To calculate the power generated by a hydropower plant, we can use the following formula:
\[ P = \eta \cdot \rho \cdot g \cdot h \cdot Q \]
where:
- \(P\) is the power (in watts),
- \(\eta\) is the efficiency of the plant (as a decimal),
- \(\rho\) is the density of water (approximately 1000 kg/m³),
- \(g\) is the acceleration due to gravity (approximately 9.81 m/s²),
- \(h\) is the height of the head (in meters),
- \(Q\) is the flow rate (in cubic meters per second).
Given:
- \(\eta = 0.65\) (65% efficiency),
- \(h = 4\) meters,
- \(Q = 2.5\) m³/s,
- \(\rho = 1000 , \text{kg/m}³\),
- \(g = 9.81 , \text{m/s}²\).
Now we can substitute these values into the formula:
\[ P = 0.65 \cdot 1000 \cdot 9.81 \cdot 4 \cdot 2.5 \]
Calculating step by step:
- \(1000 \cdot 9.81 = 9810 , \text{N/m}³\)
- \(9810 \cdot 4 = 39240 , \text{N/m³}\)
- \(39240 \cdot 2.5 = 98100 , \text{N} , (\text{or Watts})\)
- \(0.65 \cdot 98100 = 63765 , \text{Watts}\)
Therefore, the power output of the hydropower plant is approximately:
\[ P \approx 63765 , \text{Watts} , (or , 63.8 , \text{kW}) \]
So, the power of the hydropower plant is approximately 63.8 kW.