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Chapter 11

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.

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

Step-by-Step

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}