Problem 78 - Master Resource Book in JEE Main Physics

Question

A dam is situated at a height of \(550 \mathrm{~m}\) above sea level and supplies water to a power house which is at a height of \(50 \mathrm{~m}\) above sea level. \(2000 \mathrm{~kg}\) of water passes through the turbines per second. What would be the maximum electrical power output of the power house if the whole system were \(80 \%\) efficient? (a) \(8 \mathrm{MW}\) (b) \(10 \mathrm{MW}\) (c) \(12.5 \mathrm{MW}\) (d) \(16 \mathrm{MW}\)

Step-by-Step Solution

Answer

(a) 8 MW

Step 1: Determine the Height Difference

Calculate the height difference between the dam and the power house. The dam is at a height of \(550\, \text{m}\) and the power house is at a height of \(50\, \text{m}\). Thus, the height difference, \(h\), is: \[ h = 550\, \text{m} - 50\, \text{m} = 500\, \text{m} \]

Step 2: Calculate the Gravitational Potential Energy

The gravitational potential energy \(E_p\) of an object can be found using the formula: \[ E_p = m \cdot g \cdot h \] where \(m = 2000\, \text{kg}\) is the mass, \(g = 9.8\, \text{m/s}^2\) is the acceleration due to gravity, and \(h = 500\, \text{m}\) is the height. Substituting these values, we get: \[ E_p = 2000\, \text{kg} \cdot 9.8\, \text{m/s}^2 \cdot 500\, \text{m} = 9,800,000\, \text{J} \] This is the energy per second (or power in watts, since 1 J/s = 1 W) before efficiency losses.

Step 3: Account for System Efficiency

The actual power output considers the efficiency of the system. Given the system is \(80\%\) efficient, the effective power output \(P_{out}\) can be calculated using: \[ P_{out} = \text{Efficiency} \times E_p \] \[ P_{out} = 0.8 \times 9,800,000\, \text{W} = 7,840,000\, \text{W} = 7.84\, \text{MW} \]

Step 4: Choose the Closest Multiple Choice Answer

The calculated power output is \(7.84\, \text{MW}\). Looking at the provided options: (a) \(8\, \text{MW}\) (b) \(10\, \text{MW}\) (c) \(12.5\, \text{MW}\) (d) \(16\, \text{MW}\) The closest answer is \(8\, \text{MW}\) (option a).

Key Concepts

Gravitational potential energy

Gravitational potential energy is the energy stored in an object due to its position in a gravitational field, typically its height above the ground. It's a key concept for understanding how hydroelectric power is generated. In physics, the energy is given by the formula:\[ E_p = m \cdot g \cdot h \]where:

• \(m\) is the mass of the object (in kg),
• \(g\) is the acceleration due to gravity (\(9.8\, \text{m/s}^2\)),
• \(h\) is the height above the reference point (in meters).

In hydroelectric systems, the potential energy of water stored at a height is converted into electrical energy. For the dam in the exercise, with water at a height of \(500\, \text{m}\), and assuming the water flow is \(2000\, \text{kg/s}\), the gravitational energy is significant because it represents energy available to generate power.

System efficiency

System efficiency quantifies how effectively a physical system converts energy from one form to another. For systems like hydroelectric power plants, it represents the proportion of gravitational potential energy that is converted into electrical power. Efficiency is expressed as a percentage and can be calculated with:\[ \text{Efficiency} = \frac{\text{Useful output energy}}{\text{Total input energy}} \times 100\]In the exercise, the system's efficiency is \(80\%\). This means that only \(80\%\) of the gravitational potential energy of water gets converted into electrical energy. The other \(20\%\) is typically lost due to factors such as friction and heat. Understanding efficiency helps in optimizing systems, ensuring maximum energy production, and minimizing losses.

Power conversion

Power conversion in physics is the process of changing energy from one form into another—here, from mechanical to electrical energy. Power is the rate at which work is done or energy is transferred, and it's measured in watts (W). For a hydroelectric dam, potential energy is first transformed into kinetic energy as water moves, and then into electrical energy by turbines. The effective power output is calculated by multiplying the potential energy available (in joules per second) by the system efficiency:\[ P_{out} = \text{Efficiency} \times E_p \]Given that the system's efficiency is \(80\%\), the power converted is \(7,840,000\, \text{W} = 7.84\, \text{MW}\). This conversion process is crucial in power engineering, as it allows industries to harness natural processes efficiently for energy generation.

Physics problem solving

Physics problem solving involves the application of principles, formulas, and logical reasoning to find solutions. In any complex problem, such as calculating hydroelectric power output, breaking down the problem into manageable steps is essential. Here's a simple approach:

• Identify all the known values (like mass, height, and efficiency). This clarifies what you have to work with.
• Use these knowns to calculate intermediate quantities—such as gravitational potential energy.
• Apply relevant formulas systematically. For example, start calculating potential energy and then move to account for efficiency.
• Compare the calculated results with given options (for exercises with multiple choice).

Develop these problem-solving skills to become adept at tackling a range of challenges in physics and engineering.