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A centrifugal compressor takes in ambient air at \(100 \mathrm{kPa}, 17^{\circ} \mathrm{C}\) and discharges it at \(450 \mathrm{kPa}\). The compressor has an isentropic efficiency of \(80 \%\). What is your best estimate for the discharge temperature?

Short Answer

Expert verified
The discharge temperature is approximately 431.15 K.

Step by step solution

01

Analyze Given Data

First, let's identify the given information from the problem. The inlet conditions are ambient air at 100 kPa, 17°C. The discharge pressure is given as 450 kPa, and the isentropic efficiency of the compressor is 80%. We aim to determine the discharge temperature.
02

Convert Inlet Temperature to Kelvin

The given inlet temperature is 17°C. To work with thermodynamic equations, we need to convert this to Kelvin using the formula: \[ T_1 = 17 + 273.15 = 290.15 \, \text{K} \]
03

Use Isentropic Relation for Temperature

For an isentropic process, the temperature relation is given by: \[ \frac{T_{2s}}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} \]where \( \gamma \) for air is approximately 1.4. Substitute known values: \[ \frac{T_{2s}}{290.15} = \left(\frac{450}{100}\right)^{\frac{0.4}{1.4}} \]
04

Solve for Isentropic Outlet Temperature

Calculate the isentropic discharge temperature \( T_{2s} \):\[ T_{2s} = 290.15 \times \left(\frac{450}{100}\right)^{\frac{0.4}{1.4}} \]
05

Apply Compressor Efficiency

The actual discharge temperature \( T_2 \) can be found using the isentropic efficiency \( \eta \):\[ \eta = \frac{T_{2s} - T_1}{T_2 - T_1} \]Rearrange to solve for \( T_2 \):\[ T_2 = T_1 + \frac{T_{2s} - T_1}{\eta} \]
06

Calculate the Actual Temperature Increase

Substitute \( T_1 \), \( T_{2s} \), and \( \eta \) into the equation:\[ T_2 = 290.15 + \frac{T_{2s} - 290.15}{0.8} \]Complete the calculations to find the actual discharge temperature.

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Key Concepts

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

Isentropic Process
In a centrifugal compressor, an isentropic process is an ideal concept where the compression happens completely without any loss or gain of heat. This means all the work done on the air by the compressor is used to increase the air's pressure and temperature, without any wastage in the form of heat transfer. In practical terms, it serves as a benchmark to compare real processes.

For an ideal gas undergoing an isentropic process, the relation between temperature and pressure is given by:
  • \( \frac{T_{2s}}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}} \)
Where,
  • \( T_1 \) is the initial temperature
  • \( T_{2s} \) is the isentropic final temperature
  • \( P_1 \) is the initial pressure
  • \( P_2 \) is the final pressure
  • \( \gamma \) is the specific heat ratio (often 1.4 for air)
These relationships are crucial in estimating what would happen in an ideal situation without loss due to inefficiencies.

Thus, an isentropic process provides a useful theoretical model that helps in analyzing and improving the efficiency of real-world compressors.
Compressor Efficiency
Compressor efficiency is a measure of how well a compressor converts energy into useful output. It's important because practical compressors are not perfectly efficient due to losses such as friction, heat loss, and other factors. The isentropic efficiency factor, \( \eta \), is used to quantify this efficiency.

This efficiency is defined by the formula:
  • \( \eta = \frac{T_{2s} - T_1}{T_2 - T_1} \)
Where:
  • \( T_2 \) is the actual discharge temperature
  • \( T_{2s} \) is the isentropic discharge temperature
  • \( T_1 \) is the inlet temperature
To find the actual output in real conditions, we rearrange this formula to solve for \( T_2 \):
  • \( T_2 = T_1 + \frac{T_{2s} - T_1}{\eta} \)
This equation helps in adjusting the ideal temperature to account for inefficiencies, providing a more accurate representation of the compressor's performance.

Understanding compressor efficiency enables engineers to optimize compressor designs, significantly improving energy efficiency and reducing operational costs in industrial applications.
Thermodynamic Temperature Conversion
Temperature conversion is a fundamental step in thermodynamic calculations. In the context of a centrifugal compressor analysis, it's essential to convert all given temperatures into Kelvin. This is because Kelvin is an absolute temperature scale and is necessary for accurate thermodynamic computations.

The conversion from Celsius to Kelvin is straightforward:
  • \( T(K) = T(^\circ C) + 273.15 \)
For example, if the temperature is given as 17°C, then converting this to Kelvin would result in:
  • \( T(K) = 17 + 273.15 = 290.15 \, K \)
Using Kelvin for thermodynamic processes ensures that calculations remain consistent and without errors associated with a negative absolute scale.

Getting comfortable with such conversions is not just crucial for solving specific compressor problems, but also for broader applications in physics and engineering, where clarity and precision in temperature measurement can drastically impact results.

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Most popular questions from this chapter

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