/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A power plant burns \(1000 \math... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 11

A power plant burns \(1000 \mathrm{~kg}\) of coal each hour and produces \(500 \mathrm{~kW}\) of power. Calculate the overall thermal efficiency if each \(\mathrm{kg}\) of coal produces \(6 \mathrm{MJ}\) of energy.

Short Answer

Expert verified
The overall thermal efficiency is 30%.

Step by step solution

01

Determine the energy input per hour

First, calculate the total energy produced by burning coal in one hour. Given that each kilogram of coal produces 6 MJ of energy, and the power plant burns 1000 kg of coal each hour, the total energy input is:\[ \text{Total Energy Input} = 1000 \text{ kg/hour} \times 6 \text{ MJ/kg} = 6000 \text{ MJ/hour} \]
02

Convert power output to MJ per hour

Next, convert the power output from kW to MJ per hour. Since 1 kW is equivalent to 1 kJ/s, and there are 3600 seconds in an hour, we get:\[ 500 \text{ kW} = 500 \text{ kJ/s} \times 3600 \text{ s/hour} = 1800 \text{ MJ/hour} \]
03

Calculate the thermal efficiency

Finally, use the formula for thermal efficiency:\[ \text{Thermal Efficiency} = \frac{\text{Power Output}}{\text{Energy Input}} \times 100 \]Substitute the values from the previous steps:\[ \text{Thermal Efficiency} = \frac{1800 \text{ MJ/hour}}{6000 \text{ MJ/hour}} \times 100 = 30\text{ %} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

energy input
Energy input in a thermal power plant refers to the total amount of energy available from the fuel being burned. In this case, the fuel is coal, and each kilogram of coal produces a specific amount of energy. To calculate the energy input:
  • Determine the amount of fuel burned per hour.
  • Multiply this by the energy content of the fuel (in MJ per kg).
This gives you the total energy input per hour.
For example, if a power plant burns 1000 kg of coal each hour, and each kilogram of coal produces 6 MJ of energy, then the energy input per hour is:
\[ \text{Total Energy Input} = 1000 \text{ kg} \times 6 \text{ MJ/kg} = 6000 \text{ MJ/hour} \]
power output
Power output is the useful energy that a power plant produces. It is often measured in kilowatts (kW) or megawatts (MW). To convert it to the same units as energy input, you need to consider the duration for which the power is generated.
  • 1 kW = 1 kJ/s
  • There are 3600 seconds in one hour
So, converting kW to MJ per hour involves multiplying by 3600 seconds.
For instance, if a power plant produces 500 kW, you convert this to MJ per hour as follows: \[ 500 \text{ kW} = 500 \text{ kJ/s} \times 3600 \text{ seconds/hour} = 1800 \text{ MJ/hour} \]
conversion efficiency
Conversion efficiency measures how effectively a power plant converts the energy in fuel into useful power. It is a ratio of the power output to the energy input, expressed as a percentage. The formula for thermal efficiency is:
\[ \text{Thermal Efficiency} = \frac{\text{Power Output}}{\text{Energy Input}} \times 100 \]For example, if a power plant's power output is 1800 MJ/hour and the energy input is 6000 MJ/hour, then the thermal efficiency is:
\[ \text{Thermal Efficiency} = \frac{1800 \text{ MJ/hour}}{6000 \text{ MJ/hour}} \times 100 = 30\text{ %} \]This tells you that only 30% of the energy in the coal is converted into useful power, and the rest is lost as waste heat.
coal combustion energy
Coal combustion energy refers to the energy released when coal is burned. This energy is harnessed to generate electricity. Different types of coal have varying energy contents. The specific energy content is measured in megajoules per kilogram (MJ/kg).
  • Energy content of coal is crucial for calculating energy input.
  • Higher energy content means more energy is available per unit of coal burned.
Understanding coal combustion energy helps in determining the overall efficiency of the power plant. It's also essential for comparing different fuels and making decisions about fuel types for energy production.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A reversible refrigeration unit is used to cool a space to \(5^{\circ} \mathrm{C}\) by transferring heat to the surroundings, which are at \(25^{\circ} \mathrm{C}\). The same unit is then used to cool the space to \(-20^{\circ} \mathrm{C}\). Estimate the cooling rate for the second condition if the cooling rate for the first is 5 tons.

A Carnot heat engine produces \(10 \mathrm{hp}\) by transferring energy between two reservoirs at \(40^{\circ} \mathrm{F}\) and \(212^{\circ} \mathrm{F}\). Calculate the rate of heat transfer from the high-temperature reservoir. SOLUTION The engine efficiency is $$ The efficiency is also given by \(\eta=\dot{W} / \dot{Q}_{H}\). Thus, we see that $$ \dot{Q}_{H}=\frac{\dot{W}}{\eta}=\frac{(10 \mathrm{hp})(2545 \mathrm{Btu} / \mathrm{hr} / \mathrm{hp})}{0.2560}=99.410 \mathrm{Btu} / \mathrm{hr} $$ \eta=1-\frac{T_{L}}{T_{H}}=1-\frac{500}{672}=0.2560 $$

(a) What is the maximum efficiency that can result from an engine that operates on the thermal gradients in the ocean? The surface waters at the proposed location are at \(85^{\circ} \mathrm{F}\) and those at a reasonable depth are at \(50^{\circ} \mathrm{F}\). (b) What would be the maximum COP of a heat pump, operating between the two layers, used to heat an offshore oil rig?

A heat engine operates on a Carnot cycle with an efficiency of 75 percent. What COP would a refrigerator operating on the same cycle have? The low temperature is \(0^{\circ} \mathrm{C}\). The efficiency of the heat engine is given by \(\eta=1-T_{L} / T_{H}\). Hence, $$ T_{H}=\frac{T_{L}}{1-\eta}=\frac{273}{1-0.75}=1092 \mathrm{~K} $$ The COP for the refrigerator is then $$ \mathrm{COP}_{R}=\frac{T_{L}}{T_{H}-T_{L}}=\frac{273}{1092-273}=0.3333 $$

A Carnot engine operates between reservoirs at temperatures \(T_{1}\) and \(T_{2}\), and a second Carnot engine operates between reservoirs maintained at \(T_{2}\) and \(T_{3}\). Express the efficiency \(\eta_{3}\) of the third engine operating between \(T_{1}\) and \(T_{3}\) in terms of the efficiencies \(\eta_{1}\) and \(\eta_{2}\) of the other two engines.
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Dynamics

Read Explanation

Work Energy and Power

Read Explanation

Translational Dynamics

Read Explanation

Mechanics and Materials

Read Explanation

Thermodynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.