How much power generation potential is available in a run of river mini hydropower plant for flow of 25 litres/second with a head of 15 metres? Assume system efficiency of 65%.

 To calculate the power generation potential of a run-of-river mini hydropower plant, we can use the formula for hydropower:

$𝑃=𝜌\times 𝑔\times 𝑄\times 𝐻\times 𝜂$

Where:

• $𝑃$ is the power output (in watts).
• $𝜌$ is the density of water (approximately $1000{\text{kg/m}}^3$).
• $𝑔$ is the acceleration due to gravity (approximately $9.81{\text{m/s}}^2$).
• $𝑄$ is the flow rate of water (in cubic meters per second).
• $𝐻$ is the head or vertical drop (in meters).
• $𝜂$ is the efficiency of the hydropower system (expressed as a decimal).

Given:

• $𝑄=25\text{litres/second}=0.025{\text{m}}^3\mathrm{/}\text{s}$
• $𝐻=15\text{metres}$
• $𝜂=0.65$

Let's plug in the values into the formula:

$𝑃=1000{\text{kg/m}}^3\times 9.81{\text{m/s}}^2\times 0.025{\text{m}}^3\mathrm{/}\text{s}\times 15\text{m}\times 0.65$

$𝑃\approx 3.76\text{kW}$

So, the power generation potential of the run-of-river mini hydropower plant with a flow of 25 liters/second and a head of 15 meters, assuming a system efficiency of 65%, is approximately 3.76 kW (kilowatts).