12.3 Internal Combustion Engines: SI & CI Cycles

Key Takeaways

  • The ideal Otto cycle models Spark-Ignition (SI) engines with constant-volume heat addition, yielding efficiency $\eta_{Otto} = 1 - 1/r^{k-1}$ that depends strictly on compression ratio $r$ and specific heat ratio $k$.
  • The ideal Diesel cycle models Compression-Ignition (CI) engines with constant-pressure heat addition, where efficiency $\eta_{Diesel} = 1 - \frac{1}{r^{k-1}} \left[ \frac{r_c^k - 1}{k(r_c - 1)} \right]$ decreases as cut-off ratio $r_c$ increases.
  • For the same compression ratio $r$ and heat input, $\eta_{Otto} > \eta_{Dual} > \eta_{Diesel}$; however, CI engines achieve higher practical efficiencies because their knock-free compression ratios ($r = 14-22$) far exceed SI engines ($r = 8-11$).
  • Engine power metrics separate Indicated Power ($IP$, total thermodynamic power generated inside cylinders) from Brake Power ($BP$, net power available at crankshaft output flange), with mechanical efficiency $\eta_m = BP / IP$.
  • Brake Mean Effective Pressure ($BMEP$) serves as the universal benchmark for engine performance and specific torque density: $BMEP = \frac{BP \times 60}{V_d N k}$.
Last updated: July 2026

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.


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 ($q_{out} = c_v (T_4 - T_1)$).

ηOtto=1qoutqin=1T4T1T3T2=11rk1\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 ($q_{out} = c_v (T_4 - T_1)$).

ηDiesel=11rk1[rck1k(rc1)]\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: $\eta_{Otto} > \eta_{Dual} > \eta_{Diesel}$.

2. Engine Geometry & Parameter Nomenclature

ParameterSymbolDefinition / FormulaEngineering Units
Cylinder Bore$D$Internal cylinder diameter$\text{mm}$ or $\text{m}$
Piston Stroke$L$Piston travel distance between TDC and BDC$\text{mm}$ or $\text{m}$
Clearance Volume$V_c$Cylinder volume above piston when at TDC$\text{cm}^3$ or $\text{m}^3$
Displacement Volume$V_d$Volume swept by piston: $V_d = \frac{\pi}{4} D^2 L$$\text{cm}^3$, $\text{L}$, or $\text{m}^3$
Compression Ratio$r$$r = \frac{V_1}{V_2} = \frac{V_d + V_c}{V_c} = 1 + \frac{V_d}{V_c}$Dimensionless
Cut-off Ratio$r_c$$r_c = \frac{V_3}{V_2} = \text{Volume at fuel cut-off} / V_c$Dimensionless

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


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=PimiVdNncylks60×1000 (kW)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=2πTN60×1000=TN9549.3 (kW)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=IPBPFP = IP - BP ηm=BPIP=BMEPIMEP\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=BP×60VD,totalNks (kPa)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=m˙fBP (kg/kWh) or (g/kWh)BSFC = \frac{\dot{m}_f}{BP}\text{ (kg/kW}\cdot\text{h) or (g/kW}\cdot\text{h)} ηbth=BPm˙f×HV=3600BSFC(kg/kWh)×HV(kJ/kg)\eta_{bth} = \frac{BP}{\dot{m}_f \times HV} = \frac{3600}{BSFC (\text{kg/kW}\cdot\text{h}) \times HV (\text{kJ/kg})}


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 Vd=π4(0.10)2(0.12)=0.00094248 m3=0.9425 litersV_d = \frac{\pi}{4} (0.10)^2 (0.12) = 0.00094248\text{ m}^3 = 0.9425\text{ liters} VD,total=4×0.00094248=0.0037699 m3=3.77 litersV_{D,total} = 4 \times 0.00094248 = 0.0037699\text{ m}^3 = 3.77\text{ liters}

Step 2: Calculate Brake Power (BP) BP=2πTN60,000=2π×250×240060,000=3,769,91160,000=62.83 kWBP = \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=IMEPVD,totalNks60=900×0.0037699×2400×0.560=4071.4960=67.86 kWIP = \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=IPBP=67.8662.83=5.03 kWFP = IP - BP = 67.86 - 62.83 = 5.03\text{ kW} ηm=BPIP=62.8367.86=0.9259=92.6%\eta_m = \frac{BP}{IP} = \frac{62.83}{67.86} = 0.9259 = 92.6\%

Step 5: Calculate Brake Mean Effective Pressure (BMEP) BMEP=ηm×IMEP=0.9259×900 kPa=833.3 kPaBMEP = \eta_m \times IMEP = 0.9259 \times 900\text{ kPa} = 833.3\text{ kPa}

Step 6: Calculate Thermal Efficiency and BSFC Heat Input Rate Q˙in=14.0 kg/h3600 s/h×43,000 kJ/kg=167.22 kW\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} ηbth=BPQ˙in=62.83167.22=0.3757=37.6%\eta_{bth} = \frac{BP}{\dot{Q}_{in}} = \frac{62.83}{167.22} = 0.3757 = 37.6\% BSFC=m˙fBP=14.0 kg/h62.83 kW=0.2228 kg/kWh=222.8 g/kWhBSFC = \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}

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P-v Diagram Comparison: Otto vs. Diesel vs. Dual Cycles
Test Your Knowledge

An ideal Otto cycle has a compression ratio r = 8.5. Assuming an air specific heat ratio k = 1.4, what is the air-standard thermal efficiency of the cycle?

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Test Your Knowledge

A 4-stroke engine connected to a dynamometer delivers a torque of 250 N·m at 2400 RPM. What is the brake power output of the engine?

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B
C
D
Test Your Knowledge

An engine generates an indicated power of 67.86 kW and a brake power of 62.83 kW. If the total engine displacement is 3.77 liters running at 2400 RPM (4-stroke), what is the Mechanical Efficiency η_m and the Brake Mean Effective Pressure (BMEP)?

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B
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D