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=Text=
=Text=
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ANALYSIS / CALCULATION  


==SUMMARY==
==SUMMARY==
 
l t
t DISalPUNE             CON'HOL NO.             REVISON LEVEL DOCUMENT IDENTIFICATION NUMBER                               M                 96-0006         0 Tm.E                                                                                           CLASSWCATON PIECK ONE) l MUT VAPOR PRESSURE EVALUATION                                                                   % safety Related l                                                                                                                 Non safety Related MAR /SP/COWR/PEERE NUMBER l
DISalPUNE CON'HOL NO.
l                                                                                                           VENDOR DOCUuEwi NUuscR l
REVISON LEVEL DOCUMENT IDENTIFICATION NUMBER M
REVISION                                 ITEMS REVISED                                             I APPROVALS                                                                                           ,
96-0006 0
Design Engineer                   [//$ [
Tm.E CLASSWCATON PIECK ONE) l MUT VAPOR PRESSURE EVALUATION
Date Verification Engineer           hf                                                                                                       l Date/ Method
% safety Related l
Non safety Related MAR /SP/COWR/PEERE NUMBER l
l VENDOR DOCUuEwi NUuscR l
REVISION ITEMS REVISED APPROVALS Design Engineer
[//$ [
Date Verification Engineer hf Date/ Method
* R Supervisor Date
* R Supervisor Date
              " VERIFICATION METHODS: R - Design Review; A - Alternate Calculation; T - Qualfication Testing DESCRIBE BELOW IF METHOD OF VERIFICATION WAS OTHER THAN DESIGN REW!EW 1
" VERIFICATION METHODS: R - Design Review; A - Alternate Calculation; T - Qualfication Testing DESCRIBE BELOW IF METHOD OF VERIFICATION WAS OTHER THAN DESIGN REW!EW 5
l 5
PURPOSE  
PURPOSE  


==SUMMARY==
==SUMMARY==
The ourpose of this calculation is to determine the vapor pressure in MUT-1 durina draindown followiria a postulated LOCA.
The ourpose of this calculation is to determine the vapor pressure in MUT-1 durina draindown followiria a postulated LOCA.
                                                                                                                                .. a.
.. a.
1r-              .                                                                                          ;13g         , - a.y
;13g
                .s -   4, .
, - a.y 1r-
                                                                                                                              - 4. k a ., - # . $
- 4. k a., - #
            ' ._ e                   I?                                                                                   E 1. d             T2i RESULTS  
.s -
4,.
'._ e I?
E 1.
d T2i RESULTS  


==SUMMARY==
==SUMMARY==
'Thetesults of the analysis show that the vapor pressure decreases a minimum of 0.2 psia during the draindown period following a LOCA.
9604050449 960404 PDR ADOCK 0500 2
%,e3


          'Thetesults of the analysis show that the vapor pressure decreases a minimum of 0.2 psia during the draindown period following a LOCA.
DESIGN ANALYSISICALCULATION Mglorida N
I 1
Crystal River Unit 3 Page 1 of 6 CKJGUMENT IQ[NTflCATON NO.
l l
F(VL1WUN M-96-0006 0
9604050449 960404
            % ,e3 PDR        ADOCK 0500          2                                      ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
 
DESIGN ANALYSISICALCULATION Mglorida N               Crystal River Unit 3 Page 1       of 6 CKJGUMENT IQ[NTflCATON NO.                                                             F(VL1WUN M-96-0006                                                                             0


==1.0 PURPOSE==
==1.0 PURPOSE==
The purpose of this calculation is to determine the vapor pressure                                 ;
The purpose of this calculation is to determine the vapor pressure change in MUT-1 during draindown following a postulated LOCA.
change in MUT-1 during draindown following a postulated LOCA.                           The results of the calculation will provide input to the hydraulic calculation which determines the actual draindown level.
The results of the calculation will provide input to the hydraulic calculation which determines the actual draindown level.
2.0       DESIGN INPUT:
2.0 DESIGN INPUT:
2.1       The tank volume used is provided in Reference 4.1.
2.1 The tank volume used is provided in Reference 4.1.
The initial gas volume is         333.586     ft' The total tank volume is           603.669     ft 3
The initial gas volume is 333.586 ft' 3
The dished head volume             48.234     ft' / per head 2.2       The initial minimum liquid level, pressure and temperature in the MUT-1 is:
The total tank volume is 603.669 ft The dished head volume 48.234 ft' / per head 2.2 The initial minimum liquid level, pressure and temperature in the MUT-1 is:
Level = 52.3"                           Reference   4.7 Pressure = 12.08 psig                   Reference   4.7 Pressure Indicator error 1.12 psig Reference         4.9 Temperature = 135 F                     Reference   4.7 3.0       ASSUMPTIONS / PRELIMINARY DATA:
Level = 52.3" Reference 4.7 Pressure = 12.08 psig Reference 4.7 Pressure Indicator error 1.12 psig Reference 4.9 Temperature = 135 F Reference 4.7 3.0 ASSUMPTIONS / PRELIMINARY DATA:
The MUT-1 room is assumed to be at a maximum temperature of 95                           l F. This assumption is based on the maximum expected temperatures                       !
The MUT-1 room is assumed to be at a maximum temperature of 95 F.
of 95 F in adjacent EQ Zones 23 and 24 per Reference 4.5 The MUT is assumed to be stabilized at 13.2 psig pressure during normal       operation. This   includes     the   letdown     flow and recirculation flow from the make up pump.         Once the letdown and recirculation flows are terminated, the addition of heat to the MUT , terminates 'and the heat loss to the room is no longer replaced by the-system flow. Therefore af ter termination of flow the heat transfer from the tank to the room decreases system temperature. This effect will be used to detennine the change in                     y MUT vapor pressure.
This assumption is based on the maximum expected temperatures of 95 F in adjacent EQ Zones 23 and 24 per Reference 4.5 The MUT is assumed to be stabilized at 13.2 psig pressure during normal operation.
The decrease in fluid temperature is assumed to occur in the top level     of   the water in the tank.         Any heat lost through condensation on the tank walls is replaced by evaporation from the surface of the water.
This includes the letdown flow and recirculation flow from the make up pump.
If the water should completely leave the tank during drain down then the vapor pressure would drop significantly over that calculated below. This would occur since the surf ace area of the water exposed to the gas is much less than that if the water were in the tank and the vapor mass lost through condensation on the tank walls would not be replaced as rapidly.
Once the letdown and recirculation flows are terminated, the addition of heat to the MUT, terminates 'and the heat loss to the room is no longer replaced by the-system flow.
mm. , r-   m %.an       ..e.
Therefore af ter termination of flow the heat transfer from the tank to the room decreases system temperature. This effect will be used to detennine the change in y
MUT vapor pressure.
The decrease in fluid temperature is assumed to occur in the top level of the water in the tank.
Any heat lost through condensation on the tank walls is replaced by evaporation from the surface of the water.
If the water should completely leave the tank during drain down then the vapor pressure would drop significantly over that calculated below.
This would occur since the surf ace area of the water exposed to the gas is much less than that if the water were in the tank and the vapor mass lost through condensation on the tank walls would not be replaced as rapidly.
mm., r-m %.an
..e.
m L
m L


DESIGN ANALYSIS / CALCULATION                                                         .
DESIGN ANALYSIS / CALCULATION Mglorida M
Mglorida M                 Crystal River Unit 3 Page 2             of 6 oocuuon wcwcarm w                                                                           mvem M-96-0006                                                                                   0 3.0   ASSUMPTIONS / PRELIMINARY DATA Cont'd:
Crystal River Unit 3 Page 2 of 6 oocuuon wcwcarm w mvem M-96-0006 0
3.0 ASSUMPTIONS / PRELIMINARY DATA Cont'd:
This analysis does not contain any assumptions requiring later confirmation or preliminary data.
This analysis does not contain any assumptions requiring later confirmation or preliminary data.


==4.0   REFERENCES==
==4.0 REFERENCES==
4.1   FPC calculation M-96-0007 Revision 0, BWST/MUT Hytran Analysis 4.2   Buffalo Tank Drawing M-6057 (G/C drawing #58-027).
4.1 FPC calculation M-96-0007 Revision 0, BWST/MUT Hytran Analysis 4.2 Buffalo Tank Drawing M-6057 (G/C drawing #58-027).
4.3   Kreith, F.,   " Principles of Heat Transfer", second edition.
4.3
4.4   Marks, " Standard Handbook for Mechanical Engineers", eighth edition.
: Kreith, F.,
4.5   Environmental and Seismic Qualification Program Manual, Revision 8.
" Principles of Heat Transfer", second edition.
4.6   ASME Steam Tables,     Third Edition 4.7   Letter FCS-14607     Design Input dated December 19, 1995 Att. 1 4.8   Operating Procedure OP-103B, Rev. 15 4.9   Calculation DC-5515-018-26.01-ME,"BWST Suction O MUT Suction Lines Tie-In Point Pressures", Rev. 5
4.4 Marks, " Standard Handbook for Mechanical Engineers", eighth edition.
4.5 Environmental and Seismic Qualification Program Manual, Revision 8.
4.6 ASME Steam Tables, Third Edition 4.7 Letter FCS-14607 Design Input dated December 19, 1995 Att. 1 4.8 Operating Procedure OP-103B, Rev. 15 4.9 Calculation DC-5515-018-26.01-ME,"BWST Suction O MUT Suction Lines Tie-In Point Pressures", Rev. 5


==5.0 CALCULATIONS==
==5.0 CALCULATIONS==
For this analysis the tank volumes calculated in Reference 4.1 are used.                         ;
For this analysis the tank volumes calculated in Reference 4.1 are used.
The tank dimensions are provided in References 4.1 and 4.2.
The tank dimensions are provided in References 4.1 and 4.2.
Tank volume         - 590.16 ft'.             Reference 4.1 Dished head volume   = 41.48 ft'.
Tank volume
Tank OD = 8 ft.                                       Reference 4.2 Tank overall height         =  13.42 ft               Rererence 4.2 Tank cylindrical height     =  10.25 ft               Reference 4.2 The approximate surface area is as follows:
- 590.16 ft'.
Cylinder                                                   .
Reference 4.1 Dished head volume
A==*D*H=         n *28.0*   10. 25 - 2 57. 6 f t' .
= 41.48 ft'.
an                                                                   ur v. o, e,.ei ii n s ,   x.., e 2.,....ng
Tank OD = 8 ft.
Reference 4.2 Tank overall height 13.42 ft Rererence 4.2
=
Tank cylindrical height 10.25 ft Reference 4.2
=
The approximate surface area is as follows:
Cylinder A==*D*H=
n
*28.0*
: 10. 25 - 2 57. 6 f t'.
an ur v. o, e,.ei ii n s,
x..,
e 2.,....ng


l-l e
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DESIGN ANALYSiSICALCULATION MgloridaU                    Crystal River Unit 3 Page 3             of 6 OOcuutNT iDENTIM,ATm No.                                                                               AEviSON              (
DESIGN ANALYSiSICALCULATION Mglorida U
M-96-0006                                                                                             0                     l l
Crystal River Unit 3 Page 3 of 6 AEviSON OOcuutNT iDENTIM,ATm No.
l 5.0     CALCULATIONS Cont'd:
M-96-0006 0
For the dished head the area of circular plate is                                                             I conservatively used:                                                                                           l 2                2              2 A=x*D       /4 -   x*8 /4 = 50.3 ft                                                           l The heat removed from the tank (due to natural convection up                                                   l the tank walls) can be conservatively calculated using the following formula.
5.0 CALCULATIONS Cont'd:
Q = h A AT                                         Reference 4.3                                     i Where h = natural convection heat transfer coefficient h = 0.18   (AT) 333                       Reference 4.3                                     l l
For the dished head the area of circular plate is conservatively used:
A = surface area of tank.
A=x*D
AT = temperature difference (tank to air)                                                     '
/4 -
2 h = 0.18 * (40 F) 333 = 0.61 BTU /HR-FT.7 The tank upper head can be treated as a horizontal flat plate with the characteristic dimension equal to the diameter.                                 The natural convection heat transfer coeficient is calculated below:
x*8 /4 = 50.3 ft 2
h*D/k = 0.48 (Gr) Ud5 Where h = natural convection heat transfer coefficient                                               '
2 2
Gr = Grashoff Number  !
The heat removed from the tank (due to natural convection up the tank walls) can be conservatively calculated using the following formula.
                                                                = (1.44 *10'' /F- f t )
Q = h A AT Reference 4.3 i
Where h = natural convection heat transfer coefficient h = 0.18 (AT) 333 Reference 4.3 A = surface area of tank.
AT = temperature difference (tank to air) 2 h = 0.18 * (40 F) 333 = 0.61 BTU /HR-FT.7 The tank upper head can be treated as a horizontal flat plate with the characteristic dimension equal to the diameter.
The natural convection heat transfer coeficient is calculated below:
h*D/k = 0.48 (Gr) Ud5 Where h = natural convection heat transfer coefficient 3
3
* D
* D
* 45 F 3            3 Gr = 3.318 10 k = thermal conductivity of air = .00146 BTU /hr-ft-F D = Diameter = 8 Ft Therefore:
* 45 F Gr = Grashoff Number = (1.44 *10'' /F-f t )
2 h = 0.48 * (3.318 10 ) 0 25 2       *  .00146/8 = 0.08 BTU /HR-FT_p Using only the top head and cylindrical portion of the tank for surface area, the heat loss is calculated.fNote that it is assumed that the water remaining in the tank loses no heat to the environment. Heat loss from the water isiused to raise the gas temperature and replace the mass loss duento condensation on the walls through evaporation.
Gr = 3.318 10 k = thermal conductivity of air =.00146 BTU /hr-ft-F D = Diameter = 8 Ft Therefore:
Q. = (0.61*257.6 + .08*50.3)*(135 - 95) = 7250 BTU /HR The radiant temperature factor from Table 5-25 (Ref. 4.3) is taken to be 1.2 for a 95 wall temperature and a 135' tank temperature. Assuming a view factor of 1 and an emissivity of 0.8 for the tank and the walls the radiant heat transfer is:
2 h = 0.48 * (3.318 10 ) 0 25 2
3;.2 = 1/(1/.8 + 1/.8-1)     = 0.67      Reference 4.3 pg 227 h, = 1. 2
.00146/8 = 0.08 BTU /HR-FT_p Using only the top head and cylindrical portion of the tank for surface area, the heat loss is calculated.fNote that it is assumed that the water remaining in the tank loses no heat to the environment.
* 0.67     =   0.8          Reference 4.3 pg 230 Theref ore 0, = 0. 8 * (257.6 + 50.3)           *
Heat loss from the water isiused to raise the gas temperature and replace the mass loss duento condensation on the walls through evaporation.
(135-95)    = 9850 BTU /HR
Q. = (0.61*257.6 +.08*50.3)*(135 - 95) = 7250 BTU /HR The radiant temperature factor from Table 5-25 (Ref. 4.3) is taken to be 1.2 for a 95 wall temperature and a 135' tank temperature.
                                                                ~
Assuming a view factor of 1 and an emissivity of 0.8 for the tank and the walls the radiant heat transfer is:
0.67 Reference 4.3 pg 227 3;.2 = 1/(1/.8 + 1/.8-1)
=
0.8 Reference 4.3 pg 230 h, = 1. 2
* 0.67
=
(135-95)
= 9850 BTU /HR Theref ore 0, = 0. 8 * (257.6 + 50.3)
~
RET. L#e of Fent R E S P, Nuc.es En9neerisig b 95
RET. L#e of Fent R E S P, Nuc.es En9neerisig b 95


glorida       DESIGN ANALYSIS / CALCULATION ccw' TInf                     Crystal River Unit 3                                                       :
glorida DESIGN ANALYSIS / CALCULATION ccw' TInf Crystal River Unit 3 Page 4 of 6 orxu w a u m or m na rosa M-96-0006 0
Page 4                   of 6 orxu w a u m or m na                                                                                             rosa   ,
j 5.0 CALCULATIONS Cont'd:
M-96-0006                                                                                                         0       j 5.0     CALCULATIONS Cont'd:
The total heat removed from the tank by convection and radiation from the tank walls to the room is:
The total heat removed from the tank by convection and radiation from the tank walls to the room is:
Q - 7250 + 9850 - 17100 BTU /HR                                                                       )
Q - 7250 + 9850 - 17100 BTU /HR Calculation of heat loss from the water to the hydrogen gas as the level drops and the gas expands.
l Calculation of heat loss from the water to the hydrogen gas as the level drops and the gas expands.
The gas would normally cool as it expands in an adiabatic fashion assuming the level drop were fast.
The gas would normally cool as it expands in an adiabatic fashion assuming the level drop were fast.             However the level in the tank drops slowly and the heat transfer from the water surface to the gas keeps the gas in thermal equilibrium with the water.       The calculation below is an attempt to quantify the energy loss from the water as the gas expands from it's initial pressure to it's final state in the tank (assumed to be bottom dished head).
However the level in the tank drops slowly and the heat transfer from the water surface to the gas keeps the gas in thermal equilibrium with the water.
The calculation below is an attempt to quantify the energy loss from the water as the gas expands from it's initial pressure to it's final state in the tank (assumed to be bottom dished head).
Assuming the gas behaves as an isentropic (closed) system the total energy lost due to expansion can be calculated as follows:
Assuming the gas behaves as an isentropic (closed) system the total energy lost due to expansion can be calculated as follows:
T /T 2  1
(V /V ) "-"
                                        =  (V /V ) "-"
Reference 4.4 4-18 T /T
1  2                              Reference 4.4 4-18 x
=
Where; T = initial and final temperatures V = initial and final volumes                       ~
2 1
1 2
x Where; T = initial and final temperatures V = initial and final volumes
~
k = 1.41 isentropic exponent for hydrogen The energy loss for a closed system is:
k = 1.41 isentropic exponent for hydrogen The energy loss for a closed system is:
            ^~
I
c, (T2 - T)
~
        ~
^~
I u2- ui =     h-h 2   1 1            , ; -P             '          ~~
u2-ui =
h-h 2
1 c,
(T - T)
, ; -P
~~
2 1
::+
::+
Where; c, - specific heat     - 2.43 BTU /LBM *RJReference 4.3 w
Where; c, - specific heat
The initial gas volume is               333.586   ft 3 2
- 2.43 BTU /LBM *RJReference 4.3 w
The total tank volume is               603.669   ft The dished head volume                   48.234   ft' T /T 2  1
3 The initial gas volume is 333.586 ft 2
                                        =  (333.586/535.435)         = 0.8237 Assuming an initial temperature of 135'F the final temperature is:
The total tank volume is 603.669 ft The dished head volume 48.234 ft' (333.586/535.435)
T2- 0.8237 (135 + 460)         = 490 R (305F) i,'95                                                                               RET. Lde of F4nt KEEP. hweas Engmee'mg
= 0.8237 T /T
=
2 1
Assuming an initial temperature of 135'F the final temperature is:
490 R (305F)
T2- 0.8237 (135 + 460)
=
i,'95 RET. Lde of F4nt KEEP. hweas Engmee'mg


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  *        /         T     Florida           DESIGN ANALYSIS / CALCULATION l
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                      / cc ower   cwoh                        Crystal River Unit 3
5 P
              'q J                                                                                                                   l Page 5             of 6 i         DOCUMENT IDENilFCATON No.                                                                             sgymoN l         M-96-0006                                                                                             0 5.0     CALCUIATIONS Cont'd:
/ cc ower Crystal River Unit 3 l
Note that this temperature is never achieved due to heat transfer from the vessel and the water. However the temperature difference between the water and the gas may be used to determine the enthalpy change in the gas.                             The enthalpy change is made up by heat transfer from the water.                                                                                             l h-h = c. (T - T) 2    1          2    3
t cwoh
                                                              =
'q J l
2.43 * (135 - 30) = 355 BTU /Lbm of hydrogen The initial hydrogen inventory is calculated below using the perfect gas law:
Page 5 of 6 i
PV = nRT therefore m - PV/RT, R = 767 f t-lbf /lbm-R for H 2 :
DOCUMENT IDENilFCATON No.
Initial tank pressure is 13.2 psig References 4.8 & 4.9 The initial gas pressure is 13.2 psig - 2.5375 psig 2.5375 psig - vapor pressue at 135 F               Reference 4.6                                   <
sgymoN l
P H2 = 13.2 - 2.54 = 10.66 psig M= (10.66+14.696) psi *333.59f t *144in /f t / (767f t-lbf /lbm-R *595R) 3      2    2 i
M-96-0006 0
2                              m - 2.669 lbm                                                           ;          )
5.0 CALCUIATIONS Cont'd:
                    >            The energy required to replace that lost due to expansion s is:
Note that this temperature is never achieved due to heat transfer from the vessel and the water.
However the temperature difference between the water and the gas may be used to determine the enthalpy change in the gas.
The enthalpy change is made up by heat transfer from the water.
l h-h = c.
(T - T) 2.43 * (135 - 30) = 355 BTU /Lbm of
=
2 1
2 3
hydrogen The initial hydrogen inventory is calculated below using the perfect gas law:
PV = nRT therefore m - PV/RT, R = 767 f t-lbf /lbm-R for H :
2 Initial tank pressure is 13.2 psig References 4.8 & 4.9 The initial gas pressure is 13.2 psig - 2.5375 psig 2.5375 psig - vapor pressue at 135 F Reference 4.6 13.2 - 2.54 = 10.66 psig P
=
H2 3
2 2
M= (10.66+14.696) psi *333.59f t *144in /f t / (767f t-lbf /lbm-R *595R) i m - 2.669 lbm
)
2 The energy required to replace that lost due to expansion s
is:
Q = 355 BTU /LBM
Q = 355 BTU /LBM
* 2.669 LBM = 947 BTU The heat loss due to the expansion of hydrogen is much less than that lost due to heat transfer from the tank to>
* 2.669 LBM = 947 BTU The heat loss due to the expansion of hydrogen is much less than that lost due to heat transfer from the tank to>
                    >              the room.                                                                             y         4
the room.
              ~. ,m             -
y 4
                                                                                                                        .#          z KN                 The total heat lost from the water conservatively assum3.ng tank drawdown occurs in 30 minutes is; 1
z
9                                                                                                      m           I a                Q = 947 BTU       + 17100 BTU /HR
~.,m KN The total heat lost from the water conservatively assum3.ng 1
* 0.5 HR = 9497 BTU                                 w It is assumed this heat is lost from the top portion of the liquid level in the tank. Therefore the total heat loss is averaged over the volume of fluid remaining in the bottom head.
tank drawdown occurs in 30 minutes is; 9
I m
Q = 947 BTU
+ 17100 BTU /HR
* 0.5 HR = 9497 BTU w
a It is assumed this heat is lost from the top portion of the liquid level in the tank.
Therefore the total heat loss is averaged over the volume of fluid remaining in the bottom head.
Change in H O enthalpy is 9497 BTU / (48.234 f t)*61.46Lbm/f t')
Change in H O enthalpy is 9497 BTU / (48.234 f t)*61.46Lbm/f t')
2 Ah = 3.20 BTU /LBM The initial H;O enthalpy is 102.95 BTU /LBM (at 13 5 'F)                                           j
2 Ah
                                                                                                                                      )
= 3.20 BTU /LBM The initial H;O enthalpy is 102.95 BTU /LBM (at 13 5 'F) j
L 95                                                                           RE T. ufe of Piant RE SP Nuc6ese Engmeermg
)
L 95 RE T. ufe of Piant RE SP Nuc6ese Engmeermg


\       .
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Floride       DESIGN ANALYSIS / CALCULATION l
Floride DESIGN ANALYSIS / CALCULATION l
Power
Power
!                    -rm                         Crystal River Unit 3 Page 6     of 6 mmm amorm o                                                                     -
-rm Crystal River Unit 3 Page 6 of 6 mmm amorm o M-SE-0006 0
M-SE-0006                                                                       0 l
l 5.0 CALCUIATIONS Cont'd:
5.0     CALCUIATIONS Cont'd:
The final enthalpy of the water remaining in the tank is 102.95 - 3.20 - 99.75 BTU /LBM From the steam tables (Reference 4.6) the final water temperature is 131.7*F.
The final enthalpy of the water remaining in the tank is 102.95 - 3.20 - 99.75 BTU /LBM From the steam tables (Reference 4.6) the final water temperature is 131.7*F.       At this temperature the vapor pressure is 2.32 psia.
At this temperature the vapor pressure is 2.32 psia.
l The initial vapor pressure is 2.5375 psia.
The initial vapor pressure is 2.5375 psia.
Therefore the vapor pressure drop is approximately 0.2 psia. If the tank were completely empty the vapor pressure would eventually drop to room temperature or                       ;
Therefore the vapor pressure drop is approximately 0.2 psia.
approximately 0.95 psia. Therefore the vapor pressure                       i drop calculated above, due to heat transfer from the tank to the environment and expansion of the hydrogen gas is a conservative estimate of the expected decrease in vapor                     ;
If the tank were completely empty the vapor pressure would eventually drop to room temperature or approximately 0.95 psia.
pressure.                                                                   i The primary assumption that effects these results is that l
Therefore the vapor pressure drop calculated above, due to heat transfer from the tank to the environment and expansion of the hydrogen gas is a conservative estimate of the expected decrease in vapor pressure.
the heat loss from the water occurs from the top level of the water. Once the water leaves the tank the pressure in the tank would drop to approximately 0.95 psia since condensation from the gas / vapor space would not be                         ;
i The primary assumption that effects these results is that the heat loss from the water occurs from the top level of the water.
replaced by the boil off or evaporation of the water in the tank. The results are appropriate due to the other conservative assumptions used in the calculation. These are:
Once the water leaves the tank the pressure in the tank would drop to approximately 0.95 psia since condensation from the gas / vapor space would not be replaced by the boil off or evaporation of the water in the tank.
: 1.     No heat transfer from the bottom head to the environment.
The results are appropriate due to the other conservative assumptions used in the calculation.
: 2.     Water is assumed fully mixed and isothermal.
These are:
3.-     No' credit taken for pump heat which tends to raise the initial temperature in the tank before the letdown flow and pump recirculation are s       isolated.
1.
: 4.     The total tank surface area calculated does not take into account the dished head geometry.
No heat transfer from the bottom head to the environment.
: 5.     The gas / vapor region is assumed saturated at all times.
2.
: 6.     MUT-1 drawdown occurs in a period of greater than 30 minutes (Reference 4.1) .
Water is assumed fully mixed and isothermal.
6.0     RESULTS/ CONCLUSIONS:
3.-
No' credit taken for pump heat which tends to raise the initial temperature in the tank before the letdown flow and pump recirculation are s
isolated.
4.
The total tank surface area calculated does not take into account the dished head geometry.
5.
The gas / vapor region is assumed saturated at all times.
6.
MUT-1 drawdown occurs in a period of greater than 30 minutes (Reference 4.1).
6.0 RESULTS/ CONCLUSIONS:
The vapor pressure in the vapor space can conservatively be assumed to be initially at 2.54 psia. As the tank is drawn down the vapor pressure decreases linearly to 2.34 psia.
The vapor pressure in the vapor space can conservatively be assumed to be initially at 2.54 psia. As the tank is drawn down the vapor pressure decreases linearly to 2.34 psia.
3
3
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Latest revision as of 04:23, 13 December 2024

Mut Vapor Pressure Evaluation
ML20101M589
Person / Time
Site: Crystal River Duke Energy icon.png
Issue date: 04/04/1996
From:
FLORIDA POWER CORP.
To:
Shared Package
ML20101M505 List:
References
M-96-0006, M-96-0006-R00, M-96-6, M-96-6-R, NUDOCS 9604050449
Download: ML20101M589 (7)
v • d • e


Text

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Florida C. o,....e...,P, w ;

Enclosss ce 8 s

ANALYSIS / CALCULATION

SUMMARY

l t

DISalPUNE CON'HOL NO.

REVISON LEVEL DOCUMENT IDENTIFICATION NUMBER M

96-0006 0

Tm.E CLASSWCATON PIECK ONE) l MUT VAPOR PRESSURE EVALUATION

% safety Related l

Non safety Related MAR /SP/COWR/PEERE NUMBER l

l VENDOR DOCUuEwi NUuscR l

REVISION ITEMS REVISED APPROVALS Design Engineer

[//$ [

Date Verification Engineer hf Date/ Method

  • R Supervisor Date

" VERIFICATION METHODS: R - Design Review; A - Alternate Calculation; T - Qualfication Testing DESCRIBE BELOW IF METHOD OF VERIFICATION WAS OTHER THAN DESIGN REW!EW 5

PURPOSE

SUMMARY

The ourpose of this calculation is to determine the vapor pressure in MUT-1 durina draindown followiria a postulated LOCA.

.. a.

13g

, - a.y 1r-

- 4. k a., - #

.s -

4,.

'._ e I?

E 1.

d T2i RESULTS

SUMMARY

'Thetesults of the analysis show that the vapor pressure decreases a minimum of 0.2 psia during the draindown period following a LOCA.

9604050449 960404 PDR ADOCK 0500 2

%,e3

DESIGN ANALYSISICALCULATION Mglorida N

Crystal River Unit 3 Page 1 of 6 CKJGUMENT IQ[NTflCATON NO.

F(VL1WUN M-96-0006 0

1.0 PURPOSE

The purpose of this calculation is to determine the vapor pressure change in MUT-1 during draindown following a postulated LOCA.

The results of the calculation will provide input to the hydraulic calculation which determines the actual draindown level.

2.0 DESIGN INPUT:

2.1 The tank volume used is provided in Reference 4.1.

The initial gas volume is 333.586 ft' 3

The total tank volume is 603.669 ft The dished head volume 48.234 ft' / per head 2.2 The initial minimum liquid level, pressure and temperature in the MUT-1 is:

Level = 52.3" Reference 4.7 Pressure = 12.08 psig Reference 4.7 Pressure Indicator error 1.12 psig Reference 4.9 Temperature = 135 F Reference 4.7 3.0 ASSUMPTIONS / PRELIMINARY DATA:

The MUT-1 room is assumed to be at a maximum temperature of 95 F.

This assumption is based on the maximum expected temperatures of 95 F in adjacent EQ Zones 23 and 24 per Reference 4.5 The MUT is assumed to be stabilized at 13.2 psig pressure during normal operation.

This includes the letdown flow and recirculation flow from the make up pump.

Once the letdown and recirculation flows are terminated, the addition of heat to the MUT, terminates 'and the heat loss to the room is no longer replaced by the-system flow.

Therefore af ter termination of flow the heat transfer from the tank to the room decreases system temperature. This effect will be used to detennine the change in y

MUT vapor pressure.

The decrease in fluid temperature is assumed to occur in the top level of the water in the tank.

Any heat lost through condensation on the tank walls is replaced by evaporation from the surface of the water.

If the water should completely leave the tank during drain down then the vapor pressure would drop significantly over that calculated below.

This would occur since the surf ace area of the water exposed to the gas is much less than that if the water were in the tank and the vapor mass lost through condensation on the tank walls would not be replaced as rapidly.

mm., r-m %.an

..e.

m L

DESIGN ANALYSIS / CALCULATION Mglorida M

Crystal River Unit 3 Page 2 of 6 oocuuon wcwcarm w mvem M-96-0006 0

3.0 ASSUMPTIONS / PRELIMINARY DATA Cont'd:

This analysis does not contain any assumptions requiring later confirmation or preliminary data.

4.0 REFERENCES

4.1 FPC calculation M-96-0007 Revision 0, BWST/MUT Hytran Analysis 4.2 Buffalo Tank Drawing M-6057 (G/C drawing #58-027).

4.3

Kreith, F.,

" Principles of Heat Transfer", second edition.

4.4 Marks, " Standard Handbook for Mechanical Engineers", eighth edition.

4.5 Environmental and Seismic Qualification Program Manual, Revision 8.

4.6 ASME Steam Tables, Third Edition 4.7 Letter FCS-14607 Design Input dated December 19, 1995 Att. 1 4.8 Operating Procedure OP-103B, Rev. 15 4.9 Calculation DC-5515-018-26.01-ME,"BWST Suction O MUT Suction Lines Tie-In Point Pressures", Rev. 5

5.0 CALCULATIONS

For this analysis the tank volumes calculated in Reference 4.1 are used.

The tank dimensions are provided in References 4.1 and 4.2.

Tank volume

- 590.16 ft'.

Reference 4.1 Dished head volume

= 41.48 ft'.

Tank OD = 8 ft.

Reference 4.2 Tank overall height 13.42 ft Rererence 4.2

=

Tank cylindrical height 10.25 ft Reference 4.2

=

The approximate surface area is as follows:

Cylinder A==*D*H=

n

  • 28.0*
10. 25 - 2 57. 6 f t'.

an ur v. o, e,.ei ii n s,

x..,

e 2.,....ng

l-l e

DESIGN ANALYSiSICALCULATION Mglorida U

Crystal River Unit 3 Page 3 of 6 AEviSON OOcuutNT iDENTIM,ATm No.

M-96-0006 0

5.0 CALCULATIONS Cont'd:

For the dished head the area of circular plate is conservatively used:

A=x*D

/4 -

x*8 /4 = 50.3 ft 2

2 2

The heat removed from the tank (due to natural convection up the tank walls) can be conservatively calculated using the following formula.

Q = h A AT Reference 4.3 i

Where h = natural convection heat transfer coefficient h = 0.18 (AT) 333 Reference 4.3 A = surface area of tank.

AT = temperature difference (tank to air) 2 h = 0.18 * (40 F) 333 = 0.61 BTU /HR-FT.7 The tank upper head can be treated as a horizontal flat plate with the characteristic dimension equal to the diameter.

The natural convection heat transfer coeficient is calculated below:

h*D/k = 0.48 (Gr) Ud5 Where h = natural convection heat transfer coefficient 3

3

  • D
  • 45 F Gr = Grashoff Number = (1.44 *10 /F-f t )

Gr = 3.318 10 k = thermal conductivity of air =.00146 BTU /hr-ft-F D = Diameter = 8 Ft Therefore:

2 h = 0.48 * (3.318 10 ) 0 25 2

.00146/8 = 0.08 BTU /HR-FT_p Using only the top head and cylindrical portion of the tank for surface area, the heat loss is calculated.fNote that it is assumed that the water remaining in the tank loses no heat to the environment.

Heat loss from the water isiused to raise the gas temperature and replace the mass loss duento condensation on the walls through evaporation.

Q. = (0.61*257.6 +.08*50.3)*(135 - 95) = 7250 BTU /HR The radiant temperature factor from Table 5-25 (Ref. 4.3) is taken to be 1.2 for a 95 wall temperature and a 135' tank temperature.

Assuming a view factor of 1 and an emissivity of 0.8 for the tank and the walls the radiant heat transfer is:

0.67 Reference 4.3 pg 227 3;.2 = 1/(1/.8 + 1/.8-1)

=

0.8 Reference 4.3 pg 230 h, = 1. 2

  • 0.67

=

(135-95)

= 9850 BTU /HR Theref ore 0, = 0. 8 * (257.6 + 50.3)

~

RET. L#e of Fent R E S P, Nuc.es En9neerisig b 95

glorida DESIGN ANALYSIS / CALCULATION ccw' TInf Crystal River Unit 3 Page 4 of 6 orxu w a u m or m na rosa M-96-0006 0

j 5.0 CALCULATIONS Cont'd:

The total heat removed from the tank by convection and radiation from the tank walls to the room is:

Q - 7250 + 9850 - 17100 BTU /HR Calculation of heat loss from the water to the hydrogen gas as the level drops and the gas expands.

The gas would normally cool as it expands in an adiabatic fashion assuming the level drop were fast.

However the level in the tank drops slowly and the heat transfer from the water surface to the gas keeps the gas in thermal equilibrium with the water.

The calculation below is an attempt to quantify the energy loss from the water as the gas expands from it's initial pressure to it's final state in the tank (assumed to be bottom dished head).

Assuming the gas behaves as an isentropic (closed) system the total energy lost due to expansion can be calculated as follows:

(V /V ) "-"

Reference 4.4 4-18 T /T

=

2 1

1 2

x Where; T = initial and final temperatures V = initial and final volumes

~

k = 1.41 isentropic exponent for hydrogen The energy loss for a closed system is:

I

~

^~

u2-ui =

h-h 2

1 c,

(T - T)

, ; -P

~~

2 1

+

Where; c, - specific heat

- 2.43 BTU /LBM *RJReference 4.3 w

3 The initial gas volume is 333.586 ft 2

The total tank volume is 603.669 ft The dished head volume 48.234 ft' (333.586/535.435)

= 0.8237 T /T

=

2 1

Assuming an initial temperature of 135'F the final temperature is:

490 R (305F)

T2- 0.8237 (135 + 460)

=

i,'95 RET. Lde of F4nt KEEP. hweas Engmee'mg

t

,~~

'e

/

T Florida DESIGN ANALYSIS / CALCULATION

(

5 P

/ cc ower Crystal River Unit 3 l

t cwoh

'q J l

Page 5 of 6 i

DOCUMENT IDENilFCATON No.

sgymoN l

M-96-0006 0

5.0 CALCUIATIONS Cont'd:

Note that this temperature is never achieved due to heat transfer from the vessel and the water.

However the temperature difference between the water and the gas may be used to determine the enthalpy change in the gas.

The enthalpy change is made up by heat transfer from the water.

l h-h = c.

(T - T) 2.43 * (135 - 30) = 355 BTU /Lbm of

=

2 1

2 3

hydrogen The initial hydrogen inventory is calculated below using the perfect gas law:

PV = nRT therefore m - PV/RT, R = 767 f t-lbf /lbm-R for H :

2 Initial tank pressure is 13.2 psig References 4.8 & 4.9 The initial gas pressure is 13.2 psig - 2.5375 psig 2.5375 psig - vapor pressue at 135 F Reference 4.6 13.2 - 2.54 = 10.66 psig P

=

H2 3

2 2

M= (10.66+14.696) psi *333.59f t *144in /f t / (767f t-lbf /lbm-R *595R) i m - 2.669 lbm

)

2 The energy required to replace that lost due to expansion s

is:

Q = 355 BTU /LBM

  • 2.669 LBM = 947 BTU The heat loss due to the expansion of hydrogen is much less than that lost due to heat transfer from the tank to>

the room.

y 4

z

~.,m KN The total heat lost from the water conservatively assum3.ng 1

tank drawdown occurs in 30 minutes is; 9

I m

Q = 947 BTU

+ 17100 BTU /HR

  • 0.5 HR = 9497 BTU w

a It is assumed this heat is lost from the top portion of the liquid level in the tank.

Therefore the total heat loss is averaged over the volume of fluid remaining in the bottom head.

Change in H O enthalpy is 9497 BTU / (48.234 f t)*61.46Lbm/f t')

2 Ah

= 3.20 BTU /LBM The initial H;O enthalpy is 102.95 BTU /LBM (at 13 5 'F) j

)

L 95 RE T. ufe of Piant RE SP Nuc6ese Engmeermg

\\

Floride DESIGN ANALYSIS / CALCULATION l

Power

-rm Crystal River Unit 3 Page 6 of 6 mmm amorm o M-SE-0006 0

l 5.0 CALCUIATIONS Cont'd:

The final enthalpy of the water remaining in the tank is 102.95 - 3.20 - 99.75 BTU /LBM From the steam tables (Reference 4.6) the final water temperature is 131.7*F.

At this temperature the vapor pressure is 2.32 psia.

The initial vapor pressure is 2.5375 psia.

Therefore the vapor pressure drop is approximately 0.2 psia.

If the tank were completely empty the vapor pressure would eventually drop to room temperature or approximately 0.95 psia.

Therefore the vapor pressure drop calculated above, due to heat transfer from the tank to the environment and expansion of the hydrogen gas is a conservative estimate of the expected decrease in vapor pressure.

i The primary assumption that effects these results is that the heat loss from the water occurs from the top level of the water.

Once the water leaves the tank the pressure in the tank would drop to approximately 0.95 psia since condensation from the gas / vapor space would not be replaced by the boil off or evaporation of the water in the tank.

The results are appropriate due to the other conservative assumptions used in the calculation.

These are:

1.

No heat transfer from the bottom head to the environment.

2.

Water is assumed fully mixed and isothermal.

3.-

No' credit taken for pump heat which tends to raise the initial temperature in the tank before the letdown flow and pump recirculation are s

isolated.

4.

The total tank surface area calculated does not take into account the dished head geometry.

5.

The gas / vapor region is assumed saturated at all times.

6.

MUT-1 drawdown occurs in a period of greater than 30 minutes (Reference 4.1).

6.0 RESULTS/ CONCLUSIONS:

The vapor pressure in the vapor space can conservatively be assumed to be initially at 2.54 psia. As the tank is drawn down the vapor pressure decreases linearly to 2.34 psia.

3

.m m v. e.m u p w 2 ege..ees L

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