Question

Air enters an insulated compressor operating at steady state at 1 bar, \(350 \mathrm{~K}\) with a mass flow rate of \(1 \mathrm{~kg} / \mathrm{s}\) and exits at 4 bar. The isentropic compressor efficiency is \(82 \%\). Determine the power input, in \(\mathrm{kW}\), and the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), using the ideal gas model with (a) data from Table A-22. (b) \(I T\). (c) a constant specific heat ratio, \(k=1.39\).

Short answer

Calculate power input using enthalpy changes and entropy production using specific entropy changes applying isentropic efficiency correction for all three approaches: Table A-22, iterative techniques, and constant specific heat ratio.

Step-by-step solution

- Use the Ideal Gas Model

Since the air behaves as an ideal gas, use the ideal gas properties to find the specific enthalpies and specific entropies. For parts (a), (b), and (c), different approaches or data might be used, so they will be broken down individually.

- Isentropic Process for Air

Determine the isentropic exit temperature for the compressor using the given inlet temperature and pressures. Use the isentropic relation: \[ T_{2s} = T_1 \left( \frac{P_2}{P_1} \right)^{(k-1)/k} \]. Apply this formula to find the isentropic exit temperature for each part.

(a) - Data from Table A-22

Using property tables such as Table A-22, find the specific enthalpy and entropy values for initial and isentropic final states. Once the isentropic properties are found, use the isentropic efficiency to determine the actual enthalpy: \[ h_2 = h_1 + \frac{h_{2s} - h_1}{\eta_{is}} \]. Calculate the work input per unit mass: \[ W_{in} = h_2 - h_1 \]. Multiply it by the mass flow rate to get the power input.

(a) - Entropy Production Rate

Using specific entropies, determine the actual exit entropy: \[ s_2 = s_{2s} + (s_{2s} - s_1)(1 - \eta_{is}) \]. Then, calculate the rate of entropy production: \[ \dot{S}_{gen} = \dot{m}(s_2 - s_1) \].

(b) - Using IT

Similar to part (a), but use the iterative technique (IT) with known thermodynamic properties and polynomial expressions or software tools to find specific enthalpy and entropy for both states. Repeat the isentropic efficiency corrections, and calculate power input and entropy production as done in parts (3) and (4).

(c) - Constant Specific Heat Ratio

Using the constant specific heat ratio and the isentropic relation, find the isentropic exit temperature for a constant specific heat ratio (k=1.39): \[ T_{2s} = T_1 \left( \frac{P_2}{P_1} \right)^{(k-1)/k} \]. Apply the same steps to calculate the isentropic efficiency corrections, power input, and entropy production.

- Combine All Results

Combine the results from each part to summarize the power input and entropy production rates for parts (a), (b), and (c). Compare the differences observed using different methods.

Key concepts

isentropic process

An isentropic process is a thermodynamic process in which entropy remains constant. For compressors, this means that the process is ideal, with no friction, heat transfer, or other inefficiencies. In reality, real compressors aren't perfectly isentropic due to these inefficiencies, but the isentropic process serves as a useful model for understanding the theoretical limitations.

In the exercise, an isentropic process is used to find the ideal exit temperature of the compressor. The relation used is:

\[ T_{2s} = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}} \] where:

• \( T_{2s} \) is the isentropic exit temperature
• \(T_1 \) is the inlet temperature
• \(P_2 \) and \(P_1 \) are the exit and inlet pressures, respectively

This formula helps establish the baseline for what would happen in an ideal scenario, which is then adjusted for real-world inefficiencies.

ideal gas model

The ideal gas model simplifies the study of gases by considering them as collections of particles in random motion, ignoring intermolecular forces except during collisions. This model is particularly useful for high temperatures and low pressures relative to the gas's critical point.

In this exercise, air is considered an ideal gas. This allows the use of property tables or specific equations to determine enthalpies and entropies. For example, Table A-22 provides the specific enthalpy and entropy values at various temperatures and pressures useful for these calculations.

Ideal gas assumptions streamline many thermodynamic calculations, making it easier to analyze and predict system behavior. However, it’s important to remember this model has limitations, especially at high pressures or low temperatures where real gas behavior deviates from ideal.

entropy production

Entropy production occurs in any real process, producing irreversibilities due to factors like friction, unrestrained expansion, mixing of different substances, heat transfer through a finite temperature difference, etc.

For the compressor in the problem, the rate of entropy production provides a measure of how far the actual process deviates from the ideal. The entropy production rate is given by:
\[ \dot{S}_{gen} = \dot{m}(s_2 - s_1) \]

• Where \( \dot{S}_{gen} \) is the rate of entropy production
• \( \dot{m} \) is the mass flow rate
• \( s_2 \) and \( s_1 \) are the specific entropies at the exit and the inlet, respectively

The actual exit entropy is calculated considering the isentropic efficiency, capturing the real-world behavior of the compressor. This is essential for understanding the true performance and efficiency of the system.

compressor power input

The power input to a compressor is the energy required to compress the gas from the inlet state to the desired outlet state. This accounts for the work done on the gas, factoring in inefficiencies.

In the exercise, the power input is calculated using the enthalpies of the air at the inlet and outlet states through the relation:
\[ W_{in} = h_2 - h_1 \]

• \(W_{in} \) is the work input per unit mass
• \(h_2 \) is the actual specific enthalpy at the exit
• \(h_1 \) is the specific enthalpy at the inlet

The actual specific enthalpy at the exit is derived from the isentropic efficiency, ensuring a realistic calculation of the energy required. Multiplying this by the mass flow rate yields the total power input in kilowatts.

thermodynamic properties

Thermodynamic properties are essential characteristics of a system's state, including temperature, pressure, volume, internal energy, enthalpy, and entropy. In this exercise, several properties are vital:

• Specific enthalpy (\(h \)): The amount of energy per unit mass, useful for determining work done and heat transfer.
• Specific entropy (\(s \)): A measure of the disorder or randomness of a system, important for understanding irreversibilities and efficiency.

These properties can be found using ideal gas models, property tables, polynomial expressions, or software tools. Accurately managing these properties is crucial for designing and analyzing compressors and other thermodynamic systems.