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Grade: 11
        
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
5 months ago

Answers : (1)

giridharan
38 Points
							
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)
5 months ago
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