The head of water from the centre of the orifice fitted to a tank is maintained at 6 m of water. The diameter of the orifice is 150 mm. The tank is fitted with frictionless wheels at the bottom and the tank is moving with a velocity of 4 m/s due to the reaction of the jet coming out from the orifice.

Jet Propulsion of a Tank

Problem Statement

The head of water from the centre of the orifice fitted to a tank is maintained at 6 m of water. The diameter of the orifice is 150 mm. The tank is fitted with frictionless wheels at the bottom and the tank is moving with a velocity of 4 m/s due to the reaction of the jet coming out from the orifice. Determine : (i) propelling force on the tank, (ii) work done per second, and (iii) efficiency of propulsion.

Given Data & Constants

Head, $H = 6 \, \text{m}$
Orifice diameter, $d = 150 \, \text{mm} = 0.15 \, \text{m}$
Velocity of tank, $u = 4 \, \text{m/s}$
Density of water, $\rho = 1000 \, \text{kg/m}^3$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Solution

1. Calculate Jet Velocity

First, we find the theoretical velocity of the jet coming out of the orifice. Assuming an ideal orifice ($C_v = 1.0$).

$$ \text{Theoretical Velocity, } V_{th} = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 6} \approx 10.85 \, \text{m/s} $$

The relative velocity of the jet with respect to the tank is the difference between the jet's absolute velocity and the tank's velocity.

$$ \text{Relative Velocity, } V_r = V_{th} - u = 10.85 - 4 = 6.85 \, \text{m/s} $$

(i) Propelling Force on the Tank

The force is the rate of change of momentum, which is the mass flow rate multiplied by the relative velocity.

$$ \text{Area of orifice, } A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.15)^2 \approx 0.01767 \, \text{m}^2 $$ $$ \text{Mass flow rate, } \dot{m} = \rho A V_{th} = 1000 \times 0.01767 \times 10.85 \approx 191.72 \, \text{kg/s} $$ $$ \text{Force, } F = \dot{m} \times V_r = 191.72 \, \text{kg/s} \times 6.85 \, \text{m/s} \approx 1313.3 \, \text{N} $$

(ii) Work Done per Second (Power)

Work done per second is the propelling force multiplied by the velocity of the tank.

$$ \text{Power} = F \times u = 1313.3 \, \text{N} \times 4 \, \text{m/s} = 5253.2 \, \text{W} $$

(iii) Efficiency of Propulsion

Efficiency is the ratio of the useful power (work done on the tank) to the kinetic energy of the exiting jet relative to the tank.

$$ \text{Kinetic Energy of Jet per second} = \frac{1}{2} \dot{m} V_r^2 = \frac{1}{2} \times 191.72 \times (6.85)^2 \approx 4498 \, \text{W} $$ $$ \text{Efficiency, } \eta = \frac{\text{Work Done per second}}{\text{Work Done per second} + \text{KE of Jet per second}} = \frac{5253.2}{5253.2 + 4498} $$ $$ \eta = \frac{5253.2}{9751.2} \approx 0.5387 $$

Final Results:

(i) Propelling force on the tank: $ \approx 1313.3 \, \text{N} $

(ii) Work done per second: $ \approx 5.25 \, \text{kW} $

(iii) Efficiency of propulsion: $ \approx 53.9\% $

Explanation of Jet Propulsion

Propelling Force: The force that moves the tank forward is a reaction force. As the tank expels a mass of water backward (the jet), the water pushes back on the tank with an equal and opposite force, according to Newton's Third Law. This force is calculated by finding the rate of change of momentum of the water.
Work Done (Power): This is the useful output of the system. It's the propelling force multiplied by the speed at which the tank is moving.
Efficiency: The efficiency of propulsion measures how well the energy of the exiting water is converted into useful work to move the tank. The total energy expended is the sum of the useful work done and the kinetic energy that is "lost" or "wasted" in the jet of water that is left behind. The efficiency is the ratio of the useful work to this total energy.