- 2068 Magh · 6 marks
Ideal Otto cycle diagrams and efficiency
Short answer
Otto cycle: isentropic compression (1-2), constant volume heat addition (2-3), isentropic expansion (3-4), constant volume heat rejection (4-1). Compression ratio r = v1/v2. Efficiency η = 1 − 1/r^(γ−1), depending only on r and γ.
Sketch an ideal Otto cycle on P-v and T-s diagrams. Also derive an expression for its efficiency in terms compression ratio.
The ideal Otto cycle is the air-standard cycle of the spark ignition engine.
Diagrams
P-v diagram T-s diagram
P T
| 3 | 3
| |\ | /|
| | \ | / |
| | '. | / |
| | '. | .' |
| 2 '-. | 2 .' |
| \ '-. | | 4
| '. '-. | | /
| '-. '-. 4 | | .'
| '-._ | | | .'
| '-.._ | | 1.-'
| '-- 1 |
+-----------------------> v +--------------------> s
v2 v1
1-2 isentropic compression 2-3 constant-v heat addition
3-4 isentropic expansion 4-1 constant-v heat rejection
Derivation
Let r = v₁/v₂ be the compression ratio. Treat air as an ideal gas with constant cv and γ = cp/cv.
Heat added at constant volume (2-3): q_in = cv (T₃ − T₂) Heat rejected at constant volume (4-1): q_out = cv (T₄ − T₁)
η = 1 − q_out/q_in = 1 − (T₄ − T₁)/(T₃ − T₂)
Isentropic compression 1-2 and expansion 3-4, with v₂ = v₃ and v₁ = v₄:
T₂/T₁ = (v₁/v₂)^(γ−1) = r^(γ−1) T₃/T₄ = (v₄/v₃)^(γ−1) = r^(γ−1)
Hence T₂/T₁ = T₃/T₄, which gives T₄/T₁ = T₃/T₂, so
(T₄ − T₁)/(T₃ − T₂) = T₁/T₂ = 1/r^(γ−1)
η_Otto = 1 − 1/r^(γ−1)
The efficiency depends only on the compression ratio r and on γ; it rises as r increases. Example: r = 8, γ = 1.4: η = 1 − 1/8^0.4 = 56.5 %. In practice r is limited to about 10 by knocking.
✓ Answer checked by Claude Opus 5.5 (high effort)
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