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Thermodynamics
First Law of Thermodynamics
GATE 2007
Hard

For the same cycle (P-V diagram: 1→2 isobaric at 100 kPa from V1≈2.69 m³ to 1 m³, 2→3 isochoric at 1 m³ from 100 to 400 kPa, 3→1 isentropic PVᵞ=const back to state 1), if the specific heats of the working fluid are constant and the value of specific heat ratio γ is 1.4, the thermal efficiency (%) of the cycle is

  • 21
  • B40.9
  • C42.6
  • D59.7

Explanation

Using T∝PV (relative units): T1∝100×2.69=269, T2∝100×1=100, T3∝400×1=400. Heat rejected (1→2, isobaric): |Q12|=cp(T1−T2)∝cp×169. Heat added (2→3, isochoric): Q23=cv(T3−T2)∝cv×300. η=1−|Q12|/Q23=1−(169/300)×γ=1−0.564×1.4≈0.21=21%.

Source: GATE ME 2007 (2007)

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