 |
New York Codes Rules Regulations (Last Updated: March 27,2024) |
 |
TITLE 16. Department of Public Service |
|
|
Appendices |
|
|
Appendix 14-G. |
|
EXAMPLES ILLUSTRATING THE REINFORCEMENT OF WELDED BRANCH CONNECTIONS Example 1 An 8 inch outlet is welded into a 24 inch header. The header material is API 5LX 46 with /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°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: t =PD/2 SFET650 × 24/2 × 46000 × 0.60 × 1.00 × 1.00 = 0.283 inch Excess thickness in header wall (H-t) = 0.312 - 0.283 = 0.029 inch Outlet: Nominal wall thickess: tb = 650 × 8.625/2 × 35000 × 0.60 × 1.00 × 1.00 = 0.133 inch Excess thickness i outlet wall (B-tb) = 0.322-0.133 = 0.189 inch d = diameter of opening = 8.625-(2 × 0.322) = 7.981 inch Reinforcement required. AR = d × t = 7.981 × 0.283 = 2.26 square inch Reinforcement provided: A1 = (H-t)d = 0.029 × 7.981 = 0.23 square inch Effective area in outlet: Height (L) 2½ B + M (assume ¼ inch pad) = 2½ × 0.322 + 0.25 = 1.05 inches or 2½ H = 2.5 × 0.312 = 0.78 inch. Use 0.78 inch. A2 = 2(B-tb)L = 2 × 0.189 × 0.78 = 0.295 sq. in. This must be multiplied by 35000/46000 Effective A2 = 0.295 × /46000 = 0.22 sq. in. Required area A3 = AR - A1 - A2 = 2.26 - 0.23 - 0.22 = 1.81 sq. in. Use reinf. pl. ¼ inch thick (minimum practicable) × 15.5 inch diameter. Area (15.50 - 8.62) × 0.25 = 1.72 sq. in. Fillet welds (assuming two ¼ inch welds each side) 0.25 × 0.25 × 0.50 × 2 × 2 = 0.12 sq. in. Total A 3 provided 1.84 sq. in. Example 2 A 16 inch outlet is welded into a 24 inch header. The header material is API 5LX46 with /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°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: t =PD/2 SFET650 × 24/2 × 46000 × 0.60 × 1.00 × 1.00 = 0.283 inch Excess thickness is header wall (H-t) = 0.020 inch    |