THERMODYNAMIC CYCLES CLASS NOTES FOR MECHANICAL ENGINEERING
THERMODYNAMIC CYCLES CLASS NOTES
FOR MECHANICAL ENGINEERING
A thermodynamic cycle consists of thermodynamic processes in series. As a result, system returns to its initial starting point.
(a) POWER CYCLES (HEAT ENGINE CYCLES) WITH EXTERNAL COMBUSTION
Cycle 
12 process(compression) 
Process 2—3(heat addition) 
3—4 process(Expansion) 
Process 4—1(Heat rejection) 
Remarks 
CARNOT CYCLE 
ISENTROPIC 
ISOTHERMAL 
ISENTROPIC 
ISOTHERMAL 
IDEAL HEAT ENGINE CYCLE 
BELL COLEMAN CYCLE 
ADIABATIC 
ISOBARIC 
ADIABATIC 
ISOBARIC 
REVERSED BRAYTON CYCLE 
ERICSSON CYCLE 
ISOTHERMAL 
ISOBARIC 
ISOTHERMAL 
ISOBARIC 

RANKINE CYCLE 
ADIABATIC 
ISOBARIC 
ADIABATIC 
ISOBARIC 
USED IN STEAM ENGINES 
STIRRLING CYCLE 
ISOTHERMAL 
ISOCHORIC 
ISOTHERMAL 
ISOCHORIC 
USED IN STIRLING ENGINE 
(b) POWER CYCLES (HEAT ENGINE CYCLES) WITH INTERNAL COMBUSTION
BRAYTON CYCLE 
ADIABATIC 
ISOBARIC 
ADIABATIC 
ISOBARIC 
USED IN JET ENGINES 
DIESEL CYCLE 
ADIABATIC 
ISOBARIC 
ADIABATIC 
ISOCHORIC 
USED IN DIESEL ENGINES 
OTTO CYCLE 
ADIABATIC 
ISOCHORIC 
ADIABATIC 
ISOCHORIC 
USED IN PETROL AND GAS ENGINES 
© POWER CONSUMPTION CYCLES (REFRIGERATION AND AIR CONDITIONING CYCLES)
CYCLE 
COMPRESSION 
HEAT REJECTION 
EXPANSION 
HEAT ABSORPTION 
REMARKS 
VAPOR COMPRESSION REFRIGERATION CYCLE 
ADIABATIC 
ISOBARIC (ISOTHERMAL) 
ISENTHALPIC 
ISOBARIC (ISOTHERMAL) 
REVERSED RANKINE CYCLE 
VAPOR ABSORPTION REFRIGERATION CYCLE 
ADIABATIC 
ISOBARIC (ISOTHERMAL) 
ISENTHALPIC 
ISOBARIC (ISOTHERMAL) 
REVERSED RANKINE CYCLE 
GAS REFRIGERATION CYCLE 
ADIABATIC 
ISOBARIC (NONISOTHERMAL) 
ADIABATIC 
ISOBARIC(NON ISOTHERMAL) 
REVERSED BRAYTON CYCLE 
Efficiency of a cycle with one or more irreversible processes is less than the efficiency of Carnot cycle. This is due to lesser heat input at high temperature or more heat rejection at lower temperature. Mathematically express it as Clausius inequality.
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 an ideal cycle. This inequality is applicable to all heat engines in actual practice.
Prove 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)
= (Q_{H} — Q_{L}) / Q_{H}
For a reversible cycle
η = (T_{H} — T_{L}) / T_{H}
∴ (Q_{H} — Q_{L}) / Q_{H} = (T_{H} — T_{L}) / T_{H}
1– Q_{L}/Q_{H} = 1 — T_{L}/ T_{H}
Q_{L}/Q_{H} = T_{L}/ T_{H}
Q_{L}/ T_{L} = Q_{H}/ T_{H}
Clausius theorem in the integrated form around a reversible cycle becomes
∮dQ/T=0
Efficiency of an real cycle is less than that of a reversible cycle. Therefore the inequality sign comes into existence. Thus it becomes Clausius inequality.
∮dQ/T< 0
Use Van Dar Waals or an empirical equation in a real cycle.
VAN DER WAALS EQUATION
It is a real gas equation. It is not perfect gas equation. In includes two corrections.

First correction for the intermolecular attractions which are weak forces of attractions and repulsion between two molecules of the same substance.
we use p =Kinetic pressure –internal pressure =p’a/v^{2}
p = p’ –a/v^{2}
p’ = p + a/v^{2}
_{ }the value of constant ‘a’ will be different for different gases.’
Constant’ a’ represents the reduction in total pressure due to molecular attractive forces.

Second correction for the finite size of the molecules
We use vb instead v in the perfect equation
Therefore the Final Wan der Waal equation will become (p+a/v^{2})(vb) =RT
Or
p = [RT/(vb)] –a/v^{2}
p =pressure in N/m^{2}
v= molar volume = m^{3}/kg mole
T = Absolute temperature, K
R = Universal gas constant=8.314 k J/kg mole K
a = constant N m^{4}/(kg mole)^{2}
b =constant , m^{3}/kg mole
values of constants a and b will be different for different gases.
a and b are found from the critical properties of a gas under consideration.
a=(27/64) R^{2} T_{c}^{2}/p_{c}
b = RT_{c}/8p_{c}
Predicted values of constants ‘a’ and ‘b’ do not agree when found by different methods. Therefore there is a need to have a Van der Waals Equation free of constants ‘a’ and ‘b’. It was achieved with the use of reduced quantities.
REDUCED QUANTITIES are pressure, volume and temperature.

p_{r}=p/p_{c}

v_{r} = v/v_{c}

T_{r} = T/T_{c}