๐ Otto & Diesel Cycles
Spark-ignition and compression-ignition engine cycle analysis. MEP, compression ratio, cut-off ratio, and animated P-v diagram.
โก Fortran 90 Engine
Double Precision (IEEE 754)
โ ISO / ASME Validated
Thermodynamics
๐ Solver Telemetry
โ ACTIVE
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Released
Jan 2025
โฑ๏ธ Latency
< 1 ms
Reciprocating Engine Geometry
Internal combustion engines compress fuel-air mixtures between Top Dead Center (TDC) and Bottom Dead Center (BDC). The ratio of maximum volume to minimum volume defines the Compression Ratio ($r$).
โข Otto Cycle: Spark-ignition engine model with constant volume combustion.
โข Diesel Cycle: Compression-ignition model with constant pressure combustion.
โข Diesel Cycle: Compression-ignition model with constant pressure combustion.
๐ Configuration
Engine Cycle Formulas:
โข Compression Work: W_c = Cv * (T_2 - T_1)
โข Heat Input (Qin):
- Otto: Qin = Cv * (T_3 - T_2) (Constant Volume)
- Diesel: Qin = Cp * (T_3 - T_2) (Constant Pressure)
โข Expansion Work (We):
- Otto: We = Cv * (T_3 - T_4)
- Diesel: We = R * (T_3 - T_2) + Cv * (T_3 - T_4)
โข Thermal Efficiency: ฮท_th = W_net / Qin
โข Mean Effective Pressure: MEP = W_net / (v_1 - v_2)
Assumptions: Air standard ideal gas properties ($Cp = 1.005$ kJ/kgยทK, $Cv = 0.718$ kJ/kgยทK, $k = 1.4$).
โข Compression Work: W_c = Cv * (T_2 - T_1)
โข Heat Input (Qin):
- Otto: Qin = Cv * (T_3 - T_2) (Constant Volume)
- Diesel: Qin = Cp * (T_3 - T_2) (Constant Pressure)
โข Expansion Work (We):
- Otto: We = Cv * (T_3 - T_4)
- Diesel: We = R * (T_3 - T_2) + Cv * (T_3 - T_4)
โข Thermal Efficiency: ฮท_th = W_net / Qin
โข Mean Effective Pressure: MEP = W_net / (v_1 - v_2)
Assumptions: Air standard ideal gas properties ($Cp = 1.005$ kJ/kgยทK, $Cv = 0.718$ kJ/kgยทK, $k = 1.4$).
๐ Results & Visualization
Configure inputs and click Analyze Engine Cycle to view results.
๐ Calculation Methodology
Mathematical Model & Theory
Otto and Diesel cycles model spark-ignition and compression-ignition internal combustion engines. Thermodynamic efficiencies are modeled as functions of compression ratio $r$ and cutoff ratio $r_c$:
$$\eta_{th, Otto} = 1 - \frac{1}{r^{\gamma-1}}$$
$$\eta_{th, Diesel} = 1 - \frac{1}{r^{\gamma-1}} \left[ \frac{r_c^{\gamma} - 1}{\gamma(r_c - 1)} \right]$$
Assumptions & Cycle Idealizations
- Cold-air standard properties are assumed ($C_p = 1.005$ kJ/(kgยทK), $C_v = 0.718$ kJ/(kgยทK), $\gamma = 1.4$)
- No air intake friction or manifold losses
- Combustion is modeled as instant heat addition (constant volume for Otto, constant pressure for Diesel)
References & Literature
- Cengel, Y. A., & Boles, M. A., Thermodynamics: An Engineering Approach, McGraw-Hill.
- Moran, M. J., Shapiro, H. N., Boettner, D. D., & Bailey, M. B., Fundamentals of Engineering Thermodynamics, Wiley.
Worked Engineering Example
Problem Statement:
An ideal gasoline engine (Otto cycle) has a compression ratio $r = 8.5$. Find the thermal efficiency ($\gamma = 1.4$).
Step-by-step Solution:
1. Calculate Otto cycle thermal efficiency:
$$\eta_{th} = 1 - \frac{1}{r^{\gamma-1}} = 1 - \frac{1}{8.5^{0.4}} = 1 - \frac{1}{2.353} = 1 - 0.425 = 0.575 \quad (57.5\%)$$
Final Result:
The thermal efficiency is 57.5%.
An ideal gasoline engine (Otto cycle) has a compression ratio $r = 8.5$. Find the thermal efficiency ($\gamma = 1.4$).
Step-by-step Solution:
1. Calculate Otto cycle thermal efficiency:
$$\eta_{th} = 1 - \frac{1}{r^{\gamma-1}} = 1 - \frac{1}{8.5^{0.4}} = 1 - \frac{1}{2.353} = 1 - 0.425 = 0.575 \quad (57.5\%)$$
Final Result:
The thermal efficiency is 57.5%.