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Chapter 6: Problem 16

One-tenth kmol of carbon monoxide \((\mathrm{CO})\) in a pistoncylinder assembly undergoes a process from \(p_{1}=150 \mathrm{kPa}\), \(T_{1}=300 \mathrm{~K}\) to \(p_{2}=500 \mathrm{kPa}, T_{2}=370 \mathrm{~K}\). For the process, \(\mathrm{W}=-300 \mathrm{~kJ}\). Employing the ideal gas model, determine (a) the heat transfer, in kJ. (b) the change in entropy, in \(\mathrm{kJ} / \mathrm{K}\). Show the process on a sketch of the \(T-s\) diagram.

Short Answer

Expert verified
The heat transfer is -158.88 kJ and the change in entropy is -0.342 kJ/K.

Step by step solution

01

- Identify the initial and final states

The initial state of the carbon monoxide is given by: pressure, \(p_{1} = 150\, \text{kPa}\), and temperature, \(T_{1} = 300\, \text{K}\). The final state is given by: pressure, \(p_{2} = 500\, \text{kPa}\), and temperature, \(T_{2} = 370\, \text{K}\).
02

- Calculate the mass of the carbon monoxide

Using the molar mass of CO (M = 28.01 g/mol or 28.01 kg/kmol) and the amount of substance (n = 0.1 kmol), calculate the mass: \[ m = n \times M = 0.1 \times 28.01 = 2.801 \, \text{kg} \]
03

- Use the ideal gas model to find the specific heat capacities

For CO, the specific heat capacity at constant volume, \(c_{v}\), is approximately 0.718 kJ/(kg*K) and at constant pressure, \(c_{p}\), is approximately 1.007 kJ/(kg*K).
04

- Find the change in internal energy

The change in internal energy (\(\Delta U\)) can be calculated using: \[ \Delta U = m \cdot c_{v} \cdot (T_{2} - T_{1}) \] Substituting the values: \[ \Delta U = 2.801 \cdot 0.718 \cdot (370 - 300) = 2.801 \cdot 0.718 \cdot 70 = 141.12 \text{ kJ} \]
05

- Use the first law of thermodynamics to find the heat transfer

The first law of thermodynamics states that: \[ Q = \Delta U + W \] Given that work \(W = -300 \text{kJ}\) (work done on the system), and \(\Delta U = 141.12 \text{kJ}\), substitute: \[ Q = 141.12 - 300 = -158.88 \text{ kJ} \]
06

- Calculate the change in entropy

The change in entropy (\(\Delta S\)) for an ideal gas during a process can be calculated using: \[ \Delta S = m \cdot c_v \cdot \ln\left(\frac{T_2}{T_1}\right) + m \cdot R \cdot \ln\left(\frac{v_2}{v_1}\right) \] The ratio \(\frac{v_2}{v_1}\) can be replaced using the ideal gas law: \[ \Delta S = m \cdot c_v \cdot \ln\left(\frac{T_2}{T_1}\right) + m \cdot R \cdot \ln\left(\frac{p_1}{p_2}\right) \] Given \[ R = \frac{8.314}{28.01} = 0.2972 \text{kJ/(kg*K)} \] Substituting the values: \[ \Delta S = 2.801 \cdot 0.718 \cdot \ln\left(\frac{370}{300}\right) + 2.801 \cdot 0.2972 \cdot \ln\left(\frac{150}{500}\right) \] Calculating each term, \( \Delta S = 2.801 \cdot 0.718 \cdot 0.2087 + 2.801 \cdot 0.2972 \cdot (-0.9163) \) \( \Delta S \approx 0.4216 - 0.7635 = -0.3419 \) Therefore, \( \Delta S = -0.342 \text{ kJ/K} \)
07

- Plot the T-s diagram

Draw the T-s diagram with temperature (T) on the y-axis and entropy (S) on the x-axis. Mark the initial state (T1, S1) and the final state (T2, S2). Connect these two states representing the process path.

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Key Concepts

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

Thermodynamic Processes
In thermodynamics, a process involves the change of a system from one state to another. Here, we follow how carbon monoxide ostate changes under specific conditions. Usually influenced by pressure, volume, and temperature, such processes can be isothermal, isobaric, isochoric, or adiabatic.
  • Isothermal: Temperature remains constant.
  • Isobaric: Pressure remains constant.
  • Isochoric: Volume remains constant.
  • Adiabatic: No heat transfer happens.
For this problem, the carbon monoxide undergoes a non-standard thermodynamic process where both pressure and temperature change.Understanding these variations helps in calculating crucial parameters like enthalpy, work done, and heat exchange.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed, only transformed from one form to another. Mathematically, it is expressed as: \[ Q = \triangle U + W \] Where:
  • Q: Heat added to the system.
  • 螖鲍: Change in internal energy.
  • W: Work done by or on the system.
For our example, we calculate the change in internal energy (螖U) using the specific heat capacities, and then determine the heat transfer (Q) by considering the work performed on the gas. In simpler terms, the energy added to the system as heat, minus the energy lost as work, tells us how much the internal energy changes.
Entropy Change
Entropy measures the disorder or randomness in a system. In thermodynamics, understanding the entropy change (鈭哠) of a system during a process is crucial. For an ideal gas, the entropy change is given by: \[ \triangle S = m \times c_v \times \text{ln}\frac{T_2}{T_1} + m \times R \times \text{ln}\frac{v_2}{v_1} \] Here:
  • m: Mass of the gas.
  • cv: Specific heat at constant volume.
  • T: Temperature at different states.
  • R: Universal gas constant.
  • v: Specific volume at different states.
Understanding entropy allows us to predict the direction of spontaneous processes and the feasibility of energy conversions. In our solution, we use the initial and final states' temperatures and pressures to compute the entropy change, considering both temperature rise and pressure drop contributions.
Heat Transfer Calculations
Heat transfer analysis is critical in understanding energy flow in thermodynamic systems. To find the heat transfer (Q), we use the First Law of Thermodynamics, incorporating internal energy changes and work done. The heat transfer equation becomes: \[ Q = \triangle U + W \] For our exercise, we know the work done (W = -300 kJ) on the system and the internal energy change (螖U = 141.12 kJ). Substituting these values into the equation, we find: \[ Q = 141.12 \text{ kJ} - 300 \text{ kJ} = -158.88 \text{ kJ} \] This negative value signifies that heat is leaving the system. Understanding heat transfer is vital for designing efficient thermal systems and predicting their behavior during different processes.
Specific Heat Capacities
Specific heat capacities are properties of substances that indicate the amount of heat required to change their temperature. For gases, two essential forms exist: Specific heat at constant volume (cv) and at constant pressure (cp). For carbon monoxide (CO), these are:
  • cv: 0.718 kJ/(kg*K)
  • cp: 1.007 kJ/(kg*K)
The relationship between these capacities is connected through the gas constant (R): \[ c_p - c_v = R \] By knowing these values, we can calculate the changes in internal energy and entropy more accurately. In our problem, specific heat capacities help determine the internal energy change by: \[ \triangle U = m \times c_v \times (T_2 - T_1) \] Where m is the mass of CO, showcasing how these capacities enable us to relate temperature changes to energy changes effectively.

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Most popular questions from this chapter

An insulated, rigid tank whose volume is \(0.5 \mathrm{~m}^{3}\) is connected by a valve to a large vessel holding steam at 40 bar, \(500^{\circ} \mathrm{C}\). The tank is initially evacuated. The valve is opened only as long as required to fill the tank with steam to a pressure of 20 bar. Determine (a) the final temperature of the steam in the tank, in \({ }^{\circ} \mathrm{C}\), (b) the final mass of the steam in the tank, in \(\mathrm{kg}\), and (c) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\).

By injecting liquid water into superheated steam, the desuperheater shown in Fig. P6.83 has a saturated vapor stream at its exit. Steady-state operating data are provided in the accompanying table. Stray heat transfer and all kinetic and potential energy effects are negligible. (a) Locate states 1,2 , and 3 on a sketch of the \(T-s\) diagram. (b) Determine the rate of entropy production within the desuperheater, in \(\mathrm{kW} / \mathrm{K}\). $$ \begin{array}{ccccccc} \text { State } & p(\mathrm{MPa}) & T\left({ }^{\circ} \mathrm{C}\right) & v \times 10^{3}\left(\mathrm{~m}^{3} / \mathrm{kg}\right) & u(\mathrm{k}) / \mathrm{kg}) & h(\mathrm{k}) / \mathrm{kg}) & s(\mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \\ \hline 1 & 2.7 & 40 & 1.0066 & 167.2 & 169.9 & 0.5714 \\ 2 & 2.7 & 300 & 91.01 & 2757.0 & 3002.8 & 6.6001 \\ 3 & 2.5 & \text { sat. vap. } & 79.98 & 2603.1 & 2803.1 & 6.2575 \end{array} $$

A rigid, insulated tank with a volume of \(21.61 \mathrm{ft}^{3}\) is filled initially with air at \(110 \mathrm{lbf} / \mathrm{in}^{2}, 535^{\circ} \mathrm{R}\). A leak develops, and air slowly escapes until the pressure of the air remaining in the tank is \(15 \mathrm{lbf} / \mathrm{in}^{2}\). Employing the ideal gas model with \(k=1.4\) for the air, determine the amount of mass remaining in the tank, in lb, and its temperature, in \({ }^{\circ} \mathrm{R}\).

A rigid, insulated vessel is divided into two compartments connected by a valve. Initially, one compartment, occupying one-third of the total volume, contains air at \(500^{\circ} \mathrm{R}\), and the other is evacuated. The valve is opened and the air is allowed to fill the entire volume. Assuming the ideal gas model, determine the final temperature of the air, in \({ }^{\circ} \mathrm{R}\), and the amount of entropy produced, in Btu/ \(/{ }^{\circ} \mathrm{R}\) per \(\mathrm{lb}\) of air.

A cylindrical copper rod of base area A and length \(L\) is insulated on its lateral surface. One end of the rod is in contact with a wall at temperature \(T_{\mathrm{H}}\). The other end is in contact with a wall at a lower temperature \(T_{\mathrm{C}}\). At steady state, the rate at which energy is conducted into the rod from the hot wall is $$ \dot{Q}_{\mathrm{H}}=\frac{\kappa \mathrm{A}\left(T_{\mathrm{H}}-T_{\mathrm{C}}\right)}{L} $$ where \(\kappa\) is the thermal conductivity of the copper rod. (a) For the rod as the system, obtain an expression for the time rate of entropy production in terms of \(\mathrm{A}, L, T_{\mathrm{H}}, T_{\mathrm{C}}\), and \(\kappa\). (b) If \(T_{\mathrm{H}}=327^{\circ} \mathrm{C}, T_{\mathrm{C}}=77^{\circ} \mathrm{C}, \kappa=0.4 \mathrm{~kW} / \mathrm{m} \cdot \mathrm{K}, \mathrm{A}=0.1 \mathrm{~m}^{2}\), plot the heat transfer rate \(\dot{Q}_{\mathrm{H}}\), in \(\mathrm{kW}\), and the time rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), each versus \(L\) ranging from \(0.01\) to \(1.0 \mathrm{~m}\). Discuss.
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