Need Help Transposing the pressure drop formula

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Discussion Overview

The discussion revolves around the transposition of a pressure drop formula to determine the minimum diameter of a pipe, given specific parameters such as pressure drop limits and flow rates. The context includes a homework problem related to fluid dynamics in engineering applications.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the pressure drop formula: Pressure Drop = 800lQ²/Rd⁵.³¹ and seeks to transpose it to find the minimum diameter.
  • Another suggests taking the 5.31th root of both sides to isolate d.
  • There is a correction regarding the placement of the pressure drop (ΔP) in the formula, with a participant noting it should be on the right-hand side.
  • One participant calculates d using the formula and provides a numerical result, questioning if it is correct.
  • Another participant confirms the calculation looks correct but mentions not having a calculator available for verification.
  • Further calculations lead to a diameter of 68m, which raises concerns about the size being unusually large for a pipe.
  • Questions arise regarding the origin of the equation, the correctness of the value of R, and the appropriate units for the flow rate Q.
  • Clarifications are made about the definitions of variables used in the formula, including the units for flow rate and diameter.
  • A later reply suggests that 68mm is a more reasonable diameter than 68m.

Areas of Agreement / Disagreement

Participants express uncertainty about the correctness of certain values and units, and there is no consensus on the final diameter calculation or the validity of the parameters used in the formula.

Contextual Notes

There are unresolved questions regarding the units for flow rate (L/s vs m³/s) and the implications of the calculated diameter being unusually large. The discussion reflects a need for clarity on the definitions and assumptions underlying the formula.

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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