A 16 inch outlet is welded into a 24 inch header. The header material is API 5LX46 with 5/16 inch wall. The outlet is API 5L Grade B (Seamless) Sched. 20 with 0.312 wall. The working pressure is 650 PSI. The Class Location is 1. The reinforcement must be of the complete encirclement type. 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 2 in Appendix 14-G.
Header:
Nominal wall thickness:
Outlet:
Nominal wall thickness:
Excess thickness in outlet wall (B - tb) = 0.312 - 0.248 = 0.064 inch
d = diameter of opening = 16.000 - (2 x 0.312) = 15.376 inch
Reinforcement required:
AR = d x t = 15.376 x 0.283 = 4.35 sq. in.
Reinforcement provided:
A1 = (H - t)d = 0.029 x 15.376 = 0.44 sq. in.
Effective area in outlet:
Height (L) 2-1/2 B * M (assume 5/16 inch pl.) = 2.5 x 0.312 +
0.312 = 1.09 inch or 2-1/2 H = 2.5 x 0.312 = 0.78 inch.
Use 0.78 inch
A2 = 2(B-tb) L = 2 x 0.064 x 0.78 = 0.10 sq. in.
This must be multiplied by 35000/46000
Required area A3 = AR - A1 - A2 = 4.35 - 0.44 - 0.08 = 3.83 sq.
in.
Approx. required thickness of reinforcement
3.83/(30 - 16) = 0.27 inch
Use 5/16 inch pl. net reqd. length (neglecting welds)
3.83/0.312 = 12.3 inch
Use plate 29 inches long
A3 - 0.312 x (29 - 16) = 4.05 sq. in.
Two 1/4 inch welds to outlet
2 x 0.25 x 0.25 x 0.50 = 0.06 sq. in.
A3 provided 4.11 sq. in.
The use of end welds is optional.