TÀI LIỆU CUỐI KỲ MÔN THIẾT KẾ MÁY VÀ THIẾT BỊ HÓA CHẤT

I. Design for Internal Pressure
1.1 Formulas
Inside radius/diameter

Circumferential stress (Longitudinal
Cylindrical shell joints): t ≤ 0.5R; P ≤ 0.385SE
Longitudinal stress (Circumferential
joints): t ≤ 0.5R; P ≤ 1.25SE

Sphere &
Hemispherical
head

2:1 Ellipsoidal
head

Cone & Conical
section

Flanged &
Dished head

Outside radius/diameter


L/
r
M

6.5

7.0

7.5

8.0

8.5

9.0

9.5

1.3
9

1.4
1

1.4
4

1.4
6

1.4
8

1.5
0

1.5
2

10.
0
1.5
4

10.
5
1.5
6

11.
0
1.5
8

11.
5
1.6
0

12.
0
1.6
2

13.
0

1.6
5

14.
0
1.6
9

15.
0
1.7
2

16.
0
1.7
5

16.6
7
1.77

Notation
a = Half apex angle of cone, degree
D = Inside diameter, inches
Do = Outside diameter, inches
E = Efficiency of welded joints
L = Inside crown radius, inches
Lo = Outside crown radius, inches
M = Factor, see table above

P = Design pressure or maximum allowable pressure, psig
R = Inside radius, inches
Ro = Outside radius, inches
S = Stress value of material, psi
t = Thickness, inches
For Flanged & Dished head: L = Inside dish radius, inches
r = Inside corner radius, inches

Lo = Outside dish radius, inches


1.2 Example:
Cylindrical shell
Consider a vessel with an inside diameter D = 96 inches.
The design pressure P = 100 psig.
Material: SA 515-70 plate
Design Temperature: 100°F
Corrosion Allowance (c.a.): 0.125 inch (corroded I.D. = 96.250 inches.)
All circumferential and longitudinal seams are double-welded butt joints and are spot
radiographed. The vessel is to be built per ASME Code, Section VIII, Division 1.
From Table UCS-23 of the Code, for SA 515-70 at temperatures up to 650°F, S = 17,500 psi. For
spot radiographed joints, from Table UW-12, E = 0.85.
Use 0.5 inch plate.
Sphere & Hemispherical head
The calculated minimum thickness of formed heads is not rounded up to standard plate because
of the thinning that occurs in portions of the head during forming. The calculated value should be
the minimum thickness at any point on the head. Using the same vessel example as above:
The calculated thickness should be increased by corrosion allowance:
Semiellipsoidal heads
Using the same vessel as in previous examples: E = 1.0 for a seamless head.

Torispherical head
For torispherical heads where the knuckle radius is 6% of the inside crown radius, the ASME
Code equation is:
Using the same example:
L = 96 inches = D
E = 1.0 (seamless head)


II. Design for External Pressure


2.1 Cylindrical shell
2.1.1 Procedure


2.1.2 Example
Using the same vessel as in the internal pressure calculation:
• Tangent to tangent length: 36 ft 0 in = 432 in
• Two 2:1 semiellipsoidal heads
• External design pressure: 15 psig at 500°F
L = 448 inches (length of shell plus one-third of the depth of each head, 16 in.)
L/Do = 448/97 = 4.62
Do/t = 97/0.375 (corroded) = 258.67


From geometrical chart Figure 5, UGO-28.0, A = 0.00007. From Figure 5, UCS-28.2 for SA
515-70 at 500°F, the modulus of elasticity of material, E = 27,000,000 psi; and for A = 0.00007, the
value falls to the left of the applicable temperature line.
Then Pa = 2AE/3 (Do/t) = 2 x 0.00007 x 27 x 106/3 x 258.67 = 4.87 psi.
The vessel is good for only 4.87 psi external pressure, so stiffening rings are required.

Try two stiffening rings equally spaced between tangent lines.
L = 144 (length of shell between rings) + 8 (1/3 depth of head) = 152 in
L/Do = 152/97 = 1.56
A = 0.00022 from chart.
B = 2800 from Figure 5, UCS-28.2
Pa = 4B/(3(Do/t)) = (4 x 2800)/(3 x 258.67) = 14.4 psi
The vessel is still not good for 15 psi, so another ring should be added. Try three stiffening rings
equally spaced between tangent lines.
L = 108 + 8 = 116 in
L/Do = 116/97 = 1.19
A = 0.00027 from chart
B = 3700 from chart Figure 5, UCS-28.2
Pa = (4 x 3700)/(3 x 258.67) = 19.07 psi
Since Pa is greater than the design pressure, the vessel with three stiffening rings is good for full
vacuum (15 psi).

2.2 Spherical shell and Hemispherical head
2.2.1 Procedure
1. Calculate the value of A using the equation:
Where: Ro is the outside radius of the sphere.
2. Find the value of B from the Code material/temperature chart in Appendix 5.
3. Calculate the maximum allowable external pressure:
or for values of A falling to the left of the applicable temperature line:
2.2.2 Example
Using the same example:
Ro = 48.5 in
t = 0.125 in
From Figure UCS-28.2 B = 10,500

The hemispherical head is good for the external design pressure of 54 psi.

2.3 Semiellipsoidal head
According to the ASME Code, the required thickness for a semiellipsoidal head under external
pressure should be the greater of the following:
1. The thickness as calculated by the equation given for internal pressure using a design
pressure 1.67 times the external pressure and joint efficiency, E = 1.00.
2. The thickness by the equation Pa = B/(R/t) where R = 0.9Do and B to be determined for a
sphere.

2.4 Torispherical head
The required thickness is computed by the procedures given for ellipsoidal heads using a value
for R = Do.

III. Design for jacketed vessel
Case
P1 > P2 > Pa
P2 > P1 > Pa
P1 > Pa > P2
P2 > Pa > P1
Pa > P1 > P2
Pa > P2 > P1

Inner shell
Internal pressure P1, external pressure P2
Internal pressure P1, external pressure P2
Internal pressure (P1 + Pa) P1
External pressure (P2 + Pa)
Internal pressure Pa, external pressure Pa
Pa

Internal pressure
Pa, external pressure Pa
P2

Outer shell
Internal pressure P2
Internal pressure P2
External pressure Pa
Internal pressure P2
External pressure Pa
External pressure Pa


IV. Example of Internal/External Pressure Design
Determine the minimum required thickness of a cylindrical shell and hemispherical heads of a
welded pressure vessel designed for an internal pressure of 100 psi at a design temperature of
250°F. There is no corrosion. The shell, which contains a longitudinal butt weld, is also butt welded
to seamless heads. All Category A butt joints are Type (1) with full radiography (RT). E = 1.00 for
all calculations. The shell has a 5 foot 0 inch inside radius and is 30 foot 0 inch long from tangent to
tangent.
Also determine the minimum required thicknesses of the same vessel designed for an external
pressure of 15 psi at 100°F without stiffening rings. What is the stiffening ring spacing if the
required thickness of internal pressure is used?
Solution
1. For SA 515 Gr. 60, the allowable tensile stress from Table UCS-23 at 100°F is 15.0 psi, and
the external pressure chart is Figure 5-UCS-28.2.
2. As is generally the case for internal pressure on a cylinder, when E = 1.00 for all butt joints,
UG-27(c)(1) for circumferential stress (hoop stress) controls over UG-27(c)(2) longitudinal stress
by:
Check for applicability of using UG-27(c)(1):

Is t < R/2? 0.400 in. < 30

o.k.

Is P < 0.385 SE? 100v< 0.385 (15,000)(1) = 5,775

o.k.

3. For internal pressure on hemispherical heads, use UG-32(f):
Is t < 0.365 L? 0.200 < 0.365 (60) = 21.9

o.k.

Is P < 0.665 SE? 100 < 0.665 (15,000)(1) = 9,975

o.k.

4. For external pressure on cylinder, use UG-28 and Appendix 5:
For cylindrical shells with formed heads on the end the length of the shell plus 1/3 of the depth
of each head is used to determine the effective lengths (L) (see UG-28).
Determine the effective length without stiffening rings = 1/3 of each head depth plus straight
length = (1/3)(2)(60) + 360 = 400 in.
Assume tmin for internal pressure of 0.400 in and Do = 120 + 2(0.4)
L/Do = 400/120.8 = 3.31
Do/t = 120.4/0.4 = 301
a. Enter Figure 5-UGO-28.0 with L/Do = 3.31 and read across to sloping line of D o/t= 301. Read
A = 0.000075
b. Enter Figure 5-UCS-28.2 with A = 0.000075 and the modulus of elasticity E = 29.0 x 10 6
which is off the left side and cannot be read.
Following Step (7) of UG-28(c):

Pa < 15.0 psi MAWP. Increase thickness.


Assume t = 5/8 in = 0.625 in and Do = 120 + 2(0.625) = 121.25 in
L/Do = 400/121.25 = 3.30; Do/t = 121.25/0.625 = 194
a. From Figure 5-UGO-28.0, A = 0.00014
b. Recalculate Pa:
Pa < 15.0 psi MAWP. Increase thickness.
Assume t = 11/16 in = 0.6875 in and Do = 120 + 2(0.6875) = 121.375 in
L/Do = 400/121.375 = 3.30
Do/t = 121.375/0.6875 = 177
A = 0.00017
Pa > 15.0 psi MAWP:

ok.

Further calculations show that tmin = 0.64 in for 15.0 psi external pressure.
5. For external pressure on hemispherical head, use UG-33(c), UG-28(d), and Appendix 5.
First assumption, use tmin for internal pressure of t = 0.200
Assume t = 0.200 in and Ro = 0.5(120 + 2 x 0.2) = 60.2 in
a. Calculate A:
b. Enter Figure 5-UCS-28.2 with A = 0.0004 and read B = 5,800
c. Determine Pa:
Pa > 15 psi MAWP:

o.k.

Of interest is the fact that for 100 psi internal pressure the minimum required thickness of the
cylinder is 0.401, while for 15.0 psi external pressure the minimum required thickness is 0.636 in.
For the head, the minimum required thickness is only 0.200 in for internal pressure, while for

external pressure the minimum required thickness is less than 0.200 inches.
If a thickness between 0.400 in and 0.636 in is desired for the cylinder, stiffening rings are
required on the cylinder to obtain a smaller value of L to use in the calculation of P a. By “trial-anderror,” the approximate maximum stiffening ring spacing with the minimum thickness required for
internal pressure of 0.400 is 120 in as follows:
Assume t = 0.400 in and L = 120 in
L/Do = 120/120.8 = 0.993
Do/t = 302
a. Enter Figure 5-UCS – 28.0 and A = 0.00025
b. Enter Figure 5-UCS – 28.2 and B = 3500
c. Determine Pa


Pa > 15.0 psi MAWP:

o.k.

This indicates that the optimum design would be one where the shell was thickened above 0.400
in with stiffening rings being placed at a spacing larger than 120 in center-to-center. The optimum
design would be obtained by “trial-and-error.” After the “best” thickness and stiffening ring spacing
is determined, the design of the stiffening ring is developed according to UG-29.

V. Units Conversion
5.1 Length
1 in = 2.54 cm = 25.4 mm = 0.0254 m
1 ft = 12 in

5.2 Pressure
1 psi = 6.8948×103 Pa = 6.8948×10−2 bar = 7.03069×10−2 at = 6.8046×10−2 atm = 51.71493 Torr
1 psia = 1 psi
x psig = x + 14.7 psia

5.3 Temperature
x°C = (1.8x + 32)°F
x°F = ((x – 32)/1.8)°C