thermodynamics: pure substances
DESCRIPTION
An introductory lecture on Pure substancesTRANSCRIPT
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Properties of Pure Substances
Chapter 2
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Pure Substance
In Chemistry you defined a pure substance as an element or a compoundSomething that can not be separatedIn Thermodynamics we’ll define it as something that has a fixed chemical composition throughout
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Examples
Ice in equilibrium with waterAirAir in equilibrium with liquid air is not a pure substance – Why?Is sea water in equilibrium with water vapour a pure substance?
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Boiling Points of Selected Liquids
ProductBoiling Point
at Atmospheric Pressure(oC) (oF)
Acetaldehyde CH3CHO 20.8 69Ammonia -35.5 -28.1
Carbon dioxide -57 -70.6Ethanol 78.4 173
Freon refrigerant R-11 23.8 74.9Freon refrigerant R-12 -29.8 -21.6Freon refrigerant R-22 -41.2 -42.1
Helium -269 -452Nitrogen -196 -320Oxygen -183 -297Water 100 212
Water, sea 100.7 213
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Why is boiling and condensation of liquids
relevant?
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Phases of Pure Substances
We all have a pretty good idea of the what the three phases of matter are, but a quick review will help us understand the phase change process
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Consider what happens when we heat water at
constant pressure
Piston cylinder device –maintains constant pressure
Liquid Water
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T
v
1
2
5
3 4
1. Compressed liquid
2. Saturated liquid
3. Saturated mixture
4. Saturated vapor
5. Superheated vapor
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Two Phase Region
Compressed Liquid
Superheated Gas
Sometime a pure substance can decrease in temperature after critical point
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What happens at constant T?
Consider ideal gas law:PV=nRT
Applies only in gaseous state away from melting point (high volume region)Increasing pressure can condense a gas
… more later
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P-V diagram
P
v
T2>T1
T1
Critical Point
Saturated Liquid-vapour region
Superheated vapourCompressed
liquid region
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Critical Point
Above the critical point there is no sharp difference between liquid and gas!!
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Property Diagrams
So far we have sketched T – v diagram P – v diagram What about the P – T diagram?
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What is the difference between paths?
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Combine all three
You can put all three properties P T V
On the same diagram
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Contracts on Freezing Expands on Freezing
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Property Tables
P - pressureT - temperaturev – specific volumeu – specific internal energyh – specific enthalpy h = u + Pvs – specific entropy -define in Chapter 6
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Saturated Liquid and Saturated Vapor States
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Saturation Properties
Saturation Pressure is the pressure at which the liquid and vapor phases are in equilibrium at a given temperature.
Saturation Temperature is the temperature at which the liquid and vapor phases are in equilibrium at a given pressure.
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Table A-4 and A-5
A-4 - pg 830 Saturated water temperature table
A-5 - pg 832 Saturated water pressure table
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Examples
Cengel & Boles 3-1 (page 125)A rigid tank contains 50kg of saturated liquid water at 90°C. Determine the pressure in the tank and the volume of the tank.
1. How do we start?
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SolutionWater is saturated.… so we know thepressure – Psat
Use the(A-4)So Psat = 70.14kPa, & v=0.001036m3/kgTank Volume = specific volume x mass
=50x0.001036m3=0.0518m3
Water is saturated.… so we know thepressure – Psat
Use the Temperature table (A-4)
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Example 2: C&B Ex 3.2
A piston-cylinder device contains 0.06m3 of saturated water vapour at 350kPa. Determine the temperature and the mass of the vapour inside the cylinder.
All water vapour BUT saturated …
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Solution
Vapour is saturated.… so we know thetemperature – Tsat
Use the Pressure table (A-5)
V.… so we know thetemperature – TUse the Pressure table (A-5)So Tsat = 138.86K, & v=0.52422m3/kgMass of vapour = volume/specific volume
=0.06/0.52422kg=0.114kg
PresskPa300325350375400
Vapour Not liquid!
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State Variables
Once the Pressure, temperature and volume are known, other state variables are determined Internal energy Enthalpy Entropy For each phase of the pure substance
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u u uh h hs s s
fg g f
fg g f
fg g f
g stands for gasf stands for fluidfg stands for the difference between gas and fluid
Internal Energy
Enthalpy
Entropy
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Liquid-Vapour Mixtures:Quality
xmass
massm
m msaturated vapor
total
g
f g
Fraction of the material that is gas
x = 0 the material is all saturated liquid
x = 1 the material is all saturated gas
x is not meaningful when you are out of the saturation region
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X = 0 X = 1
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Example (C&B – 3-4)
A rigid tank contains 10.0kg of water at 90°C. If 8.00kg of the water is in the liquid phase, and the rest is in the vapour phase, determine:(a) the pressure in the tank(b) the volume of the tank
What is the fixed point – where we start?
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SolutionContents of tank are in themixed phase regionLiquid and vapour at 90°C, and P=PsatP=70.183kPa (from table A-4)In this case we need specific volume of liquid AND gas at these conditionsvf=0.001036m3/kg; vg=2.3593m3/kgSo V=8kgx0.001036m3/kg+2kgx2.3593m3/kg
=4.73m3
90°C
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Superheated Properties
Table A-6
Page 834
In the superheated region, there is only vapourTable looksa little different
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Linear Interpolation
A B
100 5
200 10
130 X
5105
100200100130
x
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Linear InterpolationEg: What is the saturation temperature for superheated steam at P=0.15MPa?
P1=0.1MPa, T1=99.61°CP2=0.2MPa`, T2=120.21°C… from tables
C
TTPPPPTT
TTTT
PPPP
91.109)61.9921.120(61.99
)(
1.02.01.015.0
1212
11
12
1
12
1
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Example – putting it all together
Direct Solar Steam Generation
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How do we solve?Consider just superheated steam region
0.1MPa0.5kg/s
P along 6m collectorowing to fluid dynamics!
Pinlet > 0.11MPa
Tsat>101.67°C
Texit~179.88°C Pinlet=1MPa, Tsat= 179.88°C
Compress
Tsat from interpolation: Tsat, 0.11MPa=99.61+(120.21-99.61)*(0.11-0.1)/(0.2-0.1)
Power needed?U179.88-U101.67For 0.5kg EACHEACH SECOND
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Power needed?Power = Energy / timeSo need first INTERNAL ENERGY change per kg steam from inlet to outlet temp. at 0.1MPa (ignoring friction losses)
kgkJUkgkJ
UUUU
TBOTHforTTTT
UUUU
/8.2508/9.2627
)(150200
15088.179
67.101
15020015088.179
12
1
12
1
Power = Energy / timeSo need first INTERNAL ENERGY change per kg steam from inlet to outlet temp. at 0.1MPa (ignoring friction losses)U=118.9kJ/kg0.5kg/sPower=59.9kW
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Equations of State
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Equations vs Tables
The behavior of many gases (like steam) is not easy to predict with an equationThat’s why we have tables like A-4, A-5 and A-6Other gases (like air) follow the ideal gas law – we can calculate their properties
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Ideal Gas Law
PV=nRT Used in your Chemistry class From now on we will refer to the gas
constant , R, as the universal gas constant, Ru , and redefine R=Ru/MW
PV=mRT R is different for every gas Tabulated in the back of the book
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Ideal Gas Law
Pv = RT This is the form we will use the most
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When does the ideal gas law apply?
The ideal gas equation of state can be derived from basic principles if one assumes:
1. Intermolecular forces are small 2. Volume occupied by the particles is small
These assumptions are true when the molecules are far apart – ie when the gas is not dense
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Criteria
The ideal gas law applies when the pressure is low, and the temperature is high - compared to the critical valuesThe critical values are tabulated in the Appendix
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Virial Equation of State
...5432 vTd
vTc
vTb
vTa
vRTP
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EndFireworks are beautiful Are they Exothermic or endothermic reactions.?
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Tutorial Problems
1.7c, 1.8e, 1.9, 1.12, 1.16c, 1.17c, 1.23c, 1.62, 1.65, 1.66, 2.4c, 2.88, 2.89