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