Thermodynamics Lecture Series

email: drjjlanita@hotmail.comhttp://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 DiagramsProperty Diagrams

QuotesQuotes

“You do not really understand something unless you can explain it to your grandmother.”

(Albert Einstein)

IntroductionIntroduction

Objectives:

1. State the meaning of pure substances

2. Provide examples of pure and non-pure substances.

3. Read the appropriate property table to determine phase and other properties.

4. Sketch property diagrams with respect to the saturation lines, representing phase and properties of pure substances.

FIGURE 1–5Some application areas of thermodynamics.

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1-1

Application

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

Steam Power PlantSteam Power Plant

FIGURE 1–17A control volume may involve fixed, moving, real, and imaginary boundaries.

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1-5

Open system devices

Open system devicesOpen system devices

Heat Exchanger

Throttle

CHAPTER

2

Properties of Pure Substances

Title:

Pure Substances

• Pure substancesPure substances– Substance with fixed chemical compositionSubstance with fixed chemical composition

• Can be single element: Such as, NCan be single element: Such as, N22, H, H22, O, O22

• Compound: Such as Water, HCompound: Such as Water, H22O, CO, C44HH1010,,

• Mixture such as Air, Mixture such as Air,

• 2-phase system such as H2-phase system such as H22O.O.

– Responsible for the receiving and removing dynamic Responsible for the receiving and removing dynamic energy (working fluid)energy (working fluid)

• Pure substancesPure substances– Substance with fixed chemical compositionSubstance with fixed chemical composition

• Can be single element: Such as, NCan be single element: Such as, N22, H, H22, O, O22

• Compound: Such as Water, HCompound: Such as Water, H22O, CO, C44HH1010,,

• Mixture such as Air, Mixture such as Air,

• 2-phase system such as H2-phase system such as H22O.O.

– Responsible for the receiving and removing dynamic Responsible for the receiving and removing dynamic energy (working fluid)energy (working fluid)

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

H2OSat.

liquid

Qin

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

Water interacts with thermal energyWater interacts with thermal energy

99.6

2 =

f@

100 kPa

T, C

30, m3/kg

1

H2O:C. liquid

P = 100 kPa

T = 30 C

P = 100 kPa

T = 30 C

Qin

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

H2OSat.

liquid

Qin

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

Water interacts with thermal energyWater interacts with thermal energy

H2O:Sat. Liq.

Sat. VaporSat. Vapor

Qin

99.6

2 =

f@

100 kPa

T, C

30

, m3/kg1

3

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

Water interacts with thermal energyWater interacts with thermal energy

4 =

g@

100 kPa

99.6

2 =

f@

100 kPa

T, C

30, m3/kg

1

3

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

H2O:Sat. Vapor

H2O:Sat. Vapor

Qin

H2O:Sat. Liq.

Sat. VaporSat. Vapor

Qin

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

Water interacts with thermal energyWater interacts with thermal energy

150

5

99.6

2 =

f@

100 kPa

T, C

30, m3/kg

1

4 =

g@

100 kPa

3

5 = @100 kPa, 150°C

3 = [f + x f g]@100 kPa

1 = f@T1

H2O:SuperVapor

H2O:SuperVapor

P = 100 kPa

T = 150 C

P = 100 kPa

T = 150 C

Qin

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

H2O:Sat. Vapor

H2O:Sat. Vapor

Qin

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

Water interacts with thermal energyWater interacts with thermal energy

H2O:Sat. Liq.

Sat. VaporSat. Vapor

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

Qin

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

H2O:Sat. Vapor

H2O:Sat. Vapor

Qin

P = 100 kPa

T = 150 C

P = 100 kPa

T = 150 C

H2O:SuperVapor

H2O:SuperVapor

Qin

P = 100 kPa

T = 30 C

P = 100 kPa

T = 30 C

H2O:C. liquid

Qin

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

H2OSat.

liquid

Qin

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

99.6

2 =

f@

100 kPa

T, C

30, m3/kg

1

4 =

g@

100 kPa

3

5 = @100 kPa, 150°C

3 = [f + x f g]@100 kPa

1 = f@T1

150

100

kPa

5

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.

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2-1

FIGURE 2-11T-v diagram for the heating process of water at constant pressure.

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

T, C

, m3/kg

99.6

f@

100 kPa

g@

100 kPa

100

kPa

179.9

45.8

10 k

Pa

1000

kPa

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2-2

FIGURE 2-16T-v diagram of constant-pressurephase-change processes of a puresubstance at various pressures(numerical values are for water).

99.6

45.8

179.9

T –v diagram: Multiple P

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2-3

FIGURE 2-18T-v diagram of a pure substance.

T –v diagram: Multiple P

T, C

, m3/kg

T – v diagram - Example

70

=f@70 C = 0.001023

81.3

3.240

50 kPa

P, kPa T, C

50 70

Psat, kPa Tsat, C

81.33

Phase, Y?

Compressed Liquid,

T < Tsat

, m3/kg

f@70 C

T – v diagram - Example

T, C

, m3/kgf@200 kPa

= 0.001061

200

kPa

P, kPa , m3/kg

200 1.5493

T- diagram with respect to the saturation

lines

T- diagram with respect to the saturation

lines

Phase, Why?

Sup. V., Sup. V., > >gg

Psat, kPa Tsat, C

120.2120.2

374.1

400

= 1.5493

120.23

g@200 kPa

= 0.8857

T, C

400400

T – v diagram - Example

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

T- diagram with respect to the saturation

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 – Pressure table

Pressure

P, kPa

10

50

P, MPaP, MPa

0.100

1.00

10

22.09

Specific internal energy, kJ/kg

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

191.82 2246.1 2437.9

340.44 2143.4 2483.9

417.36 2088.7 2506.1

761.68 1822.0 2583.6

1393.04 1151.4 2544.4

2029.6 0 2029.6

Specific volume, m3/kg

f, m3/kg g, m3/kg

0.001010 14.67

0.001030 3.240

0.001043 1.6940

0.001127 0.19444

0.001452 0.018026

0.003155 0.003155

Sat. temp.

Tsat, C

45.81

81.3381.33

99.6399.63

179.91179.91

311.06311.06

374.14374.14