Question

Water is flowing in a straight horizontal pipe of variable cross section. Where the cross-sectional area of the pipe is 3.70·10-2 m2, the pressure is 6.10·105 Pa and the velocity is 0.260 m/s. In a constricted region where the area is 9.50·10-4 m2, what is the velocity?

Answer 1

v = 10.1 m/s

Step-by-step explanation:

As we know that by the law of conservation of volume the rate of volume flowing through the pipe will remain conserved

so here we have flow rate given as

`Q = Area* velocity`

now we have

`A_1 v_1 = A_2 v_2`

now we have

`A_1 = 3.70 * 10^(-2) m^2`

`v_1 = 0.260 m/s`

`A_2 = 9.50 * 10^(-4) m^2`

now from above equation we have

`v_2 = (A_1)/(A_2) v_1`

`v_2 = (3.70* 10^(-2))/(9.50* 10^(-4))(0.260)`

`v_2 = 10.1 m/s`