Need Help Transposing the pressure drop formula

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Jack Mc
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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 the min diameter.

Homework Equations



compressor delivers 300 l s–1 of free air into a pipe at a pressure of 6 bar gauge.

Using the pressure drop formula: pressure drop = 800lQ2 Rd 5.31

calculate the minimum diameter of pipe if the pressure drop in a system is to be limited to 0.3 bar when is delivered through a pipe of equivalent length 160 m .

The Attempt at a Solution



d^5.31=800lQ^2/R
 
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CWatters said:
Actually I see you forgot to put the pressure drop on the right hand side. If ΔP is the pressure drop it should read..

d5.31=800lQ2/RΔP

I see thank you. Ill give this a go.
 
d^5.31=800x160x300^2/6.94x0.3
= d^5.31 = 55.331412.1
Is this the right lines?
 
I get..
d5.31 = 800x160x300^2/6.94x0.3 = 5.5 * 106
then
5.31*log(d) = Log(5.5 * 106)
so
d = 10(Log(5.5 * 106)/5.31)
= 68m
which is very large for a pipe!

Where did the equation come from?
Is the value of R correct?
Should the pressure drop be in bar or some other units?
 
I got R by...

R=p2/p1

R=6+1.01/1.01

R=6.94
 
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