- #1

- 52

- 0

**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/m

(for natural gas, [tex]\rho[/tex]= 40kg/m

^{3}, [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)

[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**

I subbed this value into the reynolds number equation;

Re = 1x10

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*10

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?

^{-6}/40 = 2.75*10^{-7}I subbed this value into the reynolds number equation;

Re = 1x10

^{-6}*0.4/2.75*10^{-7}= 16As 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*10

^{3}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?

Last edited: