Question 2:

Calculate the power of a low head 4 m hydropower plant with a turbine that can handle a flow rate of [tex]2.5 \, m^3 s^{-1}[/tex]. The overall efficiency of the power plant is [tex]65\%[/tex].

[tex]\[
\begin{array}{l}
P = \rho g h Q \eta \\
P = 1000 \, kg/m^3 \times 9.81 \, m/s^2 \times 4 \, m \times 2.5 \, m^3/s \times 0.65 \\
P = 1000 \times 9.81 \times 4 \times 2.5 \times 0.65 \\
P = \text{calculated value in Watts}
\end{array}
\][/tex]

(Note: Calculate the final value to complete the answer.)



Answer :

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]