Problem 4.93

# Problem 4.93 - Problem 4.93[Difficulty 3 V1 V2 CS p1 p2 Rx...

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Problem 4.93 [Difficulty: 3] R x y x CS V 2 V 1 p 2 p 1 Given: Data on adiabatic flow of air Find: Force of air on pipe Solution: Basic equation: Continuity, and momentum flux in x direction, plus ideal gas equation p ρ R T = Assumptions: 1) Steady flow 2) Ideal gas CV 3) Uniform flow From continuity ρ 1 V 1 A 1 ρ 2 V 2 A 2 + 0 = ρ 1 V 1 A ρ 2 V 2 A = ρ 1 V 1 ρ 2 V 2 = For x momentum R x p 1 A + p 2 A V 1 ρ 1 V 1 A () V 2 ρ 2 V 2 A + = ρ 1 V 1 A V 2 V 1 = R x p 2 p 1 A ρ 1 V 1 A V 2 V 1 + = For the air ρ 1 P 1 R air T 1 = ρ 1 200 101 + 1 0 3 × N m 2 kg K 286.9 N m × 1 60 273 + K × = ρ 1 3.15 kg
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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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