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Chapter 3: Problem 114

Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) contained in a piston-cylinder arrangement, initially at 6 bar and \(400 \mathrm{~K}\), undergoes an expansion to a final temperature of \(298 \mathrm{~K}\), during which the pressure-volume relationship is \(p V^{1.2}=\) constant. Assuming the ideal gas model for the \(\mathrm{CO}_{2}\), determine the final pressure, in bar, and the work and heat transfer, each in \(\mathrm{kJ} / \mathrm{kg}\).

Short Answer

Expert verified
Final pressure is 0.899 bar. Work done is 64.3 kJ/kg. Heat transfer is -2.8 kJ/kg.

Step by step solution

01

Determine Initial Volume

Use the ideal gas law to find the initial volume. The ideal gas law is given by \[ pV = nRT \] Since we're given the initial pressure (\[ p_1 = 6 \text{ bar} = 600 \text { kPa} \]) and temperature (\[ T_1 = 400 \text{K} \]), we'll solve for volume using the specific gas constant for \[ CO_2 \]. For \[ CO_2 \], \[ R = 0.1889 \text{ kJ/(kg K)} \]. Assuming the mass to be 1 kg,\[ p_1 \times V_1 = 1 \times 0.1889 \times 400 \] kJ. Therefore, \[ V_1 = \frac{0.1889 \times 400}{600} = 0.126 \text{ m}^3.\]
02

Use Pressure-Volume Relationship

Given the relationship \[ p V^{1.2} = \text{constant} \]), use the initial conditions to find the constant: \[ p_1 V_1^{1.2} = C \] Substitute \[ p_1 = 600 \text{ kPa} \], \[ V_1 = 0.126 \text{ m}^3 \]: \[ C = 600 \times (0.126)^{1.2} = 76.56 \]
03

Solve for Final Volume

Use the ideal gas law again at the final temperature \[ T_2 = 298 \text{K} \]: \[ p_2 V_2 = 0.1889 \times 298 \]. But we don't know \[ p_2 \] yet. We relate it using \[ p_2 V_2^{1.2} = 76.56 \] and substitute \[ V_2 = \frac{0.1889 \times 298}{p_2} \]: \[ p_2 (\frac{0.1889 \times 298}{p_2})^{1.2} = 76.56 \]. We solve for \[ p_2 \] which gives \[ 89.9 \text{ kPa} or 0.899 \text{ bar}. \]
04

Calculate Work Done

Since we are dealing with a polytropic process, the work \[ W \] is given by: \[ W = \frac{p_1 V_1 - p_2 V_2}{1 - n} \]. Using \[ n = 1.2 \], \[ V_2 = \frac{0.1889 \times 298}{0.899 \times 100} = 0.626 \text{ m}^3 \]: \[W = \frac{600 \times 0.126 - 0.899 \times 0.626}{1 - 1.2} = 64.3 \text{kJ/kg} \]
05

Determine Heat Transfer

The first law of thermodynamics for a closed system in terms of specific quantities is given by: \[ \text{Q} = \text{W} + \text{U}_2 - \text{U}_1 \]. For an ideal gas, \[ \text{U} = \text{C}_v \text{T} \]. Given \[ \text{C}_v = 0.657 \] (kJ/kg.K). Thus, \[ U_1 = 0.657 \times 400 \] and \[ U_2 = 0.657 \times 298 \]: \[ U_1 = 262.8, \text{U}_2 = 195.7 \]. Substitute into the first law, \[Q = 64.3 + (195.7 - 262.8) = 64.3 - 67.1 = -2.8 \text{kJ/kg}. \]

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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 fundamental to understanding gases' behavior under different conditions. It is represented by the equation: \[ pV = nRT \] where
  • \(p\) is the pressure,
  • \(V\) is the volume,
  • \(n\) is the number of moles,
  • \(R\) is the universal gas constant \((8.314 \text{ J/(mol路K)})\)
  • \(T\) is the temperature in Kelvin.
In many cases, especially in problems like this one, it is practical to use a specific gas constant \((R)\) for a particular gas. For carbon dioxide \((CO_2)\), the specific gas constant \(R\) is approximately \(0.1889 \text{ kJ/(kg路K)}\). The initial conditions in the problem (\[ p_1 = 600 \text{ kPa}, T_1 = 400 \text{ K}\]) allow us to use the ideal gas law to find the initial volume..
polytropic process
A polytropic process is a thermodynamic process that follows the equation \[ pV^n = k \]where
  • \(p\) is the pressure,
  • \(V\) is the volume,
  • \(n\) is the polytropic index,
  • \(k\) is a constant.
In this problem, the relationship is given as \[ pV^{1.2} = \text{constant}\] with \(n = 1.2\). First, determine the constant using initial conditions. Then, use this relationship to find the final pressure and volume given the final temperature. This approach simplifies solving problems involving work and heat transfer in polytropic processes.
first law of thermodynamics
The first law of thermodynamics, also known as the law of energy conservation, states that the total energy of an isolated system remains constant. In terms of internal energy \(U\), work \(W\), and heat transfer \(Q\), it is expressed as: \[ \text{Q} = \text{W} + \text{U}_2 - \text{U}_1 \] where
  • \(Q\) is heat added to the system,
  • \(W\) is the work done by the system,
  • \(U_1\) and \(U_2\) are the initial and final internal energies, respectively.
For ideal gases, internal energy only depends on temperature: \[ U = C_v T \]where \(C_v\) is the specific heat at constant volume. In the provided problem, the specific heat for \(CO_2\) is given as \(0.657 \text{kJ/(kg路K)}\). Using the final and initial temperatures, calculate the change in internal energy to determine the total heat transfer.
work done on gas
In thermodynamics, work done during a process is the energy transferred when a force is applied over a distance. For a polytropic process, work can be calculated using: \[ W = \frac{p_1 V_1 - p_2 V_2}{1 - n} \]where
  • \(p_1\) and \(p_2\) are the initial and final pressures,
  • \(V_1\) and \(V_2\) are the initial and final volumes,
  • \(n\) is the polytropic index.
Sometimes these pressures and volumes need to be converted to consistent units before calculation. In the provided example, the \(CO_2\) undergoes expansion with values \(p_1 = 600 \text{kPa}\), \(p_2 = 0.899 \text{bar}\), \(V_1 = 0.126 \text{m}^3\), and \(V_2 = 0.626 \text{m}^3\). We then substitute these into the equation for work done to find \(W\).
heat transfer calculations
Heat transfer in a thermodynamic system can be calculated using the first law of thermodynamics. This involves finding the difference between work done and change in internal energy. Using the internal energies derived from specific heat \(C_v\) and temperature (\(U = C_v T\)), compute: \[ \text{Q} = \text{W} + \text{U}_2 - \text{U}_1 \]Given the specific heat \(C_v\) of \(0.657 \text{kJ/(kg路K)}\) for \(CO_2\), and the temperatures provided, calculate initial and final internal energies: \[ U_1 = C_v T_1 \text{ and } U_2 = C_v T_2 \]Substitute all known values, including work \(W\), to find the heat transfer. This lets you understand energy added or removed from the system.

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Most popular questions from this chapter

Using the tables for water, determine the specified property data at the indicated states. In each case, locate the state on sketches of the \(p-v\) and \(T-v\) diagrams. (a) At \(p=2 \mathrm{MPa}, T=300^{\circ} \mathrm{C}\). Find \(u\), in kJ/kg. (b) At \(p=2.5 \mathrm{MPa}, T=200^{\circ} \mathrm{C}\). Find \(u\), in \(\mathrm{kJ} / \mathrm{kg}\). (c) At \(T=170^{\circ} \mathrm{F}, x=50 \%\). Find \(u\), in Btu/lb. (d) At \(p=100 \mathrm{lbf} / \mathrm{in}^{2}, T=300^{\circ} \mathrm{F}\). Find \(h\), in Btu/lb. (e) At \(p=1.5 \mathrm{MPa}, v=0.2095 \mathrm{~m}^{3} / \mathrm{kg}\). Find \(h\), in \(\mathrm{kJ} / \mathrm{kg}\).

The following table lists temperatures and specific volumes of water vapor at two pressures: $$ \begin{array}{lccc} {}{}{p=1.0 \mathrm{MPa}} &{}{c}{p=1.5 \mathrm{MPa}} \\ \hline T\left({ }^{\circ} \mathrm{C}\right) & v\left(\mathrm{~m}^{3} / \mathrm{kg}\right) & T(\mathrm{C}) & v\left(\mathrm{~m}^{3} / \mathrm{kg}\right) \\ \hline 200 & 0.2060 & 200 & 0.1325 \\ 240 & 0.2275 & 240 & 0.1483 \\ 280 & 0.2480 & 280 & 0.1627 \end{array} $$ Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate (a) the specific volume at \(T=240^{\circ} \mathrm{C}, p=1.25 \mathrm{MPa}\), in \(\mathrm{m}^{3} / \mathrm{kg}\). (b) the temperature at \(p=1.5 \mathrm{MPa}, v=0.1555 \mathrm{~m}^{3} / \mathrm{kg}\), in \({ }^{\circ} \mathrm{C}\). (c) the specific volume at \(T=220^{\circ} \mathrm{C}, p=1.4 \mathrm{MPa}\), in \(\mathrm{m}^{3} / \mathrm{kg}\).

In each of the following cases, ammonia contained in a closed, rigid tank is heated from an initial saturated vapor state at temperature \(T_{1}\) to the final temperature, \(T_{2}\) : (a) \(T_{1}=20^{\circ} \mathrm{C}, T_{2}=40^{\circ} \mathrm{C}\). Using \(I T\), determine the final pressure, in bar. (b) \(T_{1}=70^{\circ} \mathrm{F}, T_{2}=120^{\circ} \mathrm{F}\). Using \(I T\), determine the final pressure, in lbf/in. \({ }^{2}\) Compare the pressure values determined using \(I T\) with those obtained using the appropriate Appendix tables for ammonia.

Ammonia in a piston-cylinder assembly undergoes two processes in series. Initially, the ammonia is saturated vapor at \(p_{1}=100 \mathrm{lbf} / \mathrm{in} .^{2}\) Process \(1-2\) involves cooling at constant pressure until \(x_{2}=75 \%\). The second process, from state 2 to state 3 , involves heating at constant volume until \(x_{3}=100 \%\). Kinetic and potential energy effects are negligible. For \(1.2 \mathrm{lb}\) of ammonia, determine (a) the heat transfer and work for Process 1-2 and (b) the heat transfer for Process 2-3, all in Btu.

Five \(\mathrm{lb}\) of propane is contained in a closed, rigid tank initially at \(80 \mathrm{lbf} / \mathrm{in}^{2}, 110^{\circ} \mathrm{F}\). Heat transfer occurs until the final temperature in the tank is \(0^{\circ} \mathrm{F}\). Kinetic and potential energy effects are negligible. Show the initial and final states on a \(T-v\) diagram and determine the amount of energy transfer by heat, in Btu.
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