Question

An isentropic nozzle receives water at 10 bar, 400 C, and 20 m/s and discharges it to 5 bar. The inlet diameter of the nozzle is 0.01 m. The outlet diameter of the nozzle is 0.05m.

Determine the outlet velocity, in m/s.

Answer 1

Data provided

The Diameter at the inlet of a nozzle ( D1) =0.01 m

C/S area of inlet of nozzle ( A1) = ( 3.14 * 0.012 ) / 4 = 7.85 * 10-5 m2

Velocity at the inlet of a nozzle ( V1) = 20 m/s

The Diameter at the outlet of a nozzle ( D2) =0.05 m

C/S area of outlet of nozzle ( A2) = ( 3.14 * 0.052 ) / 4 = 1.9625 * 10-3 m2

Simply applying continuity equation between inlet and outlet i.e point 1 and point 2

A1 * V1 = A2 * V2

Here V2 is velocity at outlet of nozzle

V2 = ( A1 * V1) / A2

Putting all values

V2 = ( 7.85 * 10-5 m2 * 20 m/s ) /  1.9625 * 10-3 m2

we get

Velocity at exit of nozzle = V2 = 0.8 m / s