ML20082L101
| ML20082L101 | |
| Person / Time | |
|---|---|
| Site: | Duane Arnold  |
| Issue date: | 08/13/1993 |
| From: | IES UTILITIES INC., (FORMERLY IOWA ELECTRIC LIGHT |
| To: | |
| Shared Package | |
| ML20082L096 | List: |
| References | |
| CAL-M93-012, CAL-M93-012-R00, CAL-M93-12, CAL-M93-12-R, NUDOCS 9504200239 | |
| Download: ML20082L101 (27) | |
Text
_
Attachment to "V
NG-95-1?s2 IOWA ELECTRIC LIGHT AND POWER COMPANY DUANE ARNOLD ENERGY CENTER DESIGN CALCULATION COVERSHEET CALCULATION NO. : C AL-MU-og CALCULATION TITLE: 1 pcitE AS E IM fp pus ro UN B A LAN'C eb VGGbu) AT E R f L ot*J REFERENCE DOCUMENTS EWR NO.:
DDC NO.:
DCP NO.:
OTHER:W--80$
PREPARED BY:
4.
DATE:
T//2/f3 Responsible Design Engineer VERIFIED BY:
J DATE: hND Design Verification Engineer REVIEWED BY: Mu
.J DATE:
((I/6 Group _eader M
DATE:
d'-8 #.f APPROVED BY:
Supervising Engineer NG-007Z Rev 2 (1203.21) 950420o239 950412 DR ADOCK 0500 1
'V
^
PURPOSE The feedwater check valves (V14-1) and (V14-3) need to be tested to assure they can admit flow from HPCI and RCIC when required.
This is presently accomplished by alternately disassembling them during refueling such that each valve is evaluated every other outage.
This evaluation is a visual inspection.
It has been proposed to substitute a calculation method to support a test that would replace the disassembly process.
By looking at the simplified drawing given by Figure 1, feed flow is split into two feed headers which are connected by a cross-connect line.
The two feed headers are in turn split into two smaller lines that connect directly to the Reactor Vessel.
If one of the feed check vales were to become partially clogged or to go shut or partially shut, more water from one feed line would be redirected through the cross-connect line (given that total feed flow and power stay the same).
This would in theory, cause the pressure instrument "P1",
to read higher because of the increased DP needed to drive the increase flow down the line.
The purpose of this calculation is to demonstrate the feasibility to use the increased DP to supply " alert / action" levels to serve as the basis of a test procedure to replace the valve disassembly test.
ASSUMPTIONS AND GENERAL APPROACH Utilizing Bernoulli's Equation an expression for flow down one of the feed paths is generated.
The flow path taken will be that of the opposite of the line that contains HPCI.
By selecting this flow path, a flow split can be set equal to the flow needed by HPCI in one feed line with all other feed flow going to the other feed line.
Using plant data obtained during normal operation when flow is assumed to be split equally between the feed flow lines the equation can be set equal to the recorded DP and the remaining unknown parameters can be explicitly solved for as a lumped sum.
this is substituted into the equations for unbalanced flow and the DP is calculated.
Once a constant has been selved for that represents the unknown parameters a set of balancea and unbalanced solutions can be generated corresponding to specific power levels.
By applying appropriate instrument error adjustments to the calculated DP, appropriate
" alert / action" levels can be set that will detect a condition when either one line or the other is restricted.
The advantages of this test is that it reduces maintenance effort and Man-Rem and supplies a better evaluation of true check valve operability than a visual exam of a disassembled valve.
Iowa Electnc Light and Power Company Cedar Rapids. Iowa ET &;sLhr hired shh?
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Rev-Prepared!Date Venhed/Date Sheet #
Contmued On f 0 F 2d
The assumption that under normal operations there is balanced flow is based on the knowledge that the system is designed to give essentially balanced flow.
Pressure reading from P1, (see Figure 1) have been consistent over operating history and can be correlated to conditions when recent visual inspections have shown that valves were clear and working properly.
The cross-connect line serves to insure reasonably balanced flow.
Using the HPCI flow requirements as a means to set the action level, envelopes RCIC due to the more demanding HPCI flow requirements.
Additional assumptions are stated throughout the calculation as they come into play for use in the calculation at that point.
Iowa Electric Light and Power Company Cedar Rapids, Iowa B*
kdkTibh r/a M3 0, V B00L UlW 9 %
Doc # C A L - M 9 3 % I 7-peyg Rev Prepared'Date Venfied/Date Sheet #
. Continued On 2 oFM l
i
i,,-
REFERENCES
[1] NEDC-30603 Duane Arnold Energy Center Power Uprate
[2] ASME Steam Tables 1967
[3] CRANE TP410 Flow of Fluids through Valves, Fittings, and Pipe
[4] Technical Specification and Operating License for the Duane Arnold Energy Center
[5] FSK-8017
[6] ISO-DLA-2-3
[7] ISO-DLA-2-4
[8] BECH-M190
[9] I.PT-G080-001
[10] APED-C31-013
[11] M&TE Database
[12] DGC-Elli
[13] Irvine & Liley, Steam and Gas Tables with Computer Equations, Academic Press Orlando, FL
[14] APED-B11-008 Ds] Arrewart t, Pcm Dm lowa Electric Light and Power Company Cedar Rapids, Iowa j
g pwgQ, ya, (s &ffg yjtyg5 Doc # CAL-M 9% -Ot 2-Rev#
Rev Prepared /Date Venfied/Date Sheet #
Continued On S oeJD
L
1 2
P r
Li y
2 FW 'A' 1
i n
=
l L1 L2 16 in, lines 10 in. lines Flow path analyzed Figure 1. Portion of FW Flow Path Analyzed Constants:
8 g = 32.17-g e = 32.17:
9 A =144-Mlb = 10 Ib 2
sec lbf sec ft General equation: Ref. [3), page 1-5, equation 1-3 2
9AP 1 V
9AP 2 V
3 2
Egn.1 Z i+
+
=
Z2+
+
+h L p,_9_,
2g
- p. g 2g 9c 9c l
APs are relatively small and fluid is incompressible. Therefore: p = constant.
l l
2 i
GAP 3 GAP 2 V
V 2
1
(
Eqn.2 z 3+
z2 +
=
+
+h t j
p p _9_
2g 2g 9
Oc Oc V
V l
9A 2
1 Eqn.3
-(P 3-P)"
+(22-2)+h 2
1 L
P= g g
g 9c lowa Electric Light and Power Company j
Cedar Rapids. Iowa 4
i Doc # Chl-M41 %I 2-Rev#
l M [W.d!5 w.%hs l.Y Dbhh HIL/9 6 l
Rev Prepared /Date Venfied/Date Sheet #
Continued On or*,ld
i H (.
v.
p 9_
Eqn.4 (Pj-P)
+(22-2)+h 2
1 t
Ref. [3], pages 3-4, equations 3-5 & 3-14:
l 2
2 L
V y
L V
3 3
1 2
2 V 2 Eqn.5 h l
= h t3 + h L2
- I
+K-
+f=
+K 1D 2g 2g D 2 2g 2' 2 g 3
2 3
' [V 2h P
2 Eqn.6 P 3-P 2 2
V ge
=
3 2-2 1
+
(2 g 2 g/
gA 2
3,V L
V L
V V
3 3 +f2 2
2
+f-
+K 2
+K D
2g 3=2g 3
=
3 2 2g 2' 2 g D
l From P& ids M-107, M-114 and 7884-M-190, the 16" pipe consists of:
DLA which is schedule 80:
ID := 14.312 in A := E ID 2 2
A = 1.12 ft 4
DBD which is schedule 120: 10 := 13.562 in A := E ID2 2
A = 1 ft j
4 The 10" pipe consists of:
DLA which is schedule 80:
ID := 9.562 in A.= E ID2 2
4 A = 0.5 ft 2
2 A = 1 ft Consider two cases:
case a: Feed flow is split equally between upper and lower headers.
case b: Feed flow in upper header is reduced to that of needed HPCI flow. Then, total feed flow minus HPCI flow goes through the lower header.
Case "a" represents the nonnal and expected flow pattern. Pressure instuments read P 1 and P. Suppose the check valve in the feedline where HPCI injects were to go closed 2
to the point where feed flow in that line is reduced to HPCI Tech Spec requirements (3000 gpm). Then the remaining flow goes through the lower header. By reading the A P, this condition can be detected. By calculating this flow (case "b") based on data obtained from normal operation (case "a"), a predicted AP for case "b" can be obtained.
In doing this calculation, ignoring things which would tend to make the AP higher is conservative.
Iowa Electric Light and Power Company Cedar Rapids, Iowa 1
B W.AUG W9hk 01 U0bm *ltzl&&
Doc # OL-MA % <!L nevn Rev Prepared /Date Ventied/Date Sheet #
Continued On CoFlo 4
l For the L Itngth of 16" line, assuma all is DLA. This creates a larger ID and lower P
t.
P, i
velocity and, therefore, lower pressure drops. It is therefore cons rvative. Then:
D 3 := 14.312 in D 2 := 9.562 in 2
A 3 :=
D A 2 := 2-D 2 3
l 2
2 A 3 = 1.12 ft A 2 = 1 ft N.B. The area A represents tne area of two 10" pipes into which the flow is split 2
from the 16" pipe.
For the path taken in the lower header, use the shorter length path for the 10" line. This gives a lower & and is, therfore, conservative. However, as will be shown later, the pipe lengths do not enter into the calculation since thay are part of an constant term "C" which is evaluated from plant data.
RWCU injects into the lower feed header ordinarily. Ignoring its added flow to the feed flow decreases the calculated & and is, therefore, conservative.
The head loss coefficients due to bends, contractions, etc. (K) which are included in the l
calculation are lumped together and solved for the condition for which data has been i
obtained (i.e. case "a"). They are considered constant since they depent on piping geometry. However, the check valves in question open with increased flow thus lowering i
their K value and consequent & contribution. The check valves ponion of the total K is small and the relatively small non-conservatism of using a constant K is balanced by the other conservatisms.
l l
The friction factors (f) are weakly dependent on flow through Reynold's Number (Re).
For the flow regime we re considering, Re is high enough that the friction factor is constant.
i The flow path that is analyzed for cases "a" and "b" is circled in the diagram.
j Length of 16" DBD pipe = 29' 3" from ISO DBD-4-2 & 5.
Length of 16" DBA pipe = 37' 6" from ISO DBD-2-1 & 4.
Length of 16" DBD pipe = 60' from ISO DBD-2-4.
It turns out that the factors L and D for the pipe, which are constsnt, are lumped into and overall constant for head loss. This means that detailed knowledge of pipe lengths is not l
required to solve the problem at hand.
Elevation heads are:
2 762tt from ISO DBD-4-2 3
Z 2 810 ft + 10.5 in from ISO DLA-2-4 l
l lowa Electric Light and Power Company Cedar Rapids, Iowa Doc # CAL-u9 b -ot 2 neva ef
%DRWJ MJe 0 bk))00C' bih14 b Rev Preparec Date Venhea'Date Sheet #
Continued Cr 6 >F M
CASE"an s,
Temocrature of FW:
T m := 424 deg F from Ref. [1]
Density of FW:
Specific Volume of Saturated WateFrom Ref. [13], pages 21 - 24 DEGK(X) := 5 (X + 459.67)
TKCR := 647.3 0
3 AVF := 1 BVF := -1.9153882 VFCR := 3.15510 1
kg CVF := 1.201518610 DVF := -7.8464025 3
VFCR = 0.05 g Ib
- 3.888614 x
2.0582238 2
x
- 10829991 3
TC(TK) := TKCR - TK x
EVF.=
8.2180004 10-'
TVECTOR( x ) :=
x 4.7549742 10-'
s y
O
- g 0
x 1
5 7
YVF(TK) := AVF + BVF TC(TK)3 + CVF TC(TK)s + DVF TC(TK)e
+ EVF TVECTOR(TC(TK))
YVF(300) = 0.32 i
3 VF(TK) := VFCR YVF(TK)
VF(DEGK(T pw = 0.02 -
RHOF(TK) :=
RHOFfDEGK[Tpwh = 52.66 b VF(TK)
N A
N 3
ft Then:
p w := RHOF(DEGK(T gw p
l p w - 52.66 b and:
p 3 := p w, p 2 " P FW p
p ft Iowa Electric Light and Power Company Cedar Rapids. Iowa 6
t'KJh%s Hah 8 0. Ask01ih_ Wn/91, Doc # Ch l* fM 3 'C' i-Revn Rev Prepared /Date Venfied/Date Sheet #
Continued On 7 d t'.20
t,.
I',
Viscosity of FW:
p 0 = 1 10-7 lbf
- Avisc := 1.005885 Bvisc := 2813.416 l
ft se c16 = - 1.06607 c17 = 9.798022 centipoise =
100 Svisc 9(TF) :=
0 Avisc.e# *
- g(Tpw)
Then:
pw :=
pw = 2.43 10 Ibf
- 4 g w = 0.12 centipoise p
ft Mass Flow of FW:
l FW.100 := 7.1710 S at 100% power.
8 mdot hr mdot FW.100 mdotFWB"
- P 1 *^ 1 *V 1 P 2A 2V2 2
where: A1 = flow area of 16" pipe and A2 = flow area of two 10" pipes.
When flow splits from one 16" line into two 10" lines:
mdot FWB mdotFWB1 :=
and mdotFWB2 := mdotFWB1 2
Then:
mdotFWB 3 - 16.93 ft V 3 :=
V p1 A 3 Sec A 3 2 = 18.96 g V
V V
2 := A 3
2 sec lowa Electric Light and Power Company Cedar Rapids, Iowa e' 9$A!Y7Lwmd Plaki 0 JJ hhw */I2/9%
Doc # Ckc 9 M t2.
neva Rev Precared/Date Venfied/Date Sheet # __. _
Continued On S oPJr i
Reynolds Number of'FW Flow:
p w'V D 1 p
1 Re 3 :=
7 Re 1 - 1.36 10 4FW 1
4 mdotFWB or:
Re 3 :=
Re 3 - 1.36 10 7
x-pw D 3 P FW'V D 2 i
2 Re2:=
Re 2 - 1.02 10 7
4FW 4 mdotFWB1 or:
Re 2 :=
Re 2 - 1.02 10 7
pw D 2 x-Moodv Friction Factor:
e :=.00015 in f( c, D, Re) := f
[ c.25
)2 1
9 5.74 2 log, 3.7 D
+
Re
3 := f(c, D $, Re 3)
Then:
f f 3 - 0.0139 f2 := f c D 2Re2)
I2 = 0.0147 Evaluation of AP:
i
- f 2
2h pA Eqn.6 P 3-P 2
_2 3
V V
ge (2 g 2 g/
L V
2 gA
+f 3
3 V
L2 V 3
2 V 2 3
+K
+f-
+K' D
2g 32g D
2 2
3 2 2g 2g Iowa Electric Light and Power Company Cedar Rapids, Iowa
^
y ggm ggg QQ,9fQp gjt2,14g Doc # Chl~il42, ~Of 2-Rev#
Rev Prepared >Date Venfied/Date Sheet #
Continued On 9 oP2O
i '.
A 3
Since: V2"y'V 1 then:
Eqn.7
[A 2
3
-V g
1) 2 (A 2 V
=
3
+ Z2-2 1 2g' 2g gA
[A
[A 2
2 3
3 L
V 2
- -' V 2
l -V 3
+K-
+f2
- 1) + K ' (' 2 /
3
+f 1
3 V
L -(A 2 3
A 3
2 D 3 2g 2g D
2g' 2
2 2g' Collecting terms:
Eqn.8 AP =
2 ' [A 0
2
[g ) f N{
V 2
P L
L
-1+f 3
2 A 3
3 3 + K
1
+(22-Z) 2g A
3
+f=
+K 2
3 p
D D
2 (A 2/
(A 2/
QA j
Collect constant terms into a single constant:
[A 2
2 2
L L 2 A A
3
-1+f j + f-Eqn.9 C =
1 -
3 3+K'g
+K 3
2 (A2 D
D 2(A)
(A 2/
3 2
Then:
0 2
P FW -
Eqn.10 AP =
V 3 2 g (C) + (Z2-2) 1 gA From data :
AP g.gg := 67 psi at 100% power.
( ATT Ac r* MEMT l)
The actual driving head for the flow is found by subtracting the head due to the submergence of the FW sparger:
AP a = AP ued - 2 psi (head coITection due to submerged FW sparger; from Appendix 2) lowa Electric Light and Power Company Cedar Rapids, Iowa ft 2 90 M a /4 h{
100n,gsimqz Doc # CAL 40'4 -oI 2-Rev#
B t
Rev Prepared /Date Verified /Date Sheet #
Continued On 10 oeld
The total AP for arbitrary power levels can be calculated by correcting the AP due to flow only:
Power := 100 l
pw9 opw 9 p
67 psi - (Z2-2 f g
APa. read :=
2 - 2 )*
1
- 2 psi
+2 1
100 g
+ 2 psi (head correction due to submerged 3p a = AP a. read - 2 psi FW sparger; from Appendix 2)
AP a. read = 67 psi Then:
AP a'9 A
-(2 2-2) 1 pwg p
Eqn.11 C :=
2 V
2 3
Power 2g 100 Which yields:
C = 28.94 which depends only on geometry and is independent of flow.
Then:
f 9\\
Eqn.12 y
2 PN AP(V ) :=
+(Z2-2) 3 C-1 2g
( gA /
lowa Electric Light and Power Company Cedar Rapids Iowa Af
?L d/G K w m tM d6' 0JY00tirt. TfrzfG4 Doc # Chl W % MI 2-Rev#
Rev Prepared /Date Verified /Date Sheet #
Continued On il oe.2.4 l
\\'
To calculate AP as a function of mass flow:
m t v 3 ( mdot) := p3A 3
/
A P
V 3 ( mdot)2 M{
+ (Z 2-Z)
Eqn.12.a AP( mdot) :=
C-3 (gA /
2g and:
Re( mdot) :=
pw D 3 x-l CASE "b" 1
For the case in which all FW flow (adjusted for power level) less rated HPCI flow (1.498 Mlb/hr or about 3000 gpm) is going through one FW line:
FW.100 = 7.17. M!b mdot hr l
mdot HPCI := 1.498-hr Power l
mdatFWB = mdotFW.100 100 - mdo:HPCI Then, for Power = 100 %:
Calculate AP :
b D = AP(mdot ws)
AP F
AP b = 135.84 psi
- b. read := AP(mdotpwg) + 2 psi(head correction due to submerged but: AP FW sparger; from Appendix 2)
Then: AP b. read = 137.84. psi Iowa Electric Light and Power Company Cedar Rapids. Iowa B frOff hbd dhbs 0 20014 '0OZ/4 4 Doc # Cil-M98 t\\ l Revn Rev Prepared /Date Venfied!Date Sheet #
Continued On il or,2.d i
I
~
t..
n, For Power = 100 %:
i 1
Maximum AP for " Alert Level" If we measure a AP such that AP. read < AP < APAlert '
a then we can be 95% confident that HPCI flow will meet Tech Spec requirements.
The instmment uncertainty from Appendix 3 is:
1 AP Uncert = 40.3168 psi The maximum " Alert Level" AP is:
AP Alert := AP b. read - AP Uncert Then:
AP.lert = 97.53 psi Maximum AP for " Action Level" 1
l i
If we measure a AP such that:
AP > APAlert +
j then we can be 95% confident that the high AP is not due to instrument error.
The instrument uncertainty from Appendix 3 is:
4 AP Uncert = 40.3168 psi t
The maximum " Action Level" AP is:
AP Action := AP a. read + AP Uncert Then:
APAction = 107.32 psi Iowa Electric Light and Power Company Cedar Rapids, Iowa
.6
?n./d!41 b1mnl Binon JJ0 nrL Wh49 %
00C# -
~$A 5 l?
Rev#
Rev Prepared /Date Venfied/Date Sheet #
Continued On _I3 #20
For a range of power levels:
Power := 81,82.100 mdotpw( Power) := mdot FW.100 *
- mdot HPCI 00 AP (Power) := AP(mdatpw(Power))
b AP Alert ( Power) := AP b(Power) + 2 psi - AP Uncert pw9 p
2 AP a. read ( Power) :=
67 psi - (Z 2-2)-
1
- 2 psi 2 psi g
00 pw1 p
+(Z2-Z) 3 9A APAction( Power) := AP a. read ( Power) + AP Uncert mdot pw(Power)
Mib AP a. read ( Power)
AP Alert ( Power)
Power hr psi psi 81 4.31 50.79 47.66 82 4.38 51.56 49.95 83 4.45 52.34 52.27 84 4.52 53.13 54.63 85 4.6 53.92 57.03 86 4.67 54.73 59.47 87 4.74 55.54 61.94 88 4.81 56.37 64.45 89 4.88 57.2 67 90 4.95 58.05 69.59 91 5.03 58.9 72.21 92 5.1 59.76 74.87 93 5.17 60.63 77.57 94 5.24 61.51 80.31 95 5.31 62.4 83.08 96 5.39 63.31 85.9 97 5.46 64.21 88.75 98 5.53 65.13 91.64 99 5.6 66.06 94.56 100 5.67 67 97.53 lowa Electric Light and Power Company Cedar Rapids. Iowa M-t'.LJcr k ink 0.) 110 D nk. Y o W 4 Doc # ChL-M4%-D\\L nevn Rev Prepared /Date Verified /Date Sheet #
Continued On Id oNd
s,-
n and AP. read as a function of power:
Graphing APAlert a
100 95 90
/
85 4
^
7 80
/
AP gg( Power) 75 70 AP a.rW( Power )
/
/
P88 65
/j
/
. *
- 1..
g
/
/
=
.;/"
/
45 40 80 82 84 86 88 90 92 94 96 98 100 Power (CTO)
From the graph, this technique is impossible to apply below 83% power and difficult to apply below about 90% power.
l l
lowa Electric Ught and Power Compan)
Cedar Rapids. Iowa g
g,b % g, Qj Q'()g 6//242, Doc # OL-M 3M L.
Rev#
Rev Prepared /Date Venfied/Date Sheet #
Continued On /S oc
i
)
r i
CONCLUSIONS This calculation demonstrates the feasibility of predicting a DP that would be representative of a situation where one of the feed lines is restricted to the point where HPCI flow might be reduced to below Tech. Spec. limits.
It can be used as a basis for setting alert / action levels when a specified DP limit has been
}
reached or exceeded as long as the conditions for which the calculation are valid are adhered to.
The DP with instrument j
error included as developed by this calculation represents the i
maximum DP for alert / action levels that should be used.
The
" read DPs" must be taken at corresponding power levels as given by this calculation for the calculation to be valid.
Given the stated instrument errors this methodology becomes impractical below 90% power.
i Iowa Electric Light and Power Company Cedar Rapids, Iowa B A M M F%4 0
llY1bL Sll2 /93 Doc # Q)/ "U9 3 'Of 2-Rev#
Rev Prepared /Date
' Venfied/Date Sheet #
Continued On Ib er-34 P531750 NEW 9 89
i i.-
a Appendix 1: Calculation of Moody friction factor over range of FW flows FWB := 1 Mlbhr,1.5 Mlb. 8 Mlb mdot hr hr Re(mdotFWB) f(c,D 3, Re(mdot FWB 3.705 10_
0.016 6
5.558 13 0.015 1
6 0.015 7.411 10 6
0.014 9.263 10 0.014 7
1.112 10 0.014 7
1.297 10 0.014 7
0.014 1.482 10 0.014 7
1.667 10 0.014 7
1.853 10 0.014 7
2.038 10 0.013 7
2.223 10 7
2.408 10 0.013 7
2.594 10 7
2.779 10 7
2.964 10 o.1 hD 1, Re(mdot f
FWB
~~~~ ~...---~~. _ _..
0' 6
7 E
3 3o 31o
- 1. t o Re(mdotFWB)
Iowa Electric Light and Power Company Cedar Rapids, Iowa W,
O.L lhd < $
ul 0Yr 5lIMQ ^t, Doc # CAL-fl9 h 'O L Rev#
t Rev Prepared /Date Verified /Date Sheet #
Continued On I7 ce,20 P531750 NEW 9 89
.-,y
1 r:
Aopendix 2: Pressure correction due to submergence of FW sparger p RPV := 1020 psi at rated power.
l l
Tsat given Psat s
3 Units:
MPa 10 Pa p 0 := 1 MPa kj e 10 joule l
p0 = 145.038 psi 2
A :=.42677610 8 :=.3892710' C :=.94865410 8
TK( p) := A +
TF(p) := (TK(p) - 273.16) S + 32 l
in
+C 5
(PO
)
T T RPV = 546.92 Then:
1 RPV := RHOF(DEGK(T P
RPV P RPV = 46.174-Pressure correction:
i Normal reactor water level (RWL) elevation:
Z RWL := 537.5 in (APED-B11-008)
FW sparger elevation:
2 pws := 461 in (APED-B11-008) gwg := (Z RWL-2 FWS)'P AP RPV Then:
AP pwg = 2.044 psi Iowa Electric Light and Power Company Cedar Rapids, Iowa B
PX4 A $2, rhtk 0 11hhhrt 5W43 Doc # Chl'N0 **Cl 2-Rev#
Rev Prepare 6/Date Ventied/Date Sheet #
Continued On /f oe M
(,. o "
Aooendix 3: Instrument Uncenainties q
The use of PT1637 and FT 4563, PT4564 to determine a differential pressure is a trending activity performed under normal environmental conditions and power supply conditions. These conditions will also be consistent with the conditions during instrument calibration. Therefore, the following potential causes for instniment error are assumed to be negligible: Accuracy Temperature Effect, Overpressure Effect, Static Pressure Effect, Seismic Effect, Radiation Effect, Humidity Effect, Power Supply Effect, and effects due to RFUEMI.[12]
Instrument Accuracy Range := 2000 psi VA_PT1637 := 0.005 Range 0.5% Accuracy [9]
2-sigma j
Range := 1200 psi VA_PT4563 := 0.005 Range 0.5% Accuracy [10]
2-sigma 1
j Drift VD_PT1637_6mo := VA_PT1637 Assumption 2-sigma VD_PT4563_. no := VA_PT4563 Assumption 2-sigma I
Cal _Intvl_PT1637 := 2 yr 12 E yr Cal _Intvl_PT4563 := 1 yr 12 -
yr
' Cal _Intvl FT1637 D_PT1637 :=
VD_PT1637_6mo 6 mo 3
D_PT4563 :=
VD_PT4563_6mo 6 mo 3
4 4
Calibration H_PT1637 := 2 psi Heise Guage 0-2000 psig Accuracy (11]
F_PT1637 := 0.005 000 50 psi Fluke Voltmeter Accuracy (11]
50 j
Range = 1500 psi H_PT4563 := 0.001 Range Heise Guage 0-1500 psig Accuracy [11]
I F_PT4563 := 0.001 = 1 00 50 psi Fluke Voltmeter Accuracy [11]
50 lowa Electric Light and Power Company Cedar Rapids. Iowa 6
/X#//# r/n/tr bJ $11 hum N(2/04 Doc # fAL-4/M di 2_
Rev#
I Rev' Prepared!Date Verified /Date Sheet #
Continued On I4 4 e,Lt
i,-
Cstd_H_PT1637 := H PT1637 Calibration Standard Accuracies assumed to be at 4
least Csid_F_PT1637 := F_PT1637 4 times better than the instrument under calibration.
4 Cstd_H_PT4563 := H_4 63 4
Cstd_F_PT4563 := -
4 j
1 2
2 2
2 C_PT1637 := H_PT1637 + Cstd_H_PT1637 + F_PT1637 + Cstd_F_PT1637 C_PT4563 := )H_PT4563 - Cstd_H_PT4563 2
2 2
2
+ F_PT4563 + Cstd_F_PT4563 U := f.dVA_PT1637 2
2 2
2 2
2
+ VA_PT45G3 + C_PT1637 + C_PT4563 + D_PT1637 + D_PT4563 U = 40.317 psi 3-sigma (95% Confidence Interval for Instmment Enor)
Derived Units:
3 gpm = 0.1337 ft degF = 10 psia = psi psig = psia - 14.7 lbf min in mv = vott 10-3
- O' i
l Iowa Electric Light and Power Company Cedar Rapids. Iowa B
(TA%E 7//L/9 3 RikC 51/2K/ %
Doc # Ck~M Ab SI2-Rev#
Rev Prepared /Date Venfred/Date Sheet #
Conttr:ued On 2Cloc)4
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