The total resistance offered to the motion of the boat is 4905 N. Determine

a) Volume of watrer drawn by the pump per second

b) Efficiency of the jet propulsion

Mumbai University > Electronics and telecommunication > Sem 7 > Applied Hydraulics

Marks: 10

Years: DEC 2015

Given :

Velocity of boat, u= 5.5 m/s

Diameter of each jet d= 155 mm = 0.155 m

Area of each jet = $\dfrac {\pi}4 \times (0.155)^2 = 0.01886 m^2$

$\therefore$ Total area of the jets $a=2\times 0.01886 \\ =0.03772 m^2$

Total resistance to motion $=4905 N(500\times 9.81)$

The propelling force must be equal to the resistance to the motion

$\therefore$ Propelling force $F=4905 N$ or $(500\times 9.81 N)$

Propelling force is given by equation

$F= pa(v+u)v\\ 500\times 9.81=1000 \times 0.03772\times (v+5.5)\times v\\500= \dfrac {1000}{9.81}\times 0.03772 \times (v+5.5)\times v\\ =3.84(v+5.5)v\\ =3.84v^2+21.12v$

The above equation is quadratic and its solution is

$v= \dfrac {-21.12\pm\sqrt{21.12^2+4\times 3.84\times 500}}{2\times 3.84}\\=\dfrac {-21.12\pm 90.14}{7.68}\\ = \dfrac{90.14-21.12}{2}=34.51 m/s$

(i) volume of water drawn by the pump per second is equal to the volume of water discharged through the orifices at the back in the form of jets and this volume

$=av_e= a(v+u) \\ = 0.03772 \times (34.51+5.5)=1.509 m^3/s$

(ii) Efficiency of the jet propulsion is given by equation

$n= \dfrac {2vu }{(v+u)^2} =\dfrac {2\times 34.51\times 5.5}{(34.51+5.5)^2}= 0.2371=23.71\%$