A cylindrical pressure vessel with 1 m inner diameter is subjected to internal steam pressure of 1.5 MPa. The permissible stresses for the cylinder plate and the rivets in tension, shear and compression are 80, 60 and 120 N/mm^2 respectively. The efficiency of longitudinal joint can be taken

Question

A cylindrical pressure vessel with 1 m inner diameter is subjected to internal steam pressure of 1.5 MPa. The permissible stresses for the cylinder plate and the rivets in tension, shear and compression are 80, 60 and 120 N/mm² respectively. The efficiency of longitudinal joint can be taken as 80% for the purpose of calculating the plate thickness. The efficiency of circumferential lap joint should be at least 62%. Design the circumferential lap joint and calculate:
(i) thickness of the plate;
(ii) diameter of the rivets;
(iii) number of rivets;
(iv) pitch of rivets;
(v) number of rows of rivets; and
(vi) overlap of the plates.

Accepted Answer

[latex] ext { Given For vessel, } D_{i}=1 m \quad P_{i}=1.5 MPa [/latex].

[latex] \sigma_{t}=80 N / mm ^{2} \quad au=60 N / mm ^{2} \quad \sigma_{c}=120 N / mm ^{2} [/latex].

[latex] ext { For longitudinal joint, } \eta=80 \% [/latex].

[latex] ext { For circumferential joint, } \eta_{1}=62 \% [/latex].

Step I Thickness of plate
From Eq. (8.43),

[latex] t=\frac{P_{i} D_{i}}{2 \sigma_{t} \eta}+ CA [/latex] (8.43).

[latex] t=\frac{P_{i} D_{i}}{2 \sigma_{t} \eta}+ CA =\frac{1.5(1000)}{2(80)(0.8)}+2 [/latex].

= 13.72 or 14 mm (i)
Step II Diameter of rivets
t > 8 mm
From Eq. (8.45),

[latex] d=6 \sqrt{t} [/latex] (8.45).

[latex] d=6 \sqrt{t}=6 \sqrt{14}=22.45 \quad ext { or } \quad 23 mm[/latex].

Step III Number of rivets
From Eq. (8.58),

[latex] n=\left(\frac{D_{i}}{d} ight)^{2} \frac{P_{i}}{ au} [/latex] (8.58).

[latex] n=\left(\frac{D_{i}}{d} ight)^{2} \frac{P_{i}}{ au}=\left(\frac{1000}{23} ight)^{2} \frac{(1.5)}{(60)}=47.26 ext { or } 48 [/latex].

Step IV Pitch of rivets
From Eq. (8.59),

[latex] \eta_{1}=\frac{p_{1}-d}{p_{1}} [/latex] (8.59).

[latex] \eta_{1}=\frac{p_{1}-d}{p_{1}} \quad ext { or } \quad 0.62=\frac{p_{1}-23}{p_{1}} [/latex].

[latex] herefore \quad p_{1}=60.53 ext { or } 62 mm [/latex] (a).

From Eqs (8.48) and (8.49),

[latex] p_{\min .}=2 d [/latex] (8.48).

[latex] p_{\max }=C t+41.28 [/latex] (8.49).

[latex] p_{\min .}=2 d=2(23)=46 mm [/latex] (b).

[latex] p_{\max .}=C t+41.28=1.31(14)+41.28 [/latex].

= 59.62 mm (c)
From (a) and (c),

[latex] p_{1}>p_{\max } [/latex].

The pitch of rivets should be from 46 mm to 59.62 mm. We will assume the pitch as 55 mm and recalculate the number of rivets and diameter of rivet.

[latex] p_{1}=55 mm [/latex] (iv)
From Eq. (8.60), the number of rivets in one row is given by,

[latex] n_{1}=\frac{\pi\left(D_{i}+t ight)}{p_{1}} [/latex] (8.60).

[latex] n_{1}=\frac{\pi\left(D_{i}+t ight)}{p_{1}}=\frac{\pi(1000+14)}{55}=57.92 ext { or } 58 [/latex] (iii).

Step V Number of rows of rivets
It is assumed that the type of joint is single-riveted lap joint. The number of rows of rivets is one.
From Eq. (8.58), revised value of rivet diameter is obtained.

[latex] n=\left(\frac{D_{i}}{d} ight)^{2} \frac{P_{i}}{ au} [/latex] (8.58).

[latex] n=\left(\frac{D_{i}}{d} ight)^{2} \frac{P_{i}}{ au} \quad ext { or } \quad 58=\left(\frac{1000}{d} ight)^{2} \frac{(1.5)}{(60)} [/latex].

d = 20.76 or 21 mm (ii)
Step VI Overlap of plates
The margin m is given by,

m = 1.5d = 1.5 (21) = 31.5 or 35 mm
From Eq. (8.62),

[latex] a=p_{t}+2 m [/latex] (8.62).

[latex] a=p_{t}+2 m=0+2(35)=70 mm [/latex] (vi).

Step VII Check for design
From Eq. (8.59),

[latex] \eta_{1}=\frac{p_{1}-d}{p_{1}} [/latex] (8.59).

[latex] \eta_{1}=\frac{p_{1}-d}{p_{1}}=\frac{55-21}{55}=0.6182 ext { or } 61.82 \% [/latex].

The efficiency of the joint is very near to 62% and no changes are required in the calculations.

Question 8.23: A cylindrical pressure vessel with 1 m inner diameter is sub...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions