Need Help Transposing the pressure drop formula

AI Thread Summary
The discussion focuses on calculating the minimum diameter of a pipe to limit pressure drop to 0.3 bar over a length of 160 meters, using the formula Pressure Drop = 800lQ^2/Rd^5.31. The user initially misapplies the formula but receives clarification that the pressure drop (ΔP) should be included on the right side of the equation. After correcting the formula and performing calculations, a diameter of 68 mm is derived, which is deemed reasonable compared to the initial miscalculation of 68 meters. Questions arise regarding the source of the equation, the correctness of the value for R, and the units for flow rate. Ultimately, the discussion emphasizes the importance of unit consistency and proper formula application in engineering calculations.
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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Ok so far then take the 5.31th root of both sides.

If your calculator can't do that perhaps logarithms will help..

log(mn) = n * log(m)
 
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
 
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?
 
Looks right to me although I don't have my calculator to hand.
 
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
 
Should the flow rate Q be in L/s or m3/s ?
 
  • #10
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
 
  • #11
Ah OK 68mm makes more sense than meters.
 
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