SUMMARY TABLES

Epicurus on Importance of Summaries Epicurus

01 Title Page

Chapter 1 INTRODUCTION AND BASIC CONCEPTS

02 Thermodynamics - Basic Concepts

03 Properties, Pressure Blaise Pascal

Chapter 2 ENERGY, ENERGY TRANSFER, AND GENERALENERGY ANALYSIS

04 Energy

05 Mechanical Energy, Energy Transfer

06 Heat and Work

07 The First Law of Thermodynamics James P. Joule vonMayer Epicurus

08 Energy Conversion Efficiences

Chapter 3 PROPERTIES OF PURE SUBSTANCES

09 Pure Substance, Phase, Phase Change Process, Property Diagrams

10 Property Tables

11 The Ideal-Gas Equation of State Robert Boyle JacquesCharles Joseph Louis Gay-Lussac

12 Compressibility Factor - A Measure of deviation from Ideal-gasBehavior

Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS

14 Moving Boundary Work

15 Polytropic Process

16 Energy Balance for Closed Systems

17 Specific Heats Joseph Black Joseph Black

17 Specific Heats - Solids and Liquids

Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROLVOLUMES

18 Conservation of Mass

19 Flow Work

Energy Balance of Steady Flow

20 Steady-Flow Devices: Nozzles and Diffusers; Compressors, Turbines

Throttling Valves; Mixing Chambers; Heat Exchangers; Pipe Flow:Heaters, Condensors, Evaporators

21 Energy Analysis of Unsteady-Flow Systems

Chapter 6 THE SECOND LAW OF THERMODYNAMICS

22 Thermal Reservours, Heat Engines, Thermal Efficiency

24 Coefficient of Performance, Refrigerator, Heat Pump

23 The Second Law of Thermodynamics

25 The Carnot Cycle

Chapter 7 ENTROPY

26 Entropy

27 T-S Diagram

28 Gibbs Equation, Entropy Change

29 Reversible Steady-Flow Work

30 Isentropic Efficiences

31 Entropy Balance

Chapter 9 GAS POWER CYCLES

32 Gas Power Cycles - Basics

33 Reciprocating Engines - Overview

34 Otto Cycle

35 Diesel Cycle

36 Stirling and Ericsson Cycles

37 Brayton Cycle

Chapter 10 VAPOR AND COMBINED POWER CYCLES

38 Rankine Cycle

Chapter 11 REFRIGERATION CYCLES

39 Refrigeration Cycle

THERMODYNAMICSTHERMODYNAMICSTHERMODYNAMICSTHERMODYNAMICS

ME-321

Vladimir Soloviev

system

boundary

surroundings

THERMODYNAMICS All aspects of ENERGY and energy transformation (science of energy) Macro-thermodynamics (classical) matter is assumed to be continuum

Micro-thermodynamics (statistical) matter consists of discrete atoms

SYSTEM A region in space or a quantity of matter Open system (control volume) Closed system (control mass) Isolated system

PROPERTY Any characteristic of a system is called a property.

Intensive property mass-independent: { }T ,P, ,v,u,...ρ (lowercase letters)

Extensive property mass-dependent: { }m,V ,U ,E,... (upper case letters)

Specific property property per unit mass: { }v V m ,u U m ,e E m,...= = =

Independent properties { }T ,v ,{ }P,v ,{ }T,h ,…

STATE Set of all properties of the system defines a state { }T ,P,V ,m,u,e,v,s,h,...

EQUILIBRIUM Equilibrium is a state of balance. System can remain at equilibrium state indefinitely.

Thermodynamic equilibrium state: the system is in equilibrium regarding all possible changes of state; i.e. it maintains thermal, mechanical, phase, chemical etc. equilibria.

STATE POSTULATE The state of a simple compressible system is completely defined by two independent intensive properties PROCESS Change of a system from one state to another (transformation).

process diagram

Quasi-equilibrium process a system is close to equilibrium at any moment of process

Steady-flow process no change with time

Unsteady-flow process transient

Uniform no change with location

Iso-… process particular property remains constant (isothermal process)

cycle

heat

work

property 1

property 2

state

1State

path

2State

=1 2

heat

work

mass

Kazimir Malevich

The Black Square, 1915

PROPERTIES EXTENSIVE intensive

Mass m [ ]kg

Temperature T [ ]K

Pressure P [ ]kPa

Density m

Vρ =

3

kg

m

2H O

SGρ

ρ=

specific gravity (relative density)

Volume V 3m

V 1v

m ρ= =

3m

kg

specific volume

s

gγ ρ= 3

N

m

specific weight

Internal Energy U [ ]kJ U

um

= kJ

kg

specific internal energy

Total Energy E [ ]kJ E

em

= kJ

kg

specific total energy

Enthalpy H U PV= + [ ]kJ h u Pv= + kJ

kg

(heat content, total heat,“to heat”)

Specific heats p

c kJ

kg K

⋅ at constant pressure

v

c kJ

kg K

⋅ at constant volume

Entropy S kJ

K

s kJ

kg K

⋅

Quality vapor

liquid vapor

mx

m m=

+

quality is defined only for

saturated liquid-vapor mixtures

Blaise Pascale

( 1623 1662 )−

Pascal's Law

Manometer

h

fluid of

density ρ

oP P ghρ= +

o atmP P=

gas

Paris Museum of Measures and Arts

( ) ( )oT K T C 273.15= +

( ) ( )oT R T F 459.67= +

( ) ( )o oT F 9 5 T C 32= ⋅ +

dPg

dzρ= −

Barometer

h

atmP ghρ=

0P 0=

atmPP

1 atm 101.3 kPa=

1 atm 14.7 psi=

( )1-20 in this equation the positive direction of z is upward

=dP

gdz

ρ

gage abs atmP P P= −abs

P

atmP

vac atm absP P P= −

atmP

absP

positive direction

of z is downward

o atmP P=

oP P ghρ= +

z

h

fluid of

density ρ

P

abs atm gageP P +P=

abs atm vacP P P= −

Conversion of [psi] to [Pa]

P F

A=

[ ]psi [ ]

2

lbf

in=

[ ] ( ) 2

2

ftlbm 32.174

s

in

⋅

=

[ ] ( ) ( ) ( )

( )

2

2 222

2 2

kg ft mlbm 0.45359 32.174 0.3048

lbm s ft

1 ft min 0.3048

144 in ft

⋅ ⋅

=

⋅ ⋅

( ) ( )

( )2 2

0.45359 32.174 m 1kg

1 s m0.3048

144

⋅ = ⋅

⋅

2 2

m 16894.7 kg

s m

= ⋅

[ ]2

16894.7 N

m

=

[ ]

2

mN kg

s

= ⋅

weight F = ⋅m g

[ ]6894.7 Pa= [ ]2

NPa

m

=

ENERGY Total Energy E [ ]kJ Specific total energy E

em

= kJ

kg

(2-1)

Total energy E is the sum of all forms of energy: thermal, mechanical,kinetic, potential, electric, nuclear, etc.

Thermodynamics deals only with the change in total energy, E∆ . Microscopic related to molecular structure. Sum of all microscopic forms is called internal energy U

Macroscopic with respect to some outside frame. Mechanical energy.

Internal energy U [ ]kJ Specific internal energy u U

m=

kJ

kg

(2-1)

Kinetic energy KE2

Vm

2= [ ]kJ Specific kinetic energy ke

2V

2=

kJ

kg

(2-2), (2-3)

Potential energy PE mgz= [ ]kJ Specific potential energy pe gz= kJ

kg

(2-4), (2-5)

Total energy E U KE PE= + + [ ]kJ Specific total energy e u ke pe= + + kJ

kg

(2-6), (2-7)

E 2

VU m mgz

2= + + [ ]kJ Specific total energy e

2V

u gz2

= + + kJ

kg

(2-6), (2-7)

Equation (2-6) in SI units:

Energy flow associated with a fluid flow

Mass of volume V m ρ= V [ ]kg

Mass flow rate m� c av

A Vρ ρ= =�V kg

s

(2-8)

Total energy flow rate E� me= � kJ

kWs

=

(2-9)

E me= [ ]kJ

( ) thermal energy heat

m

z elevation=

V

amount of mass flowing through

a cross section per unit time

p.54

dmdot above a symbol indicates time rate m=

dt�

⇒ [ ] [ ][ ] [ ] [ ]

[ ] [ ]

2 22

2 2

2

2

V m m Vm kg m kg g z m m mgz2 s s 2E kJ U kJ U kJ kJ

10001 kg m1000

kJ s

⋅ + ⋅ ⋅ +

= + = + ⋅

⋅ ⋅

2

2 2

m mJ N m kg m kg

s s= ⋅ = ⋅ ⋅ = ⋅

work force dist ⋅

system moving with the velocity V

( )

( )

sensible kinetic energy of molecules

latent phase energy, binding forces

chemical

nuclear

Chapter 2

MECHANICAL ENERGY OF FLUID FLOW is the form of energy which can be converted to mechanical work completely and directly by an ideal mechanical device. It consists of energy of flowing fluid (called flow work or flow energy), kinetic energy (KE) and potential energy (PE) .

Mechanical energy of flowing fluid: mech

E ρ

= + +2

P Vm m mgz

2 [ ]kJ

Power (rate of energy transfer) mech

E� ρ

= + +

�

2P V

m gz2

kJ

kWs

=

mech

EE

t

∆

∆=� (2-11)

mech

e = + +2

P Vgz

2ρ

kJ

kg

(2-10)

Mechanical energy change of

incompressible flow ( constρ = ): mech

E∆ ( )2 2

2 1 2 1

2 1

P P V Vm g z z

2ρ

− −= + + −

[ ]kJ

mechE∆ � ( )

2 2

2 1 2 1

2 1

P P V Vm g z z

2ρ

− −= + + −

�

kJkW

s

=

(2-13)

mech

e∆ ( )2 2

2 1 2 1

2 1

P P V V g z z

2ρ

− −= + + −

kJ

kg

(2-12)

ENERGY TRANSFER

Heat Transfer Q energy transfer due to temperature difference

Work Transfer W energy interaction not caused by a temperature Mass Flow m transfer of energy carried by moving mass

Work and Heat: 1. Work and heat are recognized at the boundary (boundary phenomena). 2. Systems possess energy, not work or heat. 3. Associated with a process, not with a state. 4. Path functions (depend on path of process), not point functions which depend on state. SIGN CONVENTION

� � �flow enegy KE PE

heat

electrical

work

Examples 2-4,2-5,2-6 Heating of a Potato in an Oven

1z

1V

2V

2P , ,mρ

1P , ,mρ

State 1

State 2

2z

heating element

air

heat

+

system

-

to the system

positive

from the system

negative

HEAT energy transfer between system and surroundings due to temperature difference

Modes of heat transfer conduction cond

Q� dT

kAdx

= −

convection conv

Q� ( )surface fluidhA T T= −

radiation rad

Q� ( )4 4

surface surroundingsA T Tεσ= −

Heat Q [ ]kJ

Heat transfer per unit mass q Q

m=

kJ

kg

(2-14)

Rate of heat transfer Q� Q

t∆=

kJkW

s

=

Total heat transferred during process Q ( )2

1

t

t

Q t dt= ∫ � 2

1

t

t

Qδ= ∫ (2-15)

Q Q t∆= � (2-16)

WORK if energy crossing the boundary of closed system is not heat then it must be work

Work = energy transfer associated with force acting through the distance ( )W F s= ⋅

Work W [ ]kJ

Work done per unit mass w W

m=

kJ

kg

(2-17)

Power (work done per unit time) W� W

t∆=

kJkW

s

=

Exact differential of a property dT

Temperature change during the process: State2

State1

dT∫ 2 1

T T= −

Inexact differentials Wδ , Qδ

Work done during the process: 2

1

Wδ∫ 12

W=

Total heat transferred during process

2

1

Qδ∫ 12

Q=

heat

addition

heat

rejection

1

2

1T

2T

2 1

Change in temperature

T=T -T is the same for any

process between the states

∆

dT

1

2

Wδ

W depends on pathδ

T

12 12work W or heat Q

transferred during process

between states 1 and 2

can be different for different

paths of the process

Wδ

Energy can be neither created nor destroyed during a process.

It can only change forms.

in

out

The net change in the total energy E of the system during a process

is equal to the difference between the total energy entering E and

the total energy leaving E the system during that process

∆

.

conservation of energy

for any system and

any kind of process

inE

outE

2 1 in outE E E E− = −

James P. Joule (1818-1889)

E∆

inE

outE

1E

2E

2-6 THE 1st LAW OF THERMODYNAMICS – CONSERVATION OF ENERGY PRINCIPLE

Change in Total Energy 2 1

E E E∆ = − [ ]kJ (2-32)

in out

E E E∆ = − [ ]kJ (2-35)

in out

E E E= −� � � [ ]kW (2-36) E

Et

∆

∆=�

E∆ ( )2 2

2 1

2 1 2 1

V Vm u u g z z

2

−= − + + −

(2-33)

E� ( )2 2

2 1

2 1 2 1

V Vm u u g z z

2

−= − + + −

�

EE

t

∆

∆=�

Stationary system E∆ [ ]2 1m u u= −

Open system (energy crossing the boundary can be in the forms of heat, Q, work, W, and mass):

in out

E E− ( ) ( )mass ,net ,innet ,in net ,out

in out out in mass ,in mass ,out

EQ W

Q Q W W E E= − − − + −������������ �����

(2-34)

Closed system (energy crossing the boundary is in the forms of heat, Q, and work, W, only):

E∆net ,in net ,out

Q W= − net ,in net ,out

Q W= Q Q t∆= �

dE

dtnet ,in net ,out

Q W= −� � net ,in net ,out

Q W=� � W W t∆= �

1st Law of Thermodynamics

net ,in net ,outQ W− ( )

2 2

2 1

2 1 2 2 1 1 2 1

V Vm u u v P v P g z z

2

−= − + − + + −

[ ]kJ

net ,in net ,outQ W−� � ( )

−= − + − + + −

�

2 2

2 1

2 1 2 2 1 1 2 1

V Vm u u v P v P g z z

2 [ ]kW

net ,in net ,outQ W− ( )

2 2

2 1

2 1 2 1

V Vm h h g z z

2

−= − + + −

[ ]kJ

net ,in net ,outQ W−� � ( )

2 2

2 1

2 1 2 1

V Vm h h g z z

2

−= − + + −

� [ ]kW (5-38)

1

2

net ,inQ

net ,outW ( )E 0∆ =Cycle

enthalpy

h u Pv≡ +

1 = 2

1z

1V

2V

State1

State2

2z

fixed mass m

steady flow

m

m

2-7 ENERGY CONVERSION EFFICIENCIES

(2-41)

Combustion efficiency combustion

Q Heat released during combustion

HV Heating Value of the fuel burnedη = = (2-42)

Lighting efficacy [ ]

[ ]η =

lighting

Amount of light output lumens

Electricity consumed W

MECHANICAL AND ELECTRICAL DEVICES

mech,out mech,in mech,loss mech,loss

mech

mech,in mech,in mech,in

E E E EMechanical energy output1

Mechanical energy input E E Eη

−= = = = −

Pump efficiency mech, fluid

pump

shaft,in

EMechanical energy increase of the fluid

Mechanical energy input W

∆η = =

�

�

sys

in out

dEE E

dt= −� � (2-36)

sys shaft ,in loss ,out loss ,out

pump

shaft ,in shaft ,in shaft ,in

E W W W1

W W W

∆η

−= = = −

� � � �

� � �

2 1

sys

P PE m∆

ρ

− =

� � ( )2 1

P P= −�V (2-13)

2 1

pump

shaft ,in

P Pm

W

ρη

−

=

�

�

( )2 1

shaft ,in

P P

W

−=�

�

V

Overall efficiency

pump-motor pump motorη η η= ⋅ efficiency of pump-motor combination (2-49)

m�m�

shaft ,inW�

1 2P P<

loss,outW�

1P

2P

1

2

shaft ,inW�

outE�

Desired outputPerfomance

Required input=

3 1-4 PROPERTIES OF PURE SUBSTANCES

Pure Substance fixed chemical composition

Phases solid, liquid, gas

Phase Change Process

Phase diagram T-v diagram P-v diagram

P-T diagram

Latent Heat of Melting (Freezing)

Latent Heat of Vaporization (Condensing) fg

Q mh= [ ]kJ

fgQ mh=� � [ ]kW

attraction viscosity=

molecules arranged in lattice small attraction, random motion

1 2 3 4 5

compressed

liquid

saturated

liquid

saturated

liquid - vapor

mixture

superheated

vapor

saturated

vapor

o

P 1 atm

T 10 C

=

=o

P 1 atm

T 100 C

=

=

o

P 1 atm

T 100 C

=

=

o

P 1 atm

T 100 C

=

=o

P 1 atm

T 200 C

=

=

satP = pressure

at which phase

is changed at

given temperature

p.116Table 3 - 1

T

Vv

m=

1

2 3 4

5

p.122Table 3 - 3

P const=

( )energy absorbed during evaporation

critical

pointsatP , kPa

o

satT , C

101.4

100

liquid-vapor

saturation

curve

satT = temperature

at which phase

is changed at

given pressure

( )energy absorbed during melting

T

Vv

m=

critical

point

superheated

vaporcompressed

liquid

saturated

liquid-vapor

mixture

saturated

liquid line

saturated

vapor line

P const=

satT

critical

point

T

Vv

m=

T const=

satP

critical

point

P

Vv

m=

VAPOR

critical point

P

T

sublimation

melting evaporation

SOLID

LIQUID

triple point

subscripts

3-5 PROPERTY TABLES H U PV= + [ ]kJ enthalpy

h u Pv= + [ ]kJ kg (specific) enthalpy

f saturated liquid f

v specific volume of saturated liquid f

h enthalpy of saturated liquid

g saturated vapor g

v specific volume of saturated vapor g

h enthalpy of saturated vapor

fg difference fg g f

v v v= − fg g f

h h h= − latent heat of vaporization

Saturated Liquid or Vapor Saturated liquid-vapor mixture

Superheated Vapor: Compressed Liquid:

P

satT

T

y

gy

x 0= x 1=x

fy y

y can be v, h, u, or s

f fgy y xy= +

=g

quality:

mx

m

f

fg

y yx

y

−=o

T 25 C=

satP 3.169kPa=

oT , C

Vv

m=

fv

gv

P 5 MPa=

satT

T

Vv

m=

f @Tv v≈

T

f @T

f @T

f @T

approximations:

v v ,

u u ,

h h

≈

≈

≈

P 10kPa 0.01MPa= =

satT

T

Vv

m=

gv

fv v

T

( )f @ T f @ T sat @ Th h v P P≈ + ⋅ −

moderate pressuresfor low or

=mass of vapor

total mass

- and TABLE A - 5 Pressure table

LINEAR INTERPOLATION

BI-LINEAR INTERPOLATION

( )1

1 2 1

2 1

T T y y y y

T T

−= + ⋅ −

−

linear

interpolation1

T

y

1y

2y

2T

T

y

( )1

1 2 1

2 1

y y T T T T

y y

−= + ⋅ −

−

1T

T

2T

1y

2y

If property y is given:

If T is given:

( )

( )

11 11 12 11

2 1

12 21 22 21

2 1

T Ty y y y

T T

T Ty y y y

T T

−= + ⋅ −

−

−= + ⋅ −

−

( )11 2 1

2 1

P Py y y y

P P

−= + ⋅ −

−

( )

( )

111 1 2 1

12 11

212 1 2 1

22 21

y yt T T T

y y

y yt T T T

y y

−= + ⋅ −

−

−= + ⋅ −

−

( )11 2 1

2 1

P PT t t t

P P

−= + ⋅ −

−

If P and T are given: If P and property y are given:

bi - linear

interpolation

1 2T T T< <

1T

2T

P < <2

P1P

11y

12y

22y

21y

3-6 IDEAL GAS is a gas obeying the Ideal-Gas Equation of State

Boyle (1662) , Mariotte 1

P ~V

Charles, Gay-Lussac (1802) 1

P mRTV

= Ideal-Gas Equation of State

Ideal gas equation of state PV mRT= R uR M=

3kJ kPa m

kg K kg K

⋅=

⋅ ⋅ gas constant (Tables 1,2)

Pv RT= u

R 8.314=

3kJ kPa m

kmol K kmol K

⋅=

⋅ ⋅ universal gas constant

Clapeyron-Mendeleev u

mPV R T

M= M

kg

kmol

molar mass (Table 1)

u

PV NR T= N m

M= [ ]kmol mole* number

u

Pv R T= v V

N=

3m

kmol

molar specific volume

u U

N=

kJ

kmol

internal energy

h H

N=

kJ

kmol

enthalpy

Closed System Constant Volume

m const= 1 1 2 2

1 2

PV PV

T T= V const=

1 2

1 2

P P

T T=

Water vapor as an ideal gas Ideal Gas Tables **

A-17 Air A-18 Nitrogen N2 A-19 Oxygen O2

A-20 Carbon Dioxide CO2 A-21 Carbon Monoxide CO

A-22 Hydrogen H2 A-23 Water vapor H2O

12

( ) ( )For ideal gas, u T and h T depend on temperature only,

therefore, Tables A 17-25 are the temperature tables

∗∗

Properties per unit mole:

[ ] , KT is the absolute temperature

[ ] , kPaP is the absolute pressure

( )molecular weight

1

2

T

v

2P

1P

table ideal

table

Percentage of error

v v100

v

in assuming steam

to be an ideal gas

−×

, oT C

3,v m kg

[ ] ( )23Mole is a unit of measurement for amount of substance; 1 mol = 6.022 10 molecules Avagadro number∗ ⋅

( )

Edme Mariotte

1620 1684−( )

Robert Boyle

1627 1691− ( )

Jacques Charles

1746 1823− ( )

Joseph Gay-Lussac

1778 1850− ( )

B.P.E.Clapeyron

1799 1864− ( )

Dmitri Mendeleev

1834 1907−( )

Amadeo Avogadro

1776 1856−

3-7 COMPRESSIBILITY FACTOR – deviation from ideal-gas

Ideal Gas =i .g .

Pv1

RT For ideal gas ⇒

i .g .

RTv

P=

Compressibility factor Pv

ZRT

= For real gas ⇒ i .g .

RTv Z Z v

P= = ⋅ (3-19)

Reduced properties [ ]

[ ]R

cr

T KT

T K= reduced temperature (3-20)

R

cr

PP

P= reduced pressure

R

cr

cr

vv

TR

P

= pseudo-reduced specific volume (3-21)

Principle ( )R RZ T ,P is the same for all gases

of

corresponding Table A-15, p.932

states ( )R RZ v ,P is the same for all gases

Figure 3-49 Comparison of Z factors Table A-15 Generalized compressibility chart

Example 3-11 Find specific volume of R-134a at P 1.0 MPa= and oT 50 C 273.15K= = (superheated vapor)

Table A-1 cr

P 4.059 MPa= , cr

T 374.2 K= , R 0.08149= kJ

kg K

⋅

Table A-13 exact

v 0.021796= Exact value

Ideal gas ( )( )

( )i .g .

0.08149 323RTv 0.026333

P 1000= = = Error: 20.8 %

Z-factor ( )

( )R

cr

273.15TT 0.863

T 374.2= = = ,

( )

( )R

cr

1PP 0.246

P 4.059= = =

A 15−

⇒ Z 0.85≈

( )( )i .g .v Z v 0.85 0.026333 0.02238305= ⋅ = = Error: 2.7 %

4-1 PdV WORK – MOVING BOUNDARY WORK

Wδ F dS= ⋅ ( )

dV

P A dS= ⋅ ⋅

����

PdV= differential boundary work (4-1)

Boundary work done during the process Work done during a cycle

W δ= ∫2

1

W [ ]kJ

W

2

1

PdV= ∫ (4-2)

Direction of boundary work

Constant pressure process

Boundary work

in the 1st Law

force acting on

the moving boundary

Work is the energy transfer

between system and surroundings

during the process

Work out is used for something outside:

rotate the shaft, overcome friction, etc.

3 3

2

kNkPa m m kN m kJ

m

⋅ = ⋅ = ⋅ =

P

V

1V

2V

expansion

W 0<

−

1

2

P

V

2V

1V

compression

P

V

W 0<

−

1V

2V

21

expansion

( )2 1W P V V 0= ⋅ − >

sign will be adjusted

automatically

W 0>

+

1

2

P F PA=

A

pressure at the

inner surface

of the boundary

P

V

W 0<

−

2V

1V

2 1

compression

( )2 1W P V V 0= ⋅ − <

( )= = ⋅ −∫2

2 1

1

W PdV P V Vin

Wout

W

boundary

assume the only work

is the PdV work W

W 0>

out

dV 0>

work by system

, −net in boundary

Q W

( )

total boundary work is equal to the area

under the path (depends on path) 4-3

P

V

W

1V

2V

1

2

difference between the work done by the system

and the work done on the system

P

V

−1 2

netW 0>

P

V

−1 2

netW 0<

W 0<

work on system

dV 0<

in

dV AdS=

dS

4-1 BOUNDARY WORK FOR POLYTROPIC PROCESS OF GASES closed system, m=const

n

PV C=

n n

1 1 2 2PV PV=

n 1≠ n

PV C= polytropic process W

2

1

PdV= ∫2

n

1

C V dV−

= ∫

( )1 n 1 n

2 1

1CV CV

1 n

− −= −

− ⇐

W 2 2 1 1

PV PV

1 n

−=

−

PV mRT= 1 1 2 2

1 2

PV PV

T T=

2

1

T

T

11

n2

1

P

P

−

=

n 1

1

2

V

V

−

=

1 1 1

PV mRT=

2 2 2

PV mRT= W ( )2 1

mRT T

1 n= −

−

n 1= PV C= process W

2

1

CdV

V= ∫

2

1

VC ln

V=

W 2

1 1

1

VPV ln

V=

2

2 2

1

VPV ln

V= = 1

2 2

2

PPV ln

P

For ideal gas: PV mRT C= = ⇒ T const= (isothermal process)

W = 2

1

VmRT ln

V = 1

2

PmRT ln

P

1 1 2 2PV PV=

n 0= =P C isobaric process (constant pressure process)

W ( ) ( )2 1 2 1P V V P m v v= ⋅ − = ⋅ ⋅ −

n 0 P const isobaric

n 1 T const isothermal ideal gas

n k s const isentropic

n v const isochoric

= ⇒ =

= ⇒ =

= ⇒ =

= ∞ ⇒ =

n,C are constants

P

V

W

1V

2V

1

2

1 1PV C=

2 2P V C=

CP

V=

( )4.9

( )4.10

( )4.7

n n

1 1 2 2C P V P V= =

P

V

W

1V 2

V

1

2

n

1 1PV C=

n

2 2P V C=

n

CP

V=

ideal gas

( )4.6

P

V

W

1V

2V

1 2

P const=

lnP

lnV

lnP n lnV C= − ⋅ +

ENERGY BALANCE FOR CLOSED SYSTEMS (m=const) , (Not a flowing fluid, no flow work)

Energy can cross the boundary of a closed system only in the form of heat or work

1st Law of Thermodynamics

net ,in net ,outQ W E∆− = [ ]kJ (4-17) Q Q t∆= �

net ,in net ,out

Q W E∆− =� � � [ ]kW W W t∆= �

net ,in net ,out

Q W− ( )2 2

2 1

2 1 2 1

V VU U m mg z z

2

−= − + + −

net ,in net ,out

Q W− ( )2 2

2 1

2 1 2 1

V Vm u u g z z

2

−= − + + −

Stationary system: net ,in net ,out

Q W− 2 1

U U= − [ ]kJ

net ,in net ,out

Q W− ( )2 1m u u= − [ ]kJ

net ,in net ,out

q w− 2 1

u u= − kJ

kg

For a cycle: net ,in net ,out

Q W= [ ]kJ

net ,out net ,out

Q W=� � [ ]kW

Constant pressure process P const= Boundary work: boundary

W ( )2 1P V V= ⋅ −

( )boundary otherQ W W− +

2 1U U= −

( )2 1 otherQ P V V W− − −

2 1U U= −

other

Q W− ( )2 1

2 2 1 1

H H

U PV U PV= + − +����� �����

other

Q W− 2 1

H H= − (4-18)

other

Q W− ( )2 1m h h= ⋅ −

net,in in outQ Q Q= −

E∆

net,out out inW W W= −

=1 2

1u

2u

1

2

net ,outW

net ,inQ

KE PE 0∆ ∆= =

E 0∆ =

P 100=

2V

Q

bW

otherW

P 100=

1V

1z

1V

2V

m

1

State

2

State

2z

m

KE and PE

relative to some

reference frame

∆ ∆

4-3 SPECIFIC HEATS

Specific heat c energy required to raise the temperature

of a unit mass by one degree

v

c at constant volume

p

c at constant pressure

Energy balance of closed stationary system for process

1) without any work q 2 1

u u= −

2) with a boundary work at constant pressure, b

w , q 2 1 b

u u + w= − ( )

at constant pressure

2 1 2 1u u + P v v= − ⋅ −

�����

2 1

h h = −

Relationship between the transferred heat and the change of temperature Definition of specific heats

( )u T ,v , v const= ⇒ u u

du dT dvT v

∂ ∂= +

∂ ∂ q

uu T

T∆ ∆

∂= =

∂

v

v const

uc

T =

∂ =

∂

kJ

kg K

⋅

( )h T ,P , P const= ⇒ h h

dh dT dPT P

∂ ∂= +

∂ ∂ q

hh T

T∆ ∆

∂= =

∂

p

P const

hc

T =

∂ =

∂

kJ

kg K

⋅

IDEAL GAS Internal energy ( )u T is a function of T only (Joule’s experiment),

Pv RT= then h u Pv= +

Pv RT

=

=���

( ) +u T RT ⇒ ( )h T is a function of T only

v

duc

dT=

vdu c dT= ⇒ u∆

2 1u u= − ( )

2

v

1

c T dT= ∫ ( )v ,av 2 1 c T T≈ ⋅ − v

u c T= ⋅

p

dhc

dT=

pdh c dT= ⇒ h∆

2 1h h= − ( )

2

p

1

c T dT= ∫ ( )p ,av 2 1 c T T≈ ⋅ − p

h c T= ⋅

h ( )u Pv u T RT= + = + ⇒ dh

dT

duR

dT= +

p

c v

c R= + (4-29) p

v

ck

c= specific heat ratio (4-31)

p

c v u

c R= + (4-30) specific heats on a molar basis

kJ

kg K

⋅

for any process,

not only at constant

pressure or volume

Specific heat depends

on how the energy is

added to the system

Joseph Black

kJ

kg K

⋅

v pIdeal Gas c ,c

Table A-2 a,b,c

In Table A-17:

1v

2v

P const=v const=

T

oT 1+

T

v

P

1v

2v

1P P const=

v const=

T

oT 1+

2P

v

q − w = −2 1u u

≠

≠

2 1

2 1

P P

V V

( ) ( )p vc T ,c T in Table A-2b,c ( )av 1 2c at T T 2+

Tdu

v const=

u

dT

T P const=

h

v

du

dT

dh

dT

p

dh

dT

5-1 CONSERVATION OF MASS

CV

m∆ in out

m m= − [ ]kg (5-8) 2 1

m m− in out

m m= −

CV

m

t

∆

∆

in outm m= −� �

kg

s

(5-9) CVm

t

∂

∂CV

dVt

ρ∂

=∂ ∫

CV

dm

dt

in outm m= −∑ ∑� �

kg

s

(5-17)

CV CS

dV V ndA 0t

ρ ρ∂

+ ⋅ =∂ ∫ ∫

� �

mδ � m

t

∆

∆=

dV

t

ρ

∆

⋅=

cdA L

t

ρ

∆

⋅ ⋅=

n cV dAρ= (5-2), (5-3)

m�

cA

mδ= ∫ �

c

n c

A

V dAρ= ∫

c

n c

A

V dAρ= ∫

( )

c

avg

n c c

c A

V 5-4

1V dA A

Aρ

=

∫�������

avg cV Aρ=

Mass flow rate m� avg c

V Aρ= avg

c

VA

v=

V

v=�

Vρ= � kg

s

Volume flow rate V� avg c

V A= ⋅ vm= � m

ρ=�

3

m

s

Steady Flow (SF) ∂

= ⇒∂

CVm0

t

CVm const=

Incompressible Flow

in out

m m=∑ ∑� � (5-18)

in out

V V=∑ ∑� � (5-20)

Steady Flow Single Stream (SF SS) 1 2

m m m= =� � �

Incompressible Flow

1 1 1 2 2 2V A V Aρ ρ= (5-19)

1 1 2 2V A V A= (5-21)

c

c

Differential mass flow rate

through the differential area dA

of the inlet (exit) surface area A

( )

( )

c

c

Mass flow rate through the area A

of incompressible fluid =const

or for the case of uniform density =const over cross-section A

ρ

ρ

3m�

2m�

1m�

1 2 3m m m+ =� � �

2m�1m�1V

2V

1A 2A

Vm V

vρ= =

Vm V

vρ= =�

��

Change of mass within CV during the process

t = time interval∆

cA

inm

outm

inlets

control

volume

outlets

Mass balance

( )5.7

rate of

change

of mass

cA

cdA

Ln

V V n= ⋅� �

normal

velocity

differential volume

( )( )cdV dA L=

flow

velocityV�

n�

normal

vector

Conservation of mass

( )5-5

"Nothing comes

from nothing"

Parmenides 500BC

( )=constρ

( )=constρ

5-2 FLOW WORK – FLOW ENERGY

W F L= ⋅ ( )P A L= ⋅ ⋅ ( )P A L= ⋅ ⋅ P V= ⋅

W P V= ⋅ (5-23)

W P v m= ⋅ ⋅ Flow Work

w P v= ⋅

ENERGY TRANSPORT BY MASS Total energy of the flowing fluid

E

2mV

U PV mgz2

H

= + + +�������

e

2

h

V u Pv gz

2= + + +�������

(5-27)

e = + +2

V h gz

2 (5-27)

E

2V

m h gz2

= + +

(5-28)

E�

2V

m h gz2

= + +

� (5-29)

Energy Balance for Steady Flow Single Stream (SF SS) in a Pipe

flow work can be treated as

the energy of a flowing fluid

Total specific energy

of flowing fluid

Amount of energy transport by mass

Rate of energy transport by mass

Units for mechanical energy terms

W P v m= ⋅ ⋅� �

avg

mV

s

=� �pE mc T∆ ∆

cA

c

the rate of energy transport

by mass of flowing fluid

through the control volume

of uniform cross-section A

Internal Flow KE PE

Energy Energy+ + +

( )in out in out p avg c p p avg cE E E m h mh mc T V A c T c V A T∆ ∆ ρ ∆ ρ ∆= − = − = = =� � � � � �

equation is the same whether

the flow is to or out of the C.V.

work required to push the mass

into or out of the control volume

control volume

z

Vm, P, V

L

F PA=

A

Pm

control volume

[ ]2 2

2 2

2

2

V m mg z m

2 s s kJke + pe ,

kgm

s1000kJ

kg

+ ⋅

=

5-4 STEADY FLOW DEVICES

net ,in net ,out

Q W−� � ( )2 2

2 1

2 1 2 1

V V m h h g z z

2

−= ⋅ − + + ⋅ −�

Nozzle ( )1 2A A> Usually: Q 0=� , W 0=� , pe 0∆ = , ke 0∆ ≠

−

= − +2 2

2 1

2 1

V V0 h h

2

Diffuser ( )1 2A A<

Compressor (pump, fan) ( )1 2P P< Usually: Q 0=� , W 0≠� , pe 0∆ = , ke 0∆ =

W− � ( )2 1 m h h= −�

1 2w h h= −

W

w m

=�

�

Turbine ( )1 2P P> Usually: Q 0=� , W 0≠� , pe 0∆ = , ( )ke 0 but small∆ ≠

W− � ( )2 1 m h h= −� 1 2

w h h= −

increases velocity

at expence of pressure

increases pressure

by reduction of velocity

m�m�

W�

1 2P P<

Q� ( )if cooled

Compressor is used

to increase the

pressure of a fluid.

Requires power input.

Turbines convert energy

of flowing fluid into work.

Produce power output.

m�

m�

W�

1 2P P>

1V

2 1V V>1z 2z

( )

sat sat

Inlet: steam A-6

Outlet: sat. liq.-vapor mix. @ T and P

2 2 2

2 1

2 2

2

kJ

V V m 1 kg

2 1000s m

s

− ⋅

T

v

1P 4MPa=

2P 1MPa=

o

1T 500 C=

inlet

exit

1v .07=

o

2T 400 C=

2v .31=

1h 3435=

2h 3260=

steam

For steam turbine:

Ideal Gas

( )−

− ≈ − =2 2

1 2

p 2 1 2 1

V Vc T T h h

2

1 1 1P v RT=

2 2 2P v RT=

1 p 1h c T=

2 p 2h c T=

Energy balance for

steady state single flow

m�m�

1 2V << V

1 2P > P

m�m�

1 2V >> V

1 2P < P

T

v

1P 2000kPa=

2P 15kPa=

o

1T 500 C=

inlet

exit

1v

o

2T 54 C=

2v

T

v

2P 100kPa=

1P 90kPa=

o

2T C=

inlet

exit

o

1T 10 C=

air

1 1 2 2

1 2

V A V Am

v v= =�

1 1 2 2

1 2

V A V Am

v v= =�

1 1 2 2

1 2

V A V Am

v v= =�

1 v 1u c T=

2 v 2u c T=

1 2

often it is assumed

that V V 0≈�

5-5 ANALYSIS OF UNSTEADY-FLOW PROCESS

MASS BALANCE

CV

m∆ ( )2 1 CV m m= −

in out m m= −∑ ∑

ENERGY BALANCE

CV 2 1 2 2 1 1

E U U m u m u ∆ = − = − net ,in net ,out

Q W− ( )e e i i 2 2 1 1

out in

m h m h m u m u = − + −∑ ∑

SINGLE STREAM

2 1

m m− i e

m m= −

net ,in net ,out

Q W− ( )e e i i 2 2 1 1 m h m h m u m u = − + −

Example 5-12 Mass balance:

2 1

m m−0

CV

i e

m m= −0

⇒ 2 i

m m=

Energy balance:

in

Qout

W− e e

m h=i i 2 2 1 1

m h m u m u− + −( ) ⇒ 2 i

u h=

A-6 ⇒ i

kJh 3051.6

kg

=

⇒

2 i

kJu h 3051.6

kg

= =

A-5 ⇒ g @ P 1MPa 2

u 2583 u 3051.6 =

= < = ⇒ sup .vapor

A-6 ⇒ 2

o

@ P 1MPa and uT 456 C= =

2

3

2@ P 1MPa and u

mv 0.33

kg=

=

Find 2 2T ,m [ ]2 2

m V v 1.0 0.33 3 kg= = ≈

change in

the mass of

the system

final

state

initial

state

total mass

entering

the system

total mass

leaving

the system

The net mass transfer to the Control Volume during the process 1 2→

i e

e ,1 e ,2

e

Assumption of uniform-flow process:

properties at the inlets and outlets

h and h

are constant during the process:

h h averaged value h

2

+=

( )in other e e i i 2 2 1 1Q W m h m h m h m h − = − + −

Change in total energy:

( ) ( )= + = + − = + −other b other 2 1 2 2 1 1

If work includes the boundary work at constant pressure

W W W W P V V W P m v m v

iP 1MPa=

steam

W 0=

insulated

rigid tank

Q 0=

o

iT 300 C=

2P 1MPa=

1P 0=

3V 1 m=

outm

inm

Q

W

CVE∆

ih

eh

emim

Q

W

( )for ke pe 0∆ ∆= =

( )5-46

Charging of a

rigid tank by steam

inm

6-2 THERMAL ENERGY RESERVOIRS

6-3 HEAT ENGINES

Heat engine is a device, which converts heat into work

1. Heat is received from a source at H

T

2. Part of heat is converted to work. 3. Remaining heat is rejected to a sink at

LT

4. Heat engine operates on a complete cycle. STEAM POWER PLANT The steam power plant is an external-combustion engine

net ,out out inW W W= − (6-1)

net ,in in out H LQ Q Q Q Q= − = −

net ,out net ,inW Q =

sysE 0∆ = closed system cycle

net ,out in outW Q Q= − (6-2)

net ,out in out

W Q Q= −� ��

THERMAL EFFICIENCY th

Net work output Thermal Efficiency

Total heat inputη = =

net ,out

th

in

W

Qη = in out out L

in in H

Q Q Q Q 1 1

Q Q Q

−= = − = −

net ,out

th

in

W

Qη =

�

� in out out L

in in H

Q Q Q Q 1 1

Q Q Q

−= = − = −� � � �

� � �

outQ

SOURCE

SINKin

Q

( )p

Energy reservoir is a hypothetical body

of large heat capacitance mc 1

that can supply (source) or absorb (sink)

finite amounts of heat without any change of temperature

>>

turbine

boiler

condensor

pump

inQ

outW

outQ

inW

H

T

LT

furnace

surroundings

working

fluid

system boundary

waste energy

H L outQ Q W= +

LQ

net ,outW

L

T

H

T

Heat

Engine net,outW

HQ

LQ

energy balance for HE

THE SECOND LAW OF THERMODYNAMICS

Kelvin – Planck Statement: It is impossible for any device that operates on a cycle

to receive heat from a single reservoir and

produce a net amount of work

For heat engine to operate, the working fluid has

to exchange heat with heat sink as well with the heat source.

If L

Q 0= , then L th

H

Q1 1

Qη = − = , therefore, the 2nd Law claims that

no heat engine can be 100% efficient:

Clausius Statement: No device can operate on a cycle and produce effect

that is solely the heat transfer from

a lower-temperature body to a higher-temperature body

There are devices that can transfer heat

from lower-temperature reservoirs to higher-temperature reservoirs

but they have also to consume some energy Win

Equivalence of two statements:

If some device

violates one statement,

it also violates

the other statement,

and vice versa.

H

T

Heat

Engine net,outW

HQ

th < 1η

2) Attach heat engine to refrigerator 3) Violation of Clausius statement1) Assume that Kelvin-Planck statement is violated

L

T

H

T

H outQ =W

HQ

th

Heat

Engine

1η =

L

T

H

T

H LQ Q+

LQ

H inQ =W

HQ

th

Heat

Engine

1η =

Refrigerator

L

T

H

T

LQ

LQ

combine in a single device

Lord Kelvin

(1824-1907)

Max Planck

(1858-1947)

Rudolf Clausius

(1822-1888)

nd

Violation

of the 2 Law:

nd

Violation

of the 2 Law:

L

T

H

T

HQ

LQ

6-4 REFRIGERATOR Refrigerator transfers heat Objective: remove heat from a refrigeration space

from a low-temperature medium to a higher temperature medium

L L

R

Hin H L

L

Q Q 1COP = = =

QW Q Q1

Q

−−

L L

R

in H L H

L

Q Q 1COP = = =

W Q Q Q1

Q

−−

� �

� � � �

�

HEAT PUMP Heat pump transfers heat Objective: supply heat to a living space

from a low-temperature medium to a higher temperature medium

H H

HP

Lin H L

H

Q Q 1COP = = =

QW Q Q1

Q

−−

H H

HP

in H L L

H

Q Q 1COP = = =

W Q Q Q1

Q

−−

� �

� � � �

�

COEFFICIENT OF PERFORMANCE

6-5 PERPETUAL–MOTION MACHINES

PMM1 violates the 1st law of thermodynamics

PMM2 violates the 2nd law of thermodynamic

H L inQ Q W= +

LQ

inW

Q

LQ

inW

H L inQ Q W= +

compressor

condensor

evaporator

throttle

HQ

inW

LQ

HT

living space

LT

cold reservoir

Desired

Output

Vapor-compression

refrigeration

cycle

Desired outputCOP =

Required input

compressor

condensor

evaporator

throttle

HQ

inW

LQ

HT

LT

surroundings

refregerated space

working

fluid

refrigerant

=

Desired

Output

Vapor-compression

refrigeration

cycle

working

fluid

refrigerant

=

HEAT

Planck, p.86

§112. A process which can in no way be completely reversed is termed irreversible, all other processes reversible. That a process may be irreversible, it is not sufficient that it cannot be directly reversed. This is the case with many mechanical processes which are not irreversible (cf. 113). The full requirement is, that it be impossible, even with the assistance of all agents in nature, to restore everywhere the exact initial state when the process has once taken place. The propositions of the three preceding paragraphs, therefore, declare, that the generation of heat by friction, the expansion of a gas without the performance of external work and the absorption of external heat, the conduction of heat, etc., are irreversible processes. §115. Since there exists in nature no process entirely free from friction or heat-conduction, all processes which actually take place in nature, if the second law be correct, are in reality irreversible ; reversible processes form only an ideal limiting case. §116. The second fundamental principle of thermodynamics being, like the first, an empirical law, we can speak of its proof only in so far as its total purport may be deduced from a single self-evident proposition.

We, therefore, put forward the following proposition as being given directly by experience : It is impossible to construct an engine which will work in a complete cycle, and produce no effect except the raising of a weight and the cooling of a heat-reservoir. Such an engine could be used simultaneously as a motor and a refrigerator without any waste of energy or material, and would in any case be the most profitable engine ever made. It would, it is true, not be equivalent to perpetual motion, for it does not produce work from nothing, but from the heat, which it draws from the reservoir. It would not, therefore, like perpetual motion, contradict the principle of energy, but would, nevertheless, possess for man the essential advantage of perpetual motion, the supply of work without cost ; for the inexhaustible supply of heat in the earth, in the atmosphere, and in the sea, would, like the oxygen of the atmosphere, be at everybody's immediate disposal.

6-7 THE CARNOT CYCLE

6-8 CARNOT PRINCIPLES Efficiency of two Heat Engines operating between the same two reservoirs at L

T and H

T

C1

irreversible reversible η η<

C2 reversible 1 reversible 2

η η=

6-9 THE THERMODYNAMIC TEMPERATURE SCALE For reversible heat engine operating between L

T and H

T :

L L

H H

Q T

Q T= Kelvin Scale of Absolute Temperature

6-10 THE CARNOT HEAT ENGINE The efficiency of heat engine operating on a reversible Carnot cycle between L

T and H

T :

L L

C

H H

Q T 1 1

Q Tη = − = −

6-11 THE CARNOT HEAT PUMP Carnot Heat Pump: Carnot Refrigerator:

CHP

L L

H H

1 1COP

Q T1 1

Q T

= =

− −

CR

H H

L L

1 1COP

Q T1 1

Q T

= =

− −

H

H

Isothermal expansion

at T

heat Q is added

3

2

insulated

1

4

insulated

4

3

LQ

L

L

Isothermal compression

at T

heat Q is rejectedL

Adiabatic expansion

Q=0

Temperature dropped to TH

Adiabatic compression

Q=0

Temperature raised to T

The Carnot Cycle consists

of 4 reversible processes

IR

IR Rη η<

HT

LT

R1 R2

R1 R2η η=

(Refrigerator, Heat Pump)The Reversed Carnot Cycle

P

HT

V

net ,outW

1V

1

2

3

4

4V

2V

3V

LT

P

HT

V

net ,inW

1V

1

2

3

4

2V

4V

3V

LT

(Heat Engine)The Carnot Cycle

HQ

LQ

LT

HT

HQ

LQ

LT

HT

C

L H

is the highest possible efficiency

of the heat engine operating between

two reservoirs at T and T

η

gas

continues

to expand

gas

1V

2V

2V

3V

3V

4V

4V

1V

nd

Violation of Carnot Principles

yields violation of the

2 Law of Thermodynamics

Carnot

Efficiency

Reversible Processes

1

2

HQ

Q 0=Q 0= Q 0=Q 0=

the paths coinside•

at thermal equilibrium•

heat transfer is due to infinitely

small difference in temperature

•

1

2

Q1

2

-Q

-W W

Sadi Carnot (1822-1888)

Irreversible Processes

heat transfer is due to finite

difference in temperature

•

T 0∆ >

heat generation by friction•

INSTITUT NATIONAL DES SCIENCES APPLIQUEES, LYON

In statistical thermodynamics

entropy is treated as a measure

of molecular disorder

2nd

LAW OF THERMODYNAMICS – ramification

Q

0T

δ≤∫�

rev

Q0

T

δ =

∫�

7-1 ENTROPY Clausius: “I propose to name the quantity S the entropy of the system, after the Greek word τροπη [trope], the transformation”

dS rev

Q

T

δ =

definition of the differential of entropy

S∆ 2 1

S S= − 2

rev1

Q

T

δ =

∫ change of entropy during rev. process

S kJ

K

total entropy

S

sm

= kJ

kg K

⋅ specific entropy (entropy)

7-2 ENTROPY CHANGE AND ENTROPY GENERATION

Q

T

δ∫�

2

1

Q

T

δ= ∫

1

rev2

Q

T

δ +

∫

2

1

Q

T

δ= ∫

1

2

dS+ ∫ 0≤

2

1

Q

T

δ∫ 1 2

S S+ − 0≤

2 1

S S− 2

1

Q

T

δ≥ ∫

INCREASE OF ENTROPY PRINCIPLE

2 1

S S− 2

gen

1

Q S

T

δ= +∫

genS 0≥

isolated

S 0∆ > ( )isolated revS 0∆ =

Clausius Inequality

Carnot

Cycle

impossible

The Universe is Entropy (disorder)

an isolated of the Universe

system is increasing

⇒

gen

Entropy change S

can be negative,

but entropy generation

S cannot:

∆

( )

Entropy of irreversible

isolated system Q 0

is

during a process

=

increasing

entropy transfer with heat entropy generationentropy change

for all

cycles

for any

reversible

cycle

entropy is an extansive property:

1 2 3S S S S∆ ∆ ∆ ∆= + +

1S∆

2S∆

3S∆

T

reversible process

S

= ∫2

1

Q TdS

1S

2S

T

reversible process

dSS

Q TdSδ =

1S

2S

T

Kelvin-Planck Clausius

2 4

H L L H L

Carnot H L L L H1 3

Q Q Q Q TQ Q Q 1 0

T T T T T T Q T

δ δ δ = + = − = − =

∫ ∫ ∫�

HT

Q

LT

W

HT

Q

LT

Q

impossible

( )7-9 ( )7-11

( )7-7

( )7-10

( )7-4

Arbitrary process

Reversible1

2

lnS k p=

7-5 T-S DIAGRAM

Internally reversible process dS int rev

Q

T

δ =

( )7-4 ⇒ Qδ TdS= ( )7-14

Q 2

1

TdS= ∫ ( )7-15

q 2

1

Tds= ∫ ( )7-17

Isothermal Process

Q ( )0 2 1T S S= ⋅ − ( )7-18

q ( )0 2 1T s s= ⋅ − ( )7-19

7-4 Isentropic process ( S const= )

Example 7-6 Carnot Cycle

net ,inQ

net ,out W=

net ,out

W H L

Q Q= −

h-s diagram (Mollier diagram, p.345) is useful for analysis of steady-flow devices

T

S

Area Q=

1S

2S

1

2

Internally

reversible

process

Heat transfer during

the internally

reversible process

reversible adiabatic

isentropic process

�

isentropic process

is associated with

internally reversible

adiabatic process

w h∆=For adiabatic steady-flow devices:

nozzles, turbines, compressors, etc.

Richard Mollier (1863 1935)−

For internally

reversible process

T

S

Q 0=

1 2S S=

1

2

T

S

net ,out H LW Q Q= −

1 4S S=

2 3S S=

1 2

HT

4 3L

T

HQ

LQ

T

S

Q

1S

2S

1 2

isothermal

process is

internally

reversible0

T

q T s∆= For isothermal processes

h

s

s∆

1s

2s

1

2

h∆measure

of work

output

measure of the

irreversibilities

1h

2h

7-7 GIBBS EQUATIONS – THE Tds RELATIONS

W PdVδ = boundary work

Q Wδ δ− dU= Q TdSδ = for reversible process

⇓

TdS dU PdV= +

Tds du Pdv= +

ds 1 P

du dvT T

= +

Tds dh vdP= −

ds 1 v

dh dPT T

= −

ENTROPY CHANGE ISENTROPIC PROCESS, ∆S=0

7-8 LIQUIDS AND SOLIDS v p

c c c= = , dv 0= (incompressible substance)

ds 1

du T

=dT

c T

=

2 1

s s− 2

avg

1

T c ln

T=

2 1T T=

7-9 IDEAL GAS Pv RT= , v

du c dT= , p

dh c dT= Isentropic process

ds v

dT dv c R

T v= + 1st Gibbs Eq

ds p

dT dP c R

T P= − 2nd Gibbs Eq

Constant specific heats: v ,avgc , p,avg

c at 1 2

avg

T TT

2

+=

2 1

s s− 2 2

v ,avg

1 1

T v c ln R ln

T v= + ( )7-33

k 1

2 1

1 2

T v

T v

−

=

2 1

s s− 2 2

p ,avg

1 1

T P c ln R ln

T P= − ( )7-34

k 1

k2 2

1 1

T P

T P

−

=

k

2 1

1 2

P v

P v

=

2 1

s s− o o 2

2 1

1

P s s R ln

P= − − ( )7-39 o 0 2

2 1

1

Ps s R ln

P= +

( )T

o

p

0

dTs c T

T= ∫ , ( )

2

o o

2 1 p

1

dTs s c T

T− = ∫ 2 r 2

1 r1

P P

P P= 2 r 2

1 r1

v v

v v=

Start with closed system

energy balance

for reversible process

Gibbs equations do not depend

on the type of the process

(reversible and irreversible),

they are valid both for

closed and open systems (p.350),

because they include only properties

where, the temperature dependent part of entropy is defined by:

( )1 2 s s is an isothermal process :=Isentropic process

Isentropic process

( )1 2 s s=Isentropic process

( )−=

1 k kTP C

( )7 49− ( )7 50−

o

r r s ,P ,v

Table A-17

( )7-25st1 Gibbs Equation

( )7-26nd2 Gibbs Equation

= −p vR c c

=p

v

ck

c

= −v

Rk 1

c

Exact Analysis

Table A-2

Specific

heat ratio−

=k 1

Tv C

=k

Pv C

relative specific pressure and volume:

T

ss∆1

s2

s

1

2

1T

2T

1P

2P

v

1c R

k 1=

−

p

kc R

k 1=

−

=(polytropic process with n k )

T

s

1 2s s=

1

2

1T

2T

1P

2P

h u Pv

dh du Pdv vdP

= +

= + +

u h Pv

du dh Pdv vdP

= −

= − −

(1839-1903)

( )7-38

7-10 REVERSIBLE STEADY-FLOW WORK

rev revq w dh dke dpeδ δ− = + + Energy balance

revq Tds dh vdPδ = = − nd

2 Gibbs equation

rev

w vdP dke dpeδ = − − −

2

rev ,net ,out

1

w vdP ke pe∆ ∆= − − −∫ ( )7-51

= − ∫2

rev ,out

1

w vdP ( )7-52 turbine

= ∫2

rev ,in

1

w vdP compressor

rev ,out act ,outw w≥

rev ,in act ,inw w≤

7-11 Compression work for ideal gas (p.364)

( )

k 1

k1 2

comp ,in 2 1

1

kRT Pkw R T T 1

k 1 k 1 P

− = − = − − −

( )

n 1

n1 2

comp ,in 2 1

1

nRT Pnw R T T 1

n 1 n 1 P

− = − = − − −

2

comp,in

1

Pw RT ln

P=

Incompressible fluid ( v const= , 7-51) ⇒ ( )rev 2 1w v P P ke pe∆ ∆= − − − − ( )7-54

Bernoulli equation (steady flow of incompressible fluid through the simple pipe without friction)

( )2 2

2 1 2 1

2 1

P P V Vg z z 0

2ρ

− −+ + ⋅ − = ( )7-55

( )

Danila Bernoulov

1700 1782−

�

act act

rev rev

act rev act rev

rev act rev act

Tds

rev act act

gen

q w dh dke dpe

q w dh dke dpe

q q w w 0

w w q q

w w qds s 0

T T

δ δ

δ δ

δ δ δ δ

δ δ δ δ

δ δ δ

− = + +

− = + +

− − + =

− = −

−= − = ≥

pipe

inlet state

w 0=

outlet state

1 2

qδinlet state

wδ

outlet state

1 2

k

Isentropic process

Pv const=

Isothermal process

Pv const=

n

Polytropic process

Pv const=

−

=

1

kv cP

−

=

1

nv cP

1v cP

−=

P

v

1P

2P

( )isothermal n 1=

( )polytropic 1 n k< <

( )isentropic n k=

Flow through the pipe which involves no work interaction

Pv RT=Ideal gas

Reversible work output for

steady-flow and closed systems

( )for ke pe 0∆ ∆= =

Equations 7-57 a,b,c

outQ maximum

heat transfer

Q 0=

T const=

rev ,out act ,outw w≥

reversible

actual

turbine

compression

P

v1

v2

v

1

2

1P

2P

turbine

in derivation, we assume that both

processes are between the same states

turbine

compression

Steady-flow devices

deliver the most and

consume the least work

when the process is

reversible

≥rev ,net ,out act ,net ,out

w w

δ δ≥rev act

w w

Eq.7-9

minimum

heat transfercomp ,in

w

By cooling compressor, the requiered work input

can be minimized to achive the same pressure increase

temperature

is increased

P

v1

v2

v

1

2

1P

2P

rev ,outw

turbine

ISENTROPIC EFFICIENCIES η is a measure of the deviation of actual process from the corresponding isentropic process

Turbine

aT

s

wactual work

isentropic work wη = =

1 2aT

1 2s

h h

h hη −

=−

Compressor and pump

sC

a

wisentropic work

actual work wη = =

2s 1C

2a 1

h h

h hη −

=−

compressor

( )2 1

P2a 1

v P P

h hη

−=

− pump

Nozzle

= =2

2aN 2

2s

Vactual KE at exit

isentropic KE at exit Vη

2

2a1 2a

Vh h

2= +

1 2aN

1 2s

h h

h hη −

=−

h

s

1

1P

2P

isentropic

actual

1h

2ah

2sh

aw sw

h

s

1

2P

1P

isentropic

actual

1h

2ah

2sh

aw sw

h

s

1

1P

2P

isentropic

actual

1h

2ah

2sh

aw sw

s2

s2

s2

a2

a2

a2

7-13 ENTROPY BALANCE There is no conservation of entropy law, but we can established the entropy balance

2 1

S S− in out gen

S S S= − + kJ

K

2 1

s s− in out gen

s s s= − + kJ

kg K

⋅

system

dS

dt

in out gen S S S= − +� � �

kW

K

Entropy Transfer Entropy can be transferred with heat and with mass, and cannot be transferred with work Heat Transfer

heat ,in

S Q

T=

heatS

2

1

Q

T

δ= ∫ k

k k

Q

T=∑

heat ,out

SQ

T

−=

heatS� k

k k

Q

T=∑

�

Mass Flow

mass

S ms= s mδ= ∫ S t∆= �

mass

S i i e e

i e

m s m s= −∑ ∑

mass

S� ms= � ( )n cV A sρ=

CLOSED SYSTEMS ( ms 0= )

k

gen

k k

Q S

T+∑

2 1 S S= −

k

gen

k k

Q S

T+∑

�� cv

dS

dt=

OPEN SYSTEMS

k

i i e e gen

k i ek

Q m s m s + S

T+ −∑ ∑ ∑

2 1 S S= −

k

i i e e gen

k i ek

Q m s m s + S

T+ −∑ ∑ ∑

��� � cv

dS

dt=

kQ

kT

inlet

i im s

outlet

e em s

v

v

solids,liquids,

ideal gas:

Q mc T

Q mc T

∆

∆

=

=� �

1T

2T �

energy transfer which is not

accomponied by entropy transfer

is work

workS 0=

heat

QS

T

δ= ∫

massS ms=

surS∆

genSin

S

systemS∆

outS

by heat

by mass

by heat

by mass

vm,c

( ) ( )k

gen e i gen e i

k

QS m s s S m s s

T= − − = −∑

SF SS SF SS adiabatic

�� �� �

T Q

Q−

genS

kQ

i im s

kQ

cvS∆

kT

e em s

genS

i im s

kT

sysS∆

treated as an adiabatic system

gen sys surS = S + S system + surroundings ∆ ∆

�����������

gen sysadiabatic closed system S S∆=

T

s

system 2 1S S S∆ = −

1S 2

S

1

2

1T

2T

Q

Heat transfer at constant temperature T Heat transfer when T of a system is variable

1

2

Steady Flowk

i i e e gen

k i ek

Q m s m s + S

T+ −∑ ∑ ∑

��� � 0=

9-1 GAS POWER CYCLES – BASICS Power cycles (heat engines) produce work from heat.

Thermal efficiency: net net

th

in in

W w

Q qη = = L

th Carnot

H

T 1

Tη η≤ = −

Ideal power cycle: ● all processes are internally reversible

● no friction

● quasi-equilibrium expansion and compression

● no heat loss between devices

● negligible change in KE and PE

Air-standard cycle: ● working fluid is air (ideal gas)

● all processes are internally reversible

● combustion process is modeled by heat transfer H

Q from reservoir at H

T

● exhaust process is modeled by heat transfer L

Q to reservoir at L

T

● cold air assumption: v

c 0.718= , p

c 1.005= kJ

kg K

⋅ , k 1.4= , R 0.287=

Ideal air-standard cycle (internally reversible) is not, in general, a Carnot cycle (totally reversible)

9-2 CARNOT CYCLE ( )H H 2 1Q T S S= − ( )H H 2 1

q T s s= −

( )L L 3 4Q T S S= − ( ) ( )L L 3 4 L 2 1

q T s s T s s= − = −

Kelvin scale

H L L L

Carnot

H H H

Q Q Q T1 1

Q Q Tη

−= = − = −

( )

( )L 2 1L L L

Carnot

H H H 2 1 H

T s sQ q T1 1 1 1

Q q T s s Tη

−= − = − = − = −

−

( ) ( ) ( )L

net ,out Carnot H H 2 1 H L 2 1

H

Tw q 1 T s s T T s s

Tη

= ⋅ = − ⋅ ⋅ − = − ⋅ −

9-4 RECIPROCATING ENGINE (piston-cylinder device)

HT

T

S

net ,in H L net ,outQ Q Q W= − =

1 4S S=

2 3S S=

1 2

HT

4 3L

T

HQ

HT

LQ

net ,outW

actual

cycle

ideal

cycle HT

LT

LQ

HQ

net ,outW

( )Spark ignition engines SI

Internal combustion engines

Displacement volumemax min

V V−

Compression ratio

BDC max max

TDC min min

V V vr

V V v= = =

Mean Effective Pressure

net net

max min max min

W wMEP

V V v v= =

− −

net net

minmaxmax

max

w wMEP

1vv 1v 1

rv

= =

−−

( )net net

minmax

min

min

w wMEP

v r 1vv 1

v

= =−

−

( )Compression ignition engines CI

TDS

top dead center

bore

BDS

bottom dead center

stroke

clearance

volume

displacement

volume

MEP

P

minVmaxV

V

( )net max minW MEP V V= ⋅ −

netW

mean

effective

pressure

RECIPROCATING ENGINES – OVERVIEW

SI spark-ignition engines (combustion is initiated by a spark plug) CI compression-ignition engines (combustion is self-ignited as a result of compression)

( )Spark ignition engines SI

TDS

top

dead

center

bore

BDS

bottom

dead

center

stroke

clearance

volume

displacement

volume

MEP

P

minVmaxV

V

( )net max minW MEP V V= ⋅ −

netW

mean

effective

pressure

Internal combustion engines

Reciprocating engine (piston-cylinder device)

Displacement volumemax min

V V−

Compression ratio BDC max max

TDC min min

V V vr

V V v= = =

Mean Effective Pressure net net

max min max min

W wMEP

V V v v= =

− −

net net

minmaxmax

max

w wMEP

1vv 1v 1

rv

= =

−−

( )net net

minmax

min

min

w wMEP

v r 1vv 1

v

= =−

−

( )Compression ignition engines CI

9-5 OTTO CYCLE – SPARK-IGNITION ENGINES

Actual 4-stroke spark-ignition engine Ideal Otto cycle (air-standard cycle)

( ) ( )in out out inq q w w u∆− − − =

cycle

0= ⇒ net ,out in out

w q q= −

:→2 3 ( )in 3 2 v 3 2q u u c T T= − = −

:→4 1 ( )out 4 1 v 4 1q u u c T T= − = −

k 1k 1

1 2 3 4

k 1

2 1 4 3

T v 1 v T

T v r v T

−−

−

= = = =

⇒ 3 4

2 1

T T

T T=

k

k2 1

1 2

P vr

P v

= =

th ,Otto

η net out 4 1

in in 3 2

w q T T 1 1

q q T T

−= = − = −

−

4

11

2 3

2

T1

TT 1

T T1

T

−

= −

−

k 1

1 2

2 1

T v 1 1

T v

−

= − = −

k 11

1r

−

= −

1 k

1 r−

= −

( )Energy balance Otto cycle is executed in a closed system :

isentropic

compression

heat

addition

v const=

isentropic

expansion

heat

rejection

v const=

P

V

1

2

4

3

′4

compression

ignition

expansion

exhaust

intake

( )2 3 4 1Isentropic processes 1-2 and 3-4 with v v and v v := = under cold air assumption

( )heat transfer at v const w 0= =

( )Thermal efficiency of the ideal Otto cycle :under cold air assumption

Nikolaus Otto

(1832-1891)

Beau de Rochas

(1815-1893)http : // www.keveney.com / Engines.html

thermal efficiency

of Otto cycle under

cold air assumption

T

1 2s s=s

1

2 4

3

inq

outq

2 3 maxv v v= =

1 4 minv v v= =

3 4s s=

isentropic

expansion

isentropic

compression

p

v

ck

c=

Typical

compression

ratios for

gasoline

engines

argon

helium

air

2CO

actual engines

thermal efficiency

of Otto cycle under

cold air assumption:

1 k1 rη −

= −

compression

ratio41

2 3

vvr

v v= =

P

2 3v v= 1 4v v=

V

1

2 4

3

inq

outq

isentropic

expansion

isentropic

compression

ignition

compression

11

2 2 3

4 4

′4 ′4

( )

expansion

power

ignition

exhaust intake

9-6 CYCLE

in out 3 2q w u u− = −

in out 3 2

q w + u u = − ( )2 3 2 3 2 P v v + u u= − −

�����

3 2

h h = −

= −in 3 2

q h h ( )p 3 2 c T T= −

= −out 4 1

q u u ( )= −v 4 1 c T T

th

η ( )

( )v 4 1net out 4 1

in in p 3 2 3 2

c T Tw q T T1 1 1 1

q q c T T k T T

− −= = − = − = −

− −

4

11

2 3

2

T1

TT1 1

k T T1

T

−

= −

−

4k 1

12

1 3

2

T1

Tv1 1

k v T1

T

−

−

= −

−

( )( )

k

c

k 1

c

r 11 1 1

k r 1r−

−= −

−

( )( )

k1 kc

c

r 1r 1

k r 1

− −= − ⋅

−

k 1

1 2

k 1

2 1

T v 1

T v r

−

−

= =

41

k 1

2 3 c

TT 1

T T r−

=

k 1 k 1 k 1

4 3 3 c2

3 4 2 4

T v v rv

T v v v r

− − −

= = =

k 13 4

c

2 1

T Tr

T T

−=

3 32 2

2 3

P vP v

T T=

2 3P P=

⇒ 32

2 3

vv

T T= 3 3

c

2 2

T vr

T v= =

k 1 k 1 k4 3

c c c c

1 2

T Tr r r r

T T

− −= = =

T

1 2s s=s

1

2

4

3

inq

outq

P const=

minv v=

3 4s s=

Diesel cycle is executed in a closed system

Thermal efficiency of the ideal Diesel cycle ( ):cold -air - standard assumption

compression ratio1

2

vr

v=

3

c

2

vr

v= cutoff ratio

( )( )

k

c1 k

Diesel

c

r 11 1 r

k r 1η −

− = − ⋅ ⋅

−

Thermal efficiency of Diesel cycle

under :cold air assumption

Typical

compression

ratios for

gasoline

engines

efficiency of actual

diesel engines

efficiency of actual

gasoline engines

Otto Dieselη η>

heat transfer and boundary work at constant pressure:→2 3 :

heat transfer at constant volume:→4 1:

( )9-12

Isentropic process 1 2:→

Ideal gas

boundary work at constant pressure

Isentropic process 3 4:→

P

2v 1 4v v=

v

1

2

4

3inq

outq

kPV const

isentropic

expansion

=

isentropic

compression

of air only

2 3P P=

injection of fuel

and combustion

3v

modelling

of fresh

air intake

net,outw

net,inq

net ,ou net ,inw q=

Rudolf Diesel

( 1858 1913 )−

Constant pressure process 2 3:→

P v RT=

STIRLING CYCLE https://www.stirlingengine.com/

ERICSSON CYCLE

External combustion engine

( )

Reverend Robert Stirling

1790 1878 −

( )

John Ericsson

1803 1889 −

External combustion engine

T

4vs

1

2

4

3

inqoutq

2v

P

4s 3sv

1 2

4 3

inq

outq

P=const

HT

1s

regeneratorP=const

2s

regenerator

LT

T

1 4v v=s

1

2

4

3

inq

outq

HT const=

2 3v v=

P

4s 3sv

1 2

4 3

inq

outq

v=const

HT

1s

regeneratorv=const

2s

regenerator

LT const=

LT

L

th

H

T 1

Tη = −

Carnot

Thermal efficiency of Stirling cycles

Ericsson

-9 7

9-8 BRAYTON CYCLE

( )in 3 2 p 3 2q h h c T T= − = − ( )comp,in 2 1 p 2 1

w h h c T T= − = −

( )out 4 1 p 4 1q h h c T T= − = − ( )turb,out 3 4 p 3 4w h h c T T= − = −

net ,out in outw q q= −

3 2 4 1h h h h= − − +

( )3 4 2 1h h h h= − − −

turb ,out comp ,inw w= −

k 1k 1k 1 kk

3 32 2 kP

1 1 4 4

P TT Pr

T P P T

−−−

= = = =

32P

1 4

PPr

P P= = pressure ratio

4 3

1 2

T T

T T=

B

η net ,out

in

w

q= out

in

q1

q= − 4 1

3 2

T T1

T T

−= −

−

4

1

1 1

23

2

2

TT 1

T T1 1

TTT 1

T

−

= − = −

−

B

η k 1

kP

11

r

−= −

P

2v4v

v

1

2

4

3inq

outq

P const=

1v

s const=s const=

Gas-Turbine Engines

Open Cycle Closed Cycle

For isentropic processes 1-2 and 3-4:

( )Thermal efficiency of Brayton cycle :under cold -air assumption

( )

George Brayton

1830 1892−

⇓

T

1 2s s=s

1

2

4

3

inq

outq

P const=

P const=

3 4s s=

maxT

minT

The Ideal Bryton Cycle:

Steady-Flow

10-2 RANKINE CYCLE

turbinepump

inq

outw

outq

inw

boiler

condensor

1

2

4

3

Vapor power cycles working fluid is alternately vaporized and condensed−

1 2 Pump:→in 2 1w h h= − ( )in 1 2 1w v P P= −

11 f @Pv v=11 f @Ph h=

2 3 Boiler:→in 3 2q h h= −

3 4 Turbine:→out 4 3w h h− = −

4 1 Condenser:→out 1 4q h h− = −

2 1 inh h w= +

( )

net ,outnet ,inwq

in out out inq q w w 0− − − =

����������

Energy balance for a single stream steady-flow devices:

incompressible fluid

Ideal Rankine sycle:

T

s

inw

1 2s s=

= =2 3P P const

1

2

4

3inq

outq

3 4s s=

outw

= =1 4P P const

net out 4 1

th

in in 3 2

w q h h 1 1

q q h hη

−= = − = −

−

1 BtuHeat rate

kWh=

[ ]Amount of heat supplied in Btu

to generate 1 kWh of electricity

( )1820 1872−

Energy balance for a cycle:

Coefficient of thermal

efficiency used in US

[ ]

[ ]th

3412 Btu kWh

Heat rate Btu kWhη =

11 REFRIGERATION CYCLES

L L

R

Hin H L

L

q q 1COP = = =

qw q q1

q

−−

H H

HP

Lin H L

H

q q 1COP = = =

qw q q1

q

−−

11-2 REVERSED CARNOT CYCLE

L

R,Carnot

H Hin

L L

q 1 1COP = = =

q Tw1 1

q T− −

H

HP,Carnot

L Lin

H H

q 1 1COP = = =

q Tw1 1

q T− −

11-3 IDEAL REFRIGERATION CYCLE

1 4L

R

net,in 2 1

h hqCOP = =

w h h

−

−

2 3H

HP

net,in 2 1

h hqCOP = =

w h h

−

−

Refrigeration heat transfer from a lower temperature region to a higher temperature region−

Vapor-compression refrigeration cycle working fluid (refrigerant) is

vaporised and condensed alternately and is compressed in the vapor phase

−

H L inQ Q W= +

LQ

inW

objective:

remove heat from a refrigeration space:

−Refrigerator

objective:

supply heat to a living space:

−Heat pump

T

s

inw

s3 4s s=

P const=

1

2

4

3

L inq q=

outq

1 2s s=

P const=

s4

4s

1 2 Compressor:→in 2 1w h h= −

2 3 Condenser:→ out 2 3q h h= −

3 4 Throttle:→ 3 4h h=

4 1 Evaporator:→in 1 4q h h= −

P

h

in 2 1w h h= −

3 4h h=

1

2

4

3

in 1 4q h h= −

out 2 3q h h= −

1h 2h

T

s

inw

3 4s s=

P const=

1

2

4

3

LQ

HQ

1 2s s=

P const=

( ) ( )in out out in e iq q w w h h− − − = −SF SS

1

2

4

3

compressor

condensor

evaporator

throttle

Hq

inw

Lq

HT

LT

working

fluid refrigerant=

Vapor-compression

refrigeration

cycle

4