basic concept thermodynamics

8/9/2019 Basic Concept Thermodynamics

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Chapter 1: Basic Concepts

Introduction to Thermodynamics

Thermodynamic System and Control Volume

Properties, Processes and Cycles Units

Specific Volume and Density

Pressure Energy and Temperature

1


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Control Mass and Control Volume

Control Mass: Control Volume:

Encloses a fixed amount of mass Mass can enter or leave

Notice volume may change

Control surface

Control surface

A

B

2


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Complete system: Power Plant

Water

Loop

Heat

Loop

(Condenser)

3

Conversion of fuel energy into electricity and heat


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Complete system: Refrigerator

Refrigerant flows

in a loop through

4 devices

4

Refrigerator pushes energy out of a cold space into a warmer space (room)


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Properties, Processes and Cycles

Phases: Solid, Liquid and Vapor (gas)

State: a specific condition expressed by a unique set of property valueslike P, Tand density.

Two independent properties are needed to specify a state

Intensive properties are independent of mass (P, T,)

Extensive properties depend on mass (Mass, Volume, Energy)

Process: A change of a substance beginning at state 1 to a final state 2through a continuous variation of the state.

Example: Heat a cup of cold water to be warm water

Example: Compress air in a cylinder to a smaller volume (bike pump)

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Properties, Processes and Cycles, Continued

Process path or type: A specific succession of states is usually given aname that relates to the condition under which this happens. This is givenby the way the device behaves as a process equation or device equation.

Example: The heating of water in a cup takes place at atmosphericpressure so it is called an isobaricprocess, pressure is constant.

Example: Heating of air in a constant volume container is called anisochoricprocess, volume is constant.

Example: Air is compressed in a piston cylinder while it is maintainedat a constant temperature called an isothermalprocess.

Example: Air is compressed in a well insulated piston cylinder so it

does not have any heating or cooling called an adiabaticprocess.

The type of process is often dictated by the device behavior and couldalso be called a device equation.

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Properties, Processes and Cycles, Continued

Cycle: A process path that ends in the initial state. It can be a complexprocess or a combination of several simple processes.

Example: Water circulating in a steam power plant

Example: Refrigerant (like R-134a, R-410A or R-12) circulating in a

refrigerator or air-conditioner. Example: Air going through 4 sequential processes in a piston cylinder

similar to what happens in a car engine. This is repeated in time.

Cycles do not (net) change the working substance which keeps looping in

space (flow) or in time. However during the cycle the outside world

exchanges energy with the working substance at different conditions

during the cycle so the net effect is an energy conversion process.

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UNITS

SI or English:

SI: mass (kg), length (m), time (s), force (N)

F = ma has units 1 N = (1 kg) (1 m/s2) = 1 kg m/s2

English: mass (lbm), length (ft), time (s), force (lbf)

F = mg has units 1 lbf = (1 lbm) (32.174 ft/s2) = 32.174 lbm ft/s2

Mole: A fixed number of particles (No = 6.022*1023 particles per mol)

distinguish between gram mole (mol) and kilomole (kmol).

m = n M where m is mass , n = number of moles, M is molecular mass

Units and Conversions: See Table A.1

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Pressure

Pressure is force per unit area (normal stress)

= lim0

With uniform P, piston at rest (F

= F

)

Fext

= P Acyl

Units for P

1 Pa = 1 N/m2

1 kPa = 1000 Pa = 1 kN/m2

1 bar = 105

Pa = 100 kPa = 0.1 MPa1 atm = 101.325 kPa = 14.696 lbf/in2

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Pressure

Study 1.1 A hydraulic lift

A hydraulic cylinder has a diameter of 7.5 cm. What should the

fluid pressure inside be to create a force of 6000 N on the rod if we

neglect the outside pressure?

Solution ____ _____

From the definition of pressure as force per unit area

F = P A or P =F

A

The cross sectional area is

A = r2= D2/4 =4

0.0752m2= 0.004418 m2

so the pressure becomes

P =F

A=

6000 N

0.004418 m2 = 1 358 080 Pa = 1358 kPa

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Manometer/Barometer

Manometer

Barometer

12

Manometer

Force acting down on cross section at B:

Column mass: m = AH

F = PBA = PoA + mg = PoA + AHg

PB= PA= Po+ P = Po+ gH

Pgauge= P = gH

Barometer

The top pressure is near zero (perfect vacuum)

Column: m = AHo held up by

F = (PatmPtop) A = mg = AHog

Patm= Ptop+ P gHo


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Manometer

Manometer

13

Example: Column height and pressure difference

Find P for a 10 m column of liquid water or air, assuming

densities of 1000 and 1 kg/m3respectively, use g = 10 m/s2.

Solution ____ ______

Pliq

= gH = 1000 kg/m310 m/s210 m = 100 kPa

Patm= gH = 1 kg/m310 m/s210 m = 0.1 kPa

We can neglect P in atmospheric air for most cases.


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Energy and Temperature

A diatomic molecule, O2, N2 .

2 modes of rotation ( x, z axis)1 mode of vibration (in y axis)

A H2O molecule, 3 modes of vibration,

3 modes of rotation (not shown)

Energy on the macroscalecan be written as

E = Internal + Kinetic + Potential = U + KE + PE

= mu + mV2+ mgZ

velocity and elevation are for the center of mass.The internal energy uhas similar terms on the microscale

u = uext molecule+ utranslation+ uint molecule

Term 1: potential energy from intermolecular forces fct. of distancestrong (solid, liquid) weak for a gas (distance is large)

Term 2: molecular kinetic energy

Term 3: depends on structure of molecule

uint molecule= upotential+ urotation+ uvibration+ uatoms

Each mode of energy possesses kT of energy in average, k isBoltzmans constant. Translation always has 3 modes (x, y, z).

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Temperature

TC TF

-40 -40-17.8 0

0 3210 50

20 6825 77

30 86

37.8 10050 122

70 158100 212

Temperature is a measure of hotnesswhich is a sensible level of average energy inatoms and molecules and photons. Energy (frequency) has a distribution that is

characteristic for a single temperature. Recall the average energy per energy mode(degree freedom) is kT. In practice we use a thermometer to measure T which rely on

some other property like vor , an ohmic resistance (thermistor) or a voltage from a bi-metal junction (thermocouple) all sensitive to T.

Relative T (has offset): Celcius oC or Fahrenheit F

Absolute T ( T 0): Kelvin K or Rankine R

Ice point at 1 atm: 0.01 oC, 32 F Steam point: 100 oC, 212 F

Very cold: 40 oC, 40 F (at this point TC= TF)

TK= TC+ 273.15

TR= TF+ 459.67 = 1.8 TK

Later we show the ideal gas temperature scale, chapter 2, and the thermodynamic

temperature scale, chapter 5, and prove they are the same as the absolute T above.

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Thermocouples, thermistors

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Chapter 2

Phase Boundaries and The P-v-TSurface

The Two Phase States

The Liquid and Solid States

The Superheated Vapor States The Ideal gas States

Compressibility Factor

Equations Of State (EOS)

Computerized Tables

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Phase Boundaries and The P-v-TSurface

Complete phase diagram (P, T) separates the solid (S),

the liquid (L) and the vapor (V) phases.

Sublimation line (S above, V below)

Fusion line (S left, L right)

Vaporization curve (L above, V below)

On each curve: P = Psat(T) or T = Tsat(P)

One point on each curve is a pair (Psat, Tsat)

State a:

Superheated vapor (T > Tsat for same P)

Expanded vapor (P < Psat for same T)

State b:

Compressed liquid (P > Psat for same T)

Subcooled liquid (T < Tsat for same P)

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Vaporization curve0

Critical Point and Triple Point data

Tc,oC Pc, MPa Ttp,

oC Ptp, kPa

Water 374 22 0.01 0.6113CO2 31 7.4 -56 520

O2 -118 5.1 -219 0.15

N2 -147 3.4 -210 12.5


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The P-v-T surfaces

20

Show the (T, v) combinations following constant P processes for water

Small vs are liquid

Large vs are vapor (gas)

Liquid and vapor regions

are separated by a 2-phase

region of mixtures with a

combination of liquid andvapor.

The borders are the

Saturated liquid line and

Saturated vapor line

They meet at critical point

vapor

liquid


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Th T bl f W t


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Table B.1.3 superheated vapor along a constant P curve: Tentry, v,

Table B.1.4 compressed liquid

Subheading P (Tsat)

The Tables for Water

22

TABLE B.1.3

Superheated

Vaporv u h s

Th T bl f W t


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23

The saturated solid-vapor table and the sublimation line

Table B.1.5 Gives listing of TentryPsat, vi, vig, vg, .


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24

Example 2.1

Determine the phase for each of the following water states using table B.1 and indicate

the relative position in theP-v, T-vandP-Tdiagrams.

a. 120oC, 500 kPa b. 120oC, 0.5 m3/kg

Solution ____ _____

a. Enter B.1.1: T = 120oC: Psat= 198.5 kPa < P = 500 kPa so compressed liquid.

Enter B.1.2: P = 500 kPa: Tsat

= 151.86oC > T = 120oC so subcooled liquid

b. Enter B.1.1: T = 120oC: vf= 0.00106 < v < vg= 0.891 86 m3/kg


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The Two Phase States

25

Consider a mixture of some liquid in equilibrium

with some vapor. For the combination we can

write:V = Vf + Vg= mfvf+ mgvg

m = mf+ mg

Average specific volume

v = V/m =mf

m

vf+mg

m

vg

v = (1x) vf+ x vg

Quality x mg/ m (mass fraction: 0 x 1)

Defining vfg= vgvf we can also writev = vf+ x vfg

v is the mass fraction weighted average of vfand vg

x = mass fraction of vapor, 1x = mass fraction of liquid


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The Superheated Vapor States

26

Example 2.3Determine the phase for each of the following states using the table in Appendix B and

indicate the relative position in theP-v, T-vandP-Tdiagrams.

a. Ammonia 30oC, 1000 kPa b. R-134a 200 kPa, 0.125 m3/kg

Solution ____________________ _________

a. Enter B.2.1: T = 30oC: Psat= 1167 kPa since P < Psat it is superheated vapor.

If entry with 1000 kPa: Tsat25oC and we see it is superheated about 5oC.


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Ideal Gas Law

A different scaling can be used

P v = R T

P V = m R T = n RT

Here nis the number of moles and Mis the molecular mass.

n = m / M

From ideal gas law: 1 mol occupies same Vat a given (P, T) regardless of which substance it is.

Compare a mass at state 1 with same mass at a state 2:

m R=P2V2

T2=

P1V1T1

Differentiate the ideal gas law with time for constant state (P, T ) to get:

27

P.V =

.m R T =

.n R

_T

d l


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Ideal Gas Law

Example 2.6

28

d l G


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Ideal Gas Law

When do you have an ideal gas?

For very large v,so molecular distances are large (very small force between molecules)

To get a sense look at a T-v diagram for water.If Tis higher you can allow smaller vs and still have an ideal gas.

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The Compressibility Factor

The deviation from the ideal gas law can be expressed by a factorZso

Pv= Z RT

For an ideal gasZ = 1 and for states approaching liquidZbecomes very small. For very high pressures Z can

actually be higher than 1. A chart for nitrogen gives a general shape of a compressibility chart.

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Equations of State

The P-v-T surface is approximated in an equation of state. Several of these are extensions of the ideal gas law

as

P=RT

v b

a

v2+ cbv + db2

Here aand bare scaled to the critical properties

a= aoPcx2; b= box ; x = RTc/Pc

cand dare model constants shown in Tabel. D.1 including a correction with the accentric factor,

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Work at a Moving Boundary of a Compressible Substance

32

Work from substance to piston

W=F dx = PA dx = P dV

1W2= 1

2

PdV = area

To evaluate: P= function of V

1W2= Function(1, 2, path)

For example from figure

dV < 0 so 1W2 is negative

Magnitudes:

1W2 A< 1W2 B< 1W2 C


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Work at a Moving Boundary of a Compressible Substance

33

Example 3.4 continued

P= C1+ C2V; C2= ks/A2

1W2= 1

2

PdV = area under the process curve

1W2= 12(P1+P2)(V2V1)

=1

2m (P1+P2)(v2v1)


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Work in a Polytropic Process

34

Polytropic process definition:

PVn= constant = C or Pvn= constant

Indicated work ( n 1):

1W2= 1

2

PdV = 1

2

C VndV =C

1 n(V1 n)2

1

= C1 n(V1 n2 V1 n1 )

=1

1 n(P2V

n

2V1 n2 P1V

n

1 V1 n1 )

=1

1 n(P2V2P1V1)

Indicated work ( n = 1):

1W2= 1

2

PdV = 1

2

C VdV = C ln V21

=P1V1lnV2

V1

k l


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35

Example 3.6

An ideal gas in a piston cylinder starts out with 200 kPa, 0.04 m3. Let us look at 4

different possible processes and find the work for each case.Solution __________________________________________________

a. Process: P= C, Heat the gas to V2= 0.1 m3

1W2=

1

2

PdV =PdV=P(V2V1)

= 200 kPa (0.10.04) m3= 12.0 kJ

b. Process: T= C, Heat the gas to V2= 0.1 m3

Ideal gas: PV= mRT= C so polytropic with n= 1

1W2= 1

2

PdV =P1V1ln

V2

V1

= 200 kPa 0.04 m3 ln0.1

0.04= 7.33 kJ

To do the process the mass (load) on piston varies

k i l i


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36


c. Process: Polytropic with n= 1.3, Increase volume to V2= 0.1 m

3

P2=P1(V1/ V2)

1.3= 200 kPa (0.04 / 0.1)

1.3= 60.77 kPa

1W2=1

1 n(P2V2P1V1)

=1

11.3(60.77 0.1200 0.04) kPa m3

= 6.41 kJ

d. Process: V= C, Cool the gas toP2= 100 kPa

1W2= 1

2

PdV = 0

Comment:

To do the process (c) the mass (load) on piston varies

To do the process (d) the piston is locked in place

Th I t l E


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The Internal Energy

37

Main characteristic:

u = uext molecule + utranslation + uint molecule

uint molecule= Potential + Rotation + Vibration + Atoms

uext moleculeis sensitive to molecular distances and all parts are sensitive to T.

For liquid and solid phases this first part is significant (small distances) and it

diminishes for the vapor states going to zero for ideal gas states.

Information is in the B section tables for substances as fct(T, P).

For a two phase mixture of liquid and vapor

U = Uliq+ Uvap= mliquf+ mvapug

u = U/m = (mliq

/m) uf+ (m

vap/m) u

g

u = (1x) uf+ x ug

u = uf + x ufg

The Internal Energy


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The Internal Energy

38

Example 3.8

Determine the missing property (P, T, v, x) for water at each of the following states:

a. T= 300oC, u = 2780 kJ/kg ; b.P= 200 kPa, u = 2000 kJ/kg

Solution __________________________________________________

a) Table B.1.1 at 300oC: u > ug= 2563.0 kJ/kg, so superheated vapor,

x is undefined,

Table B.1.3 between 1600 and 1800 kPa at 300

o

C:1600 kPa, 300oC: u = 2781.03 kJ/kg; v = 0.15862 m3/kg

1800 kPa, 300oC: u = 2776.83 kJ/kg; v = 0.14021 m3/kg

linear interpolation:

P = 1600 + 20027802781.03

2776.832781.03= 1600 + 200 0.24524

= 1648 kPa

v = 0.15862 + (0.140210.15862) 0.24524 = 0.1542 m3/kg

The Internal Energy


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The Internal Energy

39


Determine the missing property (P, T, v, x) for water at each of the following states:

a. T= 300oC, u = 2780 kJ/kg ; b.P= 200 kPa, u = 2000 kJ/kg

Solution __________________________________________________

b) Table B.1.2 at 2000 kPa:

uf= 906.4 < u < ug= 2600.3 kJ/kg two-phase

u = 2000 = uf + x ufEg= 906.4 + x 1693.8

x = 0.6456

v = vf + x vfg= 0.001 177 + 0.6456 0.098 45

= 0.064 74 m3/kg

The Enthalpy


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The Enthalpy

40

Example: The constant Pprocess

Recall the work for a constant pressure process

1W2= PdV =PdV =P (V2V1)Assume changes in kinetic and potential energies are zero, the

energy equation gives the heat transfer as

1Q2= m (u2u1) + 1W2= m (u2u1) +P (V2V1)= m (u2u1) +P m (v2v1)

= m [(u2+P2v2)(u1+P1v1)]

Definition of enthalpy

H U + PV ; hu + Pv

Enthalpy has same units as u[kJ/kg], and for a two-phase state

h = (1x) hf+ x hg= hf+ x hfg

as the mass fraction [1x and x] averaged property (same as for v, u).Comment: Any combination of thermodynamic properties is another

property. Enthalpy is used because it appears often.


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The Enthalpy


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The Enthalpy

42


Work term from area in (P-v) diagram

1W2=Pm (v2v1) = 400 kPa 0.5 kg (0.65480.2) m3/kg = 91 kJ

Heat transfer from energy equation

1Q2= m(u2u1) + 1W2= m[(u2+P2v2)(u1+P1v1)]

= m(h2h1) = 0.5 kg (3066.81524.7) kJ/kg = 771.1 kJ

Comment: We could have found (u1, u2) and not used the enthalpies, same result.

The Specific Heats


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The Specific Heats

43

Specific heats

Previous cases (KE= 0 and PE = 0 ) give for an incremental process per unit mass:

dv= 0 w= 0 so q = du

dP= 0 w=P dv so q = dh

We want to express this in terms of a property we can measure: T. Due to the above

processes we define the specific heats as:

Cv=

u

T v ; CP=

h

T P

Thermodynamic properties, a function of the state. They simplify to be fct(T) only, as an

approximation, in the single phase regions: solid, liquid and ideal gas.

Cv dudT ; CP dhdT No phase change

Then

du = CvdT and u2u1= CvdT Cv(T2T1)

dh = CPdT and h2h1= CPdT CP(T2T1)

where the last expression assumes constant specific heats.

The Specific Heats: Solids and Liquids


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The Specific Heats: Solids and Liquids

44

Specific heats for solids and liquids

Matter in these phases are dense and nearly incompressible so

v= constant and v is smallThen

dh = du + d(Pv) = du + P dv + v dP du + v dPso

h2h1= u2u1+ v dP = u2u1+ v (P2P1) Cv(T2T1) + v (P2P1)

If pressures are moderate then the last term can be neglected thus changes in uand hare

the same and the two specific heats are the same.

C= Cv= CP are listed for some solids and liquids in Tables A.3, A.4, F.2 and F.3

Liquids in [kJ/kg-K]: water (4.18) fuels CxHy (2 - 2.5) metals (0.151)Solids in [kJ/kg-K]: ice (2.04) granite (0.9) metals (0.4 typ.)

If temperature varies significantly or great accuracy is required then a model for the

variation of C= Cvwith temperature is required and integral must be done

u2u1= CvdT

The Specific Heats: Ideal Gases


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The Specific Heats: Ideal Gases

45

Specific heats for ideal gases

Large intermolecular distance gives small potential energy between molecules so only the

internal (like rotation, vibration) molecular energy terms for u are present.

u = uext molecule+ utranslation+ uint molecule utranslation+ uint molecule= u(T)

Pv = RT

so

h = u + Pv = u + RT = h(T)

Cv0=du

dT ; CP0=

dh

dT= Cv0+R

u2u1 = Cv0dT Cv0 (T2T1)

h2h1 = CP0dT CP0 (T2T1)Tables: A.7-9, F.5-6 A.6 A.5, F.4

Units for Cv0, CP0andR: kJ/kg-K or Btu/lbm-R

The zero subscript indicates ideal gas, any gas whenP 0 (v ) is an ideal gas.

The Specific Heat Function of Temperature


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46

Specific heat with temperature

Monatomic gases Ar, He, ... cannot

rotate or vibrate so specific heat is

constant. Only translation active so

specific heat also low.

The complex molecules have moremodes of internal energies possible.

See Appendix C for more details.

otice CP0/R is dimensionless



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47

Example 3.13

Calculate the change of enthalpy as oxygen is heated from 300 K to 1500 K. Assume

ideal gas behavior.

Solution ________________________________________

Table A.8: h2h1= 1540.2273.2 = 1267.0 kJ/kg

Table A.6: h2h1= CP0dT = 1000 CP0d

= 1000 [0.88120.0001 2+ 130.543140.33 4]=1.5=0.3= 1241.5 kJ/kg

Table A.5: h2h1= CP0(T2T1) = 0.922 kJ/kg-K (1500300) K = 1106.4 kJ/kg

The result goes from most accurate (A.8) to least accurate (A.5).

otice how the slope of the cord is close to CP0(Tavg= 900 K)

slope = (h2h1) / (T2T1) =1267 kJ/kg

1200 K= 1.056 kJ/kg-K

CP0 900 K= 0.880.0001 0.9 + 0.54 0.920.33 0.93

= 1.077 kJ/kg-K


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The Entropy


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The Entropy

49

Entropy for water, T-sdiagramEntropy of a pure substance.

Entropy is tabulated together with otherproperties as (T, P, v, u, h, s) in the B section

of appendix. Properties (uor h) andsare

zero at a reference pointthat depends on

the substance.

In the two-phase regionsrelates to the

saturated liquid-vapor values as (v, u, h)

s = (1x)sf+ x sg

= sf+ x sfg

The Entropy


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The Entropy

50

Enthalpy-Entropy for water,

h-sdiagram, Mollier diagram

T-sdiagram from CATT3 for water.

Showing P = 1.55 MPa & v = 0.13 m^3/kg

curves with the plot option

Aqua, red separated by P = Pcritical

The Entropy


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The Entropy

51

Entropy changes in reversible processes.

In general

ds =

q

T rev and qrev= T ds

otice:

1. Change dsfollows magnitude and sign of q

2. For a given q change dsis smaller for high T

3. For a reversible process we find heat transfer

qrev= T ds 1q2= 1

2

T ds = area

These points are illustrated in the figures.

The Entropy


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The Entropy

52

T-Sdiagram


From the definition of the entropy there are two processes

for which we can evaluate the integral without furtherdevelopments.

Reversible adiabaticprocess: q = 0 so

ds =

q

T rev= 0 s = constant

1q2= 1

2

T ds = area = 0

This is called an isentropicprocess.

Reversible isothermalprocess: T = constant so

s2s1=

1

2

q

T rev=

1q2

T

1q2=

1

2

T ds = area = T (s2s1)

The two processes and the

area below the process curveshows the heat transfer.

The Carnot Cycle in a T-s diagram


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The Carnot Cycle in a T-sdiagram

53

Carnot Heat Engine

T-Sdiagram

Carnot Heat Pump

Entropy changes in the Carnot Cycle.

The Carnot cycle consists of 2 adiabatic and 2 isothermal

processes so the cycle looks like a rectangle in a T-s

diagram as shown. Notice how the heat transfer is the area

below the process curve as in the heat engine

1q2= TH(s2s1) = qHand

3q4= TL(s4s3) =qLThe two isentropic processes (rev. and adiabatic) give

s2= s3 and s4= s1

so the entropy increase in 1-2 equals the decrease in the

34 process. The net work equals the net heat transfer sothe cycle efficiency is

=wnet

qH=

qHqLqH

=area 1-2-3-4-1

area 1-2-b-a-1

The reversed cycle is a refrigerator or heat pump and it has

opposite changes in entropy and all heat transfers.

The Entropy Change in Evaporation


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The Entropy Change in Evaporation

54

Constant P process


Consider a constant pressure process of boilingsaturated liquid to saturated vapor from state 1 to state 2

in the diagram. From the energy equation

1q2= u2u1+ 1w2= h2h1= hfg

The entropy change is

s2s1= sfg= 12 qT rev=

1q

2T

=h

fgT

relating the entropy of evaporationsfgto the enthalpy of

evaporation hfg. This is a consistency requirement for

the tables of thermodynamic properties.

sfg=hfg

T

Entropy Change for a Solid or Liquid


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Entropy Change for a Solid or Liquid

55

For a solidor liquidwe have the behavior for an incompressible substance:

v constant; valso small; du = C dT

ds =duT

+PT

dv duT

=CT

dT

If specific heat is constant we can integrate

s2s1=

1

2

C

TdT = C ln

T2

T1

Since vis constant uandsare functions of temperature only. Thus in an isentropic

process the temperature does not change, like pumping a liquid to higher pressure.

Example 6.3

Liquid water is heated from 20oC to 90oC. Find the change in entropy using constant

specific heat and by using the steam tables.

Solution _____________________________________________________

Table A.4: s2s1= C lnT2

T1= 4.18 kJ/kg-K ln

363.15

293.15= 0.8951 kJ/kg-K

Table B.1.1: s2

s

1= s

f 90

s

f 20= 1.1925

0.2966 = 0.8959 kJ/kg-K

Entropy Change for an Ideal Gas


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56

ow use the two Gibbs relations for an ideal gas

Ideal gas: Pv = RT ; du = Cv0dT ; dh = Cp0dT

Gibbs: T ds = du + P dv = Cv0dT +RTv

dv

ds =Cv0

TdT +

R

vdv

s2s1=

1

2

Cv0

T dT + R ln

v2

v1 (6.13)

Gibbs: T ds = dhv dP = Cp0dT RT

PdP

ds =Cp0

TdT

R

PdP

s2s1=

1

2

Cp0

TdT R ln

P2

P1 (6.15)

Conclusions:

Ideal gas: u(T), h(T) but s(T, P) or s(T, v)

Entropy increases with T, increases with v, but decreases withP

Model for specific heat (recall Cp0= Cv0+ R) needed to evaluate the integrals



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57

Constant specific heat model:

Cp0, Cv0 assumed constant and found in Table A.5, F.4

Integrals are done to give:

s2s1=

1

2

Cp0

TdT R ln

P2

P1 s2s1= Cp0ln

T2

T1R ln

P2

P1 (6.16)

s2

s1

=

1

2

Cv0

TdT + R ln

v2

v1 s

2

s1

= Cv0

lnT2

T1 + R ln

v2

v1 (6.17)

The expressions are alternatives and consistent. Use Cp0= Cv0+ R and ideal gas law to

mathematically transform from one to the other equation.

Variable simple specific heat model:

Cp0(T) = Fct(T) polynomial, exponential or power functions

Cv0(T) = Cp0(T)RT also a function of T only

Table A.6 shows an example of polynomials used for function Cp0(T).

Entropy Change using the Standard Entropy


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Entropy Change using the Standard Entropy

58

Variable complex specific heat model:

Standard entropytabulated: s(T, P0) = s

0

T

To

T

Cp0

T dT + s

0

ref (6.18)

s2s1= s0

T2s0

T1R lnP2

P1 (6.19)

Absolute entropy: s(T, P) = s0

TR lnP

P0

Function is shown for air in A.7, F.5 and other gases in A.8, F.6 with an offset to make

the standard entropy a certain value at T0. For other Tables s0

ref= s(T0, P0) may be set to

zero. Normally we need a difference so any offset will cancel out.

otice: s0T2s

0T1=

To

T2Cp0T

dT + s0ref

To

T1

Cp0T

dTs0ref

=

To

T2

Cp0

TdT +

T1

To

Cp0

TdT =

1

2

Cp0

TdT

Entropy Change Using Different Models


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py g g

59

Example 6.4

Find the change in specific entropy of oxygen when state changes from 300 K, 200 kPa

to 1500 K, 150 kPa.

Solution _____________________________________________________

The most accurate result is using standard entropy from A.8 corrected for pressure as

s2s1= s0

T2s0

T1R lnP2

P1= 8.06496.41680.2598 ln

150

200= 1.7228 kJ/kg-K

The next best evaluation is using the equation from Table A.6 to get

s2s1= CP0/T dT = CP0/d RlnP2

P1

= [0.88 ln0.0001 + 12

0.5421

30.33 3]

=1.5=0.3Rln

P2

P1

= 1.7058 kJ/kg-K

Constant specific heat evaluation becomes

s2s1= Cp0lnT2

T1R ln

P2

P1= 0.922 ln

1500

3000.2598 ln

150

200= 1.5586 kJ/kg-K

This last evaluation is too low since Cp0is taken at 300 K and it increases with T so in

Example 3.13 we found the average should be like Cp0= 1.06 kJ/kg-K.


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Process Equation Using Entropy


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q g py

61

The Isentropic Process continued

The relations can be extended with specific volume by the ideal gas lawPv=RT

P2

P1=

T2

T1

k/(k1)=

v1

v2

k

T2

T1=

P2

P1

(k1)/k=

v1

v2

k1

These relations are known as the power relationsand are valid for an isentropic processin an ideal gas with constant specific heats. If one of the ratios is known in an isentropic

compression or expansion (typically thePor vratio) the other ratios can be found.

The pressure to volume relation can be rewritten to recognize

P vk= constant

as a polytropicprocess with polytropic exponent n = k. The different processes are:

The polytropic processes: Pvn= constant

Isobaric: n = 0, P= constant Isentropic: n = k, s= constant

Isothermal: n = 1, T= constant Isochoric: n = , v= constant

Polytropic Process in P-v and T-s diagrams


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y p g

The polytropic processes: Pvn= constant

Isobaric: n = 0, P= constant Isentropic: n = k, s= constant

Isothermal: n = 1, T= constant Isochoric: n = , v= constant

basic concept thermodynamics

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