EGR 334 Thermodynamics Chapter 3: Section 1-5 Lecture 06: Evaluating Properties Quiz Today?

EGR 334 ThermodynamicsChapter 3: Section 1-5

Lecture 06: Evaluating Properties

Quiz Today?

Today’s main concepts:• Number of properties needed to set simple compressible

system.• State properties: Temperature, Pressure, Specific Volume • Phase and Quality.• Given two properties of a pure compressible substance,

find other state properties.• Using the Thermodynamic Property tables.

Reading Assignment:

Homework Assignment:

• Read Chap 3: Sections 6-8

From Chap 3: 5, 7, 10, 29

3

Pure Substance: a substance that is uniform and invariable in chemical composition.

Sec 3.1.1: Phase and Pure Substance

Oil

Ice

Water

Honey

Air

May each of these be considered a pure substance?

Helium gas in a tank?

A contents of a glass containing water and ice?

A jar of honey, water, oil, and ice topped with air?

Atmosphericair?

Yes

Yes…maybeYes

No

4

Phase: Matter which is homogeneous throughout in both its chemical composition and physical structure.


Bottle Contents : Air (N2, O2, ect.)

Homogeneous chemical composition?

Homogeneous physical structure?Bottle : Glass


Homogeneous physical structure?

Cap : Aluminum?


Homogeneous physical structure?

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Phase: Matter which is homogeneous throughout in both its chemical composition and physical structure.


Pure substances can exist in multiple phases.

- Gaseous Phase- Liquid Phase- Solid Phase

Pure substances can undergo changes in Phase.

liquid gas: vaporization gas liquid: condensation

solid gas: sublimation

gas solid: deposition

solid liquid: melting liquid solid: freezing

6

The intensive state of a closed system at equilibrium is its condition as described by the values of its intensive thermodynamic properties.

For a simple compressible system, specifying any ___2_____ independent intensive thermodynamic properties will fix the other intensive thermodynamics properties of the system.

In this class we will be dealing with simple compressible systems.

Examples of Simple Compressible Systems:--Standard Air (Oxygen-Nitrogren mix)--Ideal Gases--Superheated Water Vapor

Intensive Properties include: Pressure Temperature Density Specific Volume Internal Energy Enthalpy Entropy

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Most important skill for today’s lecture:

•Given any two intensive properties of H20, be able to specify other intensive properties.

Given any two of Temperature Specific Volume and Pressure ...find the missing intensive property value.

( , )p p T v ( , )T T p v ( , )v v T por or

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3D p-v-T surface model: A representation of how p, v, and T take on specific values depending upon their intensive properties and phase changes.

Sec 3.2: P-v-T relation

The different surfaces of the model represent different phases of a pure substance (like H20) which depend only on the pressure, temperature, and specific volume of the state.

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Phase Diagram from the 3D p-v-T surface model: If you pull off the Pressure-Temperature projection you create a plot given as the Phase Diagram.


Triple point•

•

Critical Point

Tempera

tur

e

Pre

ssur

e

10

p-V Diagram from the 3D p-v-T surface model: The Pressure-Specific Volume projectionof the surface model is a useful toolfor showing processes involving ice-water-water vapor system.


Specific Vol.

Pre

ssur

e •

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Projections of p-v-T surface: Triple Point--the state at which solid, liquid, and gas coexist Critical Point-- the point where saturated liquid and saturated vapor lines meet.


Phase diagram p-V Diagram

Isotherms--Lines of constant temperature

Isotherms

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T-v diagram from the 3D p-v-T surface model: If you pull off the Temperature-Specific Volume projection you create a plot that can be useful for showinglines of constant pressure.


•Temperatur

e

Specific Volume

isobar: line of constant pressure

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i) Saturated Liquid Line (identified values of Tf , vf, and pf ) ii) Saturated Vapor Line (identified values of Tg , vg, and pg

iii) Critical Point (inflection point at top of dome)iv) On p-v diagram, lines of constant temperature run from high left to low right. v) On T-v diagram, lines of constant pressure run from low left to high right.vi) While traversing the liquid-vapor phase (the area under the dome), both temperature and pressure remain constant for changes in spec. volume.

Tem

pera

ture

You should be able to recognize:

Fig03_03

Consider a Constant Pressure Process

Sec 3.3: Phase Change

water water

sub-cooled liquid(compressed liquid)

vaporwater

supervaporsat.

vaporvaporwater

two phaseliquid-vapor

two phaseliquid vapor

saturated liquid

saturatedvapor

super heated vapor

(steam)

• •i j

RunAnimation

http://www.wiley.com/college/moran/0470495901/animations/liq_to_vapor/liq_to_vapor.html

Fig03_03

Sec 3.3: Phase Change

Distinguish mixture of liquid/vapor using quality.

Two Phase, Liquid-Vapor Mix

vaporliquid

vapor

mm

mx

Can have different mixtures of liquid and vapor (10% liquid, 90% vapor)

x = 1, saturated vapor (g)x = 0, saturated liquid (f)

Can also be expressed as a percentage (%)

Fig03_03

Sec 3.5: Saturation

Quality (x) is also a property. It is a way to express the relative amount of a substance that contains two different phases of material.

Two Phase, Liquid-Vapor Mix

gT

vf

T

lT v

m

mv

m

mv

For example, in a given vessel, we know that the two volumes must add to the total volume

VLT VVV

1 f gv x v xv

with

T

v

T

l

m

mx

m

mx ; 1

f g fv v x v v

17

pre

ssure

spec. vol.

*(p, v, T)

Quantitative Thermodynamic Properties:

Methods:1) Read off a p-v or a T-v diagram 2) Look up on Steam Tables: (Appendix and Handout) i) Table A.2: Prop. of Saturated H20: Temperature Table ii) Table A.3: Prop. of Saturated H20: Pressure Table iii) Table A.4: Prop. of Superheated Water Vapor iv) Table A.5: Prop. of Compressed Water v) Table A.6: Prop. of Saturated H20: Solid-vapor table3) Use a Computer Steam Application i) IT (download from www.wiley.com/college/moran) ii) http://www.dofmaster.com/steam.html

Given any two intensive properties, can you a) determine the phase (liquid, saturated, mixture, solid, or gas?)b) determine the other intensive property values at that state.

Link to IT download

Link to IT dofmaster

http://www.wiley.com/college/moran

http://www.dofmaster.com/steam.html

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Examples of Property Diagrams

Molliere chart Pyschrometric chart

Given two properties, locate and read other values.

19

Saturated Steam Table

Given two property values, look up other intensive values.

Given T= 25 deg C and saturated vapor

Read p and v.

20

Example 1: Determine property values

Determine phase or phases of H20 at the following conditions and sketch p-v and T-v diagrams showing the positions of each of the followingIdentify the specific volume, v, if possible:a) p = 5 bar T = 151.9 deg C.

b) p = 5 bar T = 200 deg. C.

c) p = 2.5 MPa T = 200 deg C.

d) p = 2.8 bar T = 160 deg C.

e) p = 1 bar T = -12 deg C.

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Determine phase or phases of H20 at the following conditions and sketch p-v and T-v diagrams showing the positions of each of the following:Identify the specific volume, v, if possible:a) p = 5 bar T = 151.9 deg C. from table A.3: at 5 bar the saturated temperature is 151.9 which meansit could be anywhere on the liq-vapor line as a two phase material.

22



b) p = 5 bar T = 200 deg. C.

c) p = 2.5 MPa T = 200 deg C.

d) p = 2.8 bar T = 160 deg C.

e) p = 1 bar T = -12 deg C.

23


Determine phase or phases of H20 at the following conditions and sketch p-v and T-v diagrams showing the positions of each of the following:b) p = 5 bar T = 200 deg. C.from table A.3: at 5 bar the temp. of 200 > 151.9 which means the substance is vapor or superheated vapor…need to consult table A.4from table A.4 at 5 bar and T = 200 deg C….v = 0.4249 m3/kg

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b) p = 5 bar T = 200 deg. C.

c) p = 2.5 MPa T = 200 deg C.

d) p = 2.8 bar T = 160 deg C.

e) p = 1 bar T = -12 deg C.

25


c) p = 2.5 MPa T = 200 deg C. from table A.3 at 2.5 MPa = 25 bar the temp of 200 is 200< 224 which means that the substance is compressed liquid. For compressed liquid we assume that the spec. vol, will be similar to vf = 1.1973x10-3. This can be checked on Table A.5 for compressed liquids which gives 1.1555x10-3 m3/kg

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b) p = 5 bar T = 200 deg. C.

c) p = 2.5 MPa T = 200 deg C.

d) p = 2.8 bar T = 160 deg C.

e) p = 1 bar T = -12 deg C.

27


d) p = 2.8 bar T = 160 deg C.from Table A.2 at Tsat = 160, psat = 6.178 > 2.8 which means the substance will be vapor or superheated…so refer to table A.4.from Table A.4 at p = 1.5 and T = 160….v = 1.317 at p = 3 and T = 160….v = 0.651

Interpolation is required:

1.317 2.8 1.5

0.651 1.317 3.0 1.5

v

0.7398v

28



b) p = 5 bar T = 200 deg. C.

c) p = 2.5 MPa T = 200 deg C.

d) p = 2.8 bar T = 160 deg C.

e) p = 1 bar T = -12 deg C.

29


e) p = 1 bar T = -12 deg C. from table A.3 at psat of 1 bar, Tsat = 99.63 C > -12 which means the substance will be compressed liquid or solid. Noticing that Table A-5 doesn’t handle pressures this low, if you assumed that the material was liquid, you would use the saturated spec. vol. at the given temperature of -12 deg. However, you recognize that -12 degrees is below the freezing point of water and therefore you expect that this is in solid phase (ice).Table A-6 also has information on saturated solid-vapor data.

Assume v = vi for a T of -12 deg. C. v = 1.0888 x 10-3 = 0.0010888 m3/kg

30

Linear Interpolation:

*

*y2

yunknown

xknown x1

y1

x2

1 1

2 1 2 1

unknown knowny y x x

y y x x

2 11 1

2 1

( )unknown known

y yy y x x

x x

You try it: What is the specific volume of a water at 40 bar and 140 oC?Linear Interpolation will be needed.

Given the following:

Sec 3.5: Evaluating P, v, T

2 11 1

2 1

( )v v

v v p pp p

@ 140°C & 25 bar, v1 = 1.0784 m3/kg@ 140°C & 50 bar, v2 = 1.0768 m3/kg

Find the spec. vol at p = 40 bar.

(1.0768 1.0784)1.0784 (40 25) 1.0774

(50 25)v

Sec 3.5: Evaluating P, v, T

@ 180°C & 40 bar, v1 = 1.1248 m3/kg

This time try it for the same pressure with different temperatures.Given: 40 bar and 180 oC, the specific volume of water is 1.1248 m3/kg. 40 bar and 140 oC, the specific volume of water is 1.0774 m3/kg.

What is the specific volume of water at 40 bar and 150 oC?

Example 2:

2 11 1

2 1

( )v v

v v T TT T

3(1.1248 1.0774)1.0774 (150 140) 1.0893 /

(180 140)v m kg

A cylinder-piston assembly initially contains water at 3 MPa and 300oC. The water is cooled at constant volume to 200 oC, then compressed isothermally to a final pressure of 2.5 MPa. Sketch the process on a T-v diagram and find the specific volume at the 3 states.

Sec 3.5.2 : Saturation Tables

Example:

TV diagram

Sec 3.5.2 : Saturation Tables

Example:

35

State 1: p1 = 3 MPa= 30 bar and T1 = 300 deg C.

i) Start with superheated vapor table A-4

ii) Interpolate between 280 and 320 deg C.

1 1( )b aa a

b a

v vv v T T

T T

1

(0.0850 0.0771)0.0771 (300 280)

(320 280)v

31 0.0811 /v m kg

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State 2: v2 = v1 = 0.0811 m3/kg and T2 = 200 deg C.

Recognize that this is in the liquid-vapor mixture range. Refer to Table A-2

i) Locate vg and vf at T = 200 deg C.

( )f g fv v x v v

31.1565 10fv

2 0.0811v

0.1274gv

ii) Determine quality, x2

f

g f

v vxv v

0.0811 0.0011565

0.633 63.3%0.1274 0.0011565

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State 3: p3 = 2.5 MPa = 25 bar and T2 = T3= 200 deg C.

Recognize that this is in the compressed liquid. Refer to Table A-5

i) Locate p3=25 bar and T3=200 deg C

3 33 1.1555 10 /v m kg

ii) Read the value of v directly.

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End of Slides for Lecture 06

EGR 334 Thermodynamics Chapter 3: Section 1-5 Lecture 06: Evaluating Properties Quiz Today?

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