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Chapter 5: Problem 46

In a few places where the air is very cold in the winter, such as \(-30^{\circ} \mathrm{C}\), it is possible to find a temperature of \(13^{\circ} \mathrm{C}\) below ground. What efficiency will a heat engine have when operating between these two thermal reservoirs?

Short Answer

Expert verified
The efficiency of the heat engine is approximately 15.04%.

Step by step solution

01

Understand the Concept of Efficiency

The efficiency of a heat engine is determined by the temperature difference between the heat reservoirs. Specifically, the efficiency is given by \( \eta = 1 - \frac{T_C}{T_H} \), where \( T_C \) and \( T_H \) are the temperatures of the cold and hot reservoirs, respectively, in Kelvin.
02

Convert Celsius to Kelvin

To apply the formula, convert the given temperatures from Celsius to Kelvin by adding 273.15. The cold temperature \(-30^{\circ} \mathrm{C}\) becomes \(243.15 \text{ K}\), and the hot temperature \(13^{\circ} \mathrm{C}\) becomes \(286.15 \text{ K}\).
03

Substitute the Temperatures into the Efficiency Formula

Substitute \( T_C = 243.15 \text{ K} \) and \( T_H = 286.15 \text{ K} \) into the efficiency formula: \[ \eta = 1 - \frac{243.15}{286.15} \].
04

Perform the Calculation

Calculate \( \frac{243.15}{286.15} \) which equals approximately 0.8496. Therefore, \( \eta = 1 - 0.8496 \approx 0.1504 \).
05

Convert Efficiency to Percentage

Lastly, convert the efficiency to a percentage by multiplying by 100. Thus, \( \eta \approx 0.1504 \times 100 = 15.04\% \).

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Key Concepts

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

Kelvin Scale
The Kelvin scale is a crucial part of understanding thermal processes, especially in physics and thermodynamics. Unlike the Celsius scale, which starts at the freezing point of water, the Kelvin scale starts at absolute zero. Absolute zero, or 0 Kelvin, is the lowest possible temperature, where particles have minimal thermal energy.
To convert temperatures from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example,
  • -30掳C converts to 243.15 K,
  • 13掳C converts to 286.15 K.

This conversion is vital when calculating the efficiency of heat engines, as the laws of thermodynamics rely on absolute temperature values.
Temperature Conversion
Temperature conversion is a fundamental step in thermodynamics calculations, particularly when transitioning from Celsius to Kelvin.
This conversion ensures that calculations remain consistent with thermodynamic laws, which require absolute temperature values defined by the Kelvin scale. The conversion formula is simple:
  • Kelvin = Celsius + 273.15

It鈥檚 essential to use this conversion when calculating the efficiency of a heat engine, as this ensures accurate computations aligned with the physical properties of the thermal systems involved.
Thermal Reservoirs
Thermal reservoirs are integral components in the study of heat engines. They refer to large bodies that can absorb or reject an unlimited amount of heat without undergoing a temperature change themselves. This property makes them ideal for modeling and understanding heat transfer processes.
In the context of the given exercise, there are two reservoirs:
  • The cold reservoir at -30掳C (243.15K)
  • The hot reservoir at 13掳C (286.15K)
These reservoirs set the temperature limits for the operation of our heat engine. They define the potential maximum efficiency by determining the temperature gradient over which the engine operates.
Heat Engine Concepts
Heat engines are fascinating devices that convert thermal energy into mechanical work. They operate between two thermal reservoirs, absorbing heat from the hot reservoir and partially converting it into work, while expelling the remaining heat to the cold reservoir.
The efficiency of a heat engine is quantified by using the Carnot efficiency formula: \[ \eta = 1 - \frac{T_C}{T_H} \]
where:
  • \( T_H \) is the temperature of the hot reservoir in Kelvin
  • \( T_C \) is the temperature of the cold reservoir in Kelvin
This relationship shows that the greater the temperature difference between the two reservoirs, the higher the potential efficiency of the engine. Understanding these concepts helps appreciate the practical limitations and potentials of real-world heat engines.

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Most popular questions from this chapter

A window-mounted air conditioner removes 3.5 Btu from the inside of a home using 1.75 Btu work input. How much energy is released outside, and what is its COP?

Hot combustion gases (air) at \(1500 \mathrm{~K}\) are used as the heat source in a heat engine where the gas is cooled to \(750 \mathrm{~K}\) and the ambient is at \(300 \mathrm{~K}\). This is not a constant-temperature source. How does that affect the efficiency?

For each of the cases below, determine if the heat engine satisfies the first law (energy equation) and if it violates the second law. a. \(\dot{Q}_{H}=6 \mathrm{~kW}, \quad \dot{Q}_{L}=4 \mathrm{~kW}, \quad \dot{W}=2 \mathrm{~kW}\) b. \(\dot{Q}_{H}=6 \mathrm{~kW}, \quad \dot{Q}_{L}=0 \mathrm{~kW}, \quad \dot{W}=6 \mathrm{~kW}\) c. \(\dot{Q}_{H}=6 \mathrm{~kW}, \quad \dot{Q}_{L}=2 \mathrm{~kW}, \quad \dot{W}=5 \mathrm{~kW}\) d. \(\dot{Q}_{H}=6 \mathrm{~kW}, \quad \dot{Q}_{L}=6 \mathrm{~kW}, \quad \dot{W}=0 \mathrm{~kW}\)

Helium has the lowest normal boiling point of any of the elements at \(4.2 \mathrm{~K}\). At this temperature the enthalpy of evaporation is \(83.3 \mathrm{~kJ} / \mathrm{kmol}\). A Carnot refrigeration cycle is analyzed for the production of \(1 \mathrm{kmol}\) of liquid helium at \(4.2 \mathrm{~K}\) from saturated vapor at the same temperature. What is the work input to the refrigerator and the COP for the cycle with an ambient at \(300 \mathrm{~K} ?\)

Air in a rigid \(1-\mathrm{m}^{3}\) box is at \(300 \mathrm{~K}, 200 \mathrm{kPa}\). It is heated to \(600 \mathrm{~K}\) by heat transfer from a reversible heat pump that receives energy from the ambient at \(300 \mathrm{~K}\) besides the work input. Use constant specific heat at \(300 \mathrm{~K}\). Since the COP changes, write \(d Q=m_{\text {air }} C_{\mathrm{v}} d T\) and find \(d W\). Integrate \(d W\) with the temperature to find the required heat pump work.
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