In a water wheel application, an overshot wheel is in use. The water in the channel flows at a speed of $1.4 \frac{\mathrm{m}}{\mathrm{sec}}$ and then reaches the wheel. The channel width is $0.5 \mathrm{~m}$, the water depth is $0.13 \mathrm{~m}$, and the wheel diameter is $3 \mathrm{~m}$. The overshot wheel converts $78 \%$ of the energy into effective work (efficiency: 0.78); how many effective horsepower is the capacity of this wheel?
Every second, $1.4 \cdot 0.5 \cdot 0.13 \cdot 1000 \mathrm{~kg}=91 \mathrm{~kg}$ of water flows onto the wheel.

The indicated power is the energy available per second, which comes from the kinetic energy ($\frac{1}{2} m v^{2}$) and the potential energy $(p \cdot s)$. Therefore,

$$
\begin{gathered}
H_{i}=\frac{1}{2} m v^{2}+p \cdot s=\left(\frac{1}{2} \cdot \frac{91}{9.81} \cdot 1.4^{2}+91 \cdot 3\right) \frac{\mathrm{m} \mathrm{kg}}{\mathrm{sec}}= \\
=282.1 \frac{\mathrm{m} \mathrm{kg}}{\mathrm{sec}}=3.76(H P)_{i}
\end{gathered}
$$

The effective power

$$
H_{e}=3.76 \cdot 0.78(H P)_{e}=2.9(H P)_{e} .
$$

The problem was solved by: Bánó L., Ehrenstein P., Frank A., Freund E., Pető L., Szilas O.