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A water-cooled air compressor takes air in at \(20^{\circ} \mathrm{C}\), \(90 \mathrm{kPa}\) and compresses it to \(500 \mathrm{kPa}\). The isothermal efficiency is \(88 \%\) and the actual compressor has the same heat transfer as the ideal one. Find the specific compressor work and the exit temperature.

Short Answer

Expert verified
Specific work: -127.34 kJ/kg; Exit temperature: 20°C.

Step by step solution

01

Define Isothermal Compressor Work

Isothermal work for compression is given by the formula \(w = \int_{v_1}^{v_2} P \, dv = P_1 V_1 \ln \left( \frac{P_2}{P_1} \right)\), where \(P_1\) and \(P_2\) are initial and final pressures, and \(V_1\) is the initial specific volume of air. Since air behaves as an ideal gas, we can use the ideal gas law to find \(V_1 = \frac{RT_1}{P_1}\). For air, \(R = 0.287 \text{ kJ/kg.K}\) and \(T_1 = 293.15 \text{ K}\). This simplifies to \(w = RT_1 \ln \left( \frac{P_2}{P_1} \right)\).
02

Calculate Ideal Work

Calculate the ideal isothermal compressor work using the formula from Step 1: \(w = 0.287 \times 293.15 \ln \left( \frac{500}{90} \right)\) kJ/kg. Simplify this to approximately \(-112.06 \text{ kJ/kg}\).
03

Adjust for Efficiency

The actual compressor work is given by \(w_{actual} = \frac{w_{ideal}}{\eta_{isothermal}}\), where \(\eta_{isothermal} = 0.88\). Thus, \(w_{actual} = \frac{-112.06}{0.88} \approx -127.34 \text{ kJ/kg}\).
04

Find Exit Temperature Assuming Isothermal Process

For an isothermal process, \(T_2 = T_1\). Hence, the exit temperature remains \(293.15 \text{ K}\) or \(20^{\circ} \text{C}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isothermal Compression
Isothermal compression is a process where the temperature of the gas remains constant while it is being compressed. In an ideal isothermal compression, heat is allowed to transfer freely between the gas and the surroundings so that the gas retains the same temperature throughout the process.

For isothermal compression in air compressors, it is important because maintaining the temperature allows us to precisely calculate the work needed for compression using the formula:
  • The work done is expressed as: \( w = P_1 V_1 \ln\left( \frac{P_2}{P_1} \right) \)
  • For an ideal gas, this becomes: \( w = RT_1 \ln\left( \frac{P_2}{P_1} \right) \)
This equation gives us the isothermal compressor work, providing a baseline for efficiency and performance analysis.
Ideal Gas Law
The Ideal Gas Law is a fundamental principle that describes the behavior of gases under various conditions. It relates pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the amount of gas in moles (\(n\)) through the equation:
  • \( PV = nRT \)
where \(R\) is the universal gas constant. For cases involving specific volume and conditions, such as in compressors, we often modify this to suit the specific gas.

For air, the specific gas constant \(R\) is 0.287 kJ/kg.K. This simplified version allows us to determine initial conditions of the gas, such as initial specific volume \(V_1\) before compression, using:
  • \( V_1 = \frac{RT_1}{P_1} \)
This equation is integral in calculating the initial conditions necessary for further analysis of compressor efficiency and performance.
Compressor Efficiency
Compressor efficiency is a measure of how effectively a compressor converts input energy into useful work. Specifically, isothermal efficiency compares actual compressor work with ideal work for isothermal compression. It is calculated using:
  • \( \eta_{isothermal} = \frac{w_{ideal}}{w_{actual}} \)
In the context of the given problem, the isothermal efficiency is given as 88%, indicating that the actual compressor work is less efficient compared to the ideal scenario. This tells us that due to losses, more work is required than in an ideal isothermally efficient process.

Understanding efficiency allows engineers to assess system performance and room for improvement in design and operation of compressors.
Specific Compressor Work
Specific compressor work refers to the work required to compress a unit mass of gas from an initial state to a final state. It gives insight into how much energy each kilogram of air requires to be compressed under specified conditions.

Calculation of this can be straightforward in an ideal situation but requires adjustments for efficiency in practical scenarios. In this exercise, the formula used was:
  • \( w_{actual} = \frac{w_{ideal}}{\eta_{isothermal}} \)
where \( w_{ideal} \) is calculated using the principles of isothermal compression and ideal gas law. Accounting for efficiency gives a realistic picture of the energy expenditure needed.

By comprehending specific work, engineers can better design systems that optimize energy usage, minimize costs, and evaluate potential improvements.

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