Recent content by Jack Mc

l = the equivalent length of pipe in metres Q = the flow through the pipe in ℓs−1 free air R = p2/p1 = the ratio of compression at the start of the pipe d = the internal diameter of the pipe in mm

I got R by... R=p2/p1 R=6+1.01/1.01 R=6.94

d^5.31=800x160x300^2/6.94x0.3 = d^5.31 = 55.331412.1 Is this the right lines?

I see thank you. Ill give this a go.

Homework Statement Hi, Iv got to find the minimum diameter of a pipe if the pressure drops in a system is to be limited to 0.3bar. when delivered through a pipe of equivalent length 160m I have to use the formula Pressure Drop = 800lQ^2/Rd^5.31 I believe I have to transpose this to find...

Was nearly right...

These are my workings... Facts; P1 = 1.013bar V1 = Unknown T1 = 15 + 273 = 288K P2 = 6 + 1.013 = 7.013 V2 = 1.2 m³-1 T2 = 30 + 273 = 303K Formula: V1= (P2 X V2 X T1) / (P1 X T2) V1= (7.013 X 1.2 X 288) / (303 X 1.013) V1= 2423.69 / 306.939 ∴ FAD of the compressor = 7.89 m³ min-1 Does this...