Determine the reinforcement requirements for a 12 -in. × 16 -in. manway opening. The 12 -in. dimension lies along the longitudinal axis of the vessel. The manway cover seals against the outside surface of the manway. The ID of the shell is 41.875 in. The allowable stress of the shell material is

Question

Determine the reinforcement requirements for a 12 -in. × 16 -in. manway opening. The 12 -in. dimension lies along the longitudinal axis of the vessel. The manway cover seals against the outside surface of the manway. The ID of the shell is 41.875 in. The allowable stress of the shell material is 20.5 ksi and of the manway material 19.4 ksi. The design pressure is 700 psi and the design temperature is 500°F. There is no corrosion. See Fig. 11.19 for details.

Accepted Answer

1 . Strength ratio is [latex]f_{r}=19.4 / 20.5=0.946[/latex].

2 . Using AD-201(a) of the ASME Code, VIII-2, the minimum required thickness of the shell is

[latex]t=\frac{P R}{S-0.5 P}=\frac{700 imes 20.938}{20,500-0.5(700)}=0.728 ext { in. }[/latex]; use 1 in.

3 . Using AD-201(a) of the ASME Code, VIII-2, the minimum required thickness of the nozzle is

[latex]t_{r n}=\frac{P r}{S-0.5 P}=\frac{700 imes 6}{19,400-0.5(700)}=0.220 ext { in. }[/latex]; use [latex]1 \frac{1}{4}[/latex] in.

4 . Examination of the longitudinal plane gives limits parallel to shell surface as:

(a) For 100% of required reinforcement area:

X = d or [latex]\left(T_{s}+T_{n}+r ight)[/latex], whichever is larger

X = 12 in. or (6 + 1 + 1.25) = 8.25 in.; use X = 12 in.

(b) For two-thirds of required reinforcement area,

[latex]X^{\prime}=r+0.5 \sqrt{R_{m} t}[/latex] or [latex]\left(T_{s}+T_{n}+r ight)[/latex] whichever is larger

[latex]X^{\prime}=6+0.5 \sqrt{(21.438)(1)}=8.315 ext { in. }[/latex] or 8.25 in.; use

[latex]X^{\prime}=8.315 [/latex] in.

(c) For limits normal to shell surface, the larger of

[latex]Y=0.5 \sqrt{r_{m} t_{n}}+K[/latex] or

[latex]Y=1.73 x+2.5 t_{p}+K \leq 2.5 T_{s}[/latex] and [latex]\leq L+2.5 t_{p}[/latex]

[latex]Y=0.5 \sqrt{6.625 imes 1.25}+0.25=1.698 ext { in. }[/latex] or

[latex]Y=0+2.5(1.25)+0.25=3.375[/latex] ≤

2.5(1) = 2.5 in. and ≤ 3 + 2.5(1.25) = 6.125 in.

use Y = 2.5 in.

(d) Reinforcement area required for

[latex]100 \%: \quad A_{r}=d t_{r} F+2 t_{n} t_{r}\left(1-f_{r} ight)[/latex]

[latex]\begin{aligned}A_{r} &=12 imes 0.728 imes 1+2 imes 1.25 imes 0.728(1-0.946) \&=8.834 ext { in. }^{2}\end{aligned}[/latex]

[latex]\frac{2}{3}: \quad A_{r}=\frac{2}{3}(8.834)=5.922 ext { in. }^{2}[/latex]

(e) Reinforcement available with 100% limit:

[latex]A_{1}=(1-0.728)(24-12)=3.264 ext { in. }{ }^{2}[/latex]

(f) Reinforcement available in nozzle wall:

outward: [latex]A_{21}=2(2.5)(1.25-0.220)(0.946)=4.872 ext { in. }{ }^{2}[/latex]

inward: [latex]A_{22}=2(2.5)(1.25)(0.946)=5.912 ext { in. }^{2}[/latex]

(g) Total reinforcement with 100% limit is

[latex]A_{t}= A _{1}+ A _{21}+ A _{22}=3.264+4.872+5.912=14.048 ext { in. }^{2}[/latex]

[latex]A_{t}=14.048>A_{r}=8.834 ext { in. }^{2}[/latex]

(h) Reinforcement available in shell with two-thirds limit:

[latex]A_{1}=(1-0.728)(2 imes 8.315-12)=1.259 ext { in. }^{2}[/latex]

(i) Total reinforcement with two-thirds limit is

[latex]A_{t}=A_{1}+A_{21}+A_{22}=1.259+4.872+5.912=12.040 ext { in. }^{2}[/latex]

[latex]A_{t}=12.040>A_{r}=5.922 ext { in. }^{2}[/latex]

5 . In examining the circumferential plane, reinforcement area required according to the ASME Code, VIII- 2 is

[latex]A_{r}=d t_{r} F=16 imes 0.858 imes 0.5=6.864 ext { in. }^{2}[/latex]

Using the arc length of 16.4 in., [latex]A_{r}=7.036 in.{ }^{2}[/latex] Either of the two areas is less than [latex]A_{r}[/latex] in item 4d and do not control. Longitudinal controls.

Question 11.10: Determine the reinforcement requirements for a 12 -in. × 16 ...

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