pure substance part1 - drjj - uitm

Thermodynamics Lecture Series

Pure substances Pure substances –– Property Property tables and Property Diagramstables and Property Diagrams

Applied Sciences Education Research Group (ASERG)Faculty of Applied SciencesUniversiti Teknologi MARA

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

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.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

FIGURE 1–5Some application areas of thermodynamics.

Application

1-1

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

SteamTurbine Mechanical Energy

to 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.


1-5

Open system devices

Open system devicesOpen system devices

ThrottleHeat Exchanger

CHAPTER

2

Properties of Pure Substances

Title:

Pure Substances

• Pure substances– Substance with fixed chemical composition

• Can be single element: Such as, N2, H2, O2

• Compound: Such as Water, H2O, C4H10,• Mixture such as Air, • 2-phase system such as H2O.

– Responsible for the receiving and removing dynamic 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, •• 22--phase system such as Hphase 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 Water

99.6ν

2 = ν

f@100 kPa

T, °C

30ν, m3/kg

ν1

H2OSat. liquid

Qin

P = 100 kPa

T = 99.6 °C

P = 100 kPa

T = 99.6 °C

H2O:C. liquid

P = 100 kPa

T = 30 °C

P = 100 kPa

T = 30 °C

Qin

Water interacts with thermal energyWater interacts with thermal energy


99.6ν

2 = ν

f@100 kPa

T, °C

30ν, m3/kgν

1

ν3

P = 100 kPa

T = 99.6 °C

P = 100 kPa

T = 99.6 °C

H2O:Sat. Liq.

Sat. VaporSat. Vapor

Qin

H2OSat. liquid

Qin




ν4 =

νg@

100 kPa

99.6ν

2 = ν

f@100 kPa

T, °C

30ν, m3/kg

ν1

ν3

H2O:Sat. Vapor

H2O:Sat. Vapor

Qin

P = 100 kPa

T = 99.6 °C

P = 100 kPa

T = 99.6 °C

H2O:Sat. Liq.


Qin



150

ν5

99.6ν

2 = ν

f@100 kPa

T, °C

30ν, m3/kgν

1

ν4 =

νg@

100kPa

ν3

ν5 = ν@100 kPa, 150°C

ν3 = [νf + x νf g]@100 kPa

ν1 = νf@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


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

H2O:Sat. Liq.


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:H2O:

Qin

P = 100 kPa

T = 150 °C

P = 100 kPa

T = 150 °C

H2O:SuperVapor

H2O:SuperVapor

Qin



99.6

ν2 =

νf@

100 kPa

T, °C

30ν, m3/kg

ν1

ν4 =

νg@

100kPa

ν3

ν5 = ν@100 kPa, 150°C

ν3 = [νf + x νf g]@100 kPa

ν1 = νf@T1

150

100 k

Pa

ν5

Compressed liquidCompressed liquid: Good : Good estimation for properties estimation for properties by taking yby taking y = = yyff@T @T where where y can be either y can be either νν, u, h or , u, h or s.s.


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

2-1


T, °C

ν, m3/kg

99.6

νf@

100 kPa

νg@

100kPa

100 k

Pa

179.9

45.8

10 kP

a

1000

kPa


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


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

T –v diagram: Multiple P

2-3

T – v diagram - Example

T, °C

ν, m3/kg

70

ν=νf@70 °C = 0.001023

81.3

3.240

50 kPaCompressed Liquid,

T < Tsat

Phase, Y?

7050

T, ° CP, kPa

81.33

Tsat, °CPsat, kPaνf@70 °C

ν, m3/kg


T, °C

ν, m3/kgνf@200 kPa

= 0.001061

200 k

Pa

T- ν diagram with respect to the saturation

lines


lines

374.1400

ν = 1.5493

120.23

νg@200 kPa= 0.8857

1.5493200

ν, m3/kgP, kPa

Sup. V., Sup. V., νν >>ννgg

Phase, Why?

120.2120.2Tsat, °CPsat, kPa

400400T, ° C


T, °C

ν, m3/kg

1,000

kPa T- ν diagram

with respect to the saturation

lines


lines

Wet Mix., Wet Mix., uuff < u << u < uugg

Phase, Why?

374.1

ννf@1,000 kPa

= 0.001127

179.9

νg@1,000 kPa= 0.19444

ν = [νf + x νf g]@1,000 kPa

2,0001,000

u, kJ/kgP, kPa

179.9179.9Tsat, °CPsat, kPa

179.9179.9T, ° C

Property TableSaturated water – Pressure table

P, P, MPaMPa

Pressure

10

50

0.100

1.00

10

22.09

P, kPa

2029.602029.6

2544.41151.41393.04

2583.61822.0761.68

2506.12088.7417.36

2483.92143.4340.44

2437.92246.1191.82

ug, kJ/kgufg, kJ/kguf, kJ/kg

Specific internal energy, kJ/kg

0.0031550.003155

0.0180260.001452

0.194440.001127

1.69400.001043

3.2400.001030

14.670.001010

νg, m3/kgνf, m3/kg

Specific volume, m3/kg

374.14374.14

311.06311.06

179.91179.91

99.6399.63

81.3381.33

45.81

Tsat, °C

Sat. temp.

pure substance part1 - drjj - uitm

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