Given: n=4 ; d=0.2  m ; S P=44 \times 10^6  W ; N=400  rpm ; Q=11.5  m ^3 / s ; H=455  m ; D=2  m ; \eta_{\text {mech }}=0.97 ; \phi=15^{\circ}

Using the definition of mechanical efficiency,

Power developed by runner =\frac{S P}{\eta_{\text {mech }}}=\frac{44 \times 10^6}{0.97}=45.36 \times 10^6  W

Flow rate per jet q=\frac{Q}{4}=\frac{11.5}{4}=2.875   m ^3 / s

Velocity of the jet  V_1=\frac{4 q}{\pi d^2}=\frac{4 \times 2.875}{\pi \times 0.2^2}=91.514  m / s

Peripheral speed of runner  u=\frac{\pi D N}{60}=\frac{\pi \times 2 \times 400}{60}=41.89  m / s

Also, from the equation for power developed per jet, we get

\begin{aligned}P &=\rho Q\left(V_1-u\right)(1+K \cos \phi) u \\11.34 \times 10^6 &=1000 \times 2.875 \times(91.54-41.89)\left(1+K \cos 15^{\circ}\right) \times 41.89\end{aligned}

Solving, we get, blade velocity coefficient, K=0.928

Relative velocity at inlet  V_{r 1}=\left(V_1-u\right)=91.514-41.89=49.624  m / s

Coefficient of velocity  C_ν=\frac{V_1}{\sqrt{2 g H}}=\frac{91.54}{\sqrt{2 \times 9.81 \times 455}}=0.9886

Head lost in the nozzle

\begin{aligned}&=H-C_ν^2 H=\left(1-C_ν^2\right) H \\&=\left(1-0.9886^2\right) \times 455=10.32  m\end{aligned}

Head lost in the buckets

\begin{aligned}&=\frac{V_{r 1}^2-V_{r 2}^2}{2 g}=\frac{V_{r 1}^2-K V_{r 1}^2}{2 g}\\&=\frac{V_{r 1}^2\left(1-K^2\right)}{2 g}=\frac{49.624^2\left(1-0.928^2\right)}{2 \times 9.81}=17.42  m\end{aligned}

Power loss in the nozzle =w Q h_n=9810 \times 11 \times 10.32=1164251  W =1164.25  kW

Power loss in the buckets  =9810 \times 11.5 \times 17.42=1965237  W =1965.237   kW.