ML19289C583
| ML19289C583 | |
| Person / Time | |
|---|---|
| Site: | Big Rock Point File:Consumers Energy icon.png |
| Issue date: | 01/16/1979 |
| From: | Charnley J, Kong Y GENERAL ELECTRIC CO. |
| To: | |
| Shared Package | |
| ML19289C578 | List: |
| References | |
| NEDO-21974-APP, NEDO-21974-APP-2, NUDOCS 7901190090 | |
| Download: ML19289C583 (33) | |
Text
.
a e
o APPENDIX II TO GENERAL ELECTRIC REPORT NED0-21974 BIG ROCK POINT CORE SPRAY SPARGER AND STEA?! BAFFLE STRESS ANALYSIS Prepared by
(
v J$E. Charnley Verified by b
Y. H. Kong Revision 1
~
~ %' ng 790119003o
1 e
I.
INTRODUCTION The purpose of this analysis is to demonstrate that the replacement core sparger and steam baffle shown on 794E830 (Reference 1) satisfies the structural aspects of design specification 22A5589, Rev.1, (Reference 2) and the appropriate repair program.
The repair program requires in kind replacement with specified exceptions. Since the existing core spray sparger and steam baffle have not structurally failed, only structural changes must be reviewed for compliance with the design specifi-cation. Therefore, the following changes must be analyzed for compliance with the design specification.
1.
Material was changed from 304 stainless steel to 316 stainless steel.
2.
Core spray sparger outlet nozzles and elbows were changed.
3.
Connector pipe was changed.
4.
Torque.n the baffle plate hold down studs was changed.
5.
Core spray sparger "J" bolt installation is now specified.
The design specification also requires that the lowest natural frequency of the core spray sparger be determined and compared to the vortex shedding frequency for the core spray sparger.
Vortex shedding frequency must be less than 2/3 of the lowest natural frequency.
The repair program requires that the reolacement core spray sparger be designed in accordance with ASA B31.1-1955 and tnat the replacement steam baffle be designed in accordance with sound engineering practices with guidance from ASME Section XI, 1974 Edition Summer 1976 Addenda, Paragraph IWA 7000. Paragraph IWA 7000 allows replacements to be designed to any code addendum after the construction code addendum. ASA B31.1-1955 does not distinguish between types of loads (primary / secondary) or types of stress (bending / membrane). Therefore, ASME Section III Subsection NG (Core Support Structure) will be used to categorize load and stress.
II.
SUMMARY
The replacement steam baffle and core spray sparger assembly satisfies the requirements of the design specification and the repair program.
J III ANALYSIS 1.
Vortex shedding frequency The vortex shedding frequency (f) can be determined from the following equation.
=.22 (Mechanical Vibrations by Den Hartog Ref. 3)
WHERE f = vortex shedding frequency d = diameter v = fluid velocity Three locations must be considered.
1.
Sparger outlet nozzles 2.
Sparger ring pipe 3.
Connector pipe Since f is increased as velocity increases, it is conservative to use the velocity in the reactor pressure vessel steam outlet nozzles which can be obtained from Q = p AV WHERE Q = total steam plus water flow rate A = total area of steam outlet nozzles V = steam velocity 6
Q = 12x10 lbs/hr 2
A=nd N T
N = number of steam outlet nozzles N=6 d = steam outlet nozzle diameter d = 12. 58 inch A = n (12.58)2 6 = 746 inch 2
T l
The fluid is a mixture of water and steam and its density is therefore between those of water and steam.
P
= 44.25 lbs/ft3 (saturated water at 575 F)
H2O P
= 2.97 lbs/ft3 (saturated steam at 575 F) steam The core exit void fraction at 100% power and 100% flow is.31.
Assume a slip ratio of 1.0.
Then 3
p = (.31) (2.97) + (.69) (44.25) = 31.5 lbs/ft 6
l=
12x10
= 20.5 ft/sec 31.5 ( g (3600)
Thus the vortex shedding frequency is related to diameter by f =.22 ( 20. 5 _f_t, )
t D
sec
= 4.50 D in feet D
D
=.84 inch sparger outlet nozzle Dconnector pipe
= 2.38 inch Osparger ring pipe
= 2.38 foutlet nozzle
= 64.3 cycles /sec.
Iconnector and ring
= 22.7 cycles /sec 2.
NATURAL FREQUENCIES The purpose of this calculation is not to precisely determine the lowest natural frequencies of the sparger, connector and outlet nozzle; but is to show that the lowest frequency of the outlet is significaatly higher than (3/2) (65) = 98.
cycles /sec and that the icwest natural frequencies of the connector and sparger are significantly higher than (3/2) (23) = 35 cycles /sec. Model the sparger outlet nozzle and elbow as follows:
5.7.f
_ 1
~ T 24 a
/
L '/E sca % ems e
f natural 3.89 (Roark Fourth Edition Reference 4)
=
outlet W13 8EI W = Weight of metal plus water in pounds 6
E = Young's modulus = 26X10 p3; 1 = length : 5.75 4
I = moment of inertia =.0171 in W = (.851 +.13) 5.75 =.47 pounds 12 Inatural 3.89
= 776 cycles /sec
=
outlet 47 (5.75)3 y
8(26x100) (.0171)
> 98 Model the sparger ring as a single span with ends simply supported:
A
_1 o
4 C
o 2.38 l fA dr 2"
SCH 40 PIPG f
3 5 (Roark Ref. 4) natural g
ring V 384EI W = (3.653 + 1.45) 39
= 16.6 lbs 12 4
I =.666 in 6
E: 26X10 p3; 1 = distance between supports 1=
W (100)
= 39 inch 8
I e
f 3.55 natural
= 130. cycles /sec
=
/ 5 16.6 39 3
ring V384(26X10")(.666)
> 35 Model the connector pipe as
/////n itiiisi
'A Stip F-A p-2" scH 40 PIPE W' 1.
/
Inatural
= 3.89 riser W = (3.653 + 1.45) 30 = 12.8 lbs.
12 6
E = 26X10 p37 4
I =.666 inch 1 = 30 inch f
" 3.89 78 cycles /sec
=
natural
--12.8 (30),
riser
> 35 g
(8) (26X106).666 Therefore since all natural frequencies are significantly higher than the vortex shedding frequency, harmonic excitation forces caused by vortex shedding will not occur and a dynamic analysis is not required.
3.
Material Change Since the Code for Pressure Piping (ASA B31.1-1955) does not give allowable stresses for either 316 or 304 another code must be used to establish the effect of a material change from 304 SS to 316 SS. The repair program is based on ASME Section XI,1971 Edition, which references ASME Section III,1971 Edition.
Therefore, tge Section III allowable stresses will be used as a basis for compari-0 son. At 600 F Sm for 304SS 's 16,400 psi at 600 F Sm for 316SS is 17,000 psi.
Therefore, the material change is an improvement and no additional analysis is required.
In addition 316 is better than 304 from a stress corrosion viewpoint.
4.
Core Spray Sparger Outlet Nozzle The only design change was to replace a bend with a mitred weld joint. The weld is full penetration thus the stiffness characteristic of the outlet nozzle are unchanged. There is no significant load on the nozzle. Therefore, there is no concern for either stress corrosion or weld efficiency factor.
Thus the design change is acceptable.
5.
Connector Pipe The original connector pipe design was an all welded assembly that was cut to fit in the field. The connector pipe was attached with marmon clamps to the thermal sleeve and to the~ sparger inlet tee.
The replacement connector pipe design incorporates a ball joint and a slip joint to facilitate installation. The replacement marmon clamp has also been modified.
The remainder of the connector pipe is unchanged.
The replacement cennector pipe design is shown schematically below:
// /,'isi I
A SFARGr R WLET n c i
V' Q'N!^ & 6 A U-To m T
%N
/
/
SLIP. Tom T V S Ll?
FtT
// /i ///
^
/ / )iv-
< <</
THCRMAL S trays.
The slip joints and ball joint will accommodate all potential loads due to installation and operation except for the hydraulic forces due to core spray flow. During core spray operation there is a 90 psi pressure drop between the thermal sleeve agd the inside cf the reactor vessel at the maximum flow rate of 477 GPM at 40 F.
The maximum hydraulic force F is equal to the sum of the pressure and momentum forces.
9 WHERE:
F = hydraulic force AP = pressure drop = 90 psi A) = thermal sleeve crossectional area :.7854 (2.89)2 = 6.56 inches 2
2 A = Pipe flow area : 3.36 inches 2
P = 62.4 lb/ft3 (40 F)
U g = 32.2 ft/sec 3
V = 477 GPM.002228 ft /sec
= 45.5 ft/sec 3.36 GPM ft 144 F = (90) 6.56 + 62.4 3.36 (45.5)2
= 684 lbs.
32.2
? ?4
/ / / / / / // / /
M r.J P
~7 1
h A s s u mc S t.r p p a-FCAN 9
TRANSmnT THC
's
- s R6Qt88 R C O
/h oM ET h
- b 2
YY F
M' F-T j
4 A
AIRz M:o M
0 f
r, f,
9 Az
,_ A 3
\\
ur
>'g
>g p
b N$
0 I
2 1
M = R f - FP) 22 "B
- kk2 9
- 0 2
1 2
RIk (Reference 4)
Ffj 92, 22l 2EI EI) j S
is not a standard text book examole and therefore must be solved j
from first principals.
R,
$=0 I J r<x11 4
$ =0 R1 A
V = Rz Shear A
m : R 1, 1
M: R1x i
Moment
(
d S dB _ M 2
dX EI dx 0 4X<1 2
$=0 @X=0 J' = 0 @X=1 2 l
e (X) =
M (X) d X =
+ C) gI 2
s a
4 R X3
[(X)=
6 (X) d X =
2
+C) X
+C2 6EI 2
@X=0
$=0 C2=0
@X=l
0 Cz
2 2
-R 2
6EI 2
$ 4 ##
2 3
[=0 X=
2 2 )2 2
2
@ X=N 6=R2 2
- R R
=
2EI 6EI 3EI 2
2 2
R X
6 (X) =
f1(X)dX =
2 2
+C EI 3
2 EI 2 R
X=
2 2 2
3EI 2 R
R
+ C 2
2 2
2 3
=
3EI EI 2
2 C
2 2
=
3 3EI 2 R
X 2R 6 (x) 2 2 2 2 EI 3EI 2
2 0 X=/ 2+
3 R
(2+ 3) - 2 R2 2 2 2 2 2 3
$ (x) = 6
=
=
3EI EI EI 3EI 2
2 2
2 a
6)
G
=
2 Flj R
A R
+ R
=
2 2 l 2 2 2 2 3 2EI)
EI) 3EI EI 2
2 F
684
=
f)=
28.49 2 = 14.75 2
1=
8.12 3
6 E
26 X 10
=
I) =.666 I2=
.777 Substitution will yield R
475 lb.
=
2 M
7000 in. Ib.
=
B M
= -12,500 in. Ib.
Maximum moment in horizontal run of pipe = R
= 7000 in. Ib.
2 2 Moment at the slip fit
" slip R
- F
=
2 2 a
fit 5.12 a =
Mslip 3490 in. Ib.
=
fit Moment at the ball joint 11 R
F5
=
ball 2 2 22.81 - 1.32 b
21.49
=
=
M
-7700 in. Ib.
=
ball I
N E
max
~
mx I
Horizontal
_ Horizontal I
7000 11,800 psi Ok
=
=
The support which is analyzed below carries all the moment at the ball joint.
Thus, the maximum moment in the vertical pipe run is either M or is ball F times the distance from the ball to the inlet tee.
li
-7700
=
ball F 7.00 = 4790 Therefore, M
7700
=
Vertical O' max 13,700 psi
=
=
Vertical (Section III subsection NG)
Moment at slip joint = 3490 in. Ib.
,/
--K f
l' N SC H 40 FIPE 3490 in. Ib.
_I
\\
b
]
\\
9.9
,1 M!d OVER LA P T 1 1
I
- d. : y 5-
~. _
~.
w_
N F The centroid of a triangle is 1/3 of its base from the end. Therefore F (2_ 4.5, 2_ 4.51 M
=
3 2
3 2 3F = M F = 3490
= 1160 lb.
3 The clearance between the pipes is small, therefore, the load will spread around the circumference of the rings. The load will be reacted by both ring action and by meridinal bending of the pipe. To simplify the problem neglect meridinal pipe bending and assume the load is entirely reacted by a 4.5 inch wide ring.
flodel the end of the pipe as a 4.5" wide ring with the 1160 lb. load distributed as shown below:
1160 i
YT i
/
)'
/
If/
1160 This loading can be obtained by superposition of three cases from Table VIII of Roark.
'" A h\\
lu 1 T._Y__
P cAss I
CASC I2 cASC 1
+
gpgg hx Igo"
!wR ly v
V=wR U
The maximum moment will occur at x = 0 and 180.
" total : M case 12 + t4
~ " case case 2
1 U
X=o x =180 x=o M
=.305 wR2 + wR2 (.02653) (1-0) - 1 wR2Z case 12 U
I x=o U = cos X e cos o = 1 Z = sin x e sin o = 0
" case
=.305 wR2 12 x=o M
=.305 wR2 + wR2 (.0269 ) (1-if) - 1 wR 2
Z case 12 2
x 180
- 1 wR2 (), 7) 3 6
U : cos 180 = -l Z = six 180 = 0 M
=.305wR2 + 2wR2 (.02653) -l 2
wR case 12 li 0
X = 180
=.1914 wR M
=.3183 WR case) x=0
" total =.305 wR2 +.1914wR2
.3183WR wR
=W
" total =.1781 WR
- max = 6 Mmax b T4
=6
(.1781) fil60) 1.11 4.5 (.154J4
- 12,900 psi f allow at 40 + 600
= 320 F is 1.5S = 1.5 (19,800) = 29,700 psi ok g
2 Moment at marmon clamp.
The marmon clamp is on the horizontal run of pipe. Therefore, the moment is equal to R I 2 2 M
7000 in. lb.
=
clamp Replacement mannon clamp:
- >T I
P.
l 7000 in. lb.
\\
j k
+x p) -p
= 90 psi 0
Po f
f
/
i P
I i
i l
l X-X The pipe and pipe flanges have not been changed. Therefore, only the clamp must be analyzed. The moment is resisted by sinusoically varying force.
l
[
F
[
R r
max
'Y L
l
~
i F = Fmax sin 9 L = R sin 9 M=2{"Rsin9Fmaxsin9d9 n
2
= 2R F max sin 9d9
= 2R Fmax
= R n F max 7
Emax = M f folNT 0.,
./
Fug _
zo The contract force must be normal to the surface 0-i Fs n_\\.
_ir I I xg\\
n g = l.ss Z = Fmax cos 20 F max = 7000
= 1401 lb 1.59n Ignoring friction X : Fmax sin 20 Z = 1401 cos 20 = 1317 lb X = 1401 sin 20 = 479 The maximum stress in the clamp will occur at point a.
-t (o.3 -418) a L,f
\\
3
,i
\\
\\ (,15t,.19'r) 1
'W' o = - 2.,e x - u. s-r,
\\
X s
O=
3G y -Y +. o916 Using the equation for the distance from a point to a line i
AX)+BY)+c[
d:
gA2+B2 d, =.364 d =.033 2
M =.364 (1317)
.033 (479) : 464 in 1b.
I=6(1)(464)
= 20,900 psi
('.365)4 allow = 1.55, 25,500 psi ok
=
The bolt preload must be larger than the total value of the x force and the pressure force.
The total value of the x force is equal to tr-X = sin ?0 F (9) d9 cos 20 o
(FmaxsinedG
=.364
=.364 (1400 (2) = 1020 lb The pressure force is P = AP2R = M (2) (1.59) = 286 lb f
1 l 1020 + 286
\\
l s
I 1020 + 286 The minimum bolt preload must be 1020 + 2EC 652 lb
=
2 e
Installation torque is 10-13 ft lb on a 7/16 - 14 UN - 2 bolt Check min. preload T =.2 Fd F: T
= 10 (12)
= 1550 lb ok
.2d
(.2).387 Check max bolt stress F = 1550 (13) = 2016 lb 10
= F/A A=bolttensilearea=j[(d')2 d = equivalent diameter
/d =.4375
.9743
=.3679 14 GT' 2016
=
.7854 (.3679)2
= 19,000 psi Based on Section III Subsection NG c7' allow = (1.2) (.9) Sy = 20,300 psi ok Sy = 18,800 psi at 600 F f
?
The moment at the ball joint (7700 in. lb.) is transferred to the baffle plate by the support as shown on Page 12.
The support is not attached to the ball joint and the hydraulic forces will cause the ball joint to more away from the support. This movement (J ) is equal to the sum of the deflection of connector pipe between the inlet tee and the ball joint (f ) plus the deflectio-Of the sparger ring (f ) plus i
2 the rotation of the sparger ring (9) times distance between inlet tee and ball joint.
[ = f) + [2
- G J2
$ 1-us.1 G, 7, _ g g, V(
)
] : 7.0 E, T (r
by3 oo The moment is reacted by the connection, therefore 3
FL 1~
3EI F = 684 L
7.0
=
6 E
26 X 10
=
I
.666
=
$)
.0045
=
[
. Fj)2 h
)1
+Y
- 3 l) ( f) +1) 2 2
1 2
=
6EI (l) + / 2)
))
14.1
=
24 2
[2
.0092
=.
)
@ = 7 (684) = 4790 in. Ib.
6 I4'I
=
8.1 G J 6
G = 9.9 X 10 J = 2I : 1.332 6=
.0032 radius
(=
.0045 +.0092 + 7 (.0032)
-f =.037 inch Thus, the total differential motion of the ball joint and the support is
.037 inches. The cending moment at the ball joint will be transferred into the support as the support will prevent rotation of the ball joint.
The contact point will be as shown below for case 1 or case 2.
j
~
~<
/
c Ass 2 gp/
f casc
= F:foEH
\\
g s
M: 7700 Q
t
\\ O' s
Case 1 d = 1.5 + $ = 1.54 The vertical force (F) needed to react the bending moment F (-
5000
=
Assume that the contact force spreads out until yield stress is reached.
Then the required contact area (A) 5000 2
A 266 in
=
=
18,800 l
[.80 d
t A
yA/2 s
/4 N
i-I R: 2.0 ' l2.lo2S = R l
20 3, $
l 1
/
/
L_ _ V Let a equal the distance between the center of the bolt and the centroid of the contact area a e 2.0 - 1.5 -(2.2 - 2.20) 2 2
.5 - j 425)
.344
=
=
Therefore, the bending moment per bolt (Mbolt) can be calculated.
5000
=
a=
(.344) = 860 in. lb.
Mbolt The tensile load in the bolt is
=f=5000 L
2500
=
Maximum bolt stress
+
I max 2500 860.375 v
4 26,400 psi
=
=
1 f (.375)4 max 7(.75) 4 4
The allowable stress is 1.2Sy 1.2 (18,800) 1.2Sy 22,600 psi
=
=
Subsection NG of Section III states the allowable primary plus secondary membrane plus bending stress in a bolt is equal to 1.2Sy or.67 L where:
L L
lower bound limit load with a psuedo Sy equal to 1.5 Se L
=
The below curve is a bending / tension interaction curve for a solid circular cross section from page 559 of Reference 6.
I.T s
C (.9 2 l.98 j
M APPt.ico 6'Asric.
/
\\
\\
\\
/
\\
17 g -
8 l
/
/
\\
i l
1
\\
//
ba.co 8
/
I
\\
Pcuastm
,23 f.0
H applh i 860 in. Ib.
=
P applied 2500 lb.
=
(1.5 Sm) f(.75)2 P elastic 11,200 lb.
=
=
(1.5 Sm) f=
(1.5) 17,000 (.78
.375)4 M elastic
= 1050 in. Ib.
=
M applied 860
- 82
=
M elastic 1050 P applied 2500 P elastic IT2gg =.22 The location of point C is (.42, 1.47) which corresponds to L '
L Length A-B is equal to A-B=[.822 +.222
.85
=
Length A-C is equal to A-C =
1.472 +.422 1.53
=
=.56 4.
67
)
Therefore, the bolts are OK.
For Case 2, the vertical force (F) occurs at the outer diameter of the ball joint.
d 2.625
=
2930 F
=
=
2.625 -
.04
- 2.00
.585 a =
=
2930
(.585) 858 in. Ib.
M
=
a =
=
bolt f=1470 L
=
M apolied 858
- 82
=
H elastic 1050 P applied 1470
- 13
=
P elastic 11200 Point D is clearly less limiting than Point B, therefore OK.
The maximua stress in the support will occur for Case 2.
(q7o (q7o E
C.s A
l 1.75 Y
--y
=
'I 2%o 1470 (.75)
M 4040 in. lb.
=
=
F
=
=
max 2
9,900 psi bT (1.75)
(.8) 6.
Baffle Plate Hold Down Studs The baffle plate assembly is bolted to brackets that are welded to the reactor pressure vessel. The hold down is shown on 137C7221 and is 3/4 - 10 UN - 2.
The installation torque is 55 ft lbs.
Installation torque and bolt axial load are related by:
(machine design-Shigley Ref. 5).
T = Fide 1 + nfdm secf3
+.625 fc 2 ndm-flsecp T = applied torque F; = axial load dm = pitch diameter 1 = thread pitch f = thread coefficient of friction B = half thread angle f = nut coefficient of friction c
Assume f = fc :.15 then T =.2 F) dm F. -
T I
.2 dm F.=
55 (12)
= 4850 lbs.
I
.2
.68
( = Fi X-A =.7854 (.75
.9743 ) 2
=.335 in 2 10 o' = 4850
= 14,500 psi <20,300 psi
.J43 OK The baffle plate hold down studs will carry additional load if the hydraulic forces are greater than the weight of the baffle plate.
QA FFL6 pt o uT4.s :-
3 so 1.
X Flotd j
i 6
Total flow rate = 12X 10 lbskr = Q Due to the large hole in the center of the baffle plate, there is no pressure drop across it. The only hydraulic force is due to change of monentum.
2 F = PAV 9
3 P = 31. lbs/ft 2
2 A ='109
.7854 = 60.1 ft 0
V = _Q = 12X10 lbs/hr 3
PA 3600 sec 60.1 ft4 31. Ibs/ft hr
= 1.8 ft/sec F = 31 (60.1) (1.8)2 = 187 lbs 32.2 Since the baffle plate weighs much more than 187 lbs, the maximum stress in the holddown studs is equal to the preload stress of 14,500 psi.
7.
Core Spray Sparger J Bolt The installation sequence is to tighten the J bolt snug and then 1/4 turn tighter.
O
.T-B o s-T
/
c S PA RGcR Rottc
'/// / /D STEAM GAFFt C (A 55unto RIGID) l
]l4-Io THRCAD 1/4 turn is 1
=.025 inch (4) (10)
The.025" deflection must be absorbed by either elongation of J bolt, bending of the J bolt, or compression of sparger ring.
7 G nT-i[
7 Goer 2 S PA R GER P
^P y
$ 1 bolt =PLf 3Ei b kbolt P1
=
AE 3
=.149 PR (Roark 4th Edition Table VIII)j sparger El
.025 =
lt + A D sparger + [2 bolt 1=4 L = 1.63 Conservatively use a constant diameter of 3/4".
A =.7854 (.75) 2 =.44 inch 2 6
E = 26X10 si Assume a.75 wide ring I =.75 (.154)3
=.000228 in#
Ib = #~(.75)4 =
4
.0155 in 12 64 R = 1.19 0.025 = P4
+.149 (1.19) P
+ P (1.63)3
=.0000463P
.44 (26X10 T 26X100 (.000228) 3(26 X 10 )(.0155) 6 P = 540 lbs.
The highest stress in the sparger will occur at the point of contact between sparger and steam baffle.
+
ROARK Table XIV b = 2.15 Y
\\ T P = 540
= 720 lbs /in
.75 b = 2.15 720 (2.38)
=.0175 inch N'
26X100 Area of contact :.75b Area =.0131 P bearing = 540
= 41,200 psi
.0131 Per Section III Subsection NG the allowable average bearing stress under the head of a bolt is 2.7 Sy (Sy 18,800 for 316 at 6000F)
= 50,800 psi. Therefore ok The highest stress is in the J bolt is equal to the sum of the bending stress plus the axial stress.
E + PLC 540 540 (1.63).375 22,600 psi
=
=
A I
.44
.0155 Based on Section III, Subsection NG, the allowable bending plus membrane stress in a bolt is 1.2 Sy = 22,600 psi. Therefore, OK.
+
IV. REFERENCES 1.
General Electric Drawing 794E830 2.
General Electric Design Specification 22A5589, Rev.1 3.
Mechanical Vibrations by Den Hartog, Fourth Edition 1956.
4.
Formulas for Stress and Strain by Roark, Fourth Edition, 1965 5.
flachine design by Shigley,1956.
6.
Advanced Mechanics of fiaterials, Sealy and Smith, Second Edition.