1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need Help Transposing the pressure drop formula

  1. Feb 25, 2017 #1
    1. The problem statement, all variables and given/known data

    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 belive I have to transpose this to find the min diameter.

    2. Relevant 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 .



    3. The attempt at a solution

    d^5.31=800lQ^2/R
     
  2. jcsd
  3. Feb 28, 2017 #2

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    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)
     
  4. Feb 28, 2017 #3

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    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
     
  5. Feb 28, 2017 #4
    I see thank you. Ill give this a go.
     
  6. Mar 1, 2017 #5
    d^5.31=800x160x300^2/6.94x0.3
    = d^5.31 = 55.331412.1
    Is this the right lines?
     
  7. Mar 1, 2017 #6

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Looks right to me although I don't have my calculator to hand.
     
  8. Mar 2, 2017 #7

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  9. Mar 2, 2017 #8
    I got R by...

    R=p2/p1

    R=6+1.01/1.01

    R=6.94
     
  10. Mar 2, 2017 #9

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Should the flow rate Q be in L/s or m3/s ?
     
  11. Mar 2, 2017 #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
     
  12. Mar 2, 2017 #11

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Ah OK 68mm makes more sense than meters.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Need Help Transposing the pressure drop formula
  1. Pressure Drop (Replies: 1)

  2. Transposing a formula (Replies: 1)

Loading...