Advanced Thermodynamics

Note 4The Second Law of Thermodynamics

Lecturer: 郭修伯

Differences between heat and work

• A conversion efficiency of 100% by elimination of friction, a dissipative process that transforms work into heat.

• All efforts to devise a process for the continuous conversion of heat completely into work or into mechanical or electrical energy has failed. Conversion efficiencies do not exceed about 40%.

• The flow of heat between two bodies always takes place from the hotter to the cooler body, and never in the reverse direction.

The second law

• No apparatus can operate in such a way that its only effect is to convert heat absorbed by a system completely into work done by the system.– It is impossible by a cyclic process to convert the heat

absorbed by a system completely into work done by the system.

• No process is possible which consists solely into the transfer of heat from one temperature level to a higher one.

• Heat engine: devices or machines that produce work from heat in a cyclic process.– Essential to all heat-engine cycles are absorption of

heat into the system at a high temperature, rejection of heat to the surroundings at a lower temperature, and production of work.

– In operation, the working fluid of a heat engine absorbs heat |QH| from a hot reservoir, produces a net amount of work |W|, discard heat |QC| to a cold reservoir, and returns to its initial state.

The first law: CH QQW

The thermal efficiency of the engine:H

C

H

CH

H Q

Q

Q

QQ

Q

W

1

• If a thermal efficiency of 100% is not possible for heat engines, what determines the upper limits?

• A heat engine operating in a completely reversible manner is called a Carnot engine (N.L.S. Carnot, 1824):– Step 1: A system at the temperature of a cold reservoir TC undergoes a reversible

adiabatic process that causes its temperature to rise to that of a hot reservoir at TH.

– Step 2: The system maintains contact with the hot reservoir at TH, and undergoes a reversible isothermal process during which heat |QH| is absorbed from the heat reservoir.

– Step 3: The system undergoes a reversible adiabatic process in the opposite direction of step 1 that brings its temperature back to that of the cold reservoir at TC.

– Step 4: The system maintains contacts with the reservoir at TC, and undergoes a reversible isothermal process in the opposite direction of step 2 that returns to its initial state with rejection of heat |QC| to the cold reservoir.

The cycle traversed by an ideal gas serving as the working fluid in a Carnot engine is shown by a PV diagram.

TH

TC

b

c

ad

P

V

|QH|

|QC|

a → b adiabatic compressionb → c isothermal expansionc → d adiabatic expansiond → a isothermal compression

For the isothermal steps b → c and d → a

b

cHH V

VRTQ ln and

a

dCC V

VRTQ ln

For adiabatic processes a → b and c → d

b

aT

T

V

V

V

T

dT

R

CH

C

ln andc

dT

T

V

V

V

T

dT

R

CH

C

ln

C

H

ad

bc

C

H

C

H

T

T

VV

VV

T

T

Q

Q

)/ln(

)/ln(

H

C

H

C

H T

T

Q

Q

Q

W 11 5.0

600

3001

Rough practical limits for ηof a Carnot engine; actual heat engines are irreversible and ηrarely exceed 0.35.

A central power plant, rated at 800,000 kW, generates steam at 585K and discards heat to a river at 295 K. If the thermal efficiency of the plant is 70% of the maximum possible value, how much heat is discarded to the river at rated power?

4957.0585

2951max 347.04957.07.0

H

C

H

C

H T

T

Q

Q

Q

W 11

347.0 kWW 800000

kWQC 1505500

• A Carnot engine:

• There exists a property called entropy S, which is an intrinsic property of a system, functionally related to the measurable coordinates which characterize the system.– If a process is reversible and adiabatic, dQrev = 0, dSt = 0. The entropy of a system is constant

during a reversible adiabatic process and the process is isentropic.– When a system undergoes an irreversible process between two equilibrium states, the entropy

change of the system ΔSt is evaluated by an arbitrarily chosen reversible process. Since entropy is a state function, the entropy changes of the irreversible and reversible processes are identical.

C

C

H

H

T

Q

T

Q 0

C

C

H

H

T

Q

T

Q

0C

C

H

H

T

dQ

T

dQ

0 T

dQrev

T

dQdS revt

T

dQS revt

Entropy changes of an ideal gas

• For one mole of fluid undergoing a mechanically reversible process in a closed system:

PdVdQdU rev

VdPPdVdUdH

VdPdHdQrev

dTCdH igP PRTV /

P

dPR

T

dTC

T

dQdS ig

Prev

0

ln0 P

P

T

dT

R

C

R

S T

T

igP

For an ideal gas with constant heat capacities undergoing a reversible adiabatic process:

0

ln0 P

P

T

dT

R

C

R

S T

T

igP

0S

1

2

1

2 lnln0P

P

T

T

R

C igP

igPCR

P

P

T

T/

1

2

1

2

RCC igV

igP ig

V

igP

CC

/)1(

1

2

1

2

P

P

T

T

The 2nd law mathematical statement

• Let a quantity of heat |Q| be transferred from the hotter (TH) to the cooler (TC) reservoir. The entropy changes of the two reservoirs are:

• For the process of irreversible heat transfer, ΔStotal is always positive, approaching zero as the process becomes reversible.

H

tH T

QS

||

C

tC T

QS

||

CH

CHtC

tHtotal TT

TTQSSS ||

0 totalS

A 40 kg steel casting (CP= 0.5 kJ/kgK) at a temperature of 450°C is quenched in 150 kg of oil (CP= 2.5 kJ/kgK) at 25°C. If there are no heat losses, what is the change in entropy of (a) the casting, (b) the oil, and (c) both considered together?

0)25)(5.2)(150()450)(5.0)(40( TT CT 25.46No heat losses:

K

kJ

T

TmC

T

dTCm

T

dQS P

Pt 33.16ln1

2 (a) the casting

K

kJ

T

TmC

T

dTCm

T

dQS P

Pt 13.26ln1

2 (b) the oil

(c) total entropy changeK

kJSSS t

oiltcasttotal 80.9

Net rate of change in entropy of flowing streams

Time rate of change entropy in control volume

Time rate of change entropy in surroundings

Total rate of entropy generation

0)(

)( G

tsurrcv

fs Sdt

dS

dt

mSdmS

0)(

)(,

Gj j

jcvfs S

T

Q

dt

mSdmS

Entropy balance for open systems

In a steady-state flow process, 1 mol/s of air at 600 K and 1 atm is continuously mixed with 2 mol/s of air at 450K and 1 atm. The product stream is at 400K and 1 atm. Determine the rate of heat transfer and the rate of entropy generation for the process. Assume that air is an ideal gas with CP = (7/2)R, that the surroundings are at 300K, and that kinetic- and potential-energy changes are negligible.

KTs

moln

A

A

600

1

C.V.

KTs

moln

400

3

KTs

moln

B

B

450

2

Q

sJTTCnTTCnHnHnHnQ BPBAPABBAA /7.8729)()(

Ks

J

T

Q

T

TCn

T

TCn

T

QSSnSSn

T

QSnSnSnS

BPB

APA

BBAA

BBAAG

446.10

300

7.8729

500

400ln)2(

600

400ln)1()314.8)(

2

7(

lnln

)()(

An inventor claims to have devised a process which takes in only saturated steam at 100 °C and which by a complicated series of steps makes heat continuously available at a temperature level of 200°C. The inventor claims further that, for every kilogram of steam taken into the process, 2000 kJ of energy as heat is liberated at the temperature level of 200°C. Show whether or not this is possible. In order to give the inventor the benefit of any doubt, assume cooling water available in unlimited quantity at a temperature of 0°C.

T’ = 200°C

kgK

kJS

kg

kJH

3554.7

2676

1

1

apparatus

kJQ 2000

Tσ = 0°C

Q

Saturated steam at 100 °C

Liquid water at 0 °C

0.0

0.0

2

2

S

H

QQQQ 2000

sWQH

QH 200026760.0

kJQ 0.676

K

kJS 3554.73554.70.0

K

kJS t 227.4

15.273200

2000

K

kJS t 4748.2

15.2730

0.676

K

kJStotal 6536.04748.2227.43554.7

Calculation of ideal work• In a process producing work, there is an absolute

maximum amount which may be accomplished as the result of a given change of state of the fluid flowing through the control volume.

• The limiting value obtains when change of state associated with the process is accomplished completely reversibly.

• The entropy generation is zero:

0)( T

QmS fs

fsmSTQ )( )()(

2

1 2 revWmSTmzguH sfs

fs

fs

fs

ideal mSTmzguHW )(2

1 2

fsfsideal mSTmHW )( STHWideal

s

idealt W

Wrequiredwork

)(

ideal

st W

Wproducedwork

)(

What is the maximum work that can be obtained in a steady-state flow process from 1 mol of nitrogen (assumed an ideal gas) at 800 K and 50 bar? Take the temperature and pressure of the surroundings as 300 K and 1.0133 bar?

mol

JEEICPHdTCH

T

T

igP 15060)5040.0,0.0,3593.0,280.3;300,800(314.8

0

For an ideal gas, enthalpy is independent of pressure, and its change is given:

mol

J

P

PR

T

dTCS

T

T

igP 042.3

50

0133.1ln314.815060ln

2

11

2

mol

JSTHWideal 15973)042.3)(300(15060

Or, Step 1, reversible, adiabatic expansion from initial state to 1.0133 bar, T’ Step 2, cooling to the final temperature

HWQ S Step 1 )( 1HHHWS

)(dQT

TTdWCarnot

Step 2 STHH

T

dQTQW

T

TCarnot )'( 2'

2

STHWWW CarnotSideal Total

An inventor claims to have devised a process which takes in only saturated steam at 100 °C and which by a complicated series of steps makes heat continuously available at a temperature level of 200°C. The inventor claims further that, for every kilogram of steam taken into the process, 2000 kJ of energy as heat is liberated at the temperature level of 200°C. Show whether or not this is possible. In order to give the inventor the benefit of any doubt, assume cooling water available in unlimited quantity at a temperature of 0°C.

T’ = 200°C

kgK

kJS

kg

kJH

3554.7

2676

1

1

apparatus

kJQ 2000

Tσ = 0°C

Q

Saturated steam at 100 °C

Liquid water at 0 °C

0.0

0.0

2

2

S

H

267626760.0 H

K

kJS 3554.73554.70.0

mol

J

STHWideal

9.666

)3554.7)(15.273(2676

kJ

TT

TWQ

7.15770200

15.2732009.666

||||

Lost work

• Work that is wasted as the result of irreversibilities in a process is called lost work: idealSlost WWW

QmzguHWfs

S

2

2

1

fs

fs

ideal mSTmzguHW )(2

1 2

QmSTW fslost )(

Surrounding temperature TσT

QmSS fsG

)(

0 Glost STW

0 Glost STW

The two basic types of steady-flow heat exchanger are characterized by their flow patterns: cocurrent and countercurrent. Consider the two cases, for each of which the following specifications apply:

The minimum temperature difference between the flowing streams is 10K. Assume that both streams are ideal gases with CP = (7/2)R. Find the lost work for both cases. Take Tσ = 300 K.

smolnKTKTKT HCHH 1300350400 121

Fig 5.9

QmzguHWfs

S

2

2

10)()( CCHH HnHn

0)()( 1212 CCPCHHPH TTCnTTCn CCHHfs SnSnnS )()()(

Negligible pressure change

1

2

1

2 lnln)(C

C

H

C

H

HPHfs T

T

n

n

T

TCnnS

QnSTW fslost )(

Case I, cocurrent:

25.1300340

350400

H

C

n

n Ks

JnS fs 667.0

300

340ln25.1

400

350ln)314.8(

2

7)1()(

s

JnSTW fslost 1.200667.0300)(

Case II, countercurrent:

5556.0300390

350400

H

C

n

n

Ks

JnS fs 356.0

300

390ln5556.0

400

350ln)314.8(

2

7)1()(

s

JnSTW fslost 7.106356.0300)(

From thermodynamic point of view, the countercurrent case is much more efficient.

The third law of thermodynamics

• The absolute entropy is zero for all perfect crystalline substances at absolute zero temperature.– When the form is noncrystalline, e.g., amorphous or

glassy, calculations show that the entropy of the more random form is greater than that of the crystalline form.

– The absolute entropy of a gas at temperature T:

T

T

gP

v

vT

T

lP

f

fTSP

v

v

f

f

dTT

C

T

HdT

T

C

T

HdT

T

CS

)()()(0