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Gas turbines are powerful engines widely used in power plants, aircraft propulsion, and industrial applications. They convert the chemical energy of fuel into mechanical work through the continuous flow of gases. The fundamental thermodynamic cycle governing gas turbines is the Brayton cycle, named after George Brayton who first described it.
The Brayton cycle is essential in engineering thermodynamics because it models the ideal operation of gas turbines, helping engineers understand and optimize their performance. It consists of four main components: a compressor, a combustion chamber (or heat exchanger), a turbine, and a heat exchanger or exhaust system. These components work together to compress air, add heat at constant pressure, expand the hot gases to produce work, and finally reject heat.
Understanding the Brayton cycle and its modifications is crucial for competitive exams and practical engineering, especially in India where gas turbines contribute significantly to power generation and aviation sectors. This chapter will guide you through the basic cycle, its thermodynamic processes, graphical representations, and ways to improve its efficiency through modifications like regeneration, reheating, and intercooling.
The ideal Brayton cycle consists of four distinct thermodynamic processes that air undergoes as it flows through the gas turbine system. These processes assume air behaves as an ideal gas, and there are no losses due to friction, heat transfer to the surroundings, or mechanical inefficiencies.
These processes can be summarized as:
Visualizing the Brayton cycle on pressure-volume (PV) and temperature-entropy (TS) diagrams helps understand the thermodynamic changes during each process.
Figure: PV and TS diagrams of the ideal Brayton cycle showing the four processes and states.
While the ideal Brayton cycle provides a useful baseline, real gas turbines incorporate modifications to enhance efficiency and performance. The three main modifications are:
Each modification affects the thermal efficiency and work output differently, often trading off complexity and cost for better performance.
graph TD A[Start: Air Intake] --> B[Compression Stage] B --> C{Intercooling?} C -- Yes --> D[Cool compressed air] D --> E[Further Compression] C -- No --> E E --> F[Heat Addition (Combustion)] F --> G{Reheating?} G -- Yes --> H[Expansion Stage 1] H --> I[Reheat Gas] I --> J[Expansion Stage 2] G -- No --> J[Expansion Stage] J --> K{Regeneration?} K -- Yes --> L[Heat Recovery from Exhaust] L --> M[Heat Added Reduced] K -- No --> M[Heat Added] M --> N[Exhaust]Figure: Flowchart illustrating Brayton cycle modifications: intercooling, reheating, and regeneration.
Key performance parameters include thermal efficiency, net work output, and specific fuel consumption. Understanding these helps evaluate and compare cycle performance.
Thermal efficiency is the ratio of net work output to heat input:
For the ideal Brayton cycle, thermal efficiency depends mainly on the pressure ratio \( r_p = \frac{P_2}{P_1} \) and the specific heat ratio \( \gamma \) of air:
The net work is the difference between turbine work and compressor work:
where \( m \) is the mass flow rate of air, and \( C_p \) is the specific heat at constant pressure.
SFC measures the fuel required to produce a unit of power, typically in kg/kWh or g/kWh. Lower SFC indicates better fuel efficiency.
Step 1: Identify given data:
Step 2: Use the thermal efficiency formula for ideal Brayton cycle:
\[ \eta_{th} = 1 - \frac{1}{r_p^{\frac{\gamma - 1}{\gamma}}} \]
Step 3: Calculate the exponent:
\[ \frac{\gamma - 1}{\gamma} = \frac{1.4 - 1}{1.4} = \frac{0.4}{1.4} \approx 0.2857 \]
Step 4: Calculate \( r_p^{(\gamma - 1)/\gamma} \):
\[ 10^{0.2857} \approx 1.93 \]
Step 5: Calculate efficiency:
\[ \eta_{th} = 1 - \frac{1}{1.93} = 1 - 0.518 = 0.482 \]
Answer: The thermal efficiency of the ideal Brayton cycle is approximately 48.2%.
Step 1: Given data:
Step 2: The temperature of air leaving the regenerator, \( T_5 \), is given by:
\[ \epsilon = \frac{T_5 - T_2}{T_4 - T_2} \]
Rearranged:
\[ T_5 = T_2 + \epsilon (T_4 - T_2) \]
Step 3: Find \( T_4 \) (turbine outlet temperature) using isentropic expansion relation:
\[ T_4 = T_3 \times \left(\frac{1}{r_p}\right)^{\frac{\gamma - 1}{\gamma}} = 1400 \times 8^{-0.2857} \]
Calculate \( 8^{-0.2857} \):
\[ 8^{-0.2857} = \frac{1}{8^{0.2857}} \approx \frac{1}{1.86} = 0.5376 \]
Therefore,
\[ T_4 = 1400 \times 0.5376 = 752.6 \, K \]
Step 4: Calculate \( T_5 \):
\[ T_5 = 600 + 0.75 \times (752.6 - 600) = 600 + 0.75 \times 152.6 = 600 + 114.45 = 714.45 \, K \]
Answer: The temperature of air leaving the regenerator is approximately 714.5 K. Regeneration reduces the fuel needed to heat air from 600 K to 1400 K by preheating it to 714.5 K, improving thermal efficiency.
Step 1: Given data:
Step 2: Calculate turbine outlet temperature after first expansion (\( T_4 \)) using isentropic relation:
\[ T_4 = T_3 \times \left(\frac{P_4}{P_3}\right)^{\frac{\gamma - 1}{\gamma}} \]
Since there are two expansion stages, assume pressure ratio across each turbine stage is \( \sqrt{r_p} = \sqrt{10} = 3.162 \).
So,
\[ \frac{P_4}{P_3} = \frac{1}{3.162} = 0.3162 \]
Calculate exponent:
\[ \frac{\gamma - 1}{\gamma} = 0.2857 \]
Calculate:
\[ T_4 = 1500 \times (0.3162)^{0.2857} = 1500 \times 0.724 = 1086 \, K \]
Step 3: Calculate turbine outlet temperature after second expansion (\( T_6 \)):
\[ T_6 = T_5 \times \left(\frac{P_6}{P_5}\right)^{\frac{\gamma - 1}{\gamma}} = 1400 \times (0.3162)^{0.2857} = 1400 \times 0.724 = 1013.6 \, K \]
Step 4: Calculate turbine work output per kg air:
\[ W_{turbine} = C_p \times [(T_3 - T_4) + (T_5 - T_6)] = 1.005 \times [(1500 - 1086) + (1400 - 1013.6)] \]
\[ = 1.005 \times (414 + 386.4) = 1.005 \times 800.4 = 804.4 \, kJ/kg \]
Step 5: Calculate compressor work:
\[ W_{compressor} = C_p \times (T_2 - T_1) = 1.005 \times (600 - 300) = 1.005 \times 300 = 301.5 \, kJ/kg \]
Step 6: Calculate net work output:
\[ W_{net} = W_{turbine} - W_{compressor} = 804.4 - 301.5 = 502.9 \, kJ/kg \]
Answer: The net work output per kg of air for the reheated Brayton cycle is approximately 503 kJ/kg.
Step 1: Given data:
Step 2: For two-stage compression with intercooling, the pressure ratio is split equally:
\[ r_{p1} = r_{p2} = \sqrt{\frac{P_3}{P_1}} = \sqrt{8} = 2.828 \]
Step 3: Calculate temperature after first compression stage:
\[ T_2 = T_1 \times r_{p1}^{\frac{\gamma - 1}{\gamma}} = 300 \times 2.828^{0.2857} = 300 \times 1.32 = 396 \, K \]
Step 4: Calculate work for each compression stage:
\[ W_{comp,1} = C_p \times (T_2 - T_1) = 1.005 \times (396 - 300) = 96.5 \, kJ/kg \]
After intercooling, temperature returns to 300 K before second stage.
Step 5: Calculate temperature after second compression stage:
\[ T_3 = T_1 \times r_{p2}^{\frac{\gamma - 1}{\gamma}} = 300 \times 2.828^{0.2857} = 396 \, K \]
Step 6: Work for second stage:
\[ W_{comp,2} = C_p \times (T_3 - T_1) = 1.005 \times (396 - 300) = 96.5 \, kJ/kg \]
Step 7: Total compressor work with intercooling:
\[ W_{total} = W_{comp,1} + W_{comp,2} = 96.5 + 96.5 = 193 \, kJ/kg \]
Answer: Intercooling reduces compressor work from 250 kJ/kg to approximately 193 kJ/kg, improving net work output.
Step 1: Given data:
Step 2: Calculate isentropic compressor outlet temperature \( T_{2s} \):
\[ T_{2s} = T_1 \times r_p^{\frac{\gamma - 1}{\gamma}} = 300 \times 12^{0.2857} \]
Calculate \( 12^{0.2857} \):
\[ 12^{0.2857} \approx 1.95 \]
Therefore,
\[ T_{2s} = 300 \times 1.95 = 585 \, K \]
Step 3: Calculate actual compressor outlet temperature \( T_2 \):
Since compressor isentropic efficiency \( \eta_c = \frac{T_{2s} - T_1}{T_2 - T_1} \), rearranged:
\[ T_2 = T_1 + \frac{T_{2s} - T_1}{\eta_c} = 300 + \frac{585 - 300}{0.85} = 300 + \frac{285}{0.85} = 300 + 335.3 = 635.3 \, K \]
Step 4: Calculate isentropic turbine outlet temperature \( T_{4s} \):
\[ T_{4s} = T_3 \times \left(\frac{1}{r_p}\right)^{\frac{\gamma - 1}{\gamma}} = 1400 \times 12^{-0.2857} = 1400 \times \frac{1}{1.95} = 1400 \times 0.513 = 718.2 \, K \]
Step 5: Calculate actual turbine outlet temperature \( T_4 \):
Turbine isentropic efficiency \( \eta_t = \frac{T_3 - T_4}{T_3 - T_{4s}} \), rearranged:
\[ T_4 = T_3 - \eta_t (T_3 - T_{4s}) = 1400 - 0.88 \times (1400 - 718.2) = 1400 - 0.88 \times 681.8 = 1400 - 599.98 = 800 \, K \]
Answer: The actual compressor outlet temperature is approximately 635 K, and the turbine outlet temperature is approximately 800 K.
When to use: When analyzing or sketching the cycle to avoid confusion.
When to use: For quick calculations during exams.
When to use: When dealing with modified Brayton cycles involving regeneration.
When to use: During temperature calculations after compression or expansion.
When to use: When solving complex modified Brayton cycle problems.
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