6-7 THE CARNOT CYCLE

6-8 CARNOT PRINCIPLES Efficiency of two Heat Engines operating between the same two reservoirs at LT and HT

C1 irreversible reversible η η< C2 reversible 1 reversible 2 η η=

6-9 THE THERMODYNAMIC TEMPERATURE SCALE For reversible heat engine operating between LT and HT :

L L

H H

Q T Q T

= Kelvin Scale of Absolute Temperature

6-10 THE CARNOT HEAT ENGINE The efficiency of heat engine operating on a reversible Carnot cycle between LT and HT :

L L C

H H

Q T 1 1 Q T

η = − = −

6-11 THE CARNOT HEAT PUMP Carnot Heat Pump: Carnot Refrigerator:

H

H

Isothermal expansion at Theat Q is added

3

2

insulated

1

4

insulated

4

3

LQ

L

L

Isothermal compression at Theat Q is rejectedL

Adiabatic expansion Q=0Temperature dropped to T H

Adiabatic compression Q=0Temperature raised to T

The Carnot Cycle consists of 4 reversible processes

IR

IR Rη η<

HT

LT

R1 R2

R1 R2η η=

(Refrigerator)The Reversed Carnot Cycle

P

HT const=

V

net ,outW

1V

1

2

34

4V 2V 3VLT const=

P

HT const=

V

net ,inW

1V

1

23

4

2V 4V 3V

(Heat Engine)The Carnot Cycle

HQ

LQ

LT

HT

HQ

LQ

LT

HT

C

L H

is the highest possible efficiency of the heat engine operating between two reservoirs at T and T

η

gas continues to expand

gas

1V

2V 2V

3V 3V

4V 4V

1V

nd

Violation of Carnot Principles yields violation of the 2 Law of Thermodynamics

CarnotEfficiency

6 - 6 Reversible Processes

1

2

HQ

Q 0=Q 0= Q 0=Q 0=

the paths coinside•

at thermal equilibrium•

heat transfer due to infinitely small difference in temperature•

1

2

Q1

2

-Q-W W

Sadi Carnot (1822-1888)

Irreversible Processes

heat transfer due to T >0ƥ

T 0∆ >

heat generation by friction•

LT const=

L C

H

T1T

η = −

HT

LT

CHPL L

H H

1 1COP Q T1 1Q T

= =− −

CRH H

L L

1 1COP Q T1 1Q T

= =− −

INSTITUT NATIONAL DES SCIENCES APPLIQUEES, LYON, FRANCE

COMMENTS AND ADDITIONAL NOTES ON CARNOT CYCLE Carnot Cycle consists of 4 reversible processes which operate between two temperature recervoirs at HT and LT .

Where HT and LT are temperatures according to not specified temperature scale (empirical). We know only that ordering of this temperature scale is such that HT > LT to be consistent with the 2nd Law of Thermodynamics (Clausius statement).

Carnot Principles CP1 any reversible η η≤ Assume the opposite, any reversible η η> , to demonstrate

that violation of the CP1 yields violation of the 2nd Law.

CP2 reversible,1 reversible,2 η η= From CP1, we have reversible,1 reversible,2 η η≤ and reversible,2 reversible,1 η η≤ Derivation of the Thermodynamic Temperature Scale (from Carnot Cycle with intermediate temperature reservoir at 0T )

H L

1 2

for any two Carnot cycles1 2between the same T and TL L

C C1 2H H

Q Q1 1Q Q

η η= − = = − , therefore,

( )LL H

H

Q f T ,TQ

= depends only on HT and LT , but not on the engine

( )H0 H

0

Q f T ,TQ

= where 0T is any such that H 0 LT T T> >

( )L0 L

0

Q f T ,TQ

=

0L L

H 0 H

QQ Q Q Q Q

= ⋅ ( )L H f T ,T= ( )( )

0 L

0 H

f T ,T f T ,T

= ( )( )

L

H

T Tθθ

=

Kelvinchosed

L

H

T T

=

Demonstrate that the temperature of the Thermodynamic Temperature Scale is the same as Absolute Temperature of the Ideal Gas equation of state PV mRT= This absolute temperature T can be measured by the Ideal Gas thermometer by measuring V of the fixed mass at constant pressure:

PT VmR

= ⋅

or by measuring pressure P in the rigid tank of constant volume V :

VT PmR

= ⋅

(formally, then zero absolute temperature corresponds to zero volume or zero pressure (no molecules in the tank))

HT

LT

0T

HQHQ

LQ

LQ

0Q

0Q

Consider reversible process without heat transfer (adiabatic) expansion/compression of ideal gas (process 4 1→ ):

Qδ PdV m du− = ⋅ differential energy balance

v1 mRTdV mc dTV

− = ⋅ from ideal gas equation of state 1P mRTV

=

vc1 1 dV dTV R T

− = ⋅

1 1 1 dV dTV k 1 T

− = ⋅−

p vc c R= + ⇒ v v

pp v

v

c c 1 1cR c c k 11c

= = =− −

−

( ) 1 1k 1 dV dT lnCV T

− ⋅ + =∫ ∫

k 1lnV lnT lnC− + =

k 1T V C− = k 1PV V CmR

− = , from ideal gas equation, PVTmR

=

kPV C= polytropic process with n k=

k

1 kPT C− = ( )11

k 1 1 1 kk 1T V C V CT CT− − −−= ⇒ = =

Carnot Cycle

1k 1

2 L

H3

V T TV

− =

Carnot Cycle 2 3→

1k 1

4 H

L1

V T TV

− =

Carnot Cycle 4 1→

1k 1

1 L 2

H4 3

V T V TV V

− = =

⇒ 3 2

4 1

V V V V

=

2H H

1

VQ mRT lnV

= 3H

4

V mRT ln

V= Carnot Cycle 1 2→

3

L L4

VQ = mRT ln

V Carnot Cycle 3 4→

Divide second equation by the first, then the Thermodynamic temperature scale of Kelvin for Carnot cycle is obtained:

L L

H H

Q T Q T

= where LT and HT are absolute temperature of ideal gas equation.

1

4 4V

1V

Q 0δ = dV

Dependence of the temperature of the atmosphere on the height above the sea level (application of adiabatic expansion of a gas) When air rises to the upper regions of lower pressure, it EXPANDS. This expansion can be considered ADIABATIC, because air is a poor conductor of heat. ideal gas eqn

dP 1g gdz v

ρ= − = − 1 Pv RT=

dP P gdz RT

= − separate variables

1 1dP gdzP RT

= −

k1 kPT C− =

k

1 kd PT 0−

=

k k

1 k 1 kT dP P T dT 0T

− − ∂

+ = ∂ , where

k k 2k 111 k 1 k 1 kk kT T T

T 1 k 1 k

−−

− − − ∂

= = ∂ − −

1 k 1dP dTP 1 k T

= −−

k 1 1dT gdz

1 k T RT− = −

−

k 1 gdT dzk R−

= −

p

gdT dzc

= − integrate, using initial temperature ( )T 0 , to solve for

( ) ( ) p

g zcT z T 0 e

−

= Then temperature gradient .o

p

dT g Cdz 9 8 dz c km

= − ≈ −

z

m

m

polytropic processfor ideal gas

then differential

use product rule

control mass

Pascal's Law

Another derivation: Consider the Carnot Cycle operating between two temperature reservoirs (for which both LT and HT are the absolute

temperature of ideal gas equation of state) for the fixed mass m of ideal gas, and show that L L

H H

Q T Q T

= .

Energy balance for Isothermal process 1 -> 2 : ( )H 12 2 1Q W m u u− = ⋅ − 2 1 HT T T const= = =

2H H

1

VQ mRT ln 0V

− = boundary work for isothermal process

2H H

1

VQ mRT lnV

=

Isothermal process 3 -> 4 : ( )L 34 4 3 Q W m u u− − = ⋅ − 3 4 LT T T const= = =

4L L

3

VQ + mRT ln 0

V= boundary work for isothermal process

4L L

3

VQ = mRT ln

V−

3L L

4

VQ = mRT ln

V

Adiabatic process 2 -> 3: Q ( )23 3 2 W m u u− = ⋅ − PdV m du− = ⋅ differential balance v PdV m c dT− = ⋅ ideal gas

v1 mRTdV mc dTV

− = ⋅ from ideal gas equation of state 1P mRTV

=

vc1 1 dV dTV R T

− = ⋅

v

v

p c

c1 1 dV dTV c c T

− = ⋅−

from p vc c R= +

p

v

1 1 1 dV dTcV T1c

− = ⋅−

1 1 1 dV dTV k 1 T

− = ⋅−

p

v

ck

c= specific heats ratio

3 3

2 2

1 1 1 dV dTV k 1 T

− = ⋅−∫ ∫

( ) 32

3 2

TV k 1 ln lnV T

− ⋅ =

k 1

32k 1

23

TV ln lnTV

−

− =

k 1

2 Lk 1

H3

V T TV

−

− = can be obtained faster for isentropic process

1k 1

2 L

H3

V T TV

− =

Adiabatic process 4 -> 1: k 1

4 Hk 1

L1

V T TV

−

− =

1

k 14 H

L1

V T TV

− =

Consider

1k 1

2 L

H3

V T TV

− =

1

14k 1

H

L

V1 VT

T

−

= =

then

2 1

43

V V VV

=

32

41

VV VV

=

Consider 2H H

1

VQ mRT lnV

= 3H

4

V mRT ln

V=

Carnot Cycle 1->2 3

L L4

VQ = mRT ln

V

Divide second equation by the first, then

L L

H H

Q T Q T

= where both LT and HT are absolute temperature of ideal gas equation of state.

Conclusion: the absolute thermodynamic scale coincides with the absolute temperature scale of the ideal gas thermometer.

Derivation of

L L C

H H

Q T 1 1 Q T

η = − = −

with the help of entropy:

( )L L 4 3Q T S S= − ⋅ −

( )H H 2 1Q T S S= ⋅ − Because adiabatic processes 2->3 and 4->1 are isentropic, 1 4S S= and 2 3S S= ,

( )L L 4 3Q T S S= − ⋅ −

( )H H 3 4Q T S S= ⋅ − Division yields

L L

H H

Q T Q T

=

COP in terms of 1 L

H

QQ

η = − :

HPL

H

1 1COP Q1Q

η= =

−

CRH

L

1 1COP 1Q 1Q

η= = −

−

Negative absolute temperature Definition of the thermodynamic temperature scale does not prohibit to choose the negative absolute temperature scale, because

L L L

H H H

Q T T Q T T

−= =

−

Show that choice of the negative absolute temperature scale will yield violation of the 2nd law of thermodynamics. Zero absolute temperature – is it allowed by this definitions?

Entropy

Equation L L

H H

Q T Q T

= looks like the …basic equation for derivation of the results for reversible machines.

Yes, this equation, in fact, is fundamental for the further development of classical thermodynamics, if rewritten as

L H

L H

Q Q T T

=

Quantity Q ST

= is called entropy. ‘loose definition of entropy”

It is equal to transferred heat per unit temperature kJK

.

Change of entropy during the Carnot cycle is zero: H L

H L

Q Q 0T T

− = .

Entropy is a property of the system and depends on its state like other properties , , ,...P T V

Nernst’s postulate: entropy of any system at zero absolute temperature is zero. Differential change of entropy during the differential part of reversible process:

Q dSTδ

= ratio of amount of heat transferred to the system

to temperature of the system during this differential reversible process 2

2 11

Q S STδ

= −∫ change of entropy does not depend on path of reversible process