How Does Pipe Configuration Affect Pressure in a Fluid System?

[teegfit]

< Mentor Note -- thread moved to HH from the technical engineering forums, so no HH Template is shown >[/color]

The system shown in the figure below consists of L = 60 m of D = 50 mm diameter cast iron pipe (k/D = 0.004), two 45° flanged long radius elbows (K = 0.25) and four 90° flanged long radius elbows (K = 0.5), a fully open flanged globe valve (K = 10) and a sharp exit (K = 1.0) into a reservoir. The elevation at point 1 Z1 = 0 m and at point 2, the free surface within the reservoir, Z2 = 5 m. The volume flow rate of water (ρ = 999 kg/m3 and µ =0.0012 Ns/m2 ) delivered by the system to the reservoir Q = 0.005 m3 /s.
Determine the gauge pressure at point 1 and:
(i) all losses within the system are neglected,
(ii) major losses only are included,
(iii) all losses within the system are included.My solution:
Firstly i calculated the mean velocity,V, by dividing the volume flow rate, Q, by the cross sectional area of the pipe which i calculated to be using the diameter,D, (1.96E-3) and therefore the velocity is 2.55 m/s.

And I calculated the pressure by using (P1-P2)=(denisty)(gravity(g))(Z2-Z1) and assumed P2 to be zero. therefore pressure 1 is 49,000.95 Pa, at loses neglected. i)

I then calculated the reynolds number of the pipe by the formula ((denistyxVxD)/(dynamic viscosity)) giving me a reynolds number of 106143.75.

Then at k/D .004 and reynolds 1E5 on the moody diagram to find a friction factor,f, of .03 which i put into the formula headloss due to friction,hfmajor, = f(L/D)((V^2)/2(g)) giving me a headloss of 11.93 m which i put into the formula P1=(density)(g)(z2+hfmajor) to give me 165917.217 pa ii)

I found the minor head loss,hfminor, by using =(k1+k2 etc.)((V^2)/2(g)) getting a headloss minor of 4.474 and plugging this back into the equation P1=(density)(g)(Z2 + hfmajor +hfminor) to get an answer of 209763.27 Pa iii)

I am posting this here as i want to ensure my methology is right as i have an exam in mechanics of fluids on monday

Thank you very much for your help

 

[SteamKing]

< Mentor Note -- thread moved to HH from the technical engineering forums, so no HH Template is shown >

The system shown in the figure below consists of L = 60 m of D = 50 mm diameter cast iron pipe (k/D = 0.004), two 45° flanged long radius elbows (K = 0.25) and four 90° flanged long radius elbows (K = 0.5), a fully open flanged globe valve (K = 10) and a sharp exit (K = 1.0) into a reservoir. The elevation at point 1 Z1 = 0 m and at point 2, the free surface within the reservoir, Z2 = 5 m. The volume flow rate of water (ρ = 999 kg/m3 and µ =0.0012 Ns/m2 ) delivered by the system to the reservoir Q = 0.005 m3 /s.
Determine the gauge pressure at point 1 and:
(i) all losses within the system are neglected,
(ii) major losses only are included,
(iii) all losses within the system are included.My solution:
Firstly i calculated the mean velocity,V, by dividing the volume flow rate, Q, by the cross sectional area of the pipe which i calculated to be using the diameter,D, (1.96E-3) and therefore the velocity is 2.55 m/s.

And I calculated the pressure by using (P1-P2)=(denisty)(gravity(g))(Z2-Z1) and assumed P2 to be zero. therefore pressure 1 is 49,000.95 Pa, at loses neglected. i)

I then calculated the reynolds number of the pipe by the formula ((denistyxVxD)/(dynamic viscosity)) giving me a reynolds number of 106143.75.

Then at k/D .004 and reynolds 1E5 on the moody diagram to find a friction factor,f, of .03 which i put into the formula headloss due to friction,hfmajor, = f(L/D)((V^2)/2(g)) giving me a headloss of 11.93 m which i put into the formula P1=(density)(g)(z2+hfmajor) to give me 165917.217 pa ii)

I agree with your calculations to this point.

I found the minor head loss,hfminor, by using =(k1+k2 etc.)((V^2)/2(g)) getting a headloss minor of 4.474 and plugging this back into the equation P1=(density)(g)(Z2 + hfmajor +hfminor) to get an answer of 209763.27 Pa iii)

I don't get the minor head loss of 4.474 m. You might want to check your arithmetic here.

 

[teegfit]

I rechecked my arithmetic and have calculated the same answer. I did this by:

(2(.25)+4(.5)+10+1)(((2.55)^2)/2(9.81)) = 4.4742 m
^(k values added) ^(velocity squared over 2 x gravity)

I could have this wrong

 

[SteamKing]

I rechecked my arithmetic and have calculated the same answer. I did this by:

(2(.25)+4(.5)+10+1)(((2.55)^2)/2(9.81)) = 4.4742 m
^(k values added) ^(velocity squared over 2 x gravity)

I could have this wrong

Nope, it was my mistake. I thought the open globe valve had a K = 1.0 instead of K = 10.0.