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Gas vs vapor cycles; Otto cycle efficiency

Short answer

Gas cycles work with a gaseous fluid without phase change (e.g., Otto, Brayton); vapor cycles involve phase change (e.g., Rankine). The Otto cycle air-standard efficiency is η = 1 - 1/r^(γ-1), derived from isentropic relations and constant-volume heat addition and rejection.

Differentiate between gas and vapor cycles. Also derive an expression for the air standard efficiency of Otto cycle in terms of compression ratio.

Gas cycles and vapour cycles

PointGas cycleVapour cycle
Working fluidA gas (air) that stays gaseous throughoutA fluid that changes phase (liquid ⇄ vapour), usually water
Phase changeNoneEvaporation and condensation occur
ExamplesOtto, Diesel, Dual, Brayton, StirlingRankine, reheat, regenerative Rankine, vapour-compression refrigeration
Heat additionCombustion in the cylinder or combustorIn the boiler
Heat rejectionExhaust to atmosphereIn the condenser
CompressionCompressor takes much of the turbine workPump handles liquid, needs very little work
EquipmentEngines, gas turbinesSteam power plants, refrigerators
Property dataIdeal gas equationsSteam tables or charts

Otto cycle efficiency

The Otto cycle is the ideal air-standard cycle of the spark-ignition engine:

  • 1-2: isentropic compression
  • 2-3: constant-volume heat addition
  • 3-4: isentropic expansion
  • 4-1: constant-volume heat rejection
 P
 |   3
 |   |\
 |   | \
 |   |  '.
 |   |    '.
 |   2      '-.
 |    \        '-.
 |     '.         '-.
 |       '-.         '-. 4
 |          '-._         |
 |              '-.._    |
 |                   '-- 1
 +-----------------------> v
    v2                   v1

Let r = v₁/v₂ be the compression ratio. For air as an ideal gas with constant specific heats:

q_in = cv (T₃ − T₂), q_out = cv (T₄ − T₁)

η = 1 − q_out/q_in = 1 − (T₄ − T₁)/(T₃ − T₂)

Isentropic processes:

  • 1-2: T₂/T₁ = (v₁/v₂)^(γ−1) = r^(γ−1)
  • 3-4: T₃/T₄ = (v₄/v₃)^(γ−1) = (v₁/v₂)^(γ−1) = r^(γ−1)

So T₂/T₁ = T₃/T₄, which gives T₄/T₁ = T₃/T₂. Then

(T₄ − T₁)/(T₃ − T₂) = T₁/T₂ = 1/r^(γ−1)

η_Otto = 1 − 1/r^(γ−1)

The Otto efficiency depends only on the compression ratio r and on γ. Example: r = 8, γ = 1.4 gives η = 1 − 1/8^0.4 = 56.5 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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