Tc

Th

heat pump

Tc

Th

heat engine

Carnot’s Theorem

We introduced already the Carnot cycle with an ideal gas

Now we show:

1 Energy efficiency of the Carnot cycle is independent of the working substance

2 Any cyclic process that absorbs heat at one temperature, and rejects heat at one other temperature, and is reversible has the energy efficiency of a Carnot cycle

W

hQ

cQ

W

WQQ ch

cQ

W

QhP

reversibleRemark:

Note:P>1Textbook:coefficient of performance

Tc

Th

heat engine X

XW

XhQ

XcQ

Let’s combine a fictitious heat engine X with CX with a heat pump

realized by a reversed Carnot cycle

Tc

Th

heat pumpChQ

CcQ

X CCW

CX WWW

We can design the engine X such that Cc

Xc QQ

Let’s calculate XW XhQ X

cQ with Xh

XX

Q

W

X

XXh

WQ

XWXcQ

X

XW

XW

Xc

X

X Q1

If X would be a Carnot engine it would produce the work CWXc

C

C Q1

with CX XWXc

X

X Q1

> CWXc

C

C Q1

T

c

Th

heat engine X

XW

XhQ

XcQ T

c

Th

heat pump

ChQ

CcQ

X CCW

CX WWW

We can design the engine X such that Cc

Xc QQ

However:

CX WWW >0

0.0 0.2 0.4 0.6 0.8 1.00

2

4

6

8

10

21

01 1

d

d

X/(1-X)

/(1

-)

C X<

C/(1-C)

CX FalseLet X be the heat pump and the Carnot cycle operate like an engine

XC False

CX

1 Energy efficiency of the Carnot cycle is independent of the working substance.

2 Any cyclic process that absorbs heat at one temperature,and rejects heat at one other temperature, and is reversible has the energy efficiency of a Carnot cycle.

Why

Because: X can be a Carnot engine with arbitrary working substance

Carnot’s theorem: No engine operating between two heat reservoirs ismore efficient than a Carnot engine.

Proof uses similar idea as before:

We can design the engine X such that

Again we create a composite device

Tc

Th

heat engine X

W

XhQ

XcQ Tc

Th

heat pumpChQ

CcQ

X C

W operates the Carnot refrigerator

My statementholds man

Let’s assume that CX

Note: this time engine X can be also work irreversible like a real engine does

CX Xh

XQ

W >

ChQ

W

Ch

Xh Q

1

Q

1

0QQ Xh

Ch

Heat transferred from the cooler to the hotter reservoir without doing work on the surrounding

Violation of the Clausius statement CX Rudolf Clausius (2.1.1822 -24.8.1888)

Applications of Carnot Cycles

Any cyclic process that absorbs heat at one temperature, and rejects heat at one other temperature, and is reversible has the energy efficiency of a Carnot cycle.

We stated:

Why did we calculate energy efficiencies for

- gas turbine

- Otto cycle

Because: they are not 2-temperature devices, but accept and reject heat at a range of temperatures

Energy efficiency not given by the Carnot formula

But: It is interesting to compare the maximum possible efficiency of a Carnot cyclewith the efficiency of engineering cycles with the same maximum and minimum temperatures

Consider the gas turbine again

0 2 40

2

4

6

P

V

2 3

41

adiabates

(Brayton or Joule cycle)

1

h

l

P

P1

Efficiency

Ph

Pl

Maximum temperature:

@ 3

Minimum temperature:

@ 1

2 3Heating the gas (by burning the fuel)

4 1 cooling

: T3

: T1

1

h

l

P

P1 with

/)1(

l

1/)1(

h

2

P

T

P

T

2

1/)1(

h

/)1(l

T

T

P

P

2

1

T

T1

Efficiency of corresponding Carnot Cycle 3

1C T

T1

With 22,33 TTT 22,3

1C TT

T1

2

1turbinegas T

T1

22,3

1cycleCarnot TT

T1

0T 2,3

Unfortunately: Gas turbine useless in the limit 0T 2,3

Because: Heat taken per cycle 0

Work done per cycle 0

Absolute Temperature

We showed: Energy efficiency of the Carnot cycle isindependent of the working substance.

Definition of temperature independent of any material property

A temperature scale is an absolute temperature scale if and only if

C2

C1

2

1

Q

Q

T

T

where

,

C1Q and C

2Q are the heats exchanged by a Carnot cycle

operating between reservoirs at temperatures T1 and T2.

Measurement of 2

1c2

C1

C T

T1

Q

Q1

Temperature ratio2

1

T

TT1

T2

W

C2Q

C1Q

As discussed earlier, unique temperature scale requires fixed point

fixC2

C1T

Q

QT or fixC T1T

Kelvin-scale: Tfix =Ttripel=273.16K

It turns out:

proportional to thermodynamic Temperature T

empirical gas temperatureV303P

3g P

Plim

Why

Because: Calculation of efficiency of Carnot cycle based on nRPV

yields2

1C 1

Ta

With fixfix T a=1 T

C2

C1

2

1

Q

Q

T

TFrom definition of thermodynamic temperature

If any absolute temperature is positive all other absolute temperatures are positive

there is an absolute zero of thermodynamic temperature

when the rejected heat C1Q 0

T=0 can never be reached, because this would violate the Kelvin statement

however