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Diesel η = 1 − (1/r^(γ−1))·(ρ^γ−1)/(γ(ρ−1))

Diesel Efficiency

📐 计算公式

η_diesel = 1 − (1/r^(γ−1))·(ρ^γ−1)/(γ(ρ−1))

输入压缩比 r、截止比 ρ 与比热比 γ，求狄塞尔循环效率。

压缩比 r

截止比 ρ

比热比 γ

Formula & Standard References

Item	Formula / Note	Source / Standard
Diesel cycle efficiency	η = 1 − (1/r^(γ−1))·(ρ^γ−1)/(γ(ρ−1))	Air-standard Diesel cycle
Compression ratio	r = V₁/V₂, diesel 14–22	IC engine principles
Cut-off ratio	ρ = V₃/V₂ cut-off ratio	Cycle parameters
vs Otto	Otto higher at same r	Cycle comparison
Actual efficiency	35–45% real engines	Engine data
Constant pressure heat	Diesel assumes constant-pressure heat	Air-standard assumption

📌 计算说明

η_diesel = 1 − (1/r^(γ−1))·(ρ^γ−1)/(γ(ρ−1))。
r=18、ρ=2、γ=1.4 → η≈63%。

📐 Physical Constants

Constants used on this page and their authoritative sources, for verification and citation

Constant	Value	Unit	Source
Heat capacity ratio γ, diatomic ideal gas	7/5 = 1.400	—	Theoretical value from equipartition (N₂, O₂ at room temperature)
γ used in air-standard cycle analysis	1.4	—	Engineering convention (page default; standard in SAE / ICE textbooks)
Gas constant of dry air R_a	287.042	J·kg⁻¹·K⁻¹	Derived from R/M (CIPM 2007 dry-air M)

Note: the air-standard diesel cycle assumes an ideal gas with constant specific heats, isentropic compression and expansion, and no friction or heat-transfer losses; real diesel engines fall well short of this ideal efficiency.

📚 深度解析：柴油机理想循环效率

💡 常见使用场景

狄塞尔循环（定压加热）热效率估算。
压缩比、截止比对效率影响。
与奥托循环对比。

压缩比 18 截止比 2

r=18, ρ=2, γ=1.4：η=1−(1/r^(γ−1))·(ρ^γ−1)/(γ(ρ−1))=1−(1/3.176)·(2.639−1)/1.4≈0.631=63.1%。

增大压缩比

r 18→22：η 升至约 65.6%。

❓ 常见问题（FAQ）

与奥托比？: 同压缩比下奥托（定容加热）效率更高，柴油机靠高压缩比取胜。
截止比？: ρ=定压加热终了体积/压缩终了体积，越大效率越低。

狄塞尔循环效率计算器