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Gas power cycles form the backbone of many mechanical engineering applications, especially in internal combustion (IC) engines and gas turbines. These cycles describe the idealized sequences of processes that convert heat energy into mechanical work using gases as the working fluid. Understanding these cycles helps engineers analyze engine performance, optimize efficiency, and design better power systems.
In this section, we will explore five important gas power cycles:
Each cycle has unique characteristics based on how heat is added and rejected, and how the working fluid is compressed and expanded. We will analyze these cycles using pressure-volume (P-V) and temperature-entropy (T-S) diagrams, derive their thermal efficiencies, and compare their performances. This knowledge is crucial for competitive exams and practical engineering applications.
The Otto cycle is the idealized thermodynamic cycle for spark ignition (SI) engines, such as petrol engines. It consists of four distinct processes involving the working gas (air-fuel mixture) inside the cylinder:
Assumptions: The cycle assumes ideal gas behavior, no friction, instantaneous combustion, and reversible adiabatic compression and expansion.
The thermal efficiency \(\eta\) of the Otto cycle depends primarily on the compression ratio \(r\) and the specific heat ratio \(\gamma\) (ratio of specific heats at constant pressure and volume, \(C_p/C_v\)). The compression ratio is defined as:
\[r = \frac{V_1}{V_2}\]
where \(V_1\) is the volume before compression and \(V_2\) is the volume after compression.
The thermal efficiency formula is derived from the first law of thermodynamics and ideal gas relations:
This formula shows that increasing the compression ratio improves efficiency, which is why SI engines strive for higher compression ratios within material limits.
The Diesel cycle models the idealized operation of compression ignition (CI) engines, such as diesel engines. It differs from the Otto cycle mainly in the heat addition process. The four processes are:
The Diesel cycle's efficiency depends on the compression ratio \(r\), cutoff ratio \(\rho\), and specific heat ratio \(\gamma\). The cutoff ratio is defined as:
\[\rho = \frac{V_3}{V_2}\]
where \(V_2\) is the volume at the start of heat addition and \(V_3\) is the volume at the end of heat addition (constant pressure process).
The thermal efficiency formula for the Diesel cycle is:
This formula shows that efficiency increases with compression ratio but is affected by the cutoff ratio, which represents the duration of fuel injection.
The Dual cycle, also known as the mixed cycle, combines features of both Otto and Diesel cycles. It models practical CI engines where heat addition occurs partly at constant volume and partly at constant pressure. The four processes are:
The dual cycle efficiency depends on compression ratio \(r\), pressure ratio \(\beta\), cutoff ratio \(\rho\), and specific heat ratio \(\gamma\). The pressure ratio is:
\[\beta = \frac{P_3}{P_2}\]
The efficiency formula is:
This cycle better represents real CI engine behavior by combining constant volume and constant pressure heat addition.
The Brayton cycle is the ideal cycle for gas turbines and jet engines. It operates on a continuous flow of air and fuel, unlike the reciprocating engines of previous cycles. The four main processes are:
The Brayton cycle efficiency depends on the pressure ratio \(r_p\) across the compressor and turbine, and the specific heat ratio \(\gamma\). The pressure ratio is:
\[r_p = \frac{P_2}{P_1}\]
The thermal efficiency formula is:
Increasing the pressure ratio improves efficiency, but practical limits arise due to material strength and temperature constraints. Regeneration (using exhaust heat to preheat compressed air) can further enhance efficiency.
The Stirling cycle is an external combustion engine cycle known for its high efficiency and quiet operation. It uses a fixed amount of working gas sealed inside the engine and operates through four processes:
The Stirling engine is an example of an external combustion engine, where heat is supplied externally rather than by internal fuel combustion.
The Stirling cycle efficiency depends only on the temperature of the hot reservoir \(T_H\) and cold reservoir \(T_L\), making it theoretically as efficient as the Carnot cycle:
This highlights the importance of maximizing the temperature difference for higher efficiency.
Step 1: Write down the formula for Otto cycle efficiency:
\[ \eta = 1 - \frac{1}{r^{\gamma - 1}} \]
Step 2: Substitute the given values:
\[ \eta = 1 - \frac{1}{8^{1.4 - 1}} = 1 - \frac{1}{8^{0.4}} \]
Step 3: Calculate \(8^{0.4}\):
\[ 8^{0.4} = e^{0.4 \ln 8} = e^{0.4 \times 2.079} = e^{0.8316} \approx 2.297 \]
Step 4: Calculate efficiency:
\[ \eta = 1 - \frac{1}{2.297} = 1 - 0.435 = 0.565 \]
Answer: The thermal efficiency is approximately 56.5%.
Step 1: Write down the Diesel cycle efficiency formula:
\[ \eta = 1 - \frac{1}{r^{\gamma - 1}} \times \frac{\rho^{\gamma} - 1}{\gamma (\rho - 1)} \]
Step 2: Calculate \(r^{\gamma - 1}\):
\[ r^{\gamma - 1} = 15^{0.4} = e^{0.4 \ln 15} = e^{0.4 \times 2.708} = e^{1.083} \approx 2.953 \]
Step 3: Calculate \(\rho^{\gamma} - 1\):
\[ 2^{1.4} - 1 = e^{1.4 \ln 2} - 1 = e^{1.4 \times 0.693} - 1 = e^{0.970} - 1 \approx 2.638 - 1 = 1.638 \]
Step 4: Calculate denominator \(\gamma (\rho - 1)\):
\[ 1.4 \times (2 - 1) = 1.4 \times 1 = 1.4 \]
Step 5: Calculate the fraction:
\[ \frac{1.638}{1.4} = 1.17 \]
Step 6: Calculate the efficiency:
\[ \eta = 1 - \frac{1}{2.953} \times 1.17 = 1 - 0.339 \times 1.17 = 1 - 0.397 = 0.603 \]
Answer: The thermal efficiency is approximately 60.3%.
Step 1: Calculate Otto cycle efficiency:
\[ \eta_{Otto} = 1 - \frac{1}{15^{0.4}} = 1 - \frac{1}{2.953} = 1 - 0.339 = 0.661 \]
Step 2: Calculate Diesel cycle efficiency (from Example 2):
\[ \eta_{Diesel} = 0.603 \]
Step 3: Compare efficiencies:
Otto cycle efficiency is higher (66.1%) compared to Diesel cycle (60.3%) for the same compression ratio.
Answer: Otto cycle is more efficient at same compression ratio, but Diesel cycle allows higher compression ratios practically.
Step 1: Write the Brayton cycle efficiency formula:
\[ \eta = 1 - \frac{1}{r_p^{\frac{\gamma - 1}{\gamma}}} \]
Step 2: Calculate efficiency for \(r_p=5\):
\[ \frac{\gamma - 1}{\gamma} = \frac{0.4}{1.4} = 0.2857 \]
\[ 5^{0.2857} = e^{0.2857 \ln 5} = e^{0.2857 \times 1.609} = e^{0.460} \approx 1.584 \]
\[ \eta = 1 - \frac{1}{1.584} = 1 - 0.631 = 0.369 \]
Step 3: Calculate efficiency for \(r_p=10\):
\[ 10^{0.2857} = e^{0.2857 \ln 10} = e^{0.2857 \times 2.303} = e^{0.658} \approx 1.931 \]
\[ \eta = 1 - \frac{1}{1.931} = 1 - 0.518 = 0.482 \]
Step 4: Effect of regeneration at \(r_p=10\):
Efficiency improves by 10%, so new efficiency:
\[ \eta_{regen} = 0.482 \times 1.10 = 0.530 \]
Answer: Efficiency increases with pressure ratio; regeneration further improves efficiency by recovering exhaust heat.
Step 1: Use the Stirling cycle efficiency formula:
\[ \eta = 1 - \frac{T_L}{T_H} \]
Step 2: Substitute the given temperatures:
\[ \eta = 1 - \frac{300}{900} = 1 - 0.333 = 0.667 \]
Answer: The thermal efficiency is 66.7%.
| Cycle | Heat Addition | Heat Rejection | Efficiency Dependence | Typical Application |
|---|---|---|---|---|
| Otto | Constant Volume | Constant Volume | Compression ratio \(r\) | Spark Ignition Engines |
| Diesel | Constant Pressure | Constant Volume | Compression ratio \(r\), Cutoff ratio \(\rho\) | Compression Ignition Engines |
| Dual | Constant Volume + Constant Pressure | Constant Volume | Compression \(r\), Pressure \(\beta\), Cutoff \(\rho\) | Practical CI Engines |
| Brayton | Constant Pressure | Constant Pressure | Pressure ratio \(r_p\) | Gas Turbines |
| Stirling | Isothermal | Isothermal | Temperature limits \(T_H, T_L\) | External Combustion Engines |
When to use: Quickly solve efficiency problems during exams.
When to use: To understand cycle behavior and prevent conceptual errors.
When to use: For conceptual questions and optimization problems.
When to use: To correctly apply formulas and avoid mixing processes.
When to use: For quick estimation or multiple-choice questions.
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