12.3 Internal Combustion Engines: SI & CI Cycles

Internal Combustion (IC) engines convert thermal energy released during fuel combustion directly into mechanical work inside engine cylinders. IC engines operate as either Spark-Ignition (SI) engines running on gasoline/gas fuels or Compression-Ignition (CI) engines running on diesel oil.

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1. Ideal Air-Standard Thermodynamic Cycles

Air-standard cycle analysis models internal combustion processes assuming the working fluid is an ideal gas (air) executing a closed, internally reversible thermodynamic cycle with constant specific heats ($k = c_p / c_v = 1.4$).

A. The Otto Cycle (Ideal SI Engine)

The ideal Otto cycle models gasoline engines where combustion occurs so rapidly at Top Dead Center (TDC) that volume remains virtually constant during heat addition.

• Process 1-2: Isentropic compression from Bottom Dead Center (BDC) to TDC.
• Process 2-3: Constant-volume heat addition ($q_{in} = c_v (T_3 - T_2)$).
• Process 3-4: Isentropic expansion (power stroke).
• Process 4-1: Constant-volume heat rejection ($q_{out} = c_v (T_4 - T_1)$).

$$\eta_{Otto} = 1 - \frac{q_{out}}{q_{in}} = 1 - \frac{T_4 - T_1}{T_3 - T_2} = 1 - \frac{1}{r^{k-1}}$$

where compression ratio $r = V_1 / V_2 = V_{BDC} / V_{TDC}$. As compression ratio $r$ increases, thermal efficiency increases. However, SI engines are limited to $r \approx 8$–$11$ to prevent autoignition (engine knock).

B. The Diesel Cycle (Ideal CI Engine)

In heavy-duty diesel engines, fuel is injected near TDC into hot compressed air and burns at approximately constant pressure.

• Process 1-2: Isentropic compression ($r = V_1 / V_2 = 14$–$22$).
• Process 2-3: Constant-pressure heat addition ($q_{in} = c_p (T_3 - T_2)$).
• Process 3-4: Isentropic expansion.
• Process 4-1: Constant-volume heat rejection ($q_{out} = c_v (T_4 - T_1)$).

$$\eta_{Diesel} = 1 - \frac{1}{r^{k-1}} \left[ \frac{r_c^k - 1}{k (r_c - 1)} \right]$$

where cut-off ratio $r_c = V_3 / V_2 = T_3 / T_2$ represents the ratio of cylinder volumes after and before fuel injection/combustion. As engine load increases, cut-off ratio $r_c$ increases, which slightly lowers ideal Diesel cycle efficiency.

C. Dual / Sabathé Cycle (Modern High-Speed CI Engine)

Modern high-speed diesel engines exhibit rapid initial combustion followed by continued burning during piston descent. Heat addition occurs partly at constant volume (2 $\rightarrow$ 2.5) and partly at constant pressure (2.5 $\rightarrow$ 3).

• Pressure ratio $r_p = P_{2.5} / P_2$.
• For identical peak pressures and heat inputs: $\eta_{Diesel} > \eta_{Dual} > \eta_{Otto}$.
• For identical compression ratio $r$ and heat input: $\eta_{Otto} > \eta_{Dual} > \eta_{Diesel}$.

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2. Engine Geometry & Parameter Nomenclature

Total engine displacement for an $n_{cyl}$-cylinder engine is $V_{D,total} = n_{cyl} \times V_d$.

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3. Power, Pressure & Efficiency Performance Metrics

A. Indicated Power ($IP$)

Indicated power is the total theoretical mechanical power produced by combustion gas expansion inside the cylinder cavity, measured via cylinder pressure indicator diagrams:

$$IP = \frac{P_{imi} \cdot V_d \cdot N \cdot n_{cyl} \cdot k_s}{60\times 1000}\text{ (kW)}$$

where $P_{imi}$ is indicated mean effective pressure (kPa), $V_d$ is single cylinder displacement (m$^3$), $N$ is engine speed (RPM), $n_{cyl}$ is number of cylinders, and $k_s$ is power stroke factor ($k_s = 0.5$ for 4-stroke; $k_s = 1.0$ for 2-stroke).

B. Brake Power ($BP$)

Brake power is the net useful mechanical power delivered by the crankshaft output flange to an external load (measured with a dynamometer brake):

$$BP = \frac{2 \pi T N}{60 \times 1000} = \frac{T N}{9549.3}\text{ (kW)}$$

where $T$ is brake torque in Newton-meters (N·m) and $N$ is rotational speed in RPM.

C. Friction Power ($FP$) & Mechanical Efficiency ($\eta_m$)

Friction power represents total energy consumed by piston ring sliding friction, bearing friction, pumping losses, and auxiliary drives (water pump, oil pump, alternator):

$$FP = IP - BP$$ $$\eta_m = \frac{BP}{IP} = \frac{BMEP}{IMEP}$$

D. Brake Mean Effective Pressure ($BMEP$)

$BMEP$ is a hypothetical constant pressure that, if acting on the piston throughout the power stroke, would produce the measured brake power output:

$$BMEP = \frac{BP \times 60}{V_{D,total} \cdot N \cdot k_s}\text{ (kPa)}$$

E. Specific Fuel Consumption ($BSFC$)

Brake Specific Fuel Consumption measures engine fuel conversion economy (mass of fuel burned per unit of brake energy output):

$$BSFC = \frac{\dot{m}_f}{BP}\text{ (kg/kW}\cdot\text{h) or (g/kW}\cdot\text{h)}$$ $$\eta_{bth} = \frac{BP}{\dot{m}_f \times HV} = \frac{3600}{BSFC (\text{kg/kW}\cdot\text{h}) \times HV (\text{kJ/kg})}$$

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4. Worked Step-by-Step IC Engine Calculation

Problem Statement: A 4-cylinder, 4-stroke diesel engine has a bore $D = 100\text{ mm}$ ($0.10\text{ m}$) and stroke $L = 120\text{ mm}$ ($0.12\text{ m}$). Operating at $2400\text{ RPM}$, the engine records a dynamometer torque $T = 250\text{ N}\cdot\text{m}$, fuel consumption $\dot{m}_f = 14.0\text{ kg/h}$ ($HV = 43,000\text{ kJ/kg}$), and an indicated mean effective pressure $IMEP = 900\text{ kPa}$. Calculate:

1. Total engine displacement volume $V_{D,total}$
2. Brake power $BP$ and Indicated power $IP$
3. Friction power $FP$ and Mechanical efficiency $\eta_m$
4. Brake Mean Effective Pressure $BMEP$
5. Brake thermal efficiency $\eta_{bth}$ and $BSFC$

Step-by-Step Solution

Step 1: Calculate Displacement Volumes

$$V_d = \frac{\pi}{4} (0.10)^2 (0.12) = 0.00094248\text{ m}^3 = 0.9425\text{ liters}$$ $$V_{D,total} = 4 \times 0.00094248 = 0.0037699\text{ m}^3 = 3.77\text{ liters}$$

Step 2: Calculate Brake Power (BP)

$$BP = \frac{2 \pi T N}{60,000} = \frac{2 \pi \times 250 \times 2400}{60,000} = \frac{3,769,911}{60,000} = 62.83\text{ kW}$$

Step 3: Calculate Indicated Power (IP) For a 4-stroke engine, $k_s = 0.5$:

$$IP = \frac{IMEP \cdot V_{D,total} \cdot N \cdot k_s}{60} = \frac{900 \times 0.0037699 \times 2400 \times 0.5}{60} = \frac{4071.49}{60} = 67.86\text{ kW}$$

Step 4: Calculate Friction Power (FP) and Mechanical Efficiency

$$FP = IP - BP = 67.86 - 62.83 = 5.03\text{ kW}$$ $$\eta_m = \frac{BP}{IP} = \frac{62.83}{67.86} = 0.9259 = 92.6\%$$

Step 5: Calculate Brake Mean Effective Pressure (BMEP)

$$BMEP = \eta_m \times IMEP = 0.9259 \times 900\text{ kPa} = 833.3\text{ kPa}$$

Step 6: Calculate Thermal Efficiency and BSFC

$$\text{Heat Input Rate } \dot{Q}_{in} = \frac{14.0\text{ kg/h}}{3600\text{ s/h}} \times 43,000\text{ kJ/kg} = 167.22\text{ kW}$$ $$\eta_{bth} = \frac{BP}{\dot{Q}_{in}} = \frac{62.83}{167.22} = 0.3757 = 37.6\%$$ $$BSFC = \frac{\dot{m}_f}{BP} = \frac{14.0\text{ kg/h}}{62.83\text{ kW}} = 0.2228\text{ kg/kW}\cdot\text{h} = 222.8\text{ g/kW}\cdot\text{h}$$