47617_cive1400-200102 - This question paper consists of 3...

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Turn Over 1 This question paper consists of 3 printed pages, each of which is identified by the Code Number CIVE 140001 ª UNIVERSITY OF LEEDS May/June 2002 Examination for the degree of BEng/ MEng Civil Engineering FLUID MECHANICS Time allowed: 2 hours Attempt 4 questions Useful formulae: Parallel axis theorem 2 x A I I GG oo + = , 2 nd moment of area for a rectangle, 12 3 bd I GG = , and for a triangle 36 3 bd I GG = 1. (a) A “U”-tube manometer containing mercury of density 13600 kg/m 3 is used to measure the pressure drop along a horizontal pipe. If the fluid in the pipe has a relative density of 0.8 and the manometer reading is 0.6m, what is the pressure difference measured by the manometer? [8 marks] (b) A tank with vertical sides is filled with water to a depth of 4.0m. The water is covered with a layer of oil 0.5m thick. If the relative density of the oil is 0.8, find the resultant force (per unit width) and its line of action on the wall of the tank. [9 marks] (c) Water is being fired at 20 m/s from a hose of 80mm diameter into the atmosphere. The water leaves the hose through a nozzle with a diameter of 25mm at its exit. Find the pressure just upstream of the nozzle and the force on the nozzle. [8 marks] 1.a
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Turn Over 2 datum A B a b c d e Manometric fluid density ρ Fluid density ρ w Hg h Figure of manometer setup density of mercury ρ = 13000 kg/m 3 pressure at C and D is equal: p C = p D p A + ρ w g a = p B + ρ w g ( b - h ) + ρ Hg g h p A - p B = ρ w g b - ρ w g h - ρ w g a + ρ Hg g h = ρ w g ( b - a ) + hg ( ρ Hg - ρ w ) As horizontal a = b p A - p B = hg ( ρ Hg - ρ w ) = 0.6 × 9.81 × ( 13600 - 800 ) = 75 340 N/m 2 = 75.34 kN/m 2 1.b ρ o = 800 kg/m 3 ρ w = 1000 kg/m 3 h 1 = 0.5m, h 2 = 4.0m Disatnce of centroids of pressure diagram from surface L 1 = 0.5 *2/3 = 0.333m L 2 = 2.5m L 3 = 0.5 + 4*2/3 = 3.1667m Areas of pressure diagram A 1 = 800 × 0.5 × 9.81 × 0.5 × 0.5 = 981 A 2 = 1000 × 4.0 × 9.81 × 4.0 × 0.5 = 78480 A 3 = 800 × 0.5 × 9.81 × 4.0 = 15696 Total area = resultant focre per unitwidth = R = 95157 N Take moment about surface to find point of action L R
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Turn Over 3 R L R = A 1 L 1 + A 2 L 2 + A 3 L 3 L R = 2.587 m 1.c 1 2 u 1 = unknown d 1 = 0.08m u 2 = 20 m/s d 2 = 0.025m Use continuity to calculate the unknown velocity s m d d u u u d u d u A u A / 953 . 1 08 . 0 025 . 0 20 4 4 2 2 1 2 2 1 2 2 2 1 2 1 2 2 1 1 = = = = = p p Area of the jet 2 2 2 2 00049 . 0 4 025 . 0 14159 . 3 4 m d A = × = = p Force on the water: ( 29 ( 29 ( 29 jet) the to direction opposite in the ( 86 . 176 ) 953 . 1 0 . 0 2 ( 0 . 20 00049 . 0 1000 1 2 1 1 2 1 2 N u u Au u u Q u u m F T = - × × = - = - = - = r r Force on the wall = R = -F = -176 N (in the direction of the jet) Pressure force Use Bernoulli to calculate the unknown pressure 2 2 2 2 1 2 1 1 2 2 z g u g p z g u g p + + = + + r r As horizontal z 1 = z 2 p 2 = atmospheric = 0.0
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Turn Over 4 ( 29 ( 29 2 2 2 2 2 1 2 2 1 2 2 2
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47617_cive1400-200102 - This question paper consists of 3...

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