Question

Air enters a compressor operating at steady state with a volumetric flow rate of \(8 \mathrm{~m}^{3} / \mathrm{min}\) at \(23^{\circ} \mathrm{C}, 0.12 \mathrm{MPa}\). The air is compressed isothermally without internal irreversibilities, exiting at \(1.5 \mathrm{MPa}\). Kinetic and potential energy effects can be ignored. Evaluate the work required and the heat transfer, each in \(\mathrm{kW}\).

Short answer

The work required is approximately \(-0.391 \mathrm{kW}\) and the heat transfer is \( -0.391 \mathrm{kW} \) as well.

Step-by-step solution

Identify the Process and Given Data

This is an isothermal compression process. Given data are: \( \dot{V_{in}} = 8 \mathrm{~m^3 / min} \), since 1 \( \mathrm{min} = 60 \mathrm{s} \), it translates to \( \dot{V_{in}} = \frac{8 \mathrm{~m^3}}{60 \mathrm{s}} = \frac{2}{15} \mathrm{~m^3 / s} \) Initial temperature \( T_1 = 23^{\circ} \mathrm{C} = 296 \text{ K} \) Initial pressure \( P_1 = 0.12 \mathrm{~MPa} \) Final pressure \( P_2 = 1.5 \mathrm{~MPa} \)

Apply the Ideal Gas Law to Determine Initial Mass Flow Rate

Using the ideal gas law, \( PV = nRT \), where, \( n = \frac{m}{M} \), therefore, \( V = \frac{mRT}{P} \)Rearranged, we have, \( m = \frac{PV}{RT} \) Since \( V_{in} = \dot{V_{in}} \times t \), the mass flow rate is: \( \dot{m} = \frac{P_1 \dot{V_{in}}}{R T_1} \), where, \( R = 287 \mathrm{J/(kg \cdot K)} \) Substitute the values to get mass flow rate, \( \dot{m} \): \( P_1 = 0.12 \mathrm{MPa} = 120 \mathrm{kPa} = 120000 \mathrm{Pa} \), \( \dot{V_{in}} = \frac{2}{15} \mathrm{m^3/s} \), \( T_1 = 296 \mathrm{K} \)\( \dot{m} = \frac{120000 \times \frac{2}{15}}{287 \times 296} \approx 0.00288 \mathrm{~kg/s} \)

Calculate Work Required for Isothermal Compression

For isothermal processes, work required is given by: \[ W = -nRT \ln \left(\frac{P_2}{P_1} \right) \] since \( n = \frac{m}{M} \) \[ W = - \dot{m} R T_1 \ln \left(\frac{P_2}{P_1} \right) \] Substitute values: \( \dot{m} = 0.00288 \mathrm{~kg/s} \), \( R = 287 \mathrm{J/(kg\cdot K)} \), \( T_1 = 296 \mathrm{K} \) \( P_1 = 0.12 \mathrm{MPa} \), \( P_2 = 1.5 \mathrm{MPa} \) \[ W = -0.00288 \times 287 \times 296 \ln \left(\frac{1.5}{0.12} \right) \approx -2.88 \times 287 \times 296 \times 1.610 \approx -390.6 \mathrm{W} \] Convert to \( \mathrm{kW} \): \[ W \approx -0.391 \mathrm{kW} \]

Calculate Heat Transfer for Isothermal Process

Using the first law of thermodynamics for an isothermal process: \[ \Delta U = Q - W = 0 \] (because internal energy change for isothermal process is zero). Therefore, \( Q = W \).Substitute the value of work calculated, thus: \[ Q \approx -0.391 \mathrm{kW} \]

Key concepts

steady state operation

In a steady-state operation, the properties of the system do not change with time. This means that the mass flow rate, pressure, and temperature at any given point remain constant over time.
In the context of the problem, the compressor is operating in a steady state. This implies that the air entering and leaving the compressor does so continuously without any accumulation inside the system. Steady states simplify analysis because we can assume that any changes in properties are due to the flow rather than time-dependent fluctuations.

ideal gas law

The Ideal Gas Law is a fundamental relation between pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas. It is expressed as: \( PV = nRT \)Here, \(R\) is the universal gas constant. For air, we often use \( R = 287 \mathrm{J/(kg \bullet K)} \).
In this exercise, the Ideal Gas Law helps determine the mass flow rate. By knowing the volumetric flow rate, pressure, and temperature of the air entering the compressor, we can rearrange the Ideal Gas Law to: \( m = \frac{PV}{RT} \)This formula allows us to calculate the mass \(m\) of the air entering per unit time.

mass flow rate calculation

Mass flow rate is the mass of a substance passing through a cross-section per unit of time. It is denoted as \( \dot{m} \). For the given problem, to calculate \( \dot{m} \):1. Convert volumetric flow rate to compatible units: \( \dot{V}_{in} = \frac{8 \ m^3}{60s} = \frac{2}{15} \mathrm{m^3/s} \)2. Use the Ideal Gas Law to find the mass of air, and then the mass flow rate: \( \dot{m} = \frac{P_1 \bullet \dot{V}_{in}}{R \bullet T_1} \)Substitute the known values:\( P_1 = 120 \mathrm{kPa} = 120000 \mathrm{Pa} \)\( T_1= 296 K \)\( R = 287 \mathrm{J/(kg \bullet K)} \)\( \dot{m} = \frac{120000 \bullet \frac{2}{15}}{287 \bullet 296} \approx 0.00288 \ kg/s \)This tells us how much air mass enters the compressor each second.

work required for compression

In an isothermal compression process, the temperature remains constant. The work required (W) for compressing an ideal gas is given by:\[ W = - \dot{m} R T \text{ln} \left(\frac{P_2}{P_1}\right) \]Given that temperature is constant, use: \( T = T_1 \).
Now substitute the known values:\( \dot{m} = 0.00288 \ kg/s, \, R = 287 \mathrm{J/(kg \bullet K)}, \, T_1 = 296 K, \, P_1 = 0.12 MPa, \, and \, P_2 = 1.5 MPa \).\[ W = -0.00288 \bullet 287 \bullet 296 \bullet \text{ln} \left(\frac{1.5}{0.12}\right) \]\[ W \approx -390.6 \mathrm{W} \text{ or } -0.391 \mathrm{kW} \]This negative sign indicates work is done on the gas (input).

heat transfer calculation

To find the heat transfer (Q) during an isothermal process, we use the first law of thermodynamics: \( \Delta U = Q - W \)Since the internal energy change \( \Delta U \) is zero for an isothermal process (temperature is constant), the equation simplifies to:\( Q = W \)From the previous section, the work done on the gas \( W \) is -0.391 kW.
Therefore, the heat transferred also equals -0.391 kW but with a direction opposite to the work input (indicating heat is released to the surroundings).