ML20084E350
| ML20084E350 | |
| Person / Time | |
|---|---|
| Site: | Ginna  |
| Issue date: | 03/15/1973 |
| From: | ROCHESTER GAS & ELECTRIC CORP. |
| To: | |
| Shared Package | |
| ML20084E344 | List: |
| References | |
| NUDOCS 8304150027 | |
| Download: ML20084E350 (18) | |
Text
,
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RESP 0!4SE TO AEC QUESTIONS
SUBJECT:
RGE - Pressurizer Safety Valve Discharge Piping Time History Dynamic Analysis MAR 15 EH 8304150027 730315 gDRADOCK 05000244 PDR
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Question: What is the critical percentage damping used in the analysis?
Answer:
The pressurizer safety valve discharge piping system is analyzed by using 0.5 percent critical damping.
Question: How many degrees of freedom are considered in the analysis?
Answer:
Each lumped mass is presented by three translatory degrees of freedom.
FIGURE 1 (PCV 435) shows the 25 lumped mass model.
The system is analyzed for (25x3) 75 degrees of freedom.
FIGURE 2 (PCV 434) shows the 22 lumped mass model.
The system is analyzed for 22x3 66 degrees of freedom.
Question: What method is used for the normal mode -solution?
Answer:
The modified Jacobi method is used for the normal mode solution, where w represents the system frequencies and 4 represents the mode shapes of the system.
Question:
Furnish the time history plots of the most highly stressed node point in each system.
Answer:
The Figure 3 shows the maximum stress from the time history analysis for the PCV 435 (FIGURE 1).
The Figure 4 shows the maximum stress from the time history analysis for the PCV 434 (FIGURE 2).
Question:
Furnish the plots of the time history displacements for each line at or near the most highly stressed point.
Answer:
The figures 5, 6, and 7 show the tine history displacecents in X, Y, and Z directions for line PCV 435 (FIGUD.E 1).
The Figures S, O, and 10 show the time history displactnents in X, Y, and Z directicns for line PCV 434 (FIGURE 2).
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j Question:
How are the stresses combined?
I Answer:
The stresses due to deadweight, pressure, seismic and time history dynamic analyses are calculated separately.
It is j
conservatively assumed that the maximum stress around the pipe circumference occurs at the same point for all load cases considered.
These stresses are added absolutely and compared with the code allowable stress limit of 1.2 x Sa' where S, = stress allowable.
l 1
Question:
Describe the method used in the time history analysis.
Answer:
FIGURES 1 and 2 show the lumped mass mathematical model used
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in the time history analysis.
Support and restraints are represented as springs in the model.
l The time history dynamic solution employs the following computer j
codes. A brief description of the function of each code is also provided.
j WESTDYH:
Normal mode solution (w,0) l FLASH:
Calculates the time history transient hydraulic forces in the system.
FIXFM:
Calculates the dynamic time history displacements at each mass point in the system using the time history hydraulic forces.
WESDYN-2: Time history dynamic displacements are applied as 4
external boundary conditions to evaluate the piping forces and stresses at all node points in the system
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(not only mass points). Time history forces in i
supports and piping restraints are also calculated.
Question:
Which mode contributes most to the highest stresses point in the system?
i Answer:
Although the time history analysis employs modal coordinates in solving the equations of motion, the individual modal contributions are not saved but immediately combined to provide the total i
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response for all modes. We, therefore, cannot ascertain which mode contributes most significantly to the maximum stress.
Question:
List a few fundamental frequencies for the dynamic analysis.
Answer:
Tables 1 and 2 show the fundamental frequencies and their predominant direction of motion.
TABLE 1 PCV 435 (FIGURE 1)
Frequency Predominant Direction cps of liotion 23.19 Z
23.22 X
23.51 Y
29.97 Y
30.84 X
36.91 Y
43.75 X
46.75 Z
54.48 Y
61.59 Z
TABLE 2 PCV 434 (FIGURE 2) 16.81 Y
29.72 X
29.78 Z
33.44 X
39.99 Y
41.03 X
51.87 Y
58.17 Z
79.24 Y
81.67 X
L
O O
Question:
Prepare a combined stress summary table including the node j
points representing elbows and straight runs.
Answer:
Tables 3 and 4 show the stress summary for the combined load cases.
Thermal Expansion Stress, KSI S
=
E Deadweight Stress, KSI S
=
D Pressure Stress, KSI S
=
p Thrust Forces Stress, KSI S
=
T Seismic Stress, KSI S
=
3 T + S ' ESI
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SD+SP+S S
=
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CLASS I MATERIAL - SA 376 Stainless Grade 316 l
(1)
B31.1 Allowable Stress - Sustained Mechanical Load l
S,
= 1.2 x S l
H
= 1.2 x 17.05 = 20.45 ksi S
=SD+
Sp+ ST+ S S S, > S (2)
B31.1 Allowable Expansion Stress S,
= f (1.25 S + 0.25 S )
c h
= 1 (1.25 x 18.75) + (0.25 x 17.05) 27.66 ksi
=
S, > SE f
= Stress range reduction factor for cyclic conditions S
= Allowable stress of material at maximum hot temperature H
S
= All wable stress of material at ambient temperature c
CLASS II MATERIAL - SA 106B Carbon Steel (1)
B31.1 Allowable Stress - Sustained Mechanical Load S
= 1.2 x S a
H
=1.2(15.0)
= 18.0 ksi S, >
S (2)
B31.1 Allowable Expansion Stress S,
=.F(1.25 S + 0.25S) c h
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=(1)
(1.25[15] + 0.25[15])
l.
= 22.5 ksi S,
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V TABLE 3 v
PCV 435 (FIGURE 1) i b
b S
S b
b
!! ode E
D P
T S
Point ksi ksi ksi ksi ksi ksi 104 3.65 0.31 3.1 0.27 0.24 3.92 110 2.24 0.41 3.1 0.15 0.17 3.83 4 120 2.98 1.28 3.1 1.07 0.32 5.77 ?
130 3.32 0.21 3.1 0.17' O.10 3.58 ?h 150 7.98 1.77 3.1 0.46 0.38 5.71 160 1.36 0.89 3.1 0.59 0.32 4.90k 170 6.61 0.58 3.1 0.46 0.36 4.50$
4.03$
180 4.27 0.17 3.1 0.58 0.18 190 5.74 1.18 3.1 1.16 0.43 5.87 O 200 4.02 0.36 3.1 0.32 0.07 3.85 g 210 6.65 2.05 3.1 1.12 0.30 6.57 M
220 2.42 1.0 3.1 0.63 0.26 4.99 %T 230 2.75 1.91 3.1 1.84 0.52 7.37
.J 234 2.56 2.5 3.1 2.62 0.77 8.99 242 20.48 0.41 3.1 0.43 0.16 4.10 260 24.10 0.28 3.1 1.24 0.11 4.73 4 270 22.82 0.26 3.1 1.06 d.06 4.48Q 280 8.41 0.31 3.1 0.53 0.13 4.07 M 290 10.39 0.22.
3.1 0.74 0.22 4.28 300 10.85 0.32 3.1 0.77 0.07 4.26 '
80 310 10.48 0.62 3.1 0.88
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'{6.69 y 315 17.69 1.42 3.1 1.89 0.28
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V TABLE 4 PCV 434 (FIGURE 2)
S S
S S
b S
flode E
D P
T S
Point ksi ksi ksi ksi ksi ksi I
103 3.87 0.36 3.1 0.18 0.33 3.97g l
110 2.76 0.56 3.1 0.29.
0.23 4.18&
120 8.38 1.93 3.1 0.85 0.54 6.42 "
140 5.60 0.45 3.1 0.63 0.24 4.42 150 7.28 0.57 3.1 0.45 0.34 4.46 4 160 4.29 0.19 3.1 0.61 0.61 4.51}
170 9.73 0.56 3.1 0.92 0.16 4.74 180-6.17 0.51 3.1 0.24 0.32 4.17 M 6.45h 190 8.71 1.50 3.1 0.86 0.99 I
200 2.63 0.57 3.1 0.58 0.11 4.36}
210 3.00 0.61 3.1 0.90 0.26 4.87 O 223 19.03 0.49 3.1 0.65 0.22 4.46 e
2qq 23.42 0.36 3.1 0.50 0.14 4.10 250 24.78 0.26 3.1 0.44 0.11 3.91
-260 8.19 0.54 3.1 0.35 0.10 4.09 D 270.
13.36 0.82 3.1 0.35 0.09 4.36 '
280 14.85 0.61 3.1 0.55 0.17 4.43 3 286 22.83 0.77 3.1
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