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Finite Volume Analysis: Application of Conservation of Energy

Example 1 (continued)

Step 2b

(continued)

Simplify for steady, uniform, incompressible flow with one

stream entering and one stream exiting the control surface.

For steady flow ∂/∂t ∫ e ρ dV = ∂/∂t ∫ [ u + p/ρ + ½ V² + gz ] ρ dV = 0

cv cv

For no heat transfer across the control surface Q _{net in =} 0. The energy equation

reduces to: .

∫ e ρ V ● n dS = W _{shaft net in} where e = u + p/ρ + ½ V² + gz

∫ e ρ V₁ ● n₁ dS + ∫ e ρ V₂● n₂ dS = W _{shaft net in}

cs1 cs2

Now V₁ ● n₁ = - V₁ and V₂● n₂ = V₂

Since uniform, incompressible flow with one stream entering the control surface and

one stream leaving the control surface

∫ ρ V₁ dS = ρ V₁ A₁ = ρ Q and ∫ ρ V₂ dS = ρ V₂ A₂ = ρ Q = dm/dt
cs1 cs2

Here dm/dt is the mass flow rate in slugs/sec

- [u₁ + p₁/ρ + ½ V₁² + gz₁ ] dm/dt + [u₂+ p₂/ρ + ½ V₂² + gz₂ ] dm/dt = W _{shaft net
in}

where terms on the left side are (ft lb / slug) (slugs/sec) = ft lb / sec

and the net shaft power in is W _{shaft net in} is in ft lb / sec