Pipeflow Example (Type 1 conditions – with a pump)                Click here for another example.  

 

Example 1  At a ski resort, water at 40o F is pumped through a 3 in. diameter, 2000 ft long steel

pipe from a pond at an elevation of 4286 ft to a snow-making machine at an elevation of 4623 ft

at a rate of 0.26 ft3/sec.  It is necessary to maintain a pressure of 180 psi at the snow-making

machine.  See the figure below.  Find the horsepower added to the water by the pump.  Neglect

minor losses.

                         

 

Strategy:  Apply the energy equation between stations 1 and 2.     (all terms in ft)

 

                     P1/γ  +  V12/2g  +  z1  +  WP  =  P2/γ  +  V22/2g  +  z2  +  HL 

 

At station 1:  P1  =  0  psfg,  V1  =  0  ft/sec,  z1  =  4286 ft

 

So           WP  =  P2/γ  +  V22/2g  +  z2 - z1  +  HLmajor  +  HLminor  ,            HLminor  =   0

 

At station 2:    P2  =  (180 psi)(144 in2/ft2)  =     25,920 psfg,  γ = 62.4 lb/ft3  P2 /γ = 415.4 ft

 

  V2  =  Q/A  =  V  = 0.26 ft3/sec) / [π(1/4)2/4] = 5.2964 ft/sec,  V2/2g = 0.43 ft

 

                       z2  - z1 =  337 ft,   HL  =  HLmajor  = [f L/D] (V2/2g)

 

Use Reynolds number, relative roughness, and Moody chart to determine the friction factor, f.   

Rey  =  VD/ν  =  (5.2964)((1/4)/(1.664x10-5)  =   79600   

For steel pipe  ε/D  =  0.00015/(1/4) = 0.0006   From the Moody Chart,    f  =  0.0215  

 

Thus  WP  =  415.4 + 0.43  +  337  +  0.0215(2000/0.25)(0.43)  =  826.8 ft

 

Now   WP (γQ)  =  (826.8 ft)(62.4 lb/ft3) x 0.26 ft3/sec) (1 hp/ 550 ft lb/sec)  =  24.4 hp (result)

 

Return to Notes on Fluid Mechanics


Copyright © 2019 Richard C. Coddington
All rights reserved.