Interactive Lecture
Otto, Diesel, Carnot, Rankine, and Brayton Cycle Analysis
by Dr Arnaldo Delli Carri
A thermodynamic cycle is a series of thermodynamic processes that returns a system to its initial state. Engines, refrigerators, and heat pumps all run on such cycles. The system undergoes a sequence of processes (heating, cooling, expansion, compression) and ultimately returns to its starting point, ready to repeat the cycle.
Over one complete cycle, every state property returns to its initial value. The net change in internal energy is zero (\(\Delta U_{cycle} = 0\)), so the net work output equals the net heat input: \(W_{net} = Q_{in} - Q_{out}\)
Thermodynamic cycles can be classified into two main categories:
The Carnot cycle is the most efficient thermodynamic cycle operating between two temperature limits. It consists of four reversible processes:
Carnot Cycle Processes:
Carnot Efficiency:
\[\eta_{Carnot} = 1 - \frac{T_C}{T_H}\]
Where \(T_H\) and \(T_C\) are absolute temperatures (Kelvin)
The Otto cycle is an idealised model for spark-ignition (gasoline) engines. It consists of four processes:
Otto Cycle Processes:
Otto Cycle Efficiency:
\[\eta_{Otto} = 1 - \frac{1}{r^{\gamma - 1}}\]
Where:
The Diesel cycle models compression-ignition engines where fuel is injected and burns at constant pressure:
Diesel Cycle Processes:
Diesel Cycle Efficiency:
\[\eta_{Diesel} = 1 - \frac{1}{r^{\gamma - 1}} \cdot \frac{r_c^{\gamma} - 1}{\gamma(r_c - 1)}\]
Where \(r_c = \frac{V_3}{V_2}\) is the cutoff ratio
The Rankine cycle is used in steam power plants and involves phase changes of the working fluid:
Rankine Cycle Processes:
Rankine Cycle Efficiency:
\[\eta_{Rankine} = \frac{W_{turbine} - W_{pump}}{Q_{in}} = \frac{(h_3 - h_4) - (h_2 - h_1)}{h_3 - h_2}\]
For any power cycle, thermal efficiency is defined as:
\[\eta_{th} = \frac{W_{net}}{Q_{in}} = \frac{Q_{in} - Q_{out}}{Q_{in}} = 1 - \frac{Q_{out}}{Q_{in}}\]
Where:
Explore different thermodynamic cycles on P-V and T-S diagrams. Select a cycle type, adjust parameters, and press Play to animate the cycle through each process.
| Concept | Equation | Description |
|---|---|---|
| Thermal Efficiency | \(\eta = \frac{W_{net}}{Q_{in}} = 1 - \frac{Q_{out}}{Q_{in}}\) | General efficiency for power cycles |
| Carnot Efficiency | \(\eta_{Carnot} = 1 - \frac{T_C}{T_H}\) | Maximum possible efficiency between two temperatures |
| Otto Efficiency | \(\eta_{Otto} = 1 - \frac{1}{r^{\gamma-1}}\) | Ideal spark-ignition engine efficiency |
| Diesel Efficiency | \(\eta_{Diesel} = 1 - \frac{1}{r^{\gamma-1}} \cdot \frac{r_c^{\gamma}-1}{\gamma(r_c-1)}\) | Ideal compression-ignition engine efficiency |
| Compression Ratio | \(r = \frac{V_{max}}{V_{min}}\) | Ratio of maximum to minimum cylinder volume |
| Specific Heat Ratio | \(\gamma = \frac{C_p}{C_v}\) | Ratio of specific heats (≈1.4 for air) |
| Isentropic Process | \(PV^{\gamma} = \text{const}\) or \(TV^{\gamma-1} = \text{const}\) | Adiabatic reversible process |
| Ideal Gas Law | \(PV = nRT\) or \(PV = mR_sT\) | Equation of state for ideal gases (\(n\) = moles, \(R = 8.314\) J/mol·K; or \(m\) = mass, \(R_s = R/M\) = specific gas constant) |
| Work (Isothermal) | \(W = nRT\ln\frac{V_2}{V_1}\) | Work during constant temperature process |
| Heat (Constant Volume) | \(Q = nC_v\Delta T\) | Heat transfer at constant volume |
An Otto cycle engine has a compression ratio of 9:1 and operates with air as the working fluid (γ = 1.4). Calculate the thermal efficiency of the engine.
Show Solution
Solution:
Given:
Formula:
\[\eta_{Otto} = 1 - \frac{1}{r^{\gamma-1}}\]
Calculation:
\[\eta_{Otto} = 1 - \frac{1}{9^{1.4-1}} = 1 - \frac{1}{9^{0.4}}\]
\[\eta_{Otto} = 1 - \frac{1}{2.408} = 1 - 0.415 = 0.585\]
Answer: The thermal efficiency is 58.5%
A Carnot engine operates between a hot reservoir at 800 K and a cold reservoir at 300 K. Calculate:
Show Solution
Solution:
Given:
Formula:
\[\eta_{Carnot} = 1 - \frac{T_C}{T_H}, \qquad \frac{Q_{out}}{Q_{in}} = \frac{T_C}{T_H}, \qquad Q_{in} = \frac{W}{\eta}\]
Calculation (a) — Thermal Efficiency:
\[\eta_{Carnot} = 1 - \frac{300}{800} = 1 - 0.375 = 0.625 = 62.5\%\]
Calculation (b) — Heat Rejected Ratio:
\[\frac{Q_{out}}{Q_{in}} = \frac{300}{800} = 0.375 = 37.5\%\]
Calculation (c) — Heat Absorbed:
\[Q_{in} = \frac{W}{\eta} = \frac{100}{0.625} = 160 \text{ kJ}\]
Answers:
Compare the thermal efficiency of an Otto cycle and a Diesel cycle, both operating with:
Show Solution
Solution:
Given:
Formula:
\[\eta_{Otto} = 1 - \frac{1}{r^{\gamma-1}}, \qquad \eta_{Diesel} = 1 - \frac{1}{r^{\gamma-1}} \cdot \frac{r_c^{\gamma}-1}{\gamma(r_c-1)}\]
Calculation (Otto):
\[\eta_{Otto} = 1 - \frac{1}{18^{0.4}} = 1 - \frac{1}{3.179} = 0.685 = 68.5\%\]
Calculation (Diesel):
\[\eta_{Diesel} = 1 - \frac{1}{18^{0.4}} \cdot \frac{2^{1.4}-1}{1.4(2-1)} = 1 - \frac{1}{3.179} \cdot \frac{2.639-1}{1.4}\]
\[\eta_{Diesel} = 1 - 0.3147 \cdot 1.171 = 1 - 0.368 = 0.632 = 63.2\%\]
Comparison: At the same compression ratio the Otto cycle is more efficient, because Diesel heat addition at constant pressure pushes part of the combustion to larger volume where less work is recovered. Diesel engines win in practice by running much higher compression ratios (14:1–25:1) than spark ignition allows.
Answer: \(\eta_{Otto} = 68.5\%\), \(\eta_{Diesel} = 63.2\%\)