Awood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in.2 the allowable stress for the hoops is σallos=11.4 ksi. determine the maximum spacing s along the pipe so that the pipe can resist an internal gauge pressure of 4 psi.

31.67 in

Explanation:

Given:

Diameter of the pipe, D = 3ft = 36 in

cross-sectional area of the steel = 0.2 in²

Note: Refer to the figure attached

From the free body diagram represented in the figure, we have

ΣFx = 0

or

pressure × projected area = 2 × Force in steel

Now, the projected area = spacing (s) × diameter of the wood pipe

force in steel = stress in steel (σ) × cross-sectional area of the steel

on substituting the values we get

4 ksi × (s × 36 in) = 2 × σ × 0.2 in²

also, allowable hoop stress = 11.4 ksi

thus,

σ = 11.4 ksi = 11.4 × 10³ psi

therefore, we have

4 psi × (s × 36 in) = 2 × 11.4 × 10³ psi × 0.2 in²

thus,

s = 31.67 in

hence the maximum spacing is 31.67 in

v=ir = 3x15 =45 v

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat