Textbook Question
A heat engine does 200 J of work per cycle while exhausting 400 J of waste heat. What is the engine's thermal efficiency?
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Ch 21: Heat Engines and Refrigerators
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 21, Problem 7
The power output of a car engine running at 2400 rpm is 500 kW. How much (a) work is done and (b) heat is exhausted per cycle if the engine's thermal efficiency is 20%? Give your answers in kJ.
Verified step by step guidance1
Step 1: Understand the given data. The engine's power output is 500 kW, the rotational speed is 2400 rpm, and the thermal efficiency is 20% (0.20). The goal is to calculate the work done and the heat exhausted per cycle. Start by converting the rotational speed from rpm to cycles per second (frequency). Use the formula: \( f = \frac{\text{rpm}}{60} \).
Step 2: Calculate the work done per second (useful power output). Since power is the rate of doing work, the work done per second is equal to the power output. Use the formula: \( W_{\text{useful}} = P \times t \), where \( P \) is the power output (500 kW) and \( t \) is 1 second. This gives the work done per second in kilojoules.
Step 3: Determine the work done per cycle. Divide the work done per second by the frequency (cycles per second) calculated in Step 1. Use the formula: \( W_{\text{cycle}} = \frac{W_{\text{useful}}}{f} \).
Step 4: Relate the thermal efficiency to the heat input. Thermal efficiency is defined as \( \eta = \frac{W_{\text{useful}}}{Q_{\text{in}}} \), where \( Q_{\text{in}} \) is the heat input. Rearrange this formula to find \( Q_{\text{in}} = \frac{W_{\text{useful}}}{\eta} \).
Step 5: Calculate the heat exhausted per cycle. The heat exhausted is the difference between the heat input and the work done. Use the formula: \( Q_{\text{exhausted}} = Q_{\text{in}} - W_{\text{useful}} \). Divide this value by the frequency to find the heat exhausted per cycle: \( Q_{\text{exhausted, cycle}} = \frac{Q_{\text{exhausted}}}{f} \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Power and Work
Power is the rate at which work is done or energy is transferred over time. In the context of engines, power output indicates how much work the engine can perform in a given time frame. The relationship between power (P), work (W), and time (t) is given by the equation P = W/t, which can be rearranged to find work done as W = P * t.
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Power
Thermal Efficiency
Thermal efficiency is a measure of how effectively an engine converts the energy from fuel into useful work. It is defined as the ratio of the useful work output to the total energy input, expressed as a percentage. In this case, with a thermal efficiency of 20%, only 20% of the energy consumed by the engine is converted into work, while the remaining 80% is lost as waste heat.
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Heat Exhaustion
Heat exhaustion refers to the energy that is not converted into useful work and is instead released into the environment as waste heat. In an engine, this is calculated by determining the total energy input and subtracting the useful work output. Understanding heat exhaustion is crucial for evaluating engine performance and efficiency, as it highlights energy losses that could be minimized for better fuel economy.
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Related Practice
Textbook Question
A gas following the pV trajectory of FIGURE EX21.11 does 60 J of work per cycle. What is Vmax?
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Textbook Question
What are (a) the thermal efficiency and (b) the heat extracted from the hot reservoir for the heat engine shown in FIGURE EX21.16?
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Textbook Question
What are (a) Wout and QH and (b) the thermal efficiency for the heat engine shown in FIGURE EX21.14?
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Textbook Question
A Boeing 777 jet engine, the world's largest, has a power output of 82 MW. It burns jet fuel with an energy density of 43 MJ /kg. What is the engine's fuel consumption rate, in kg/s, if its efficiency is 30%?
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