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	1-2 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 (NON-ISOTHERMAL)	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.

1. First correction for the inter-molecular 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/v²

p = p’ –a/v²

p’ = p + a/v²

 the value of constant ‘a’ will be different for different gases.’

Constant’ a’ represents the reduction in total pressure due to molecular attractive forces.

2. Second correction for the finite size of the molecules

We use v-b instead v in the perfect equation

Therefore the Final Wan der Waal equation will become (p+a/v²)(v-b) =RT

Or

p = [RT/(v-b)] –a/v²

p =pressure in N/m²

v= molar volume = m³/kg mole

T = Absolute temperature, K

R = Universal gas constant=8.314 k J/kg mole K

a = constant N m⁴/(kg mole)²

b =constant , m³/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) R² T_c²/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

Put the values in terms of reduced quantities, the Van der Waals equation becomes

(p_r+ 3/V_r²) (V_r—1/3) =(8/3) T_r

Now this Van der Waals Equation can be applied to any gas

COMPRESSIBILITY FACTOR ‘Z’ OR COMPRESSION FACTOR

At high pressures and low temperatures there is a big deviation in the ideal gas equation if used. It is because of that the volume of the gas becomes very large under high pressures and low temperatures. This deviation is expressed in terms of compressibility factor Z = pV/RT

Z = 1 for a perfect gas

Z>1 or Z<1 for a real gas.

At low pressures, Z =1 even for a real gas and there is no compressibility factor at low pressures.

This compressibility factor is due to inter molecular attractive and repulsive forces. These forces decrease with the increase of temperature and decrease of pressure. This factor ‘Z’ helps to find the difference between real gas and ideal gas behavior. It is an empirical factor. It is different for different gases. Z>1 at high pressures. The role of Z is more predominant near the phase change condition because of large variations in the resulting volumes.