393
D
Sample of Pressure Vessel Design Data Sheets
Pressure Vessel Design Data Sheet
1) Requirements:
Working Pressure: 525 psig
Design Code: ASME Section VIII – Div. 1
Working Temperature: 500 ∘ F
Construction: All Welded
corrosion allowance: 0.063 in.
2) Material Specifications:
Material
Material specifications
Maximum allowable stress
Reference
P no.
Grp. no.
Plate
SA-204, Grade B
20,000 psi
Tbl. 1A, II-D
3
2
Castings
SA-216, Grade WCA
16,300 × 0.8 = 13,040
Tbl. 1A, II-D UG-24(a)
1
1
Forgings
SA-181
16,300
Tbl. 1A, II-D
1
1
Bolting
SA-193, Grade B16
20,000
Tbl. 1A, II-D
Piping
SA-106, Grade B
17,100
Tbl. 1A, II-D
1
1
Gaskets
1/16 in. mineral
y = 3700; m = 2.75
Tbl. 2–5.1
3) Sketch and General Dimensions of Vessel
4-in. outlet with welding neck flange
Seamless dished head, pressure on
concave side, unstayed
8-in. outlet with lap joint flange
8-in. manway
36 in.
nominal inside
diameter
Special cast steel welding
neck flange
Special forged steel blind
flange
1-1/2 in. blowdown outlet with welding
neck flange
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
D Sample of Pressure Vessel Design Data Sheets
Design of Head and Shell
Shell Thickness: UG-27 (c)(1)
Minimum required thickness, t =
1)
PR
(SE) − (0.6P)
(estimate E = 0.70, Tbl UW − 12)
(double welded butt joint)
(525)(18)
= 0.691
(20,000 × 0.70) − (0.6 × 525)
add corrosion allowance ∶ 0.691 + 0.063 = 0.753 in.
2) Check postweld heat treatment and radiograph requirements.
UCS-56, Tbl UCS-56 Note (3)(a): Postweld heat treatment required if thickness exceeds 5/8 in; because
0.753 > 0.625, postweld heat treatment required UW-11 (a), UCS-57 (note: cf. UW-52): radiograph required
P-3, Gr 1, 2, 3 if t > 0.750 in.; because 0.753 > 0.750, radiography is required;
3)
recalculate thickness required using E = 0.85(Tbl UW − 12, spot examined)
525 × 18
t+c=
+ 0.063 = 0.566 + 0.063 = 0.629
(20,000 × 0.85) − (315)
use plate thickness = 0.750
4) Check applicability of formula: UG-27 (c)(1)
0.385 SE = (0.385) (20,000) (0.85) = 7700; 525 < 7700 formula OK
0.5R = (0.5) (18) = 9.0;0.750 < 9.0 formula OK
Head Thickness: (UG-32)
1) A dished head will be used, UG-32 (j) requires that the inside knuckle radius shall be not less than 6% of the outside
diameter of the skirt, nor less than three times the head thickness. Also, the inside crown radius is to not exceed
the outside diameter of the skirt, thus try:
inside crown radius, L = 27 in. < 37.375 in.
inside knuckle radius, r = 4 in. > 0.06 × 37.375 = 2.24 in.
2) Thickness required: UA-4 (d)
PLM
t=
(2SE) − (0.2P)
t+c =
E = 1.00 for seamless head
for L∕r = 27∕4 = 6.75,
M = 1.40(Tbl UA − 4.2)
(525)(27)(1.40)
+ 0.063
(2 × 20,000 × 1) − (0.2 × 525)
= 0.500 + 0.063
t + c = 0.563 therefore use head plate thickness of 0.688 in.(11∕16)
3) Check minimum thickness: UG-32 (b).
For a joint efficiency of 0.85 between the head and shell, postweld heat-treated double-butt weld for the circumferential joint: UG-32 (f ).
[
]
1
PL
tmin =
(0.85) (2SE) − (0.2P)
[
]
(525)(18)
1
tmin =
; tmin = 0.237
0.85 (2 × 20,000 × 1) − (0.2 × 525)
0.237 < 0.688, thickness OK.
from UCS-16 (b): 3/32 < 0.688, thickness OK.
4) Check minimum inside knuckle radius: UG-32 (j)
(3)(0.688) = 2.063 4.0 > 2.063 knuckle radius OK
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394
5) Cylindrical skirt: UW - 13 (a) 0.688 < (1.25)(0.750) = 0.937
A cylindrical skirt is optional for the head butt-welded to the shell, a tangent of 1 1/2 in. length will be used to
provide a bevel for the butt weld; this will avoid cutting the bevel into the knuckle.
Sketch of the head and shell showing principal dimensions and tolerance.
11
16
3
4
7± 1
35 8 8
[UG–80]
1
1
a) Vessel to be postweld heat-treated at 1100 ∘ F for 3/4 h.
b) Maximum permissible offset of circumferential joints = 3/16′′ (UW-33).
c) Spot examination of welded joints per UW-52.
1 3/32′′ reinforcing bead: UW-35 (a), UW-51 (a)(1).
No offset preparation necessary between the head and shell because the thicknesses do not differ by more than 1/8 in.
UW-9(c).
Inside surface of head shell not deviate
more than 7/16 in. from 27 in. radius UG–81 (a)
4 in. R
27 i
1–1/2 in., bevel 30″ for welding
n. R
1
18 in.
4
14 in.
Data Sheet for Reinforcement Calculations (UG-37, −40)
1) Shell
Description: (longitudinal butt joint, double-welded, postweld heat-treated, spot radiography).
Design pressure
P
525 psig
Joint efficiency
E
0.85
Maximum allowable stress
S
20,000 psi
Corrosion allowance
c
0.063 in.
Inside radius of shell (specify) before corrosion allowance is added R, L, K 1 , D
Nominal thickness, exclusive of corrosion allowance
t
0.687 in.
Minimum required thickness (reference: UG-27)
tr
0.480 in.
tr =
(525)(18)
PR
=
(SE) − (0.6P) (20,000 × 1) − (0.6 × 525)
Excess thickness
t – tr
0.207
2) Nozzle
Material used: seamless steel pipe, 5A-106, Grade B (Schedule 80).
18 in.
395
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Sample of Pressure Vessel Design Data Sheets
D Sample of Pressure Vessel Design Data Sheets
Maximum allowable stress
S
17,100 psi
Inner diameter of finished opening in corroded condition
d
7.750 in.
Nominal thickness exclusive of corrosion allowance
tn
0.438 in.
Thickness required for hoop stress (UG-27)
t rn
0.117 in.
trn =
(525)(7.50∕2)
PR
=
(SE) − (0.6P) (17,100 × 1) − (0.6 × 525)
3) Sketch of reinforcement with dimensions and welding detail (UW-15, −16, −18)
8.625
0.063
1 in.ϕ
Tell tale
4
Hole (UW-15c)
0·207
0.687
tr n = 0.117
xs = 0.321
14.625 DIA
50
1.592
tn = 0.438
d = 7.750
15.50
Reinforcement Calculations for 8 in. Nozzle
1) Reinforcement required: UG-37
A = 3.720 sq. in.
A = (d)(t r )(F) = (7.750) (0.480)(1)
2) Metal considered to have reinforcing value: UG-40(d).
A1 = metal in the shell and in the nozzle wall within the shell wall thickness available for reinforcement:
Shell: (15.50–8.625)
(0.207)
= 1.423 sq. in.
Nozzle: (8.625–7.750)
(0.207)(17,100/20,000)
= 0.155 sq. in.
(cf UG-41a)
A1 = 1.578 sq. in.
A2 = metal in the nozzle wall outside the shell thickness available for reinforcement:
(0.438–0.117)(2)(h)
A2 = 0.642 h sq. in.
(0.438–0.117)(2)(1.595) see below for h
= 1.024 sq. in.
A3 = metal added as reinforcement and metal in attachment welds:
Added reinforcement (1/2 in. plate, 8–5/8 ID by 14–5/8 OD).
(0.500)(14.625–8.625)
= 3.000 sq. in.
One-half in. full fillet welds around outside of reinforcing plate and around nozzle wall:
(0.50)(0.50)(0.50)(4)
= 0.500 sq. in.
A3 = 3.500 sq. in.
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396
3) Width of area of reinforcement normal to vessel wall:UG-40(c)
(2.5)(0.687)
= 1.720 in.
or
(2.5)(0.438)
= 1.095.
Added reinforcement
= 0.500.
h = 1.595 (this governs)
4) Summary:
Reinforcement used:
A1 : shell and nozzle
1.578
A2 : nozzle
1.024
A3 : added reinforcement and welds
3.500
6.102 sq. in. used >3.72 required.
Attachment Welding for 8 in. Outlet (UW-15, -16)
1) Allowable stresses in welds: (Tbl. UW-15)
a) Combined end and side loading stress in butt welds
(20,000) (0.74) =
14,800 psi
b) Combined end and side loading stress in fillet welds
(20,000) (0.49) =
9800 psi
2) Strength of Welds:
a) Outer fillet weld between vessel wall and reinforcing plate
(π) (14.625/2) (0.500) (9800) =
112,600 lb.
b) Butt weld between vessel wall and nozzle wall
(π) (8.625/2) (0.687) (14,800) =
137,800 lb.
c) Fillet weld between nozzle and reinforcing plate
(π) (8.625/2) (0.500) (9800) =
66,400 lb.
d) Butt weld between nozzle wall and reinforcing plate
(π) (8.625/2) (0.500) (14,800) =
100,300 lb.
3) Loads:
a) Strength in tension of the plate removed (UG-41(b)(2)
(20,000) (0.480) (8.625) =
82,800 lb.
b) Strength of metal in vessel wall available for reinforcement
(20,000) (0.207) (15.50–8.625) =
28,460 lb.
c) Reinforcement load carried by the nozzle wall
(17,100) (2) (0.438) (0.687) =
10,290 lb.
+ (17,100) (2) (0.321) (1.595) =
17,510 lb.
27,800 lb.
397
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Sample of Pressure Vessel Design Data Sheets
D Sample of Pressure Vessel Design Data Sheets
d) Reinforcement load carried by reinforcing plate
82,800 − (28,460 + 27,800) =
26,540 lb.
4) Summary:
a) Load to be carried by attached reinforcement
82,800–28,460 =
54,340 lb.
Strength of attachment = 112,600 + 137,800 =
250,400 lb.
b) Load to be carried by nozzle wall =
17,510 lb.
Strength of attachment =
137,800 lb.
c) Load to be carried by reinforcing plate =
26,540 lb.
Strength of attachment =
112,600 lb.
Attachment welding is satisfactory.
8 in. Lap-Joint Flange Attachment and Selection
1) Check welding between the nozzle and shell and reinforcing for strength to withstand hydrostatic end force.
Permissible shear stress in butt welds (Tbl UW-15)
(20,000) (0.60) =
12,000 psi.
Shear strength of butt weld (cf sheet 4 for dimensions)
(π) (8.625) (0.687) (12,000) =
223,400 lb.
Hydrostatic end force to inner diameter of nozzle
(π/4) (7.750)2 (525) =
24,700 lb.
Thus, the butt weld alone is safe for hydrostatic end force
223,400 > 24,700 lb.
2) From ANSI B16.5-2017 Steel Pipe Flanges and Fittings, use 400 lb.
Rated at 665 psi at 500 ∘ F.
From vendor’s catalog:
Number of bolts
12
Size of bolts
1 in.
Bolt circle diameter
13 in.
3) Use lap-joint stub end, 8.625 OD by 7.625 ID by 8 in. long
4) From vendor’s catalog, use 1/16 in. gasket – 12 in. OD by 8 in. ID.
Preliminary Design of 18 in. Integral Flange
A = (27.125)
C = (24.625)
W
critical section
R
R
g1
B = 16.875
t
W
g1/2
3
h
1
Stress
g0 = 0.563
18 in. O.D.
Design pressure =
525 psi
Flange design stress = 12,000 psi
Bolt design stress =
20,000 psi
Try 1.250 in. Diameter bolts
Root area = 0.943 sq. in. (8 thrd)
Estimate g0 = 0.563 in.
Section for
Section for
Radial
M longitudinal
hub stress
Flange
M
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398
Trial Dimensions
1) Hub length, h ≅
√
√
Bg. = (16.875)(0.563) = 3.08 in.
try h = 3.00 in.
2) Center line bolt circle to hub ≅ (1.5)Db = (1.5)(1.25) = 1.875
try R = 2.00 in.
3) Minimum bolt circle diameter for nut clearance = 18 + (2)(1 + 2)
try C = 24.0 in.
4) Bolt spacing = 3 in. or (2.25 × Db ) = (2.25)(1.25) = 2.81 in.
5) Number of bolts = π(24.0)/2.81 = 26.8
try 24 bolts
6) Minimum flange OD = 24.0 + 2(1.25) = 26.5
try A = 26.5 in.
Flange Thickness Calculations
Assume that the hydrostatic end force, H, is effective to 22 in. diameter, the assumed location of the gasket reaction,
and use a factor of safety of 2 to assure a tight joint.
1) Design hydrostatic end force, H = (π/4)(222 )(525)(2) H = 400,000 lb
2) Required number of bolts to carry this load:
= (400,000)/(20,000)(0.943) = 21.2 (this compares favorably with 24)
3) g 1 /2 = (1/2)(0.563 + 1.0) = 0.782
4) For the critical section at 16.875 + (2)(0.782) = 18.438 diameter, the arc length of a sector containing one bolt will
be: arc length = (π)(18 438)/24 = 2.42 in.
5) Allowable bolt load on sector = (20,000)(0.443) W = 18,860 lb/bolt
6) Gasket reaction = 18,860 – (1/24)(π/4)(222 )(525) = 18,860 – 8330 = 10,530 lb/bolt
7) Bending moment at critical section
= (18,860)(1/2)(24.0 – 18.438) – (10,530)(1/2)(22.00 – 18.438)
= 52,500 – 18,750 = 33,850 lb in.
8) Calculated thickness for radial flange stress: Sf = Mc/I; (UA-52)
12,000 = (33,850)/(1/6)(2.42)(t 2 );t 2 = 7.00
t = 2.65 in.
try t = 2.75 in.
9) Calculated thickness for longitudinal hub stress (UA-52)
(12,000)(1.5) = (33,850)/(1/6)(2.42)(g 1 2 ) ∴ g 1 2 = 4.66 and g 1 = 2.16 in.
10) Revise estimated flange dimensions using:
g 1 = 2.00 in
C = 16.875 + 2(2.00 + 1.875)= 24.625 in.
R = 1.875 in
A = 24.625 + 2(1.25)
= 27.125 in.
h = (3)(2 – 0.563) = 4.31, say 4.50 in.
Preliminary Design of 18 in. Integral Flange
1) A 1/16 in. flat asbestos gasket will be used, 19 in. ID by 22 in. OD
2) From Table 2-5.2:
Basic gasket seating width, bo = (1/2)(22√
– 19)/2 = 0.750 in.
Effective gasket seating width, b = (1/2) 0.750 = 0.433 in.
3) Location of gasket load reaction, Table 2-5.2
G = 22.0 – 2(0.433) = 22.0 – 0.866
G = 21.133 in.
399
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Sample of Pressure Vessel Design Data Sheets
Calculation for 18 in. Flange, 2–4 to 2–8
1) From UG-34 (c)(2): t = d CP∕S + 1.9WhG ∕Sd3 , Figure UG-34(j).
2) From Calculation sheet above: W = 453,000 lb.
hG = (1∕2)(24.625 − 21.133) = 1.746 in.
d = G = 21.133 in.
H = 184,000 lb
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D Sample of Pressure Vessel Design Data Sheets
400
Blind Flange for 18 in. Manway√
Flange
3) From sheet 1: S = 16,300 psi (SA-181 forging).
4) C = 0.30 UG-34(d)
√
(0.3)(525) 1.9(184 000)(1.746)
= 2.467
t = (21.133)
+
16,300
(16,300)(21.133)3
adding corrosion allowance: 2.467 + 0.063 = 2.530
use 2–5/8 thickness.
5) Sketch of blind flange
2–1/4 in. dia. spot face for nuts
Bolt holes,
evenly spaced,
24 at
1–5/8 in. dia.
2–5/8 in.
63
22 in. outside diameter of raised face
1/4 in.
raised face
24–5/8 bolt circle diameter
27–1/8 in. dia. flange outside diameter
Data Sheet for Reinforcement Calculations (UG-37, -40)
1) Shell
Description: longitudinal butt joint, double-welded, postweld heat-treated, spot radiography.
Design pressure
P
525 psig
Joint efficiency
E
0.85
Maximum allowable stress
S
20,000 psi
Corrosion allowance
c
0.063 in.
Inside radius of shell before corrosion allowance is added
R, L, K 1 , D
18 in.
Nominal thickness, exclusive of corrosion allowance
t
0.687 in.
Minimum required thickness (reference UG-27) see sheet 3
tr
0.480 in.
Excess thickness
t – tr
0.207 in.
2) Nozzle
Material used: seamless steel pipe, SA-106, Grade B (Schedule 40).
Maximum allowable stress
S
17,100
Inside diameter of finished opening in corroded condition
d
17.0 in.
Nominal thickness exclusive of corrosion allowance
tn
0.500 in.
Thickness required for hoop stress (UG-27)
t rn
0.266 in.
trn =
(525)(17.0 × .5)
PR
=
(SE) − (0.6P) (17,100 × 1) − (0.6 × 525)
401
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Sample of Pressure Vessel Design Data Sheets
D Sample of Pressure Vessel Design Data Sheets
3) Sketch of reinforcement with dimensions and welding detail (UW-15, -16, -18).
18.0
tn = 0.500
1.72
trn = 0.266
27.5
0·207
Tell-tale hole,
1/4 IN-D/A
0.75
0.480
0.687
0.063
d = 17.00
34.00
Reinforcement Calculations for 18 in. Manway Alternate Method: Load Calculations Based on Nozzle
and Vessel O. D
1) Total load to be carried within area of reinforcement
(525) (37.375/2) (34)
= 333 000 lb.
2) Strength of vessel walls within area of reinforcement
(0.687) (34.0–18.0) (20,000)
= 219,800 lb.
3) Strength of nozzle wall within vessel wall thickness
(0.687) (18.0–17.0) (17,100)
= 11,700 lb.
4) Load carried by nozzle wall due to pressure in nozzle
(525) (18) (h)
(est. h = 1)
(525) (18) (1.72)
= 9450 h lb.
= 16 260 lb. corr.
5) Strength of nozzle wall outside of vessel wall thickness
(0.500) (2) (h) (17,100)
= 17,100 h lb.
(0.500) (2) (1.72) (17,100)
29,400 lb. corr.
6) Strength of nozzle wall available for reinforcement
(11,700) + (17,100 – 9450) (h)
= 19,350 lb. est.
(11,700) + (7650 × 1.72) = 11,700 + 13,200
24,900 lb. corr.
7) Load to be carried by added reinforcement
333 000 – 219,800 – 19,350
= 93,850 lb. est.
333 000 – 219,800 – 24,900 (h = 1.72 in.)
70,400 lb. corr.
8) Area required of added reinforcement
93,850/20,000
= 4.693 in.2 est.
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402
9) Area of reinforcement used
3/4 in. plate, 18 in. ID by 27.5 in. OD
= 7.12 in.2
Two 3/4 by 3/4 fillet welds, (0.75) (0.75) (2)(1/2)
= 0.56 in.2
= 7.68
10) Strength of reinforcement added
(7.68) (20,000)
= 153,600 lb.
11) Width of area of reinforcement, h
(2.5)(0.687)
=
1.72 in. (this governs)
(2.5)(0.500)
=
1.25
Added reinf.
=
0.75
h = 1.72 in.
OR
2.00
Summary
12) Load calculated to have existed in the metal removed:
(525) (37.375/2)(18)
= 176,200
13) Strength of metal in vessel wall available for reinforcement:
(0.207)(34 – 18)(20,000)
= 66,200 lb.
14) Strength of nozzle wall available for reinforcement
= 24,900 lb.
15) Strength of added reinforcement
= 153,600 lb.
= 244,700 lb.
Total strength
Because the total strength exceeds the load calculated to have existed in the metalremoved, the design is
satisfactory.
Data Sheet for Reinforcement Calculations (UG-37, -40)
1) Head
Description: dished, 27 in. crown radius, 4 in. knuckle radius.
Design pressure
P
525 psig
Joint efficiency
E
1.0
Maximum allowable stress
S
20,000 psi
Corrosion allowance
c
0.063 in.
Inside radius of shell, or inner crown radius, or equivalent spherical radius (specify)
before corrosion allowance is added
R, L, K 1 , D
27.0 in.
Nominal thickness, exclusive of corrosion allowance
t
0.625 in.
Minimum required thickness (reference: UG-37(b), UA-4(d)
tr
0.355
E = 1, M = 1
t=
Excess thickness
(525)(27)
PLM
=
(2SE) − (0.2P) (2 × 20,000) − (0.2 × 525)
t − tr
0.270
403
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Data Sheet for Reinforcement Calculations (UG-37, -40)
D Sample of Pressure Vessel Design Data Sheets
2) Nozzle
Material used: seamless steel pipe, SA-106, Grade B (Schedule 80).
Maximum allowable stress
s
17,100 psi
Inner diameter of finished opening in corroded condition
d
3.951 in.
Nominal thickness exclusive of corrosion allowance
tn
0.274 in.
Thickness required for hoop stress (UG-27)
t rn
0.0618 in.
trn =
(525)(3.951 × .5)
PR
=
(SE) − (0.6P) (17,100 × 1) − (0.6 × 525)
3) Sketch of reinforcement with dimensions and welding detail (UW-15, -16, -18).
4.500
tn = 0.274
0.686
trn = 0.0618
5
8
0.355
0.688
11
16
0.063
d = 3.951
7.902
Reinforcement Calculation for 4 in. Nozzle in Dished Head
1) Total load to be carried within area of reinforcement:
(525)(1/2)(27 + 1.25)(7.902)
= 58 600 lb.
2) Strength of head within area of reinforcement
(0.625)(7.902 – 4.50)(20,000)
= 42,500 lb.
3) Strength of nozzle wall within head wall thickness
(0.625)(4.50 – 3.951)(17,100)
= 5870 lb.
4) Load carried by nozzle wall due to pressure in nozzle
(525)(4.50)(h)
= 2360 h lb.
5) Strength of nozzle wall outside of head thickness
(0.274)(h)(2)(17,100)
= 9370 h lb.
6) Strength of nozzle wall available for reinforcement
5870 + (9370 – 2360)(h)
(est. h = 1)
5870 + 4810
= 12,880 lb. est.
= 10,680 lb. corr.
7) Load to be carried by added reinforcement
58 600 – (42,500 + 12,880)
= 3220 lb. est.
58 600 – (42,500 + 8840)
= 7260 lb. corr.
10.1002/9781119311515.app4, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/9781119311515.app4 by Nat Prov Indonesia, Wiley Online Library on [21/01/2025]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License
404
8) Area required of added reinforcement
3220/17,100
= 0.188 sq. in. est.
9) Area of reinforcement used: two 5/8 in. fillet welds inside and out plus extension on nozzle inside of head
Welds: (0.625)(0.625)(0.5)(4)
= 0.781 sq. in.
Nozzle: (0.625)(0.274)(2)
= 0.342 sq. in.
10) Strength of reinforcement added
(0.781)(20,000) + (0.342)(17,100)
= 21,500 lb.
11) Width of area of reinforcement, h
(2.5)(0.274)
h = 0.685 in.
Summary
12) Load calculated to have existed in the metal removed
(525)(1/2)(27 + 1.25)(4.50)
33 400 lb.
13) Strength of metal in head available for reinforcement
(0.270)(7.902 – 4.50)(20,000)
18,370 lb.
14) Strength of nozzle wall available for reinforcement
10,680 lb.
15) Strength of added reinforcement
21,500 lb.
16) Total strength
50,550 lb.
Because the total strength of 50,550 lb. exceeds the load of 33 400 lb. the design is satisfactory.
405
10.1002/9781119311515.app4, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/9781119311515.app4 by Nat Prov Indonesia, Wiley Online Library on [21/01/2025]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License
Summary