Question

Calculate the power of a low head of 4 meter hydropower plant with a turbine that can handle a flow rate of 2.5 cubic meters per second. The overall efficiency of the power plant is 65 percent

Answer 1

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:

1. \(1000 \cdot 9.81 = 9810 \, \text{N/m}³\)
2. \(9810 \cdot 4 = 39240 \, \text{N/m³}\)
3. \(39240 \cdot 2.5 = 98100 \, \text{N} \, (\text{or Watts})\)
4. \(0.65 \cdot 98100 = 63765 \, \text{Watts}\)

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**.