Problem 4.99

# Problem 4.99 - Problem 4.99 [Difficulty: 4] Given: Data on...

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Problem 4.99 [Difficulty: 4] Given: Data on flow in wind tunnel Find: Mass flow rate in tunnel; Maximum velocity at section 2; Drag on object Solution: Basic equations: Continuity, and momentum flux in x direction; ideal gas equation p ρ R T = Assumptions: 1) Steady flow 2) Uniform density at each section From continuity m flow ρ 1 V 1 A 1 = ρ 1 V 1 π D 1 2 4 = where m flow is the mass flow rate We take ambient conditions for the air density ρ air p atm R air T atm = ρ air 101000 N m 2 kg K 286.9 N m × 1 293 K × = ρ air 1.2 kg m 3 = m flow 1.2 kg m 3 12.5 × m s π 0.75 m () 2 4 × = m flow 6.63 kg s = Also m flow A 2 ρ 2 u 2 d = ρ air 0 R r V max r R 2 π r d = 2 π ρ air V max R 0 R r r 2 d = 2 π ρ air V max R 2 3 = V max 3m flow 2 π ρ air R 2 = V max 3 2 π 6.63 × kg s m 3 1.2 kg × 1 0.375 m 2 × = V max 18.8 m s = For x momentum R x p 1 A + p 2 A V 1 ρ 1 V 1 A A 2 ρ 2 u 2 u 2 d + = R x p 2 p 1 A V 1 m flow 0 R r ρ air V max r R 2 2 π r
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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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