To calculate the power output of a hydropower plant, you need to use the following formula for power (P):
[tex]\[ P = \rho \times g \times h \times Q \times \eta \][/tex]
where:
- [tex]\(\rho\)[/tex] is the density of water ([tex]\(1000 \, \text{kg/m}^3\)[/tex])
- [tex]\(g\)[/tex] is the acceleration due to gravity ([tex]\(9.81 \, \text{m/s}^2\)[/tex])
- [tex]\(h\)[/tex] is the head or height through which the water falls (4 meters)
- [tex]\(Q\)[/tex] is the flow rate ([tex]\(2.5 \, \text{m}^3/\text{s}\)[/tex])
- [tex]\(\eta\)[/tex] is the efficiency (65%, or 0.65 in decimal form)
Let's break down the calculation step-by-step:
1. Substitute the known values into the formula:
[tex]\[
P = 1000 \times 9.81 \times 4 \times 2.5 \times 0.65
\][/tex]
2. Multiply the values together:
- First, multiply the densities and gravity:
[tex]\[
1000 \times 9.81 = 9810
\][/tex]
- Then, multiply by the head:
[tex]\[
9810 \times 4 = 39240
\][/tex]
- Then, multiply by the flow rate:
[tex]\[
39240 \times 2.5 = 98100
\][/tex]
- Finally, apply the efficiency:
[tex]\[
98100 \times 0.65 = 63765
\][/tex]
So, the power output of the hydropower plant is [tex]\(63765 \, \text{Watts}\)[/tex] or [tex]\(63.765 \, \text{kW}\)[/tex].
Thus, the power of a low head of 4 m hydropower plant with a turbine that can handle a flow rate of 2.5 [tex]\(m^3/s\)[/tex] and an overall efficiency of 65% is:
[tex]\[ 63765 \, \text{Watts} \][/tex]
