Determine the reinforcement requirements of an 8-in. ID nozzle that is centrally located in a 2 : 1 ellipsoidal head. The inside diameter of the head skirt is 41.75 in. The allowable stress of the head material is 20.5 ksi and of the nozzle material is 21.6 ksi. The design pressure is 700 psi and

Question

Determine the reinforcement requirements of an 8-in. ID nozzle that is centrally located in a 2 : 1 ellipsoidal head. The inside diameter of the head skirt is 41.75 in. The allowable stress of the head material is 20.5 ksi and of the nozzle material is 21.6 ksi. The design pressure is 700 psi and the design temperature is 500°F. There is no corrosion. See Fig. 11.18 for details of nozzle.

Accepted Answer

1 . The minimum required thickness of the 2 : 1 ellipsoidal head using Fig. AD-204.1 of the ASME Code, VIII-2, is determined as follows:

[latex]\frac{P}{S}=\frac{700}{20500}=0.034[/latex] which gives [latex]\frac{t}{L}=0.021[/latex]

L = 0.9D = 0.9(41.75) = 37.575 in.

t = 0.021L = 0.021(37.575) = 0.789 in.;

use 1.0-in. thickness

2 . 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 4}{21,600-0.700}=0.132 in.[/latex]; use [latex]1 \frac{1}{8}[/latex] in. thickness

3 . Limits parallel to head surface are:

(a) For 100% or required reinforcement area:

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

X = 8 in. or (4 + 1 + 1.125) = 6.125 in.; use X = 8 in.

(b) For two-thirds required reinforcement area is the larger of:

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

[latex]X^{\prime}=4+0.5 \sqrt{(38.075)(1)}=7.085[/latex] in. or 6.125 in.; use [latex]X^{\prime}=7.085[/latex] in.

4 . Limits normal to head surface, the larger of

[latex]\begin{aligned}Y=& 0.5 \sqrt{r_{m} t_{n}}+K ext { or } 1.73 x+2.5 t_{p}+K \leq \& 2.5 T_{n} ext { and } \leq L+2.5 t_{p}\end{aligned}[/latex]

[latex]Y=0.5 \sqrt{(4.563)(1.125)}+0.25=1.383[/latex] in. or

[latex]=0+2.5(1.125)+0.25=3.063 ext { in. } [/latex] ≤

2.5(1) = 2.5 in. and ≤ 4 + 2.5(1.125) = 6.813 in.; use Y = 2.5 in.

5 . Reinforcement area required for

[latex]100 \%: \quad A_{r}=d t_{r} F=8(0.789)(1)=6.312 ext { in. }^{2}[/latex]

[latex]\frac{2}{3}: \quad A_{r}=\frac{2}{3}(6.312)=4.208 ext { in. }^{2}[/latex]

6 . Reinforcement available with 100% limit is

[latex]A_{1}=\left(T_{s}-t_{r} ight)(2 X-d)=(1.0-0.789)(16-8)=1.688 ext { in. }^{2}[/latex]

7 . Reinforcement available in nozzle wall is

[latex]A_{2}=2 Y\left(T_{n}-t_{r n} ight)=2(2.5)(1.125-0.132)=4.965 ext { in. }{ }^{2}[/latex]

8 . Total reinforcement available form head and nozzle within 100% reinforcement limit is

[latex]A_{t}=A_{1}+A_{2}=1.688+4.965=6.654 ext { in. }^{2}>6.312 ext { in. }^{2}[/latex]

9 . Reinforcement available with two-thirds limit is

[latex]\begin{aligned}A_{1} &=\left(T_{s}-F t_{r} ight)\left(2 X^{\prime}-d ight)=(1-0.789)(2 imes 7.085-8) \&=1.302 ext { in. }^{2}\end{aligned}[/latex]

10 . Total reinforcement available from head and nozzle using two-thirds limit is

[latex]A_{t}=A_{1}+A_{2}=1.302+4.965=6.267 ext { in. }^{2}>4.208 ext { in. }^{2}[/latex]

Question 11.9: Determine the reinforcement requirements of an 8-in. ID nozz...

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