Question

1. For thick spheres, a y factor is used to determine thickness, t, if P is known. The equations are:

t = R(y1/3 – 1) and y = 2(SE+P) 2SE-P

Where: t=minimum required thickness of sphere (inch); R=inside diameter of sphere (inch); S=allowable stress (lbf/in2); E=joint efficiency (%); P=allowable pressure. Assuming S=14,000 lbf/in2, P=3000 psi and E=100%, the thickness of a sphere with a 10 inch inside diameter to be used with a 3000 psi allowable pressure is                        .

1. 8/16 inch                 b.            5/7 inch                   c.           9/16 inch                 d.           7/15 inch

Answer 1

Ang :- Given t=R (yy-2) & y= 2(SE+A 2SE -P where to minimum required thickness R = 10 inch ¿ S = 14000 lbf/m2 p=3000 psi E-100,1.0 I lbf/in2 14000 lbf/in2 / 1 Psi = 1400 psc y = 20" 14000 x1 + 3000) = 2X17000 28000-3000 2X Orox1 300 y= 34000 = 34 = 1.36 25 25000 = 10 3 t= R(yY3-1) 10 (11-36)*3 -) to 10 (1-10793165-1) t= 10 X0.10793165 ta = 1.0793165 inch CS Scanned with CamScanner