Figure 11.10 shows a 66-in. ID steam drum containing five different diameters and two types of nozzles. What are the nozzle reinforcement requirements? The design data are Design pressure = 2875 psi. Design temperature = saturation at design pressure approximately 689°F. Materials 70,000 psi UTS

Question

Figure 11.10 shows a 66-in. ID steam drum containing five different diameters and two types of nozzles. What are the nozzle reinforcement requirements? The design data are

Design pressure = 2875 psi.

Design temperature = saturation at design pressure approximately 689°F.

Materials 70,000 psi UTS drum plate.

Allowable stress at saturation temperature = 16,800 psi.

Weld joint efficiency is E = 0.95.

Nozzles are [latex]3 \frac{1}{2}[/latex] in., 4 in., 5 in., [latex]6 \frac{7}{8}[/latex] in., and 24-in. ID.

Accepted Answer

1 . Minimum required thickness of shell at E = 0.95 is

[latex]t=\frac{P R}{S E-0.6 P}=\frac{2875 imes 33}{16,800 imes 0.95-0.6 imes 2875}=6.665 [/latex] in .

use [latex]6 \frac{3}{4}[/latex] in. plate

2 . Minimum required thickness of seamless shell is

[latex]t_{r}=\frac{P R}{S E-0.6 P}=\frac{2875 imes 33}{16,800-0.6 imes 2875}=6.29 [/latex] in.

3 . Minimum required thickness of nozzle is

[latex]t_{r n}=\frac{P r}{S E-0.6 P}=\frac{2875 r}{16,800-0.6 imes 2875}=0.190 \ r[/latex]

4 . Maximum diameter of a single unreinforced opening is determined as

[latex]D_{o}=66+2(6.75)=79.5 [/latex] in.

K from Eq. 11.12 is

[latex]\begin{aligned}\frac{P D_{o}}{1.82 S T_{ s }}=\frac{2875 imes 79.5}{1.82 imes 16,800 imes 6.75} &=1.107 \K_{\max }&=0.99\end{aligned}[/latex]

[latex]K=\frac{p D_{o}}{1.82 S T_{s}}[/latex] (11.12)

Using Eq. 11.10, the maximum diameter is

[latex]d_{\max }=2.75 \sqrt[3]{D_{o} T_{s}(1-K)}[/latex] (11.10)

with the limits of 0.5 ≤ K ≤ 0.99.

p = internal design pressure of maximum allowable working pressure (psi)

[latex]d_{\max }[/latex] = maximum allowable diameter of opening (in.)

[latex]D_{o}[/latex] = outside diameter of shell (in.)

[latex]T_{s}[/latex] = nominal thickness of shell (in.)

S = allowable tensile stress (psi)

[latex]d_{\max }=2.75 \sqrt[3]{79.5 imes 6.75 imes(1-0.99)}=4.814 [/latex] in.

5 . The only single openings are 5 in. and [latex]6 \frac{7}{8}[/latex] in.; others are not considered single. Consequently, all nozzles have the reinforcement area calculated (see Table 11.2).

Table 11.2 Reinforcement Calculations for [latex]3 \frac{1}{2}[/latex] in., 4 in., 5 in., and [latex]6 \frac{5}{8}[/latex] in. Nozzles on a 66 in. ID Steam Drum
X = parallel Y = normal
d	[latex]A_{r}[/latex]	[latex]T_{n}[/latex]	d	[latex]\begin{aligned}&6.75+ \&T_{n}+ r\end{aligned}[/latex]	[latex]2.5T_{n}[/latex]	[latex]2.5T_{s}[/latex]	[latex]t_{r n}[/latex]	[latex]A_{1}[/latex]	[latex]A_{2}[/latex]	[latex]A_{t}[/latex]
3.5	22.03	1.875	3.5	10.375	4.688	16.875	0.334	7.87	14.44	22.31
4.0	25.18	2.125	4.0	10.875	5.313	16.875	0.381	8.09	18.53	26.62
5.0	31.47	2.5	5.0	11.75	6.25	16.875	0.477	8.44	25.29	33.73
6.875	43.27	3.0	6.875	13.188	7.5	16.875	0.656	8.89	35.16	44.05
Area available is greater than area required and values of [latex]T_{n}[/latex] that were assumed are correct.

6 . All nozzles except the 24 -in. ID nozzle are "set-on" type and calculated together.

7 . Reinforcement area required by Eq. 11.15 is

[latex]A_{r}=d t_{r} F=d(6.294)(1.0)=6.294 \ d[/latex]

8 . Limit parallel to shell is greater of

d or [latex]T_{s}+T_{n}+0.5 d=6.75+T_{n}+r[/latex]

9 . Limit normal to shell is smaller of

[latex]2.5 T_{s}=2.5(6.75)=16.875[/latex] in. or [latex]2.5 T_{n}[/latex]

10 . Area available in shell wall is

[latex]\begin{aligned}A_{1}=&(2 X-d)\left(T_{s}-t_{r} ight)=(2 X-d)(6.75-6.294) \& A_{1}=(2 X-d)(0.456)\end{aligned}[/latex]

11 . Area available in nozzle wall is

[latex]A_{2}=2 Y\left(T_{n}-t_{r n} ight)[/latex]

12 . A 24-in. ID downcomer has "set-in" nozzle and calculates d/D = 24 / 66 = 0.36 exceeds limit of 0.33 D for vessels over 60 in. ID. Alternate rules are also recommended.

13 . Minimum required thickness of a downcomer nozzle is

[latex]t_{r n}=0.190(12)=2.289 ext { in. }[/latex] use [latex]T_{n}=5 \frac{1}{4} ext { in. }[/latex]

14 . Reinforcement area required is

[latex]A_{r}=6.294(24)=151.06 ext { in. }^{2}[/latex]

15 . Limit parallel to shell is greater of

d = 24 in. or [latex]T_{s}+T_{n}+0.5 d=6.75+5.25+12=24 ext { in. }[/latex]

16 . Limit normal to shell is smaller of

16.875 in. or 2.5(5.25) = 13.125 in.

17 . Area available in shell wall is

[latex]A_{1}=(2 imes 24-12)(0.456)=5.47 ext{ in.} ^{2}[/latex]

18 . Area available in nozzle wall; wall extends inward for [latex]6 \frac{1}{2}[/latex] in. as full reinforcing limit outward:

[latex]A_{21}=2(13.125)(5.25-2.289)=77.73 ext { in. }^{2}[/latex]

[latex]A_{22}=2(6.5)(5.25-0)=68.25 ext { in. }^{2}[/latex]

19 . Total area available is

[latex]A_{t}=A_{1}+A_{21}+A_{22}=151.45 ext { in. }^{2} \geq A_{r} ext { of } 151.06 ext { in. }^{2}[/latex]

20 . Also, check "close-in" limit. Determine limit parallel to shell as the greater of

0.75 d = 0.75(24) = 18 in. or [latex]T_{s}+T_{n}+r=24 ext { in. }[/latex]

Because the parallel limit is the same for the "close-in" limit, the area required of [latex]0.67 A_{r}[/latex] is also satisfied without further calculations.

Question 11.4: Figure 11.10 shows a 66-in. ID steam drum containing five di...

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