Question

Water discharges from a horizontal, cylindrical pipe at the rate of 5.00x10^-3 m^3/s. At a point in the pipe where the cross-section is 1.000x10^-3 m^2, the absolute pressure is 1.60x10^5 Pa. What is the pipe radius at a constriction if the pressure there is reduced to 1.20*10^5 Pa?

I know that we are to find the velocity first, which I got 8.944 m/s approximately. Then after I found the radius and the answer I got was 0.0000236, is this correct ?

Answer 1

As per theorem of continuity

Av = constant = Rate of flow

now v = Rate/Area

v=5.00 x 10–³/1 x 10–³ = 5m/s at point I

also by Bernoulli theorem

P + 1/2dv² + dgh = constant

hence P2 - P1 = 1/2d(v2²- v1²)

1.60 x 10^5 - 1.2 x 10^5 = 1/2 x 10³ (v2²- v1²)

8 x 10 = v2²- 25

v2 = 105sq root

now A1v1 = A2v2 again we get A² as

0.0000236