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A diesel engine performs J of mechanical work and discards J of heat each cycle. How much heat must be supplied to the engine in each cycle?
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Ch 20: The Second Law of Thermodynamics
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552
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TOC.tableOfContents
- Ch 01: Units, Physical Quantities & Vectors37
- Ch 02: Motion Along a Straight Line57
- Ch 03: Motion in Two or Three Dimensions45
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws54
- Ch 06: Work & Kinetic Energy11
- Ch 07: Potential Energy & Conservation6
- Ch 08: Momentum, Impulse, and Collisions15
- Ch 09: Rotation of Rigid Bodies37
- Ch 10: Dynamics of Rotational Motion44
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics27
- Ch 13: Gravitation34
- Ch 14: Periodic Motion26
- Ch 15: Mechanical Waves22
- Ch 16: Sound & Hearing28
- Ch 17: Temperature and Heat28
- Ch 18: Thermal Properties of Matter35
- Ch 19: The First Law of Thermodynamics29
- Ch 20: The Second Law of Thermodynamics18
- Ch 21: Electric Charge and Electric Field28
- Ch 22: Gauss' Law21
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF24
- Ch 26: Direct-Current Circuits29
- Ch 27: Magnetic Field and Magnetic Forces16
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction22
- Ch 30: Inductance14
- Ch 31: Alternating Current20
- Ch 33: The Nature and Propagation of Light17
- Ch 34: Geometric Optics29
- Ch 35: Interference13
- Ch 36: Diffraction19
- Ch 37: Special Relativity19
- Ch 38: Photons: Light Waves Behaving as Particles12
- Ch 39: Particles Behaving as Waves25
- Ch 40: Quantum Mechanics I: Wave Functions20
- Ch 41: Quantum Mechanics II: Atomic Structure21
- Ch 42: Molecules and Condensed Matter17
- Ch 43: Nuclear Physics13
- Ch 44: Particle Physics and Cosmology17
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Non è quello che usi tu?Cambia libro di testo
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Young & Freedman Calc 15th EditionCh 20: The Second Law of ThermodynamicsProblema 8a
Young & Freedman Calc 15th EditionCh 20: The Second Law of ThermodynamicsProblema 8aCapitolo 20, Problema 8a
Calculate the theoretical efficiency for an Otto-cycle engine with and .
Guida verificata passo dopo passo1
Understand the Otto cycle: The Otto cycle is a thermodynamic cycle that describes the functioning of a typical spark ignition piston engine. It consists of two isochoric processes (constant volume) and two adiabatic processes (no heat exchange).
Identify the formula for the efficiency of an Otto cycle engine: The efficiency \( \eta \) of an Otto cycle engine is given by the formula \( \eta = 1 - \frac{1}{r^{\gamma - 1}} \), where \( r \) is the compression ratio and \( \gamma \) is the specific heat ratio.
Substitute the given values into the formula: You are given \( \gamma = 1.40 \) and \( r = 9.50 \). Substitute these values into the efficiency formula: \( \eta = 1 - \frac{1}{9.50^{1.40 - 1}} \).
Simplify the expression: Calculate \( 1.40 - 1 \) to get \( 0.40 \). Then, compute \( 9.50^{0.40} \) to find the value of the denominator.
Calculate the efficiency: Subtract the result from 1 to find the theoretical efficiency of the Otto-cycle engine.
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Otto Cycle
The Otto cycle is a thermodynamic cycle that describes the functioning of a typical spark-ignition piston engine. It consists of two adiabatic and two isochoric processes. Understanding the Otto cycle is crucial for calculating the efficiency of engines that operate on this cycle, such as gasoline engines.
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10:59
The Otto Cycle
Specific Heat Ratio (γ)
The specific heat ratio, denoted as γ (gamma), is the ratio of the specific heat at constant pressure (Cp) to the specific heat at constant volume (Cv). It is a critical parameter in thermodynamics, affecting the efficiency of cycles like the Otto cycle. For air, γ is typically around 1.40, which influences the theoretical efficiency calculation.
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06:50Specific Heat & Temperature Changes
Compression Ratio (r)
The compression ratio, denoted as r, is the ratio of the maximum to minimum volume in the cylinder of an internal combustion engine. It is a key factor in determining the efficiency of the Otto cycle, as higher compression ratios generally lead to higher efficiencies. In this question, r is given as 9.50, which is used in the efficiency formula.
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17:42Finding the Compression Ratio
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A Carnot engine is operated between two heat reservoirs at temperatures of K and K. If the engine receives kJ of heat energy from the reservoir at K in each cycle, how many joules per cycle does it discard to the reservoir at K?
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