Finite Volume Analysis:   Application of Conservation of Linear Momentum     

 

Example 3  (continued)

 

Use the dot product et  =  cos θ i  +  sin θ j   with (2B).    

 

   ˗  ρ  Vj2 A1 i  +  ρ  Vj2 A2 ( cos θ i  +  sin θ j ) ˗ ρ  Vj2 A3 ( cos θ i  +  sin θ j ) = R n    (2B)

                     

  The result is:                ˗  ρ  Vj2 A1 cos θ + ρ  Vj2 A2  ˗  ρ  Vj2 A3  =  0                            (3)

 

Note:   dm/dt  =  ρ  Vj A

 

So                                             ˗  dm1/dt cos θ +  dm2/dt ˗ dm3/dt  =  0

                                                                           

And from conservation of mass:   dm1/dt  =  dm2/dt  +  dm3/dt    or

 

                                                                   dm2/dt   =  dm1/dt   ˗  dm3/dt

 

    dm1/dt cos θ =  dm1/dt   ˗  dm3/dt ˗ dm3/dt  = 

 

                          dm1/dt   ˗ dm1/dt cos θ  =  2 dm3/dt 

 

                           dm3/dt  =  (1/2) dm1/dt [ 1 ˗ cos θ ]                                               (result)

 

       Now     dm2/dt  =  dm1/dt  ˗  dm3/dt  =  dm1/dt ˗ (1/2) dm1/dt [ 1 ˗ cos θ ]    

 

 Or                     dm2/dt  =  (1/2) dm1/dt [ 1 + cos θ ]                                                (result)

 

Note:          As expected     dm2/dt  >  dm3/dt

 

Also note:   If   θ  =  90o ,   dm2/dt  =  dm3/dt  =  (1/2) dm1/dt    as expected.

 

And also    If  θ  =  0o ,   dm2/dt  =  dm1/dt    and   dm3/dt  =  0  as expected.

                        

 

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