ML20101M580
| ML20101M580 | |
| Person / Time | |
|---|---|
| Site: | Crystal River  |
| Issue date: | 04/04/1996 |
| From: | FLORIDA POWER CORP. |
| To: | |
| Shared Package | |
| ML20101M505 | List:
|
| References | |
| M-96-0004, M-96-0004-R00, M-96-4, M-96-4-R, NUDOCS 9604050447 | |
| Download: ML20101M580 (9) | |
Text
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ANALYSIS / CALCULATION
SUMMARY
l DISCIPUNE CONT 50L NO.
WSION LEVEL DOCUMENT IDENTIFICATION NUMBER M
96-0004 0
ym CLASSiFCATION (CHECK ONE)
MUT DISSOLVED GAS EVOLUTION ANALYSIS
$ safety Rei ted O Non sar tr R.iat.o MAR /SP/CGWR/PEERE NUMBER VENDOR DOCUMENT NUMBER N
ITEMS REVISED APPROVALS Design Engineer ul;/7~[jd Date Verification Engineer ((d Dete/ Method
- R Supervisor Date
- VERIFICATION METHODS: R - Design Review; A - Alternate Calculation; T - Qualfication Testing DESCRIBE BELOW IF METHOD OF VERIFICATION WAS OTHER THAN DESaGN REVIEW q
PURPOSE
SUMMARY
The purpose of this calculation is to determine the effect of the dissolved aas in the MUT water on final tank pressure after -
the MUT is drawndown durina a LOCA.The results of this analysis will be used in the MUT/BWST HYTRAN analysis (M96-00071
~r
,- d RESU(TS
SUMMARY
The results of the analysis show that the maximum H pressure increase due to the draindown is less than 0.20 psi. The 2
realistic asessment of pressure increase due to reevolution is approximately 0.09 psi.
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9 Florida DESIGN ANALYSIS / CALCULATION Power Crystal River Unit 3 tm Page 1 of 8 oc;c we%1 a.~recom eo.
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i Table of Contents Section Description Page 1.0 P u rpo s e....................................
2 2.0 Design in puts................................
2 3.0 Assu m ptions.................................
2 4.0 References..................................
3 5.0 Calculations................
3 6.0 R e s u lts.....................................
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Crystal River Unit 3 l
Page 2 of 8 DOCU8AENT OEtwTWsCATON NQ HEvi3ON M-96-0004 0
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1.0 Purpose
The purpose of this calculation is to determine the effect of the dissolved gas in the water on final tank pressure after the MUT is drawndown during a LOCA. This effect will be the increase in pressure resulting from the evolving of the dissolved gas from the water space to the gas space within the MUT. This re-evolution of the gas takes place because the partial pressure of the hydrogen in the gas space is decreased as water is removed from the MUT during this event.
2.0 Design inputs:
The MUT volume is taken from Ref. 4.1.
The starting pressure / temperature for the MUT contents is obtained from Ref. 4.3 and pressure gage instrument error is obtained from Ref. 4.7 The value for Henry's constant was obtained from reference Ref. 4.2
3.0 Assumptions
3.1 The contents of the tank remain at a constant temperature during the entire 3
drawdown. This assumption is based on the amount of heat that would be transfer i
during the event as being very small compared to the total heat contained by the r
tank contents. As the level in the MUT drops to the lower region of the MUT below zero level indication, the contents of the tank will start to lose a larger percentage of u
its heat but this value will still be very small compared to the total heat still contained within the MUT.
. 3.2 The state point properties of the gas within the MUT can be determined by use e
of the perfect gas law. The pressures and temperature of the gas during this event-
- are within the boundaries for application of this law.
1 4
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3.3 Transient is sufficiently slow that the dissolved gas in the water in the MUT can "re-evolve into the gas space. Henry's law is valid for static conditions. This transient is considered to proceed at a rate which will allow the hydrogen in the water and the gas to achieve equilibrium with each other because of the decrease in hydrogen partial pressure of the during the drawdown.
3.4 The overgas in the MUT will be considered all hydrogen. Plant records show that the gas space in the MUT will have hydrogen and nitrogen in the gas space.
However, hydrogen has greater solubility in water than nitrogen and therefore will have a greater effect on the resulting pressure increase after drawdown, i.e., more gas is available to re-evolve.
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DESIGN ANALYSIS / CALCULATION Mglorida N
Crystal River Unit 3 Page 3 of 8 DOCUMtNT OtNN LATIUN NQ fgvlSON M-96-0004 0
Other assumptions are contained within the body of the calculation and require no later confirmation.
4.0
References:
4.1 Crystal River Unit 3 Calculation No. M-96-0007 4.2 Water Chemistry Manual for 177FA Plants, B&W Nuclear Technologies, BAW-1385, Rev. 6, Dec.1992 4.3 Crystal River Unit 3, Operating Procedure, OP-1038, Rev 15 4.4 Crystal River Unit 3 Piping Isometrics 4.5 Crane Technical Paper 410 4.6 ASME Steam Tables 4.7 Calculation DC-5515-018-26.01-ME, "BWST Suction & MUT Suction Unes Tie-in Point Pressures", Rev 5
5.0 Calculations
Start with the tank level at 52.3" and the overgas pressure at 13.2 psig (Ref. 4.3, page 33 &
34 and Ref. 4.7).
The MUT tank contents are at 135 *F and remain in that condition through-out the transient, i.e., isothermal expansion (see assumption 3.1). The pipe from the tank to the MU pump suction header. is 4" and has approximately 85 feet of horizontal run. There is approximately 16 feet of vertical run. There is 14.5 feet of vertical run from the tank to the horizontal section (see Ref. 4.4).
The following are the volumes of the MUT (see Ref. 4.1):
TOP & BOTTOM HEADS, TEACH 48.234 ft' VOLUME OF SHELL 507.202 ft' TOTAL TANK VOLUME 603.669 ft The vertical 14.5 feet of 4" schedule 40 pipe has a volume of 0.088 X 14.5=1.276 ft' The horizontal 85 feet of 4" schedule 40 pipe has a volume of 0.088 X 85=7.48 ft The remaining 1.5 feet of 4" schedule 40 pipe has a volume of 0.088 X 1.5=0.132 ft (Pipe volume obtained from Ref. 4.5)
Total volume of the pipe is 8.888 ft' Because assumption 3.1 maintains the MUT temperature constant, the partial pressure of the water vapor remains constant at 2.5375 psia (Ref. 4.6) during the transient. Hydrogen, H, will be used as the cover gas ( see assumption 3.4) yielding a conservative estimate of 2
off gassing as the total pressure decays due to drawdown.
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DESIGN ANALYSIS / CALCULATION Mglorida N
Crystal River Unit 3 Page 4 of_8 uxwm uwwm,<3 mi3.
M-96-0004 0
Use a Henry's Law Constant for H of 0.9245 Kg H 0-psia /cc-H (STP)( see Ref. 4.2, page 2
2 2
1-54) for the temperature of interest. The reference has no corrections for the influence of pressure on this value. For this calculation, an assumption is that the influence of the gas's partial pressure on it's solubility is negligible over the range of interest and that the B&W constant will remain constant over this range.
The total pressure is 13.2 psig. Using an atmospheric pressure of 14.7 psia, the absolute total pressure at the start of drawdown is pr = 13.2 + 14.7 = 27.9 psia Of the total pressure in the tank, the vapor pressure will remain constant at 2.5375 psia.
Therefore, at the start of the drawdown, the partial pressure of the H is 2
ps = 27.9 -2.5375 = 25.3625 psia The mass of H in the gas space at a level of 52.3" can be determined by the perfect gas 2
law ms = pn(144)V,/(2T)
Where: ps is the partial pressure of H,25.3625 psia 2
V, = 333.586 ft'(Ref. 4.1) 2 = 767.0 lbf-ft/lbm "R (Ref. 4.5)
T is the absolute gas temperature, 135 + 460 = 595 R ms = 2.6696236 lbm At the start of the tank drawdown, the following is the amount of water in the tank and i
P pe, V, = Vwr + V, = 270.083 (Ref. 4.1) + 8.888 V, = 278.971 ft' The amount of water in Kg in this water volume is M = V, p,/2.205 Where: p, = 61.46282 lbm/ft, water density at 135 F, (Ref. 4.6)
M = 278.971 X 61.46282/2.205 = 7776.04 Kg m
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DESIGN ANALYSIS / CALCULATION Mglorida N
Crystal River Unit 3 Page 5 of 8 cumur u.nrwurm uo.
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The maximum concentration of H dissolved in the water is 2
Cs = ps/C,w i
Where: C,w = 0.9245 Kg H 0-psia /cc-H (STP) (Ref. 4.2) 2 2
Cs = 25.3625/0.9245 = 27.433748 cc-H (STP)/Kg H O 2
2 Therefore, the total volume of H dissolved in the water is 2
Vs.w = M X Cs Vs.w = 7776.04 X 27.433748 = 213325.92 cc-H (STP) 2 l
The mass of H dissolved on the water is 2
m.w = RW)V.w/E s
s Where: P is standard pressure,14.7 psia T is standard temperature,68 F or 528 R 2 is as defined before
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m.w = 14.7 X 144 X (213325.92/28317.02)/((767) X 528) s m.w = 0.0393774 lbm s
When the MUT is completely drawndown to the entrance to the 6" header, the volume for the gas is now
. ~..
Vooo = V, + V, = 333.586 + 278.971
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= 612.557 ft'
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Now, assuming the partial pressure of the water remains at 2.5375 psia and if the mass of the H is still the same, 2.6696236 lbm, the partial pressure of the H is 2
2 psoo = m H/Vooo = 2.MMM X 99) X m/04 X M2.W) s
= 13.811838 psia Therefore, the total pressure is now proo = psoo + p, = 13.811898 + 2.5375 = 16.349398 psia in u s. o, s.m...~...,e....,
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Mglorida DESIGN ANALYSIS / CALCULATION N
Crystal River Unit 3 Page 6 of 8 DOCUMLNI OLNt5CATON NQ.
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if the total mass of H dissolved in the water at the start of drawdown is re-evolved into the 2
cover gas at the end of drawdown, the total increase in total pressure would be msrs = 2.6696236 + 0.0393774 = 2.709001 lbm Again, using the perfect gas law, the partial pressure of the H can be determined 2
psoo.,, = m isENooo = 2.709001 X (767) X 595/(144 X 612.557) n
= 14.015626 psia And the total pressure would be pmo = Psoo a + p, = 14.015626 + 2.5375 = 16.553126 psia The above result is a worst case estimate of the increase in pressure due to potential gas re-evolution from the decrease in H partial pressure at the end of drawdown.
2 Since the total H contained in the MUT water at the start of drawdown can not physically 2
be re-evolved, a better estimate of the amount of gas re-evolved is determined by the following H mass balance on the tank:
2 srs = m,, + rn, + m i, m
s s
a where: m
= 2.709001 lbm srs m,, = mass of H in the gas volume at the end of drawdown s
2 m,, = mass of H dissolved in the water volume at the end of s
2 drawdown vatt m,, = mass of H which left the tank dissolved in the water s
2 tynW removed during the drawdown. The H concentration!through-2 out the drawdown is assumed to be the same H,. concentration
~r when the last of the water is removed, i.e., the c'oncentration of i(
H in the water at the final partial pressure of the H -
2 2
Conservative assumption since more H is left in the tank.
2 The m,, term can be determined by the perfect gas law:
s m,, = ps,14 Vooo # T s
where: ps,is the final H partial pressure 2
Vooo, R & T are as given before
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DESIGN ANALYSIS / CALCULATION Mglorida rU Crystal River Unit 3 Page 7 of 8 ca, m r w e orm,o rosa M-96-0004 0
The m,, term can be determined from:
s m, = C ps, V.,
s where: C is a constant which converts the water volume, V.,, into the amount of water mass in kg and determines the concentration of H dissolved in the 2
water in cc/kg of water at STP. The numbers used in C are as follows:
C = 61.46282(14.7)144/(28317.02(2.205)2(528)0.9245) 61.46282, density of water, Ibm /ft 14.7. standard pressure, psi 2
2 144, conversion, in to ft 28317.02, conversion, cc to ft' 2.205, conversion, Ibm to kg 528, standard temperature, *R 0.9245 kg-psia /cc(STP) for H (Ref. 4.2) 2 C = 0.0042687/2 The m, term can be determined from:
s msw = C ps, ( V., - V.,)
where the terms are as given before and V,is the volume of water left when drawdown is completed.
Substituting these terms into the mass balance equation yields the following:
2.709001 = ps,144 Vooo /2 T + C ps, V, + C ps, ( V. - V,)
Collectinhierms and solving for the partial pressure of H after water has been removed 2
from the tank yields the following:
ps, = 2.70900M / (0.New Vooo + 0.0042687 V. )
Substituting for Vooo 612.557 ft and for V 278.971 ft' yields:
ps, = 2.709001(767)/(0.2420168(612.557) + 0.0042687(278.971))
= 13.903938 psia The total pressure in the tank is P, = 13.903938 + 2.5375 = 16.441438 psia b 95 R E T. L/e of F.atti RE5P Pesear Eng<neenng
Mglorida DESIGN ANALYSIS / CALCULATION M
Crystal River Unit 3 Page 8 of 8 N_* "ENI OLNiiG(,A YON NO FtE(SeON M-96-0004 0
6.0 Results
The following table presents the results of the final pressure when starting at a level of 52.3' and a pressure of 16.22 psig.
l Results Table Method Final Pressure in MUT, psia A pressure from no gas re-
)
evolution, psi i
No Gas Re-evolution 16.349398 0.0 All Gas Dissolved in Water 16.553126 0.203728 released to MUT l
Gas Mass Accounting 16.441438 0.0920397 The very conservative approach of assuming all the gas dissolved in the water is re-evolved yields a maximum A pressure increase of 0.203728 psi. A mass accounting method which is still conservative, since the water leaving through-out the event is assumed saturated with dissolved H at the MUT event ending pressure, yields a A 2
pressure increase of 0.0920397 psi. Therefore, effects of re-evolving H within the MUT will 2
not exceed 0.203728 psi additional pressure.
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