Question

A compressor operating at steady state takes in atmospheric air at \(20^{\circ} \mathrm{C}, 1\) bar at a rate of \(1 \mathrm{~kg} / \mathrm{s}\) and discharges air at 5 bar. Plot the power required, in \(\mathrm{kW}\), and the exit temperature, in \({ }^{\circ} \mathrm{C}\), versus the isentropic compressor efficiency ranging from 70 to \(100 \%\). Assume the ideal gas model for the air and neglect heat transfer with the surroundings and changes in kinetic and potential energy.

Short answer

Calculate theoretical exit temperature, adjust for efficiency, then determine power required. Plot results for efficiency range 70-100%.

Step-by-step solution

Define Given Data and Assumptions

List all known values and assumptions: \( T_1 = 20 \degree C = 293.15 \; K \), \( P_1 = 1 \; bar \), \( m = 1 \; kg/s \), \( P_2 = 5 \; bar \), Ideal gas model for air, neglect heat transfer, kinetic and potential energy changes.

Isentropic Process Calculations

Calculate the theoretical (isentropic) exit temperature: \[ \frac{T_{2s}}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}} \] where \( \gamma \) (ratio of specific heats) for air is typically 1.4. Solve for \( T_{2s} \): \[ T_{2s} = 293.15 \times \left( \frac{5}{1} \right)^{\frac{1.4 - 1}{1.4}} \]

Actual Exit Temperature with Efficiency

The actual exit temperature can be calculated using the efficiency: \[ T_2 = T_1 + \frac{T_{2s} - T_1}{\eta} \] where \( \eta \) is the isentropic efficiency.

Calculate Power Required

The power required can be found using: \[ W = m \cdot c_p \cdot (T_2 - T_1) \] where \( c_p \) (specific heat capacity of air at constant pressure) is typically 1.005 \( kJ/kg \cdot K \).

Plotting Results

Vary \( \eta \) from 70% to 100% and calculate corresponding \( T_2 \) and \( W \). Use these results to create two plots: Power required vs. efficiency and Exit temperature vs. efficiency.

Key concepts

Isentropic Process

An isentropic process is a thermodynamic process where entropy remains constant. This implies that there is no heat exchange with the surroundings, making the process reversible and adiabatic. For air compression, an isentropic process can be theoretical, helping us understand the limits of efficiency. The formula \(\frac{T_{2s}}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}}\) helps in calculating the theoretical exit temperature (\(T_{2s} \)) under these ideal conditions. Here, \(\gamma \) is the ratio of specific heats (usually 1.4 for air). This theoretical scenario aids in establishing a benchmark for real-world compressor performance.

Ideal Gas Model

The ideal gas model simplifies the behavior of gases by assuming no interactions between gas molecules and that the volume of the gas molecules themselves is negligible. This model works well for many practical applications, including air compression. Using the ideal gas law, \(PV = nRT \), allows for straightforward calculations of pressure, volume, and temperature relationships. For the given exercise, the assumptions include treating air as an ideal gas and simplifying the thermodynamic equations, making it easier to predict system behavior under various conditions.

Compressor Power Calculation

Calculating the power required by a compressor involves understanding both the ideal and actual performance scenarios. The power calculation follows the steps:
1. Calculate the actual exit temperature using efficiency: \(T_2 = T_1 + \frac{T_{2s} - T_1}{\eta} \), where \(\eta \) is the isentropic efficiency.
2. Determine the power required using: \(W = m \cdot c_p \cdot (T_2 - T_1) \), where \(c_p \) is the specific heat capacity of air at constant pressure. The specific heat, roughly 1.005 \(\frac{kJ}{kg \cdot K}\), helps approximate the required energy input accurately. By calculating the power for varied efficiencies, one can then plot the power required versus the efficiency range.

Thermodynamic Efficiency

Thermodynamic efficiency evaluates how well a compressor converts input energy into useful work, essentially comparing real systems to ideal ones. The isentropic efficiency \(\eta \) is a key metric, calculated as the ratio of work for an ideal isentropic process to the actual work required. Higher efficiency means closer performance to the ideal isentropic process, minimizing energy losses. Efficiency is vital in optimizing energy consumption and operational costs in compressor systems. By plotting efficiency from 70% to 100%, students can see how closer efficiency to the ideal (100%) considerably reduces the energy required.

Air Compression

Air compression involves compressing atmospheric air, increasing its pressure and temperature. It’s used in various applications, from industrial machinery to HVAC systems. The process, ideally modeled by assuming no heat loss and perfect behavior (isentropic and ideal gas assumptions), simplifies analysis. Calculations begin with known conditions (initial temperature, pressure, and mass flow rate) and end at new conditions after compression. Understanding these basics helps in designing efficient compressors, crucial for optimizing performance in real-world scenarios with minimal energy wastage. Students often find visualizing the plots for required power and exit temperature against efficiency useful in grasping the practical impacts of improved efficiency.