Problem 4.76 - ⋅ − 12 13 p 4 ⋅ A 4 ⋅ V 1 2 ρ − V...

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Problem 4.76 [Difficulty: 3] Given: Flow into and out of CV Find: Expressions for rate of change of mass, and force Solution: Basic equations: Mass and momentum flux Assumptions: 1) Incompressible flow 2) Uniform flow For the mass equation dM CV dt CS ρ V A () + dM CV dt ρ V 1 A 1 V 2 A 2 V 3 A 3 + V 4 A 4 + () + = 0 = dM CV dt ρ V 1 A 1 V 2 A 2 + V 3 A 3 V 4 A 4 () = For the x momentum F x p 1 A 1 2 + 5 13 p 2 A 2 + 4 5 p 3 A 3 5 13 p 4 A 4 0 V 1 2 ρ V 1 A 1 () + 5 13 V 2 ρ V 2 A 2 () + 4 5 V 3 ρ V 3 A 3 () 5 13 V 3 ρ V 3 A 3 () + + ... = F x p 1 A 1 2 5 13 p 2 A 2 4 5 p 3 A 3 + 5 13 p 4 A 4 + ρ 1 2 V 1 2 A 1 5 13 V 2 2 A 2 4 5 V 3 2 A 3 + 5 13 V 3 2 A 3 + + = For the y momentum F y p 1 A 1 2 + 12 13 p 2 A 2 3 5 p 3 A 3
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Unformatted text preview: ⋅ − 12 13 p 4 ⋅ A 4 ⋅ + V 1 2 ρ − V 1 ⋅ A 1 ⋅ ( ) ⋅ + 12 13 V 2 ⋅ ρ − V 2 ⋅ A 2 ⋅ ( ) ⋅ − 3 5 V 3 ⋅ ρ V 3 ⋅ A 3 ⋅ ( ) ⋅ 12 13 V 3 ⋅ ρ V 3 ⋅ A 3 ⋅ ( ) ⋅ − + ... = F y p 1 A 1 ⋅ 2 − 12 13 p 2 ⋅ A 2 ⋅ + 3 5 p 3 ⋅ A 3 ⋅ + 12 13 p 4 ⋅ A 4 ⋅ − ρ 1 2 − V 1 2 ⋅ A 1 ⋅ 12 13 V 2 2 ⋅ A 2 ⋅ − 3 5 V 3 2 ⋅ A 3 ⋅ + 12 13 V 3 2 ⋅ A 3 ⋅ − ⎛ ⎝ ⎞ ⎠ ⋅ + =...
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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