PRESSURE VESSEL SAMPLE PROBLEMS:
1. Determine the bursting pressure of a steel shell with diameter of 10 inches and made of ¼” thick steel plate.
The joint efficiency is at 70% and the tensile strength is 60 ksi.
1 Given:
Sol'n:
Di=
10 in
t/d=
0.025 (vessel is thin-walled since t/d <0.07)
t=
0.25 in
t=
pD/(4St*E)
e=
0.7
p=
4200 psi
St=
60000 psi
Req'd:
P=
?
2. Determine the safe wall thickness of a 30 inches steel tank with internal pressure of 7.82 MPa. The yield stress
of the material is at 275.48 MPa. The factor of safety to use is 2.
2 Given:
Sol'n:
D=
30 in
t=
pD/(2St*E)
p=
7.82 Mpa
t=
0.851604 in
Sy=
275.48 MPa
t/d=
0.028387 (vessel is thin-walled since t/d <0.07)
FS=
2
Req'd:
t=
?
3. A cylindrical tank with 10 inches inside diameter contains Oxygen gas at 2500 psi. Calculate the required wall
thickness in (mm) under stress of 28,000 psi.
3 Given:
Sol'n:
Di=
10 in
t=
pD/(2St*E)
p=
2500 psi
t=
0.446429 in
S=
28000 psi
t/d=
0.044643
Req'd:
t=
11.33929 mm
t=
?
4. Determine the thickness of a steel air receiver with 30 inches diameter and pressure load of 120 psi,
design stress of 8,000 psi.
4 Given:
Sol'n:
Di=
30 in
t=
pD/(2St*E)
p=
120 psi
t=
0.225 in
S=
8000 psi
t/d=
0.0075
Req'd:
t=
?
5. A steel cylindrical air receiver with 5 feet diameter and pressure load of 180 psi, design stress of 9,500 psi
maximum. The pressure vessel is to be provided 1-1/2” diameter drain valve installed at the bottom of the vessel
and safety pressure relief valve installed either at the top most or the side with pop out rating of 200 psi. Assume
of 100% weld joint efficiency. Determine the bursting pressure of this air receiver if the lap welding tensile strength
is 65,000 psi.
5 Given:
Sol'n:
D=
Pw=
Sd=
Dv=
Pr=
E=
St=
Req'd:
Pb=
5f
180 psi
9500 psi
1.5 in
200 psi
100%
65000 psi
t=
t=
pD/(2St*E)
0.568421 0.009474
t=
t=
pD/(2St*E)
0.631579
t=
Pb=
pD/(2St*E)
1368.421 psi
?
6. The work cylinder of a hydraulic system is acted by a hydraulic pressure of 370 psi while the maximum load
of the piston is 5500 lbs. If the allowable tensile stress is 2000 psi, what is the required wall thickness of the
cylinder.
6 Given:
Sol'n:
p=
730 psi
t=
pD/(2St*E)
F=
5500 lb
p=
F/A
St=
2000 psi
A=
7.534247 in2
Req'd:
D=
3.097237 in
t=
?
t=
0.565246 in
checking of pressure vessel type: t/D=
0.1825 not thin-walled
Use thick walled-cylinder formula: t=
D/2*{ [ (St+P) / (St - O) ] ^ 0.5 -1 }
t=
0.721894 in
checking of pressure vessel type: t/D=
0.233077 thick walled
7. A cylindrical pressure vessel is fabricated from steel plating that has a thickness of 15 mm. The diameter
of the pressure vessel is 400 mm and its length is 2.0 m. Determine the maximum internal pressure that can
be applied if the longitudinal stress is limited to 140 MPa, and the circumferential (tangential) stress is limited to 60 MPa.
7 Given:
Sol'n:
t=
15 mm
St=
pD/(2t*E)
D=
400 mm
p=
4.5 Mpa
safe pressure
Sl=
140 Mpa
Sl=
pD/(4t*E)
St=
60 MPa
p=
21 MPa
Pmax=
?
8. A cylindrical steel pressure vessel 16 inches in diameter with a wall thickness of 1 inch, is subjected to an
internal pressure of 650 psi.
(a) Calculate the tangential and longitudinal stresses in the steel in psi.
(b) To what value may the internal pressure be increased if the stress in the steel is limited to 18,000 psi?
8 Given:
Sol'n:
D=
16 in
a) St=
pD/(2t*E)
t=
1 in
St=
5200 psi
p=
650 psi
b) Sl=
pD/(4t*E)
a) St=
?
Sl=
2600 psi
b) Sl=
?
c) p=
(2t*St)/D
c) p=
?
p=
2250 psi
p @ St=
18000 psi
thick steel plate.
d since t/d <0.07)
Pa. The yield stress
d since t/d <0.07)
the required wall
s of 9,500 psi
ottom of the vessel
of 200 psi. Assume
ding tensile strength
maximum load
ckness of the
he diameter
ssure that can
stress is limited to 60 MPa.
bjected to an
1.A cast iron flywheel is rotated at a speed of 1600 rpm and having a rim diameter of 2 feet. If the weight
of the rim is 50 lbs, What is the centrifugal force.
1 N=
D=
W=
Fc=
1600 rpm
Fc=
mV2/R
2f
50 lbs
Fc=
V=
V=
Fc=
WV2/(g*R) (R= rim radius)
πDN
167.552 f/sec
43,593 lbf
?
(Fc= centrifugal force)
2.What pressure is required to punch a hole 1.5 inches diameter through a 3/8” inch steel plate?
2 D=
1.5 in
P=
D*t*80
tons
t=
0.375 in
P=
45 tons
P=
?
3.A flywheel for a punching press must be capable of furnishing 3800 Nm of energy during the 25% revolution
while the hole is being punched. The flywheel maximum speed is 250 rpm and the speed decreases by 8.7%
during the load stroke. The mean radius of the rim is 1025 mm and contributes 90% of the energy requirements.
Approximate total weight of the flywheel to be 1.2 that of the rim. Compute the total weight of the flywheel.
3 KEf=
3800 Nm
KEr=
0.9 *KEf
%Rev=
0.25
KEr=
3420 Nm
N1=
N2=
250 rpm
0.087 *N1
V1=
V1=
πDN1
Rm=
V2=
πDN2
KEr=
1025 mm
0.9 *KEf
Wf=
1.2 Wr
KEr=
[Wr*(V1 -V22)]/2g
Wr=
Wr=
Wf=
Wf=
[KEr*2g] / (V12-V22)
559.9 N
1.2 Wr
671.9 N
Wf=
?
26.8 m/sec
V2=
24.5 m/sec
2
4. A shearing machine requires 150 kg-m of energy to shear steel sheet, and has a normal speed of 3 rev/sec,
slowing down to 2.8 rev/sec during the shearing process. The flywheel has a mean diameter of 80cm and
weighs 0.018 kg/cm3. The width of the rim is 20cm. If the hub and the arms of the flywheel account for 15%
of its total weight, find the thickness of the rim.
4 KEf=
150 kg-m
V1=
πDN1
N1=
N2=
Dm=
ρ=
0.018 kg/cm3
V1=
V2=
V2=
KEf=
b=
20 cm
Wf =
[KEf*2g] / (V12-V22)
0.15 Wf
Wf =
401.65 kg
Wr+Wah
Wha=
t=
3 rev/s
2.8 rev/s
80 cm
?
Wf=
7.54 m/sec
πDN2
7.04 m/sec
[Wf*(V1 -V22)]/2g
2
Wr=
Wr=
Wr=
Wr=
t=
Wf-Wah
Wf - 0.15*Wf
341.40 kg
πD*b*t*ρ
3.77 cm
5. A flywheel weighing 1100 kg has a radius of gyration of 1524mm. The shaf journals are 304.8mm in diameter
and have a coefficient of friction of 0.06. Afer the wheel reaches 100 rpm, the driving force is withdrawn and
the wheel slows to 50 rpm. How much energy does the wheel loss?
5 m=
1100 kg
w1=
2πN1
(w is angular velocity)
k=
1524 mm
w1=
10.47 rad/sec
D=
304.8 mm
w2=
2πN2
f=
0.06
w2=
5.24 rad/sec
N1=
N2=
KE=
100 rpm
50 rpm
?
KE=
KE=
KE=
mk2*(w1-w2)
105,064 Nm or J
105.1 kJ
6. A punch punches a 1-in diameter hole in a steel plate ¾” inch thick every 10 sec. The actual punching
takes 1.5 sec. The ultimate shear strength of the plate is 60ksi. The flywheel of the punch press has a mass
moment of inertia of 500 in-lb-sec2 and rotates at mean speed of 180 rpm. (a) What is the horsepower
required for the punch operation.
b)What is the total speed fluctuation of the flywheel in rpm?
Given:
Sol'n:
D=
1 in
power= E/Ta
E=
1/2*I*(w1^2-w2^2)
t=
0.75 in
E=
F*t/2
E=
1/2*I*(w1-w2)(w1+w2)
T=
Ta=
Ssu=
Im=
Im=
Nm=
Req'd:
P=
(N1-N2)
Cf
10 sec
1.5 sec
60 ksi
500 in-lbm-s2
41.666667 f-lb-s2
180 rpm
?
?
hp
RPM
Ssu=
F/(πDt)
wm=
F=
F=
E=
P=
P=
Ssu*(πDt)
141,372.00 Lbs
4417.875 lb-f
2945.25 lb-f/sec
5.355 hp
(w1+w2)=
37.6992 rad/sec
(W1-w2)=
5.625 rad/sec
(N1-N2)= 53.714668 rpm
Cf=
(w1-w2)/wm
Cf=
0.2984
Note:
lbm=
lbf=
(w1+w2)/2
32.2 slugs
1 slug-ft/s2
7. A flywheel rotating at 120 rpm is slowed down to 102 rpm during a punching operation that requires
¾ second for the punching portion of the cycle. What angular deceleration does the flywheel experience in radians per sec per
Given:
Sol'n:
N1=
120 rpm
α=
(W2-w1)/t
N2=
102 rpm
w1=
2πN1=
12.57 rad/sec
t=
0.75 sec
w2=
2πN2=
10.68 rad/sec
Req'd:
α=
α=
-2.51 rad/s2
(negative value denotes decceleration)
?
8. A flywheel has diameter of 1.5m and a mass of 1000 kg. What torque is needed to produce an angular acceleration of 120 re
Given:
Sol'n:
D=
1.5 m
T=
1/2mk^2*α
m=
1000 kg
α=
120 rev/min/sec
α=
120 rev/min/sec α=
12.57 rad/s2
(α=angular acceleration)
Req'd:
T=
3534.3 J
T=
9. A flywheel rotates at 220 rpm slowed down to 65% of its revolution during the three-fourth second punching portion of the
Compute the angular acceleration of the flywheel in rad per sec2.
Given:
Sol'n:
N1=
220 rpm
α=
(w2-w1)/t
N2=
0.65 *N1
w1=
2πN1
N2=
143 rpm
w1=
23.04 red/sec
t=
0.75 or 3/4 sec
w2=
2πN2
Req'd:
w2=
14.97 red/sec
α=
?
α=
-10.75 rad/s2
(α=angular acceleration)
10) A punching machine flywheel accelerates from rest with 7 radian per sec2. In 10 sec, how many radians are achieved?
Given:
Sol'n:
V1*t + 1/2at2 (distance formula)
w1=
0
S=
α=
t=
Req'd:
ϴ=
7 rad/s2
10 sec
ϴ=
ϴ=
w1*t+1/2αt2
(Angular formula; similar to distance formula)
350 radians
11) A sheet metal working company purchased a shearing machine from a surplus dealer without a flywheel. It is calculated th
will use 2380 Joules of energy to shear a 1.2mm thick sheet metal. The flywheel to be used will have a mean diameter of 91.44
The normal operating speed is 180 rpm and slows down to 160 rpm during the shearing process. Assuming that the arms and
of the rim weight concentrated at the mean diameter and the material density is 0.26 lb per cubic inches.
a)Determine the weight of the flywheel.
b)Determine the thickness of the rim.
Given:
Sol'n:
E=
2380 J
KE=
Wf/2g*(V1^2-V2^2)
tm=
1.2 mm
V1=
πDN1=
8.62 m/s
Dm=
91.44 cm
V2=
πDN2=
7.66 m/s
b=
25.4 cm
Wf=
2995.6794268 N
N1=
180 rpm
Wr=
ρVr
N2=
160 rpm
Wr=
ρ*(πDm*br*tr)
Wah=
0.12 Wr
Wf=
Wr+Wah
d=
0.26 lb/in3
Wf=
Wr+.12Wr
d=
0.0071955 kg/cm3
Wr=
2674.713774 N
Req'd:
Wf=
Tr=
?
?
Wr=
Tr=
272.65176085 kg
5.19 cm
12) A punching machine requires 1800 f-lb of energy to punch a certain hole. The speed of the machine is reduced from 200 r
The weight of the arms and hub are considered to be 12% of the rim weight, and the mean diameter is 28in. What is the coeffi
Given:
Sol'n:
KE=
1800 f-lb
KE=
Wf/2g*(V1^2-V2^2)
N1=
200 rpm
V1=
πDN1=
24.43 fps
N2=
180 rpm
V2=
πDN2=
21.99 fps
Wah=
0.12 Wr
v=
23.21 fps
Dm=
28 in
Cf=
(V1-V2)/V
Req'd:
Cf=
0.1053
Cf=
?
Wf=
1021.86 lbs
Wf=
Wr+Wah
Wf=
Wr+.12Wr
Wr=
912.38 lbs
revolution
equirements.
m in diameter
-w2)(w1+w2)
ience in radians per sec per sec?
6
es decceleration)
ngular acceleration of 120 revolution per minute per second?
nd punching portion of the cycle.
y radians are achieved?
flywheel. It is calculated that the machine
e a mean diameter of 91.44cm with a width of 25.4 cm.
ssuming that the arms and the hub will account for 12%
chine is reduced from 200 rpm to 180 rpm during the punching operation.
er is 28in. What is the coefficient of fluctuation and the weight of the rim?
1 A three extension coil springs are hooked in series that support a single weight of 100 kgs.
The first spring is rated at 0.4 kg/mm and the other 2 lower springs are rated at 0.64 kg/mm.
Compute the total deflection.
W=
100 kg
y1=
W/k1
k1=
0.4 kg/mm
y1=
250 mm
k2=k3
0.64 kg/mm
y2=y3= W/k2
Yt=
?
y2=y3=
156.25 mm
Yt=
y1+y2+y3
Yt=
562.5 mm
2 Compute the maximum deflection of a 20 coil helical spring having a load of 75 kgs.
The spring is squared and ground ends with modulus of elasticity in shear of 79.84 Gpa,
outside diameter of 101.6mm, wire diameter of 9.525mm.
Nt=
20 total coils Nt=
n+2
for square & ground ends
W=
75 kg
n=
Nt-2
G=
79.84 Gpa
n=
18 active coils
D
=
Do=
101.6 mm
Do-d
m
dw=
y=
9.525 mm
?
Dm=
C=
C=
y=
y=
92.075 mm
Dm/dw
9.6667
(8Fc n) / (G*dw)
3
125.85 mm
3 Compute the deflection of a 18 coils helical spring having a load of 100 kg. The modulus of elasticity in
shear of spring is 96.92 Gpa, inside diameter of 9.525cm and wire diameter of 9.525mm. The spring is square and ground
Nt=
18 total coils Nt=
n+2
for square & ground ends
W=
100 kg
n=
Nt-2
G=
96.92 Gpa
n=
16 active coils
Dm/dw
Di=
9.525 cm
c=
dw=
y=
9.525 mm
?
c=
y=
y=
11
(8Fc n) / (G*dw)
3
181.04 mm
4 A high alloy spring having squared and ground ends has a total of 16 coils and modulus of elasticity in shear of 85 Gpa.
Compute the Wahl Factor. The spring outside diameter is 9.66 cm and wire diameter is 0.65cm.
Nt=
16 total coils K=
(4C-1)/(4C-4)+0.615/C
G=
85 Gpa
C=
Dm/d
Do=
9.66 cm
C=
13.8615
d=
0.65 cm
K=
1.1027
K=
?
5 A helical spring having a squared and ground ends has a total of 18 coils and its material has modulus of elasticity in shear
If the spring has an outside diameter of 10.42cm and a wire diameter of 0.625cm, compute the maximum deflection that
Nt=
18 total coils Nt=
n+2
for square & ground ends
G=
78.91 Gpa
n=
Nt-2
Do=
12 cm
n=
16
active coils
Dm/dw
d=
0.65 cm
c=
W=
Y=
40 kg
?
c=
y=
y=
17.46
(8Fc3n) / (G*dw)
521.36 mm
6 The load on a helical spring is 1600-lb and the corresponding deflection is to be 4-in. Rigidity modulus is 11 million psi and
maximum intensity of safe torsional stress is 60,000 psi. Design the spring for the total number of turns if the wire is circul
cross section with a diameter of 5/8-in and a center line radius of 1 1/2 in. The spring is squared and ground ends.
Dm/dw
W=
1600 lbs
C=
y=
G=
Ss=
d=
Rm=
sq&gr
Nt=
3.5 in
11000000 psi
60000 psi
0.625 in
1.5 in
?
C=
K=
K=
y=
n=
n=
Nt=
Nt=
4.8
(4C-1)/(4C-4)+0.615/C
1.3255
3
(8Fc n) / (G*dw)
n+2
16.998 coils
17 active coils
for square & ground ends
19 total coils
7 All four compression coil spring support one load of 700 kg. All four springs are arranged in parallel and rated same at 0.60
Kt=
F/Yt
n=
4 springs
W=
k1=k2=k3=k4=
Yt=
?
900 kg
0.609 kg/mm
Kt=
k1+k2+k3+k4
Kt=
Yt=
Yt=
2.436 kg/mm
F/Kt
369.5 mm
8 A safety valve spring having 9.5 active coils has the ends squared and ground. The outside diameter
of the coil is 115mm and the wire diameter is 13mm. It has a free length of 203mm. The length of which this
spring must be initially compressed to hold a boiler pressure of 1.38MPa on the valve seat of 32mm diameter.
Modulus of rigidity is taken as G=80GPa. Find the ff:
a.Force acting on the valve seat, N
P=
F/A
b.Mean diameter, mm
F=
1109.8644 N
c.Spring Index
Dm=
Do-d
d.Spring deflection, mm
Dm=
102 mm
e.Operating length, mm
C=
Dm/d
f.Solid Length, mm
C=
7.8461538
g.Deflection of solid length, mm
Y=
8Fc^3n/Gd
h.Solid force
Y=
39.176051 mm
i.Wahl factor
OL=
FL-Y
n
n=
Do=
d=
FL=
P=
Dv=
G=
9.5 coils
115 mm
13 mm
203 mm
1.38 Mpa
32 mm
80 Gpa
OL+
SL=
SL=
Ysolid=
Ysolid=
k=
Fs=
Fs=
K=
K=
163.82395 mm
(n+2)*d
149.5 mm
FL-SL
53.5 mm
F/Y=Fs/Ys
F*Ys/Y
1515.6645 N
(4C-1)/(4C-4)+0.615/C
1.1879329
9 A spring is designed to fire 2 kg projectile. The outside diameter of the coil is 160mm with an 18mm wire and a total of 22
The spring has a free length of 650mm. When set or loaded, the spring is compressed to a length of 450mm. Determine th
a)Spring index
C
C=
Dm/d
b)Active number of coils n
C=
6.5
c)Spring Deflection
Y
n=
Nt-2
d)Spring rate
k
n=
20
d)Wahl Factor
K
Y=
FL-CL
W=
2 kg
Y=
110
Do=
150 mm
k=
F/y
d=
20 mm
k=
0.0181818182 kg/mm
Nt=
22 sq&gr
K=
(4C-1)/(4C-4)+0.615/C
FL=
560 mm
K=
1.230979021
CL=
450 mm
10 A crate of mass 1800 kg is moving at a speed of 1.2 m/s. It is brought to rest by 2 helical compression springs. In stopping t
If the spring has a spring index of 6 and allowable stress of 360MPa, determine the ff:
a.Wahl Factor
K
K=
(4C-1)/(4C-4)+0.615/C
b.Wire Diameter
d
K=
1.2525
F=
ma
a=
(Vf2-Vo2)/2S
W=
1800 kg
S=y=
200
V=
1.2 m/s
a=
3.6 m/s2
N=
2
F=
6480 N
Y=
200 mm
Ss=
8FKDm/Pid3= 8FKC/Pid2
C=
6
K=
(4C-1)/(4C-4)+0.615/C
Ss=
360 Mpa
d2=
344.46
d=
18.559671897 mm
11. Find the number of active coils of a No. 8 wire Helical Spring with index of 6, steady load with spring rate of 42.5 lb/in . Ma
Dm/dw
Mean Radius is 0.486 in
C=
C=
k=
6
42.5 lb/in
Dm=
Dm=
C*dw
0.972 in
Ss=
n=
dw=
60 ksi
?
G=
12000000 psi
0.162 in
k=
K=
K=
Ss=
F=
y=
Y=
n=
n=
F/y
(4C-1)/(4C-4)+0.615/C
1.2525
8FKDm/Pidw^3
82.283583234 lb
1.9360843114 in
8Fc3n/Gdw
26.47 coils
27 active coils (rounded up value)
ring is square and ground ends.
ty in shear of 85 Gpa.
ulus of elasticity in shear of 78.910 Gpa.
maximum deflection that can be produced in the spring due to a load of 50 kgs.
dulus is 11 million psi and the
f turns if the wire is circular in
and ground ends.
el and rated same at 0.609 kg/mm. Compute the deflection in mm.
f which this
mm diameter.
mm wire and a total of 22 coils squared and ground.
of 450mm. Determine the ff:
sion springs. In stopping the crate, the springs are compressed 200mm.
ng rate of 42.5 lb/in . Maximum allowable stress is 60 ksi.