THERMODYNAMIC CYCLES

A Thermodynamic Cycle consists of thermodynamic processes in series and such that system returns to its initial starting point.

(a)    POWER CYCLES (HEAT ENGINE CYCLES) WITH EXTERNAL COMBUSTION

CycleProcess 1—2(compression)Process 2—3(heat addition)Process 3—4(Expansion) Process 4—1(Heat rejection)Remarks
CARNOT CYCLEISENTROPICISOTHERMALISENTROPICISOTHERMALIDEAL HEAT ENGINE CYCLE
BELL COLEMAN CYCLEADIABATICISOBARICADIABATICISOBARICREVERSED BRAYTON CYCLE
ERICSSON  CYCLEISOTHERMALISOBARICISOTHERMALISOBARIC
RANKINE CYCLEADIABATICISOBARICADIABATICISOBARICUSED IN STEAM ENGINES
STIRRLING CYCLEISOTHERMALISOCHORICISOTHERMALISOCHORICUSED IN STIRLING ENGINE

(b)   POWER CYCLES (HEAT ENGINE CYCLES) WITH INTERNAL COMBUSTION

BRAYTON CYCLEADIABATICISOBARICADIABATICISOBARICUSED IN JET ENGINES
DIESEL CYCLEADIABATICISOBARICADIABATICISOCHORICUSED IN DIESEL ENGINES
OTTO CYCLEADIABATICISOCHORICADIABATICISOCHORICUSED IN PETROL AND GAS ENGINES

© POWER CONSUMPTION CYCLES (REFRIGERATION AND AIR CONDITIONING CYCLES)

CYCLECOMPRESSIONHEAT REJECTIONEXPANSIONHEAT ABSORPTIONREMARKS
VAPOR COMPRESSION REFRIGERATION CYCLEADIABATICISOBARIC   (ISOTHERMAL) ISENTHALPICISOBARIC   (ISOTHERMAL) REVERSED RANKINE CYCLE
VAPOR ABSORPTION REFRIGERATION CYCLEADIABATICISOBARIC (ISOTHERMAL)ISENTHALPICISOBARIC   (ISOTHERMAL) REVERSED RANKINE CYCLE
GAS REFRIGERATION CYCLEADIABATICISOBARIC (NON-ISOTHERMAL)ADIABATICISOBARIC(NON -ISOTHERMAL)REVERSED BRAYTON CYCLE

A cycle with one or more irreversible processes will have efficiency less than that of Carnot cycle working between the same temperature limits. This may be due to lesser heat input at high temperature or more heat rejection at lower temperature. Mathematically it can be expressed as Clausius inequality as given below:Clausius Inequality

∮dQ/T≤0

But ds = dQ/T

The equal sign is applicable only to the ideal cycle (Carnot cycle). In this, integral is equal to net change in entropy in one complete cycle. Therefore change of entropy is zero in the most efficient cycle. This inequality is applicable to all heat engines in actual practice.

Prove of Clausius Inequality Theorem

∮dQ/T ≤0

Proof

For an heat engine
η  = ((Heat supplied at high temp–Heat rejected at low temp))/(Heat supplied at high temp)

= (QH — QL) / QH

For a reversible cycle

η = (TH — TL) / TH

∴  (QH — QL) / QH  = (TH — TL) / TH

1– QL/QH = 1 — TL/ TH

QL/QH = TL/ TH

QL/ TL = QH/ TH

Clausius theorem in the integrated form around a reversible cycle becomes

∮dQ/T=0

For a real cycle with an irreversible process, its efficiency will be less than that of a reversible cycle. Therefore the equality sign comes into existence and it becomes Clausius inequality.

∮dQ/T< 0