Finite Volume Analysis:   Application of Conservation of Linear Momentum   

 

Example 1  (continued)    

                                  

 

Thus for conservation of linear momentum   ∫  V ρ V ● n  dS   =   Σ Fx i

                                                                      cs

 

      ∫  V ρ V ● n  dS   =   ∫  V1 ρ V1 ● n1  dS1  +    ∫  V ρ V ● n2  dS2       where  V = (3/2) V1  i

    cs                           cs1                               cs2                                 and  dS2 = 2π r dr

 

                                                           r = d/2

     ∫  V ρ V ● n  dS  =  - ρ V12 π (d/2)2  +  ∫ (3/2)V1[(r/d/2)] i ρ (3/2)V1[(r/d/2)] i ● i  2 π r dr

    cs                                                     r = 0

 

     ∫  V ρ V ● n  dS  =   [ - ρ V12 π (d/2)2  +  ρ(9/32) V12  π d2  ] i    

 

                 Recall  V1  =  100 ft/sec     and   d = 2 ft    ( area = π ft2 )

 

and           Σ Fx i  =  ( P1 A1 – P2 A2 - FD ) i  =  [ (0.2)(144)π – (0.15)(144)π  -  FD  ] i

 

                  - ρ 1002 π +  ρ(36/32) 1002  π   =  (0.2)(144)π – (0.15)(144) π -  FD

 

Assume  ρ = constant = 2.38 x 10-3  slugs/ft3   (Taken from table on properties for air

at standard atmospheric pressure. )

 

                FD  =  13.3 lb  or  FD  =  - 13.3 i  lb  (result)

 

 

Click here for another example.

 

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