### Carnot cycle

```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)
90
80
70
60
50
p
Isotherm, pV = constant = NkBT
40
30
20
10
0
10
20
30
40
50
60
V
70
80
90
100
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
Carnot cycle
pA, VA, TA
Isotherm 1
Th
p
Qh
pB, VB, TB
pD, VD, TD
Isotherm 2
pC, VC, TC
V
Carnot cycle
pA, VA, TA
Isotherm 1
Th
p
Qh
pB, VB, TB
pD, VD, TD
Isotherm 2
pC, VC, TC
V
Carnot cycle
pA, VA, TA
Isotherm 1
Tl
p
Qh
pB, VB, TB
pD, VD, TD
Ql
Isotherm 2
pC, VC, TC
V
Carnot cycle
pA, VA, TA
Isotherm 1
Tl
p
Qh
pB, VB, TB
pD, VD, TD
Ql
Isotherm 2
pC, VC, TC
V
Why such a strange engine?
Will discuss in class
Efficiency of a Carnot engine
,  ,
Isotherm 1 ⇒  =  = ℎ
p
,  ,
,  ,
Isotherm 2 ⇒  =  =
V
,  ,
Efficiency of a Carnot engine
,  , ℎ
Isotherm 1
p
⇒ Δℎ = ℎ ln

,  , ℎ
⇒ Δ = 0
−1
−1
⇒   = ℎ
Adiabat 1 ⇒ Δ = 0
−1
−1
⇒ ℎ
=
,  ,
Isotherm 2

⇒ Δ =  ln

V
,  ,
Efficiency of a Carnot engine

Δℎ = ℎ ln

−1
ℎ
−1
=

Δ =  ln

−1

(1) From isotherm 1
(3) From isotherm 2
−1
= ℎ
From the first law of thermodynamics: Δ = Δ + Δ
For the complete Carnot cycle Δ = 0
since  is a state variable
Efficiency of a Carnot engine

Δℎ = ℎ ln

−1
ℎ
−1
=

Δ =  ln

−1

(1) From isotherm 1
(3) From isotherm 2
−1
= ℎ
From the first law of thermodynamics: Δ = Δ + Δ
For the complete Carnot cycle Δ = 0 since  is a state variable
⇒ Δ = −Δ
Δis the work done on the engine (system), let  be the work done by the engine
⇒  = −Δ = Δ

From (1) and (3): Δ = Δℎ +Δ = ℎ ln  +  ln

Efficiency of a Carnot engine

Δℎ = ℎ ln

−1
ℎ
(1) From isotherm 1
−1
=

Δ =  ln

−1

(3) From isotherm 2
−1
= ℎ

From (1) and (3): Δ = Δℎ +Δ = ℎ ln  +  ln  =

Efficiency is defined as:

Output
Input
Output is the work done by the engine i.e.  and input is the
heat absorbed by the engine i.e. Δℎ
⇒  efficiency =

Δℎ +Δ
Δ
=
=1+
Δℎ
Δℎ
Δℎ
Efficiency of a Carnot engine

Δℎ = ℎ ln

−1
ℎ
(1) From isotherm 1
−1
=

Δ =  ln

−1

(3) From isotherm 2
−1
= ℎ
⇒  efficiency =

Δℎ +Δ
Δ
=
=1+
Δℎ
Δℎ
Δℎ

ln

⇒ = 1+
=1−

ℎ ln
ℎ ln

ln
ℎ

⇒
=
(2)

−1
and (4)
ℎ

⇒
=

(from (1) and (3))
−1

⇒
=

⇒
=

Efficiency of a Carnot engine

⇒ =1−
ℎ
Laws of thermodynamics
0. There is a game
1. You can never win
2. You cannot break even, either
3. You cannot quit the game
Carnot engine: Schematic representation
ℎ
ℎ = Δℎ
= Δℎ +Δ = ℎ −
Carnot
= Δ = −Δ

Carnot engine: Schematic representation
ℎ
ℎ
= ℎ −
Carnot

Carnot engine is reversible
ℎ
ℎ
= ℎ −
Carnot

Carnot engine is reversible (refrigerator)
ℎ
ℎ
= ℎ −
Carnot

Carnot’s theorem
reversible
Of all heat engines working between two given
temperatures, none is more efficient than a Carnot engine
Carnot engine is reversible (refrigerator)
ℎ
ℎ′
ℎ
′ = ℎ′ − ′
= ℎ −
R
Carnot
′

Adjust the cycles so that  =  ′
Carnot engine is reversible (refrigerator)
ℎ
ℎ′
ℎ
= ℎ −  = ℎ′ − ′
R
Carnot
′

Carnot engine is reversible (refrigerator)
ℎ
ℎ′
ℎ

>
ℎ′ ℎ
= ℎ −  = ℎ′ − ′
⇒ ℎ′ < ℎ ⇒ ℎ − ℎ′ > 0
R
Carnot
If ′ >  then:
Also: ℎ −  = ℎ′ − ′
′

⇒ ℎ − ℎ′ =  − ′ > 0
Is this possible?
ℎ
ℎ
ℎ − ℎ′
ℎ′
⇒ ℎ − ℎ′ =  − ′ > 0
= ℎ −  = ℎ′ − ′
R
Carnot

−
′

⇒ ℎ > ℎ′   > ′
′
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.
⇒ ′ ≯
Carnot’s theorem
reversible
Of all heat engines working between two given
temperatures, none is more efficient than a Carnot engine
<
⇒  <
For reversible engines
ℎ
ℎ′
ℎ
= ℎ −  = ℎ′ − ′
R
Carnot
′

⇒ ′ ≯    ≯ ′ ⇒  = ′
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 Carnot
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.
Carnot refrigerator and Kelvin violator
ℎ
ℎ′
ℎ
⇒ ℎ −ℎ′ =  > 0
= ℎ′ = ℎ −
Kelvin
violator
Carnot

⇒ ℎ − = ℎ′
Carnot engine and Claussius violator
ℎ
ℎ

= ℎ −
Claussius
violator
Carnot

```