ML20101M561
ML20101M561 | |
Person / Time | |
---|---|
Site: | Crystal River |
Issue date: | 02/13/1996 |
From: | Harrison D MPR ASSOCIATES, INC. |
To: | |
Shared Package | |
ML20101M505 | List:
|
References | |
102075DHH06, 102075DHH06-R00, 102075DHH6, 102075DHH6-R, NUDOCS 9604050441 | |
Download: ML20101M561 (9) | |
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- Print page 1 l V EVALUATION OF MAKEUP TANK MAXIMUM PRESSURE PURPOSE:
1 The purpose of this calculation is to evaluate the maximum pressure in the makeup tank as a function of levelin the tank. The expressions are derived in hand calculations. Some parameters are also established by separate hand calculations.
CASE COVERED:
This is the nominal reference case except that the temperature in the makeup tank is assumed to be 100*F instead of 135'F. An expression for the difference in allowable pressure for the two temperatures is derived and evaluated, both as pressure and head.
PARAMETERS:
Vapor pressure of water in the makeup tank at 100*F: Psat := 0.94924 b Psat = 136.691 b n h*
Density of water in the makeup tank at 100 F: P m := 61.9963 Ib l
Density of water in the BWST at 100 F: p , := 61.9963 5 ft3 Acceleration of gravity: g := 32.17- h tec,
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Minimum levelin the BWST: Lm'm := 5.0 ft Minimum elevation of the surface in the BWST: E s:= Lmin + 119.67 ft E , = 124.67 ft Minimum margin between the top of the pipe at the tie-in and
- i h and the gas in the makeup tank:
Elevation of the top of the 6-in pipe at the tie-in point: Ed := 104.5 ft + 3 in Ed = 104.75 ft
' Minimum elevation of the levelin the pipe from the MUT to the top of the 6 in pipe at the tie-in point: Ex := Ed+Xm Ex = 106.75 ft 1
Mole ratio of Nitrogen to Hydrogen in the MUT cover gas: y := 0.1 Pipe flow velocity (in the 6-in pipe) at tie-in point: u c := 6.6641 see Gage pressure in the BWST: Psg := 1.0 ft 62.4 E ft3 l Absolute pressure in the BWST: Ps := 14.7 b + Psg Ps = 2.054 10 3 *b
' 2 f;2 in Total head loss from the BWST to the tie-in point at the centerline of the 6-in pipe: AHac '= ll.8704 ft Combined Henry's Law and other constants for Hydrogen and Nitrogen, respectively:
- i O DH := 0.018003 DN := 0.013609 The number of calculational steps to reduce the liquid volume in the makeup tank to zero: k := 20 I
Mathcad PLUS 6.0 EVAL 4RR.MCD 11:48 AM on 2/12/96 4
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,. i MPR Associates,Inc. Pagef 320 King Street, Alexandria, VA 22314 Calc 102075DHH06 Preparer:
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EXPRESSIONS USED:
The expressions below for the maximum pressure in makeup tank are derived in hand calculations:
vo t (j - 1) 6v t DH-(1 - v -j 6v) vo t (j - 1) 6v t D g(1 - vo- j 6v)
R H(j,v ,6v) = RN(I' V0'0V) ;*
vo + j 6v + D g(1 - vo- j 8v) vo +j 6v t D y(1 - vo- j.6v) k k !
RH j,vo, +t RN l'Y '
T(1,vo,k) :=
"' 5* .
l 1+7 i The maximum pressure in the makeup tank as a function of the initial gas volume fraction, vo,is as follows: ,
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-p m g Ex + p , g E ,- (p ,- p m) g Ed
- P s g- - AHac - - Psat (psg 2g j Pmo(vo,y,k) := Psat + y T(Y,vo,k) {
If we now define another quantity, Pxmo, as the allowable pressure at the maximum temperature in the makeup tank of 135'F, we can get the difference between the two pressures.
p The saturation pressure at 135'F is: Pxsat := 2.5375- Pxsat = 365.4 b k " II The density at 135'Fis: p ,x := 61.4628-ft The non-dimensional constants (from hand calculations) are: Dx H := 0.01892 Dx N := 0.01246 Now determine the new terms for the maximum temperature condition as above:
vo t (J - 1) 6vt Dx g(1 - vo- j 6v) vo + (j - 1) 6v+ Dx g(1 - vo- j 6v) l Rx H(j,vo,6v) ': Rx N(I'V0'0V) ;*
vo +J 6v+ Dx g(1 - vo- j 6v) vo +j 6v+ Dx g(1 - vo-j 6v) k k f
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j,vo, ti Rx N V0' I r k i k j j=1 Tx(y,vo,k) := j = 1 1+7 The maximum allowable pressure in the makeup tank, Pxmo, as a function of the initial gas volume fraction, vo, is as follows for the 135'F makeup tank temperature is:
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-p mx g Ex + p , g E ,- (p ,- p mx) g Ed + p s g- - AHac - - Pxsat (psg 2g j Pxmo(vo,Y,k) := Pxsat +
Tx(7,vo,k)
. The gas volume fraction, vo, is related to the measured level in the makeup tank, in inches, by the following relation (obtained from hand calculations):
vo(xm) := 0.9089- 0.0068355 m in' x Mathcad PLUS 6.0 EVAL 4RR.MCD 11:48 AM on 2/12/96
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,O If we now define the difference between the two allowable pressures as AP (where Pmo is the allowable ;
pressure at a makeup tank temperature of 100'F and Pxmo is the allowable pressure for a tank temperature of 135'F):
AP(vo,y,k) := Pmo(vo,y,k) - Pxmo(vo,y,k)
If AP is greater than zero, the allowable pressure at the makeup tank temperature of 100*F is greater than the ,
allowable pressure at the 135'F temperature used in the original base calculation and it would show that the assumption of the higher (135'F) temperature is conservative because it results in a lower allowable pressure.
Range of level: x,:= 0 in,5 in 100 in 10 .
l l AP(vo(x m).7,k) 6 l tbri l j l l 4 I" l l l /
M 0 20 40 60 80 100 As shown by the curve, the allowable pressure for 135'F is always lower than the allowable pressure at l 100*F. This confirms that the use of the higher temperature in the analysis. Furthermore, it shows that the use of the higher temperature involves a substantial additional margin in the calculation. The curve of pressure difference can also be expressed in terms of feet of head as follows:
15 AP(vo(x m).T.k) 10 5Pm ft 5 0 20 40 60 80 100 l
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