Pipeflow Example              Type 3 Application      

 

Example 3  Water flows downward through a vertical, smooth pipe as shown in the figure below. 

When the flowrate is 0.5 ft3/sec, there is no change in pressure along the pipe.  Find the diameter

of the pipe.

                                          

 

Strategy:  Apply conservation of mass and of energy across stations 1 and 2 shown above.

 

For conservation of mass:  (Assume steady, uniform, incompressible flow.)

 

                     ρ1 A1 V1  = ρ2 A2 V2     which gives    V1  =  V2

 

For conservation of energy between stations 1 and 2 : 

 

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

 

For a vertical pipe:   z1  = L,   z2  =  0  and since no pressure change,  P1  =  P2

 

                                             L  =  HL  =  f (L/D) V2/2g

 

Now  V  =  Q/A  =  4Q/πD2   Cancel L.     So   1  =  f [ 4Q/πD2 ]2 / 2gD  =  8f Q2/ π2gD5

 

Giving           f  =  π2gD5 / 8Q2    Here there are two unknowns,  f  and  D.

 

Click here to continue with this problem.

 

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