Pipeflow Example     Type 1 Application      (continued)

 

Example  

                        

 

 

Next evaluate the head loss, HL, both the major and the minor head loss.

 

 HL  =  ( f L/D) (V2/2g)  +  KL (V2/2g)  =  [ f L/D + KL ] (V2/2g) 

 

where  f  = 0.03, L = 60 ft,  D = 0.1 ft,  KL = 12

 

Note: Since the friction factor, f, is given, there is no need for iteration.

 

Now   V  =  Q/A  =  Q / ( π D2/4 )  =  4 Q / (π D2)  =  400 Q / π

 

and    HL  =  [ (0.3)(60)(1/10) + 12 ] [400 Q / π ]2  [ 1/64.4 ]

 

Recall from conservation of energy:       HP  =    HL  

 

        So           1.7628/Q  =  [ (0.3)(60)(1/10) + 12 ] [400 Q / π ]2 

 

       or           Q3  =  ( 1.7628) / ( 7551.889 )

 

                                Q  =  0.0615 ft3 / sec     (result)

 

Click here to return to section on Pipe Flow Types.

 

 

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