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Finite Volume Analysis: Application of Conservation of Energy

Example 1 (continued)

Group like terms in the energy equation from the previous screen.

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

Now (u₁ – u₂) dm/dt + [(p₁ – p₂)/ρ] dm/dt + ½ (V₁² - V₂²) dm/dt .

+ g(z₁– z₂) dm/dt = - W _{shaft net in}

[(u₂ – u₁)dm/dt = [(p₁ – p₂)/ρ] dm/dt + ½ (V₁² - V₂²) dm/dt .

+ g(z₁– z₂) dm/dt /+ W _{shaft net in}

dm/dt = ρQ = 1.94 (slugs/ft³) (150) (ft³/sec) = 291 slugs/sec

p₁ = 60 (psi) 144 in²/ft²) = 8640 lb/ft² and

p₂ = - 10 (in Hg) (13.6 sg)(1ft/12 in) = - 11.33 ft of water

So p₂ = - 11.33 ft (62.4 lb/ft³) = -707.2 lb/ft²

V₁ = Q/A₁ = 150 (cfs) / (π)(3²)/4 (ft²) = 21.22 ft/sec

V₂ = Q/A₂ = 150 (cfs) / (π)(4²)/4 (ft²) = 11.94 ft/sec

For the datum selected z₂ = 0 ft. So g z₂ = 0 and gz₁ = 32.2(10) = 322 (ft/sec²) ft

W _{shaft net in} = - 2500 hp (note: net shaft work by turbine is “out”)

Now (u₂ - u₁) dm/dt = power lost between section 1 and 2.