Question

A large water jet with a discharge of 2m^3 /s rises 90m above the ground. The exit nozzle diameter to achieve this must be. (a) 0.246m (b) 0.318m (c) 1.3m (d) 0.052m (e) None of the above

Answer 1

Answer:

The correct answer is option 'a': 0.046 meters.

Explanation:

We know that the exit velocity of a jet of water is given by Torricelli's law as

$v=\sqrt{2gh}$

where

'v' is velocity of head

'g' is acceleration due to gravity

'h' is the head under which the jet falls

Now since the jet rises to a head of 90 meters above ground thus from conservation of energy principle it must have fallen through a head of 90 meters.

Applying the values in above equation we get the exit velocity as

$v=\sqrt{2\times 9.81\times 90}=42.02m/s$

now we know the relation between discharge and velocity as dictated by contuinity equation is

$Q=V\times Area$

Applying values in the above equation and solving for area we get

$Area=\frac{Q}{v}=\frac{2}{42.02}=0.0476m^{2}$

The circular area is related to diameter as

$Area=\frac{\pi D^{2}}{4}\\\\\therefore D=\sqrt{\frac{4\cdot A}{\pi }}=\sqrt{\frac{4\times 0.0476}{\pi }}=0.246m$

Thus the diameter of the nozzle is 0.246 meters