thermodynamics: pure substances
DESCRIPTION
An introductory lecture on Pure substancesTRANSCRIPT
Properties of Pure Substances
Chapter 2
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
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?
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
Why is boiling and condensation of liquids
relevant?
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
Consider what happens when we heat water at
constant pressure
Piston cylinder device –maintains constant pressure
Liquid Water
T
v
1
2
5
3 4
1. Compressed liquid
2. Saturated liquid
3. Saturated mixture
4. Saturated vapor
5. Superheated vapor
Two Phase Region
Compressed Liquid
Superheated Gas
Sometime a pure substance can decrease in temperature after critical point
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
P-V diagram
P
v
T2>T1
T1
Critical Point
Saturated Liquid-vapour region
Superheated vapourCompressed
liquid region
Critical Point
Above the critical point there is no sharp difference between liquid and gas!!
Property Diagrams
So far we have sketched T – v diagram P – v diagram What about the P – T diagram?
What is the difference between paths?
Combine all three
You can put all three properties P T V
On the same diagram
Contracts on Freezing Expands on Freezing
Property Tables
P - pressureT - temperaturev – specific volumeu – specific internal energyh – specific enthalpy h = u + Pvs – specific entropy -define in Chapter 6
Saturated Liquid and Saturated Vapor States
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.
Table A-4 and A-5
A-4 - pg 830 Saturated water temperature table
A-5 - pg 832 Saturated water pressure table
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?
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)
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 …
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!
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
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
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
X = 0 X = 1
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?
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
Superheated Properties
Table A-6
Page 834
In the superheated region, there is only vapourTable looksa little different
Linear Interpolation
A B
100 5
200 10
130 X
5105
100200100130
x
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
Example – putting it all together
Direct Solar Steam Generation
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
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
Equations of State
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
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
Ideal Gas Law
Pv = RT This is the form we will use the most
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
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
Virial Equation of State
...5432 vTd
vTc
vTb
vTa
vRTP
EndFireworks are beautiful Are they Exothermic or endothermic reactions.?
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