For the same turbine as Sample Prob. 6.6, find the head converted into mechanical work.
Eq. (6.32):
\begin{aligned}h_M & =\left|\frac{\Delta\left(u V_u\right)}{g}\right|=\frac{u_1 V_{u 1}-u_2 V_{u 2}}{g} \\\\& =\frac{15.708 \times 17.833-9.425 \times 4.477}{32.2}=7.39 \mathrm{ft} \\\\P & =\frac{\gamma Q h_M}{550}=\frac{62.4(7.5) 7.39}{550}=6.29 \mathrm{hp}\end{aligned}Note:
This agrees very well with Part (c) of Sample Prob. 6.6.
h_M=\frac{u_1 V_1 \cos \alpha_1-u_2 V_2 \cos \alpha_2}{g} (6.32)