Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine ? (g=10 m/s²)

First, find the useful power supplied by the falling water (the input power to the turbine system).

$P\_\{\backslash text\{in\}\}\; =\; \backslash dot\; m\; \backslash ,\; g\; \backslash ,\; h$

where

$\backslash dot\; m\; =\; 15\backslash \; \backslash text\{kg/s\}$ (mass flow rate)

$g\; =\; 10\backslash \; \backslash text\{m/s\}^2$ (acceleration due to gravity)

$h\; =\; 60\backslash \; \backslash text\{m\}$ (height of fall)

Substituting the values:

$P\_\{\backslash text\{in\}\}\; =\; 15\; \backslash times\; 10\; \backslash times\; 60\backslash \; \backslash text\{W\}$

$P\_\{\backslash text\{in\}\}\; =\; 15\; \backslash times\; 600\backslash \; \backslash text\{W\}\; =\; 9000\backslash \; \backslash text\{W\}$

Thus the water supplies 9000 W, i.e., 9 kW of mechanical energy.

The turbine experiences frictional losses equal to 10 % of this input energy. Therefore the power that actually reaches the turbine (the useful power) is 90 % of the input:

$P\_\{\backslash text\{out\}\}\; =\; (1\; -\; 0.10)\backslash ,\; P\_\{\backslash text\{in\}\}\; =\; 0.90\; \backslash times\; 9000\backslash \; \backslash text\{W\}$

$P\_\{\backslash text\{out\}\}\; =\; 8100\backslash \; \backslash text\{W\}$

Converting to kilowatts:

$P\_\{\backslash text\{out\}\}\; =\; 8.1\backslash \; \backslash text\{kW\}$

Hence the turbine generates 8.1 kW of power.