Fluid-III-solutions

Fluid-III-solutions - le (txl65) Fluid-III mohanty (81440)...

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le (txl65) – Fluid-III – mohanty – (81440) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. 001 10.0 points A vessel is flled with a liquid o± density 1300 kg / m 3 . There are two holes (one above the other) in the side o± the vessel. Liquid streams out o± both o± these holes in a hori- zontal direction. The upper hole is a distance 90 m below the sur±ace o± the liquid and the lower hole is a distance 107 m below the sur- ±ace o± the liquid. The acceleration o± gravity is 9 . 8 m / s 2 . At what horizontal distance ±rom the vessel does the stream ±rom the upper hole collide with the stream ±rom the lower hole? Correct answer: 196 . 265 m. Explanation: Let : y 1 = 90 m , y 2 = 107 m , and g = 9 . 8 m / s 2 . x h 90 m 107 m The fgure is not drawn to scale. Since the pressure on the sur±ace o± the liquid and at the exit holes are approximately the same, Bernoulli’s equation gives us 1 2 ρ v 2 = ρ g y , so the velocity o± the liquid as it emerges ±rom the hole is v x = r 2 g y , where y is the distance o± the hole below the sur±ace o± the liquid. For each hole, we have x = v t , so h y = 1 2 g t 2 = 1 2 g x 2 v 2 = 1 2 g x 2 2 g y 4 y ( h y ) = x 2 . Thus y 1 ( h y 1 ) = y 2 ( h y 2 ) ( y 2 y 1 ) h = y 2 2 y 2 1 = ( y 2 + y 1 ) ( y 2 y 1 ) h = ( y 2 + y 1 ) and x = r 4 y 1 ( y 2 + y 1 y 1 ) = 2 y 1 y 2 = 2 r (90 m) (107 m) = 196 . 265 m . 002 10.0 points Water comes down a spillway o± a dam ±rom an initial vertical height o± 174 m . What is the velocity o± the water at the end o± the spillway? (Neglect ±rictional losses.) The acceleration o± gravity is 9 . 8 m / s 2 . Correct answer: 58 . 3986 m / s. Explanation: Let : h = 174 m and g = 9 . 8 m / s 2 . Applying Bernoulli’s equation, P + 1 2 ρ v 2 + ρ g y = constant , so P 1 + 1 2 ρ v 2 1 + ρ g y 1 = P 2 + 1 2 ρ v 2 2 + ρ g y 2 .
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le (txl65) – Fluid-III – mohanty – (81440) 2 Since P 1 = P 2 , v 1 = 0, v 2 = v , and y 1 y 2 = h , ρ g y 1 = 1 2 ρ v 2 2 + ρ g y 2 v 2 = 2 g ( y 1 y 2 ) v = r 2 g h = R 2 (9 . 8 m / s 2 ) (174 m) = 58 . 3986 m / s . 003 10.0 points Consider an airplane of wing area 20 m 2 (both wings combined) ±ying through air of density 0 . 462 kg / m 3 . From the plane’s point of view, the air ±owing above the plane’s wings has speed 242 m / s , but the air ±owing below the wings has speed 210 m / s . What is the net upward lift force on the plane? Assume laminar air ±ow both above and below the plane’s wings. The plane is ±ying horizontally. Correct answer: 66 . 8237 kN. Explanation: Let : ρ = 0 . 462 kg / m 3 , v (1) wings = 242 m / s , v (2) wings = 210 m / s , and A wings = 20 m 2 .
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This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.

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Fluid-III-solutions - le (txl65) Fluid-III mohanty (81440)...

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