Problem 4.95 - Problem 4.95 [Difficulty: 3] V1 V2 CS p1 p2...

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Problem 4.95 [Difficulty: 3] R x y x CS V 2 V 1 p 2 p 1 V 3 ρ 1 2 Given: Data on heated flow of gas Find: Force of gas on pipe Solution: Basic equation: Continuity, and momentum flux in x direction p ρ R T = Assumptions: 1) Steady flow 2) Uniform flow From continuity ρ 1 V 1 A 1 ρ 2 V 2 A 2 + m 3 + 0 = V 2 V 1 ρ 1 ρ 2 m 3 ρ 2 A = where m 3 = 20 kg/s is the mass leaving through the walls (the software does not allow a dot) V 2 170 m s 6 2.75 × 20 kg s m 3 2.75 kg × 1 0.15 m 2 × = V 2 322 m s = For x momentum R x p 1 A + p 2 A V 1 ρ 1 V 1 A
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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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