/*! 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 9 A gasoline engine has a power ou... [FREE SOLUTION] | 91Ó°ÊÓ

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A gasoline engine has a power output of \(180 \mathrm{~kW}\) (about \(241 \mathrm{hp}\) ). Its thermal efficiency is \(28.0 \%\). (a) How much heat must be supplied to the engine per second? (b) How much heat is discarded by the engine per second?

Short Answer

Expert verified
Heat supplied per second: 642,857.14 J/s. Heat discarded per second: 462,857.14 J/s.

Step by step solution

01

Understand the Problem

We are given a power output of the engine, which is the useful work done over time, as 180 kW, and the thermal efficiency of the engine as 28%. We need to find how much heat is supplied to the engine per second (part a) and how much heat is discarded (part b).
02

Recall Efficiency Formula

The thermal efficiency \(E\) of an engine is given by \[E = \frac{W}{Q_{in}},\]where \(W\) is the useful work done per second (power output), and \(Q_{in}\) is the heat energy supplied to the engine per second.
03

Calculate Heat Supplied

Rearrange the efficiency formula to find \(Q_{in}\): \[Q_{in} = \frac{W}{E}\]Substitute \(W = 180 \text{ kW} = 180,000 \text{ J/s}\) and \(E = 0.28\):\[Q_{in} = \frac{180,000}{0.28} \approx 642,857.14 \text{ J/s}.\]
04

Calculate Heat Discarded

The heat discarded \(Q_{out}\) is the difference between the heat supplied and the useful work: \[Q_{out} = Q_{in} - W.\]Substitute the known values:\[Q_{out} = 642,857.14 - 180,000 \approx 462,857.14 \text{ J/s}.\]
05

Summarize Findings

From our calculations, we find that the heat supplied to the engine per second is approximately 642,857.14 J/s, and the heat discarded per second is approximately 462,857.14 J/s.

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics and is simply the movement of thermal energy from one substance or system to another. It can occur through conduction, convection, or radiation. In the context of a heat engine, like the gasoline engine described in the exercise, heat transfer involves the engine receiving heat energy from a source, such as burning gasoline, which then facilitates mechanical work.
To understand how heat is utilized in engines, we first consider the heat input (\(Q_{in}\)), which is the thermal energy supplied to the engine. This energy is not fully converted into useful work due to inherent inefficiencies, which result in some heat being expelled or wasted. This concept aids in comprehending the thermal efficiency of engines.
Heat not used for work output is often released through exhaust gases or dissipated by cooling mechanisms. Heat transfer efficiency is crucial as it impacts the amount of fuel required and also the environmental footprint.
Power Output
Power output refers to the rate at which a machine, like our internal combustion engine, performs work. It signifies how much energy is converted into mechanical work per unit of time. In the exercise example, it is clearly defined as 180 kW, which is equivalent to about 241 horsepower. This figure represents the engine's ability to convert heat energy from fuel into motion or force. It's also an essential component in calculating engine efficiency, as we consider it while determining how much of the supplied heat is ultimately useful.
A greater power output implies that the engine can perform more work over the same period, but it’s also necessary to consider the efficiency, as a less efficient engine might use more fuel to achieve this power output. Therefore, understanding the balance between power output and heat input is crucial for evaluating engine performance.
Thermodynamics
Thermodynamics is the science of energy and entropy, encompassing the laws that govern heat transfer and energy conversion. It plays a critical role in understanding the operation of engines. The first law of thermodynamics, also known as the conservation of energy, states that energy cannot be created or destroyed but only transformed from one form to another.
In engines, mechanical work is produced by converting heat energy as per the thermodynamic principles. The efficiency of this process is explained by the second law of thermodynamics, which involves entropy. This law posits that no process is perfectly efficient, meaning some energy will invariably be lost as waste heat.
  • First Law: Energy conservation - Used to track energy flow in engines.
  • Second Law: Entropy increase - Justifies energy loss in systems.
These thermodynamic laws underpin our calculations for thermal efficiency and emphasize the importance of minimizing heat loss for better fuel economy and performance.
Heat Engine Calculations
Heat engine calculations provide a structured approach to evaluating the performance of engines, like determining how much heat must be supplied or discarded. With these calculations, we better understand the balance between energy input, output, and losses.
In our exercise, the engine's thermal efficiency helps compute the heat supplied (\(Q_{in}\)) using the formula \(E = \frac{W}{Q_{in}}\). By rearranging the formula, we calculated the heat supplied as \(Q_{in} = \frac{180,000}{0.28} \approx 642,857.14 \text{ J/s}\). This shows us how much energy is required for the engine to perform its work. Additionally, calculating the heat discarded (\(Q_{out}\)) is crucial to understanding energy losses: \(Q_{out} = Q_{in} - W\). This yields \(462,857.14 \text{ J/s}\), indicating the portion of energy not converted into work. Accounting for these calculations gives valuable insights into an engine’s fuel efficiency, helping optimize both its environmental impact and economic use of resources.

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Most popular questions from this chapter

An ice-making machine operates as a Carnot refrigerator. It takes heat from water at \(0.0^{\circ} \mathrm{C}\) and exhausts the heat into a room at \(24.0^{\circ} \mathrm{C}\). Suppose that it converts \(85.0 \mathrm{~kg}\) of water at \(0.0^{\circ} \mathrm{C}\) into ice at \(0.0^{\circ} \mathrm{C}\). (a) How much heat must be removed from the water? (b) How much work energy must be supplied to the refrigerator?

A typical coal-fired power plant generates \(1000 \mathrm{MW}\) of usable power at an overall thermal efficiency of \(40 \%\). (a) What is the rate of heat input to the plant? (b) The plant burns anthracite coal, which has a heat of combustion of \(2.65 \times 10^{7} \mathrm{~J} / \mathrm{kg} .\) How much coal does the plant use per day, if it operates continuously? (c) At what rate is heat ejected into the cool reservoir, which is the nearby river? (d) The river's temperature is \(18.0^{\circ} \mathrm{C}\) before it reaches the power plant and \(18.5^{\circ} \mathrm{C}\) after it has received the plant's waste heat. Calculate the river's flow rate, in cubic meters per second. (e) By how much does the river's entropy increase each second?

Premium gasoline produces \(1.23 \times 10^{8} \mathrm{~J}\) of heat per gallon when it is burned at a temperature of approximately \(400^{\circ} \mathrm{C}\) (although the amount can vary with the fuel mixture). If the car's engine is \(25 \%\) efficient, three-fourths of that heat is expelled into the air, typically at \(20^{\circ} \mathrm{C}\). If your car gets 35 miles per gallon of gas, by how much does the car's engine change the entropy of the world when you drive 1.0 mile? Does it decrease or increase it?

An engineer is working with a Carnot engine that has an unknown cold-reservoir temperature \(\left(T_{\mathrm{C}}\right)\) but a known and controllable hot- reservoir temperature \(\left(T_{\mathrm{H}}\right) .\) He measures the efficiency of the engine as a function of the hot-reservoir temperature and produces the following data set: $$ \begin{array}{cc} \hline T_{\mathrm{H}}(\mathrm{K}) & e \\ \hline 300 & 0.133 \\ 370 & 0.298 \\ 425 & 0.390 \\ 483 & 0.461 \\ 535 & 0.513 \\ \hline \end{array} $$ Produce a linearized plot of the engine efficiency as a function of the hot- reservoir temperature. Using a "best fit" to the data, determine the cold- reservoir temperature.

If the proposed plant is built and produces \(10 \mathrm{MW}\) but the rate at which waste heat is exhausted to the cold water is \(165 \mathrm{MW}\), what is the plant's actual efficiency? A. \(5.7 \%\) B. \(6.1 \%\) C. \(6.5 \%\) D. \(16.5 \%\)

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