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 3
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%?
Verified step by step guidance1
Step 1: Understand the relationship between power, energy density, and efficiency. The power output of the engine is related to the rate of energy consumption and the efficiency of the engine. Efficiency is defined as the ratio of useful power output to the total energy input.
Step 2: Write the formula for efficiency: \( \text{Efficiency} = \frac{\text{Useful Power Output}}{\text{Total Energy Input Rate}} \). Rearrange this formula to find the total energy input rate: \( \text{Total Energy Input Rate} = \frac{\text{Useful Power Output}}{\text{Efficiency}} \).
Step 3: Substitute the given values into the formula. The useful power output is \( 82 \text{ MW} = 82 \times 10^6 \text{ W} \), and the efficiency is \( 30\% = 0.30 \). Calculate the total energy input rate using \( \text{Total Energy Input Rate} = \frac{82 \times 10^6}{0.30} \).
Step 4: Relate the total energy input rate to the fuel consumption rate. The energy density of the fuel is given as \( 43 \text{ MJ/kg} = 43 \times 10^6 \text{ J/kg} \). The fuel consumption rate can be calculated using \( \text{Fuel Consumption Rate} = \frac{\text{Total Energy Input Rate}}{\text{Energy Density}} \).
Step 5: Substitute the values for total energy input rate and energy density into the formula for fuel consumption rate: \( \text{Fuel Consumption Rate} = \frac{\text{Total Energy Input Rate}}{43 \times 10^6} \). This will give the fuel consumption rate in \( \text{kg/s} \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Power Output
Power output refers to the rate at which energy is produced or consumed. In this context, the Boeing 777 jet engine has a power output of 82 MW, which means it converts energy at a rate of 82 megajoules per second. Understanding power output is crucial for calculating fuel consumption, as it directly relates to how much energy the engine requires to operate efficiently.
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Energy Density
Energy density is a measure of how much energy is stored in a given mass of fuel, expressed in joules per kilogram (J/kg). For jet fuel, the energy density is 43 MJ/kg, indicating that one kilogram of jet fuel can produce 43 megajoules of energy when burned. This concept is essential for determining how much fuel is needed to achieve the desired power output of the engine.
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Efficiency
Efficiency in this context refers to the ratio of useful power output to the total power input, expressed as a percentage. The engine's efficiency of 30% means that only 30% of the energy from the fuel is converted into useful work, while the rest is lost as waste energy. This concept is critical for calculating the actual fuel consumption rate, as it affects how much fuel is needed to produce the required power output.
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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
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.
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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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