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3.1 The Pure substance A pure substance is one that has a

homogeneous and invariable chemical composition.Ex. water, steam (a mixture of liquid water

and water vapor) Air (X)mixture of two pure substances

Chapter 3 Pure Substance Behavior

3.2 Vapor-liquid-solid-phase equilibrium in a pure substanceExplain Fig. 3.1 (p.45)

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Saturation temp. (Tsat): the temp. at which vaporization takes place at a given pressure.Exat 1 atm, Tsat = 99.6 (0.1MPa)

Saturation pressure (Psat): the pressure at which vaporization takes place at a given temperature.ExAt 99.6 , Psat = 1 atm (101.35 kPa)

At 20 , Psat = 2.3385 kPa

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For a pure substance, there is a definite relation between Tsat and PsatVapor-pressure curve (Fig. 3.2, p. 45)

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Saturated liquid: a substance existed as liquidat Tsat and Psat.

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Exwater at 1 atm, 20 T Tsat

Psat = 1 atm, Tsat = 99.6 20 < 99.6 (Subcooled liquid)

Tsat = 20 , Psat = 2.3385 kPa P > Psat(101.35kpa) (2.3385kpa)

(Compressed liquid)

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Quality (x)only at Tsat and Psat, a substance existed as part liquid and part vapor (Fig. 3.1b, Psat = 1 atm, Tsat = 99.6 )then x is defined asx = mass of vapor/total mass

= mass of vapor/(mass of vapor + mass of liquid)

Exmvap = 0.2 kg, mliq = 0.8 kg x = 0.2 or 20%

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x = 0 % saturated liquidx = 100 % saturated vapor (dry

saturated vapor) Superheated () vapor : vapor at a

temperature greater than that of the saturated vapor. (Tvap > Tsat)

For superheated vapor, its temp. and pressure are independent.Gas highly superheated vapor

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For water(Fig. 3.3, P.46)

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ABCD ( = 0.1MPa)AT = 20 Bsaturated liquid state Tsat = 99.6 , x = 0 %Csaturated vapor state Tsat = 99.6 , x = 100%BCphase change (x varied) T = Tsat = const.CDsteam is superheated at const. pressure

T, VEFGH ( = 1 MPa) MNO ( = 22.09MPa)

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Ncritical point saturated liquid state and saturated vapor state are identical.

Temperature at critical pt. critical temp. Pressure at critical pt. critical pressure Specific volume at critical pt. critical

specific volumeExsee Table 3.1 (p.46)Waterat critical pt.

T = 374.14 , P = 22.07 MPa, = 0.003155 m3/kg

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QuestionIf choose constant-pressure process at P = 40MPa (PQ) > Pcrit

(super critical pressure)What phase can we observe?

AnsFluid ()

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Pressure > critical pressureNo equilibrium state can exist

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At a given pressure, the solid starts to melt and the temperature remains constant. In this state, the solid is called a saturated solid.

In general, solid liquid for most substanceBut for the ice melting process

Sublimation(): solid phase becomes vapors phase directly.

ExIceIce VaporSuperheated vapor (T=20, P=0.26kPa) (-10 , P=0.26kPa) (-10 , P=0.26kPa)

Triple point: three phases (solid, liquid and vapor) are in equilibrium. Ex: see Table 3.2 (P.48)

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melting at const. temp.

Heat inHeat inHeat in

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Saturated solid At a given pressure, the solid starts to melt and the temperature remains constant in this state, the solid is called a saturated solid.(the cross-section of EF and fusion line)

Sublimationsolid phase becomes vapor phase directly

Triple pointthree phases (solid, liquid and vapor) are in equilibrium

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3.3 Independent properties of a pure substance

T, P, For pure substance, two of them are enough to

describe the state, provided that they are independent.Note that Psat and Tsat are related. (quality or

must be specified)

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3.4 Tables of thermodynamic properties Table Given T and P (state) (specific

volume), u (internal energy), h (enthalpy), s (entropy)

All of them are independent Two problemsIf T and P are not given at same time troubleInterpolation

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Table B.1.1 Saturated water (liquid): Temp. table P.702~P.705

Table B.1.2 Saturated water (liquid): Pressure table P.706~P.709

Table B.1.3 Superheated vapor water: P.710~P.715 ()

Table B.1.4 Compressed liquid water: P.716~P.717

Table B.1.5 Saturated solid saturated vapor water P.718~P.719

Table B.1.5

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f specific volume of saturated water (liquid)gspecific volume of saturated vapor

Both f & g are given in Tables B.1.1 & B.1.2fg = g f

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Quality (x)(In a saturation state) V = Vliq + Vvap

m = m liqf + m vapg = (1 x) f + xg

= f + x(g f)= f + xfg

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ExFind the specific volume of saturated steam at 200 having a quality of 70%

Check Table B.1.1 (704)Tsat = 200, Psat = 1.5538MPa

f = 0.001156 m3 / kg, g = 0.12736 m3 / kg = 0.001156 + (0.70) (0.12736 0.001156)

= 0.08950 m3 / kg

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Table B.1.3 superheated vapor (states on the right-hand-side of vaporization line)

Remind the definition: vapor at a temp. greater than that of saturated vapor. (Tvap > Tsat) at a given pressure

T and P are independent (p. 710)

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Ex (0.5MPa=500kPa, 200) = 0.42492 m3/kg

critical pt. (P > 22.09 MPa, P.675)

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Table B.1.4 (P.716) compressed liquid (states on the left-hand-side of vaporization line but in liquid phase)

Table B.1.3 (P.710) vaporization line

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For compressed liquidAt same temp., compressed liq. f (At saturation state)

Ex:At 100, Psat = 0.10135 MPa, f = 0.001044 m3/kg

(p.702)

At 100, P = 10 Mpa, = 0.001039 m3/kg (p.717)

For compressed liquid, const. at same temp.

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Table B.1.5. saturated solid-saturated vapor water (p. 718)

Temp. must be below the temp. of triple pt. states on the sublimation line similar to

Table B.1.1 or B.1.2.

Ex3.5 and 3.6 (p.58), for fluid other than water

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3.5 Thermodynamic surfaceP - T

Fig. 3.18 (p.59): a substance such as water in which the specific volume increases during freezing

Fig.3.19 (p.60):

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Solid, liquid and vapor phases are curved surface

Triple pt P - T Triple pt. line P T and P project area ( 3.19; p.60)

critical isotherm has a point of inflection

3.6 The P--T behavior of low-and-moderate-density gases

For gases at low densityEquation of state(for ideal gas)

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P1V1/T1 =P2V2 /T2

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Ideal-gas equation of state (at low-density)Questions1)What is the range of density for which ideal-gas

equation of state can be applied with accuracy? 2)For given T and P, how far does the real gas

deviate from ideal gas?

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Introduce the compressibility factor, Z

1)For ideal gas, Z = 1.02)Z is a measurement of deviation of actual gas

from ideal gas

RTPZ

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ExN2 (Fig. 3.22; p.64)(1) P 0, > Z 1(2) T 300 K, P 10MPa > Z 1(3) P = 4 MPa, T < 300K > Z < 1 real > ideal

Reason: intermolecular force(4) P > 30MPa > Z > 1 real < ideal

Reason: repulsive force

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To quantify Z with respect to P and T, define the reduced temperature as TrTr = T / Tcrit

And the reduced pressure as PrPr = P / Pcrit

Tcrit and Pcrit are the temperature and pressure at critical pt.

Tr , Pr generalized compressibility chart

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Fig. D.1 (P.755) simple spherical molecular

At what conditions, actual gas ideal gas

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From Fig. D.11) Pr