Question

Air enters an insulated compressor operating at steady state at \(0.95\) bar, \(27^{\circ} \mathrm{C}\) with a mass flow rate of \(4000 \mathrm{~kg} / \mathrm{h}\) and exits at \(8.7\) bar. Kinetic and potential energy effects are negligible. (a) Determine the minimum theoretical power input required, in \(\mathrm{kW}\), and the corresponding exit temperature, in \({ }^{\circ} \mathrm{C}\). (b) If the exit temperature is \(347^{\circ} \mathrm{C}\), determine the power input, in \(\mathrm{kW}\), and the isentropic compressor efficiency.

Short answer

Minimum theoretical power input: 290.09 kW. Exit temperature for isentropic process: 286.85°C. Actual power input: 357.6 kW. Isentropic efficiency: 81.1%.

Step-by-step solution

- Identify Given Values

List all the given values from the problem: Initial pressure: \(P_1 = 0.95\) bar Initial temperature: \(T_1 = 27^{\text{°C}} = 300.15\) K Mass flow rate: \(\text{ṁ} = 4000\) kg/h = \(\frac{4000}{3600}\) kg/s Final pressure: \(P_2 = 8.7\) bar

- Convert Mass Flow Rate to SI Units

Convert the mass flow rate from kg/h to kg/s: \( \text{ṁ} = \frac{4000}{3600} \text{ kg/s} \approx 1.111 \text{ kg/s} \)

- Use Isentropic Relations

For an isentropic process in an ideal gas, the relation between pressures and temperatures is: \[ \frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} \] For air, \(\gamma \approx 1.4\). Substitute the known values: \[ T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{0.4}{1.4}} \] \[ T_2 = 300.15 \left( \frac{8.7}{0.95} \right)^{\frac{0.4}{1.4}} \]

- Calculate Exit Temperature for Isentropic Process

Calculate the exit temperature: \[ T_2 \approx 300.15 \times (9.158)^{0.286} \approx 300.15 \times 1.866 \approx 560 \text{ K} \] Thus, the exit temperature is \(560 \text{ K} - 273.15 \approx 286.85^{\text{°C}}\)

- Calculate Work Required for Isentropic Process

Use the relation for work done for an isentropic process: \[ W = \text{ṁ} \cdot c_p (T_2 - T_1) \] \[ c_p \approx 1.005 \text{ kJ/kg K} \] Substitute the values: \[ W = 1.111 \cdot 1.005 \cdot (560 - 300.15) \] \[ W \approx 1.111 \times 1.005 \times 259.85 \approx 290.09 \text{ kW} \]

- Calculate Actual Work for Given Exit Temperature

If the exit temperature is \(347^{\text{°C}} = 620.15 \text{ K}\), calculate the actual work done: \[ W_{\text{actual}} = \text{ṁ} \cdot c_p (T_{2\text{actual}} - T_1) \] Substitute the values: \[ W_{\text{actual}} = 1.111 \cdot 1.005 \cdot (620.15 - 300.15) \] \[ W_{\text{actual}} \approx 1.111 \times 1.005 \times 320 \approx 357.6 \text{ kW} \]

- Calculate Isentropic Efficiency

Isentropic efficiency \( \eta \) is given by: \[ \eta = \frac{W_{\text{isentropic}}}{W_{\text{actual}}} \] Substitute the values: \[ \eta = \frac{290.09}{357.6} \approx 0.811 \text{ or } 81.1\% \]

Key concepts

Isentropic Process

An isentropic process is an idealized thermodynamic process in which entropy remains constant. This type of process is also referred to as a reversible adiabatic process. Being adiabatic means there is no heat transfer into or out of the system. In the context of compressors, an isentropic process allows us to determine the maximum efficiency and minimum work input required. To analyze an isentropic process, we employ the relationship between pressure and temperature. For instance, the specific formula used in this process is given by: \[ \frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} \] where \( T \) represents temperature, \( P \) represents pressure, and \( \gamma \) (gamma) is the specific heat ratio. For air, \( \gamma \) is approximately 1.4. This equation allows us to determine the final temperature \( T_2 \) when given initial conditions and the pressure ratio.

Ideal Gas Relations

Ideal gas relations are fundamental in thermodynamics for simplifying the analysis of gases under various conditions. The behavior of an ideal gas is described by the equation: \[ PV = nRT \] where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature. This relation implies that if you know the amount of gas, volume, and temperature, you can easily determine the pressure, and vice versa. In the scenario of a compressor, ideal gas relations help us estimate the changes in temperature and pressure that occur during the compression process. For an isentropic process in particular, the change can be understood using the relation between pressures and temperatures as shown. Understanding these relations helps in determining the power requirements and efficiency accurately.

Isentropic Efficiency

Isentropic efficiency is a measure of how close a real process comes to its ideal counterpart. In the context of compressors, it tells us how efficiently the compressor operates versus the isentropic (ideal) condition. The efficiency is calculated using the formula: \[ \eta = \frac{W_{\text{isentropic}}}{W_{\text{actual}}} \] Here, \( W_{\text{isentropic}} \) is the work done in an isentropic process, and \( W_{\text{actual}} \) is the actual work input. For example, if a compressor's actual exit temperature is higher than the theoretical isentropic exit temperature, the efficiency will be less than 100%, reflecting losses due to irreversibilities. By understanding and calculating its efficiency, engineers can evaluate the performance of compressors and identify opportunities for improving energy efficiency.

Power Input Calculation

Calculating the power input for a compressor requires understanding both the ideal and actual scenarios. The work done during an isentropic process (ideal) is calculated using: \[ W = \dot{m} \cdot c_p (T_2 - T_1) \] In this equation, \( \dot{m} \) is the mass flow rate of the air, \( c_p \) is the specific heat at constant pressure, and \( T_2 \) and \( T_1 \) are the final and initial temperatures, respectively. For the actual work input, the same formula can be used with the actual final temperature. For instance, if the actual exit temperature is higher than the isentropic one, this indicates higher energy consumption and lower efficiency. Calculating power input correctly helps in determining energy usage and operational costs for the compressor.