:...b~- 2 2 16*m2 a -(m+ l)+b *(m-1) a4-(3*m+ 1)-b4*(m-1)- 2*a2 *b2-(m+ 1)-S*m*a2*b2 *In (a)b -4*a *b -(m+ l)*ln (a)b 2 2 2 2 2 a -(m+ t)+b *(m-1) h ell '¥ Ghc:=--- Gpc *-
..- - - 3 Gsc:=-- Hub Ghc =7 .025* l 0-g *in* lb(
1 x*b2*Edc Edc*t Edc*t3 1 Disk, Pressure Gpc =2.446-101 *in*psf dw 2 Ksc :- _ . dw 1 4*:c-(a + tb)*tb*Ebc Kpc:---- Disk, Seat force Gsc=l.373*101 *in*lb( 4*tb*Ebc 1 Body, Seat forces Ksc =4.013*10-g *in*lbf 1 Body, Bonnet Pressure: Kpc =5.043*101 *in*psf Wedging Interference at Reference Conditions
*1r := 2*Ghc*Hc + (Gsc+ Ksc)-(Suc + Sdc) + Gpc-(Puc +Pde- 2*Pbc) + 2*Kpc*Pbc ...
-4
+-dw*adc-(Tdc- Tr)+ dw*abc*(Thc- Tr) wr =7.47758*10 *in QEIERMINATION OF OPENING EQRCES flexibilities for Opening a6*(7*m+3)+b 6-{m- 1)- a4 *b 2*(m+7)-a2*b" -{7*m- 5) ...
- m. *- 2 vdo 222 2
+-4*a *b *[a *(S*m- 1) + b *(m+ l)].In ba - 42 t6*a *b *(m+ 1)-ln ba 3*(m2 - 1) 4> :=-'---""""'
( ) ( ) 2 2 2 16*m a *(m+t)+b *(m- l)
'l':= *(m2 - t). 4 4 2 2 a *(3*m+ 1)- b -(m- l)-2*a *b -(m+ 1)- 8*m*a *b *In 2 2 (a)b - 4*a *b *(m+ l)*ln (a)b 2 2 2
3 2 [ 2 2 4*m *it a *(m + 1) + b *(m- 1) h ell 'l' 1 Gho:=--- Gpo:=-- Gso:=-- Hub Gho = 6.942* I 0....., *in* lbf
- n:*b2*Edo Edo*t3 Ed~*t3 Disk, Pressure Gpo =2.417* 10-8 *in*ps( 1 1
dw dw 2 Disk, Seat Force Gso = 1.357* l0-8 *in*lbf Kso :=--~--- Kpo:=--- 4*:n:*(a+ tb)*tb*Ebo 4*tb*Ebo 1 Body, Seat Forces Kso =3.965*10....., *in*lbf 1 1 Body, Bonnet Pressure: Kpo = 4.983* 10 *in*psf Wedging Interference at Opening Conditions
-4 wo :=wr+ dw*ado*(Tdo-Tr)- dw*abo*(Tbo-Tr) WO =7.47758*10 *in
-4 Define: W :=wo- Opo-(Puo + Pdo- 2*Pbo)- 2*Kpo*Pbo W = 5.97822* 10 *in Malhcad 5.0 - EJM01.MCD 5:05:21 PM on &'29196
MP'R Associates, Inc., eX'andria, VA 22314 e I i ....v I . Sheet4 Checked by: /! !.:. :C-v:vv.,.,_......,.....,.......__. 2
. 2 Pbo-n:*b
-n:*a *(Puo- Pbo)-
cos(O) 1.213* 103 O*lbf 0 2 Pbo*n*b *tan( 8) 243.815 O*lbf 0 Po:= 2 Po= *!bf Pbo*n*b *tan(O) 243.815 1 3 Pbo*n:*b 8.724*10
-n*a2*(Pdo- Pbo)-
cos(0) 0 l l 4 O*lbf 3.41* 10 _ w Gso+ Kso
-1 0 0 0 0 0 0
µo*sin(9) cos(0) 0 0 0 0 -1 0
- µo*cos(0) sin(B) 0 0 0 0 0 0 0 0 - µo*sin(0) - cos(0) 0 0 Mo:= 0 0 0 -µo*cos(0) sin(B) 1 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 -1 2*Gho*cos(0) 0 0 0 0 0 Gso+ Kso The forces are:
Solving for the opening forces, So: 9.878*1a3
-1 Suo :=So 0 Suo = 9878 *!bf So :=Mo *Po 8.664*10 3
Huo :=So 1 Huo =8664*lbf 3.818* lct Fuo :=So2 Fuo =3818*lbf 4 1.711* 10 So= *)bf Sdo :=So3 Sdo = 171lO*lbf 3 8.386-10 Hdo := So4 Hdo = 8386 *!bf 7.013*103 Fdo :=So5 Fdo =7013*1bf 3 9.01*10 Ho :=So6 Ho =9010*lbf 1.083* 104 Fo := So7 Fo = 10831 *Ibf Opening Forces Calculated Disk Opening Force Fo =10831 *lbf Pbo =2235 *psi F th.stem =4298*1bf 3 Packing load F i>ack =3.827* 10
- lbf F th.bonn =:: 363 *lbf Stem rejection force 2 7t F rej.o :=Pbo-D *4 F rej.o =2222 *!bf Stem opening Force F open :=Fo+ F pack- Frej.o F open =12437 *lbf Mathcad 5.0 - EJM01.MCD 5:0S:44 PM on 8129196
MPR Associates, Inc., Calculation No. 108*173-EJM-01 EVALUATION OF GATE VALVE OPENING THRUST PURPOSE: This calculation evaluates the equations for disk opening force derived in calculation 002-156-DHH-816. VALVES EVALUATED: Salem Pressurizer Block Valves 1 PR7, 1 PR6, 2PR7, 2PR6 Dimensions M~teria1 er212~rtl~~ Closing Opening Disk Thickness: t :=0.761-in 6 6 Disk Elastic Modulus.: Ede := 25.0* 10 *psi Edo :=25.3*10 *psi Disk Seating Radius: a:= 1.281-in Disk Poisson's Ratio: vdc :=0.3 vdo :=0.3 Disk Hub Radius: b :=0.630*in -6 -I -6 -I Disk Coef. of Expan.: adc :=9.61*10 *deg ado:= 9.53* 10 *deg Hub Length from Center: h :=0.219*in 6 6 Body Elastic Modulus: Ebe := 25.0* 10 *psi Ebo := 25.3* 10 *psi Disk Total Width at Center: dw := 1.959*in Body Poisson's Ratio: vbc :=0.3 vba :=0.3 Body Wall Thickness: tb :=0.76J.in -6 -1 -6 -I Body Coef. of Expan.: abe:=9.61*10 *deg abo :=9.53*10 *deg Half Disk Wedge Angle: 0 :=5*deg Stem Diameter: D := l.125*in Stroke L :=2.688*in Reference Temperature: Tr :=70*deg Ambient Temperature: T amb := 120*deg
- onditlons Closing Opening Upstream Pressure: Puc :=2235*psi Puo := 1457*psi Bonnet Pressure: Pbc :=2235*psi Pbo :=2235* psi Downstream Pressure: Pde := 2235*psi Pdo :=O*psi Disk Temperature: Tde := 653*deg Tdo := 593*dcg Body Temperature: Tbc :=653*deg Tba :=593*deg Friction Coefficient: µc := 0.40 µo :=0.44 Closing Forces Closing Stem Force: F stem : =10966* lbf Packing Force: F pack :=3827*lbf Thermal Binding, Stem lbf F th.stem :=3*-.-*L*(Tbc- T amb) F th.stem =4298 *!bf Expansion after Closing: m*deg F fu.stem := O*lbf
!bf Thermal Binding, bonnet Fth.bonn :=2.25*-.-*L*(Tbc-Tbo) F th.bonn = 363 *lbf cooling: m*deg F th.boJ'!n :=O*lbf 2
Stem Rejection Force: F reJ.c
* :=Pbc*D *.!:.
4 F rej.c =2222*lbf Closing Disk Force: Fe : = F stem+ F th.stem+ F th.bonn - F pack- F rej.e Fe =4917*lbf Mathcad 5.0
- EJM01.MCD 5:06:25 PM on 6129196
.MPR Associates, Inc., 9o King Street xandrla, VA 22314 Sheet 2
~ 'i ./
,f""J(1 ..., /
v1':;,!,, / .l'....-n-- DL, i:RMINATION OF CLOSING FORCES 2 2 Pbc*1t*b
- n:*a -(Puc - Pbc) - - - -
cos(&) 3
-2.797*10 O*lbf 0
2 Pbc*?t*b *tan( 0) 243.815 Pc:= O*lbf Pc= 0 *lbf 2 243.815
.Pbc*n*b *tan(0) 2 Pbc*n:*b 2 -2.797* 103
- it*a *(Pde - Pbc)- - - -
cos(B) 4.917*Hf Fe I -1 0 0 0 0 0
-µc*sin(e) cos(e) O 0 0 0 -1
µc*cos(0) sin{0) -1 0 0 0 0 Mc:= 0 0 0 µe*sin(9) -cos(9) 0 1 Solving for the force vector, Sc, 0 0 0 µe*cos(&) sin(B) -1 0 -l Sc :=Mc *Pc 0 0 0 1 -l 0 0 0 1 0 0 0 The forces are:
0 5.063* Ht Sue :=Se0 Sue = 5063 *!bf 3 7.86*10 Hue :=Sc 1 Hue =7860*lbf 3 2.459-10 Fuc = 2459 *lbf Fuc :=Sc2 3 *!bf Sc= 5.063* 10 Sdc :=Sc3 Sdc = 5063 *!bf 3 7.86*10 Hdc :=Sc4 Hde =7860*Jbf 2.459* 10 3 Fdc :=Sc5 Fdc =2459*lbf 7.654* ta3 He:= Sc6 He =7654*1bf 5:08:43 PM en 8129196 Mathcad 5.0 * 'EJMD1.MCD
- MPR ~ssociates, Inc., ndria, VA 22314 ERMINAJION OF WEQG!NG INTERFERENCE Flexibilities for Closing 6 6 4 2 l 4 a *(7*m+ 3) + b *(m- 1) - a *b *(m + 7)- a '.b *(7*m- 5) .**
2 m:=- 1 2 ) 2 2 2 2
+-4*a *b *[ a *(5*m- 1) + b *( m + 1) ].in(~)- 16*a4-b 2*(m + 1)*ln(~)
vdc cp:=3* ( m -1. --------------.!..b-=---*------~b_,__ 2 2 16*m2 a *(m+ l)+ b *(m- 1) 2
)
(a) a4 *(3*rn+ 1)- b4 '(m-1)- 2*a 2*b2*(m+ 1)- 8*m*a2*b2*ln - - 4*a2*b2*(m+ l)oln - (a) 2 (
'I':= 3* m - 1 . b b 2 [ 2 2 4*m '1t a *(m+ l)+b *(m-1)
CJl 'I' ""9 -1 h Gpc:=-- Gsc:=-- Hub Ghc = 7.025* 10 *in* !pf Ghc: Edc*t3 Edc*t3 ' -a -I Disk, Pressure Ope =2.446-10 *in* psi dw 2 -s -I Ksc : = - - - - - - Kpc : = - - - dw Disk, Seat Force Gsc =1.373* 10 *in* lbf 4*n*(a+ tb)*tb*Ebc 4*tb*Ebc
'""9 -I Body, Seat Forces Ksc =4.013*10 *in*lbf 1
Body, Bonnet Pressure: Kpc =5.043*10-s *in*psC Wedglng lnlederence at Reference Conditions
...r :=2*Ghc*Hc+ (Gsc+ Ksc)*(Suc + Sdc) + Gpc*(Puc +Pde - 2*Pbc) + 2*Kpc*Pbc ... -4
+-dw*adc*(Tdc- Tr)+ dw*cxbc*(Thc- Tr) wr = 5.12634* 10 *in PETEBMINATION OE OPENING FORCES Flexibilities for Openjng 6 4 2 2 4 a6*(7*m+3)+b -(m- !)- a *b *(m+7)- a *b *(7*m- 5) ...
1 2 m:=- vdo 2 ) 2 2[ 2 2 (8) 42
+-4*a *b
- 8 *(5*m- l}+b *(m+ 1) J*In - -16*8 *b *(m+ l)*ln - (8) 3* ( m - l b b cp:- . ---------------'---'---------..!......-'-2 16*m2 a2*(m+l)+b *(m-l) 4 4 qt:= J*(m2 _ l). a *(3*m+ 1)- b *(m- 1 )- 2*a *b *(m+ 1)-
2 2 2 2 (a) 2 2 8*m*a *b *In b - 4*a *b -(m+ l)*ln b (a) 2 2*n 2 2 4*m r a *(mt- 1) + b *(m- 1) CJl '¥ 1 h Gso:=.-- Hub Gho=6.942*10-g *in*lb( Gho:=--- Gpo:=-- it*b2*Edo Edo*t3 Edo*t3 1 Disk, Pressure Gpo = 2.417* 10-s *in*psi" dw Kpo := dw 2 Disk, Seat Force Gso =1.357* l 0"11 *in* lb( 1 Kso : = - - - - - - 4*tb*Ebo Kso =3.965* 10 *in* !bf 4*n*(a+ tb)*tb-Ebo ""9 -I Body, Seat Forces Body, Bonnet Pressure: Kpo =4.983* 10-s *in* psi" 1 Wedging Interference at Opening Conditions
-4 10 := wr + dw*ado*(Tdo - Tr) - dw*cxbo-(Tbo- Tr) WO =5.)2634*10 *in
-4 Define: W := wo - Gpo-(Puo + Pdo - 2*Pbo) - 2*Kpo*Pbo W =3.62698* 10 *in 5:07:52 PM on 8129196 Mathcad 5.0
- EJM01.MCO
- MPR Associates, Inc.,
Calculation No.108-173-EJM-01 2
*2 Pbo*'lt*b
- Jt*a *(Puo- Pbo) -
cos(9) 1.213* 103 O*lbf 2 0 Pbo*'lt*b *tan(9) 243.815 O*lbf 0 Po:= 2 . Po= *lbf Pbo*'lt*b *tan(9) 243.815 2 2 Pbo*n*b 8.724*103
- Jt*a *(Pdo- Pbo)-
cos(9) 0 O*lbf 2.069"10 4 l w Gso+ Kso
-1 0 0 0 0 0 0
µo*sin(9) cos(9) 0 0 0 0 -l 0
-µo*cos(0) sin(9) 0 0 0 0 0 0 0 0 -µo*sin(B) - cos(O) 0 1 0 Mo:= 0 0 0 -µo*cos(0) sin( 9) 1 0 0 0 0 0 I -1 0 0 0 0 0 0 0 1 0 -1 2*Gho*cos(9) 0 0 0 0 0 Gso+Kso The forces are:
Solving for the opening forces, So: SJ 16-103 1 So :=Mo- *Po Suo :=So 0 Suo = 5116*lbf 3.902*Ia3 Huo :=So 1 Huo = 3902*lbf 2.146" 103 4 Fua :=So2 Fuo =2146*lbf 1.235* 10 So= *lbf Sdo :=So 3 Sdo =12349 *lbf 3.624*103 Hda :=So4 Hdo =3624*lbf 3 5.341*10 Fdo :=Sos Fdo =5341 *Ibf 4.084* Ut Ho :=So6 Ho =4084*lbf 3 7.487*10 Fo :=So7 Po =7487*1bf Opening Forces Calculated Disk Opening Force Fo = 7487 *lbf Pbo =2235*psi F th.stem =0 *lbf Packing load F pack= 3.827* la3 *lbf F th.bonn ':"O*lbf Stem rejection force F lej.O
* :=Pbo*D2 -~
4 F rej.o -2222*1bf 3tem opening Force F open:= Fa+ F pack- F rej.o F open= 9092*lbf Mathcad 5.0
- EJM01.MCO 5:08:14 PM on 8129196
MPR Associates, Inc., e 320Ki t? Alexandria, VA 22314
- i Sheet 1 Calculation No. "108-173-EJM-01 Prepared by:__.,~--'---------- Checked by:-.A ~___.'~'""Y'""'il"V='-=-~~------
....f.l_*....
EVALUATION OF GATE VALVE OPENING THRUST PURPOSE: This calculation evaluates the equations for disk opening force derived in calculation 002-156-DHH-816. VALVES EVALUATED: Salem Pressurizer Block Valves 1 PR7, 1 PR6, 2PR7, 2PR6 Plmenslons Mll~clal ec212!i!tll11 Closing Opening Disk Thickness: t := 0.761-in 6 6 Disk Elastic Modulus.: Ede:= 25.0* 10 -psi Eda := 25.3* 10 *psi Disk Seating Radius: a := 1.281-in Disk Poisson's Ratio: vdc :=0.3 vdo :=0.3 Disk Hub Radius: b := 0.630-in -6 *I -6 -I Disk Coef. of Expen.: ade :=9.61*10 *deg ado :-9.53*10 *deg Hub Length from Center: h := 0.219*in 6 6 Body Elastic Modulus: Ebe := 25.0* 10 *psi Ebo :=25.3* l0 *psi Disk Total Width at Center: dw := 1.959*in Body Poisson's Ratio: vbc :=0.3 vbo :=0.3 Body Wall Thickness: tb := 0.761*in . -6 -I -6 -1 Body Coef. of Expan.: abc :=9.61*10 *deg abo :=9.53*10 *deg Half Disk Wedge Angle: 9 := S*deg Stem Diameter: D := 1.125*in Stroke L :=2.688*in Reference Temperature: Tr :=70*deg Ambient Temperature: T amb := 120*deg
.:ondltions Closing Opening Ups1ream Pressure: Puc :=2235*psi Pua:= 1457*psi Bonnet Pressure: Pbc := 2235*psi Pbo := 1457*psi Downstream Pressure: Pde := 2235*psi Pdo :=O*psi Disk Temperature: Tde := 653*deg Tdo := 593*deg Body Temperature: The :=653*deg Tbo := 593-deg Friction Coefficient µc := 0.40 µa :=0.44 Closing Forces Closing Stem Force: F stem := 10966* lbf Packing Force: F pack := 3827-lbf Thermal Binding, Stem Expansion after Closing:
F th.stem :=J*~*L*(Tbc-T m*deg amb) F th.stem =4298 *!bf F th.stem :=O* lbf
!bf Thermal Binding, bonnet Fthb
. onn
- =2.25*--*L*(Tbc-Tbo) in*deg F th.bonn =363 *!bf cooling:
F th.boI!n :=O*lbf 2 Stem Rejection Force: F reJ.C
* :=Pbc*D -~
4 F rej.c =2222 *lbf Closing Disk Force: Fe := F stem + F th.stem+ F th.bonn - F pack - F rej.c Fe =4917*lbf J.1athcad 5.0 - EJM01.MCD 5:09:11 PM on 8129/96
MPR As~ociates, Inc., andria, VA 22314 e f/il i/ Sheet2 Checked by: /J/:.,,l 9//-./J?-.-- DE: 1 cRMINATION OF CLOSING FORCES 2 2 Pbc*n*b
-1t*a *(Puc - Pbc)- - - -
cos(B}
-2.797*1Cf O*lbf 0
2 Pbc*tt*b *tan(9) 243.815 Pc:= O*lbf Pc"' 0 *!bf 2 Pbc*n*b *tan(9). 243.815 2 3 2 Pbc*?t*b -2.797*10
-n*a *(Pde- Pbc)- - - -
cos(9) 4.917* Hf Fe
-1 0 0 0 0 0
-µc*sin(0) cos(0) 0 0 0 0 -1
µc*cos( 0) sin( 9) - 1 0 0 0 0 Mc:= 0 0 0 µc*sin(9) -cos(9) 0 Solving for the force vector, Sc, 0 0 0 µc*cos(e) sin(B) -1 0 -l Sc :=Mc *Pc 0 0 0 1 -1 0 0 0 0 0 0 0 The forces are:
5.063*103 Sue :=Sc0 Sue= 5063*1bf 3 7.86*10 Hue:= Sc 1 Hue =7860*1bf 2.459* 103 Fuc :=Sc2 Fuc =2459*lbf Sc= 5.063* 103 *!bf Sdc = 5063 *lbf Sdc :=Sc3 7.86* 1<>3 Hdc :=Sc4 Hdc = 7860 *!bf 2.459' 103 Fdc :=Scs Fdc = 2459 *lbf 3 7.654* 10 He :=Sc6 He =7654*lbf Mathcad 5.0 - EJMOUACD 5:09:29 PM on 8129196
MPR Associates, Inc., -2~KingStr~eet,Ale~ria, VA 22314 e ,// : /;,. "i I . Sheet3
. ! .'/ _)..',.
Caleulation No.108-173-EJM-01 Prepared by: _ ___,,~-+-"--=---- Checked by: ,-,,;-_,, ~ * ~*////rr_. ERMINATION OF WEDGING INTERFERENCE Flexibilities for Closing 6 6 4 2 2 4 a *(7*m+ 3) + b *(m- 1)- a *b *(m+ 7)- a *b *(7*m- 5) ... J (a) b - 16*a4*b2*(m+ I )*In (a) I 2 m::::- vdc
+-4*a2*b2 *[ a2 *(S*m- 1) + b2*(mt- 1) *In b 2 . 2 a *Cm+ 1) + b *(m- l) 4 4 2 2
'P :=J*(m2 - l). a *(3*m+ 1)-b *(m-1}- 2*a *b *Cm+ 1)- S*m*a *b *In 2 2 (a)b - 4*a 2 2
*b *(m+ lHn (a)b 2 2 [ 2 2 4*m *'It a -(m+ l)+b *(m-1) h cJ> 'o/ 1 Ghc :=--- Gpc:=-- Gsc:=-- Hub Ghc = 7 .025* 10-9 *in* lb(
n:*b2*Edc Edc*t3 Edc*t3 Disk, Pressure Gpc =2.446-*10-s *in*psf 1 dw 2 Ksc : = - - - - - - Kpc:=--- dw Disk, Seat Force Gsc =1.373*10-8 *in*lbf 1 4*7t*(a + tb)'tb*Ebc 4*tb*Ebc 1 Body, Seat Forces Ksc =4.013*10-9 *in*lb( 1 Body, Bonnet Pressure: Kpc = 5.043* 10-11 *in*ps( Wedging Interference at Reference Conditions
* :=2*Ghc-Hc+ (Gsc+ Ksc)*(Suc + Sdc) + Gpc*(Puc +Pde- 2*Pbc) + 2*Kpc*Pbc ...
-4
+- dw*adc*(Tdc - Tr)+ dw*abc*(Tbc - Tr) wr = 5.12634* IO *in PETEBMINATION OE OPENING FORCES Flexibilities for Opening 6 6 4 2 2 4 a *(7*m+ 3) + b *(m- I)- a *b *(m+ 7)- a *b *(7*m- 5) ...
l m *- J (ba ) - 16*a42*b *(m+ l)*ln (b)2 a vdo 222
.[ 2
+-4-a *b
- a *(5*m- 1)+ b -(m+ 1) *ln 3*(m - I)
$:=-"'~-~*
2 2 2 2 16*m a -(m+ l)+b -(m- l)
'o/ := *(m2 - i). 4 4 2 2 a *(3*m+ 1)- b *(m- 1)- 2*a *b *{m+ 1)- B*m*a *b *ln 2 2 (a)b - 4*a *b *(m+ l)*ln (a)b 2 2 2
3 2 [ 2 2 4*m *l'C a -(m+ l)+b *(m-1) h 11> 'l' 1 Gho:=--- Gpo:=-- Gso:=-- Hub Gho =6.942*10-9 *in*lbf Tt*b2*Edo Edo*t3 Edo*t3 1 Disk, Pressure Gpo =2.417*10-8 *in*psf 2 Gso =1.357* 10-s *in* lb( 1 dw dw Disk, Seat Force Kso : - - - - - - - Kpo:=--- 4*n*(a+ tb)*tb*Ebo 4*tb*Ebo -1 Body, Seat Forces Kso =3.965* 10-g *in- lbf 1 Body, Bonnet Pressure: Kpo =4.983" 101 *in*ps( Wedging lntedereoce at Opening Conditions wo := wr + dw*ado-(Tdo - Tr) - dw*abo*(Tbo - Tr) WO =5.12634* lQ-4 *in Define: W :=wo-_ Gpo*(Puo+ Pdo- 2*Pbo) - 2*Kpo*Pbo W =4.02634*10-4 *in Mathcad 5.D - EJMD1.MCO ' 5:10:36 PM on 6129/96
' MPR Associates, Inc., Calculation No.108-173-EJM-01 Prepared by:
- --..p.....,,e...,F-L-"--dn~a-,_v_A_2231c4hecke9d by: ;./
... {;_*{./_~
..... ~
., ._,. ,,,n
~ ~
sheet 4 2 2 Pbo*it*b
-it*a *(Puo- Pbo) -
cos(0) O*lbf
-1.s24* 1rr 0
2 Pbo*n*b *tan(e) 158.943 O*lbf 0 Po:= 2 Po= *lbf Pbo*n*b *tan( 0) 158.943 2 3 2 Pbo*7t*b 5.688*10
-7t*a *(Pdo- Pbo)-
cos( 0) 0 l O*lbf 2.296* 104 w Gso-r Kso
-1 0 0 0 0 0 0
µo*sin(0) cos(0) 0 0 0 0 -1 0
- µo:cos( e) sin( a) 0 0 0 0 0 0 0 0 -µo*sin(0) -cos(8) 0 0 Mo:= 0 0 0 - µo*.cos( a) sin( e) 0 0 0 0 0 I -1 0 0 0 0 0 1 0 0 0 -l 2*Gho*cos(0) 0 0 0 0 0 Gso+ Kso The forces are:
Solving for the opening forces, So: 3 S.077*10 1 3 . Suo :=So 0 Suo = 5077 *lbf So :=Mo- *Po 6.901*10 Huo :=So 1 Huo = 6901 *lbf 3 1.783* 10 Fuo :=So 2 Fuo = 1783*lbf 4 1.231*10 So= *!bf Sdo := So 3 Sdo =12310*lbf 3 6.622*10 Hdo := So 4 Hdo =6622 *lbf 4.977*1a3 Fdo :=Sos Fdo = 4977 *!bf 7.069-lrf Ho :=So6 Ho = 7069 *!bf 6.76* la3 Fo :=So7 Fo =6760*lbf Opening Forces. Calculated Disk Opening Force Fo =6760*lbf Pbo =1457*psi F th.stem= 0 *lbf 3 F th.bonn :;:: 0 *!bf Packing load F pack= 3.827* 10 *Jbf Stem rejection force 2 re F rej.o := Pbo-D *4 F rej.o =1448*lbf
.em opening Force F open := Fo + F pack - F rej.o F open =9139*lbf 5:10:59 PM on 8129/96 Mathcad 5.0
- EJM01.MCD
~-----~-------------=- ~...:...--- .. ~. :r .. S-l-llC-MDC-U9(001) Page 11 of31 l'(C.DE-PS.ZZ-0033(Q)-A8 Rev .4 EXHIBIT 1 MOV No: 1PR6
'.3_1.8 Assumption Verification and Setpoint Determination A6/Al7 AS Baseline A14 Controlled Source Assumptions Capablllty Swic Ti:st DP Test Setpaints Rev:~ Rev:..1.. No: J_ No: fiM_
11-~~~~~~-t-= Work Onicr Date J.L (Stem) ROI.. ROL MRT Target 10244 I Thrust 18000 1 Test C14 11388 3 Thrust Cl6 Thrust/Torque Margin Min Req Torque Torque C14 IBS JBS I Limit Cl6 Test Cl4 Torque C16 M'm TSS 2.0 1 Max TSS 3.0 1 As-Left TSS A4 12 ser: 8.9 12 Ut: 8.59 8.59 12 Stt! Page S of 7 Assessment Rev ._4_
""-'-"l~\..*l\1 Ul.:..Q90.:i {002)
, h;c 11 or.CJ eL" ~.DE-PS.ZZ-0033 (Q)-A8 Rev .4 EXHIBIT 1 MOV No: 2PR7 3.1.8 Assumption Vcrification and SetpOint Determination A6/A17 AS Baseline A14 Controlled Source Assumptions Capability Static Test DP Test Secpoints Rev:~ Rev:~ No:_L No:~ Work Order
' Date Fv Fl
µ.(Stem)
ROI.. ROL MRT Target 10244 1 Thrust Thrust Limit 18000 1 Test C14 9441 3 Thrust C16 Thrust/Torque Margin Min Req Torque l Torque Limit Cl4
--- 185 185 1 l C16 Test Cl4 I; Torque C16 Min TSS. 2.00 I I Max TSS As-LeftTSS 3.00 1 A4 STROKE OPEN JOsec 1.94 1.94 10 sec TIMES CLOSE 10 sec 1.95 7.95 10 sec Page S of-7 Assessment Rev ...-!._
S*l*RC-¥DC-0905(001) Page 11 of 41 t NC.DE-PS.ZZ-0033(Q)-A8 Rev.4 EXHIBIT l MOV No: 2PR6 l 3.1.8 Assumption Verification and Setpoint Determination A6/Al7 Assumptions Rev:~ AS Capability Rev:..1., Baseline Static Test No: 4_ Al4 DP Test No: NIA Controlled Setpoints Source l Work Order Date Fv f Fs -*---* .: q
µ.(Stem) .;
i ROI...., ROL MRT Target 10244 1 I* Thrust I I Thrust Limit 18000 I l ! Test Cl4 10298 3
----- ! 1I Thrust C16 11768 ; JI:
Thrustrrorque Margin 130/o , di , ' Min Req i l
- I Torque I I
)i Torque __Cl4
..____ 185 185 1 I I I I 1:
i' Limit Cl6 i I I. Test Cl4 I Torque C16 1. I Min TSS 2.00 1 I Max TSS 3.00 1 As*Left TSS STROKE OPEN NIA 8.9 8.35 NIA 3 TIMES CLOSE
-----*4---------4---*--*'-----
10 s~c 8.59 8.60 IO sec 3 Page S of 7 Assessment Rev._L
S-!-RC-MOC--i89(002J Pa;e l1 ons f.t(:.DE-PS.ZZ-0033 (Q)-A8 Rev.4 EXHIBIT 1 MOVNo: 1PR7
- 3.1.8 Assumption Vefification and Sctpoint Determination A6/A17 A8 Baseline A14 Conrrolled Source Assumptions Capability Static Test DP Test Setpaints Rev~_!!_ Rcv:J_ No:J_ No: .11.UA..
Wotlc: Order Due F,,
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