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Air enters the compressor of a simple gas turbine at 100 \(\mathrm{kPa}, 300 \mathrm{~K}\), with a volumetric flow rate of \(5 \mathrm{~m}^{3} / \mathrm{s}\). The compressor pressure ratio is 10 and its isentropic efficiency is \(85 \%\). At the inlet to the turbine, the pressure is \(950 \mathrm{kPa}\), and the temperature is \(1400 \mathrm{~K}\). The turbine has an isentropic efficiency of \(88 \%\) and the exit pressure is \(100 \mathrm{kPa}\). On the basis of an air- standard analysis, (a) develop a full accounting of the net exergy increase of the air passing through the gas turbine combustor, in \(\mathrm{kW}\). (b) devise and evaluate an exergetic efficiency for the gas turbine cycle. Let \(T_{0}=300 \mathrm{~K}, p_{0}=100 \mathrm{kPa}\).

Short Answer

Expert verified
The net exergy increase of the air passing through the gas turbine combustor can be calculated using enthalpies and irreversibilities. Exergetic efficiency is found by comparing net work output to the exergy input.

Step by step solution

01

Determine Compressor Exit State

Calculate the pressure and temperature at the compressor exit. Given the pressure ratio is 10, the exit pressure is \[P_{2} = 10 \times P_{1} = 10 \times 100 \text{ kPa} = 1000 \text{ kPa}.\]Using the isentropic efficiency and the temperature relation for an ideal gas, \[T_{2s} = T_{1} \left(\frac{P_{2}}{P_{1}}\right)^{\frac{k-1}{k}},\]where \(k\) is the specific heat ratio (typically \(k = 1.4\) for air). Thus, \[T_{2s} = 300 \left(\frac{1000}{100}\right)^{\frac{0.4}{1.4}} \approx 579 \text{ K}.\]Using the isentropic efficiency of 85%, \[\eta_{c} = \frac{T_{2s} - T_{1}}{T_{2} - T_{1}},\]\[0.85 = \frac{579 - 300}{T_{2} - 300},\]\[T_{2} \approx 626 \text{ K}.\]
02

Determine Turbine Exit State

Calculate the pressure and temperature at the turbine exit. Given the exit pressure is 100 kPa, using the isentropic efficiency, \[T_{4s} = T_{3} \left(\frac{P_{4}}{P_{3}}\right)^{\frac{k-1}{k}},\]\[T_{4s} = 1400 \left(\frac{100}{950}\right)^{\frac{0.4}{1.4}} \approx 890 \text{ K}.\]Using the isentropic efficiency of 88%, \[\eta_{t} = \frac{T_{3} - T_{4}}{T_{3} - T_{4s}},\]\[0.88 = \frac{1400 - T_{4}}{1400 - 890},\]\[T_{4} \approx 974 \text{ K}.\]
03

Determine Work done by Compressor and Turbine

Use the temperatures to find work done. For compressor, \[W_{c} = \dot{m} C_p (T_{2} - T_{1}),\]where \( C_p \approx 1.005 \text{ kJ/kg.K} \) for air. For turbine, \[W_{t} = \dot{m} C_p (T_{3} - T_{4}).\]
04

Exergy Analysis

The exergy increase through the combustor can be found using: \[\Delta e = (h_{3} - h_{2}) - T_{0}(s_{3} - s_{2}),\]where \(h\) and \(s\) are enthalpy and entropy, respectively. For a more straightforward approach, \[\Delta E = \dot{m} \cdot \Delta e.\]
05

Calculate Exergetic Efficiency

The exergetic efficiency is given by the ratio of net work output to the exergy input. \[\eta_{ex} = \frac{W_{net}}{\Delta E} \approx \frac{W_{t} - W_{c}}{\Delta E}.\]

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Key Concepts

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

Isentropic Efficiency
Isentropic efficiency measures how efficiently a compressor or turbine performs compared to an ideal, frictionless device. It is a crucial metric in gas turbine cycle analysis to evaluate the real performance.
The formulae are as follows:
For a compressor: \( \eta_{c} = \frac{T_{2s} - T_{1}}{T_{2} - T_{1}} \)
For a turbine: \( \eta_{t} = \frac{T_{3} - T_{4}}{T_{3} - T_{4s}} \)
Where:
  • \( T_{2s} \) and \( T_{4s} \) are the ideal temperatures after compression and expansion, respectively.
  • \( T_{2} \) and \( T_{4} \) are the actual temperatures after compression and expansion, respectively.
  • \( T_{1} \) and \( T_{3} \) are the intake temperatures.

Isentropic efficiency helps diagnose where the real-world process deviates from the ideal process. Lower efficiencies indicate higher irreversibilities due to friction, heat losses, or other inefficiencies.

Consider this when analyzing the state changes in gas turbine cycles, as deviations impact both performance and energy transfer. Therefore, accurately accounting for isentropic efficiency is essential for a complete gas turbine cycle analysis.
Exergy Analysis
Exergy analysis is a powerful tool in thermodynamics to measure how much useful work can be extracted from a system. Unlike energy, which is conserved, exergy can be destroyed due to irreversibilities in a process.
For calculating the exergy increase through a gas turbine combustor, use:
\( \Delta e = (h_{3} - h_{2}) - T_{0}(s_{3} - s_{2}) \)
  • \( h \) represents enthalpy.
  • \( s \) represents entropy.
  • \( T_0 \) is the reference temperature, usually the ambient temperature.

Exergy analysis helps in identifying where and how energy losses occur, enabling better optimization of the cycle.
It's essential to calculate both the net exergy and the exergetic efficiency for a complete picture:
\( \Delta E = \dot{m} \cdot \Delta e \)
\( \eta_{ex} = \frac{W_{net}}{\Delta E} \)
  • \( \dot{m} \) is the mass flow rate.
  • \( W_{net} \) is the net work output.

By understanding exergy, engineers can pinpoint inefficiencies and make design improvements.
Compressor and Turbine Work Calculations
Calculating the work done by compressors and turbines is crucial in determining the performance of a gas turbine cycle.

For the compressor, the work required is given by:
\( W_{c} = \dot{m} \cdot C_p \cdot (T_{2} - T_{1}) \)
  • \( C_p \) is the specific heat capacity at constant pressure (usually \( 1.005 \text{kJ/kg.K} \) for air).
  • \( \dot{m} \) is the mass flow rate.
  • \( T_{2} \) and \( T_{1} \) are the exit and entry temperatures of the compressor, respectively.

For the turbine, the work output is given by:
\( W_{t} = \dot{m} \cdot C_p \cdot (T_{3} - T_{4}) \)
  • \( T_{3} \) and \( T_{4} \) are the entry and exit temperatures of the turbine, respectively.

These calculations help us understand how much energy is required to compress air and how much useful work is extracted by the turbine, impacting the overall efficiency of the gas turbine cycle.
Ideal Gas Relationships
Ideal gas relationships are fundamental in thermodynamics for simplifying the analysis of gas turbine cycles. The ideal gas law is expressed as:
\( PV = nRT \)
  • \( P \) is the pressure.
  • \( V \) is the volume.
  • \( n \) is the number of moles.
  • \( R \) is the universal gas constant.
  • \( T \) is the temperature.

For specific scenarios in gas turbines, additional relationships are used:
  • Isentropic Relations: \( T_{2s}/T_{1} = (P_{2}/P_{1})^{(k-1)/k} \), which help in calculating exit temperatures.
  • Constant Pressure and Volume Processes: Applying specific heat capacities, \( C_p \) and \( C_v \), to calculate changes in energy.

Understanding these relationships helps in accurately determining state properties, such as temperature, pressure, and volume, thereby simplifying the overall analysis and ensuring accurate results.

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Most popular questions from this chapter

A combined gas turbine-vapor power plant (Fig. 9.23) has a net power output of \(100 \mathrm{MW}\). Air enters the compressor of the gas turbine at \(100 \mathrm{kPa}, 300 \mathrm{~K}\), and is compressed to \(1200 \mathrm{kPa}\). The isentropic efficiency of the compressor is \(84 \%\). The conditions at the inlet to the turbine are \(1200 \mathrm{kPa}\) and \(1400 \mathrm{~K}\). Air expands through the turbine, which has an isentropic efficiency of \(88 \%\), to a pressure of \(100 \mathrm{kPa}\). The air then passes through the interconnecting heat exchanger, and is finally discharged at \(480 \mathrm{~K}\). Steam enters the turbine of the vapor power cycle at \(8 \mathrm{MPa}, 400^{\circ} \mathrm{C}\), and expands to the condenser pressure of \(8 \mathrm{kPa}\). Water enters the pump as saturated liquid at \(8 \mathrm{kPa}\). The turbine and pump have isentropic efficiencies of 90 and \(80 \%\), respectively. Determine (a) the mass flow rates of air and steam, each in \(\mathrm{kg} / \mathrm{s}\). (b) the thermal efficiency of the combined cycle. (c) a full accounting of the net exergy increase of the air passing through the combustor of the gas turbine, \(\dot{m}_{\text {air }}\left[\mathrm{e}_{\mathrm{f} 3}-\mathrm{e}_{\mathrm{f} 2}\right]\), in MW. Discuss. Let \(T_{0}=300 \mathrm{~K}, p_{0}=100 \mathrm{kPa}\).

If the inlet state and the exit pressure are specified for a two-stage turbine with reheat between the stages and operating at steady state, show that the maximum total work output is obtained when the pressure ratio is the same across each stage. Use a cold air-standard analysis assuming that each compression process is isentropic, there is no pressure drop through the reheater, and the temperature at the inlet to each turbine stage is the same. Kinetic and potential energy effects can be ignored.

Using the ideal gas model, determine the sonic velocity in \(\mathrm{m} / \mathrm{s}\) of steam at \(600 \mathrm{~K}\) and 50 bar.

A two-stage air compressor operates at steady state, compressing \(0.15 \mathrm{~m}^{3} / \mathrm{min}\) of air from \(100 \mathrm{kPa}, 300 \mathrm{~K}\), to \(1100 \mathrm{kPa}\). An intercooler between the two stages cools the air to \(300 \mathrm{~K}\) at a constant pressure of \(325 \mathrm{kPa}\). 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.

An engine working on the air standard Otto cycle is supplied with air at \(0.1 \mathrm{MPa}, 27^{\circ} \mathrm{C}\). The compression ratio is 8 . The heat supplied is \(1400 \mathrm{~kJ} / \mathrm{kg}\). Calculate the maximum pressure and temperature of the cycle, the cycle efficiency, and the mean effective pressure.

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