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Chapter 20: Problem 3

A Gasoline Engine. A gasoline engine takes in \(1.61 \times\) \(10^{4} \mathrm{J}\) of heat and delivers 3700 \(\mathrm{J}\) of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of \(4.60 \times 10^{4} \mathrm{J} / \mathrm{g} .(\mathrm{a})\) What is the thermal efficiency? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?

Short Answer

Expert verified
The thermal efficiency is 22.98%, 12400 J of heat is discarded per cycle, the mass of fuel burned is 0.350 g per cycle, and the power output is 222 kW (or 297.19 hp).

Step by step solution

01

Calculate Thermal Efficiency

The thermal efficiency of a heat engine is given by the formula \( \eta = \frac{W}{Q_{in}} \), where \( W \) is the work done (3700 J) and \( Q_{in} \) is the heat input (\( 1.61 \times 10^4 \) J). Plug in the values to get:\[ \eta = \frac{3700}{1.61 \times 10^4} \approx 0.2298 \text{ or } 22.98\% \].
02

Calculate Heat Discarded

The heat discarded \( Q_{out} \) can be calculated using the formula \( Q_{out} = Q_{in} - W \). Substitute \( Q_{in} = 1.61 \times 10^4 \) J and \( W = 3700 \) J:\[ Q_{out} = 1.61 \times 10^4 - 3700 = 1.24 \times 10^4 \text{ J} \].
03

Calculate Mass of Fuel Burned

The mass of the fuel can be found using the heat input formula: \( Q_{in} = m \times c \), where \( c \) is the heat of combustion per gram (\( 4.60 \times 10^4 \text{ J/g} \)). Solving for \( m \), we get:\[ m = \frac{Q_{in}}{c} = \frac{1.61 \times 10^4}{4.60 \times 10^4} \approx 0.350 \text{ g} \].
04

Calculate Power Output in Kilowatts and Horsepower

Power (\( P \)) is the work done per unit time. With 60 cycles per second, the work done per second is \( W \times 60 \). Thus:\[ P = 3700 \times 60 = 222000 \text{ W} = 222 \text{ kW} \].To convert to horsepower, use the conversion \( 1 \text{ hp} = 746 \text{ W} \):\[ P = \frac{222000}{746} \approx 297.19 \text{ hp} \].

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Key Concepts

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

Thermal Efficiency
Thermal efficiency is a measure of how well a heat engine converts the heat it receives into useful work. It's expressed as a percentage, indicating the proportion of the heat energy input (\(Q_{in}\)) that is converted into work (\(W\)). In the context of our gasoline engine example, we can derive the thermal efficiency with the formula:
  • \[\eta = \frac{W}{Q_{in}} \]
  • Insert the values for the cycle given: Work output (\(3700 \, J\)) and the heat input (\(1.61 \times 10^4 \, J\)).
  • This yields a thermal efficiency of approximately 22.98%.
The closer this percentage is to 100%, the better the engine is at converting heat into work. However, no engine can achieve perfect efficiency; losses like heat dissipation always occur.
Understanding this concept helps in improving designs to enhance engine performance and energy conservation.
Heat Engine
A heat engine, like the gasoline engine in our example, is a device that converts heat energy into mechanical work. It operates through a thermodynamic cycle, often involving the following steps:
  • Heat \(Q_{in}\): Absorbs heat energy from burning fuel.
  • Work \(W\): Converts part of this heat into work.
  • Heat \(Q_{out}\): Discards leftover heat to a cooler environment.
For our gasoline engine, each cycle involves an intake of heat energy and some discarded heat. To maintain continuous operation, a heat engine must always discard some heat since not all of the input heat is converted into work. This disposal typically occurs through the exhaust system in a vehicle. In engineering terms, understanding the cycle helps in optimizing engines for better power output and lower fuel consumption.
Power Output
Power output reflects how much work an engine can produce per unit of time, usually measured in watts (W) or horsepower (hp). For an engine running at 60 cycles per second, it's crucial to calculate how effectively it performs:
  • Calculate total work done in one second, multiplying the work per cycle (\(3700 \, J\)) by the number of cycles per second (60).
  • This results in a total power output of \(222,000 \, W\), which converts to \(222 \, kW\).
  • To express this in horsepower, use the conversion rate \(1 \, hp = 746 \, W\), yielding approximately 297.19 hp.
Power output is vital for real-world applications like how fast a car can accelerate or how much weight a machine can lift鈥攈ighlighting the practicality of engine efficiency.
Mass of Fuel
The mass of fuel relates directly to the energy needs of the engine. In our example, it is determined by the heat of combustion \(c\) and the heat input per cycle \(Q_{in}\). The process is as follows:
  • Use formula:\[m = \frac{Q_{in}}{c}\]
  • Here, \(Q_{in} = 1.61 \times 10^4 \, J\) and \(c = 4.60 \times 10^4 \, J/g\).
  • This computes to a fuel mass of about \(0.350 \, g\) per cycle.
Understanding fuel mass is crucial for assessing how long an engine can run before refueling and the overall economic and environmental footprint of the engine. Lower fuel mass necessary per cycle signifies a more effective use of resources, resulting in cost savings and reduced emissions.

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Most popular questions from this chapter

A Carnot engine has an efficiency of 59\(\%\) and performs \(2.5 \times 10^{4} \mathrm{J}\) of work in each cycle. (a) How much heat does the engine extract from its heat source in each cycle? (b) Suppose the engine exhausts heat at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) What is the temperature of its heat source?

A 15.0 -kg block of ice at \(0.0^{\circ} \mathrm{C}\) melts to liquid water at \(0.0^{\circ} \mathrm{C}\) inside a large room that has a temperature of \(20.0^{\circ} \mathrm{C}\) . Treat the ice and the room as an isolated system, and assume that the room is large enough for its temperature change to be ignored. (a) Is the melting of the ice reversible or irreversible? Explain, using simple physical reasoning without resorting to any equations. (b) Calculate the net entropy change of the system during this process. Explain whether or not this result is consistent with your answer to part (a).

The coefficient of performance \(K=H / P\) is a dimension-less quantity. Its value is independent of the units used for \(H\) and \(P\) as long as the same units, such as watts, are used for both quantities. However, it is common practice to express \(H\) in Btu/h and \(P\) in watts. When these mixed units are used, the ratio \(H / P\) is called the energy efficiency rating \((\) EER). If a room air conditioner has a coefficient of performance \(K=3.0,\) what is its EER?

A room air conditioner has a coefficient of performance of 2.9 on a hot day and uses 850 \(\mathrm{W}\) of electrical power. (a) How many ioules of heat does the air conditioner remove from the room in one minute? (b) How many joules of heat does the air conditioner deliver to the hot outside air in one minute? (c) Explain why your answers to parts (a) and (b) are not the same.

An aircraft engine takes in 9000 \(\mathrm{J}\) of heat and discards 6400 \(\mathrm{J}\) each cycle. (a) What is the mechanical work output of the engine during one cycle? (b) What is the thermal efficiency of the engine?
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