Shaft power = 12000 KW, head = 350 m, speed =750 rpm, overall eficiency = 86 %, jet diameter is not to exceed one sixth of the wheel diameter. Determine

(i) Wheel diameter

(ii) Number of jets required

(iii) Diameter of jet

Take coefficient of velocity is 0.985 and speed ratio 0.45

Given

Shaft power S.P= 12000 KW

head H = 350 m

speed N =750 rpm

overall eficiency %= 86 % or 0.86

Ratio of jet dia to wheel dia $\dfrac dD =\dfrac 16$

Coefficient of velocity $K_{v1}=C_v=0.985 $

Speed ratio $K_{u1} = 0.45 $

Velocity of jet = $V_1 = C_v\sqrt{2gH} \\ =0.985\sqrt{2\times9.81\times 350}\\ =81.62 m/s$

The velocity of wheel $u=u_1=u_2\\ \text {speed ratio }\times \sqrt{2gH}\\ =0.45\times \sqrt{2\times 9.81\times350}\\ =37.29 m/s$

But $u=\dfrac {\pi DN}{60}\therefore 37.29=\dfrac {\pi DN}{60}\\ D=\dfrac {60\times 37.29}{\pi \times N} =\dfrac {60\times 37.29}{\pi \times 750} =0.9495 m$

But $\dfrac dD= \dfrac 16 $

Dia of jet $d=\dfrac 16\times D =\dfrac {0.949}6 = 0.158 m $

Discharge of one jet q = Area of jet $\times $ velocity of jet

$=\dfrac {\pi}4 d^2 \times v_1 \\ \dfrac \pi4\times (0.158)^2 \times 81.62 \\ = 1.600 m^3/s \\ n_0 =\dfrac {S.P}{W.P} =\dfrac {12000}{\dfrac {Pg\times Q\times H}1000}\\ 0.86 =\dfrac {12000\times 1000}{1000\times 9.81\times Q\times 350} \\ Q =\dfrac {12000\times {1000}}{1000\times 9.81 \times 350 \times 0.86} \\ =4.063 m^3/s $

Number of jets = $\dfrac {\text {Total discharge}}{\text {Discharge of one jet}} = \dfrac Qq = \dfrac {4.063}{1.600} = 2.53 \approx $ 2 jets