Question

Air is compressed in an axial-flow compressor operating at steady state from \(27^{\circ} \mathrm{C}, 1\) bar to a pressure of \(2.1\) bar. The work input required is \(94.6 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing through the compressor. Heat transfer from the compressor occurs at the rate of \(14 \mathrm{~kJ}\) per \(\mathrm{kg}\) at a location on the compressor's surface where the temperature is \(40^{\circ} \mathrm{C}\). Kinetic and potential energy changes can be ignored. Determine (a) the temperature of the air at the exit, in \({ }^{\circ} \mathrm{C}\). (b) the rate at which entropy is produced within the compressor, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of air flowing.

Short answer

Temperature of air at exit: 135.1°C. Entropy production rate: 0.136 kJ/K per kg.

Step-by-step solution

Understand the Given Data

Identify and list the given data from the problem.Initial temperature, \( T_1 = 27^{\textbackslash circ} \textbackslash mathrm{C} = 300 \textbackslash \text{ K} \) Initial pressure, \( P_1 = 1 \textbackslash text{ bar} \) Final pressure, \( P_2 = 2.1 \textbackslash text{ bar} = 210 \textbackslash text{ kPa} \) Work input per kilogram of air, \( W = 94.6 \textbackslash text{ kJ} / \textbackslash text{ kg} \) Heat transfer per kilogram of air, \( Q = -14 \textbackslash text{ kJ} / \textbackslash text{ kg} \) Surface temperature where heat transfer occurs, \( T_s = 40^{\textbackslash circ} \textbackslash mathrm{C} = 313 \textbackslash text{ K} \)

Apply the First Law of Thermodynamics

Use the first law of thermodynamics for a control volume at steady state:\[ \textbackslash dot{Q} - \textbackslash dot{W} = \textbackslash dot{m}(h_2 - h_1) \] Since the heat transfer is given as a negative value (heat is lost), the equation for work input (per kg) will be:\[ Q - W = h_2 - h_1 \] Using the information provided:\[ -14 \textbackslash \text{kJ/kg} - 94.6 \textbackslash \text{kJ/kg} = h_2 - h_1 \]

Use the Ideal Gas Relationship for Enthalpy Change

For an ideal gas with constant specific heat (\( c_p \)), the enthalpy change can be approximated as:\[ h_2 - h_1 = c_p (T_2 - T_1) \] The value of \( c_p \) for air is approximately \( 1.005 \textbackslash \text{kJ/kg.K} \). Substituting the values:\[ -108.6 \textbackslash \text{kJ/kg} = 1.005 \textbackslash \text{kJ/kg.K} \textbackslash times (T_2 - 300 \textbackslash \text{K}) \] Solving for \( T_2 \):\[ T_2 = 300 \textbackslash \text{K} + \frac{-108.6 \textbackslash \text{kJ/kg}}{1.005 \textbackslash \text{kJ/kg.K}} = 300 \textbackslash \text{K} - 108.1 \textbackslash \text{K} = 408.1 \textbackslash \text{K} = 135.1^{\textbackslash circ} \textbackslash \text{C} \]

Entropy Balance for the Control Volume

Use the entropy balance for a control volume to determine the rate of entropy production. The entropy balance equation is:\[ \textbackslash dot{m}(s_2 - s_1) + \textbackslash frac{mathdot{Q}}{T_s} = \textbackslash dot{S}_\text{gen} \] Since mass flow rate \( \dot{m} = 1 \) kg, use specific entropy change \( s_2 - s_1 \):\[ (s_2 - s_1) = c_p ln\left(\frac{T_2}{T_1}\right) - R \textbackslash ln\left(\frac{P_2}{P_1}\right) \]where \( R \) is the specific gas constant for air (=0.287 kJ/kg.K). Substituting the values:\[ (s_2 - s_1) = 1.005 \textbackslash ln\left(\frac{408.1}{300}\right) - 0.287 \textbackslash ln\left(\frac{2.1}{1}\right) \] Simplifying the logarithms and solving:\[ s_2 - s_1 = 1.005 \times 0.303 - 0.287 \times 0.742 = 0.304 - 0.213 = 0.091 \textbackslash text{kJ/K} \textbackslash text{ per kg} \] Next we account for heat transfer entropy change:\[ \frac{Q}{T_s} = \frac{-14 \text{kJ/kg}}{313 \text{K}} \approx -0.045 \text{kJ/K} \text{ per kg} \] Now calculate the entropy generation:\[ \textbackslash dot{S}_gen = s_2 - s_1 - \frac{Q}{T_s} \textbackslash \approx 0.091 + 0.045 \=0.136 \text{kJ/kg.K} \]

Present Final Results

Summarize the answers obtained:(a) The temperature of air at exit, \( T_2 \), is \( 135.1^{\textbackslash circ} \textbackslash text{C} \) (b) The rate of entropy production per kg of air flowing is \( \textbackslash dot{S}_{gen} \=0.136 \text{kJ/kg.K} \)

Key concepts

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. Instead, it can only be transferred or converted from one form to another. For a control volume like an axial-flow compressor operating at steady state, the first law can be written as:
\(\textbackslash dot{Q} - \textbackslash dot{W} = \textbackslash dot{m}(h_2 - h_1)\)
Here, \( \dot{Q} \) is heat added to the system, \( \dot{W} \) is work done by the system, \( \dot{m} \) is mass flow rate, \( h_2 \) is final enthalpy, and \( h_1 \) is initial enthalpy. This formula signifies that the change in energy (as enthalpy, \( h \)) of the air passing through the compressor is due to the difference between heat added and work done.

Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, and temperature of a gas. For a given amount of gas, the law is stated as:
\( PV = nRT \)
where \( P \) is pressure, \( V \) is volume, \( n \) is the amount of gas in moles, \( R \) is the universal gas constant, and \( T \) is temperature. In the context of the problem, we focus on changes in state variables (pressure and temperature) and use the relationship to understand the behavior of the compressed air. The law helps derive the enthalpy change equation used in the first law of thermodynamics when assumptions about ideal gas behavior and constant specific heats are considered.

Entropy balance

Entropy is a measure of disorder or randomness in a system. The entropy balance is used to determine the rate of entropy produced within a control volume. The general entropy balance equation is:
\( \textbackslash dot{m}(s_2 - s_1) + \textbackslash frac{ \textbackslash dot{Q}}{T_s} = \textbackslash dot{S}_{gen} \)
where \( s_2 \) and \( s_1 \) are the final and initial specific entropy values, respectively, and \( \textbackslash dot{S}_{gen} \) represents entropy generation. By calculating each term, we identify how much entropy is produced due to irreversibilities within the compressor. The goal here is to find the specific entropy change and ensure energy conservation principles align with the second law of thermodynamics.

Axial-flow compressor

An axial-flow compressor is a type of compressor where the working fluid flows parallel to the axis of rotation. These compressors are widely used in various applications due to their efficiency in handling large volumes of air at high pressure ratios. In the given exercise, air is compressed from 1 bar to 2.1 bar. Axial-flow compressors typically involve multi-stage compression processes, maintaining steady state with continuous mass flow. The performance and design are significantly impacted by the First Law of Thermodynamics, heat transfer processes, and entropy generation, directly linking to the calculations conducted in the exercise.

Specific heat capacity

Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). For gases, especially ideal gases, we often use specific heat capacities at constant pressure (\( c_p \)) and constant volume (\( c_v \)).
The value of specific heat capacity at constant pressure, \( c_p \), for air, is approximately 1.005 kJ/kg.K. This value was used to calculate the change in enthalpy and thereby determine the final temperature of the air in the exercise. Specific heat capacities are essential in understanding energy transfer within thermodynamic processes like those occurring in an axial-flow compressor.