ML19316A493
| ML19316A493 | |
| Person / Time | |
|---|---|
| Site: | Oconee  |
| Issue date: | 06/09/1970 |
| From: | Cady K US ATOMIC ENERGY COMMISSION (AEC) |
| To: | Long C US ATOMIC ENERGY COMMISSION (AEC) |
| References | |
| NUDOCS 7912200747 | |
| Download: ML19316A493 (11) | |
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UNITED STATES 3
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ATOMIC ENERGY COMMISSION l
.__L- } - l WASHINGTON. D.C. 20*,45 l
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i June 9, 1970 Charles G. Long, Chief, PWR Prc. :t Branch 2 Division of Reactor Licensing REVIEW OF LOSS-OF-FLOW TRX;SIENT ON OCONEE
References:
1)
B&W Loss-of-Flow Transients, letter to A. Schwencer from K. B. Cady, May 11, 1970.
2)
Duke Oconee FSAR, Docket Nos. 50-269, 50-270, and 50-287 I have reviewed the ability of Duke-Oconee to handle the four-pump loss-of-flow transient without reaching an unacceptable DNB ratio. The applicant, in Reference 2, gives an acceptance criterion that DNBR shall not go below a DNER of 1.55 which is the hot channel DNBR at 114% rated power.
Because of errors and non-conservative assumations in the Oconee analysis, it. is likely the B&W design cannot meet this criterion.
This is based on a comparison with the Turkey Point, three-loop plant which reaches a DNER l
of_l.31 on the loss of all coolant pumps.
I have attached a detailed dis-cussion of this comparison and a description of the pump coastdown model whi ch <!nv r:: l'conee's calculuLien to be in error.
I met with W. R. Smith and J. Mallay of B&W on May 7 or 8 and told them of some of my concerns and gave them a copy of the enclosed at tachmen ts.
As a result'they said that B&W, through Duke, would correct the analyses of loss-of-flow transients using:
(1) a more correct, four-pump coastdown ti=e constant of 8 or 9 seconds, in place of the 12-second time constant used in the Duke FSAR; (2) a corrected discussion of the trip delay times; (3) a corrected abscissa scale on Figure 3.6; (4) more conservative initial power and temperature assumptions to reflect
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deadband in addition to instrument error.
They-also indicated they would provide a more complete discussion of the
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two-pump coastdown transient which forms the basis for the power-to-flow reactor trip setting.
They indicated they were not planning to provide a discussion of why no low pump frequency trip was provided, and suggested that if we had concerns about the possibility of underfrequency from the grid, we should address these to Duke..
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J82229747
Charles G. Long 2
June 9, 1970 I was told that the Duke /B&W response to question 3.1.4 (February 14, 1970) about sene enrichments was a whitewash which was provided in order not to leave the question unanswered in the public record.
They said the zone enrichments were proprietary and DRL would receive the requested information in a letter.
if* oda/s
~
/ L K. B. Cady PWR Project Branch 2 Division of Reactor Licensing
Enclosures:
1.
Comparisen of B&W & W Pump Coastdown 2.
Pump Coastdown Model cc:
R. C. DeYoung A. Schwencer D. F. Ross R. R. Povell R. A. Birkel J. A. Murphy T. M. Novak Docket (3)
PWR-2 Reading 6
May 7, 1970 COMPARISON OF B&W AND WESTINGHOUSE PUMP C0ASTDOWN I am concerned that the Oconee design is less able to accommodate loss-of-flow transients than typical Westinghouse designs such as the three-loop Turkey Point plant.
The power-to-flow reactor trip signal setting uses loss-of-flow transients as their Tech Spec basis.
My basic concern is that the higher system pressure drop in the B&W design results in a shorter coastdown time for reactor coolant flow in the event of loss of power to 'one or more reactor coolant pumps.
The attached table shows a comparison of the Westinghouse (Turkey Point) and E&W (Oconee) designs.
As an example, I take the case of all pump coastdown with reactor trip occurring by a loss of pump pcwer signal. The B&W analysis shows a mini-mum DNBR of 1.78 compared to a minimum DNBR of 1.31 for the Westinghouse design.
This comparison is misleading because the Westinghouse analysis is more conservative due to the following assumptions:
1)
Westinghouse uses a +4* F initial temperature error compared to
+2* F for B&W.
The Westinghouse 4' F number has 2* for deadband and 2* for ins trument error.
The B&W analysis has only instrument error.
Two degrees Fahrenheit is wertH about 0.05 cn DNBR.
2)
Westinghouse uses a 4% initial power error compared to 2% for B&W.
This 2% difference is worth about 0.08 on DNBR.
3)
Westinghouse uses a delay time of 1.6 seconds (time to initial rod motion) while B&W uses 0.5 second.
4)
Westinghouse uses a coastdown time constant of 11 seconds (my calcu-lation is 12.6 seconds) whereas B&W uses a coastdown time constant of 12 seconds (my calculation is 9.4 seconds).
With respect to the general ability of the B&W design to accommodate 1 css-of-flow transients (loss of ac power, loss of all pumps, loss of one pump)
I expect their design to be less capable than &stinghouse design for the following reasons:
a 1)
The coastdown times according to my calculations are 12.6 secends for Westinghouse and only 9.4 seconds for B&W due primarily to the greater system pressure drop. The Westinghouse analysis uses a conservative time constant of 11 seconds.
The B&W analysis-uses a-very non-conservative time constant of 12 seconds.
They have. prob-ably made an error.
1 k
,~--s_-----_.------,e--
2 2)
The minimum steady state pressure of the B&W system is 65 psi below nominal pressure and the Westinghouse minimum steady state pressure is only 30 psi below nominal.
This 35 psi difference is equivalent to about 0.05 on DNBR.
3)
The vent valve design for B&W poses the problem of a stuck or leaking valve which can reduce core flow and aggravate loss-of-flow transients.
4)
The Westinghouse design has an under-frequency pump motor trip; none is provided for B&W.
5)
The Wes tinghouse design has a reactor coolant pump circuit breaker trip for loss of a single pump.
The B&W design has no pump breaker trip for loss of a single pump and must rely on the power-to-flow comparator trip which is slower acting.
There is one of fsetting reason that the B&W design is more conservative than the Westinghouse design.
The hot channel DNBR is 1.81 in the Westinghouse design and 2.0 in the B&W design.
Turkey Point Oconee-wncrinennuce Dr.W thermal power 2200 MWt 2568 flow rate 101.5 x 106 lb/hr 131.3 x 106 avg mass velocity 2.32 x 106 lb/hr-ft2 2.53 x 106 I
nominal pressure 2250 psia 2200 minimum steady state pressure 2220 psia 2135 nominal hot channel DNBR 1.81 2.0 design overpower 112%
114%
DNBR G overpower 1.3 1.55' power density 83 kW/t 83 kW/t average linear heating 5.5 kW/ft 5.656 inlet temperature 546.2* F 554' F vessel outlet temperature 602.1* F
- 604.7* F hot channel outlet temperature 642' F-642.8 4
All pump loss-of-flow transient:
-30 psi-
.-65
+4' F
.+2' F
ATin 104%
102%
overpower initial DNBR 1.63 time to rod motion 1.6 sec.
0.5 sec.
E assumeducoastdown time constant 1 11 sec.
12 sec.
my calculated time censtant 12.6 sec.
9.4 sec.
time to 80% flux 2.25 sec.-
1.2 sec.
time to minimum DNBR 2.25 sec.
9 4
4
May 1970 PUMP COASTDOWN MODEL Summary Using a simple model, described below, I find that the all-pump flow coast-down for a PWR gives a relative reactor coolant flow rate as a function of time after' loss of pump power, 9 " l + t/:
where : is a time constant for the coolant loop with its pump and motor.
This model well describes the time behavior assumed by applicants in loss-of-flow transients.
A comparison of the calculated time constants for Oconee cnd Turkey Point are given in the following table:
Duke (B&W)
Turkey Point (W)
Calculated time constant 9.4 see 12.6 see Applicant's assumed time 12 see 11 sec constant The Turkey Point application assumes a time constant of 11 seconds which is conservative for the sy: tem which' has a time constant of 12.6 seconds.
They use high estimates lof the locp pressure drops.
By comparison, the.
Duke application uses a non-conservative time constant of 12 seconds which is considerably above the more likely estimate of 9.4 seconds.
An es ti-mate of about 8 seconds (instead of 12) would have made Duke's loss-of-flow transient comparable to the Westinghouse calculation.
Starting at 102% rated power, the FSAR's of the two applications indicate the loss-of-flow transient is worse for the Westinghouse three-loop plant (minimum DNBR = 1.3 at 2.25 seconds) than the B&W analysis (minimum DNBR =
1.78).
In reality the B&W plant coasts down f aster than the Westinghouse plant and is less able to sustain -loss-of-flow transients.
We should consider careful verification during pre-op testing of the flow coastdown' characteristics of Oconee in order to verify the model' they have. assumed.
Model Description
. The simplest coastdown model for a fluid. loop is
' Time rate of. change'
' Power dissipated I
' Pumping power added'
,of K.E. of= fluid.
,in-loop resistance)
,to fluid by pump i
1 h
7 M
4 2
Because the KE of the fluid is proportional to the flow rate squared, and assuming the power dissipated in the loop is proportional to flow rate cubed, we obtain f+Q = H; Q(o) = 1 (1)
To where the time constant t = 2E /P is twice the ratio of the initial o
o o KE of the fluid and the initial hydraulic power, Q is the ratio of flow rate to initial flow rate, and H is the ratio of the pump head to the initial pump head.
In a similar manner, the pump equation is
' Time rate of chr ae Power delivered by
' Power delivered to '
,of KE of pump & motor, the pump shaft
, pump shaf t by motor; During coastdown, the power delivered by the motor is zero.
Because the KE of the pump and motor is proportional to the shaf t speed squared, and the pcwcr delivered by the pump shaf t (to the water) is the pr; duct of the shaf t speed and the torque, we get dQ(t) + T = 0 ; G(o) = 1, (2)
Tp dt where the time constant 2E P
7 -
P P /n o o is twice the initial KE of the pump and motor divided by the initial power delivered by the pump shaft, G is the ratio of the pump speed to the initial speed, and T is the ratio of the shaf t torque to the initial torque.
The initial hydraulic ef ficiency of the pdmp is n.
~
o 1
Centrifugal pumps are characterized by two independent parameters, usually taken to be speed and flow rate.l>
Writing the torque. and head in terms of these-variables gives equations of the form T =-T(Q,0), H = H(Q,0),
(3)
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1)
A.
J'." Stepanoff, Centrifugal and Axial-Flow Pumps, Wiley 1957 (see Chap ter :13) '
H N
...m
i 3
which together with the l' op and pump differential equations above, give o
enough information. to solve the coas tdown problem. Typical pump curves look like i
l 1 - -..,.
T A ' ' ~ ~ ~
T I
I N
2 N
I
- /1. !
'N :
m N
G
\\1.
G \\
w, 3 A
\\
A'o f
_n o
\\
-.1 m r o
-z.o The simplest model is to assure that the pump torque and head are propor-rt on.11 T=H (4) and that the flow rate and pump slow down together I
Q = 0.
(5).
This gives 0; Q(o).= 1,
' (6)
T
+Q
=
where the time _. constant of the loop-pump sys tem is 2(E
+n'E )'
P
.T = T +T (7)-
=
op
. P, the only number needed to characterize the system coastdown.. The quantity Eo +'n E is, the sum of the kinetic energy of - the, loop. fluid oP plus the fraction of the initial' pump KE which will be dissipated ~in the
- resistance of the fluid loop cutside the pump.
The' remaining fraction
-(1-n )E will be dissipated as heat..as the water flows through the pump.
P
.In. this model the pump ef ficiency Ldefina.d 'as.
-4 s
4
' f'
- I--
e
,e
4 nE'n,ff (8) is constant at the initial efficiency n.
The solution to Eq. (6) is 0 " l + t/T'
(
and the time constant T is the' time for the ficw to fall to half of -
zits initial value.
The~above model doesn't seem to appear in the literature (it must, but I didn' t find it).
Yokomura ). has a related model where the time constant is given by 2
2n (E +E)
P T=
(10) y o
apparently duc to an arror.
Burgreen3) has a much more complicated model which aJsumes that the pump characteristics, Eqs. (3) are 2
H = 0',
T=0 (11) which result is a pump equation given by dn p gg- + 0 = o; Q(o) = 1 (12)
T whose sclution is 0=1+
77 (13) p 2)
T. Yokomura, Flow Coastdown in' Centrifugal-Pump Systems, Nuclear Engineering)and Design, 10,, 250 (1969).
3)' : D. Burgrcen, Flow Coastdown -in a Looo Af ter Pumping Power Cutof f,
. Nuclear. Science. Engineering,. 6.1306 (1959).
b r
i
5 This gives a flow equation, 1
-f+Q Q(o) = 1 (14)
T
=
y + gj7 o
i pi 4
whose solution is (S+1)(S-1+ S-)(aT+1)0 - (S-1) (S+1 S-)
Q = 2(cT+1)
(S+1 2)+(S-1 + 1') (aT+1) S (15) a o
a where a, 8 and T are defined as a 5 -[, SE $ - + 1, T E t/T (16)
\\ 3 9
P a
Thic ic a tt.'o paraneter t'T
&T ) solution and is much ecre cc=clex than the one parameter solution $ Eq. 9.
Bo th ' Eq. (9) and Eq. (16) have the identical behavior for large and small a and direct calcula-tion shows that they are never different by more than 6% at a = 1.
This is well within the accuracy of either model. -Therefore, Eq _9, because it is much simpler is preferable, unless the actual pump characteristics are used.
The Oconee FSAR shows-a design head of 365 feet (Figure 4-8) and a flow rate of 131.3 x 106 lbm/hr resulting in a hydraulic power of 6
, 365 x 131.3 x 10
- 6
= 13.32 x'10 ft-lb/sec,
.p o
3600 g
db e
The kinetic energy.of the pumps and motors at a speed of 1190 rpm, and 2
a moment of inertia' of 70,000 f t -lbm per pump is 9
1 2, 4 x 70,000 gl190 x 22)
= 67.5 x 10 ft-lb.
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6 E
Ju
=
p 2g 64.4 60 c
Y e
I m
p
6 This gives a time constant 'for the pump and motor, using an hyiiraulic efficiency of 85%,
P 2x0 5 x 67.5
8.6 seconds, T
p p
o The kinetic energy of the fluid in the primary system is small and is made up of:
3 a) 4 cold legs G 370 ft of 554*F water moving 46.6 ft/sec.
b) 2 hot legs @ 490 f t of 604*F water moving 61.5 ft/sec.
3 c) reactor core of 650 f t of 579*F water moving at 15.7 ft/sec.
3 d) 2 steam generators @ 1500 f t of 579'F water moving 16.4 f t/sec.
This is a total flow energy of 6
E = 5.37 x 10 ft-lb, with a resulting time constant of 6
2 x 5.37 x 10 0.8 seconds, T =
=
a o
13.32 x 10 The resulting, ' combined time constant is 1
T=-t
+T
= 9.4 seconds, o
P which compares with the non-conservative value of 12 seconds given in Figure 14-16.
A more proper, calculation using conservatively high system pressure drops should be about 8 seconds.
The Turkey Point FSAR shows a ~ design head of 254 feet (Figure 4.2-6) and a flow' rate of 101.5 x 106 lbm/hr resulting in a hydraulic power of 6
254 x 10.5 x 10 7.16x-lbb ft-lb/sec.
P.=
x c-S 7
L A-
~
7 The kinetic energy.of the pumps and motors at a speed of 1180 rpm, and a 2
moment of inertia of 70,000 f t -lbm per pump is (1180 x 21)2 1
2 3 x 70,000 6
E =
Ju 49.8 x 10 ft-lb.
=
=
2g 64.4 60 C
)
This gives a time constant for the pump and motor (using an hydraulic.
efficiency of 85%)
i 2nE i
p-2 x 0.85 x 49.8 11.8 seconds.
T =
=
=
p 7.16 o
i The kinetic energy of the fluid in the primary system is small and is made up of 3
a) 3 suction side cold legs @ 176 ft of 546*F water moving 38.7 ft/sec.
3 j
b) 3 dicchcrge cide cold icg: 0 175 ft of 546*F water _ moving 49.2 ft/ccc.
c).
3 hot legs @- 100 f t of 602*F water, moving 48.6 f t/sec.
t' 3
i d) reactor core of 650 'f t of 574*F water moving 14.3 f t/sec.
e) 3 steam generators of 928 ft of 574*F water moving 19.9 f t/sec.
I This is a total of l
6 E = 2.79 x 10 ft-lb, i
o 5
giving a time constant of -
2E T, = p
= 0.8 seconds, o
The _resulting, combined -time constant is i
-T =.T
+T
= 12.6 seconds, O.
. _p
-- which compares with the value1of jll seconds L given'on F'igure 14.1.9-1.
L 1Their value is obtained using high estimates.of loop pressure drops _in order ' to be: conservative.
e I
T i
F T 4