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

1 answer

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.