Problem 94 - University Physics with Modern Physics

Question

The Grand Coulee Dam is 1270 \(\mathrm{m}\) long and 170 \(\mathrm{m}\) high. The electrical power output from generators at its base is approximately 2000 \(\mathrm{MW}\) . How many cubic meters of water must flow from the top of the dam per second to produce this amount of power if 92\(\%\) of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of water has a mass of 1000 \(\mathrm{kg.})\)

Step-by-Step Solution

Answer

Approximately 1304 cubic meters of water must flow per second.

Step 1: Understanding the Problem

We need to find the cubic meters of water flowing over the dam each second to produce 2000 MW of electricity. We know that 92% of the potential energy is converted into electrical energy.

Step 2: Determine Potential Energy per Cubic Meter

Every cubic meter of water has a mass of 1000 kg. The potential energy (U) for each cubic meter of water falling from the height (h) of 170 m is calculated as:\[U = m \, g \, h = 1000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 170 \, \text{m}\]This gives us the potential energy in joules for each cubic meter of water.

Step 3: Calculate Potential Energy

Solving for potential energy, we get:\[U = 1000 \times 9.8 \times 170 = 1666000 \, \text{Joules}\]Thus, each cubic meter of water has 1,666,000 Joules of potential energy.

Step 4: Adjust for Energy Conversion

Since only 92% of this potential energy is converted to electrical energy, we have:\[E_{converted} = 0.92 \times 1666000 \]Calculate the energy converted per cubic meter.

Step 5: Calculate Converted Energy

Substitute the values to find the energy converted:\[E_{converted} = 0.92 \times 1666000 = 1532720 \, \text{Joules}\]So, 1,532,720 Joules of energy per cubic meter is used for electricity generation.

Step 6: Relate Power to Energy

The power output in watts (which is joules per second) is given as 2000 MW:\[P = 2000 \times 10^6 \, \text{J/s}\]We need to calculate how many cubic meters of water per second correspond to this power.

Step 7: Calculate Water Flow Rate

The flow rate of water in cubic meters per second (Q) can be determined by dividing the total power output by the energy per cubic meter:\[Q = \frac{P}{E_{converted}} = \frac{2000 \times 10^6}{1532720}\]Solve for Q to get the flow rate.

Step 8: Compute Flow Rate

Solving the above equation:\[Q \approx \frac{2000 \times 10^6}{1532720} \approx 1304 \, \text{m}^3/\text{s}\]Thus, approximately 1304 cubic meters of water need to flow from the top of the dam per second.

Key Concepts

Potential Energy

When we talk about hydropower, potential energy is a fundamental concept. It's the energy stored in an object due to its position. In the case of dams, it's all about the height the water falls from.
Each cubic meter of water at the top of a dam like the Grand Coulee Dam contains potential energy because of its elevation above the base. The formula used to calculate this energy is \[ U = mgh \] where:

• \( m \) is the mass of the water in kilograms.
• \( g \) is the acceleration due to gravity, approximately \( 9.8 \,\text{m/s}^2 \).
• \( h \) is the height in meters.

Therefore, for every cubic meter of water weighing 1000 kg falling from a height of 170 meters, the potential energy calculates to 1,666,000 joules.
Understanding this concept helps appreciate how much energy water can store before it even starts generating electricity.

Energy Conversion Efficiency

In any energy conversion process, some energy is always lost. This is where conversion efficiency comes into play.
For the Grand Coulee Dam, not all of the potential energy of the falling water is converted into electricity. Instead, only 92% of this potential energy is harnessed for electricity production.
The rest might be lost due to factors like friction within the pipes and the mechanical parts of the generators. The effective energy conversion can be expressed as: \[ E_{\text{converted}} = \eta \times U \] where \( \eta \) is the efficiency (0.92 in this case) and \( U \) is the potential energy.
For each cubic meter at the Grand Coulee Dam, this means that only around 1,532,720 joules are converted for use.
Having an efficient energy conversion process is crucial for optimizing power output and minimizing energy waste.

Flow Rate Calculation

Flow rate in the context of hydropower refers to the volume of water moving through the system each second.
This is important because it determines how much potential energy is available to be converted into electrical energy. To find out how many cubic meters of water need to move over the dam to meet the required electrical output of 2000 MW, we use the formula: \[ Q = \frac{P}{E_{\text{converted}}} \] Where:

• \( Q \) is the flow rate in cubic meters per second.
• \( P \) is the power output in watts (joules per second), which is \( 2000 \times 10^6 \, \text{J/s} \) in this case.
• \( E_{\text{converted}} \) is the converted energy per cubic meter.

By substituting the known values, we find that approximately 1304 cubic meters of water must flow every second to generate the desired power output.
This concept highlights how integral calculating the right flow rate is for achieving efficient and sustainable hydropower production.