Thermodynamics Lecture Series

Thermodynamics Lecture Series

email: [email protected]://www5.uitm.edu.my/faculties/fsg/drjj1.html

Applied Sciences Education Research Group (ASERG)

Faculty of Applied SciencesUniversiti Teknologi MARA

Pure substances – Pure substances – Property tables and Property tables and Property Diagrams & Property Diagrams &

Ideal GasesIdeal Gases

Submitted-Post-assessmentSubmitted-Post-assessment

Unreadable Self-assessmentUnreadable Self-assessment

CHAPTER

2

Properties of Pure Substances- Part 2

Pure substance

Send self-assessments to:[email protected]@salam.uitm.edu.my

Send self-assessments to:[email protected]@salam.uitm.edu.my

QuotesQuotes

"Education is an admirable thing, but it is well to remember from time to time that nothing that is nothing that is worth knowing can be taughtworth knowing can be taught." - Oscar Wilde

"What we have to learn to do, we learn by doingwe learn by doing." -Aristotle "What we have to learn to do, we learn by doingwe learn by doing." -Aristotle

IntroductionIntroduction

1. Choose the right property table to read and to determine phase and other properties.

2. Derive and use the mathematical relation to determine values of properties in the wet-mix phase

3. Sketch property diagrams with respect to the saturation lines, representing phases, processes and properties of pure substances.

Objectives:Objectives:

IntroductionIntroduction

4. Use an interpolation technique to determine unknown values of properties in the superheated vapor region

5. State conditions for ideal gas behaviour

6. Write the equation of state for an ideal gas in many different ways depending on the units.

7. Use all mathematical relations and skills of reading the property table in problem-solving.

Objectives:Objectives:

Example: A steam power cycle.Example: A steam power cycle.

SteamTurbine

Mechanical Energyto Generator

Heat Exchanger

Cooling Water

Pump

Fuel

Air

CombustionProducts

System Boundaryfor ThermodynamicAnalysis

System Boundaryfor ThermodynamicAnalysis

Steam Power Plant

Phase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure Change

Water when pressure is reducedWater when pressure is reduced

2 =

f@

30 °C4.246

T = 30 C

P = 100 kPa

T = 30 C

P = 100 kPa

H2O:C. liquid

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

H2O:Sat.

liquid

100

P, kPa

, m3/kg

1

30 C

2 = f @ 30 °C

1 = f@ 30 °C

TTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPaTTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPa


H2O:Sat. Liq.

Sat. VaporSat. Vapor

3

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

H2O:Sat.

liquid


2 =

f@

30 °C

, m3/kg1

100

P, kPa

4.24630 C

3 = [f + x f g]@ 30 °C

2 = f@30 °C

1 = f@ 30 °C



4 =

g@

30 °C

H2O:Sat. Vapor

H2O:Sat. VaporH2O:

Sat. Liq.


T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa


1

3

P, kPa

, m3/kg2 =

f@

30 °C

100

4.24630 C 4 = g@ 30 °C

3 = [f + x f g]@ 30 °C

1 = f@ 30 °C

2 = f@ 30 °C



H2O:Sat. Vapor

H2O:Sat. Vapor

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

2

5

H2O:SuperVapor

H2O:SuperVapor

T = 30 C

P = 2 kPa

T = 30 C

P = 2 kPa


4 =

g@

30 °C

2 =

f@

100 kPa

, m3/kg1

3

2 =

f@

30 °C

1

3

100

P, kPa

4.24630 C

3 = [f + x f g]@ 30 °C

4 = g@ 30 °C

1 = f@ 30 °C

2 = f@ 30 °C

5= @2kPa, 30 °C



H2O:Sat. Vapor

H2O:Sat. Vapor

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

H2O:SuperVapor

H2O:SuperVapor

T = 30 C

P = 2 kPa

T = 30 C

P = 2 kPa

H2O:Sat. Liq.


T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

T = 30 C

P = 100 kPa

T = 30 C

P = 100 kPa

H2O:C. liquid

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

H2O:Sat.

liquid



Phase Change of Water- Pressure ChangePhase Change of Water- Pressure ChangePhase Change of Water- Pressure ChangePhase Change of Water- Pressure Change

Compressed liquidCompressed liquid: Good : Good estimation for properties estimation for properties by taking yby taking y = y = yf@T f@T where where

y can be either y can be either , u, h or , u, h or s.s.

2 =

f@

30 °C

4.246

3

2

5

4 =

g@

30 °C

, m3/kg

1

100

P, kPa

30 C 3 = [f + x f g]@ 30 °C

4 = g@ 30 °C

1 = f@ 30 °C

2 = f@ 30 °C

5= @2kPa, 30 °C


Phase Change of WaterPhase Change of WaterPhase Change of WaterPhase Change of Water

1,553.8

200 C

1.2276 10 C

g@100 C

P, C

, m3/kg

101.35 100 C

f@

100 C

Phase Change of WaterPhase Change of WaterPhase Change of WaterPhase Change of Water

P, C

, m3/kg

101.35

g@

100 C

1,553.8

1.2276

200 C

10 C

100 C

f@

100 C

22,090

P- diagram with respect to the saturation

lines


lines

T, C

, m3/kg

T – v diagram - Example

70

=f@70 C = 0.001023

50 kPa

P, kPa T, C

50 70

Phase, Y?

Compressed Liquid,

T < Tsat or P > Psat

, m3/kg

f@70 C

Tsat, C

81.33

Psat, kPa

31.19

81.3

3.2400.001030

P, kPa

, m3/kg

P – v diagram - Example

P, kPa T, C

50 70

Phase, Y?

Compressed Liquid,

P > Psat or T < Tsat

, m3/kg

f@70 C

70 C

31.19

5.0420.001023

Tsat, C

81.33

Psat, kPa

31.19

=f@70 C = 0.001023

50

400 C

P – v diagram - Example

P, kPa T, C

200 400


lines


lines

Phase, Why?

Sup. Vap., T >TSup. Vap., T >Tsatsat

Psat, kPa Tsat, C

NA 120.2120.2

P, kPa

, m3/kg

22,090.0

= 1.5493

, m3/kg

1.5493

200

f@200 kPa

= 0.001061

g@200 kPa

= 0.8857

120.2 C


T, C

, m3/kg

1,00

0 kP

a

P, kPa u, kJ/kg

1,000 2,000

T- diagram with respect to the saturation

lines


lines

Phase, Why?

Wet Mix., uWet Mix., uf f < u < u< u < ugg

Psat, kPa Tsat, C

179.9179.9

374.1

f@1,000 kPa

= 0.001127

179.9

g@1,000 kPa

= 0.19444

T, C

179.9179.9

= [f + x f g]@1,000 kPa

Property TableSaturated water – Temperature table

Temp

Tsat, C

10

50

100

200

300

374.14

Specific internal energy, kJ/kg

uf, kJ/kg ufg, kJ/kg ug, kJ/kg

42.00 2347.2 2389.2

209.32 2234.2 2443.5

418.94 2087.6 2506.5

850.65 1744.7 2593.3

1332.0 1231.0 2563.0

2029.6 0 2029.6

Specific volume, m3/kg

f, m3/kg g, m3/kg

0.001000 106.38

0.001012 12.03

0.001044 1.6729

0.001157 0.13736

0.001404 0.02167

0.003155 0.003155

Sat. P.

P, kPa

1.2276

12.34912.349

P, MPaP, MPa

0.102350.10235

1.55381.5538

8.5818.581

22.0922.09

Saturated Liquid-Vapor Mixture

H2O:Sat. Liq.


gf VVV t

g

m

mx

Given the pressure, P, then T = Tsat, yf < y <yg


Vapor Phase:, VVapor Phase:, Vgg, m, mgg, , gg, u, ugg, h, hgg

Liquid Phase:, VLiquid Phase:, Vff, m, mff, , ff, u, uff, h, hff

Mixture:, V, m, Mixture:, V, m, , u, h, x, u, h, x

gft mmm

ggffavgt mmm Specific volume of Specific volume of mixture?? Since mixture?? Since

V=m V=m

Specific volume of Specific volume of mixture?? Since mixture?? Since

V=m V=m

Mixture’s qualityMixture’s qualityMixture’s qualityMixture’s quality

More vapor, higher More vapor, higher qualityquality

x = 0 for saturated x = 0 for saturated liquidliquid

x = 1 for saturated x = 1 for saturated vaporvapor

More vapor, higher More vapor, higher qualityquality

x = 0 for saturated x = 0 for saturated liquidliquid

x = 1 for saturated x = 1 for saturated vaporvapor


H2O:Sat. Liq.


t

g

m

mx



Mixture:, V, m, Mixture:, V, m, , u, h, x, u, h, xMixture’s qualityMixture’s qualityMixture’s qualityMixture’s quality

Divide by Divide by total mass, total mass,

mmtt

Divide by Divide by total mass, total mass,

mmtt

gft

gavg x

m

m

1

ggffavgt mmm

fgfg wherewherewherewhere

fgf xx fgf x

ggfgt mmm




H2O:Sat. Liq.


t

g

m

mx



Mixture:, V, m, Mixture:, V, m, , u, h, x, u, h, xMixture’s qualityMixture’s qualityMixture’s qualityMixture’s quality



fgfg wherewherewherewherefgf x

fgfg yyy wherewherewherewherefgf xyyy

If x is known or has been determined, use If x is known or has been determined, use above relations to find other properties. If above relations to find other properties. If

either either , , u, h are known, use it to find quality, x. h are known, use it to find quality, x.

If x is known or has been determined, use If x is known or has been determined, use above relations to find other properties. If above relations to find other properties. If

either either , , u, h are known, use it to find quality, x. h are known, use it to find quality, x.

y can be y can be , u, h, u, hy can be y can be , u, h, u, h

Interpolation: Example – Refrigerant-134a

T, C

, m3/kg

TH

L

TL

H

T = ??

m1

m2

21 mslopemslope

LH

LH

L

L TTTT

LH

LLHL

TTTT

P, kPa , m3/kg

200 0.10600

Phase, Why?

Sup. Vap., Sup. Vap., > > gg

Psat, kPa Tsat, C

- -10.09-10.09

T, C

??

Assume properties are linearly dependent. Perform interpolation in superheated vapor phase.

Assume properties are linearly dependent. Perform interpolation in superheated vapor phase.


P, kPa , m3/kg

200 0.10600

Assume properties are linearly dependent.

Perform interpolatio

n in superheate

d vapor phase.

Assume properties are linearly dependent.

Perform interpolatio

n in superheate

d vapor phase.

Phase, Why?


Psat, kPa Tsat, C

- -10.09-10.09

T, C

??

T, C , m3/kg u, kJ/kg

TL= 0 L = 0.10438 uL = 229.23

0 < T0 < TLL < T < THH = 0.10600 0 < u0 < uLL < u < uHH

TH= 10 H = 0.10922 uH = 237.05

LH

LLHL

TTTT

LH

LLHL

uuuu


P, kPa , m3/kg

200 0.10600


lines


lines

Phase, Why?


Psat, kPa Tsat, C

- -10.09-10.09

P, kPa

, m3/kg

22,090.0

T, C

3.353.35

f@200 kPa

= 0.0007532

g@200 kPa

= 0.0993

200

-10.09 C

u, kJ/kg

231.85231.85

= 0.10600

T = 3.35 CPsat ??

Properties of Pure Substances- Ideal Gases Equation of State

Ideal Gases

Ideal Gases

• Equation of StateEquation of State– An equation relating pressure, temperature and An equation relating pressure, temperature and

specific volume of a substance.specific volume of a substance.

– Predicts P- Predicts P- -T behaviour quite accurately-T behaviour quite accurately

– Any properties relating to other propertiesAny properties relating to other properties

– Simplest EQOS of substance in gas phase is Simplest EQOS of substance in gas phase is ideal-gasideal-gas ((imaginary gasimaginary gas) equation of state) equation of state

Ideal Gases

• Equation of State for ideal gasEquation of State for ideal gas– Boyle’s Law: Pressure of gas is inversely Boyle’s Law: Pressure of gas is inversely

proportional to its specific volume P proportional to its specific volume P VP /1

• Equation of State for ideal gasEquation of State for ideal gas– Charles’s Law: At low pressure, volume is Charles’s Law: At low pressure, volume is

proportional to temperatureproportional to temperature TV

Ideal Gases

• Equation of State for ideal gasEquation of State for ideal gas– Combining Boyles and Charles laws:Combining Boyles and Charles laws: kgkJRTP / ,

M

RR u

and where and where RRuu is universal gas constant is universal gas constant RRuu = 8.314 kJ/kmol.K = 8.314 kJ/kmol.Kand where and where RRuu is universal gas constant is universal gas constant RRuu = 8.314 kJ/kmol.K = 8.314 kJ/kmol.K

and where M is molar massand where M is molar massand where M is molar massand where M is molar mass

RTm

VP So,So,So,So, kJmRTPV ,

TM

RMNPV uEQOSEQOS: : since the masssince the mass m = MNm = MN

where N iswhere N is number of moles:number of moles:EQOSEQOS: : since the masssince the mass m = MNm = MN where N iswhere N is number of moles:number of moles:

So,So,So,So, kJTNRPV u ,

where gas constant R iswhere gas constant R iswhere gas constant R iswhere gas constant R is

EQOSEQOS: : Since the total volume Since the total volume is V = mis V = m, so : , so : = V/m = V/m

EQOSEQOS: : Since the total volume Since the total volume is V = mis V = m, so : , so : = V/m = V/m

Ideal Gases

• Equation of State for ideal gasEquation of State for ideal gas– Real gases with low densities behaves like an ideal gasReal gases with low densities behaves like an ideal gas

kgkJRTP / ,

M

RR u

P << PP << Pcrcr, T >> T, T >> TcrcrP << PP << Pcrcr, T >> T, T >> Tcrcr

where,where,where,where,

kJmRTPV , kJTNRPV u ,

Hence real gases satisfying conditionsHence real gases satisfying conditionsHence real gases satisfying conditionsHence real gases satisfying conditions

RRuu = 8.314 kJ/kmol.K, = 8.314 kJ/kmol.K,RRuu = 8.314 kJ/kmol.K, = 8.314 kJ/kmol.K, V = m andV = m andm = MNm = MNm = MNm = MN

Obeys EQOSObeys EQOSObeys EQOSObeys EQOS

Ideal Gases

Gas Mixtures – Ideal GasesGas Mixtures – Ideal Gases

Low density (mass in 1 m3) gases

Molecules are further apart

Real gases satisfying condition

PPgasgas << P << Pcritcrit; T; Tgasgas >> T >> Tcrit crit , have low density and can be treated as ideal gases

High density

Low densityMolecules far apart

Ideal Gases


Equation of StateEquation of State - P--T behaviour

PP=RT=RT (energy contained by 1 kg mass) where is the specific volume in m3/kg, RR is gas constant, kJ/kgK, TT is absolute temp in Kelvin.

High density

Low densityMolecules far apart

Ideal Gases



PP=RT=RT , since = V/m then,P(V/m)=RT. So,

PV =mRTPV =mRT, in kPam3=kJ.Total energy of a system.

Low density

High density

Ideal Gases



PV =mRTPV =mRT = NMRT = N(MR)THence, can also write PV = PV = NRNRuuTT whereNN is no of kilomoles, kmol,MM is molar mass in kg/kmole and

RRuu is universal gas constant; RRuu=MR=MR.

RRuu = 8.314 kJ/kmol = 8.314 kJ/kmolKK

Low density

High density


T, C

, m3/kg

1,00

0 kP

a

P, kPa u, kJ/kg

1,000 2,000


lines


lines

Phase, Why?

Wet Mix., uWet Mix., uf f < u < u< u < ugg

Psat, kPa Tsat, C

179.9179.9

374.1

f@1,000 kPa

= 0.001127

179.9

g@1,000 kPa

= 0.19444

T, C

179.9179.9

= [f + x f g]@1,000 kPa

Thermodynamics Lecture Series

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