The Classical Second Law of Thermodynamics - 서강대학교 …home.sogang.ac.kr/.../Lists/b17/Attachments/27/Chapter5.pdf · 2017-05-04 · Fundamentals of Thermodynamics Chapter

Prof. Siyoung Jeong

Thermodynamics I

MEE2022-02

Fundamentals of Thermodynamics

Chapter 5

The Classical Second Law of

Thermodynamics

Thermal Engineering Lab. 2

• First law : no restriction on the direction of heat and

work flow

- 열역학 1법칙만 생각하면, 따뜻한 커피가 온도가 낮은 주변 공기에서

열을 받아 가열되는 것도 가능

- 그러나 실제로 이러한 현상은 일어나지 않음

- Process의 direction에 관한 법칙 필요

1212 UUQ

• Second Law : 열역학적 Process의 direction에 관한

법칙

Chapter 5. The Classical Second Law of Thermodynamics


5.1 Heat engines and refrigerators


• Heat engine

- Operating in a cycle

- Net positive heat transfer & net positive work (Qin & Wout)


• Refrigerator (Heat Pump)

- Operating in a cycle

- Heat transfer from the low-temperature system to the high-temperature system

(TL→TH, Win)


QL







• Thermal efficiency of a heat engine

H

LH

H

thermalQ

QQ

Q

W

costs)at (energy th

sought)(energy

H

L

Q

Q1


Ex. 5.1 An automobile engine produces 136 hp on the output shaft with a thermal

efficiency of 30%. The fuel it burns gives 35000 kJ/kg as energy release.

Find the total rate of energy rejected to the ambient and the rate of fuel

consumption in kg/s.




• A simple vapor-compression refrigeration cycle



• Coefficient of Performance (COP)

HLLH

HH

LHLH

LL

QQQQ

Q

W

Q

QQQQ

Q

W

Q

1

1'

1

1 (COP)

* Thermal reservoir

- 온도 변화 없이 열교환

- Source 또는 Sink로 작용

e.g.) Atmosphere, ocean


Ex. 5.2 The refrigerator in a kitchen shown in Fig. 5.7 receives electrical input

power of 150 W to drive the system, and it rejects 400 W to the kitchen

air. Find the rate of energy taken out of the cold space and the COP of the

refrigerator.



5.2 The second law of thermodynamics


• The Kelvin-Planck statement

It is impossible to construct a device that will operate in a cycle and produce no effect other than the raising of a weight and the exchange of heat with a single reservoir.

(M. Planck, 1897)



• The Clausius statement

It is impossible to construct a

device that operates in a cycle

and produces no effect other

than the transfer of heat from

a cooler body to a warmer

body.

(R. Clausius, 1854)



• Equivalence of the two statements



• Perpetual-motion machine of the second kind



• Reversible process (가역 과정)

- A process that once having taken place can be reserved

and in so doing leave no change in either system or

surroundings.

5.3 The reversible process



• Irreversible process



• Reversible process



• Reversible process

- 마찰이 없는 mechanical process

- 마찰이 없는 단열 상태 변화

5.4 Factors that render process irreversible

• Irreversible process

: 어느 방법으로도 원상태 불가능 (모든 자연 Process)

- Friction

- Unrestrained expansion

- Heat transfer through a finite temperature difference

- Mixing of two different substances

- Other factors



• Friction



• Unrestrained expansion



• Mixing



Reversible Internally reversible

Externally irreversible



- 주어진 두 온도 사이에서 가장 효율적인 cycle

5.5 The Carnot cycle

1 : rev. isothermal process, QH is transferred to the system

2 : rev. adiabatic process, Working fluid : TH → TL

3 : rev. isothermal process, QL is rejected from the system

4 : rev. adiabatic process, Working fluid : TL → TH







• First proposition

5.6 Two propositions regarding the efficiency of a Carnot cycle

• Second proposition

revirr ηη

),( LHrev TTfη





5.7 The thermodynamic temperature scale

1

1 ,

Lth

H

L H

Q

Q

T T

H

L

H

Lth

H

L

H

L

T

T

Q

Q

Tf

Tf

Q

Q

Tf

TfTT

Tf

TfTT

Tf

TfTT

TTTTTTQ

Q

Q

Q

Q

Q

TTQ

QTT

Q

QTT

Q

Q

11 (

)(

)(

)(),( ,

)(

)(),( ,

)(

)(),(

),(),(),(

),( , ),( , ),(

)

3

131

3

232

2

121

313221

3

2

2

1

3

1

31

3

132

3

221

2

1



5.8 The ideal-gas temperature scale

PtPtPt P

PT

P

P

T

T

vRTPv

,,,

16.273

const




Ex. 5.3 In a certain constant-volume ideal-gas thermometer, the measured

pressure at the ice point (see Section 1.11) of water, 0℃, is 110.9 kPa and

at the steam point, 100℃, it is 151.1 kPa. Extrapolating, at what Celsius

temperature does the pressure go to zero (i.e., zero absolute temperature)?






Example Ptp(1)=1 bar P(1)=1.36587 bar Ti(1)=373.10 K Ptp(2)=2 bar P(2)=2.73127 bar Ti(2)=373.05 K T=373.15 K



0

0

212

1

reversible work :

Ideal Gas :

0 ln

v

v

H H

w Pdv

Pv RT

du C dT

q du w

RTC dT dv

v

vq q RT

v

①→②



4

3

1

2

1

4

2

3

1

4

2

3

4

1

4

3

4

3

3

434

2

3

lnln

ln0

ln

lnln0

ln0

0

0

v

v

v

v

v

v

v

v

v

v

v

v

v

vRdT

T

C

v

vRTq

v

vRT

v

vRTqq

v

vRdT

T

C

T

q

H

L

L

H

T

T

v

LL

LLL

T

T

v

②→③

③→④

④→①

0 3

2

lnH

L

T v

T

C vdT R

T v*

3

4

2

1

ln

ln

L

L L

H H

H

vRT

vq T

q TvRT

v

0 4

1

lnH

L

T v

T

C vdT R

T v*



5.9 Ideal versus real machines

LH

L

H

L

TT

T

QQ

Qβ

L

L

H H

H H L

Q T

Q Q T T

H

L

H

L

T

T

Q

Qη 11 thermalreal

LH

L

H

L

TT

T

QQ

Qβ

L

real

L

realH H

H H L

Q T

Q Q T T


Ex. 5.4 Let us consider the heat engine, shown schematically in Fig. 5.25, that

receives a heat-transfer rate of 1 MW at a high temperature of 550℃ and

rejects energy to the ambient surroundings at 300 K. Work is produced at

a rate of 450 kW. We would like to know how much energy is discarded

to the ambient surroundings and the engine efficiency and compare both

of these to a Carnot heat engine operating between the same two

reservoirs.



Ex. 5.5 As one mode of operation of an air conditioner is the cooling of a room

on a hot day, it works as a refrigerator, shown in Fig. 5.26. A total of 4

kW should be removed from a room at 24℃ to the outside atmosphere at

35. We would like to estimate the magnitude of the required work. To do

this we will not analyze the processes inside the refrigerator, which is

deferred to Chapter 9, but we can give a lower limit for the rate of work,

assuming it is a Carnot-cycle refrigerator.




5.10 Engineering applications





• Power output and the Carnot cycle

- Internally reversible

b

L

a

H

HthLH

LbLL

aHHH

T

Q

T

Q

QQQW

TTCQ

TTCQ

)(

)(



1

2

( ) ( )

1 1

1 1

1

( ) 1 0

H H a L b L

a b

H LH L

a b

H Hb L

L a

bth H H H a

a

bb a

a bH H a H

a a a

b bH

a a

C T T C T T

T T

T TC C

T T

C TT T

C T

TW Q C T T

T

TT T

T TWC T T C

T T T

T T TC

T T

1 1 0bH

H

a a

TC

T T



2

21 1

1 1

1

b H H H HL

a L a L a

H La H H L

H L H L

L Hb L H L

H L H L

L L H H Lbth

a H H L H L

Lth

H

T C T C TT

T C T C T

C CT T T T

C C C C

C CT T T T

C C C C

C T C T TT

T C T C T T

T

T



H

L

LHL

LLth

HL

H

L

HH

LHH

th

LH

LHLHH

LHHLL

th

T

T

TTC

TC

CC

T

T

TC

TTC

CC

TTCTC

TTCTC

11

11

1

만약

만약



Homework Problems 2017: 33, 36, 51, 63, 69, 81

The Classical Second Law of Thermodynamics - 서강대학교 …home.sogang.ac.kr/.../Lists/b17/Attachments/27/Chapter5.pdf · 2017-05-04 · Fundamentals of Thermodynamics Chapter

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