For the vessel described in Example 16.1, determine the total longitudinal stress in the cylindrical shell above and below the support line that is at the lower shell-to-head junction. The value of SE = 15,000 psi.

Question

Accepted Answer

Assume the internal pressure is set by Eq. 16.17 for a value of SE = 15,000 psi. Rearranging the terms gives

[latex]S E=P\left(\frac{R}{t}+0.6 ight)[/latex] (16.17)

where S = allowable tensile stress (psi)

E = weld joint efficiency (E = 1.0 for seamless )

P = internal design pressure (psi)

R = inside radius (in.)

t = minimum required thickness (in.)

[latex]P=\frac{S E t}{R+0.6 t}=\frac{(15,000)(0.5)}{(30)+0.6(0.5)}=245 \ psi[/latex]

The dead load of vessel above the support line is

Shell 9,700

Upper head [latex] \underline{815} [/latex]

10,515 lb

The longitudinal stress using Eq. 16.19 is

[latex]\sigma_{L}=+P\left(\frac{R}{2 t}-0.2 ight) \pm \frac{W}{\pi D_{m} t}[/latex] (16.19)

[latex]\sigma_{L}=+(245)\left(\frac{30}{2 imes 0.5}-0.2 ight)-\frac{(10,515)}{\pi(60.5)(0.5)}[/latex]

[latex]\sigma_{L}[/latex] = +7300 - 110 psi

[latex]\sigma_{L}[/latex] = 7190 psi tension with internal pressure

[latex]\sigma_{L}[/latex] = 110 psi compression without internal pressure

The dead load of vessel and contents below the support line is

Lower head 815

Shell fluid 20,620

Head fluid [latex] \underline{2,290} [/latex]

23,725 lb

The longitudinal stress using Eq. 16.19 is

[latex]\sigma_{L}=+(245)\left(\frac{30}{2 imes 0.5}-0.2 ight)+\frac{(23,725)}{\pi(60.5)(0.5)}[/latex]

[latex]\sigma_{L}[/latex] = +7300 + 250 = 7550 psi tension with internal pressure

Question 16.6: For the vessel described in Example 16.1, determine the tota...

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