Fluids PIPE FLOW RATE Q

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1. The pressure at the start of a 180km stainless steel natural gas pipeline is 4Mpa and at the end is 2Mpa. If the diameter of the pipe is 0.4m, what will be the gas flow rate through the system? (ignore losses other than friciton)
(for natural gas, [tex]\rho[/tex]= 40kg/m3, [tex]\mu[/tex]= 11x10-6 Pas



2.

[tex]\nu[/tex]=[tex]\mu[/tex]/[tex]\rho[/tex] (kinematic velocity)

Re = [tex]\mu[/tex]*D/[tex]\nu[/tex] (Reynolds number)

Q= ([tex]\pi[/tex]*D^4/ 128*[tex]\mu[/tex]*L)*(P1-P2) (Flow rate)



3. First I found the Kinematic velocity; [tex]\nu[/tex]=11x10-6/40 = 2.75*10-7

I subbed this value into the reynolds number equation;

Re = 1x10-6*0.4/2.75*10-7 = 16

As the Re number is less than 2100 the flow is laminar and I can use the above equation, I also used the moody diagram to establish the friction factor; The pipe is stainless steel so the relative rougness was equal to .045/.4*103 this part I wasn't quite sure of but I found the friction factor value to be (.002)

Now all I have to do is sub into this equation:
Q= ([tex]\pi[/tex]*D^4/ 128*[tex]\mu[/tex]*L)*(P1-P2) correct?

But how do I incorporate the friction factor? Also Is my analysis of the problem correct?
 
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  • #2
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Anyone have any help?
 
  • #3
dlgoff
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Moody diagrams are used to find pressure drops I think. Here you know the pressure drop.
 
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  • #5
dlgoff
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I don't see any "friction factor" terms in your Q equation. And you do know P1-P2. So...
 
  • #6
dlgoff
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If you were wanting to determine the pressure drop due to the roughness of the pipe then yes you would need to know the friction factor. But you already know the pressure drop.
 
  • #7
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Ah ok, If its a laminar flow no need to account for friction. Just seems a bit too easy, was one of the questions with the most marks
 
  • #8
dlgoff
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Ah ok, If its a laminar flow no need to account for friction. Just seems a bit too easy, was one of the questions with the most marks
That's the way I see it. :smile:
 
  • #9
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I've just noticed something, My Re equation is wrong it should be (u) Mean velocity instead of [tex]\mu[/tex] Viscosity, And to find u = Q/A , Now i'm completely confused.
 

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