Overview of thermodynamics of fuel-air cycles and real cycles

Derivation of Otto cycle efficiency

The Otto cycle is the ideal air-standard cycle of a spark-ignition engine. It has four processes: 1-2 isentropic compression, 2-3 constant-volume heat addition, 3-4 isentropic expansion and 4-1 constant-volume heat rejection.

 p
 |        3
 |       /|\
 |      / | \         1-2 compression
 |     /  |  \        2-3 heat in (v = c)
 |    2   |   \       3-4 expansion
 |   /    |    \      4-1 heat out (v = c)
 |  /      \    4
 | /        \__/
 |/_______1__/______ v
     v2        v1

For 1 kg of air:

$$\begin{aligned} Q_{in} &= c_v (T_3 - T_2) \\ Q_{out} &= c_v (T_4 - T_1) \\ \eta &= 1 - \frac{Q_{out}}{Q_{in}} = 1 - \frac{T_4 - T_1}{T_3 - T_2} \end{aligned}$$

For the isentropic processes, with $r = v_1/v_2 = v_4/v_3$:

$$\frac{T_2}{T_1} = r^{\gamma-1}, \qquad \frac{T_3}{T_4} = r^{\gamma-1}$$

So $T_3/T_2 = T_4/T_1$, which gives $(T_4 - T_1)/(T_3 - T_2) = T_1/T_2$. Therefore

$$\eta_{Otto} = 1 - \frac{T_1}{T_2} = 1 - \frac{1}{r^{\gamma-1}}$$

The efficiency depends only on $r$ and $\gamma$. It rises as $r$ increases.

Numerical

$$\begin{aligned} T_2 &= 300 \times 8^{0.4} = 689.2\ \text{K} \\ p_2 &= 100 \times 8^{1.4} = 1837.9\ \text{kPa} \\ T_3 &= T_2 + \frac{Q_{in}}{c_v} = 689.2 + \frac{1500}{0.718} = 2778.4\ \text{K} \\ p_3 &= p_2 \frac{T_3}{T_2} = 1837.9 \times \frac{2778.4}{689.2} = 7409\ \text{kPa} \\ \eta &= 1 - 8^{-0.4} = 0.5647 \\ W_{net} &= \eta\, Q_{in} = 0.5647 \times 1500 = 847.1\ \text{kJ/kg} \end{aligned}$$

Check: $T_4 = T_3/8^{0.4} = 1209.3$ K, so $Q_{out} = 0.718(1209.3 - 300) = 652.9$ kJ/kg and $1500 - 652.9 = 847.1$ kJ/kg.

Specific volumes:

$$\begin{aligned} v_1 &= \frac{RT_1}{p_1} = \frac{0.287 \times 300}{100} = 0.861\ \text{m}^3/\text{kg} \\ v_2 &= v_1/8 = 0.1076\ \text{m}^3/\text{kg} \\ \text{mep} &= \frac{W_{net}}{v_1 - v_2} = \frac{847.1}{0.861 - 0.1076} = 1124\ \text{kPa} \end{aligned}$$

Answer: $T_{max} = 2778$ K, $p_{max} = 7.41$ MPa, $W_{net} = 847$ kJ/kg, mep = 11.24 bar (efficiency 56.5%).

The Diesel cycle has isentropic compression (1-2), constant-pressure heat addition (2-3), isentropic expansion (3-4) and constant-volume heat rejection (4-1).

 p
 |      2________3
 |     /         \
 |    /           \
 |   /             \
 |  /               4
 | /              _/
 |/____1________/____ v

Temperatures

$$\begin{aligned} T_2 &= T_1 r^{\gamma-1} = 300 \times 18^{0.4} = 953.3\ \text{K} \\ T_3 &= \rho\, T_2 = 2.2 \times 953.3 = 2097.3\ \text{K} \\ T_4 &= T_3 \left(\frac{\rho}{r}\right)^{\gamma-1} = 2097.3 \times \left(\frac{2.2}{18}\right)^{0.4} = 904.7\ \text{K} \end{aligned}$$

Heat and efficiency

$$\begin{aligned} Q_{in} &= c_p (T_3 - T_2) = 1.005(2097.3 - 953.3) = 1149.7\ \text{kJ/kg} \\ Q_{out} &= c_v (T_4 - T_1) = 0.718(904.7 - 300) = 434.2\ \text{kJ/kg} \\ W_{net} &= 1149.7 - 434.2 = 715.5\ \text{kJ/kg} \\ \eta &= \frac{715.5}{1149.7} = 0.622 \end{aligned}$$

Check with the formula:

$$\eta = 1 - \frac{1}{18^{0.4}} \times \frac{2.2^{1.4} - 1}{1.4 \times 1.2} = 1 - \frac{1}{3.178} \times \frac{2.0157}{1.68} = 0.622$$

Mean effective pressure

$$\begin{aligned} v_1 &= \frac{0.287 \times 300}{100} = 0.861\ \text{m}^3/\text{kg}, \quad v_2 = \frac{0.861}{18} = 0.0478 \\ \text{mep} &= \frac{715.5}{0.861 - 0.0478} = 880\ \text{kPa} \end{aligned}$$

Comparison with Otto

For $r = 18$: $\eta_{Otto} = 1 - 18^{-0.4} = 0.685$, which is higher than the Diesel value of 0.622.

At the same compression ratio the Otto cycle is more efficient. In practice an SI engine cannot use $r = 18$ because of knock, while a CI engine compresses only air and can. So real Diesel engines (r = 16-22) are more efficient than petrol engines (r = 8-10).

Answer: $T_2 = 953$ K, $T_3 = 2097$ K, $T_4 = 905$ K; $Q_{in} = 1150$ kJ/kg, $Q_{out} = 434$ kJ/kg; $\eta = 62.2\%$; mep = 880 kPa.

Air-standard assumptions

• The working fluid is air only, treated as a perfect gas with constant specific heats.
• Combustion is replaced by heat addition from an outside source, and exhaust by heat rejection.
• All processes are reversible; compression and expansion are isentropic.
• No heat loss, no leakage, no valve flow loss.

Fuel-air cycle

The fuel-air cycle uses the real composition of the charge (air + fuel + residual gas) and the following:

• Specific heats of the gases increase with temperature ($c_v$ rises, so $\gamma$ falls).
• Dissociation: at high temperature CO$_2$ breaks into CO and O$_2$, and H$_2$O into H$_2$ and O$_2$. This absorbs heat, so the peak temperature falls.
• The number of moles changes during combustion.
• Fuel quantity affects the heat released (rich mixtures cause incomplete combustion).

Because of these, the peak temperature and pressure of the fuel-air cycle are lower than those of the air-standard cycle, and the efficiency is about 80-85% of it.

Composition of cylinder gases

Burnt gases contain CO$_2$, H$_2$O, N$_2$ and excess O$_2$ (lean mixture), or CO, H$_2$ and unburnt hydrocarbons (rich mixture), plus residual gas from the previous cycle.

Losses in the actual cycle

1. Time loss (finite combustion time): combustion is not at constant volume, so peak pressure is lower and falls after TDC.
2. Heat loss to cylinder walls and head during compression, combustion and expansion.
3. Exhaust blowdown loss: the exhaust valve opens before BDC, so some expansion work is lost.
4. Pumping loss in the induction and exhaust strokes.
5. Incomplete combustion, leakage past rings and friction.

The actual indicated efficiency is only 80-90% of the fuel-air cycle.

Brayton cycle efficiency

The Brayton (Joule) cycle is the ideal cycle of a gas turbine. It has isentropic compression 1-2 in a compressor, constant-pressure heat addition 2-3 in a combustor, isentropic expansion 3-4 in a turbine and constant-pressure heat rejection 4-1.

 Air --> [Compressor] --> [Combustor] --> [Turbine] --> Exhaust
             1-2             2-3             3-4        4-1
              \_______________shaft__________/ --> load

$$\begin{aligned} Q_{in} &= c_p (T_3 - T_2), \qquad Q_{out} = c_p (T_4 - T_1) \\ \eta &= 1 - \frac{T_4 - T_1}{T_3 - T_2} \end{aligned}$$

Since $p_2 = p_3$ and $p_1 = p_4$, $T_2/T_1 = T_3/T_4 = r_p^{(\gamma-1)/\gamma}$. Hence $T_4/T_1 = T_3/T_2$ and

$$\eta_{Brayton} = 1 - \frac{T_1}{T_2} = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}}$$

For $r_p = 8$, $\gamma = 1.4$:

$$\eta = 1 - 8^{-0.2857} = 1 - 0.552 = 0.448$$

Answer: $\eta = 44.8\%$. It depends only on pressure ratio, not on the maximum temperature.

Atkinson cycle

• Same as the Otto cycle (isentropic compression, constant-volume heat addition) but the expansion goes on until the gas reaches atmospheric pressure. The expansion ratio is larger than the compression ratio.
• Heat is rejected at constant pressure.
• It extracts more work from the same fuel, so the efficiency is higher than Otto's.
• It is achieved with a long expansion stroke (original linkage design) or, in modern engines, by closing the inlet valve late (Miller/Atkinson-type timing). The cost is a lower power density, so it is used in hybrid cars.

Stirling cycle

• A closed cycle with an external heat source and a regenerator. Processes: isothermal compression, constant-volume heating (regenerator), isothermal expansion, constant-volume cooling (regenerator).
• With a perfect regenerator its efficiency equals the Carnot efficiency, $1 - T_L/T_H$.
• Heat is supplied from outside, so any fuel or solar heat can be used and combustion is clean and quiet.
• Drawbacks: large size, slow response and sealing problems, so it is rare in vehicles.