homework2_solns - Homework 2 solutions Fluid Mechanics CE...

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Homework 2 solutions Fluid Mechanics CE 3502 Fall, 2008 (1) The velocity of water flow in the nozzle shown is given by the following expression: where V= velocity in meters per second, t = time in seconds, x = distance along the nozzle in m, and L = length of the nozzle in m. Find: What is the total acceleration and what is the pressure gradient dp/dx when t = 2 s, at x= 0.25 L ? Solution: t v 0 0 x v v t v z v v y v v x v v dt dv a x x x x x z x y x x x x + + + = + + + = = () 2 x L x 5 . 0 1 t 2 v = ; 2 x L x 5 . 0 1 2 t v = ; ( ) L 5 . 0 L x 5 . 0 1 2 t 2 x v 3 x = when t = 2 s, at x= 0.25 L () ( ) s / m 224 . 5 49 / 256 8 / 7 4 25 . 0 5 . 0 1 2 2 v 2 2 x = = = = 2 2 2 x s / m 612 . 2 49 / 128 8 / 7 2 25 . 0 5 . 0 1 2 t v = = = = ( ) s / 1 985 . 2 7 / 8 * 2 2 5 . 0 8 / 7 2 2 2 x v 3 3 x = = = ( ) 2 x 2 2 2 x s / m 2 . 18 a s / m 207 . 18 s / m 9851 . 2 s / m 985 . 2 224 . 5 a = = + = Euler: x a z p x ρ = γ + ; Since we are not given the temperature, I will assume T =4°C. 0 x z = ; ( ) m / kPa 2 . 18 x p s / m 2 . 18 m / kg 1000 0 p x 2 3 = = + (2) The velocity in the outlet pipe from this reservoir is 5 m/s and h = 15 m. Because of the rounded entrance to the pipe, the flow is assumed to be irrotational. Find: Under these conditions, what is the pressure at A ? Please state any other assumptions you make to solve this problem. Solution: Consider point B, on the same streamline as A. Bernoulli’s equation: 2 v z p 2 v z p 2 B B B 2 A A A ρ + γ + = ρ + γ + (1) : s / m 5 v A = ; m 15 z z A B = Assume the cross sectional area of the tank is much less than the cross sectional area of the pipe. Continuity means if A B A A >> then A B v v << , so we neglect the velocity at B: 0 v B Plugging this information in (1): () () ( ) 0 m 15 m / N 9810 p 2 / s / m 5 m / g k 1000 p
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homework2_solns - Homework 2 solutions Fluid Mechanics CE...

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