10 20 30 40 50 60 70 80 90 1000

10

20

30

40

50

60

70

80

90

p

V

Isotherm, pV = constant = NkBT

Adiabat, pV = constant

v = 10:1:100;t = 100;r = 8.314;gamma = 1.67;p = r*t./v;k = (10^(gamma-1)).*r*t;pa = k./v.^gamma;plot(v,pa,v,p)

Carnot cycle v = 10:1:100;th = 100;tl = 50;r = 8.314;gamma = 1.67;p1 = r*th./v;k1 = (30^(gamma-1)).*r*th;pa1 = k1./v.^gamma;p2 = r*tl./v;k2 = (30^(gamma-1)).*r*tl;pa2 = k2./v.^gamma;plot(v,p1,v,pa1,v,p2,v,pa2)

Carnot cycle

p

V

Qh

Th

Isotherm 1

Adiabat 1

Adiabat 2

Isotherm 2

pA, VA, TA

pB, VB, TB

pC, VC, TC

pD, VD, TD

Carnot cycle

p

V

Qh

pA, VA, TA

pB, VB, TB

pC, VC, TC

pD, VD, TD

Th

Carnot cycle

Isotherm 1

Adiabat 1

Adiabat 2

Isotherm 2

p

V

Qh

pA, VA, TA

pB, VB, TB

pC, VC, TC

pD, VD, TD

Tl

Ql

Carnot cycle

Isotherm 1

Adiabat 1

Adiabat 2

Isotherm 2

p

V

Qh

pA, VA, TA

pB, VB, TB

pC, VC, TC

pD, VD, TD Ql

Tl

Carnot cycle

Isotherm 1

Adiabat 1

Adiabat 2

Isotherm 2

Why such a strange engine?

Will discuss in class

p

V

Efficiency of a Carnot engine

⇒𝑇 𝐴=𝑇 𝐵=𝑇 h

⇒𝑇𝐶=𝑇 𝐷=𝑇 𝑙

𝑝𝐴 ,𝑉 𝐴 ,𝑇 𝐴

𝑝𝐵 ,𝑉 𝐵 ,𝑇 𝐵

𝑝𝐶 ,𝑉 𝐶 ,𝑇𝐶

𝑝𝐷 ,𝑉 𝐷 ,𝑇 𝐷

Isotherm 1

Adiabat 1

Adiabat 2

Isotherm 2

Efficiency of a Carnot engine

𝑝𝐴 ,𝑉 𝐴 ,𝑇 h

𝑝𝐵 ,𝑉 𝐵 ,𝑇 h

𝑝𝐶 ,𝑉 𝐶 ,𝑇 𝑙

𝑝𝐷 ,𝑉 𝐷 ,𝑇 𝑙

⇒ Δ𝑄h=𝑅𝑇 h ln𝑉 𝐵

𝑉 𝐴p

V

⇒ Δ𝑄𝑙=𝑅𝑇 𝑙 ln𝑉 𝐷

𝑉 𝐶

⇒ Δ𝑄=0⇒𝑇h𝑉 𝐵

𝛾−1=𝑇 𝑙𝑉 𝐶𝛾−1

⇒ Δ𝑄=0⇒𝑇 𝑙𝑉 𝐷

𝛾−1=𝑇h𝑉 𝐴𝛾−1

Isotherm 1

Adiabat 1

Adiabat 2

Isotherm 2

Efficiency of a Carnot engine

Δ𝑄h=𝑅𝑇 h ln𝑉 𝐵

𝑉 𝐴

Δ𝑄𝑙=𝑅𝑇 𝑙 ln𝑉 𝐷

𝑉 𝐶

𝑇 h𝑉 𝐵𝛾− 1=𝑇 𝑙𝑉 𝐶

𝛾 −1

𝑇 𝑙𝑉 𝐷𝛾 −1=𝑇 h𝑉 𝐴

𝛾 −1

(1) From isotherm 1

(3) From isotherm 2

(2) From adiabat 1

(4) From adiabat 2

From the first law of thermodynamics: Δ𝑈=Δ𝑄+Δ𝑊

For the complete Carnot cycle since is a state variable

Efficiency of a Carnot engine

Δ𝑄h=𝑅𝑇 h ln𝑉 𝐵

𝑉 𝐴

Δ𝑄𝑙=𝑅𝑇 𝑙 ln𝑉 𝐷

𝑉 𝐶

𝑇 h𝑉 𝐵𝛾− 1=𝑇 𝑙𝑉 𝐶

𝛾 −1

𝑇 𝑙𝑉 𝐷𝛾 −1=𝑇 h𝑉 𝐴

𝛾 −1

(1) From isotherm 1

(3) From isotherm 2

(2) From adiabat 1

(4) From adiabat 2

From the first law of thermodynamics: Δ𝑈=Δ𝑄+Δ𝑊

For the complete Carnot cycle since is a state variable

⇒ Δ𝑄=− Δ𝑊

From (1) and (3): +

is the work done on the engine (system), let be the work done by the engine

⇒𝑊=− Δ𝑊=Δ𝑄

Efficiency of a Carnot engine

Δ𝑄h=𝑅𝑇 h ln𝑉 𝐵

𝑉 𝐴

Δ𝑄𝑙=𝑅𝑇 𝑙 ln𝑉 𝐷

𝑉 𝐶

𝑇 h𝑉 𝐵𝛾− 1=𝑇 𝑙𝑉 𝐶

𝛾 −1

𝑇 𝑙𝑉 𝐷𝛾 −1=𝑇 h𝑉 𝐴

𝛾 −1

(1) From isotherm 1

(3) From isotherm 2

(2) From adiabat 1

(4) From adiabat 2

From (1) and (3): +

Efficiency is defined as: OutputInput

Output is the work done by the engine i.e. and input is the heat absorbed by the engine i.e.

⇒𝜂 ( efficiency )= 𝑊Δ𝑄h

=Δ𝑄h+ Δ𝑄 𝑙

Δ𝑄h

=1+Δ𝑄𝑙

Δ𝑄h

Efficiency of a Carnot engine

Δ𝑄h=𝑅𝑇 h ln𝑉 𝐵

𝑉 𝐴

Δ𝑄𝑙=𝑅𝑇 𝑙 ln𝑉 𝐷

𝑉 𝐶

𝑇 h𝑉 𝐵𝛾− 1=𝑇 𝑙𝑉 𝐶

𝛾 −1

𝑇 𝑙𝑉 𝐷𝛾 −1=𝑇 h𝑉 𝐴

𝛾 −1

(1) From isotherm 1

(3) From isotherm 2

(2) From adiabat 1

(4) From adiabat 2

⇒𝜂 ( efficiency )= 𝑊Δ𝑄h

=Δ𝑄h+ Δ𝑄 𝑙

Δ𝑄h

=1+Δ𝑄𝑙

Δ𝑄h

⇒𝜂=1+𝑅𝑇 𝑙 ln

𝑉 𝐷

𝑉 𝐶

𝑅𝑇 h ln𝑉 𝐵

𝑉 𝐴

=1−𝑇 𝑙

𝑇h

ln𝑉 𝐷

𝑉 𝐶

ln𝑉 𝐴

𝑉 𝐵

(from (1) and (3))

(2) ⇒𝑇 h

𝑇 𝑙

=(𝑉 𝐶

𝑉 𝐵)𝛾 −1

and (4) ⇒𝑇 h

𝑇 𝑙

=(𝑉 𝐷

𝑉 𝐴)𝛾− 1

⇒𝑉 𝐶

𝑉 𝐵

=𝑉 𝐷

𝑉 𝐴

⇒𝑉 𝐴

𝑉 𝐵

=𝑉 𝐷

𝑉 𝐶

Efficiency of a Carnot engine

⇒𝜂=1−𝑇 𝑙

𝑇 h

0. There is a game

Laws of thermodynamics

 

 

 

1.You can never win

2. You cannot break even, either

3. You cannot quit the game

𝑇 h

𝑇 𝑙

Carnot

𝑄h=|Δ𝑄h|

𝑄 𝑙=|Δ𝑄 𝑙|=− Δ𝑄 𝑙

𝑊=Δ𝑄h+ Δ𝑄𝑙=𝑄h−𝑄𝑙

Carnot engine: Schematic representation

𝑇 h

𝑇 𝑙

Carnot

𝑄h

𝑄𝑙

𝑊=𝑄h−𝑄𝑙

Carnot engine: Schematic representation

𝑇 h

𝑇 𝑙

Carnot

𝑄h

𝑄𝑙

𝑊=𝑄h−𝑄𝑙

Carnot engine is reversible

𝑇 h

𝑇 𝑙

Carnot

𝑄h

𝑄𝑙

𝑊=𝑄h−𝑄𝑙

Carnot engine is reversible (refrigerator)

Carnot’s theorem

Of all heat engines working between two given temperatures, none is more efficient than a Carnot engine

reversible

𝑇 h

𝑇 𝑙

Carnot

𝑄h

𝑄𝑙

𝑊=𝑄h−𝑄𝑙

Carnot engine is reversible (refrigerator)

R

𝑊 ′=𝑄h′ −𝑄 𝑙

′

𝑄h′

𝑄 𝑙′

Adjust the cycles so that

𝑇 h

𝑇 𝑙

Carnot

𝑄h

𝑄𝑙

𝑊=𝑄h−𝑄𝑙=𝑄h′ −𝑄𝑙

′

Carnot engine is reversible (refrigerator)

R

𝑄h′

𝑄 𝑙′

𝑇 h

𝑇 𝑙

Carnot

𝑄h

𝑄𝑙

𝑊=𝑄h−𝑄𝑙=𝑄h′ −𝑄𝑙

′

Carnot engine is reversible (refrigerator)

R

𝑄h′

𝑄 𝑙′

If then:

𝑊𝑄h′ >

𝑊𝑄h

⇒𝑄h′ <𝑄h⇒𝑄h−𝑄h

′ >0

Also:

⇒𝑄h−𝑄h′ =𝑄 𝑙−𝑄𝑙

′ >0

𝑇 h

𝑇 𝑙

Carnot

𝑄h

𝑄𝑙

𝑊=𝑄h−𝑄𝑙=𝑄h′ −𝑄𝑙

′

Is this possible?

R

𝑄h′

𝑄 𝑙′

⇒𝑄h>𝑄h′ 𝑎𝑛𝑑𝑄𝑙>𝑄𝑙

′

⇒𝑄h−𝑄h′ =𝑄 𝑙−𝑄𝑙

′ >0

𝑄h−𝑄h′

𝑄𝑙−𝑄𝑙′

The Second Law of Thermodynamics

•Clausius’ statement: It is impossible to construct a device that operates in a cycle and whose sole effect is to transfer heat from a cooler body to a hotter body.

⇒𝜂 ′≯� 𝜂

Carnot’s theorem

Of all heat engines working between two given temperatures, none is more efficient than a Carnot engine

reversible

𝑊 𝑖𝑟𝑟<𝑊 𝑟𝑒𝑣

⇒𝜂 𝑖𝑟𝑟<𝜂𝑟𝑒𝑣

𝑇 h

𝑇 𝑙

Carnot

𝑄h

𝑄𝑙

𝑊=𝑄h−𝑄𝑙=𝑄h′ −𝑄𝑙

′

For reversible engines

R

𝑄h′

𝑄 𝑙′

⇒𝜂 ′≯� 𝜂𝑎𝑛𝑑𝜂≯� 𝜂′⇒𝜂=𝜂′

Carnot’s theorem

Of all heat engines working between two given temperatures, none is more efficient than a Carnot engine

All reversible engines working between two temperatures have the same efficiency as

The Second Law of Thermodynamics

•Clausius’ statement: It is impossible to construct a device that operates in a cycle and whose sole effect is to transfer heat from a cooler body to a hotter body.

•Kelvin-Planck statement: It is impossible to construct a device that operates in a cycle and produces no other effect than the performance of work and the exchange of heat from a single reservoir.

𝑇 h

𝑇 𝑙

Carnot

𝑄h

𝑄𝑙

𝑊=𝑄h′ =𝑄h−𝑄𝑙

Carnot refrigerator and Kelvin violator

Kelvin violator

𝑄h′

⇒𝑄h−𝑄𝑙=𝑄h′

⇒𝑄h−𝑄h′ =𝑄 𝑙>0

𝑇 h

𝑇 𝑙

Carnot

𝑄h

𝑄𝑙

𝑊=𝑄h−𝑄𝑙

Carnot engine and Claussius violator

Claussius violator

𝑄𝑙

𝑄𝑙