To calculate the power output for a hydropower site, you can use the following formula:

$P = \eta \cdot \rho \cdot g \cdot H \cdot Q$

Where:

• $P$ is the power output in watts (W).
• $\eta$ is the efficiency of the turbo-generator (in decimal form).
• $\rho$ is the density of water (approximately $1000 \, \text{kg/m}^3$).
• $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$).
• $H$ is the effective head (in meters).
• $Q$ is the flow rate (in cubic meters per second, m³/s).

Now, let's substitute the given values:

• $\eta = 0.8$
• $\rho = 1000 \, \text{kg/m}^3$
• $g = 9.81 \, \text{m/s}^2$
• $H = 15 \, \text{m}$
• $Q = 0.04 \, \text{m}^3/\text{s}$

Substituting these into the formula:

$P = 0.8 \times 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 15 \, \text{m} \times 0.04 \, \text{m}^3/\text{s}$

$P = 0.8 \times 1000 \times 9.81 \times 15 \times 0.04$

Now, calculating the value:

$P = 0.8 \times 1000 \times 9.81 \times 15 \times 0.04 = 4708.8 \, \text{W}$

Since 1 kW = 1000 W, the power output in kilowatts is:

$P = \frac{4708.8}{1000} \, \text{kW} = 4.7088 \, \text{kW}$

So, the power output is 4.7088 kW.

Would you like any further details or have any questions?

Here are 5 related questions for further exploration:

1. How would the power output change if the effective head increased to 20 m?
2. What happens to the power output if the efficiency of the turbine decreases to 70%?
3. How would the power output differ if the flow rate was doubled to 0.08 m³/s?
4. What is the impact of water density variations on the power output calculation?
5. How would seasonal changes in water flow affect the annual power production?

Tip: Always convert the final power output to kilowatts (kW) for practical applications, as this is the standard unit used in power generation discussions.