Engr Help

Finite Volume Analysis: Application of Conservation of Energy

Example 1 (continued)

Step 3

Solve for the energy lost between sections 1 and 2.

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

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

Now (u₂ - u₁) dm/dt = power lost = PL

Conversion to hp: (1 hp = 550 ft lb/ sec)

[(p₁ – p₂)/ρ] dm/dt = { 8640 – (- 707.2)] / 1.94 } 291 = 1402080 ft lb/sec = 2549.2 hp

½ (V₁² - V₂²) dm/dt = ½ ( 21.22² – 11.94²) 291 = 44789.8 ft lb / sec = 81.44 hp

gz₁ dm/dt = 322 (291) = 93702 ft lb / sec = 170.37 hp

W _{shaft net in} = - 2500 hp

[(u₂ – u₁)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 = 2549.2 + 81.44 + 170.37 – 2500 = 301 hp (result)