Question

A gas mixture having a molar analysis of \(60 \% \mathrm{O}_{2}\) and \(40 \% \mathrm{~N}_{2}\) enters an insulated compressor operating at steady state at 1 bar, \(20^{\circ} \mathrm{C}\) with a mass flow rate of \(0.5 \mathrm{~kg} / \mathrm{s}\) and is compressed to \(5.4\) bar. Kinetic and potential energy effects are negligible. For an isentropic compressor efficiency of \(78 \%\), determine (a) the temperature at the exit, in \({ }^{\circ} \mathrm{C}\). (b) the power required, in \(\mathrm{kW}\). (c) the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\).

Short answer

The exit temperature is calculated using compressor efficiency and isentropic relations, power via mass flow rate, cp mixture, and entropy production with entropy change.

Step-by-step solution

- Calculate Specific Heat Ratio

Determine the mole fractions of oxygen and nitrogen. Since the molar analysis shows 60% O₂ and 40% N₂, we can represent them in decimal form as: \[ y_{O_2} = 0.60 \] \[ y_{N_2} = 0.40 \]The specific heat ratio (\[ \frac{c_p}{c_v} \]) for a mixture can be found from the ratios and mole fractions of the components. Given \[ \frac{c_p}{c_v} \] values: \[ \frac{c_p}{c_v}_{O_2} = 1.40 \] \[ \frac{c_p}{c_v}_{N_2} = 1.39 \], the mixture specific heat ratio becomes: \[ \frac{c_p}{c_v}_{mix} = y_{O_2} \times \frac{c_p}{c_v}_{O_2} + y_{N_2} \times \frac{c_p}{c_v}_{N_2} \]

- Calculate isentropic exit temperature

Use the isentropic relation for an ideal gas to find the exit temperature (\[ T_{2s} \]). The relation is \[ \left(\frac{T_{2s}}{T_1}\right) = \left(\frac{P_2}{P_1}\right)^{\frac{(k-1)}{k}} \]Where \[ T_1 \] is the initial temperature, \[ P_1 \] is the initial pressure, \[ P_2 \] is the final pressure, and \[ k \] is the specific heat ratio.Given \[ T_1 = 20^\text{°C} = 293.15 \text{K}, P_1 = 1 \text{ bar}, P_2 = 5.4 \text{ bar}, k = 1.395 \] (approx), calculate \[ T_{2s} \].

- Calculate Actual Exit Temperature

Since the compressor efficiency is \[ 78\% \], the actual exit temperature \[ T_2 \] can be found from: \[ T_2 = T_1 + \eta_{comp} (T_{2s} - T_1) \]Where \[ \eta_{comp} \] is the compressor efficiency. Substituting the known values, solve for \[ T_2 \].

- Calculate Power Required

Use the power formula and the mass flow rate to find the power required: \[ W_{comp} = \dot{m} c_p (T_2 - T_1) \].Find the specific heat at constant pressure (\[ c_p \]) for the gas mixture: \[ c_p = y_{O_2} c_{p_{O_2}} + y_{N_2} c_{p_{N_2}} \] and use the obtained temperatures from step 3. The known values are: \[ \dot{m} = 0.5 \text{ kg/s} \]. Given \[ c_{p_{O_2}} = 0.918 \text{ kJ/kg.K}, c_{p_{N_2}} = 1.040 \text{ kJ/kg.K} \], calculate \[ c_p \], then the compressor power.

- Calculate Rate of Entropy Production

The rate of entropy production (\[ \frac{S}{K} \]) in an insulated system is given by the change in entropy of the gas mixture: \[ \dot{S}_{gens} = \dot{m} c_p \ln \left(\frac{T_2}{T_1}\right) \]. Given \[ T_2 \] from the previous step and \[ T_1 = 293.15 K, c_p \] obtained, substitute the values and calculate the rate of entropy production.

Key concepts

specific heat ratio

In thermodynamics, the specific heat ratio, denoted as \( \frac{c_p}{c_v} \), is a critical parameter. It is the ratio of the specific heat at constant pressure (\( c_p \)) to the specific heat at constant volume (\( c_v \)). This ratio influences how substances respond to compressive heating or expansion cooling. For gas mixtures like the one in our exercise, we calculate the specific heat ratio by considering each component's proportion and their respective specific heat ratios. In the example, the mixture is made of 60% oxygen (\( O_2 \)) and 40% nitrogen (\( N_2 \)). Hence, we calculate the mixture’s specific heat ratio as: \[ \frac{c_p}{c_v}_{mix} = y_{O_2} \times \frac{c_p}{c_v}_{O_2} + y_{N_2} \times \frac{c_p}{c_v}_{N_2} \] where \( y_{O_2} \) and \( y_{N_2} \) are the mole fractions of oxygen and nitrogen, respectively. This method ensures we accurately reflect the behavior of the entire mixture.

isentropic process

An isentropic process is a pivotal concept in thermodynamics, implying a process where entropy remains constant—no entropy change. In compressors, an isentropic process ideally means no energy losses. This theoretical premise helps in calculating the temperature changes under ideal conditions. The relation used to find the exit temperature \(T_{2s}\) from an isentropic process is: \[ \frac{T_{2s}}{T_1} = \bigg( \frac{P_2}{P_1} \bigg)^{\frac{(k-1)}{k}} \] Here, \( T_1 \) is the initial temperature, \( P_1 \) is the initial pressure, \( P_2 \) is the final pressure, and \( k \) is the specific heat ratio of the mixture. This calculation gives us the ideal, or theoretical, exit temperature without considering real-world inefficiencies.

compressor efficiency

Compressor efficiency is a measure of how effectively a compressor converts input energy into useful work. For this exercise, we consider the isentropic efficiency, which is the ratio of the ideal work to the actual work done by the compressor. Given as \( \frac{78\text{\textpercent}} \) in this instance, the actual exit temperature \( T_2 \) can be calculated using: \[ T_2 = T_1 + \text{\texteta_{comp}} (T_{2s} - T_1) \] In this formula, \( T_1 \) is the initial temperature, \( T_{2s} \) is the ideal exit temperature, and \( \text{\texteta_{comp}} \) is the compressor efficiency. Understanding this efficiency is crucial, as it accounts for the real-world departures from the ideal isentropic process due to factors like friction and other irreversible processes.

power calculation

To determine a compressor's power requirement, we use the specific heat of the mixture and temperature changes. The formula to calculate the power required is expressed as: \[ W_{comp} = \frac{\text{\textdot{m}} \times c_p \times (T_2 - T_1)}{1000} \] Here, \( \text{\textdot{m}} \) represents the mass flow rate, \( c_p \) is the specific heat at constant pressure, \( T_1 \) is the initial temperature, and \( T_2 \) is the actual exit temperature from the compressor. In our problem, we need to first calculate the specific heat \( c_p \) of the mixture by proportionally blending the specific heat values of oxygen and nitrogen. These must be converted into kilojoules per kilogram Kelvin (kJ/kg.K) for consistency.

entropy production

Entropy production is an important measure of irreversibilities within a system, showing how much energy disperses rather than contributing to useful work. For a compressor in an insulated system, this is given by: \[ \text{\textdot{S}}_{gens} = \frac{\text{\textdot{m}} \times c_p \times \text{In} \bigg( \frac{T_2}{T_1} \bigg)}{1000} \] This equation uses the same terms as before: \( \text{\textdot{m}} \) for mass flow rate and \( c_p \) for specific heat. But now, it considers the natural logarithm of the ratio of the final and initial temperatures. Calculating the rate of entropy production helps us understand how much energy is lost due to system inefficiencies and can guide improvements in compressor performance.