Chapter 3 Lecture Thermodynamics

7/27/2019 Chapter 3 Lecture Thermodynamics

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Chapter 3

PROPERTIES OF PURESUBSTANCES


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Objectives

Introduce the concept of a pure substance.

Discuss the physics of phase-change processes.

Illustrate the P-v, T-v, and P-Tproperty diagrams and P-v-T

surfaces of pure substances.

Demonstrate the procedures for determining thermodynamic

properties of pure substances from tables of property data.

Describe the hypothetical substance ideal gas and the

ideal-gas equation of state.

Apply the ideal-gas equation of state in the solution of typical

problems.

Introduce the compressibility factor, which accounts for the

deviation of real gases from ideal-gas behavior.


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PURE SUBSTANCE

Pure substance: A substance that has a fixed (not

necessarily single) chemical composition throughout.Any homogeneous mixture is considered a pure substance.

Non-homogeneous mixture (oil and water) is not a pure

substance.

Nitrogen and gaseous air arepure substances. A mixture of liquid water and

gaseous water is a pure

substance, but a mixture of liquid

air and gaseous air is not.

AIR


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PHASES OF A PURE SUBSTANCE

A phase has a distinct molecular arrangement.

Three principal phases

Solid

Liquid

Gas

Phase change processes are Melting: solid liquid

Solidification: Liquid solid

Vaporization: Liquid gas

Condensation: Gas

liquid Internal energy of the substance changes during phase

change.


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PHASES OF A PURE SUBSTANCE

In many applications, we have a mixture of the two

phases (phase change).

Boiler, condenser, refrigerator, etc.

The properties we are interested in:

P Pressure

specific volume T Temperature

U internal energy

Cv, Cp specific heats

X quality (% of mass converted to vapor) We will learn to determine these properties during

thermodynamic processes.


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PHASE-CHANGE PROCESSES OF PURE

SUBSTANCES

Compressed liquid (subcooled liquid): A substance that it

is not about to vaporize. Saturated liquid: A liquid that is about to vaporize.

At 1 atm and 20C,

water exists in the

liquid phase(compressed l iqu id).

At 1 atm pressure

and 100C, water

exists as a liquid

that is ready to

vaporize

(saturated l iqu id).

Due to increased T, v also increases slightly as P remains constant


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Saturated vapor: A vapor that is about to condense.

Saturated liquidvapor mixture:The state at which the liquid andvapor phases coexistin equilibrium.

Superheated vapor: A vapor that is not about to condense (i.e., not a

saturated vapor).

As more heat is transferred,

part of the saturated liquid

vaporizes (saturated liqu idvapor mix ture).

At 1 atm pressure, the

temperature remainsconstant at 100C until the

last drop of liquid is vaporized

(saturated vapo r).

As more heat is

transferred, thetemperature of the

vapor starts to rise

(superheated vapor).

v5 > v4 > v3 > v2 > v1


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Saturation Temperature and Saturation Pressure The temperature at which water starts boiling depends on the pressure;

therefore, if the pressure is fixed, so is the boiling temperature.

Saturation temperature Tsat: The temperature at which a pure substance

changes phase at a given pressure.

Saturation pressure Psat: The pressure at which a pure substance changes

phase at a given temperature.


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Exercise

What is the saturation pressure for water at 270oC?

From Table A-4 (page 915)

Psat = 5503.0 kPa

Consider water at 270oC and 4000 kPa, is it liquid,

gas, or solid

Tsat = 250.35oC at P = 4000 kPa (A-5) so it is

vapor or gas state.

10


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PROPERTY DIAGRAMS FOR PHASE When water is kept under different pressures in a piston-cylinder

system, upon heating, water follows different process paths

As the pressure is increased, the saturation line continues to shrinkand it becomes a point, called critical point.

T-vdiagram of constant-

pressure phase-change

processes of a pure substance

at various pressures


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T-vdiagram of a pure substance.


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At supercritical pressures (P >Pcr),

there is no distinct phase-change

process.

Critical point: The point at

which the saturated liquid

and saturated vapor states

are identical.

For water: Table A-1 forother pure substances

Pcr= 22.06 MPa

Tcr= 373.95oC

Vcr= 0.003106 m3/kg


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P-vdiagram of a pure substance.

T-vdiagram of a pure substance.


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Exercise Plot T-v diagrams for water going through the following process

Isothermal process. T = 200oC, P1 = 1 MPa, P2 = 5 MPa

For T = 200oC, Psat = 1.5549 MPa (Table A-4) P1 < Psat, so initially (1) we have vapor, but P2>Psat, so at (2), we

have compressed liquid

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Exercise contd Plot P-v diagrams for water going through the following process

Isothermal process. T = 200oC, P1 = 1 MPa, P2 = 5 MPa

For T = 200oC, Psat = 1.5549 MPa (Table A-4) P1 < Psat, so initially (1) we have vapor, but P2>Psat, so at (2), we

have compressed liquid

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Exercise contd What are the initial and final specific volumes, v1 and v2

At state 1, we have superheated vapor then use Table A-6, page

918 Use T1 = 200

oC and P1 = 1 MPa to read v1 = 0.20602 m3/kg

At state 2, we have compressed liquid, then use Table A-7, page

922 Use T2 = 200

oC and P2 = 5 MPa to read v2 = 0.0011531 m3/kg

17


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Extending the

Diagrams to Include

the Solid Phase

P-vdiagram of a substance thatcontracts on freezing.

P-vdiagram of a substance that

expands on freezing (such as water).

At triple-point pressureand temperature, a

substance exists in three

phases in equilibrium.

For water,

Ttp = 0.01C

Ptp = 0.6117 kPa


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Sublimation: Passing from

the solid phase directly into

the vapor phase.

At low pressures (belowthe triple-point value),

solids evaporate without

melting first (sublimation).

P-Tdiagram of pure substances.

Table 3.3 for Triple Point temperatures and

pressures

Phase Diagram


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P-v-Tsurface of a substancethat contracts on freezing.

P-v-Tsurface of a substance thatexpands on freezing (like water).


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PROPERTY TABLES For most substances, the relationships among thermodynamic properties are too

complex , therefore, properties are frequently presented in the form of tables.

Some thermodynamic properties can be measured easily, but others cannot and

are calculated by using the relations between them and measurable properties.

EnthalpyA Combination Property

The combination

u +Pvis

frequently

encountered in

the analysis of

control volumes.

The productpressure

volume has energy units.


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Saturated Liquid and Saturated Vapor States

Table A4: Saturation properties of water under temperature.

Table A5: Saturation properties of water under pressure.

A partial list of Table A4.

Enthalpy of vaporization, hfg(Latent

heat of vaporization): The amount of

energy needed to vaporize a unit mass

of saturated liquid at a given

temperature or pressure.

E l A i id t k t i 50 k f


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Examples: A rigid tank contains 50 kg ofsaturated liquid at 90oC. Determine the

pressure in the tank and the volume of the

tank.

Solution:

m= 50 kg, saturated liquid water

T = 90oC

Find pressure from Table A-4

At T = 90oC Psat = 70.183 kPa

Find vfat 90oC: vf= 0.001036 m

3/kg

V = vf.m = (0.001036)(50) = 0.0518 m3

E amples A f 200 f t t d li id


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Examples: A mass of 200 g of saturated liquidwater is completely vaporized at a constant

pressure of 100 kPa. Determine (a) the volume

change and (b) the amount of energy transferred

to the water.

Solution: Table A-5 at 100 kPa

vfg = vg vf= 1.6941 0.001043 = 1.6931 m3/kg

Change in volume = m.vfg = 0.3386 m3

The amount of energy needed to vaporize a unitmass is the enthalpy of vaporization at that

pressure, which is hfg = 2257.5 kJ/kg for water at

100 kPa.

Thus, total energy: H = m.hfg

= (0.2 kg)(2257.5

kJ/kg) = 451.5 kJ


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Saturated LiquidVapor MixtureQuality, x: The ratio of the mass of vapor to the total mass of the mixture.

Quality is between 0 and 1 0: sat. liquid; 1: sat. vapor.

A two-phase system can be treated as a

homogeneous mixture for convenience.


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Quality is related

to the horizontal

distances on P-v

and T-v

diagrams.

The vvalue of a saturated

liquidvapor mixture lies

between the vfand vgvalues at

the specified TorP.

Vf= Specific volume of saturated liquid

Vg = Specific volume of saturated vapor

Vavg = Specific volume of saturated liquid-

vapor mixture


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y v, u, orh.

We can use quality x to determine the

enthalpy, h, and internal energy, u, of the

saturated liquid-vapor mixture

Review examples 3-4 page 128


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A rigid tank contains 10 kg of water at 90oC. If 8 kg of the water is in

the liquid form and the rest is in the vapor form, determine (a) the

pressure in the tank and (b) the volume of the tank.

Review examples 3-4, page 128

Review examples 3-5 page 129


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An 80L vessel contains 4 kg of refrigerant-134a at a pressure of 160

kPa. Determine (a) the temperature, (b) the quality, the enthalpy of

the refrigerant, and (d) the volume occupied by the vapor phase

Review examples 3 5, page 129


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Superheated VaporIn the region to the right of thesaturated vapor line and at

temperatures above the critical

point temperature, a substance

exists as superheated vapor.In this region, temperature and

pressure are independent

properties.

A partial

listing of

Table A6.

At a specified

P, superheated

vapor exists at

a higherh than

the saturatedvapor.

Compared to saturated vapor,

superheated vapor is characterized by

The compressed liquid properties


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Compressed LiquidCompressed liquid is characterized by

yv, u, orh

A more accurate relation forh

A compressed liquid

may be approximated

as a saturated liquid at

the given temperature.

At a given P

and T, a puresubstance will

exist as a

compressed

liquid if

The compressed liquid properties

depend on temperature much more

strongly than they do on pressure.

Table A-7 is the only table exists in

this text.

Interpolation


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Interpolation

In most applications, we cannot read the properties of water directly off the tables.

Interpolation Example


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Interpolation Example

Determine the temperature of water at P = 0.5 MPa and h = 2890 kJ/kg.

From Table A-5, we determine that we have superheated water vapor. Why?

Use Table A-6, superheated vapor at P = 0.5 MPa

T1 = 200oC h1 = 2855.8 kJ/kg

T2 = 250oC h2 = 2961.0 kJ/kg

(h-h1)/(h2-h1) = (T-T1)/(T2-T1)

h2 h1 = 105.2 kJ/kg

h h1 = 2890 2855.8 = 34.2 kJ/kg

(h-h1)/(h2 h1) = 34.2/105.2 = 0.3250

T2-T1 = 250oC-200oC = 50oC

(h-h1)/(h2-h1).(T2-T1) = 0.3250 (50) = 16.44oC

T = T1 + (h-h1)/(h2-h1).(T2-T1) = 200oC = 16.44oC = 216.44oC

Problem


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Problem

A piston-cylinder device contains 0.1m3 liquid water an 0.9m3 of watervapor in equilibrium at 800 kPa. Heat is transferred at constant pressureuntil T= 350oC.

(a) What is the initial temperature?

(b) Determine the total mass of water.

(c) Calculate the final volume.

(d) Show the process on a p-v diagram with respect to saturation lines.

P1 = 800 kPa P2 = 800 kPa

T1 = ? T2 = 350oC

Vapor

Liquid

(a) Initially two phases coexist, thus we have a saturated liquid-vapormixture. Then the temperature in the tank must be the saturationtemperature

T = Tsat@800kPa = 170.41oC Table A-5 Page 917

Problem


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Problem

(b) To find total mass we can add the mass of each phase

mf= Vf/vf= 0.1 m3/0.001115 m3/kg = 89.69 kg

mg= Vg/vg = 0.9 m3/0.2404 m3/kg = 3.74 kg

mTotal = mf+ mg = 93.43 kg

(c) At final state, water is superheated vapor. P2 = 800 kPa and T2 = 350oC

Using Table A-6, v2 = 0.3544 m3/kg, thus, V2 = mv2 = (93.43) (0.3544) =

0.33.1 m3


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Problem 3-26

Complete this table for H2O

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T, C P, kPa h, kJ / kg x Phase description200 0.7

140 1800

950 0.0

80 500

800 3162.2


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Problem 3-26

Complete this table for H2O

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T, C P, kPa h, kJ / kg x Phase description120.21 200 2045.8 0.7 Saturated mixture

140 361.53 1800 0.565 Saturated mixture

177.66 950 752.74 0.0 Saturated li quid

80 500 335.37 - - - Compressed liquid

350.0 800 3162.2 - - - Superheated vapor

THE IDEAL GAS EQUATION OF STATE


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THE IDEAL-GAS EQUATION OF STATE

Instead of using tables to obtain properties, using simple equations

(if we have them available) would be easier.

Equation of state: Any equation that relates the pressure, temperature,and specific volume of a substance.

The simplest and best-known equation of state for substances in the gas

phase is the ideal-gas equation of state. This equation predicts the P-v-T

behavior of a gas quite accurately within some properly selected region.

R: gas constant

M: molar mass (kg/kmol)

Ru: universal gas constant

Ideal gas equation of state

Different substances have differentgas constants.

P: absolute pressure

T: absolute temperature in Kelvin (K)V: Specific volume


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The ideal-gas relation often is

not applicable to real gases;

thus, care should be exercised

when using it.

Mass = Molar mass Mole number

Various expressions of idealgas equation

Ideal gas equation

at two states for afixed mass

Real gases behave as an ideal gas at low

densities (i.e., low pressure, high

temperature).

Is Water Vapor an Ideal Gas?


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Is Water Vapor an Ideal Gas?

At pressures below 10 kPa, watervapor can be treated as an ideal

gas, regardless of its temperature,with negligible error (less than 0.1percent).

At higher pressures, however, theideal gas assumption yieldsunacceptable errors, particularly in

the vicinity of the critical point andthe saturated vapor line.

In air-conditioning applications, thewater vapor in the air can betreated as an ideal gas. Why?

In steam power plant applications,

however, the pressures involvedare usually very high; therefore,ideal-gas relations should not beused.


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OF DEVIATION FROM IDEAL-GAS BEHAVIOR

The compressibility factor is

unity for ideal gases.

Compressibility factorZ

A factor that accounts for

the deviation of real gases

from ideal-gas behavior at

a given temperature and

pressure.

The farther away Zis from unity, the more the

gas deviates from ideal-gas behavior.

Gases behave as an ideal gas at low densities

(i.e., low pressure, high temperature).

Question: What is the criteria for low pressure

and high temperature?

Answer: The pressure or temperature of a gas

is high or low relative to its critical temperatureor pressure.

At very low pressures, all gases approach

ideal-gas behavior (regardless of their

temperature).

How to determine Z


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42Figure A-15 (page 932) Comparison ofZfactors for various gases.

Reduced

temperature

Reduced

pressurePseudo-reduced

specific volume

Zcan also be determined from

a knowledge ofPRand vR.

o to dete e

Deviation from Ideal Behavior


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At very low pressure, PR>1)

The deviation from ideal gas

behavior is greatest in the vicinity

of the critical point

Deviation from Ideal Behavior

Example 3-11


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Determine the specific volume v, of refrigerant -134a at P = 1 MPa and at

50oC.

(a) Use ideal gas equation

(b) Use compressibility chart

Solution

(a) We need to find gas constant, R, for refrigerant.

From Table A-1, R = 0.08149 kJ/kgK

Ideal gas equation, Pv = RT

v = RT/P = 0.08149 (273+50)/1000 = 0.02632 m3/kg

(b) To use compressibility chart, we need to calculate PR and TR

PR

= 1 MPa/4.059 MPa = 0.246

TR = 323K/374.2K = 0.863

Using the chart (Fig. 3.49), we find Z = 0.84

Then v = Z videal = 0.0221 m3/kg

Note that more accurate value of v is obtained from table A-13 asv = 0.02179 m3/k


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OTHER EQUATIONS

OF STATE

Several equations have been proposed torepresent the P-v-Tbehavior of substances

accurately over a larger region with no

limitations.

Van der Waals

Equation of State Critical isothermof a pure

substance has

an inflection

point at the

critical state.

This model includes two effects not considered

in the ideal-gas model: the intermolecular

attraction forces and the volume occupied by the

molecules themselves. The accuracy of the van

der Waals equation of state is often inadequate.

S


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Summary Pure substance

Phases of a pure substance

Phase-change processes of pure substances

Compressed liquid, Saturated liquid, Saturated vapor, Superheated vapor

Saturation temperature and Saturation pressure

Property diagrams for phase change processes

The T-vdiagram, The P-vdiagram, The P-Tdiagram, The P-v-Tsurface

Property tables

Enthalpy

Saturated liquid, saturated vapor, Saturated liquid vapor mixture,Superheated vapor, compressed liquid

Reference state and reference values

The ideal gas equation of state

Is water vapor an ideal gas?

Compressibility factor

Other equations of state

Problem


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Problem A piston-cylinder contains 50 L liquid water at 25oC and 300 kPa. Heat is

added at constant pressure until entire liquid is vaporized.

(a) What is the mass of water?

(b) What is the final temperature?

(c) Determine the total enthalpy change

T1 = 25oC T1 < Tsat at 300 kPa

P1 = 300 kPa Initially we have compressed

water water

For this case, P1 = 300 kPa this is too low to use Table A-7

Then, v1 = vf@25C = 0.001003 m3/kg

h1 = hf@25C = 104.83 kJ/kg

(a) m = V1/v1 = 0.05 m3 / 0.001003 m3/kg = 49.85 kg

(b) At the final state, the cylinder contains saturated vapor

Final temperature must be saturation temperature. For 300 kPa, T = 133.52oC

(c) Final enthalpy is h2 = hg@300kPa = 2724.9 kJ/kg

Chapter 3 Lecture Thermodynamics

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