Water flows in a horizontal tube as shown in figure. The pressure of water changes by `600 N//m^(2)` between `x` and `y` where the areas of cross-section are `3cm^(2)` and `1.5cm^(2)` respectively. Find the rate of flow of water through the tube.

Explanation

Let the velocity at `x = v_(x)` and that at `y = v_(y)`.
By the equation of continuity, `(v_(y))/(v_(x)) = (3cm^(2))/(1.5cm^(2)) = 2`.
By Bernoulli's equation,
`P_(x) + (1)/(2) rhov_(x)^(2) = P_(y) + (1)/(2)rhov_(y)^(2)` or, `P_(x) - P_(y) = (1)/(2)rho(2V_(y))^(2) - (1)/(2)rhov_(y)^(2) = (3)/(2)rhoV_(y)^(2)`
or, `600 (N)/(m^(2)) = (3)/(2)(1000(kg)/(m^(3)))v_(x)^(2)`
or, `v_(x) = sqrt(0.4 m^(2) // s^(2)) = 0.63 m//s`.
The rate of flow `= (3 cm^(2)) (0.63 m//s) = 189 cm^(3)//s`.