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Chapter 5: Problem 19

A cyclic heat engine operates between a source temperature of \(700^{\circ} \mathrm{C}\) and a sink temperature of \(25^{\circ} \mathrm{C}\). What is the least rate of heat rejection per \(\mathrm{kW}\) net output of the engine?

Short Answer

Expert verified
0.442 kW

Step by step solution

01

Convert temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula: \[ T(K) = T(掳C) + 273.15 \]. For the source temperature: \[ T_1 = 700 + 273.15 = 973.15 \text{ K} \]. For the sink temperature: \[ T_2 = 25 + 273.15 = 298.15 \text{ K} \].
02

Calculate the efficiency of the Carnot engine

The efficiency of a Carnot engine is given by: \[ \text{Efficiency} = 1 - \frac{T_2}{T_1} \]. Substitute the values: \[ \text{Efficiency} = 1 - \frac{298.15}{973.15} \approx 0.6933 \text{ or } 69.33\text{\%} \].
03

Relate efficiency to heat rejected

The efficiency of the engine is also given by: \[ \text{Efficiency} = \frac{W}{Q_1} \], where \( W \) is the work output and \( Q_1 \) is the heat input. From this, \[ Q_1 = \frac{W}{\text{Efficiency}} \]. The heat rejected \( Q_2 \) is given by: \[ Q_2 = Q_1 - W \].
04

Substitute values to find heat rejected

Using the efficiency as a decimal (0.6933), and \( W = 1 \text{ kW} \): \[ Q_1 = \frac{1 \text{ kW}}{0.6933} \approx 1.442 \text{ kW} \]. Then, \[ Q_2 = 1.442 \text{ kW} - 1 \text{ kW} \approx 0.442 \text{ kW} \].

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Key Concepts

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

heat engine
A heat engine is a device that converts thermal energy (heat) into mechanical work. It operates by taking heat from a high-temperature source, converting part of this energy into work, and then releasing the remaining heat to a low-temperature sink. The critical function of a heat engine is to harness the energy from thermal gradients.

Heat engines are found in various applications, including internal combustion engines in cars and power plants.

Key points about heat engines:
  • They operate between two reservoirs: a high-temperature source and a low-temperature sink.
  • They convert part of the input heat into useful work.
  • The remaining heat is expelled to the sink.
Understanding the operation of heat engines is essential for optimizing their efficiency and developing sustainable energy solutions.
thermodynamic cycles
Thermodynamic cycles are sequences of processes that involve heat and work transfer, returning a system to its initial state. These cycles are fundamental to understanding how heat engines operate. In each cycle, the system absorbs heat, performs work, and rejects heat.

Important aspects of thermodynamic cycles:
  • They consist of distinct stages such as compression, heat addition, expansion, and heat rejection.
  • The net work output is the difference between the work done during expansion and the work required for compression.
  • Carnot and Stirling cycles are examples of idealized thermodynamic cycles.
Thermodynamic cycles are crucial for analyzing the performance of heat engines and refrigeration systems, focusing on maximizing work output or minimizing energy input.
Carnot cycle efficiency
The Carnot cycle represents an idealized heat engine with maximum possible efficiency operating between two temperatures. Its efficiency sets an upper limit for real-world engines.

Key formula for Carnot efficiency:
\text{Efficiency} = 1 - \frac{T_2}{T_1}

Here, \( T_1 \) and \( T_2 \) are the absolute temperatures of the source and sink, respectively, measured in Kelvin.

Steps to calculate Carnot engine efficiency:
  • Convert temperatures from Celsius to Kelvin using \( T(K) = T(掳C) + 273.15 \).
  • Use the Carnot efficiency formula to find the ratio of heat conversion into work.
For example, a heat engine with a source temperature of 700掳C (973.15 K) and a sink temperature of 25掳C (298.15 K) has an efficiency of 69.33%.

Understanding Carnot cycle efficiency helps in assessing and improving real engine performance.
temperature conversion
Temperature conversion is critical when dealing with thermodynamic processes, as equations often require temperatures in Kelvin rather than Celsius.

Formula for converting Celsius to Kelvin:
  • \( T(K) = T(掳C) + 273.15 \)
This straightforward addition ensures temperatures are in the absolute scale (Kelvin), providing a common basis for thermodynamic calculations.

Example conversions:
  • For 700掳C: \( 700 + 273.15 = 973.15 \) K
  • For 25掳C: \( 25 + 273.15 = 298.15 \) K
Understanding temperature conversion is essential for accurately applying thermodynamic principles and performing efficiency calculations.

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Most popular questions from this chapter

A power cycle operating at steady state receives energy by heat transfer at a rate \(\dot{Q}_{\mathrm{H}}\) at \(T_{\mathrm{H}}=1000 \mathrm{~K}\) and rejects energy by heat transfer to a cold reservoir at a rate \(Q_{\mathrm{C}}\) at \(T_{C}=300 \mathrm{~K}\). For each of the following cases, determine whether the cycle operates reversibly, operates irreversibly, or is impossible. (a) \(\dot{Q}_{\mathrm{H}}=500 \mathrm{~kW}, \dot{Q}_{c}=100 \mathrm{~kW}\) (b) \(\dot{Q}_{\mathrm{H}}=500 \mathrm{~kW}, \dot{W}_{\text {cycle }}=250 \mathrm{~kW}, \dot{Q}_{\mathrm{C}}=200 \mathrm{~kW}\) (c) \(\dot{W}_{\text {cyde }}=350 \mathrm{~kW}, \dot{Q}_{\mathrm{C}}=150 \mathrm{~kW}\) (d) \(\dot{Q}_{\mathrm{H}}=500 \mathrm{~kW}, \dot{Q}_{\mathrm{c}}=200 \mathrm{~kW}\)

An inventor claims to have developed a refrigerator that at steady state requires a net power input of \(0.54 \mathrm{~kW}\) to remove \(12,800 \mathrm{~kJ} / \mathrm{h}\) of energy by heat transfer from the freezer compartment at \(-20^{\circ} \mathrm{C}\) and discharge energy by heat transfer to a kitchen at \(27^{\circ} \mathrm{C}\). Evaluate this claim.

The U.S. Food and Drug Administration (FDA) has long permitted the application of citric acid, ascorbic acid, and other substances to keep fresh meat looking red longer. In 2002 , the FDA began allowing meat to be treated with carbon monoxide. Carbon monoxide reacts with myoglobin in the meat to produce a substance that resists the natural browning of meat, thereby giving meat a longer shelf life. Investigate the use of carbon monoxide for this purpose. Identify the nature of myoglobin and explain its role in the reactions that cause meat to brown or, when treated with carbon monoxide, allows the meat to appear red longer. Consider the hazards, if any, that may accompany this practice for consumers and for meat industry workers. Report your findings in a memorandum.

Insulin and several other pharmaceuticals required daily by those suffering from diabetes and other medical conditions have relatively low thermal stability. Those living and traveling in hot climates are especially at risk by heatinduced loss of potency of their pharmaceuticals. Design a wearable, lightweight, and reliable cooler for transporting temperature-sensitive pharmaceuticals. The cooler also must be solely powered by human motion. While the long-term goal is a moderately priced consumer product, the final project report need only provide the costing of a single prototype.

Experiments by Australian researchers suggest that the second law can be violated at the micron scale over time intervals of up to 2 seconds. If validated, some say these findings could place a limit on the engineering of nanomachines because such devices may not behave simply like miniaturized versions of their larger counterparts. Investigate the implications, if any, for nanotechnology of these experimental findings. Write a report including at least three references.
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