Question

Through a horizontal pipe of varying cross section at any two places the diameter of the tuber 4 cm and 2 cm if the pressure difference between these two places be equal to 4.5 cm water then determine the rate of flow of water in a tube

Answer 1

The rate of flow of water is 0.0003 m^3 / s

Explanation:

Let the velocity at first place  =  v1

Velocity at second place =  v2

So ρv1^2 / 2 = ρv2^2 / 2 + ρgh

So v1^2 =  v2^2 + 2gh--------(1)

Now

π d1^2 / 4 v1 = π d2^2 / 4 v2

π d2^2 / 4 v2 = π d2^1 / 4 v1

Or v2 = v1 (d1 / d2)^2

Substituting the values in equation 1 we get

So v1^2 = v1^2 (d1 / d2)^4 + 2gh

V1^2(1 – (d1/d2)^4) = 2gh

Now v1^2 = 2gh / 1 – (d1/d2)^4

Rate of flow of water = d v/dt

= πd1^2 / 4 v1

= πd1^2 / 4√2gh / 1 – (d1/d2)^4

d v/dt = π (0.02)^2 / 4 √ 2(9.8)(0.045) / 1 – (2/4)^4

= 0.0003 m^3 / s

Thus the rate of flow of water is  0.0003 m^3 / s

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