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Finite Volume Analysis: Application of Conservation of Linear Momentum

Example 3 (continued)

The equation for conservation of linear momentum becomes:

∂/∂t ∫ ρ V dV + ∫ V ρ V • n dS = ∑ F

cv cs

V₁ ρ V₁ • n₁ A₁ + V₂ ρ V₂ • n₂ A₂ + V₃ ρ V₃ • n₃ A₃ = R = R n (2A)

Now V₁ = V₂ = V₃ = V_j = speed of the air jet

and recall n₁ = ˗ i , n₂ = cos θ i + sin θ j , n₃ = ˗ cos θ i ˗ sin θ j

˗ ρ V_j² A₁ i + ρ V_j² A₂ ( cos θ i + sin θ j ) ˗ ρ V_j² A₃ ( cos θ i + sin θ j ) = R n (2B)

Take the dot product with the unit normal, n, on both sides to get the normal component

of the linear momentum. Recall n = ˗ sin θ i + cos θ j

˗ ρ V_j² A₁ sin θ + ρ V_j² ( ˗ sin θ cos θ + sin θ cos θ ) A₂

˗ ρ V_j² ( ˗ sin θ cos θ + sin θ cos θ ) A₃ = R

The result is: R = ρ V_j² A₁ sin θ

To obtain the mass rate of air exiting the control surface, calculate the tangential component

of the linear momentum again using the dot product. In this case use the unit vector in

the tangential direction (along the plate is e_t = cos θ i + sin θ j .