Fluid-III-solutions

# Fluid-III-solutions - le(txl65 Fluid-III mohanty(81440 This...

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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 – find all choices before answering. 001 10.0points A vessel is filled with a liquid of density 1300 kg / m 3 . There are two holes (one above the other) in the side of the vessel. Liquid streams out of both of these holes in a hori- zontal direction. The upper hole is a distance 90 m below the surface of the liquid and the lower hole is a distance 107 m below the sur- face of the liquid. The acceleration of gravity is 9 . 8 m / s 2 . At what horizontal distance from the vessel does the stream from the upper hole collide with the stream from 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 figure is not drawn to scale. Since the pressure on the surface of 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 of the liquid as it emerges from the hole is v x = radicalbig 2 g y , where y is the distance of the hole below the surface of 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 = radicalbig 4 y 1 ( y 2 + y 1 y 1 ) = 2 y 1 y 2 = 2 radicalbig (90 m) (107 m) = 196 . 265 m . 002 10.0points Water comes down a spillway of a dam from an initial vertical height of 174 m . What is the velocity of the water at the end of the spillway? (Neglect frictional losses.) The acceleration of 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 = radicalbig 2 g h = radicalBig 2 (9 . 8 m / s 2 ) (174 m) = 58 . 3986 m / s . 003 10.0points Consider an airplane of wing area 20 m 2 (both wings combined) flying through air of density 0 . 462 kg / m 3 . From the plane’s point of view, the air flowing above the plane’s wings has speed 242 m / s , but the air flowing below the wings has speed 210 m / s . What is the net upward lift force on the plane? Assume laminar air flow both above and below the plane’s wings. The plane is flying 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 . In a laminar flow, the Bernoulli equations apply to the individual streamlines. Work in the plane’s frame of reference, and consider two separate airstreams: stream (1) flowing just over the wings, and stream (2) flowing just below the wings. Both streams are horizontal, so the Bernoulli equations for the two streams do not have ρ g Δ h terms, but only P (1) wings + ρ 2 parenleftBig v (1) wings parenrightBig 2 = P (1) ahead + ρ 2
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