1. A cylindrical shell is subjected to internal pressure such that Hoop’s Hoop’s and axial stresses in the wall reach 150 and 75 MPa respectively. respectively. Two small circular holes are punctured into the shell wall such that the line joining the centres of the two circles is parallel to the axis of the cylinder. The centres of the two circles are separated by distance 8x. One circular hole is of diameter 2x and the other of 4x. Estimate the Hoope’s stress value in t he wall at the midpoint of the line joining the centres of the two holes.
Solution:
σH = 150 MPa σL = 75 MPa θ = 90° Since the point of calculating the stress is midpoint of the line joining the centers of the two holes, r = 4x for both the holes. For radius (a) = x,
σt (x, 90°) =
150
2
x4
x 1 + (4x) − 1502 1 + 3 × 2
2
(4x)4
Cos180°
= 155.56 MPa
For radius (a) = 2x,
σt (2x, 90°) =
150 2
2
4x 1 + (4x) − 1502 1 + 3 × 2
16x 4 (4x)4
Cos180°
= 75 × 39/16 = 182.8 MPa
Stress intensification at that point along circumference circumference due to axial stress :
σt (x, 0°) =
2
x 75 − 1 + 3 × 1+ 2 (4x) 2
75
2
x4 (4x)4
Cos0°
= 1.9 MPa
σt (2x, 90°) =
2
4x 75 − 1 + 3 × 1+ 2 (4x) 2
75
2
16x 4 (4x)4
Cos0°
= 2.34 MPa
St at the midpoint Hoops stress will be = 150 + 55.56 + 82.8 + 1.9 + 2.34 = 292.6 MPa
1
2. A 1200 mm (48 inch) NB XS pipe has the following dimensional standards. Nominal Bore
Schedule
Average wall thickness
48”
XS
0.500”
For this pipe, recommend the maximum internal working pressure that can be used safely. If such a pressure is applied, what would be the values of Hoop’s stress and axial stress in the pipe and what would be their nature?
If the pipe is to withstand net external pressure with same magnitude as the above design pressure without provision of stiffening rings, what would be the maximum length of the pipe that you would allow? What would be the values of Hoop’s and axial stresses and their nature in this case?
The design temperature is 500 C and the allowable stress of material at design temperature is 140 MPa. For external pressure service, use applicable material chart for SA-l06-B supplied to you. Recommended corrosion allowance is nil and 12.5% negative mill tolerance is recommended.
Solution:
Do = 48 in. = 1219 mm = 1.219 m t = 0.5 in. = 0.0127 m m = 12.5% Sa = 140 MPa = 140000000 Pa E = 0.9 (assumed)
Removing mill tolerance, t = (1 - 12.5/100) × 0.0127 = 0.011 m t=
or,
PD o 2Sa E 0.4P
P 1.219 2140000000 0.9 0.4P
= 0.011
or, P = 2290530 Pa = 22.9 bar = 332.2 psi
Hoop's stress =
PD i 2t
=
−
P(D o 2t) 2t
=
−
2290530 × (1.219 0.022) 0.022
2
= 124625655 Pa = 124.6 MPa
Axial stress = 124.6/2 = 62312827.5 Pa = 62.3 MPa
Part II Do / t = 1.219/0.011 = 110.8 =111
Pa =
B=
or,
4B 3Do /t 3P a (D o /t) 4
=
3×332.2×111 4
= 27656
From chart, L/Do = 5.5 or, L = 5.5 × 48 = 264 in. = 22 ft
3. An ellipsoidal head with a = 1.2 m and b = 0.6 m and with thickness of 6 mm is subjected to an internal pressure. Determine the MAWP if the MoC allowable is 100 MPa.
Solution:
Given, Sa = 100 MPa = 100000000 Pa t = 6 mm = 0.006 m This is an ellipsoidal closure (2:1) with radius = 1.2 m Therefore, Do = 2.4 m E = 0.9 (assumed)
t=
or, P
PD o
2 S a E+0.9P
=
2S a Et
−
D o 1.8t
=
2 ×100000000 × 0.9 × 0.006 2.4
− 1.8 × 0.006
= 452034 Pa =
4.5 bar
4. Find: a.
The largest size of a self compensating, non-protruding 40 SCH nozzle that can be provided on a 24” SCH 40 pipe of the same material. Design pressure is 7.5 bar.
b.
The lightest 16” NB pipe that can withstand internal vacuum over its 10 m unstiffened run without the possibility of buckling failure. Design temperature is 400 C. Assume the MoC to be SA-106-B. 3
Take the allowable stress for the MoC as 120 MPa. Corrosion allowance as 1.5 mm, all welds fully radiographed, zero mill tolerance.
Solution:
P = 7.5 bar T = 400 °C Sa = 120 MPa = 12000000 Pa L = 10 m C = 1.5 mm E=1
5. A vessel with the following design specifications is available. Recommend a safe operating pressure for the same.
Cylindrical body :
OD - 3 m, Wall Thickness - 12 mm, Tangent to Tangent length - 10 m
Top Closure
: 2:1 ellipsoidal, 12 mm wall thickness, flanged connection with shell
Bottom Closure
: Conical with 30 degrees half cone angle, 12 mm wall thickness, welded to shell
Nozzles: 1. Nozzle on cylindrical body with its axis at 30 cm above bottom tangent, 4” NB, 40 SCH, seamless, protruding 3 cm inside the vessel, not reinforced 2. Nozzle on top closure at its crown, 6” NB, 40 SCH, non-protruding, seamless, not reinforced 3. Nozzle on conical bottom, at half depth of the cone, oriented normal to cone wall, 4” OD, 40 SCH, non-protruding, seamless, not reinforced
Solution:
For top 2:1 ellipsoidal closure: ltop = 1/4 of Do = 3/4 = 0.75 m
For bottom conical closure: 4
l bottom = (Do/2)/tan30° = (3/2)/0.5773 = 2.598 m
For cylindrical body: Do = 3 m t = 12 mm = 0.012 m L = 10 m 1/3rd length of conical closure = 0.866 m 1/3rd length of ellipsoidal closure = 0.25 m Effective Length (Le) of shell along with 1/3rd length of two different closures = 10 + 0.25 + 0.86 = 11.11 m Le/Do = 11.11/3 = 3.7 Do/t = 3/0.012 = 250 The value of Factor A = 0.000085 E = 23800000 psi (assumed)
Since, value of Factor A is falling to the left of t he temperature line Pa can be calculated using the formula below: Pa = 2 AE / 3(Do / t) = 2 × 0.000085 × 23800000 / (3 × 250) = 5.39 psi
Internal Pressure Design:
For cylindrical body: Do = 3 m = 118 in. t = 12 mm = 0.012 m =0.47 in. L = 10 m tc =
PDo 2Sa E 0.4P
0.47
P 118 223800000 0.9 0.4P
or, P =
Nozzles: 5
6. Find the adequacy of the pressure vessel with design as in Question 5 for vacuum service. Assume that the nozzles and the conical closure are unlikely to fail because of buckling. If you find the cylindrical body to be inadequate, suggest suitable number of stiffeners. If top closure is inadequate to withstand vacuum, suggest a suitable wall thickness for it.
7. A 150 mm (6”) NB pipe needs to be designed for an external design pressure of 400 psi and a design temperature of 400 C. The MoC of the seamless pipe is austenitic stainless to ASTM A 312 TP 304 L. The corrosion allowance is nil. Negative mill tolerance of 12.5 % should be used on nominal thickness. Select a proper pipe schedule for a pipe segment of 7.5 m with flanged connections at both the ends. Recommend maximum operating pressure if the pipe designed by you is also to be used for internal pressure service occasionally. Allowable stress of the MoC at design temperature is 40000 psi.
Solution:
P = 400 psi (external design pressure) T = 400°C L = 7.5 m Sa = 40000 psi Do = 168.3 mm = 6.62 in.
Let us assume a thickness of 0.2 in. or 5.08 mm L/Do = 7.5/0.168 = 44.64 Do/t = 0.168/0.005 = 33.6 Factor A = 0.022 Factor B = 11000 P=
4B Do 3( ) t
=
4 × 11000 3 × 33.6
= 436.5 psi
Since this pressure is more than design pressure assumed thickness is ok. Mill tolerance = 0.125 × 5.08 = 0.635 mm After adding mill tolerance, t = 5.08 + 0.635 = 5.715 mm Based on available pipe schedule, 40 SCH can be recommended. 6
8. Design a spherical vessel for storage of a chemical for the f ollowing data.
Design Pressure (Internal)
: 100 psig
Design Temperature
: 300°C
Yield Stress at room temperature
: 60000 psi
Weld Joint Efficiency
: 0.85
Outer Diameter
:2m
Corrosion Allowance
: 3 mm
Negative mill tolerance on nominal thickness of 5% is acceptable.
Two nozzles (one protruding, another non-protruding) are to be provided on this vessel. Both nozzles are made out of 6” NB 40 SCH welded pipe with weld joint efficiency of 0.85. Check the nozzles for reinforcement requirement and suggest reinforcement pad thickness if the pad is required. For suitable support design, the Civil department needs to be provided with the weight of the tank if filled with water. Calculate the same.
Solution:
Design Pressure (Internal)
: 100 psig = 114.7 psi
Design Temperature
: 300 C
Yield Stress at room temperature
: 60000 psi
Weld Joint Efficiency
: 0.85
Outer Diameter
: 2 m = 78.7 in.
Corrosion Allowance
: 3 mm = 0.118 in.
t=
=
PD o 4Sa E 0.4P
114.7 78.7 460000 0.85 0.4 114.7
0.044 in. 1.12 mm
Negative mill tolerance = 5/100 × 1.12 = 0.056 Adding corrosion allowance and negative mill tolerance we get, t = 1.12 + 3 + 0.056 = 4.18 mm
7
Nozzle:
9. A pressure vessel has a diameter of 1500 mm. It is to be provided with a flanged closure at its top. The design pressure in the vessel is 35 bar (internal). Allowable stress for the MoC is 100 MPa. Determine the wall thickness of:
1. Formed hemispherical cover 2. 2:1 ellipsoidal closure with weld joint efficiency of 0.85 3. Formed Dished closure with crown radius same as vessel OD and knuckle radius 10% of the crown radius. 4. Conical head with cone angle of 45 degrees, fully radiographed.
Solution:
Diameter = 1500 mm = 1.5 m Internal Design Pressure = 35 bar = 3500000 Pa Allowable stress = 100 MPa
1. For hemispherical closure: t=
=
PD o 4Sa E 0.4P
3500000 1.5 4100000000 0.85 0.4 3500000
0.015 m 15 mm
2. For 2:1 ellipsoidal closure: t=
PD o 2Sa E 0.9P
3500000 1.5 2100000000 0.85 0.9 3500000
3. For dished closure: t=
PDcrown M o 2Sa E 0.5P(M - 0.2)
Crown radius (L) = 1.5 m 8
0.03 m 30 mm
Knuckle radius (r) = (10/100) × 1.5 = 0.15 m Therefore, L/r = 10 M (from chart) = 1.54
t=
=
PDcrown M o 2Sa E 0.5P(M - 0.2)
3500000 1.5 1.54 2100000000 0.85 0.5 3500000 (1.54 - 0.2)
0.046 m 46 mm
4. For conical closure: Half cone angle = 45° E=1 t=
=
PD o 2cosαSa E 0.4P 3500000 1.5
2 0.707 100000000 1 0.4 3500000
0.036 m 36 mm
10. Design the lightest cylinder with welded hemispherical closures to store 15 kg air at 100 bar pressure. Take allowable stress of MoC as 40000 psi, weld joint efficiency as 1 for all welds. Make suitable assumptions if necessary. Corrosion allowance and negative mill tolerance are zero. Fabrication is done using sheet material. Consider two cases: (A) When all parts must be fabricated using sheet of same thickness. (B) When different parts could be fabricated using sheets of different thicknesses.
Solution:
Internal design pressure(P) = 100 bar Allowable stress (Sa) = 40000 psi = 2758 bar Weld joint efficiency (E) = 1 Diameter (Do) = 1.5 m ( assumed)
Case A - When all parts could be fabricated using same sheet thickness: 9
Spherical shape always experiences less stress compared to cylindrical shape. Therefore,
Case B - When all parts could be fabricated using sheet of different thickness: For hemispherical closure: t=
=
PD o 4Sa E 0.4P
100 1.5 42758 0.85 0.4 100
0.0148 m 14.8 mm
For cylindrical vessel: t=
=
PD o 4Sa E 0.4P
100 1.5 22758 0.9 0.4 100
0.0297 m 29.7 mm
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