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

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

Step-by-step explanation:

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

`v=√(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=√(2* 9.81* 90)=42.02m/s`

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

`Q=V* Area`

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

`Area=(Q)/(v)=(2)/(42.02)=0.0476m^(2)`

The circular area is related to diameter as

`Area=(\pi D^(2))/(4)\\\\\therefore D=\sqrt{(4\cdot A)/(\pi )}=\sqrt{(4* 0.0476)/(\pi )}=0.246m`

Thus the diameter of the nozzle is 0.246 meters