Question

A two-stage air compressor operates at steady state, compressing \(10 \mathrm{~m}^{3} / \mathrm{min}\) of air from \(100 \mathrm{kPa}, 300 \mathrm{~K}\), to \(1200 \mathrm{kPa}\). An intercooler between the two stages cools the air to \(300 \mathrm{~K}\) at a constant pressure of \(350 \mathrm{kPa}\). The compression processes are isentropic. Calculate the power required to run the compressor, in \(\mathrm{kW}\), and compare the result to the power required for isentropic compression from the same inlet state to the same final pressure.

Short answer

The power required for the two-stage compressor is found by summing the power for each stage. Compare this with the power calculated for an isentropic single-stage compressor.

Step-by-step solution

- Determine air properties at different states

Identify the initial, intermediary, and final states of the air. State 1: Initial state where pressure, \(P_1 = 100 \text{kPa}\), and temperature, \(T_1 = 300 \text{K}\). Intermediary State 2: After first-stage compression to \(P_2 = 350 \text{kPa}\), final state where \(P_3 = 1200 \text{kPa}\) after second-stage compression.

- Calculate the compression ratios for each stage

Find the compression ratios \(r_{12} = \frac{P_2}{P_1} \) and \(r_{23} = \frac{P_3}{P_2}\). This gives \(r_{12} = \frac{350}{100} \approx 3.5\) and \(r_{23} = \frac{1200}{350} \approx 3.43\).

- Use isentropic relations to find final temperatures

For isentropic processes, we have \(T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}}\) and \(T_3 = T_2 \left( \frac{P_3}{P_2} \right) ^{\frac{\gamma - 1}{\gamma}}\). With \(\gamma = 1.4\) for air, calculate \(T_2\) and \(T_3\).

- Determine the specific work of each compression

The specific work \(w\) for an ideal gas undergoing isentropic compression is \(w = c_p (T_{out} - T_{in})\), where \(c_p = 1005 \text{J/kg.K}\) for air. Calculate the specific work for both stages.

- Calculate the mass flow rate

Convert the volumetric flow rate to mass flow rate using \(\dot{m} = \frac{ \dot{V} \cdot P }{ R \cdot T }\). Here, \dot{V} = 10 \text{m}^3/\text{min}\, \(R = 287 \text{J/kg.K}\), \(P = 100 \text{kPa}\), and \(T = 300 \text{K}\).

- Calculate the total power required for the two stages

Multiply mass flow rate by the specific work for each stage and sum the values: \(W_{total} = \dot{m} \cdot (w_{12} + w_{23}) \). Convert this to kW.

- Compare with power for a single-stage isentropic compression

Calculate the total specific work for a direct isentropic compression from 100 kPa to 1200 kPa. Compare with the two-stage compression.

Key concepts

Isentropic Process

An isentropic process is a thermodynamic process during which entropy remains constant. It is an idealized process and often used as a standard for comparison.
During compression in an isentropic process, no heat is transferred to or from the gas, leading to maximum efficiency. Mathematically, for an ideal gas undergoing isentropic compression, we use the relation: oindent \[T_2 = T_1 \bigg(\frac{P_2}{P_1}\bigg)^{\frac{(\gamma - 1)}{\gamma}}\] where \ T \ is the temperature, \ P \ is the pressure, and \ \gamma \ is the heat capacity ratio (typically \ \gamma = 1.4 \ for air). This relationship allows us to determine the final temperature after compression, given initial conditions and the compression ratio.
In the exercise, the air undergoes two isentropic compressions separated by an intercooler. The intercooler maintains the temperature at 300 K between stages, preventing the cumulative temperature rise, thereby improving efficiency.

Compression Ratio

The compression ratio is the ratio of the final pressure to the initial pressure during a compression process. For multiple stages, this ratio is critical in splitting the total compression work efficiently between stages.
The compression ratio for each stage is given as: \[ r = \frac{P_{out}}{P_{in}} \]
In our exercise, we have two stages:

• The first stage compresses from 100 kPa to 350 kPa, yielding \ r_{12} = 3.5\.
• The second stage compresses from 350 kPa to 1200 kPa, giving \ r_{23} = 3.43\.

Using these ratios, we can calculate intermediate temperatures and specific works more accurately, optimizing energy expenditure for the compressor.

Power Calculation

Power calculation involves determining the specific work done during compression and multiplying it by the mass flow rate. The specific work of compression for an ideal gas in an isentropic process is given by: \[ w = c_p (T_{out} - T_{in}) \] where \ T \ is temperature and \ c_p \ (specific heat at constant pressure) is typically 1005 J/kg.K for air.
To find the total power required, we first convert the volumetric flow rate (10 m³/min) into a mass flow rate using the ideal gas law: \ \dot{m} = \frac{\dot{V} \cdot P}{R \cdot T} \text{ where }\ \dot{V}\ is volumetric flow rate, \ R\ is the specific gas constant, and \ P \ and \ T \ are initial conditions. \ R = 287 J/kg.K \ for air.
Then, multiply the mass flow rate by the specific work of each stage and sum the results: \ W_{total} = \dot{m} \cdot (w_{12} + w_{23}) \. This will give the power requirement in watts, which can be converted to kilowatts (kW).

Ideal Gas Properties

Understanding ideal gas properties is fundamental for accurately solving thermodynamic problems. The behavior of an ideal gas is given by the equation of state: \[ PV = nRT \], where \ P \ is pressure, \ V \ is volume, \ n \ is the number of moles, \ R \ is the universal gas constant, and \ T \ is the temperature.
For engineering applications, we often use the specific gas constant, \ R \, in the form \[ \ R = \frac{\bar{R}}{M} \] where \ \bar{R}\ is the universal gas constant (8.314 J/(mol·K)) and \ M \ is the molar mass of the gas (28.97 g/mol for air).
For the exercise, air is treated as an ideal gas, simplifying calculations: \[ \dot{m} = \frac{ \dot{V} \times P}{ R \times T } \]. Here, \ \dot{V} = 10 \text{m}^3/\text{min}, \ P = 100 kPa, \ R = 287 \text{J/kg.K}, \text{ and } T = 300 K \.
This understanding provides the basis for mass flow rate and specific work calculations, critical for determining the overall power requirement for the compressor.