At a certain waterfall, 8 m³ of water falls per second to a depth of 5 m. The energy from this needs to be converted into electrical energy and transmitted to a nearby city for lighting and driving machines. The efficiency of the turbines is 80%, the efficiency of the generators is 90%, and the average loss in the transmission line from the production site to the consumption site is 5%. In the city, 5 motors, each with 10 horsepower, need to be driven (with an efficiency of 85%) and 24 incandescent lamps, each 40 volts and 10 amperes, need to be powered. How many 55-watt incandescent lamps can still be powered with the remaining energy?
At the water turbine, the energy generated per second:

$$
\begin{gathered}
E=8000 \cdot 5 \frac{\mathrm{kg} \mathrm{m}}{\mathrm{sec}}=\frac{8000 \cdot 5}{75} \mathrm{HP}=\frac{8000 \cdot 5 \cdot 736}{75} \text { watt. } \\
E=392500 \text { watt. }
\end{gathered}
$$

Taking all losses into account, we have available

$$
\begin{gathered}
H=392500 \cdot 0.8 \cdot 0.9 \cdot 0.95 \text { watt. } \\
H=268500 \text { watt. }
\end{gathered}
$$

From this, the motors consume

$$
\frac{5 \cdot 10 \cdot 736}{0.85}=43290 \text { watt }
$$

the arc lamps

$$
24 \cdot 40 \cdot 10=9600 \text { watt. }
$$

Leaving for the incandescent lamps

$$
268500-43290-9600=215610 \text { watt }
$$

thus we can power

$$
215610: 55=3920 \text { incandescent lamps. }
$$

(P̈̈zel Tibor, Budapest.)