Finite Volume Analysis: Application of Conservation of Linear Momentum

Example 1 (continued)

The axisymmetric velocity profile at section 2 is

V [ ( r / d/2) ] i and n₂ = i and where V is currently unknown.

Also note the expression for the element of area, dS = 2π r dr so that

r = d/2

∫ ρ V₂ ● n₂ dS = ∫ ρ V [ ( r / d/2) ] 2π r dr = (1/6) V π ρ d²

cs₂ r = 0

Thus for conservation of mass - ρ V₁ (π d²/4) + (1/6) V π ρ d² = 0 and

V = (3/2) V₁ V₁ = 100 ft/sec so V = 150 ft/sec (result)

Continue with the solution by:

Step 3

Write the equation for conservation of linear momentum and simplify for known values. Check for steady flow. Include pressure forces and the drag force by the screen.

d/dt ∫ V ρ dV + ∫ V ρ V ● n dS = Σ F_x i

cv cs

For steady flow the first term is zero leaving ∫ V ρ V ● n dS = 0

cs

Click here to continue with this example.

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