Question

A horizontal pipe is fitted with a nozzle. The inlet diameter of the nozzle is 40 mm and the outlet diameter is 20 mm. The flow rate in the pipe is 1.2 m3 /min and water density is 1000 kg/m3 . Determine the force exerted by the nozzle on the water.

Answer 1

Answer:

969.68N

Explanation:

d₁=0.04 m      A₁=$\frac{\pi d^2_{1} }{4}$

$A_{1} =\frac{\pi \times .04^2}{4}= 0.00125m^{2}$

d₂=0.02 m      A₂=$\frac{\pi d^2_{2} }{4}$

$A_{2} =\frac{\pi \times .02^2}{4}= 0.00031m^{2}$

Q=1.2m³/min        Q=1.2/60=0.02m³/s

using continuity equation

Q₁=A₁v₁

v₁=Q₁/A₁=0.02/0.00125=16m/s

Q₂=A₂v₂

v₂=Q₂/A₂=0.02/0.00031=64.5m/s

$F_{inlet}=\rho A_{1}v_1^{2}$

$F_{inlet}=1000\times 0.00125\times16^{2}=320N$

$F_{outlet}=\rho A_{2}v_2^{2}$

$F_{outlet}=1000\times 0.00031\times64.5^{2}=1289.68N$

Force on the nozzle=F_{outlet}-F_{inlet}

= 1289.68-320

=969.68N