Problem 79 - University Physics with Modern Physics

Question

A hydroelectric dam holds back a lake of surface area \(3.0 \times 10^{6} \mathrm{m}^{2}\) that has vertical sides below the water level. The water level in the lake is 150 \(\mathrm{m}\) above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is converted into electrical energy with 90\(\%\) efficiency. (a) If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the water in the lake? The density of water is 1000 \(\mathrm{kg} / \mathrm{m}^{3} .(\mathrm{b})\) What volume of water must pass through the dam to produce 1000 kilo-watt-hours of electrical energy? What distance does the level of water in the lake fall when this much water passes through the dam?

Step-by-Step Solution

Answer

(a) 4.41 x 10¹² J; (b) 2.72 x 10⁴ m³ water causes a fall of 0.00907 m.

Step 1: Calculate the Volume of the Top Meter of Water

The surface area of the lake is given as \(3.0 \times 10^6 \, \text{m}^2\). To find the volume of the top meter, multiply the surface area by 1 meter (the height). Therefore, the volume is \(V = 3.0 \times 10^6 \, \text{m}^2 \times 1 \, \text{m} = 3.0 \times 10^6 \, \text{m}^3\).

Step 2: Calculate the Mass of the Top Meter of Water

The density of the water is given as \(1000 \, \text{kg/m}^3\). Multiply the volume (from Step 1) by the density to find the mass: \(m = 3.0 \times 10^6 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 = 3.0 \times 10^9 \, \text{kg}\).

Step 3: Calculate the Potential Energy of the Top Meter of Water

The potential energy stored in this mass of water at a height of 150 m can be calculated using \(PE = mgh\), where \(m = 3.0 \times 10^9 \, \text{kg}\), \(g = 9.8 \, \text{m/s}^2\), and \(h = 150 \, \text{m}\). This results in \(PE = 3.0 \times 10^9 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 150 \, \text{m} = 4.41 \times 10^{12} \, \text{J}\).

Step 4: Calculate the Volume of Water for Required Energy

First, convert 1000 kilo-watt-hours to joules: \(1000 \, ext{kWh} = 1000 \, \times 3.6 \times 10^6 \, ext{J} = 3.6 \times 10^9 \, ext{J}\). Since the conversion efficiency is 90%, multiply the required energy by the inverse of the efficiency: \(E = \frac{3.6 \times 10^9 \, \text{J}}{0.9} = 4.0 \times 10^9 \, \text{J}\).

Step 5: Calculate the Mass and Volume of Water Needed

The potential energy of water is given by \(PE = mgh\). Rearranging for mass gives \(m = \frac{E}{gh}\), where \(E = 4.0 \times 10^9 \, \text{J}\), \(g = 9.8 \, \text{m/s}^2\), and \(h = 150 \, \text{m}\). This calculates to \(m = \frac{4.0 \times 10^9 \, \text{J}}{9.8 \, \text{m/s}^2 \times 150 \, \text{m}} = 2.72 \times 10^7 \, \text{kg}\). Dividing this mass by the density gives the volume: \(V = \frac{2.72 \times 10^7 \, \text{kg}}{1000 \, \text{kg/m}^3} = 2.72 \times 10^4 \, \text{m}^3\).

Step 6: Calculate the Drop in Water Level

The total surface area of the lake is \(3.0 \times 10^6 \, \text{m}^2\), and the volume of water removed is \(2.72 \times 10^4 \, \text{m}^3\). The drop in water level, \(d\), can be calculated by dividing the volume by surface area: \(d = \frac{2.72 \times 10^4 \, \text{m}^3}{3.0 \times 10^6 \, \text{m}^2} = 0.00907 \, \text{m}\).

Key Concepts

Gravitational Potential Energy

When water is held high above the ground, like in a dam, it possesses gravitational potential energy. This energy is due to its height above a reference point, in this case, the base of the dam. The gravitational potential energy (PE) of an object can be calculated using the formula: \[ PE = mgh \] 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 above the reference point in meters.

For example, in our hydroelectric dam scenario, we calculate the energy stored in the top meter of water using this formula. Understanding this form of energy is crucial because it represents the potential to do work, such as generating electricity.

Efficiency in Energy Conversion

In the context of a hydroelectric dam, energy conversion efficiency refers to how much of the water's mechanical energy is successfully converted into electrical energy. This efficiency is significant because not all input energy is transformed into useful output energy due to losses, predominantly through heat and friction. To express conversion efficiency, we use:\[ \text{Efficiency} = \frac{\text{Useful Output Energy}}{\text{Total Input Energy}} \]This value is often expressed as a percentage. In our dam example, the turbines convert the water's mechanical energy to electrical energy with 90% efficiency. This means only 90% of the potential energy is turned into electricity. Knowing the efficiency helps in calculating the actual amount of energy we can expect to produce and plan accordingly.

Mass and Volume Calculations

To find the amount of energy stored in a specific volume of water, we first need to know both the mass and the volume of that water. The relationship between mass, volume, and density is given by:\[ m = \rho V \]where:

• \( m \) is the mass,
• \( V \) is the volume,
• \( \rho \) is the density, which for water is 1000 kg/m³.

In our problem, the volume for one meter of depth in the lake is calculated by multiplying the surface area by the depth. This helps determine the mass, which then is used to compute the potential energy. Understanding these calculations allows us to assess the system's energy potential accurately.

Energy Units Conversion

Handling different energy units is crucial for accurately calculating and understanding energy systems. In the field of hydroelectric power, energy is typically measured in joules, while electricity often appears in kilowatt-hours (kWh). The conversion between these units is essential when performing calculations.The conversion factor is:\[ 1 \, \text{kWh} = 3.6 \times 10^6 \, \text{J} \]This means that to convert kilowatt-hours to joules, we multiply by 3.6 million. For instance, in the dam scenario, to calculate how much energy in joules is equivalent to 1000 kWh, we use the conversion factor. Understanding conversions is important because it allows us to translate energy requirements and outputs into a consistent form for analysis and engineering.