Question

A dam 50 meters high is constructed on a river and is delivering \(180 \mathrm{MW}\) at some moment in time. What is the flow rate of water, in cubic meters per second, if the facility converts gravitational potential energy into electricity at \(90 \%\) efficiency?

Short answer

Answer: The flow rate of water in this scenario is approximately 0.32997 cubic meters per second.

Step-by-step solution

Calculate the Gravitational Potential Energy

To find the flow rate, we will first calculate the gravitational potential energy (GPE) of the water as it falls through the dam. The GPE can be calculated using the formula: GPE = m * g * h where, m = mass of water (kg) g = acceleration due to gravity (9.81 m/s^2) h = height of the dam (50 meters) However, since we don't have the mass of water directly, we will use the power generation capacity of the dam given as 180 MW.

Utilize the relation between Power, Potential Energy, and Time

We have the power generation capacity P given by 180 MW (180 * 10^6 W) and the potential energy calculated above. Since power is defined as the work done per unit time, we can relate the potential energy and power as follows: P = GPE / t Where t is time in seconds. Now, using the fact that the dam's energy conversion efficiency is 90%, the energy conversion can be represented as: Total_GPE = GPE * Efficiency

Calculate the Mass Flow Rate of Water

Now we have the total gravitational potential energy and the relation between power and time. We can find the mass flow rate (mass per unit time) using these variables: Mass_Flow_Rate = Total_GPE / (g * h) Substituting the values for power, efficiency, g, and h: Mass_Flow_Rate = (180 * 10^6 * 0.9) / (9.81 * 50) Calculating the mass flow rate, we get: Mass_Flow_Rate ≈ 329.97 kg/s

Calculate the Volume Flow Rate

Now that we have the mass flow rate of water, we can convert this into volume flow rate. To do this, we need to divide the mass flow rate by the density of water (ρ), which is approximately 1000 kg/m^3: Volume_Flow_Rate = Mass_Flow_Rate / ρ Substitute the value of the mass flow rate and the density of water: Volume_Flow_Rate = 329.97 kg/s / 1000 kg/m^3 Calculating the volume flow rate, we get: Volume_Flow_Rate ≈ 0.32997 m^3/s So, the flow rate of water is approximately 0.32997 cubic meters per second.

Key concepts

Gravitational Potential Energy

Gravitational Potential Energy (GPE) is the energy possessed by an object due to its position relative to the Earth. In the case of hydropower, it refers to the energy stored in water held at a height in a dam. This stored energy can be used to produce electricity.
The GPE of the water can be calculated using the formula:\[ \text{GPE} = m \cdot g \cdot h \]where:

• \(m\) is the mass of the water in kilograms (kg),
• \(g\) is the acceleration due to gravity, which is approximately \(9.81 \, \text{m/s}^2\),
• \(h\) is the height of the water above the ground, in this instance the height of the dam (50 meters).

When water is released from the dam, this potential energy is converted into mechanical energy and subsequently into electricity. Understanding GPE is crucial since it forms the basis for calculating other factors such as power, energy efficiency, and flow rates in hydropower systems.

Mass Flow Rate

The mass flow rate refers to the amount of mass passing through a point per unit of time. In hydropower, it is the rate at which water mass is flowing through the system, and it's key to determining the amount of energy that can be converted to electricity.
To find the mass flow rate, we use the formula:\[ \text{Mass Flow Rate} = \frac{\text{Total GPE}}{g \cdot h} \]For our problem, the dam provides a power of 180 MW with an energy conversion efficiency of 90%, so:

• The total GPE available is: \(180 \times 10^6 \times 0.9 \) (Watts).
• Substituting into the formula gives \( \approx 329.97 \, \text{kg/s} \) for the mass flow rate.

This indicates the mass of water flowing through the dam per second, which is crucial to further calculations for volume flow rate and ensuring optimal energy production.

Volume Flow Rate

Volume Flow Rate gives us the amount of water volume passing through a point per unit of time. It is derived from the mass flow rate by considering the density of water. In hydropower, it's essential as it helps design the plant's capacity and efficiency.
The relationship between mass flow rate and volume flow rate is given by:\[ \text{Volume Flow Rate} = \frac{\text{Mass Flow Rate}}{\rho} \]where \(\rho\) is the density of water

• The density of water is usually \(1000 \, \text{kg/m}^3\).
• Substituting the mass flow rate calculated (\(329.97 \, \text{kg/s}\)) results in a volume flow rate of approximately \(0.32997 \, \text{m}^3/\text{s}\).

Understanding volume flow rate is important for constructing the right capacity of turbines and for ensuring smooth operation of hydropower plants.

Energy Conversion Efficiency

Energy Conversion Efficiency in hydropower systems measures how well the energy stored as GPE is converted to electrical energy. It is expressed in percentage and determines the overall performance of the dam.
It is calculated using the formula:\[ \text{Efficiency} = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100 \]For our hydropower plant:

• The useful energy output is 180 MW, and the efficiency is given as 90%.
• Meaning, 10% of the potential energy is lost due to factors like turbine friction and other mechanical losses.

By understanding efficiency, one can improve the design and management of hydropower facilities, ensuring maximum electricity is produced from the available water. Increasing the efficiency of energy conversion means harnessing more power from natural resources while minimizing losses, which is vital for sustainable energy production.