Finite Volume Analysis:   Application of Conservation of Linear Momentum   

 

Example 2  (continued)   For the given data (as listed below):

            

                    Vr = 20 ft/sec    r = 8 inches = 2/3 ft,     t = 0.5 inches  =  1/24 ft 

 

And for water    ρ  =  1.94 slugs/ft3

 

 

Step 2

Write the equation for conservation of linear momentum and simplify for known values.  Check for steady flow.  Include the anchoring force, Fy.

Step 3

Solve for desired anchoring force.

 

                              

For conservation of linear momentum:   d/dt  ∫ ρ V dV  +   ∫ ρ V (V●n) dS  =  Σ F

                                                                        cv               cs

Since the flow is steady the first integral is zero.

 

Now  V  =  Vr er  = Vr ( sin θ  i – cos θ j ) where  n = er      so  V ● n  =  Vr er ● er   =  Vr

 

Also    Σ F  =  Fx i +  Fy j    and   dS  =   r dθ t

 

                                      π/2                                                                                  π/2

      ∫ ρ V (V●n) dS  =  ρ ∫ Vr ( sin θ  i – cos θ j ) Vr  r dθ t  =  ρ (Vr )2 (- sin θ )| r t  j

     cs                             -π/2                                                                                 -π/2

 

Therefore   Fx  = 0  (note symmetry)  and  Fy  =  - 2 ρ (Vr )2 r t

 

            Fy  =  - 2 (1.94)  (20 )2 (2/3) (1/24)  =  - 43.11 lb      (result)

 

Click here for another example.

 

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