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Chapter 6: Problem 148

Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) at 1 bar, \(300 \mathrm{~K}\) enters a compressor operating at steady state and is compressed adiabatically to an exit state of 10 bar, \(520 \mathrm{~K}\). The \(\mathrm{CO}_{2}\) is modeled as an ideal gas, and kinetic and potential energy effects are negligible. For the compressor, determine (a) the work input, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of \(\mathrm{CO}_{2}\) flowing, (b) the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of \(\mathrm{CO}_{2}\) flowing, and (c) the isentropic compressor efficiency.

Short Answer

Expert verified
a) 186.12 \text{ kJ/kg} b) 0.027 \text{ kJ/K鈰卥g} c) 29.5\text{%}

Step by step solution

01

- Identify given data

Identify the initial and final states of the carbon dioxide: Initial state: - Pressure, \( P_1 = 1 \text{ bar} = 100 \text{ kPa} \) - Temperature, \( T_1 = 300 \text{ K} \) Final state: - Pressure, \( P_2 = 10 \text{ bar} = 1000 \text{ kPa} \) - Temperature, \( T_2 = 520 \text{ K} \)
02

- Apply the first law of thermodynamics

The process is adiabatic, so \( Q = 0 \). The first law of thermodynamics gives: \( \text{Work input per kg of } CO_2 \ \rightarrow W_{in} = \frac{{h_2 - h_1}}{{m}} \) Use the ideal gas model to get specific enthalpies, where \( h = c_p T \).
03

- Specific heat capacity for CO2

Assume specific heat capacity at constant pressure \( c_p = 0.846 \text{ kJ/kg鈰匥} \). Calculate the change in enthalpy: \( \triangle h = c_p \times (T_2 - T_1) \) Substituting values, we get: \( \triangle h = 0.846 \times (520 - 300) \) \( \triangle h = 0.846 \times 220 \) \( \triangle h = 186.12 \text{ kJ/kg} \)
04

- Work input

Since it's an adiabatic process, work input \( W_{in} = \triangle h \). Therefore, \( W_{in} = 186.12 \text{ kJ/kg} \).
05

- Entropy change

Entropy change in adiabatic process: Using ideal gas assumptions, \( \triangle s = c_p \times \text{ln} \frac{T_2}{T_1} - R \times \text{ln} \frac{P_2}{P_1} \). Given \( R_{CO2} = 0.1889 \text{ kJ/kg鈰匥} \), \( \triangle s = 0.846 \times \text{ln} \frac{520}{300} - 0.1889 \times \text{ln} \frac{1000}{100} \) \( \triangle s = 0.846 \times 0.5465 - 0.1889 \times 2.3026 \) \( \triangle s = 0.462 \text{ kJ/K鈰卥g} - 0.435 \text{ kJ/K鈰卥g} \) \( \triangle s = 0.027 \text{ kJ/K鈰卥g} \)
06

- Entropy production rate

Since entropy production is always positive in irreversible processes, Rate of entropy production: \( \triangle S_{gen} = \triangle s = 0.027 \text{ kJ/K鈰卥g} \).
07

- Isentropic efficiency

For isentropic efficiency, use the formula: \( \text{Isentropic efficiency} = \frac{W_{is}}{W_{in}} \) Isentropic process follows: \( T_{2,s} = T_1 \times \bigg( \frac{P_2}{P_1} \bigg)^{\frac{(\text{纬}-1)}{\text{纬}}} \), where \( \text{纬} = \frac{c_p}{c_v} \text{ (ratio of specific heats)} \), \( \text{纬} = 1.289 \). Calculate \( T_{2,s} \): \( T_{2,s} = 300 \times \bigg( \frac{1000}{100} \bigg)^{\frac{(1.289 - 1)}{1.289}} \) \( T_{2,s} = 300 \times 2.303^0.224 \) \( T_{2,s} = 300 \times 1.2163 \) \( T_{2,s} = 364.9 \text{ K} \). Now, isentropic work \( W_{is} \) uses enthalpy change: \( W_{is} = c_p \times (T_{2,s} - T_1) \) \( W_{is} = 0.846 \times (364.9 - 300) \) \( W_{is} = 0.846 \times 64.9 \) \( W_{is} = 54.9 \text{ kJ/kg} \). Therefore, \( \text{Isentropic efficiency} = \frac{54.9}{186.12} = 0.295 = 29.5\text{%} \)

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is essential for understanding energy changes in thermodynamic systems. It states that the energy added to a system as heat, minus the work done by the system, equals the change in the system's internal energy. Mathematically, it is expressed as:
\[ \text{螖U} = Q - W \] For an adiabatic process like the one in our exercise, there is no heat exchange (\(Q = 0\)). Therefore, the work done on the system (\
Ideal Gas Model
The ideal gas model simplifies the study of gases under various conditions. In this model, gas molecules do not interact except during elastic collisions, and the volume of the molecules themselves is negligible compared to the container. The relationship among pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) is given by the ideal gas equation:
\[ PV = nRT \] where \(R\) is the universal gas constant, and \(n\) is the number of moles of gas. This model allows us to calculate the specific enthalpy (\(h\)) and specific entropy (\(s\)) precisely using the specific heat capacities.
In the given exercise, we treat CO\(_2\) as an ideal gas to simplify the calculations for work input and entropy changes.
Entropy Production
Entropy measures the randomness or disorder in a system. During an adiabatic compression, although no heat is exchanged, entropy can still be produced due to irreversibilities. Entropy change (\(\Delta s\)) in an adiabatic process for an ideal gas is given by:
\[ \Delta s = c_p \ln\left( \frac{T_2}{T_1} \right) - R \ln\left( \frac{P_2}{P_1} \right) \] In our exercise, CO\(_2\)'s entropy change is calculated, showing a small but positive value indicating entropy production. Positive entropy production implies the real process is dissipative, not perfectly efficient.
Isentropic Efficiency
Isentropic efficiency measures how close a real process comes to an ideal isentropic process (a process with no entropy change). For a compressor, isentropic efficiency (\(\eta\)) is defined as:
\[ \eta = \frac{\text{Isentropic work}}{\text{Actual work}} \] In our solution, we determined the ideal exit temperature (\(T_{2,s}\)) of an isentropic process first. Then we used it to compute the isentropic work input. Finally, the isentropic efficiency of the compressor tells us how efficiently the real process was carried out compared to the ideal case. It is a critical performance parameter for compressors and other thermodynamic devices.

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Most popular questions from this chapter

Air within a piston-cylinder assembly, initially at \(30 \mathrm{lbf} /\) in. \({ }^{2}, 510^{\circ} \mathrm{R}\), and a volume of \(6 \mathrm{ft}^{3}\), is compressed isentropically to a final volume of \(1.2 \mathrm{ft}^{3}\). Assuming the ideal gas model with \(k=1.4\) for the air, determine the (a) mass, in lb, (b) final pressure, in lbf/in. \({ }^{2}\), (c) final temperature, in \({ }^{\circ} \mathrm{R}\), and (d) work, in Btu.

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.

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.

Air at \(400 \mathrm{kPa}, 970 \mathrm{~K}\) enters a turbine operating at steady state and exits at \(100 \mathrm{kPa}, 670 \mathrm{~K}\). Heat transfer from the turbine occurs at an average outer surface temperature of \(315 \mathrm{~K}\) at the rate of \(30 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing. Kinetic and potential energy effects are negligible. For air as an ideal gas with \(c_{p}=1.1 \mathrm{~kJ} /\) \(\mathrm{kg} \cdot \mathrm{K}\), determine (a) the rate power is developed, in kJ per \(\mathrm{kg}\) of air flowing, and (b) the rate of entropy production within the turbine, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of air flowing.

Nitrogen \(\left(\mathrm{N}_{2}\right)\) undergoes an internally reversible process from 6 bar, \(247^{\circ} \mathrm{C}\) during which \(p v^{1.20}=\) constant. The initial volume is \(0.1 \mathrm{~m}^{3}\) and the work for the process is \(121.14 \mathrm{~kJ}\). Assuming ideal gas behavior, and neglecting kinetic and potential energy effects, determine heat transfer, in \(\mathrm{kJ}\), and the entropy change, in \(\mathrm{kJ} / \mathrm{K}\). Show the process on a \(T-s\) diagram.
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