Short Version : 19. 2nd Law of Thermodynamics

19.1. Reversibility & Irreversibility

Block slowed down by friction:

irreversibleBouncing ball:

reversible

Examples of irreversible processes:

• Beating an egg, blending yolk & white

• Cups of cold & hot water in contact Spontaneous process:

order disorder

( statistically more probable )

19.2. The 2nd Law of Thermodynamics

Heat engine extracts work from heat reservoirs.

• gasoline & diesel engines

• fossil-fueled & nuclear power plants

• jet engines

Perfect heat engine: coverts heat to work directly.

Heat dumped

2nd law of thermodynamics ( Kelvin-Planck version ):There is no perfect heat engine.

No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.

EfficiencyNet WorkHeat Input

e

h c

h

Q QeQ

1 c

h

QQ

(any engine)

(Simple engine)

0U W Q (any cycle)

Hero Engine

Stirling Engine

Carnot Engine

1. isothermal expansion: T = Th , W1 = Qh > 0

2. Adiabatic expansion: Th Tc, W2 > 0

3. isothermal compression: T = Tc , W3 = Qc < 0

Adiabatic compression : Tc Th , W4 = W2 < 0

Ideal gas:

Adiabatic processes:

1 1h B c CT V T V

1 1c D h AT V T V

C B

D A

V VV V

1 1c ccarnot

h h

Q TeQ T

3 ln2

Bh AB h

A

VQ Q n R TV

ln Dc CD c

C

VQ Q n R TV

AB: Heat abs.

C D: Heat rejected:

B C: Work done BC BC c hW U nR T T

D A: Work done

32

DA DA

h c

W U

nR T T

Engines, Refrigerators, & the 2nd Law

Carnot’s theorem:

1.All Carnot engines operating between temperatures Th & Tc have the same efficiency.

2.No other engine operating between Th & Tc can have a greater efficiency.

Refrigerator: extracts heat from cool reservoir into a hot one.

work required

2nd law of thermodynamics ( Clausius version ):

There is no perfect refrigerator.

perfect refrigerator: moves heat from cool to hot reservoir without work being done on it.

No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature.

Perfect refrigerator Perfect heat engine

Clausius Kelvin-Planck

Carnot engine is most efficient

eCarnot = thermodynamic efficiency

eCarnot erev > eirrev

Carnot refrigerator, e = 60%

Hypothetical engine, e = 70%

19.3. Applications of the 2nd LawPower plant

fossil-fuel : Th = 650 K

Nuclear : Th = 570 K

Tc = 310 K

3101650fossil

KeK

52 %

3101570nuclear

KeK

46 %

Actual values:

efossil ~ 40 % enuclear ~ 34 % ecar ~ 20 %

Prob 54 & 55

Heat source

Boiler

Turbine

Generator

Electricity

Condenser

Waste water

Cooling water

Application: Combined-Cycle Power Plant

Turbine engines: high Th ( 1000K 2000K ) & Tc ( 800 K ) … not efficient.

Steam engines : Tc ~ ambient 300K.

Combined-cycle : Th ( 1000K 2000K ) & Tc ( 300 K ) … e ~ 60%

Example 19.2. Combined-Cycle Power Plant

The gas turbine in a combined-cycle power plant operates at 1450 C.

Its waste heat at 500 C is the input for a conventional steam cycle, with its condenser at 8 C.

Find e of the combined-cycle, & compare it with those of the individual components.

273 81273 1450combinede

84 %

273 5001273 1450turbinee

55 %

273 81273 500steame

64 %

Refrigerators

Coefficient of performance (COP) for refrigerators :

cQCOPW

c

h c

QQ Q

c

h c

TT T

COP is high if Th Tc .

Max. theoretical value (Carnot)

1st law

W = 0 ( COP = ) for moving Q when Th = Tc .

Example 19.3. Home Freezer

A typical home freezer operates between Tc = 18C to Th = 30 C.

What’s its maximum possible COP?

With this COP, how much electrical energy would it take to freeze 500 g of water initially at 0

C?

5.31 273 18

30 18

c

h c

TCOP

T T

Q L m 334 / 0.5kJ kg kg 167 kJ Table 17.1

QWCOP

167

5.31kJ

31 kJ

2nd law: only a fraction of Q can become W in heat engines. a little W can move a lot of Q in refrigerators.

Heat Pumps

ccooling

QCOPW

c

h c

QQ Q

in

out in

TT T

Heat pump as AC :

hheating

QCOPW

h

h c

QQ Q

in

in out

TT T

Heat pump as heater :

Ground temp ~ 10C year round (US)

Heat pump: moves heat from Tc to Th .

19.4. Entropy & Energy Quality

Energy quality measures the versatility of different energy forms.

2nd law:Energy of higher quality can be converted completely into lower quality form.

But not vice versa.

Entropy

lukewarm: can’t do W,

Carnot cycle (reversible processes):

c c

h h

Q TQ T

0c h

c h

Q QT T

Qh = heat absorbed

Qc = heat rejected

Qh , Qc = heat absorbed

0C

d QT

C = any closed path

2

1

d QST

S = entropy [ S ] = J / K

Irreversible processes can’t be represented by a path.

Entropy

lukewarm: can’t do W,

Carnot cycle (reversible processes):

c c

h h

Q TQ T

0c h

c h

Q QT T

Qh = heat absorbed

Qc = heat rejected

Qh , Qc = heat absorbed

0C

d QT

C = any closed path

2

1

d QST

S = entropy [ S ] = J / K

Irreversible processes can’t be represented by a path.

C = Carnot cycle

Contour = sum of Carnot cycles.

Entropy change is path-independent.

( S is a thermodynamic variable )

S = 0 over any closed path

S21 + S12= 0

S21 = S21

Entropy in Carnot CycleIdeal gas：

ln Bh AB h

A

VQ Q n R TV

ln Dc CD c

C

VQ Q n R TV

Adiabatic processes：1 1

h B c CT V T V

1 1c D h AT V T V

C B

D A

V VV V

ABC ABQ Q

ADC DCQ Q

Heat absorbed:

Heat rejected:

ln Bh

A

Vn R TV

ln Cc

D

Vn R TV

ABABC

h

QST

ln B

A

Vn RV

DCADC

c

QST

ln C

D

Vn RV

ABCS

Irreversible Heat Transfer

Cold & hot water can be mixed reversibly

using extra heat baths.

f

c

T

c T

dQST

1

cQT

f

h

T

h T

dQST

2

hQT

c hS S S 1 2

c hQ QT T

1 2

c cQ QT T

1 2

1 1c T T

Q

0

0S Actual mixing, irreversible processes

reversible processes

T1 = some medium T.

T2 = some medium T.

2 1T T

Adiabatic Free Expansion

. exp. 0adW 0vacuump

Adiabatic Qad.exp. = 0

. exp. 0adU

. exp. 0adT

S can be calculated by any reversible

process between the same states.

dQST

1 dQT

QT

2

1

lnVn RV

0

WT

2

1

1 V

V

n R T dVT V

p = const.Can’t do work

degraded.

isothermal

Entropy & Availability of Work

Before adiabatic expansion, gas can do work isothermally

2

1

lnVW Q n R TV

After adiabatic expansion, gas cannot do work, while its entropy increases by

QST

2

1

lnVn RV

unavailableE T S

In a general irreversible process minunavailableE T S

Coolest T in system

Example 19.4. Loss of

A 2.0 L cylinder contains 5.0 mol of compressed gas at 300 K.

If the cylinder is discharged into a 150 L vacuum chamber & its temperature remains at 300 K,

how much energy becomes unavailable to do work?

2

1

lnVn R TV

minunavailableE T S

150 25.0 8.314 / / 300 ln2.0L Lmol J K mol K

L

54 kJ

A Statistical Interpretation of Entropy

Gas of 2 distinguishable molecules occupying 2 sides of a box

Microstates Macrostates probability of macrostate

1/4

2 ¼ = ½ 1/4

Gas of 4 distinguishable molecules occupying 2 sides of a box

Microstates Macrostates probability of macrostate

1/16 = 0.06

4 1/16 = ¼ =0.25

4 1/16 = ¼ =0.25

1/16 = 0.06

6 1/16 = 3/8 = 0.38

Gas of 100 molecules Gas of 1023 molecules

Equal distribution of molecules

lnBS k Statistical definition of entropy : # of micro states

Entropy & the 2nd Law of Thermodynamics

2nd Law of Thermodynamics : 0S in any closed system

S can decrease in an open system by outside work on it.

However, S 0 for combined system.

S 0 in the universe

Universe tends to disorder

Life ?