Appendix 14-G Example 1

 

An 8 inch outlet is welded to a 24 inch header. The header material is API 5LX 46 with 5/16 inch wall. The outlet is API 5L Grade B (Seamless) Sched. 40 with 0.322 inch wall. The working pressure is 650 PSI. The Class location is 1. The joint efficiency is 1.00. The temperature is 100 deg F. Design Factors F=0.60, E=1.00, T=1.00. For dimensions see Figure for Example 1 , Appendix 14-G.

 

Header:

Nominal wall thickness:

Excess thickness in header wall (H-t) = 0.312 - 0.283 = 0.029 inch

 

Outlet:

Nominal wall thickness:

Excess thickness in outlet wall (B-tb) = 0.322 - 0.133 = 0.189 inch

d = diameter of opening = 8.625-(2 x 0.322) = 7.981 inch

 

Reinforcement required.

AR = d x t = 7.981 x 0.283 = 2.26 square inch

 

Reinforcement provided.

A1 = (H - t)d = 0.029 x 7.981 = 0.23 square inch

 

Effective area in outlet:

Height (L) 2-1/2 B + M (assume 1/4 inch pad) = 2-1/2 x 0.322 + 0.25 = 1.05 inches

or

2-1/2 H = 2.5 x 0.312 = 0.78 inch.

Use 0.78 inch.

 

A2 = 2( B-tb)L = 2 x 0.189 x 0.78 = 0.295 sq. in.

This must be multiplied by 35000/46000

 

Required area A3 = AR - A1 - A2 = 2.26 - 0.23. - 0.22 = 1.81 sq. in.

 

Use reinf. pl. 1/4 inch thick (minimum practicable) x 15.5 inch diameter

 

Area (15.5 - 8.62) x 0.25 = 1.72 sq. in.

 

Fillet welds (assuming two 1/4 inch welds each side)

 

0.25 x 0.25 x 0.50 x 2 x 2 = 0.12 sq. in

 

Total A3 provided 1.84 sq. in.