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(II) A certain power plant puts out 580 MW of electric power. Estimate the heat discharged per second, assuming that the plant has an efficiency of 32\(\%\).

Short Answer

Expert verified
The plant discharges approximately 1232.5 MW of heat per second.

Step by step solution

01

Understand the Concept

To find the heat discharged by the power plant, we need to use the concept of efficiency in power plants, which is defined as the ratio of the useful electric power output to the total power input.
02

Identify Given and Required Information

We are given that the electric power output is 580 MW and the efficiency is 32%. We need to find the total heat power input and the heat discharged per second.
03

Convert Efficiency to Decimal

Efficiency is given in percentage form, so we convert it to a decimal by dividing by 100. Therefore, the efficiency is \(0.32\).
04

Calculate Total Power Input

Use the formula for efficiency: \( \text{Efficiency} = \frac{\text{Useful Power Output}}{\text{Total Power Input}} \). Rearranging gives \( \text{Total Power Input} = \frac{\text{Useful Power Output}}{\text{Efficiency}} \). Substitute in the known values: \( \text{Total Power Input} = \frac{580 \text{ MW}}{0.32} \approx 1812.5 \text{ MW} \).
05

Calculate Heat Discharged

The heat discharged per second is the difference between the total power input and the electric power output. So, \( \text{Heat Discharged} = \text{Total Power Input} - \text{Useful Power Output} \). Substitute the values: \( \text{Heat Discharged} = 1812.5 \text{ MW} - 580 \text{ MW} = 1232.5 \text{ MW} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Discharge Calculation
Calculating the heat discharged by a power plant involves understanding how efficiently the plant converts raw energy into usable electric energy. In the context of our exercise, the heat discharged is the leftover energy after useful electricity has been produced. When a power plant operates, it cannot convert all the fuel energy into electrical energy due to inherent inefficiencies. This unconverted energy is usually expelled as waste heat.

To find the heat discharged per second, we first need to determine the **total power input**, which includes both the useful power output and the waste power discharge. Using the efficiency equation:
  • Efficiency ( \(\eta\) ) = \( \frac{\text{Electric Power Output}}{\text{Total Power Input}} \)
Once we find the total power input, heat discharge is calculated by subtracting the electric power output from this total power input. This gives us:
  • Heat Discharged per Second = Total Power Input - Electric Power Output.
In this example, the heat discharged is 1232.5 MW, which highlights how much energy is lost as heat in the power generation process.
Electric Power Output
Electric power output is the measure of usable energy produced by a power plant. This is the energy that is supplied to homes, businesses, and other facilities. In our exercise, the power plant generates an electric power output of 580 MW. This number represents the effective power being delivered for practical use.

Electric power output is crucial for calculating efficiency. Efficiency tells us how well a power plant converts the fuel it consumes into useful electricity. The formula to determine efficiency is:
  • \( \text{Efficiency} = \frac{\text{Electric Power Output}}{\text{Total Power Input}} \)
Given the efficiency and knowing either the total power input or the power output, you can solve for the missing variable. In practice, maintaining a high electric power output with minimal loss is essential for economic and environmental considerations.
Total Power Input
The total power input of a power plant is the initial energy in the form of fuel or other sources that the plant uses to eventually generate electricity. It encompasses all energy that the plant processes, not just the electricity produced. This concept is central when evaluating how efficient a power plant is.

Using the efficiency formula, the total power input can be calculated from the known efficiency and electric power output:
  • \( \text{Total Power Input} = \frac{\text{Electric Power Output}}{\text{Efficiency}} \)
For our example, the plant's efficiency is 32%, and the output is 580 MW. Replacing these values into the formula gives us a total power input of approximately 1812.5 MW. This indicates the total amount of energy the plant takes in to operate, a large portion of which does not end up as usable electricity, showing the inefficiencies inherent in energy conversion.

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Most popular questions from this chapter

An ideal heat pump is used to maintain the inside temperature of a house at \(T_\mathrm{in} = 22^\circ\)C when the outside temperature is \(T_\mathrm{out}\). Assume that when it is operating, the heat pump does work at a rate of 1500W. Also assume that the house loses heat via conduction through its walls and other surfaces at a rate given by (650 W/C\(^\circ\))(\(T_\mathrm{in} - T_\mathrm{out}\)). (\(a\)) For what outside temperature would the heat pump have to operate all the time in order to maintain the house at an inside temperature of 22\(^\circ\)C? (\(b\)) If the outside temperature is 8\(^\circ\)C, what percentage of the time does the heat pump have to operate in order to maintain the house at an inside temperature of 22\(^\circ\)C?

(II) A Carnot engine's operating temperatures are 210\(^\circ\)C and 45\(^\circ\)C. The engine's power output is 910W. Calculate the rate of heat output.

(I) Solar cells (Fig. 15\(-\)26) can produce about 40 W of electricity per square meter of surface area if directly facing the Sun. How large an area is required to supply the needs of a house that requires 24 kWh/day? Would this fit on the roof of an average house? (Assume the Sun shines about 9 h/day.)

It has been suggested that a heat engine could be developed that made use of the temperature difference between water at the surface of the ocean and water several hundred meters deep. In the tropics, the temperatures may be 27\(^\circ\)C and 4\(^\circ\)C, respectively. (\(a\)) What is the maximum efficiency such an engine could have? (\(b\)) Why might such an engine be feasible in spite of the low efficiency? (\(c\)) Can you imagine any adverse environmental effects that might occur?

(II) A four-cylinder gasoline engine has an efficiency of 0.22 and delivers 180 J of work per cycle per cylinder. If the engine runs at 25 cycles per second (1500 rpm), determine (\(a\)) the work done per second, and (\(b\)) the total heat input per second from the gasoline. (\(c\)) If the energy content of gasoline is 130 MJ per gallon, how long does one gallon last?

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