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## Q. 11.9

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. ## Verified Solution

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:

$\frac{P}{S}=\frac{700}{20500}=0.034$        which gives $\frac{t}{L}=0.021$

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

$t_{r n}=\frac{P R}{S-0.5 P}=\frac{700 \times 4}{21,600-0.700}=0.132 in.$;              use $1 \frac{1}{8}$ in. thickness

3 . Limits parallel to head surface are:

(a) For 100% or required reinforcement area:

X = d       or      $\left(T_{s}+T_{n}+r\right)$      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:

$X^{\prime}=r+0.5 \sqrt{R_{m} t}$       or      $\left(T+T_{n}+0.5 d\right)$

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

4 . Limits normal to head surface, the larger of

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

$Y=0.5 \sqrt{(4.563)(1.125)}+0.25=1.383$ in.       or

$=0+2.5(1.125)+0.25=3.063 \text { in. }$

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

$100 \%: \quad A_{r}=d t_{r} F=8(0.789)(1)=6.312 \text { in. }^{2}$

$\frac{2}{3}: \quad A_{r}=\frac{2}{3}(6.312)=4.208 \text { in. }^{2}$

6 . Reinforcement available with 100% limit is

$A_{1}=\left(T_{s}-t_{r}\right)(2 X-d)=(1.0-0.789)(16-8)=1.688 \text { in. }^{2}$

7 . Reinforcement available in nozzle wall is

$A_{2}=2 Y\left(T_{n}-t_{r n}\right)=2(2.5)(1.125-0.132)=4.965 \text { in. }{ }^{2}$

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

$A_{t}=A_{1}+A_{2}=1.688+4.965=6.654 \text { in. }^{2}>6.312 \text { in. }^{2}$

9 . Reinforcement available with two-thirds limit is

\begin{aligned}A_{1} &=\left(T_{s}-F t_{r}\right)\left(2 X^{\prime}-d\right)=(1-0.789)(2 \times 7.085-8) \\&=1.302 \text { in. }^{2}\end{aligned}

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

$A_{t}=A_{1}+A_{2}=1.302+4.965=6.267 \text { in. }^{2}>4.208 \text { in. }^{2}$