ML20049H270

App 3D to Gessar, Analysis of Recirculation Motor & Pump Under Accident Conditions.

ML20049H270
Person / Time
Site:	05000447
Issue date:	02/12/1982
From:	GENERAL ELECTRIC CO.
To:
References
NUDOCS 8202230026
Download: ML20049H270 (102)

Text

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 i

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l APPENDIX 3D i O ANALYSIS OF RECIRCULATION MOTOR AND i

PUMP UNDER ACCIDENT CONDITIONS }

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8202230026 820212 PDR ADOCK 05000447 K pyg

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0

() APPENDIX 3D CONTENTS Section Title Page 3D APPENDIX 3D - ANALYSIS OF RECIRCULATION MOTOR AND PUMP UNDER ACCIDENT CONDITIONS 3D.1 INTRODUCTION 3D.1-1 3D.2 PUMP SPEED UNDER LOCA CONDITIONS 3D.2-1 3D.2.1 Discharge Line Breaks 3D.2-2 3D.2.2 Suction Line Breaks 3D.2-2 3D.2.2.1 Longitudinal Breaks 3D.2-2 3D.2.2.2 Guillotine Breaks 3D.2-3 3D.3 MOTOR GENERATED MISSILES 3D.3-1 3D.3.1 Rotor Failure 3D.3-4 3D.3.2 Fan Failure 3D.3-6 3D.3.3 Retaining Ring Failure 3D.3-7 3D.3.4 The Motor as a Potential Missile 3D.3-9

(^N

() 3D.4 PUMP IMPELLER-GENERATED MISSILES 3D.4-1 3D.4.1 Impeller Penetration of the Pump Case 3D.4-1 3D.4.2 Potential Pump Impeller Missiles from the Pipe Break 3D.4-3 3D.4.2.1 Characteristics of Potential Pump Impeller Missiles 3D.A-3 3D.4.2.1.1 Missile Velocity in the Piping System 3D.4-6 3D.4.1.2 Escape Pattern 3D.4-8 3D.4.2.2 Damage Potential of Escaped Missiles 3D.4-8 3D.4.2.2.1 Formulas for Missile Perforation of Targets 3D.4-8 3D.4.2.2.2 Steel Targets 3D.4-9 3D.4.2.2.3 Concrete Targets 3D.4-ll 3D.4.2.2.4 Damage Probability to Piping and Critical Equipment 3D.4-12 3D.4.2.2.5 Containment Perforation Probability 3D.4-12 3D.5 PUMP SPEED ANALYSIS UNDER LOCA CONDITIONS 3D.5-1 3D.5.1 Flow Loss Coefficient 3D.5-2 3D.5.2 Combining of Loss Coefficients 3D.5-5

( 3D.5.3 Two-Phase Flow Rates 3D.5-7 3D-i

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O CONTENTS (Continued)

Section Title Page 3D.5.4 Determination of Critical Plow 3D.5-7 3D.5.5 Pump Volume Flow Rate 3D.5-8 3D.5.6 Pump Windmilling Speed 3D.5-10

, 3D.5.7 Pump Maximum Torque 3D.5-12

.i 3D.6 REFERENCES 3D.6-1 3D.A ATTACIIMENT A TO APPENDIX 3D 3D.A.1 O

O 3D-li

i 1

GESSAR II 22A7007 1 l

4 238 NUCLEAR ISLAND Rev. 0 l l

() APPENDIX 3D TABLES t

Table Title Page

~

i 3D.2-1 Recirculation Pump Data 3D.2-5 3D.2-2 Discharge Line Guillotine Breaks 3D.2-5

3D.2-3 Suction Line Longitudinal Breaks 3D.2-6

) 3D.2-4 Suction Line Guillotine Breaks 3D.2-6 i

ILLUSTRATIONS Figure Title Page 3D.2-1 Complete Pump Characteristics - Double-i Suction Pump (N g = 1800) 3D.2-7 3D.2-2 Singl'e-Phase Full Performance Map for ,

Test Pump 3D.2-8 i

3D.2-3 Zero Torque Lines 3D.2-9

() 3D.2-4 3D.2-5 Double Volute Pump Casing Recirculation Loop Schematic Diagram 3D.2-10

! with Break Locations 3D.2-ll 3D.4-1 Recirculation Loop Break Designations 3D.4-15 3D.5-1 Head versus Flow (N s = 1800) 3D.5-13 3D.5-2 Head versus Flow (N s = 4200) 3D.5-14 3D.5-3 Flow Loading 3D.5-15 3D.5-4 Vessel Equivalent Total Pressure as a 3D.5-16 Function of Flow Loading 3D.5-5 Torque versus Reverse Flow - Pump at i Zero Speed 3D.5-17 3

3DA-1 Critical Flow at Longitudinal Break 3D.A-33

! 3DA-2 Pressure at Pump Cut Water 3D.A-34 3DA-3 , Complete Pump Characteristics - Double-

. Suction Pump (N g = 1800) 3D.A-35 3DA-4 i Pump Flow Volume 3D.A-36 2 i.

O i

3D-ii/3D-iv

GESSA A II 22A7007 238 NUCLEA!1 ISLAND Rev. 0

/~N ANALYSIS OF RECIRCULATION MOTOR AND

()

PUMP UNDER ACCIDENT CONDITIONS 3D.1 INTRODUCTION Given a major recirculation line break which produces high volume flow, the recirculation pump will operate as a hydraulic turbine with speed proportional to flow rate. Evaluations of the theoret-ical pump speed for various recirculation break types and conse-quences of predicted speed are presented in this appendix. This appendix also presents a aiscussion and analytical explanation of pump speed under loss-of-coolant accident (LOCA) conditions, the potential for motor-generated missiles, and the analyses necessary to support a conclusion that missiles will not be generated or the impact of potential missiles is acceptable.

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l v

3D.1-1/3D.1-2

GESSAR II 22A7007 238 NUCLEAR ISLAND Rsv. 0 3D.2 PUMP SPEED UNDER LOCA CONDITIONS

)

Complete operating maps or four-quadrant pump curves, for specific speeds of 1800 and 4200 are shown in Figures 3D.2-1 and 3D.2-2, respectively. Of particular interest when considering pump maxi-mum spe( d are the zero torque characteristic lines in the first (forward flow and speed) and the third (reverse flow and speed) quadrants of Figures 3D.2-1 and 3D.2- 2. These are shown in Figure 3D.2-3 for both the 1800 and 4200 specific speed pumps. All boiling water reactor (BWR) recirculation pumps have specific speeds between 1800 and 4200. The highest speed curve per volume flow rate of Figure 3D.2-3 has been used to predict pump speed

, during postulated pipe breaks.

The analytical method used to predict pump speed involves the calculation of maximum two-phase mass flow rate through the pump and the specific volume of the two-phase mixture in the pump for

N postulated breaks. The product of mass flow rate and specific

(%- volume results in maximum two-phase volume flow rate through the pump. This flow rate is used in conjunction with Figure 3D.2-3 to calculate pump maximum speed.

The two-phase flow rate is dependent on recirculation potential choke points. These are:

1. recirculation pump cutwater area (Figure 3D.2-4);

2. recirculation pump impeller eye area (Figure 3D.2-4); and

3. jet pumps.

Table 3D.2-1 summarizes BWR/6 recirculation pump characteristics and Figure 3D.2-5 shows a recirculation loop schematic with break locations.

O O

3D.2-1

GESSAR II 22A7007 238 NU" LEAR ISLAND Rev. 0 3D.2.1 Discharge Line Breaks Discharge line breaks can be either longitudinal or circumferential and in sizes up to and including a guillotine break. All discharge line breaks result in forward flow of a two-phase mixture through the pump, causing forward speed of the pump assuming loss of power to the recirculation pump motor. The guillotine breaks result in the highest flow rate and potential pump speed.

Calculated pump flow and speed for discharge line guillotine breaks are given in Table 3D.2-2. All other break configurations (longitudinal and connecting branches) in the recirculation dis-charge line will result in speeds less than those shown in Table 3D.2-2 for the guillotine breaks.

3D.2.2 Suction Line Breaks Suction line breaks can fall into several categories: longitudinal and guillotine breaks in the main line and longitudinal and guillotine breaks in connecting lines. Analysis of connecting line breaks is not required since the connecting line flow areas will be smaller than the areas associated with main line maximum longitudinal and guillotine breaks.

3D.2.2.1 Longitudinal Breaks In the event of a longitudinal break, the flow path will be from the reactor vessel to the break through the suction line and by reverse flow through the jet pumps and the recirculation pump. The latter path is a high-flow resistance / low-flow area path compared to the suction line. Thus, the low-flow resistance /high-flow area of the suction line path provides a high percentage of the flow to the break resulting in proportionately less reverse flow through the recirculation pump.

O 3D.2-2

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.2.2.1 Longitudinal Breaks (Continued)

Table 3D.2-3 gives calculated pump flow and reverse speed for the largest suction line longitudinal break (equal to an area that will produce critical flow in the suction line just upstream of the a

break).

I 3D.2.2.2 Guillotine Breaks In the event of a guillchine break,.the pump will experience reverse flow through the jet pumps, discharge headers, an'd revir-culation piping and out the break. Flow will also occur from the suction nozzle, down the suction line, and out the other end of the broken pipe. However, this flow will have no effect on reverse flow through the pump as the broken ends of the break are con-servatively assumed sufficiently separated to preclude jet inter-action and reduction of reverse flow through the pump.

O Table 3D.2-4 displays calculated pump torque and reverse speed for the suction line guillotine break. These values are based on

steady-state analysis assuming that full reactor pressure / flow is maintained long enough to fully sccelerate the pump to windmill speed associated with reverse flow.

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3D.2-3/3D.2-4

. - . . , - - - _ - _ - - - -- . . , _ . . . - . . - - __ - _= . . -

O O O Table 3D.2-1 RECIRCULATION PUMP DATA Impeller Rated Rated Number of Product Cutwater Eye Loop Pump Pump Pump Nozzle V2 Ant Line & Area Area Diameter Head Flow Nozzles Diameter

'ize Vendor (in.2) (in.2) (in.) (ft) (qpm) Per Loop (in) 218 BWR 6 - B 135 87 20 815 32,500 50 1.13 218 BWR 6 - A 120 250 20 815 32,500- 50 1.13 238 BWR 6 - A 175 275 24 780 42,000 50 1.30 238 BWR 6 - A 175 300 24 780 42,000 50 1.30 251 BWR 6 - A 175 280 24 835 47,100 60 1.24 $

Eo

.a Table 3D.2-2 @s yy DISCHARGE LINE GUILLOTINE BREAKS **

E 5U Product Max. Flow Ratio Speed Ratio 5 Plant Line Q/Q Rated N/N Rated $

Size & Vendor Slip Homogeneous Slip Homogeneous 218 BWR 6 - B 3.55 2.50 1.74 1.22 218 BWR 6 - A 4.84 3.48 2.37 1.71 238 BWR 6 - A 5.21 3.72 2.55 1.82 238 BWR 6 - A 5.26 3.76 2.58 1.85 251 BWR 6 - A 4.67 3.32 2.29 1,63 wU

%5 Pump Efficiency Considerations Not Included for Conservatism *8

Table 3D.2-3 SUCTION LINE LONGITUDINAL BREAKS Product Max Flow Ratio Speed Ratio Plant Line Q/Q Rated N/N Rated Size & Vendor Slip Homogeneous Slip Homogeneous 218 BWR 6 - B 1.0 0.9 1.8 1.5 218 BWR 6 - A 1.0 0.9 1.8 1.5 238 BWR 6 - A 1.0 0.8 1.7 1.5 238 BWR 6 -A 1.0 0.8 1.7 1.5 251 BWR 6 - A 1.0 0.8 1.7 1.4 m w

co Pump Efficiency Considerations Not Included for Conservatism g u o tn N

E$

4 Table 3D.2-4 xx SUCTION LINE GUILLOTINE BREAKS y[

C Product $

Plant Line Torque Ratio T/T Rated Speed Ratio N/N Rated Size & Vendor Slip Homogeneous Slip Homogeneous 218 BWR 6 - B 2.74 1.61 6.12 5.38 218 BWR 6 - A 4.18 2.47 8.35 9.01 238 BWR 6 - A 4.41 2.57 9.24 9.19 238 BWR 6 - A 4.54 2.66 9.31 9.52 251 BWR 6 - A 3.88 2.26 8.24 8.21 M g

$5 Pump Efficiency Considerations Not Included for Conservatism *$

o -a O O O

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O r

(

x s gi 1.,75%

V f .c

%i

? .,,f 7 N

\ f *'* .,,,

M;'a

'l*9

.*C

.,m _  ? K.E.90 sector, no missile penetration of the stator and stator frame will occur.

3D.3.2 Fan Failure W = fan weight = 78 lbs.

Rg = (outer radius of fan) = 18.00 inches Rg =

(inner radius of fan) = 13.25 inches I = b M (R 2 +R 2) total 2 o s 1

I g

=

3 2 8

2

+ 13.29) 144

4.2 lb ft sec 2 K.E. g = Iw 2

(4.2)(550)2

= 6.35 x 10 5 ft-lbs.

• 5

= ft-lbs = 1,59 x 10 ft-lbs K.E.90o sector 4 (assumes all energy is translational).

3D.3-6

GESSAR II 22A7007 238 NUCLEAR ISLAND Rnv. 0 3D.3.2 Fan Failure (Continued)

Calculating the energy for the fan to penetrate the overhanging  !

4 ends of the stator coils and the stator frame:

( = 6.17 inches (missile diameter)-

D =

1

3. h5 3 where:

missile height = 3.375 inches, missile width = 36 inches E = ( . )( 000) 16000(3.5)2 s-stator coils 46500 .

+ 1500 !

!39.77) (3.5) y 4 ,I _

= 1.05 x 10 8 ft lbs O E s-stator frame

= (6.17)(7000 0 46500 16000(1)2

~

+ 1500 1 (39. 77)lgy)

(4 j ,

8

= 0.29 x 10 ft-lbs E = 1.34 x 10 8 ft-lbs s-total Since E s-toul > K.E.90' sector, no penetration will occur.

3D.3.3 Retaining Ring Failure I

• total o s

=f 22l l144 (19.82 2 + 19,11 2)

(

I = 9.65 lb ft sec 2 total 3D.3-7

GESSAR II 22A7007 238 NUCLEAR ISLAND Rav. 0 3D.3.3 Retaining Ring Failure (Continued)

K.E.

o al Iw = (9.65)(550)2

= 1.56 x 10 6 ft-lbs

= = 3.65 x 10 5 ft-lbs K.E.90 sector where weight of retaining ring = 118 lbs outside radius of retaining ring = 19.82 in.

inside radius of retaining ring = 19.11 in.

Calculating the energy to penetrate overhanging ends of stator coils and stator frame:

D = (2) (39.64) (5) = 8*88 O

39.65 + 5

( ( }

E =

46500 16000(1)2 s-stator frame _

39.64

+ 1500 (1)

E = 412595 ft-lbs s-stator frame

~

h, " (8.88)(32000) 16000(3.5)2 s-stator coils 46500 .

39.64

+ 1500 (3.5)

E .

= 1515688 ft-lbs s-stator coils 6

= 1.5 x 10 6 + 0.4 x 10 s-total 1.9 x 10 ft-lbs.

3D.3-8

GESSAR II 22A7007 238 NUCLEAR ISLAND R2v. O 3D.3.3 Retaining Ring Failure (Continued)

O Since Eg _ g y > K.E.90 sector' no Penetration by the retaining 4 ring is possible.

3D.3.4 The Motor as a Potential Missile Since pump shaft failure decouples the motor from the driving blowdown force, only those cases with peak torques less than that required to fail the pump shaft (five times rated) will have the capability to drive the motor to overspeed. An analysis showing the load carrying capacity of pump shafts follows.

Conservatively taking the following pump shaft cross section below the pump half coupling and with no allowance for stress con-centration factors associated with keyways, the pump shaft will fail as follows:

O S s

= 16T(12)

D3 (70,000)(0.5777) = (16) T(12)

(68 )

(70,000)(0.577) (6 8)

T =

(12) (16)

T = 142,750 ft-lbs f= ,f-f = 5.6 times rated torque where I

I Sg = 0.577 x ultimate tensile strength (accepted ratio of ultimate shear strength to ultimate tensile strength)

A Shaft torque (ft/lbs)

T =

l

()

3D.3-9 i

i_ . __. , . _ _ _ _ - . _ _ _ _ _ _ _ . _

GESSAR II 22A7007 238 NUCLEAR ISLAND Rov. 0 3D.3.4 The Motor as a Potential Missile (Continued)

D = Shaft diameter (inches)

T = Rated torque (ft-lbs).

r Taking the pump cross section at the midpoint of the coupling and allowing for stress concentration, we have:

S s

D3 1 (

(70,000)(0.577) =

(5.5)'

T = 109,954 ft-lbs I = 4.31 times rated T

r 2.2 SHEAR STRESS 12 -

1.6 -

l 8

h/r 0.1 1.4 -

0.2 h

b '

+h! N 0.5 1.0 - r 44 n 08 -

MAXIMUM SHE AR STRESS 06 O 0.2 0.4 0.6 OB 1.0 R ATIO y/h 3D.3-10

7 _

.s GESSARdI .22A7007 238 NUCLEAR ISLAND j :Rsv. O s

3D.3.4 The Motor as a Potential Missile (Continued) (

Including stress concentration factor-m h ~

~ ~

y = 0.75 - = 0.27 h = 0.75 k = 1.25 r = 2.75 h1 = 1.00 ~

h=

r 1, = 3.45 times rated.

Since the load-carrying capability of the pump-to-motor bolts is -

very much greater than the load-carrying capability of the pump _ .

shaft, it is clear that no mechanism occurring in the pump can -

transmit sufficient load to cause the bolts holding the pump to X .

the motor to fail. -

4 ,

A possibility does exist, however,'that a seizure of the~ motor

1. rotor to the motor stator could p,roduce a failure of the bolts holding the pump to the motor. The rotational kinetic energy of ,

4 the moter rotor would upon seizure be converted into; strain energy '

in these bolts.

s.

Any discussion about the capability of these bolts'to wiEhs. tan'd a l sudden seizure will depend upon the rotational speed at which g seizure occurs and the duration of time required for deceleration.'

It should be understood that since torque equals rotational ine'rtia times angular acceleration, any change in speed must be associated with some finite time thus making an instantaneous seizure a ,

• ~

physical impossibility. -

.'s u g %.

i The following calculation provided demonstrates that the bolt.ings s .

, 1 that secures the motor to the pump is sufficient to survive an \

7Q "

incredible deceleration withqttt failure. When the argument t$at_

the bolting can survive a near instantaneoun deceEeration is '

1

, s g

. .W 3D.3.-11 i  % ,

S.

N '%, g .

• s R '*

% t

. . _ _-. ._. ._ ..__._ ____ u

GESSAR II 22A7007 238 NUCLEAR ISLAND R3v. 0 3D.3.4 The Motor as a Potential Missile (Continued) taken together with the fact that the pump shaft will fail with more than 5 times rated torque, there is sufficient evidence to show that these bolts will not fail. Furthermore, for the motor to escape as a missile, three seismic shock suppressors and the motor shaft would also be required to fail, thus further adding to the conservatism of these calculations.

Typical Large BWR Recirculation Pump:

OD of rotor = 42 in.

Weight of rotor = 12,300 lbs Rated torque of motor = 23,402 ft-lbs WK 2 = 17,875 lb/ft 2 motor WK 2 =

825 lb/ft 2

12-1 1/2"-6 mounting bolts on 87 in bolt circle NOTE This data represents that, associated with motors having the highest possible inertial forces of our product lines,

, this should present the most conservative picture of load carrying capability.

To calculate the maximum torque carrying capability of pump-to-motor bolts:

=

2T T

d "^#

b d "A# '

T =

b 0

3D.3-12

GESSAR II 22A7007 238 NUCLEAR ISLAND R:v. 0 3D.3.4 The Motor as a Potential'Mi3sile (Continued) i T = (87)(12)(1.392) T 2 ,

in in.2 lbs in.2 Tbolts = 726 T in-lbs for A193B7 bolts min ult. = 125,000 psi tensile strength T = (0.577)(125,000)(726) bolts

, = 5.236 x 10 7 in. lbs where 0.577 is the ratio of ultimate shear strength to ultimate tensile strength

- \m Maximum pump torque that could continue to drive motor with-out failing pump shaft (torque required to fail pump shaft is estimated at five times rated torque)

T pump

=

5(23492)(12) ft-lbs in/ft.

T pump

= 1.404 x 10 6 in-lbs The torque available to decelerate the motor without failing the pump to motor bolts is:

T = T bolts

-T pump

= (5.236 x 10 7 - 1.404 x 10') in-lbs

! T = 5.095 x 10 7 in. lbs i

f O

3D.3-13/3D.3-14 9

GESSAR II 22A7007 238 NUCLEAR ISLAND Rav. 0 l

3D.4 PUMP IMPELLER-GENERATED MISSILES This section assesses the consequences of impeller-generated missiles. As in the case of the motor-generated missiles dis-cussed in Section 3D.3, the bursting speed of the pump impeller is taken to be equal to the 5252 rpm bursting speed calculated for the motor rotor.

Assuming the pump impeller bursts, there are two ways in which a missile can become external to the pump casing: impeller penetra-tion of the pump casing or escape of the missile through the break in the recirculation piping. Both of these possibilities are examined in this section.

3D.4.1 Impeller Penetration of the Pump Case The analysis provided in this subsection calculates the energy of a 90* section of a complete impeller (this missile possesses the maximum translational kinetic energy). The translational energy in the missile is compared to that required to penetrate the pump case.

Missile Kinetic Energy:

2 WK 2 for impeller = 900 lb/ft I = = = 27.95 lb ft sec g 32 2 K.E. TOTAL

= Iw = (27.95)(550)2

= 4.23 x 10 8 ft-lbs

= = 1.06 x 10 8 ft-lbs K.E.90' SECTOR O

3D.4-1

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.4.1 Impeller Penetration of the Pump Case (Continued)

Penetration Energy:

I* h (2) l I (E)

D = tl , (4 / = 20.84 in, t+R n_d_ , g 4

where D = effective missile diameter (in.)

d = impeller diameter (in.)

E = height of impeller (in.)

Using the Stanford missile equation:

E =

46500 16000 (3.5)2 + 1500 l (3.5)

S _ 4/ ,

E g

= 6. M x 10 6 Nh Since E g > K.E., no penetration of the pump case is possible.

O 3D.4-2

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O p

Q 3D.4.1 Impeller Penetration of the Pump Case (Continued)

+ 5/8 5/8 TYP - d j(

V v wxc:ox . n V

'/

h a e 5

V s I 17.14 J( 7 -

+331/2 0 Y,- l 9

e 1, '

V

, 231/2 dia

, 5 7/8 dia

) WT = 1550 lbs P

AVG = 0.15U lbs/in.3 T.

IMPE LLER CROSS SECTION 1

l 3D.4.2 Potential Pump Impeller Missiles from the Pipe Break In order to assess the consequences of a pump impeller missile exiting the pipe break, it is necessary to characterize the potential missiles and then examine their damage potential.

l 3D.4.2.1 Characteristics of Potential Pump Impeller Missiles l

l Break locations must be examined in the determination of missile l

characteristics. No breaks are postulated to occur at the b)

\s i

3D.4-3 1

l

GESSAR II 22A700/

238 NUCLEAR ISLAND Rev. 0 3D.4.2.1 Characteristics of Potential Pump Impeller Missiles (Continue )

terminal ends of the piping and branch runs. In addition, breaks are postulated to occur between terminal ends

• of each piping run or branch run:

(1) when the maximum stress range exceeds 2.4 S as calculated between any two load sets (including zero load set) according to Subarticle NB-3600 ASME Code Section III for upset plant conditions and an Opera-tional Basis Earthquake (OBE) load at locations where:

(a) The stress range is calculated using Equation (10) of the ASME Subarticle [if the calculated maximum stress range exceeds 2.4 S but is not greater than 3 S , no breaks will be postulated unless the cumulative usage factor exceeds 0.1 as calculated from Equation 14}, or (b) The stress ranges as calculated by Equations (12) and (13) exceed 2.4 S m r if the cumulative usage factor exceeds 0.1 as calculated from Equation (14) where Equation (10) exceeds 3S ; or (2) If two or more intermediate locations cannot be determined by stress or usage factor limits, a total of

• Terminal-ends are extremities of piping runs that connect to structures, components, or pipe anchors that in the piping stress analysis are assumed to act as rigid constraints to piping thermal expansion. A branch connection to a main piping run is a terminal-end for a branch run except when the branch and main run is modeled as a common piping system during the piping stress analysis.

3D.4-4

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O n

(b' ) 3D.4.2.1 Characteristics of Potential Pump Impeller Missiles (Continued) two intermediate lccations shall be identified on a reasonable basis

• for each piping run or branch run.

Break locations identified using these criteria are shown in Figure 3D.4-1. These locations are based on a representative ASME Code Section III piping analysis for a plant using standard-plant envelope seismic-spectra for the BWR/6 recirculation piping system in a Mark III containment.

Any instantaneous circumferential break in the suction line, some of the instantaneous circumferential breaks in the discharge line and some longitudinal breaks will generate destructive recircula-tion pump impeller overspeed that will generate missiles from the impeller failure. However, only instantaneous circumferential breaks in the suction line and in the discharge line before the

{~)

\- flow control valve will result in destructive overspeed of the recirculation pump. All longitudinal breaks and instantaneous circumferential breaks in the discharge line after the flow-control valve may result in overspeed, but not destructive overspeed.

• Reasonable basis shall be:

(1) Fitting locations, and/or (2) Highest Stress or Usage Factor Locations.

Where more than two such intermediate locations are possible, using the application of this reasonable basis, two locations possessing the greatest damage potential will be used.

f i

N3),

3D.4-5

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.4.2.1 Characteristics of Potential Pump Impeller Missiles f (Continued)

If a design basis loss-of-coolant accident occurs as an instanta-neous circumferential break in the suction line, the recirculation pump impeller will overspeed destructively and may break up in just about any way with various-size pieces being formed. The most likely breakup will result in five main blade segments, five main shroud segments, and numerous smaller pieces. Of these, the large shroud segments are the only ones of sufficient mass to cause significant damage. The size and shape of these shroud seg-monts are such that it would be very difficult for a shroud seg-ment to leave the pump and enter the piping system. To enter the piping the missile must be ejected with its longitudinal axis within a few degrees of a specific orientation and it must over-come the centrifugal force in the metal / fluid mixture; therefore the probability that one of the five possible large shroud seg-ments would escape into the pipe is very small. Assigning a con-servative probability of 0.10 to this occurrence, and assuming (in the worst case) that each of the five large shroud segments exist and have an independent opportunity to exit from the pump, the result is a total probability of 0.4 (which is derived from 1-0.9 5) for missiles of sufficient energy to enter the piping sys-tem after a design basis LOCA break.

3D.4.2.1.1 Missile Velocity in the Piping System The missile velocity is calculated using the following time-d pendent numerical integration formula:

(V g -V g)2 Vg = Vg +P cat f 2'd DM o

O 3D.4-6

GESSAR II 22A7007 238 NUCLEAR ISLAFD Rev. 0

.3D.4.2.1.1 Missile Velocity in the Piping System (Continued) where W = weight of the missile, lb f Pg = density of water-steam mixture, lb/ft 3 V = fluid velocity,.ft/sec; f

V = initial missile velocity, ft/sec; M

o C = drag coe M cient, dimensionless 9

A = m ssile reference area, Pd ;

M t = time, s.

This equation is valid for straight pipes only. When a missile passes through a 90-degree elbow, most of the velocity is lost due to impact and friction. The velocity after passing through the elbow is conservatively estimated to be half of the velocity of the velocity before the elbow. No velocity loss is ascribed to passage through a tee.

The following typical values are used in the calculation of missile velocity.

W = 67 lb (weight of missile) 3 8 I P f

=

2.222 lb/ft for suction line, and 12.5 lb/ft for discharge line V = 441 ft/sec for suction line, and 208 ft/sec for dis-f charge line ,

C = 0.484 D

A

• M Due to the lack of velocity in the direction of the pipe, the initial missile velocity when escaping into the suction line is zero. For exit into the discharge line, the initial velocity of the missile is conservatively estimated to be 104 ft/sec, which is one-half of V g.

3D.4-7

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.4.1.2 Escape Pattern The only possible escape route is through the break, since any missile escaping into the broken pipe does not have enough energy to perforate the piping system when it hits an elbow, a tee, or another change-of-direction fitting.

If the gap between the two adjacent ends of pipe is less than or equal to one pipe diameter, the missile will either hit the pipe end opposite the end from which it is escaping and be stopped or it will penetrate the other pipe end and its direction will be reversed within two feet of the broken end. The missile will ricochet about the region of the break until it exists. The escape velocity will be very low because the exit point is not preceded by a long run of straight pipe.

If the gap is more than one pipe diameter, the missile will escape the confines of the pipe with a flight path which lies within a gg trajectory impact cone with a half angle of 14.5 degrees. It is furthermore estimated that the possibility for escape is equally distributed over the cone. The escape velocity will remain rela-tively unchanged until the missile impacts.

3D.4.2.2 Damage Potential of Escaped Missiles Possible consequences of missiles being formed ar.d expelled are:

(1) perforation of the primary containment; (2) perforation of a main steamline; (3) perforation of a feedwater line; (4) perforation of an ECCS line; or (5) destruction of an inboard main steamline isolation valve operator.

3D.4-8

. . . - - - . . . ~ _ _ - . . .. .-_ . . .. .. . -

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O i

s 3D.4.2.2 Damage Potential of Escaped Missiles (Continued) 1 The possibility of each of these occurrences is estimated in the event of a design basis LOCA occurring at each of the break loca-tions from which a missile could be expelled.

3D.4.2.2.1 Formulas for Missile Perforation of Targets In assessing the damage potential, only the velocity at impact perpendicular to the surface of the material is' considered when

! calculating the minimum thickness needed to-prevent perforation.

The following symbols are used in the perforation analysis.

A Impa t angle, target surface to velocity vector 2

(degrees)

A * "I" 3 ^2 B = 8D"  ;

bg = 4 inches"  !

D = Missile equivalent impact-point diameter (in.)

Di = Missile minimum perforation-cross-section diameter (in.)

E = Energy required for perforation (in./lb)

I T = Minimum thickness of steel target needed to prevent 2

perforation (back of target unsupported)

, U = Ultimate tensile strength of steel target (1b/in 2) j V3 = Missile velocity (ft/sec)

W = Missile weight (lb) l X = Minimum thickness of 4000 psi concrete target needed to prevent perforation (inches) 1 3D.4.2.2.2 Steel Targets Stanford Research Institute (SRI) has developed an empirical per-foration equation for the case of small, tool-steel, right, circular cylinders fired against a low-strength carbon-steel plate.

I l

3D.4-9 I

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.4.2.2.2 Steel Targets (Continued)

This equation is not fully applicable to the present case of a tumbling missile of complex shape made of soft, fully-annealed stainless steel. However, as discussed later, the SRI equation is considered to be conservative. The SRI equation is:

U l 16,000 B I E/D = (T 500 T 46500 1 2)2 + B

\ S 2}I

• Setting the length of a square side between rigid supports (B) equal to 8D and the length of a standard width (Bg) to 4 inches and solving the equation for T 2 considering that:

E =

32 2 ( A) 3 3 , yi lds 0.04514W (V3A )2 3

2 T = +

-0.09372D+\0.008783D 2 DU NOTE The plate used by SRI was rigidly held and the missiles used were made of tool steel. The recirculation missile is made of relatively soft stainless steel and the pipes are flexible relative to the SRI target plate; this makes the calculated minimum material thickness of the steel targets conservative.

The calculated escape velocity of a 67-pound missile passing through a 60-foot straight pipe (which is longer than the length of the total suction line including elbows) is 240 ft/sec when its initial velocity is zero. The fluid velocity in the discharge line is 208 ft/sec which is the maximum velocity the missile can attain. In both cases, a 1.3-inch-thick steel target will resist perforation (this is a conservative value).

3D.4-10

1 GESSAR II 22A7007 g

238 NUCLEAR ISLAND Rev. 0

() 3D.4.2.2.3 Concrete Targets The three empirical performation equations considered for 4000-psi-ultimate-strength concrete targets are the Petri, the Ballistic l

Research Lab (BRL), and the Amman. All three equations are based I on armor-piercing projectiles, not on annealed stainless-steel mis-siles, which makes the use of any one of the three equations conservative:

Petri Equation:

~

~

[ (V3A3) 2) 6.072436 W Log X = 2.0 10 f+215000j (Dt )2 ,

BRL Equation 2 o

C

~

! (V3 A3 \*

6.7515 W X = 1.25 (1000/

(D 1)l* .

i Amman Equation:

/v3As\

• 3.76025 W i

X = 2.0 (1000)

- * ~

l (D 1) l l

, NOTE In all three equations, the first number to the right of the equal sign is a safety factor (2 for Petri and Amman, and i 1.25 for BRL).

l 3D.4-ll l

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.4.2.2.4 Damage Probability to Piping and Critical Equipment f Only the identical postulated breaks RSlA and RSlB can generate recirculation pump missiles (Figure 3D.4-1). The calculated missile velocity is 142 ft/sec. If the pipe ends are aligned, the missile can penetrate only 0.83 ft against the flow of fluid; if the pipe ends are not aligned, a minimum of 0.73 inch thickness of steel is required to prevent perforation.

For the identical breaks RSlA/B, the pipes are secured with pipe-whip restraints that restrict a broken pipe end to less than one pipe diameter movement which is a completely safe condition.*

If the pipe end could move enough to allow a missile to escape, the breaks would still be completely safe, as the only possible target is the reactor vessel with a 5-inch thick wall which is considerably thicker than the 1.3 inch value required to prevent perforation. The conclusion is that, given the breaks postulated for the BWR/6 Mark III power plants, no pipe or critical equipment will be damaged by a recirculation pump missile.

3D.4.2.2.5 Containment Perforation Probability The BWR/6 Mark III Reactor Building has a free-standing steel con-tainment which contains the drywell. To perforate the steel con-tainment, a missile generated within the drywell must either pierce the drywell with sufficient energy remaining to perforate the steel containment or generate a secondary missile powerful enough to perforate the steel containment.

For a 90-degree impact (A =1) of a 67-pound missile having a 3

minimum cross-section diameter of 3.6 inches, the minimum required

• A break is considered safe if the postulated missile is contained within the piping; leaves the piping system at a velocity insuffi-cient to perforate the containment or an essential piping system; or will impact a nonessential target which does not escalate the consequences of the accident.

3D.4-12

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O i ,

i

() 3D.4.2.2.5 Containment Performation Probability (Continued) j- thickness (x inches) of 4,000 psi reinforced concrete has been computed using all three equations for various missile velocities (V3 ).

i i

X (inches)

Vs(ft/sec) Petri BRL Amman 550 24.0 25.5 17.2

. 500 21.1 22.5 14.5 4

450 18.1 19.5 12.0 400 15.2 16.7 9.7 350 12.3 14.0 7.6 300 9.6 11.4 5.8 250 7.0 9.0 4.2 200 4.7 6.7 2.8 l

n\' The calculated escape velocity of a 67-pound missile passing through a 60-foot straight pipe is 240 ft/nec, when the initial velocity is zero. The suction line has a total length of less than

! 60 feet. The fluid velocity in the discharge line is 208 ft/sec, which is the maximum velocity the missile can attain there. In both cases, a 9.0-inch, 4000-psi reinforced-concrete wall resists i perforation. Furthermore a 21-inch thick, 4,000-psi, reinforced-concrete wall will resist spalling.3 These calculations are conservative.

The minimum strength required for BWR/6-Mark III reinforced dry-i well concrete is 4,000 psi and the maximum missile velocity is less than 250 ft/sec. To avoid perforation at that velocity, the dry-well wall, the drywell ceiling, and the drywell weir wall thickness must be at least 9.0 inches using the most conservative equation.

}

This is well below the wall thickness of even the weir wall which is 22 inches thick.

Os 3D.4-13 1

..--_._~.....__.-~_,-..-.--_,...__...._---.-..____.._._--.._,--_.-.._._...-4. -,-----.-__.,-m- , - -

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.4.2.2.5 Containment Perforation Probability (Continued)

The dome above the reactor cannot be reached by a missile due to the dome size and placement. The dome has the same diameter as the shield wall and is placed above the shield wall. The missile attempting to enter the dome would hit the shield wall or the ceiling and be stopped.

The conclusion is that the containment cannot be perforated by any recirculation-pump-generated missile.

O O

3D.4-14

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O O lRosAiaI r AndAlsj 2 O* RISER Ao,a 0

(F 140' 90

.. RD3 A/8l f RS1 A/8 RISER NO,4 001 A/S I l RO2 A/8l zf s - -IRoe AiB l

~

C MRD7 A/8l roser NO. , N RISER NO.3 lRO9 A/B si g , ,

0 - N 3 ,

N HEAOgg

([ DISCHARGE RHR LOOP 8 ONLY M V V t k~ ~

NOTES

^ BREAK RSt A on TH L Op o pgg ON THE 8 LOOP Figurc 30,4_3- Recirculation L P Break Designations 3D.4-15/30,4_3g

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.5 PUMP SPEED ANALYSIS UNDER LOCA CONDITIONS The analysis made for this appendix is a limiting speed calculation assuming full reactor pressure /two-phase flow exists long enough to accelerate the pump and motor to full windmilling speed. This is conservative since large recirculation line breaks cause rapid vessel depressurization and time-dependent windmilling speed will not be achieved.

The general approach to the analysis is:

(1) The flow loss coefficients of the piping, fittings, jet pumps, and the recirculation pump in the flow path from the reactor to the point of the postulated break in the recirculation line were calculated.

(2) Using the two-phase mass flow rate versus loss coef-4

' ficients for both slip flow models and homogeneous flow

( -models, the flow loadings at potential critical flow points in the path were calculated. This is either in the recirculation pump cut water, impeller eye area, the jet pumps for the guillotine breaks, or at the break area for longitudinal breaks.

(3) The critical mass flow ratc is used to determine the pressure in the pump and the void fraction in the pump.

The mass flow rate and average specific volume based on static pressure are used to determine the volume flow rate in the pump.

(4) The zero torque curves of the reported Ns = 1800-rpm and Ns = 4200-rpm four-quadrant pump curves are used i to predict runaway speed.

O U

3D.5-1

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.5 PUMP SPEED ANALYSIS UNDER LOCA CONDITIONS (Continued)

NOTE The analysis conservatively calculates the highest speed based on:

(a) two-phase flow rate and void fraction based on the slip flow or the homogeneous flow model (b) the worst case of runaway speed for Ns = 1800 rpm or Ns = 4200 rpm from reported four quadrant pump curve.5

(5) For the special case of the guillotine suction line break, the pump loss coefficient at zero speed is cal-culated, and (1), (2), and (3), are recalculated to determine peak torque applied by the impeller to the shaft. This permits analysis for the potential of shearing the pump shaft and calculating the rate of acceleration of the pump versus the vessel depressuriza-tion transient.

3D.5.1 Plow Loss Coefficient The use of normal loss coefficients for two-phase analysis is considered a6 equate since the single-phase characteristics predominate over the region where most of the hydraulic losses occur. Furthermore, steam quality never gets very high making this assumption even better.

Hydraulic losses in a pipe, fittings, or valves are characterized by the equation:

fL V 2 KV 2 0"

D- 2g 2g 3D.5-2

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0

() 3D.5.1 Flow Loss Coefficient (Continued) where-AH = loss, ft of fluid f = Friction factor, dimensionless

= equivalent length to diameter ratio of the component, dimensionless V = fluid velocity, ft/sec g = gravitational constant, 32.2 ft/sec 2 f

K = loss coefficient = gL .

For the pump speed / torque analysis, the following loss coefficients were used:

elbows = 0.2

() gate valve = 0.2 jet pump = 0.2.

The elbow and valve coefficients are taken as 80% of the values reported by Crane 7. The jet pump coefficient is for reverse flow under LOCA conditions and is based on 0.5 for sudden contractions (vessel into the jet pump nozzle) and 0.1 for diffuse-type expansion into the jet pump piping system.

t The recirculation pump is also a fitting which produces a loss coefficient whether the pump is windmilling or at zero speed.

Figures 3D.2-1 and 3D.2-2 are four-quadrant pump curves for specific pump speeds for 1800 and 4200 rpm, respectively '". A 3

replot of Figures 3D.2-1 and 3D.2-2 show head loss across the pump versus flow through the pump for several speed conditions gives Figures 3D.5-1 and 3D.5-2. These figures closely approximate l

the hydraulic loss coefficients and can be represented in the

() following form:

I 3D.5-3 i

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.5.1 Flow Loss Coefficient (Continued) l AHp = KP IYPI 2g where AHp = pump loss (ft)

Kp = pump loss coefficient in terms of piping system flow area (Attachment A, Section A, paragraph II.3)

D Vp = velocity in pump in terms ,f piping system flow area (ft/sec)

Now D Op vp =

DL where Op = pump flow (ft 3/sec)

A DL

= scharge piping flow area Ut2)

NOTE All BWR/6 pumps have suction piping diameters equal to discharge piping diameter.

Therefore D

2g AHpA D

Kp = .

2 Op 0

3D.5-4

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.5.1 Flow Loss Coefficient (Continued)

The method of analysis used for pump speed is to:

(1) Determine the value of 100H/hr at 300% rated flow from Figures 3D.5-1 and-3D.5-2 to minimize. the effects of Figures 3D.4-1 and 3D.5-1.to deviate slightly 1 m 2

a0 or V 2 equation; (2) Calculate AHp for this equation-as Hp = Hr(100H/Hr);

(3) Use the lesser of AHp determined from Figures 3D.5-1 and 3D.5-2 to minimize loss and maximum flow rate; and (4) Using the calculated flow rate, apply iterative pro-cedures and recalculate pump loss coefficient and then flow rate until calculated and assumed flow rates equal each other.

3D.S.2 Combining of Loss Coefficients Subsection 3D.4.1 notes that the pressure drop across a component is characterized by:

=

KV 2 AH 2g where V = the velocity in the component.

In a piping system, flow areas may_not be constant along the entire flow path. A series of translations and summations can be made O

3D.5-5

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.5.2 Combining of Loss Coefficients (Continued) along the flow path to determine total flow loss coefficient by recognizing that:

/A B V = V I 1 A B(Agj where V = velocity at point A Vg = Velocity at point B A = flow area of point A A

B

= fl W r f Point B.

As an example of the method used to combine loss coefficients, consider the flow system shown below:

FLOW DIR ECTION 7 g l Y D1 D2 D3 V I v 4

ja xi

>l< x2

-l: x3

>-l A B C D g _

fL v _ KV 2 D 2g 2g D1 K = K1 (loss coefficient at B with respect to flow area of D1)

O 3D.5-6

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 7w 3D.5.2 Combining of Loss Coefficients (Continued) b

K = K =K (loss coefficient a,t B with respect to flow area of D2)

K = K + K2 (loss coefficient from A to C with res-A pect to flow area of D2)

D3 K = K (loss coefficient from A to D D2 (/D3 " +-K3 with respect to flow area of D3).

3D.S.3 Two-Phase Flow Rates Figure 3D.5-3 is a plot of maximum flow load (lbm/sec/ft2) as a function of loss coefficient of the piping system for the reactor vessel to a check point of interest in the piping system or to the break point. Both the slip flow model and the homogeneous model g- g curves are based on initial reactor conditions prior to postulated

(, / recirculation pipe breaks of 1020 psia and saturated water enthalpy of 545 BTU /lb.

3D.5.4 Determination of Critical Flow The method of analysis used is to determine critical flow at potential choke points in the recirculation flow path using Figure 3D.5-3 to define the critical flow rate as a function of loss coefficient from the reactor vessel to the point being l investigated. The following potential choke points exist.

l ll Discharge Line Breaks 1

(1) Pump impeller eye (2) Pump cut water area l

O 3D.5-7

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.5.4 Determination of Critical Flow (Continued)

Suction Line Breaks (1) Jet pump nozzle (2) Pump impeller eye (3) Pump cut water area See Figures 3D.2-4 and 3D.2-5 for location of these areas.

The applicable loss coefficient (K) to be used with Figure 3D.5-3 is calculated using the methods of (Attachment A, Section A, paragraph II.3) considering the loss coefficients in path to the potential points (Figure 3D.2-5).

The maximum mass flow rate in the piping system is equal to the mass flow rate associated with the smallest or critical volume flow rate at a choke point. Flow loading at any point becomes critical mass flow rate times specific volume times area taken at the point of interest.

3D.5.5 Pump Volume Flow Rate Figure 3D.5-4 shows total pressure at the entrance to a flow restriction in the line as a function of flow loading at this point for both the slip model and the homogeneous model.

The volume flow rate during two-phase blowdown is a function of mass flow rate at the point where flow chokes and the specific volume is a function of the quality under isenthalpic expansion from saturated water at reactor. vessel enthalpy to pressure condi-tions in the pump and the two-phase flow model used, slip or homogeneous.

O 3D.5-8

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0

) 3D.5.5 Pump Volume Flow Rate (Continued)

kJ Pressure conditions in the pump are calculated as follows:

4 (1) Average static presoure in the' pump is calculated by averaging the static pressure at.the pump cut water and the pump _ impeller eye. Static pressure is taken to be equal to 60% of the stagnation pressure for the slip flow case and 80% of the stagnation pressure for homogeneous flow.

(2) Flow rates for discharge line breaks are based on the cut water area.

(3) Flow rates for suction line breaks are based on the impeller eye area.

Two-phase quality in the pump is calculated by using isenthalpic

[}

' expansion of saturated water at initial reactor pressure-to-pressure conditions in the pump determined from Figure 3D.5-4.

Specific volume of the two-phase mixture in the pump depends on the model used. For slip flow model:

i 1-X "I

I I "SI + I (1 - X 7 ) v py] X7+ K I

where K

7

=

(v g7 /vpy) i For homogeneous flow model:

4

() "I FI + I ("SI ~ "FI}

! 3D.5-9 I

ex- -

e -

--e-. - .w<, - - - - - - . - -

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 3D.5.5 Pump Volume Flow Rate (Continued) where O

X = two-phase steam t,uality in pump 7

v py

= saturated water specific volume at pump pressure 3

(ft /lb) v g7 = saturated steam specific volume at pump pressure 3

(ft /lb).

After the two-phase specific volume has been calculated, the two-phase flow volume is calculated by:

Op = MpV x 7.4 7

where Op = pump volume flow rate (gpm)

Mp = piping system flow rate (lb/sec) v = specific volume of two-phase flow mixture (ft 3/lb) 7 3D.5.6 Pump Windmilling Speed The previous subsections show the method used to calculate the volume flow rate and specific volume in the recirculation uump under postulated recirculation line breaks. It can be shown analytically that the maximum impeller blade velocity under zero torque windmilling operation will be a linear function of pump volume flow rate. Figures 3D.2-1 and 3D.2-2 repo:.-ted liquid phase four-quadrant curves for pump specific speeds of 1800 and 4200 rpm, respectively. A replot of Figures 3D.2-1 and 3D.2-2 showing maxi-mum pump speed under zero torque (windmilling) conditions versus pump volume flow rate is presented in Figure 3D.2-3.

O 3D.5-10

. +

.- g GESSAR II 22A7007 j

238 NUCLEAR ISLAND \ .

Rev ?. O i J

. i's , .;

3D.5.6 Pump Windmilling Speed '(Continued) .

~ , .s i -

% s z.y ( '

)

,~ .

Several remarks are in order concerning Figure 3D.2-3. -

y p s . . - ., .

A. '

\< <cs  ; -

(1) The curves are based on a'mo,delihaf >- single-phase 1] quid

\.

flow through the pump. Tne'analysi's used in this replot assumed a two-phase f1'ow a'nalysls%alth behavior as a I single-phase liquid flow Qhile, in fact, blade tolocity ,

l under two-phase flow ic~expecded to be less t$ddj'predic- [

ted blade velocity under_equiv51enh singlo PNaae flow. ,

i \ -

g ,

. N '.

(2) The analysis assumed BWR pumpc Will respond in accord- ,

ance with the curve of N g (pumyspecificspeed) =

4200 rpm; while in fact,..all BWR pumps have specific ,

' ~

speeds less than 4200 rpm.y '

n

. t With the conservatism noted above, the zero torque' (freewhdelincj)

N 4 speeds have been predicated on conservative basis of \ \ l' N = 4200 rpm for discharge line breaks and Ns = 1800{rphfor; l s

suction line breaks. The N =~4200-rpm curve more accurately s s

accounts for the reluctance of the pump to run'at near zero flow s l

! due to shaft and motor bearing friction. -

s 'T- ..

] Therefore, the analysis uses:  !

s- N S  %

(1) For forward flow (discharge line breaks) -

,' ,5 *,

N/N R "

OR 0/2 W h",,

.i l ,

(2) For reverse flow (suction line breaks)

l . i  % ,

. ,- , l N/N R =

(0/QR ) G00 7 5). 7 ,[, y

- u

,s.

y 4 - r

i. i ..f 7  !-

n%

q .

s ,

4 -

d ,-  %'***

3D.5-11 ..s -

* * \ u .. < .

• ~

i t

=_m-.,__m,___- _ . _ _ , _ _ _ , , . _ . , _ _ _ . . . . - , - _ - . , _ . . _ m.. _ , _ . _ . , , , , , . _ . . . _ . - , = , . , , , .

s GESSAR Il 22A7007 238 NUCLEAR ISLAND Rev. 0

). -

3D.5.7 Pump Maximum Torque in tne evont that pump speed, based on a steady state, freewheeling upeed exceeds acceptable limits, it is desirable to calculate torgue applied to the impeller by two-phase blowdown from two viewpoints:

(1) The torque may be sufficiently high to shear the impeller / shaft / motor train to free the motor from high acceleration torques.

(2) The torque may be sufficiently low to limit pump / motor acceleration to less than zero torque /windmilling speed before reactor depressurization occurs.

Figure 3D.5-5 is a replot of Figures 3D.2-1 and 3D.2-2 showing torque under reverse flow (suction line break) conditions. How-ever, this figure is based on single-phase flow (liquid). Applied torque, especially under reverse flow, is calculated (for both single and two-phase flow) using the following equation:

V T =

Cq Op .

Then applied torque at the impeller under postulated recirculation line breaks, assuming the recirculation pump is at zero spced, has been calculated.

O 3D.5-12 V

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 4

10 3 -

10 TURBINE AT T = 0 1 TURBINE AT N = 0 y .

5 6- a PUMP AT T = 0 s

2E ~

2@I O-O 5

~

t 2 -

10 H = 2.1 Q '

I '

H = 0.72241 O

• ISOII H = 0.2847 O 1 t I t 1l t

,f f l t t I f FLOW THROUGH PUMP (% reted)

(tooo/Og l

~

O Figure 3D.5-1. Head versus Flow (N s = 1800) 3D.5-13

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 10 10 TURBINE AT T = 0

^c -

TURBINE AT N = 0 x

?

PUMP AT T = 0 g_e a

A 3.

8I o8 h=

z 10

, i , , I i i , i I i i i i 3

10 0 1 2 3 10 39 10 10 FLOW (% Og l O

Figure 3D.5-2. Head versus Flow (N s = 4200) 3D.5-14

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O I

i 1

O 9000 sooo -

f l 7000 sooo -

5000 4000 -

'*2 x

g 3000 -

' O a E

3 5

d 2000 -

SLIP HOMOGENEOUS l

I I I I I I o 1o 20 30 40 So 60 70 l KaFUD Figure 3D.5-3. Flow Loading 3D.5-15

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O O

10 -

3 ~

E R

E i

!s O E

O 10 -

E 0N5 o - HOMOGEN EOUS (0.993076 3 e

SLIP (0.29345 GOM,

' ' ' ' i ' ' ' '

2 2 3 4 10 10 10 FLOW LOADING (Ibm /sec/f t )

Figure 3D.5-4. Vessel Equivalent Total Pressure as a Function of Flow Loading 3D.5-16

i GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 10*

l

- l 10 -

N, = 4200 N,= 1800 F- ~

6 i

5 O i -

l 2

10 T/TR= 1.37(Q/QR I' '

T/TR= 02(Q/OR I'

, e i i i l i i i i I i i e i 1

10 2 10 3 gg 4 10 FLOW (% Og )

I Figure 3D.5-5. Torque versus Reverse Flow - Pump at Zero Speed 3D.5-17/3D.5-18 1

' - - - - --- - - _ - - - ,-,o-_-,._-_y,

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O 3D.6 REFERENCES O

1. . R.C. Gwaltney, " Missile Generation and Protection in Light-Water-Cooled Power Reactor Plants", ORNL-NSIC-22, Oak Ridge National Laboratory, September 1968.

2. R.J. Roark, " Formulas for Stress and Strain", Fifth Edition, 1975.

3. Bechtel Corporation, San Francisco, California, Design of Structures for Missile Impact #BC-TOP-9, October 1972 and July 1973.

4. F.J. Moody, APED 4827, " Maximum Two-Phase Vessel Blowdown from Pipes".

5. A.J. Stepanoff, " Centrifugal and Axial Flow Pumps- Theory, Design, and Application", John Wiley & Sons, Second Edition.

6. EPRI-NP-128, "Two Phase Performance Program, Preliminary Test Plan", September 1975.

7. " Flow of Fluids Through Valves, Fittings, and Pipe", Crane Company, Technical Paper No. 410 0969).

O lO l

l 3D.6-1/3D.6-2 l

1 -_ _

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O O

ATTACIIMENT A TO APPENDIX 3D O SAMPLE CALCULATIONS O

- .. _ _ - - ._. - . - - . _ . _ _ . ~ . . . . - -. __

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O i

i Attachment A SAMPLE CALCULATIONS 4

NOMENCLATURE Definition Dimensions Computer A

CW Pump cut water flow area in.2 gy

A Pump cut water flow area ft 2 31 CW A

IE Pump impeller eye area in.2 A2 A Pump impeller eye area ft 2 A2 IE A gg Suction line flow area ft 2 A3 A Suction line friction factor --- A5 LOOPF l A Discharge line flow area ft: 34 DL D gL Suction line inside diameter inches D

() D DL Discharge line inside diameter inches 2

D1 G2 G Maximum cut water flow loading lb/sec/ft CW 2

G Maximum impeller eye flow loading lb/sec/ft G1,G2 IE 2

G Actual flow loading at cut water lb/sec/ft G3 FL

, hg Saturated water enthalpy at BTU /lb H3 i reactor pressure I

hf Saturated water enthalpy at pump BTU /lb H4 impeller pressure j h Enthalpy of evaporation at pump BTU /lb HS f9 impeller pressure H Pump rated head ft l R i K Suction line loss coefficient ---

K1 8 (vessel to pump suction) l l

K gyg Sucti9, n 1tne friction loss --- F5 l

coefficient K Suction line loss ;oefficient in --- K2 IE terms of impeller eye area l

3DA-1 r , - , - , -

., - _ ..- .,,.a ,, , , , , , . , -. -..- . .,.. ,.,.,..,: .. .n , ..n., , - - , . , ..,r

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 NOMENCLATURE (Continued)

Definition Dimensions Computer K Pump loss coefficient in terms of K3 P suction line area K

CW oss efficient, jet pump to K pump for suction line breaks.

Vessel suction nozzle through pump for discharge line breaks K. Jet pump nozzle loss coefficient K7 3

K.D Jet pump nozzle loss coefficient K8 3

in terms of discharge line flow area K

Loss coefficient of jet pump 180 K9 bend K

D Loss coefficient of main dis- K6 charge line K

IR Jet pump internal riser loss K (2) coefficient D

K pamp n rn r ser loss K Q)

IR coefficient in terms of discharge line flow area K

ER pymp x ernal riser loss K M) coefficient D

K ER et pump external riser loss K(5) coefficient in terms of discharge line flow area K" Recirculation header loss K(6) coefficient K"D Recirculation header loss K(7) coefficient in terms of discharge line flow area K yp Loss coefficient under reverse K(8) flow due to suction line break -

jet pumps too but not including recirculation pump 3DA-2

GESSAR II 22A7007 238 NUCLEAR ISLAND Rsv. O NOMENCLATURE (Continued)

Definition Dimensions Computer D

K Loss coefficient - recirculation K(9)

P pump K

T Loss coefficient - under reverse K(10) flow due to suction line break -

jet pumps through and including recirculation pump K uction line loss coefficient in K4 CW terms of cut water area N. Number of jet pumps N1 3

P Pressure at cut water psi P1 CW P Pressure in impeller psi P1 7

QR Rated pump flow GPM Q1

._ OR Rat ed pump flow cfs Q2

'\ - 'O Calculated pump volume flow rate GPM 03 3

v g7 Specific volume of steam at pumr. ft /lb V4 impeller pressure 3

v Specific volume of saturated ft /lb V3 f7 water at pump impeller pressure v Specific volume of two phase ft'/lb V6 7

mixture in pump impeller X Quality of fraction steam in X3 7

two phase blowdown flow in pump impeller l

l 3

3DA-3/3DA-4

~

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev.'O Attachment A O SECTION A EQUATIONS AND SAMPLE CALCULATION DISCilARGE LINE GUILLOTINE BREAK I. Sample Calculation Input Data A

CW

= 00 in.2 D gg = 25.53 in.

A IE

= 0 in.2 ^LOOPF II R

= 0h K SUC QR = 45,200 GPM II. Calculations

1. Areas and Suction Piping Losa I " A "*

ACW( CW Example: A CW = 200/144 = 1.3889 ft 2

= + l44 AIE(ft) AIE(i"*)

Example: A  ! "

• IE i

^LOOPF

+K 0.0113j (0. 8 )

K gg = 1.22 SUC Example: K gg = 1.22 x x 0.8 0.

+ 0.184 = 1.16 A "

SL 576 SL}

Example: A gL =

576 (

O 3DA-5

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0

2. Maximum Possible Flow Loading nt Impeller Eye

/A IE IE SL A gLj Example: K

• IE "

• O 3.5 5 G ~

gure 3DA-U IE IE' o' o Example: (slip flow model) G = 7755 ie-s (homogeneous model) G ie-Il = 5177 Using Figure 3DA-2 to calculate the pressure at the pump impeller eye:

P = (slip flow slip 0.293448 (G.le-s) model)

995 psi P

homogeneous 0.9930732 (Gie-H)

= 974 psi (homogeneous model)

Steam quality (545.6 - h f974) 545.6 - 538.4 IIomogeneous: X ie-II 1193.4 - 538.4 (h -hf974) 9974

0.0110 Slip: X.

(545.6 - hf995) _ (545.6 - 541.6)

(h g995 1192.6 - 541.6 le-s

-hf995)

= 0.0061 0

3DA-6

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0

2. Maximum Possible Flow Loading at Impeller Eye (Continued)

Specific volume llomogeneous: v "

ie-h "f974 + Xie-h I"s974 - "f974) i

= 0.0251 + 0.0110 (0.4593 - 0.0251)

= 0.0299 Slip: /v s995 I 0.4485 )1/3

" = 3. M le-s v 0.01257j 3

f995/

"le-s ie-s "s995

(

T

+K ie-s II-Xie-s) "f995)'

/ l-Xie-s) 1 X

( ic-s

• K ie-s )

(

= I (0.0061 (0.4485)

(

)

+ (3.29) (1-0.0061) (0.01257) I

)

1 0.0061 + 1 ~ 3 29

= 0.0135

3. Maximum Possible Flow Loading at the Pump Cut Water Pump equivalent loss coefficients Through iterative procedure it was determined that pump

loss coefficients should be determined at a pump flow of 3.78 Q R f r the homogeneous flow case and 5.37 Q R #

the slip flow case. The head developed by the pump i

3DA-7

GESSAR II 22A7007 238 HUCLEAR ISLAND Rev. 0

(

3. Maximum Possible Flow Loading at the Pump Cut Water (Continued) would be 2.69 II R f r the homogeneous case and 4.95 II R

for the slip flow case. The pump loss coefficients are calculated as follows.

P 2g allp Ap 2 Kp _

2 09 For homogeneous flow:

Kp = (64.4)(2.69)(710)(3.555)2 = 10.73 (3.78 x 100.7)2 For slip flow:

K = (64.4) (4. 5((710)(3.555)2 = 9.78 P

(5.37 x 100.7)2 Combined loss coefficient at pump cut water D (Acw}

K cw IKSL

• p)

(Agg)2 flomogeneous flow:

K =

1. 889)2 = 1.815 (1.16 + 10.73) cw (3.555)2 Slip flow:

K = Il.16 + 9.78)

(1.3889)2 = 1.670 cw -

(3.555)2 3DA-8

GESSAR II 22A7007 238 NUCLEAR ISLAND R3v. 9

's 3. Maximum Possible Flow Loading at the Pump Cut Water (Continued)

Flow loading at pump cut water I G CW

= f(KCW' Po, ho) (Figure 3DA-1)

Example: (Slip Flow Model) G = 7244 by CW S Figure 3DA-3 j

Example: (Homogeneous Model) G " 4044 DY CW H Figure

3DA-3 Using Figure 3DA-2 to calculate the pressure at the

, pump cut water:

Homogeneous flow model:

P homogeneous

= 0.9930732 G CW-H homogeneous = 923 l

Slip flow model:

P gygg

=

0.293448 G CW-S '

P "

slip l Steam quality I

Slip:

545.6 - h

" f935 545.6 - 532.2 i

(} h s935 -h f935 1194.8 - 532.2 = 0.0202 3DA-9 l

{

l. - . . - - _ _ ._ _ _ . _ _ _ . _ _ _ . _ _ - - - _ - - -- - - - - -- - - - -

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0

3. Maximum Possible Flow Loading at the Pump Cut Water (Continued)

K "

I"s935 II 0.48045  !

= 2.823 CW-S (vf935/ 0.02135j Ilomogeneous:

545.6 - h

" f923 545.6 - 530.3 CW-li h -h = 0.023 s923 f923 1195.3 - 530.3 Specific volume llomogeneous:

"923 "f923

• I"s923 - "f923}

=

0.02131 + 0.023 (0.49912 - 0.02131) g

= 0.0323 W Slip:

"s935 I 935 "s935 + cws (1-X935}"f935 3 I-*935 935 K cws -

=

[0.0202(0.4805) + (2.823)(1-0.0202)0.02135]

~

0.0202 +

2.823

= 0.0253 0

3DA-10

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0

4. Location of Critical Flow If G IE A

IE cw

^ cw critical flow is at the pump cut l water, and

^

O fl =C, c I

If GIE ^IE < cw ^cw critical flow is at the pump impeller eye And, flow loading at cut water is total flow at impeller eye times the area of the impeller eye divided by the i

cut water area 1

i C "

FL IE ^IE ! ^CW Example: (slip flow model)

GIE ^IE * * "

G A = 7244 x 1.3889 = 10061 CW CW Critical flow in pump cut water G

F -s = 10061 i Example: (homogeneous model)

GIE ^IE

• GCW ^CW i

critical flow in pump cut water G

F -H 6728

O i

3DA-ll i

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0

5. Average Static Pressure in Pump Slip p = IE + cw

_ (995 + 935)0.6 = 579 avg 2 2 Horaogeneous

( IE + cw

• (974 + 923) 0.8 avg 2. 2

6. Pluid Properties in Impeller Based on Pava, Isenthalpic Expansion from Vessel Pressure and Saturated Enthalpy ho - h fgp avg X =

fg @ P avg "SI "S avg "fI "f avg Example: (slip flow) hg = 467.4 @ 579 psi ho = 545.6 h = 737.0 @ 579 psi f

v f

= 0.0201 @ 579 psi 545. 467.4 X = = 0.1061 v = 0.7976 @ 579 psi 7 s Example: (homogeneous flow) h = 502.4 @ 758.8 psi f

ho = 545.6 h = 698.2 @ 758.8 psi f

v = 0.0207 @ 758.8 psi f

545.6 502.4 Xy = = 0.0619 v = 0.6028 @ 758.8 psi s

O 3DA-12

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O

,O 7. For Slip Flow, Determine Fluid Specific Volume in Pump Impeller

=

fvg,)1/3 K l l I

\"fIl Slip -

1-X, "I

I "SI + I II -

I) "fil _

I+ K I -

Example: K = "

• 7 01 v7 = [0.1061 x 0.7976

+ 3.4109(1 - 0.106D x 0.0201]

x 0.1061 + 1 -3.4109 0.1061 .,

= 0.0537 Homogeneous v7 v f7 + X7 (v g7 -v f7)

=

Example: v7 = 0.0207

+ 0.0619 (0.6028 - 0.0207) '

= 0.0567

. 8. Pump Flow Volume Rate and Speed Q = V 7

G pg A CW (cfs) = v 7 G pL A CW x 7.48 x 60 (gpm)

O 3DA-13

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0

8. Pump Flow Volume Rate and Speed (Continued) 0 N/N R ~~ IIO/O ), where $ = fr m Figure 3DA-4 R 5 Example: (slip flow)

Q = 0.537 x 7244 x 1.3889 x 7.48 x 60 = 242480 O__\ = 242480 45,200 = 5.36 (OR)

= x 5.36 = 2.62 Example: (homogeneous flow)

> Q = 0.0567 x 4844 x 1.3889 x 7.48 x 60 = 171203 IQT 171203 45,200 = 3.79

=

Q-p N__ 10 g

x 3.79 = 1.85 0

1 l

O 3DA-14 1

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0

' ' SECTION B EQUATIONS AND SAMPLE CALCULATION SUCTION LINE BREAK - PUMP AT ZERO SPEED I. Additional Input Data D = 25.53 K "

• DL J Dy = 3.14 K = 0.42 n

Ny = 10 K = 2.8 + 0.40 + 0.73 0.8 = 3.424 D (0.01 9_ 1 1

II. Calculation  ;

+

D DL 1)2

. J

= Ky xlD (yj {j Example: = 0.6 x ,y ph = 26.220 D 2 D "

DL 1 n n (7j j

/ ) Ih Example: K

[j *

(7 n )

I i

l NOTE All BWR/6 jet pumps use a jet pump 180* bend with

! a 7-inch internal diameter.

4 D "

.I DL I 1 )2 IR IR D IR/ J  !)

Example: K = 0.35 IR (00 13) (0.8)

+ 0.1341 = 0.4637 a

3DA-15

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O II. Calculation (Continued)

K 7p = (0.4637) 6 1 = 0.9417 D

• DL I1 ER ER D ER/ ("J!

Example: K * *

• ER . 13

+ 0.1840 = 1.2009 9

• K * *

• ER 1. 6 Since N y = 10, F g = 0.4 D "

DL

(

U H \Dg ) S Example: K = 0.06 l l0119 (0. 8 )

H

+ 0.0720 = 0.0120 2

Kg = 0.120 yg,20 0.0 = 0.0%

= K + + +

Ky_p D J n IR ER H Example: Kg_p = 3.424 + 26.22 + 0. m + 0. m

+ 1.219 + 0.060 = 32.607 2

n DL .

A 1" DL 576 O

3DA-16

GESSAR II 22A7007 I

238 NUCLEAR ISII.ND Rev 0 p II. Calculation (Continued)

Example: A " "

• DL 76 7

Through iterative process it was determined that if the pump loss coefficient was calculated for a flow of 4.92 Q f r the R homogeneous flow case and 5.67 QR f r the slip flow case, the calculated flow would equal the assumed flow. The following analysis presents these results:

Homogeneous flow model 1

i Pump head at 4.92 Q R H

p

= 0.72241 (4.92)2.05404(730) = 13913.3 l

Pump loss coefficient:

l KP ,

p ^D ,

(64.4)(13913.3)(3.555)2 = 46.13 P Op 2

(4.92 x 100.7)2 l'

Slip flow model I

Pump head at 5.67 Og i

H = 0.72241 (5.67)2.05404(730) = 18621 Pump loss coefficient:

K P = (64.4)(18621)(3.555)2 = 46.49 E

i (5.67 x 100.7)2 3DA-17 l

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O II. Calculation (Continued)

Check for critical flow at jet pump 2 2 3

Ay =

(10) = 0.538 ft 777 (3.14)

Ky = 0 (no losses at jet pump nozzle entrance)

Gy -

fW y = 0, Po, ho) (Figure 3D.A-1)

Example: G y - 5300 (homogeneous) Gy = 8000 (slip)

Check _for critical flow at pump cut water IA CW CW J-P A DL/

G que 3DA-1)

O CW CW, ,

Example: K I 200 2

= 4.98 CW l511.9

\

G geneous)

CW l

G = 4500 (slip)

CW Check for critical flow at impeller c'e D

K. = K. + K le ]-p p l A.

le K. = K.

le le A DL O

3DA -18

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 -

1

II. Calculation '(Continued)

Example:

1 K "

I 250\2 (32.607 + 46.49) (slip) le

, l511.9)i

\

I 250 h2 K " (32.607 + 46.13) (homogeneous) ie l511.9j

(

K ie

= 18.87 (slip) K ie 18.78 (homogeneous)

G ie

= f(K ie, Po, ho) (Figure 3DA-1) i G = 2420 (slip) Gie = 1990 (homogeneous) ie i

i- Location of Critical Flow i

Homogeneous Flow Model:

GA = (5300)(0.538) = 2851.4 (jet pump) 33 i

I G ^

cw cw

= (3230)(1.3889) = 4486.1 (pump cut water)

G b ie ie

= (1990)(1.7361) = 3454.8 (pump impeller eye)

Since G A <G ie Aie < Gcw Acw, hoke flow is at the jet

pumps.

I Slip Flow Model:

G.A. = (8000)(0.538) = 4304.0 (jet putap)  !

] ]

G cw^cw (4500)(1.3889) = 6250.0 (pump cut water)

^

O 1 1

~

3DA-19

. . ~ . , - _ _ . _ _ . _ _ . . - . . . _ _ _ . _ . _ _ , - _ . _ _ _ _ _ . , _ . - _ . _

GESSAR II 22A7007 238 11UCLEAR ISLAf1D Rev. O II. Calculation (Continued)

G A ic ie

=

(2420)(1.7361) = 4201.4 (pump impeller eye)

<G A choke flow is at the pump Since Gie^ie < G A ,

impeller eye.

Calculation of Steam Properties in Pump Stagnation pressure at pump cut water:

P = f(G) Figure 3DA-2 P = f(3230) = 666 psi (homogeneous)

P = f(4500) = 607 psi (slip flow)

Stagnation pressure at pump impeller eye:

P. =

f (G) Figure 3DA-2 le P

yg

= f(1990) = 451 psi (homogeneous)

P = f (2420) = 346 psi (slip flow)

Static Pressure (Average Static Pressure in Pump):

(666 + 451 p

avg

= l 2

(0.8) = 446.8 psi (homogeneous)

(

607 + 346 P

avg

=

2 (0.6) = 285.9 psi (slip flow)

(

(static pressure equals 0.8 times stagnation pressure for homogeneous flow and 0.6 times stagna-tion pressure for slip flow) 3DA-20

.__e

$2 GESSAR'II 22A7007 x 238 NUCLEAR ISLAND Rev. 0

+

,,q\'l't i n -

l ,

, b,

,,4 1 ,

II. Calculation (Continued) N 9

t 4 .,, ~.

i *s j Steam Quality: .3 4

, 3r L

X avg

=

545.6 = 436.57 769.03

= 0.142 (homogeneous) * )

2

, ,n .

545.6 = 389.3.

X avg

=

814.21

=

0.192 (slip)< '

g .. . r r 6

4 , ,

Specific Volume: ~

\  ;

v avg

=

0.01953 + (0.142[1.0402 - 0.01953]) 'e  ;

= 0.1645 (homogeneous)' ,

i K = [1.6188 l\ I! = 4.413 I

avg l{0.01883 j

. + .t .

g s .

'1 v avg

= 0.192(1.6188) + (4.413)(1 - 0.192)(0.01883)

' \ "

O.192 + 1 -4.413 0*192 I = 0.1418 (slip)

• 3 ( j i

I Flow Rate:

i

n. J ,.

O = (vg) (Gj ) (A3 ) (7.48) (60) = (0.1645) (5300) (0.538Ii ' "

(7.48) (60) = 210512 (homogeneous) ,

O/Q R = 4.66 , (homogeneous)  !

, ~\

\

Q u (vs) (G.le) (A.le) (7.48) (60) '= (0.1418) (2429(0.17361)

(7. 48) (60) = 267374 (slip) l .

I O/Q R = 5.92 - (slip) 3DA-21'

---,_-_,.z_ - . . - _ _ _ , _ . - . _ _ , _ . _ _ , , , , _ _ , _ . . . , . . - . _ , . . . . . . . _ - , . . . , _ . . - . _,,4.,,__,, $__,,.._,-_,_,.. - , _ . . - _ , , . . . , . . . - , , , _ - _ . . , , , - - , _ . . ,

'. GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O II. Calculat. ion (Continued)

Torque:

1.35 0 1 90535

! R

• v 62.4 T/T R = 2.47 (homogeneous)

T/T R = 4.52 (slip) l l

\

i O

I 1

1 l

O 3DA-22

i GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O t

i SECTION C EQUATIONS AND SAMPLE CALCULATION SUCTION LINE GUILLOTINE BREAK - PUMP AT ZERO TORQUE i

(WINDMILLING SPEED) ,

I. Additional Input Data None II. Calculation l Through iterative process it was determined that if the pump loss coefficient was calculated for a flow of 4.92 Q R f r the homogeneous case and 4.96 Qg for the slip flow case, the calculated flow would equal the assumed flow. The following calculations present the results:

Homogeneous Flow Model

. Pump head at 4.92 Q R AH = 2.1 (4.92)l.95081 (710) = 33,371 p

Pump Loss Coefficient:

K = 9 "9 ^ = (64.4)(33371)(3.555)2 = 110.65 P Op 2

(4.92 x 100.7) 2 Slip Flow Model i

Pump head at 4.96 Q

• R AH = 2.1-(4.96)l.95081 (710) = 33902 j 3DA-23

, -- , --.. . - . _ , _ _ . .. - - , , - . . - . _ - . . e. - . ._

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O II. Calculation (Continued)

Pump Loss Coefficient:

K D , (64.4)(33902)(3.555)2 P = 110.61 (4.96 x 100.7)2 The calculations in Section B for flow at the jet pump and pump cut water still apply to this case; the calculation for flow through the pump is performed using the preceding pump loss coefficients.

K. = h. .

P ^te Yl le le A DL/

K ie

= K P + Kp K = 32.607 + 110.65 = 143.26 (homogeneous) ie K = 34.16 (homogeneous) ie l

K[i

= 32.607 + 110.61 = 143.22 (slip)

K = 34.16 (slip) ie G = f(K, P , ho) (Figure 3DA-1) ie G ig = 1620 (homogeneous) G = 1870 (slip) e Calculating steam properties at the impeller eye Pressure at impeller eye:

P = f(G) Figure 3DA-2 3DA-24 1

L

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 II. Calculation (Continued)

, P ie

= 0.9930732 (1620)0.8054544 = 387 psi (homogeneous)

P ie

=

0.293448 (1870)C.9076 = 274 psi (slip)

Average steam quality and specific volume in pump.

i

!!omogeneouis :

(P ie p cw ( }

(0. 8) = (0. 8) = 421.2 psi h

f

= 429.4 h = 1205.5 9

v = 0.01942 f

v = 1.10886 g

545.6 - 429.4 '

1205.5 - 429.4 = 0.1497 X =

v = 0.01942 + 0.1497 (1.10886 - 0.01942) = 0.1825 Slip:

p , ( ie + Pcw) (0.6) - (274 + 607) (0.6) = 264.3 avg 2 2 I

h =- 381.8 f

I, h = 1202.7 9

O j 3DA-25 l-

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O II. Calculation (Continued) vg = 0.018728 v = 1.7432 9

9 f! 1.7432 h l/3 kI = 3 1 = 1 = 4.5319 v fj 0.018728j

.6 - h

• f 454.6 - 381.8 h -h

" - 0.1995 I

g 1207.7 - 381.8

/ (1 - X.)h v

7

=

[X_v g +kg(1 - X7)vf] l

\ X7+ kg )

Location of Critical Flow liomogeneous Flow Model:

G.A. = (5300)(0.538) = 2851.4 (jet pump)

J J G A = (3230)(1.3889) = 4486.1 (pump cut water)

G A (1645)(1.7361) = 2855 (pump impeller eye) ie ie GA <G A and choke flow is at the jet cAie < G w pumps.

Slip Flow Model:

G.A.

J J

=

(8000)(0.538) = 4304.0 (jet pumps)

O 3DA-26

._~ . . .__ .- . . - __ .. .- - _ _ _ . __ -

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0

() II. Calculation (Continued)

G A = (4500)(1.3889) = 6250.0 (pump cut water) cw G A = (1870)(1.7361) = 3247 (pump impeller eye) ie ie G

cA ie ' j^j < cw cw

^ and hoke flow is at the pump impeller eye.

i Calculate Pump Flow Rate and Speed Q =

(vg) (Gy) (A7) (7.48) (60)

= (0.1825) (5300)(0.538)

(7. 4 8) (60) 0 = 233547 (homogeneous) l

() Q/Q g = 5.17 (homogeneous)

N/N R =

(5.17)(100/57.5) = 8.99 (homogeneous)

Q =

(vy) (Gy ) (A7) (7.48) (60.0) =

(0.1564)(1870)

(1.7361)(7.48)(60.0)

O = 227880 (slip) i l O/Q R = 5.04 (slip)

N/N R =

(5.04)(100/57.5) = 8.77 I

O l

l 3DA-27/3DA-28!

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 SECTION D EQUATIONS AND SAMPLE CALCULATION 4

SUCTION LINE LONGITUDINAL BREAK - PUMP AT ZERO TORQUE (WINDMILLING)

I. Additional Input Data None II. Calculation The analysis is based on a longitudinal break in the line close to the pump nozzle. The break will have flow from the i

suction line and reverse flow through the recirculation pump. The combined flow coefficient in:

K D

= K

\^DL/

O

[ 1 }2 K

C +

SL K T )

where K gg and K T are calculated in Sections A and B of this Attachment respectively.

i 2

1 ,

Example: K

• 1 C 1 143.2 k 1.16 i

j Now, the largest longitudinal break will result in critical flow just upstream in the suction line or:

/K T2

^LBM ^SL K j l

l 3DA-29 i . _ _ _ . .

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O II. Calculation (Continued)

Example: A LBM

0. 6 (3.555) = 5.022 Determine critical flow at longitudinal break:

G C, , o gure 3DA-U LB Example: G omogeneous)

LB G = 4500 (slip)

LB Determine fluid pressure at break (conservatively assume this pressure also exists in pump impeller):

P = gure 3DA-2) 7 f (GLB)

Example: P = 670 (homogeneous) P = 6G7 (slip)

O 7 7 Now, use the methods of the previous sections to calculate the fluid properties in the pump:

545 - 485.8 = 0.0834 (homogeneous)

Example: X 7 =

545 - 473.1 = 0.0992 (slip)

X =

7 K = = 3.355 (slip) 7 0.0 016 v = 0.0760 (homogeneous) 7 v = 0.0502 (slip) 7 9

3DA-30

GESSAR II 22A7007

, 238 NUCLEAR ISLAND Rev. O i

II. Calculation (Continued) d Now, the reverse flow rate through the pump is a fraction of the total flow rate defined by:

4 0.5

/KC \

P ^LBM LB j 6

Example: Mp = (5.022)(3250) (01 2

1347.5 (homogeneous) 4 0.5 Mp

(5.022)(4500) (O.

= 1865.7 (slip) r i The pump volume flow rate is Q = Mp7V x 7.48 x 60 and IN I =-

100 Q i l

gy 7 )l (R R

! Example: Q/Q R = 1.02 (homogeneous)

/ I 1 = 1.77 (homogeneous)

I[N--

l R)

Q/Q R = 0.93 (slip) i i \

i

= 1.62 (slip).

l R)

!O 3DA-31/3DA-32 i

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 0 9000 8000 7000 6000 -

4000 -

E 5

sE a-3 3000 -

O i

=

2000 -

SLIP HOMOGENEOUS i i I I I I 1000 o to 20 30 40 50 60 70 K,, LOSS COEF FICIENT Figure 3DA-1. Critical Flow at Longitudinal Break 3DA-33

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. O O

10 -

3 E

5 P

9 E

[

E 3

u

< \

E R

O 3 -

E 10 ld _

HOMOGENEOUS (09930732 G OM4W 3

[O SLIP (0.29348 GOS076 ,

2 i i i t l i e t i 3g 3 4 10 10 10 FLOW LOADING (Ibm /sec/f t2)

O Figure 3DA-2. Pressure at Pump Cut Water 3DA-34

GESSAR II 22A7007 l 238 NUCLEAR ISLAND Rev. O O

\ , N.m v g L *200 y

.iD A *175 [ *i f/es N x \ s J'"* m lf?pp' *$

.,m_

p -

M ,,,, n ,

a lIi .s0 xx N

! \ \ j

,! ~

I ss e f //

I llI l+ , 20 t

-=

l

,llll ll l

+10 s - f y .20

1. /?_,

f I l  % ,

'9

[ --- CONSTANT HE AD LINES lI Jl

-- CONSTANT TOROVE LINES

~" '

.- ,Y^"M1 O "a^<"'

+ f OPER ATION N

I l / / -"

s .T jIffI '  :' 6"

~

lh,~ ,/ g /p D SS PATION

/

l f

,f ,

f

, , / /, j j j,, gyAL /

N l,j'/, "

?*f

~'" '

l I E"Ar.o l I

)l I g

f l/

~

h h ) C Hoo Fh

} l' l

I I I I m / ///  ;' " " ,2 l EPAmN s

2 (lI', f I;

\

( '[/ "#'""@

1' I ENERG,

• N y,} ff  ! DISStPA TION PWP jQ

g Q51

, g'. f; f / g E uPL AN ATORY CHART

'S NS ~ ' 9pm. ENTERING SUCTION NO22L E

,j a -. f j

g

• p ff/j f '

- som. ENTERING DISCHARGE N0ZZLE

+ H. HIGHER HE AD AT DISCHARGE NOZ2LE

{

j f f Ikg I j "1 - H.HtGMER HE AD AT SUCTION NOZZLE o ,

I t' ,E /

• rpm. NORMAL PUMP. - epm. NORM AL TUR8tNE b epmi

~

l l p/e N 8 4'8 C + T - Ct+ h +e OR g g

/ gf[f .g g

- T = Ct - $ + epm) OR Cl* W - epm) 100% -50% 0 +50% '100% +150% *200% + 250% +300%

- 200% -150%

- epm

• epm O Figure 3DA-3. Complete Pump Characteristics - Double Suction Pump (Ns = 1800) 3DA-35

GESSAR II 22A7007 238 NUCLEAR ISLAND Rev. 0 ZERO TORQUE O

(WINDELLING)

LINES

+100% - --

N, = 4200 N, = 1800

+50% - --

~a:

0  ! l f.

N, = 4200

_m - -.-

N, = 1800 I

l i  !

oog 0 +100% +200% +300%

FLOW. % OF Og O

l Figure 3DA-4. Pump Flow Volume

{

3DA-36

_- -. -- . _ - .