Jet propulsion of ship when the inlet orifices face the direction of motion of the ship.

Figure shows, a ship which is having the inlet orifices facing the direction of motion on the ship.

In this case, the expression for propelling force and work done per second will be the same as in the 1$^{st}$ case in which inlet orifices at right angles to the ship. But the energy required to be supplied by the pump will be different, as in this case, the water enters in with a velocity equal to the velocity of the ship that is with a velocity u.

$ = \dfrac 12 \text{(Mass of water supplied per sec)}\times[v_r^2-u^2]$

$ = \dfrac12(\rho av_r)\times [v_r^2-u^2]$

$v_r = (v+u)$, as in the previous case.

Therefore K.E supplied by the pump = $\dfrac 12\rho a(v+u)[(v+u)^2-u^2].............(1)$

Efficiency of propulsion, $\eta = \dfrac{\text{work done per sec}}{\text{energy supplied}}$

$\therefore, \eta = \dfrac{\rho av_r\times v\times u}{\dfrac 12\rho av_r[v_r^2-u^2]}$

$\eta = \dfrac{2vu}{v_r^2-u^2}$

$\eta = \dfrac{2vu}{(v+u)^2-u^2}$

$\eta = \dfrac{2vu}{v^2+u^2+2vu-u^2}$

$\eta = \dfrac{2vu}{v^2+2vu}$

$\eta = \dfrac{2u}{v+2u}...................(2)$

Q1) The water in a jet-propelled boat is drawn through inlet openings facing the direction of motion of the ship. The boat is moving in sea-water with a speed of 30 km/hour. The absolute velocity of the jet of water discharge at the back is 20 m/s and area of the jet of water is 0.03m$^2$. Find the propelling force and efficiency of the propulsion.

Solution: Given:

i) Speed of boat, u = 30km/hour

$u = \dfrac{30\times 1000}{60\times 60} = 8.33m/s$

ii) Absolute velocity of jet, v = 20m/s

iii) Area, a = 0.03m$^2 $

To find:

a) Propelling force (F)

$F = \rho av_r\times v$

$F = \rho a(v+u)\times v$

$F = 1000\times0.03\times (20+8.33)\times 20$

$F = 16997.98N$

b) Efficiency of propulsion ($\eta$)

$\eta = \dfrac{2u}{v+2u}$

$\eta = \dfrac{2\times 8.33}{20+2\times 8.33}$

$\eta = 0.4544 = 45.44\%$