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water is flowing through a horizontal pipe of varying cross section. At any two places the diameter of tube are 4 cm and 2 cm,if pressure difference between these 2 palaces is 4.5 cm (water) ,then determine rate of flow
let , r1= 2cm r1= 2x 10^-2 m, r2= 1x 10^-2 m, rho = 1000 kg/m^3, pressure = rho x g x h => 10^3 x 10 x 4.5 x10^-2, = 450 Pa . according to equation of continuty, A1V1= A2V2, (22/7 x 4 x 10^-4) x V1=(22/7 x 1 x 10^-4) x V2, V2/V1=4. (i) Now using bernoulli’s equation: P1 + ½ x rho x (V1)^2 = P2 + ½ x rho x (V2)^2 P1 – P2 = ½ x16x (V1)^2 – ½ x rho (V1)^2 [ V1 = 4xV2 ] ( using (i) equation) 450 = 15 x rho x (V1)^2 / 2 900 = 3x 10^3 x V1^2 (900/15000??) ^ ½ = V1 V1=24.4 x 10^-2 m/s The rate of flow = V x A = 24.4 x 10^-2 x 22/7 x (2 x 10^-2)^2 = 306.464 x 10^-2 cm^3/s(approx)
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