A tall vessel is constructed with a cylindrical shell and two hemispherical heads. The vessel is designed for full vacuum (15 psi external pressure) at a design temperature of 550° F. The material is SA-516 Grade 60 . There is no corrosion allowance required. The vessel is 10 ft 0 in. inside

Question

A tall vessel is constructed with a cylindrical shell and two hemispherical heads. The vessel is designed for full vacuum (15 psi external pressure) at a design temperature of 550° F. The material is SA-516 Grade 60 . There is no corrosion allowance required. The vessel is 10 ft 0 in. inside diameter and 116 ft 8 in. from tangent to tangent. The vessel is to be installed in the vertical position and supported at the bottom tangent line. It contains a fluid weighing 50 lb / ft³. Three stiffening rings are evenly spaced at 30 ft 0 in. with 28 ft 4 in. from each tangent line. What are the longitudinal stresses above the support line?

Accepted Answer

1 . Determine a preliminary thickness based on the external pressure alone using the procedure of the ASME Code, VIII-1, UG-28(c)(1). Assume a thickness of t = 0.75 in. [latex]D_{o}[/latex] = 120 + 2(0.75) = 121.5 in. [latex]D_{o}/t[/latex] = 121.5 / 0.75 = 162, and [latex]L/D_{o}[/latex] = 360/121.5 = 2.9630. From Fig. 5-UGO-28.0, A = 0.00022.

[latex] P_a = \frac{2(0.00022)(26.38 imes 10^6)}{3(162)} = 23.9 \ psi ; \ \ t=3/4 ext{ in. OK} [/latex]

2 . Determine the weight of the vessel and contents when

Shell-weight [latex]=\pi\left(60.75^{2}-60^{2} ight)(1400)(490 / 1728)=112,950[/latex]

Heads-weight [latex]=(4 / 3) \pi\left(60.75^{3}-60^{3} ight)(490 / 1728)=9,740[/latex]

Contents-shell [latex]=\pi(60)^{2}(1400)(50 / 1728) =458,150[/latex]

Contents-head [latex]=(4 / 3) \pi(60)^{3}(50 / 1728)=26,180[/latex]

3 . Determine the maximum compressive load, (lb/in.) using Eq. (16.23) as follows:

[latex]P_{v}=\frac{W}{\pi D_{m}}[/latex] (16.23)

[latex]P_{v}=\frac{W}{\pi D_{m}}=\frac{(112,950)+0.5(9740)}{\pi(120.75)}=310.6 ext{ lb / in.}[/latex]

[latex]P_{o}=15.0 \ psi[/latex]

[latex]\alpha=\frac{P_{v}}{P_{o} D_{o}}=\frac{(310.6)}{(15)(121.5)}=0.1704[/latex]

[latex]m=\frac{(1.23)}{\left(L / D_{o} ight)^{2}}=\frac{(1.23)}{(2.963)^{2}}=0.1401[/latex]

n = 3.0 from Fig. 16.6 for [latex]L / D_{o}[/latex] = 2.963 and [latex]D_{o} / t[/latex] = 162

[latex]P_{o}^{\prime}=\frac{9-1+0.14+0.14(0.1704)}{9-1+0.14}(15)=15.04 \ psi[/latex]

As determined in step 1 , the maximum allowable external pressure based on t = 0.75 in. is [latex]P_{o}[/latex] = 23.9 psi at 550° F. Because the required pressure of 15.04 psi is less than the permissible pressure of 23.9 psi , t = 0.75 in. is satisfactory.

Because [latex]\alpha[/latex] is less than 1.0, no cantilever beam check is needed.

Question 16.10: A tall vessel is constructed with a cylindrical shell and tw...

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