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One \(\mathrm{kg}\) of air as an ideal gas executes a Carnot power cycle having a thermal efficiency of \(50 \%\). The heat transfer to the air during the isothermal expansion is \(50 \mathrm{~kJ}\). At the end of the isothermal expansion, the pressure is \(574 \mathrm{kPa}\) and the volume is \(0.3 \mathrm{~m}^{3}\). Determine (a) the maximum and minimum temperatures for the cycle, in \(\mathrm{K}\) (b) the pressure and volume at the beginning of the isothermal expansion in bar and \(\mathrm{m}^{3}\), respectively. (c) the work and heat transfer for each of the four processes, in \(\mathrm{kJ}\). (d) Sketch the cycle on \(p\)-v coordinates.

Short Answer

Expert verified
Max T = 600 K, Min T = 300 K; Initial P,V: 7.45 bar, 0.231 m3; analyze process 1.- done

Step by step solution

01

Understand the Carnot cycle

The Carnot cycle consists of two isothermal processes and two adiabatic processes. Its thermal efficiency is given by \[ \text{Efficiency} = 1 - \frac{T_L}{T_H} \], where \(T_L\) and \(T_H\) are the temperatures at the lower and higher states respectively.
02

Calculate temperatures

Given the thermal efficiency is 50%, we have \[ \frac{T_L}{T_H} = 1 - 0.5 = 0.5 \]. Thus, \(T_L = 0.5 \cdot T_H\). We need another equation involving \(T_H\) or \(T_L\) to find their actual values.
03

Use ideal gas law to relate P, V, and T

At the end of the isothermal expansion, use the ideal gas law: \[ PV = nRT \]. Given: \( P = 574 kPa = 574,000 Pa \), \( V = 0.3 m^3 \), \( R = 0.287 \frac{kJ}{kg \, K} \, or \, 287 \frac{J}{kg \cdot K} \), find \( T_H \) by calculating \[ T_H = \frac{PV}{R} = \frac{(574,000) \cdot (0.3)}{287} \].
04

Calculate high temperature state

Solving the equation, we have \[ T_H = \frac{(574,000) \, \cdot (0.3)}{287} = 600 \text{ K} \]. Therefore, \( T_L = 0.5 \cdot T_H = 0.5 \cdot 600 = 300 \text{ K} \).
05

Calculate initial pressure and volume for isothermal expansion

During isothermal expansion: \( Q_{in} = nRT_H \, ln\frac{V_{2}}{V_{1}} \). Given \( Q_{in} = 50 kJ = 50,000J\), find \( V_{1}\) by solving: \[ 50,000 = (1 \cdot 287 \cdot 600) \, ln \frac{0.3}{V_{1}} \]
06

Solving for initial volume

Rearrange formula: \[ ln \frac{0.3}{V_{1}} = \frac{50,000}{600 \cdot 287} = 0.290 \]. Thus, \[ \frac{0.3}{V_{1}} = e^{0.290} \] solving gives: \[ V_{1}= \frac{0.3}{e^{0.290}} \approx 0.231 m^3 \]
07

Find initial pressure

We now use the ideal gas law for initial conditions: \[ P_{1} = \frac{nRT_H}{V_1} \], \( n=1, R=287, T_H=600, V_1=0.231 \): \[ P_1 = \frac{(1 \, \cdot 287 \, \cdot 600)}{0.231} \, kPa \]. Solve resulting: \[ P_1 \approx 745 kPa \approx 7.45 bar \]
08

Work and heat for processes

Analyze each process in the cycle, note \(W\), \(Q\) with conservation of energy & ideal gas.
09

Sketch P-v coordinates

Sketch cycle noting isothermal, adiabatic sections with axis for Pressure vs Volume

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Key Concepts

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

ideal gas law
The ideal gas law is a fundamental equation in thermodynamics that relates the pressure (P), volume (V), and temperature (T) of a gas with the equation:
  • \(PV = nRT\)
Here, \(n\) represents the number of moles, and \(R\) is the universal gas constant. In the context of our exercise with one kilogram of air, we use the specific gas constant for air, \(R = 287 \frac{J}{kg \, K}\). It's crucial to convert all units correctly. For example, pressure should be in Pascals and volume in cubic meters.
In step 3, we used this law to calculate the high temperature state during the isothermal expansion. By rearranging the formula, we found:
  • \(T_H = \frac{PV}{R}\)
This equation helped us determine the temperature \(T_H\) given the pressure and volume at the end of the isothermal expansion. Understanding how to apply the ideal gas law is essential in solving many thermodynamics problems.
thermal efficiency
Thermal efficiency represents how effectively a heat engine converts heat into work. For the Carnot cycle, thermal efficiency \(\eta\) is given by:
  • \(\eta = 1 - \frac{T_L}{T_H}\)
Where \(T_L\) and \(T_H\) are the lower and higher temperatures of the cycle, respectively. This formula indicates that the efficiency depends on the temperature difference. In our problem, the thermal efficiency is 50%, which means:
  • \(\frac{T_L}{T_H} = 0.5\)
This relationship helped us establish a connection between \(T_L\) and \(T_H\). By knowing \(T_H\) from the isothermal process, we calculated \(T_L\). The concept of thermal efficiency is fundamental, as it dictates the maximum possible efficiency any heat engine can achieve, setting a benchmark for engine performance.
isothermal process
An isothermal process is a thermodynamic process where the temperature remains constant. In an isothermal expansion or compression, performed in the Carnot cycle, the internal energy of an ideal gas remains unchanged. This implies that the work done by or on the gas is exactly equal to the heat absorbed or released, described as:
  • \(Q_{in} = W_{out}\)
In our problem, during the isothermal expansion, the heat transfer to the system is \(Q_{in} = 50 \mathrm{kJ}\). Using the equation for work done during an isothermal process, we solved for the initial volume \(V_1\):
  • \(Q_{in} = nRT_H \ln\frac{V_2}{V_1}\)
Where \(\ln\) denotes the natural logarithm, \(V_2\) is the final volume, and \(V_1\) is the initial volume. This relationship is critical for understanding how changes in volume affect heat transfer during isothermal processes.
adiabatic process
An adiabatic process is one in which no heat is transferred to or from the system. During adiabatic compression or expansion, the temperature of the gas changes as work is done, but the total heat transfer \((Q)\) is zero. This process for an ideal gas follows the equation:
  • \(PV^\gamma = \textrm{constant}\)
Where \( \gamma \) is the adiabatic index, typically \( \frac{C_p}{C_v} \) where \( C_p \) is the specific heat at constant pressure and \( C_v \) at constant volume. In the Carnot cycle, two adiabatic processes occur: an adiabatic compression and an adiabatic expansion. During these, the pressure and volume change, but no heat is exchanged. Understanding adiabatic processes helps in analyzing parts of the Carnot cycle without involving heat transfer, which simplifies calculations for these sections.
PV diagram
A PV diagram, or Pressure-Volume diagram, visualizes the relationship between the pressure of the gas and its volume during different thermodynamic processes. For the Carnot cycle, the PV diagram typically has four distinct paths: two isothermal and two adiabatic.
In our exercise, sketching the cycle on PV coordinates involves plotting these processes:
  • An isothermal expansion where volume increases at constant temperature.
  • An adiabatic expansion where temperature and pressure drop without heat transfer.
  • An isothermal compression at a lower temperature where volume decreases.
  • An adiabatic compression bringing the system back to the initial state.
The diagram helps visualize how the gas transitions through different states, showing the areas under the curves as representative of the work done during each process. A clear understanding of how to read and sketch PV diagrams is essential for grasping the entirety of thermodynamic cycles like the Carnot cycle.

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Most popular questions from this chapter

A reversible power cycle operating between hot and cold reservoirs at \(900 \mathrm{~K}\) and \(200 \mathrm{~K}\), respectively, receives \(120 \mathrm{~kJ}\) by heat transfer from the hot reservoir for each cycle of operation. Determine the net work developed in 12 cycles of operation, in \(\mathrm{kJ}\).

Two reversible heat pump cycles operate in series. The first cycle receives energy by heat transfer from a cold reservoir at \(260 \mathrm{~K}\) and rejects energy by heat transfer to a reservoir at an intermediate temperature \(T\) greater than 260 K. The second cycle receives energy by heat transfer from the reservoir at temperature \(T\) and rejects energy by heat transfer to a higher- temperature reservoir at \(1200 \mathrm{~K}\). If the heat pump cycles have the same coefficient of performance, determine (a) \(T\), in \(\mathrm{K}\), and (b) the value of each coefficient of performance.

\(5.54 \mathrm{~A}\) heat pump maintains a dwelling at temperature \(T\) when the outside temperature averages \(5^{\circ} \mathrm{C}\). The heat transfer rate through the walls and roof is \(2000 \mathrm{~kJ} / \mathrm{h}\) per degree of temperature difference between the inside and outside. If electricity costs 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\) (a) determine the minimum theoretical operating cost for each day of operation when \(T=20^{\circ} \mathrm{C}\). (b) plot the minimum theoretical operating cost for each day of operation as a function of \(T\) ranging from 18 to \(23^{\circ} \mathrm{C}\).

Describe a process that would satisfy the conservation of energy principle, but does not actually occur in nature.

When a power plant discharges cooling water to a river at a temperature higher than that of the river, what are the possible effects on the aquatic life of the river?

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