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Ideal diesel cycle: diagrams and efficiency

Short answer

Efficiency = 1 - (1/r^(gamma-1)) * (r_c^gamma - 1) / (gamma (r_c - 1)). Derived from q_in = cp(T3-T2), q_out = cv(T4-T1), using isentropic relations: T2 = T1 r^(gamma-1), T3 = T1 r^(gamma-1) r_c, T4 = T1 r_c^gamma.

Sketch an ideal Diesel cycle on P-v and T-s diagram. Also derive an expression for its efficiency.

The ideal diesel cycle is the air-standard cycle of a compression-ignition engine: two isentropic processes, one constant-pressure heat addition and one constant-volume heat rejection.

Diagrams

 P-v diagram                      T-s diagram
 P                                T
 |   2 _______ 3                  |               3
 |   |          \                 |             .'|
 |   |           '.               |           .'  |
 |   |             '.             |         .'    |
 |    \              '-.          |    2 _.'      |
 |     \                '-. 4     |    |          4
 |      '.                  |     |    |        .'
 |        '-.               |     |    |      .'
 |           '-._           |     |    |    .'
 |               '--.._     |     |    1.-'
 |                     '--- 1     |
 +-------------------------> v    +--------------------> s
    v2                      v1
 1-2 isentropic compression       2-3 constant-P heat addition
 3-4 isentropic expansion         4-1 constant-v heat rejection

Derivation

Let r = v₁/v₂ (compression ratio) and r_c = v₃/v₂ (cut-off ratio). Air is an ideal gas with constant specific heats; γ = cp/cv.

Heat added (2-3, constant P): q_in = cp (T₃ − T₂) Heat rejected (4-1, constant v): q_out = cv (T₄ − T₁)

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

Express all temperatures in terms of T₁:

  • 1-2 isentropic: T₂ = T₁ r^(γ−1)
  • 2-3 constant pressure: T₃/T₂ = v₃/v₂ = r_c, so T₃ = T₁ r^(γ−1) r_c
  • 3-4 isentropic, with v₄/v₃ = v₁/v₃ = r/r_c: T₄ = T₃ (r_c/r)^(γ−1) = T₁ r_c^γ

Substituting:

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

η = 1 − T₁ (r_c^γ − 1)/(γ T₁ r^(γ−1) (r_c − 1))

η_Diesel = 1 − (1/r^(γ−1)) × (r_c^γ − 1)/(γ (r_c − 1))

Comparison with the Otto cycle (η = 1 − 1/r^(γ−1)): the extra bracket (r_c^γ − 1)/(γ(r_c − 1)) is always greater than 1, so for the same compression ratio the diesel cycle is less efficient. Diesel engines, however, work at higher compression ratios and so reach higher practical efficiency. As r_c → 1, the bracket → 1 and the diesel cycle becomes the Otto cycle.

✓ Answer checked by Claude Opus 5.5 (high effort)

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