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Chapter 11

Q. 11.11

A steam pipe has a 24-in. inside diameter with a design pressure of 2500 psi and an allowable stress of 14,500 psi at the design temperature. A branch pipe with an inside diameter of 8 in. connects at an angle of α = 75°. The branch is attached by a full-penetration weld that is radiographed so that E = 1.0. Determine the thickness and reinforcement requirements.

Step-by-Step

Verified Solution

1 . Determine the minimum required thickness of the run pipe as follows:

t_{m h}=\frac{P R}{S E-0.6 P}=\frac{2500 \times 12}{14,500 \times 1-0.6 \times 2500}=2.308 in.;          use T_{s}=2.5 in.

2 . Determine the minimum required thickness of the branch pipe as follows:

t_{m b}=\frac{P r}{S E-0.6 P}=\frac{2500 \times 4.0}{14,500 \times 1-0.6 \times 2500}=0.769 in.;          use T_{n}=2.0 in.

3 . Area required for reinforcement according to Eq. 11.29 is

A_{r}=1.07 t_{m h} d_{1}(2-\sin \alpha)                         (11.29)

A_{r}=1.07(2.308)(8)\left(2-\sin 75^{\circ}\right)=20.430 \text { in. }{ }^{2}

4 . Horizontal limits of reinforcement are the larger of

d = 8 in.       or      T_{s}+T_{n}+r=2.5+2.0+4=8.5 \text { in. };           use 8.5 in.

5 . Perpendicular limit of reinforcement is as follows: Assume a 0.75 in. thick pad is added and attached by full penetration welds that are examined.

t_{e}=0.75 \text { in. }

L=2.5 T_{n}+t_{e}=2.5(2)+0.75=5.75 \text { in. }

6 . Area available for reinforcement is

A_{1}=(2 \times 8.5-8)(2.5-2.308)=2.880 \text { in. }^{2}

A_{2}=\frac{2(5.75)(2-0.769)}{\sin 75^{\circ}}=14.656 \text { in. }{ }^{2}

A_{3}=(0.5)^{2}=0.250 \text { in. }^{2}

A_{4}=(2 \times 8.5-8-4)(0.75)=3.750 \text { in. }{ }^{2}

A_{t}=A_{1}+A_{2}+A_{3}+A_{4}=21.536 \text { in. }^{2}>A_{r}=20.430 \text { in. }{ }^{2}

Shell requirement:  24 -in. ID × 2.5 in. thick

Nozzle requirement:  8 -in. ID × 2.0 in. thick

Fillet weld requirements:    2 with 0.5 in. legs

Pad requirement:    17 -in. OD ring × 0.75 in. thick

Nozzle attached to shell and pad by full-penetration welds.