Problem 86 - University Physics with Modern Physics

Question

The Grand Coulee Dam is 1270 m long and 170 m high. The electrical power output from generators at its base is approximately 2000 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 kg.)

Step-by-Step Solution

Answer

To produce 2000 MW, approximately 373.7 cubic meters of water must flow each second.

Step 1: Understand the Problem

We need to find out how many cubic meters of water per second must flow to generate 2000 MW of power if 92% of gravitational work is converted to electrical energy. We know the height of the dam is 170 meters and each cubic meter of water has a mass of 1000 kg.

Step 2: Calculate Gravitational Potential Energy

The gravitational potential energy (GPE) for a volume of water is given by the formula: \( \,GPE = m \cdot g \cdot h \,\), where \( m \,\) is the mass (in kg), \( g \,\) is the acceleration due to gravity (9.8 m/s²), and \( h \,\) is the height in meters. Here, \( h = 170 \,\) meters. We'll express the energy per cubic meter: \[ GPE_{per\,m^3} = 1000 \cdot 9.8 \cdot 170 \]

Step 3: Calculate Energy Converted to Electrical Power

Out of the total gravitational potential energy, only 92% is converted to electrical power. Therefore, the energy available per cubic meter for conversion is \( 0.92 \times GPE_{per\,m^3} \). Calculate this value.

Step 4: Relate Power to Flow Rate

Power is the amount of energy transferred per unit time. We know the required power output is 2000 MW, or \( 2000 \times 10^6 \, W \). Using \( P = \frac{E}{t} \), where \( E = \text{energy converted per second} \) and \( t = 1 \) second. Relate it to flow rate: \[ 2000 \times 10^6 = \dot{V} \times (0.92 \times GPE_{per\,m^3}) \]

Step 5: Solve for Flow Rate (\( \dot{V} \))

Rearrange the equation to solve for the flow rate, \( \dot{V} \): \[ \dot{V} = \frac{2000 \times 10^6}{0.92 \times 1000 \times 9.8 \times 170} \]Calculate \( \dot{V} \) to find out how many cubic meters of water are needed each second.

Key Concepts

Gravitational Potential Energy

When water is stored at a height, it possesses energy due to its position. This is called gravitational potential energy (GPE). It's the energy that can be harnessed to do work when the water falls. The formula to calculate it is:\[ GPE = m \cdot g \cdot h \]Where:

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

The Grand Coulee Dam is 170 meters high. If we consider each cubic meter of water has a mass of 1000 kg, then this energy calculation shows us how much potential work is available from the water just because of its height. Understanding GPE is crucial for recognizing how water can be a powerful resource for generating electricity.

Energy Conversion Efficiency

In real-world applications, not all energy can be converted into the desired form. The efficiency of energy conversion tells us how effectively we can convert input energy to output energy. For the Grand Coulee Dam, the problem specifies that 92% of the gravitational work done by the water is converted into electrical energy. This means not all of the GPE becomes electricity; some energy is lost, perhaps due to friction or other factors, which is common in energy transformation processes.The calculation of useful energy involves multiplying the total potential energy by the conversion efficiency:\[ \text{Energy Converted} = \text{Efficiency} \times \text{GPE} \]Where the efficiency here is 92%, or 0.92 as a decimal. This highlights the importance of efficiency in maximizing the output that can be gained from a given input energy.

Power Output Calculation

Power refers to the rate at which energy is transferred or converted. In hydroelectric power generation, it indicates how much energy can be produced in a certain time. The Grand Coulee Dam needs to generate a power output of 2000 MW.To relate this to water flow, we use the formula for power, which is energy per time:\[ P = \frac{E}{t} \]With water flow, energy per second is crucial because it tells us how much water must move each second to meet the power requirement. By substituting the energy converted from GPE and trying to match it with the power output, we solve for the required water flow rate:\[ \dot{V} = \frac{2000 \times 10^6}{0.92 \times 1000 \times 9.8 \times 170} \]This equation allows us to determine \( \dot{V} \), the necessary flow rate in cubic meters per second, ensuring that the dam’s output meets energy demands efficiently. Power output calculation links the physical movement of water and the technical specifications of generating large-scale electricity.