Download - Thermodynamics Lecture Series
Thermodynamics Lecture Series
email: [email protected]://www5.uitm.edu.my/faculties/fsg/drjj1.html
Applied Sciences Education Research Group (ASERG)
Faculty of Applied SciencesUniversiti Teknologi MARA
Pure substances – Pure substances – Property tables and Property tables and Property Diagrams & Property Diagrams &
Ideal GasesIdeal Gases
Submitted-Post-assessmentSubmitted-Post-assessment
Unreadable Self-assessmentUnreadable Self-assessment
CHAPTER
2
Properties of Pure Substances- Part 2
Pure substance
Send self-assessments to:[email protected]@salam.uitm.edu.my
Send self-assessments to:[email protected]@salam.uitm.edu.my
QuotesQuotes
"Education is an admirable thing, but it is well to remember from time to time that nothing that is nothing that is worth knowing can be taughtworth knowing can be taught." - Oscar Wilde
"What we have to learn to do, we learn by doingwe learn by doing." -Aristotle "What we have to learn to do, we learn by doingwe learn by doing." -Aristotle
IntroductionIntroduction
1. Choose the right property table to read and to determine phase and other properties.
2. Derive and use the mathematical relation to determine values of properties in the wet-mix phase
3. Sketch property diagrams with respect to the saturation lines, representing phases, processes and properties of pure substances.
Objectives:Objectives:
IntroductionIntroduction
4. Use an interpolation technique to determine unknown values of properties in the superheated vapor region
5. State conditions for ideal gas behaviour
6. Write the equation of state for an ideal gas in many different ways depending on the units.
7. Use all mathematical relations and skills of reading the property table in problem-solving.
Objectives:Objectives:
Example: A steam power cycle.Example: A steam power cycle.
SteamTurbine
Mechanical Energyto Generator
Heat Exchanger
Cooling Water
Pump
Fuel
Air
CombustionProducts
System Boundaryfor ThermodynamicAnalysis
System Boundaryfor ThermodynamicAnalysis
Steam Power Plant
Phase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure Change
Water when pressure is reducedWater when pressure is reduced
2 =
f@
30 °C4.246
T = 30 C
P = 100 kPa
T = 30 C
P = 100 kPa
H2O:C. liquid
T = 30 C
P = 4.246 kPa
T = 30 C
P = 4.246 kPa
H2O:Sat.
liquid
100
P, kPa
, m3/kg
1
30 C
2 = f @ 30 °C
1 = f@ 30 °C
TTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPaTTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPa
Phase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure Change
H2O:Sat. Liq.
Sat. VaporSat. Vapor
3
T = 30 C
P = 4.246 kPa
T = 30 C
P = 4.246 kPa
H2O:Sat.
liquid
Water when pressure is reducedWater when pressure is reduced
2 =
f@
30 °C
, m3/kg1
100
P, kPa
4.24630 C
3 = [f + x f g]@ 30 °C
2 = f@30 °C
1 = f@ 30 °C
TTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPaTTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPa
Phase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure Change
4 =
g@
30 °C
H2O:Sat. Vapor
H2O:Sat. VaporH2O:
Sat. Liq.
Sat. VaporSat. Vapor
T = 30 C
P = 4.246 kPa
T = 30 C
P = 4.246 kPa
Water when pressure is reducedWater when pressure is reduced
1
3
P, kPa
, m3/kg2 =
f@
30 °C
100
4.24630 C 4 = g@ 30 °C
3 = [f + x f g]@ 30 °C
1 = f@ 30 °C
2 = f@ 30 °C
TTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPaTTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPa
Phase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure Change
H2O:Sat. Vapor
H2O:Sat. Vapor
T = 30 C
P = 4.246 kPa
T = 30 C
P = 4.246 kPa
2
5
H2O:SuperVapor
H2O:SuperVapor
T = 30 C
P = 2 kPa
T = 30 C
P = 2 kPa
Water when pressure is reducedWater when pressure is reduced
4 =
g@
30 °C
2 =
f@
100 kPa
, m3/kg1
3
2 =
f@
30 °C
1
3
100
P, kPa
4.24630 C
3 = [f + x f g]@ 30 °C
4 = g@ 30 °C
1 = f@ 30 °C
2 = f@ 30 °C
5= @2kPa, 30 °C
TTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPaTTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPa
Phase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure Change
H2O:Sat. Vapor
H2O:Sat. Vapor
T = 30 C
P = 4.246 kPa
T = 30 C
P = 4.246 kPa
H2O:SuperVapor
H2O:SuperVapor
T = 30 C
P = 2 kPa
T = 30 C
P = 2 kPa
H2O:Sat. Liq.
Sat. VaporSat. Vapor
T = 30 C
P = 4.246 kPa
T = 30 C
P = 4.246 kPa
T = 30 C
P = 100 kPa
T = 30 C
P = 100 kPa
H2O:C. liquid
T = 30 C
P = 4.246 kPa
T = 30 C
P = 4.246 kPa
H2O:Sat.
liquid
Water when pressure is reducedWater when pressure is reduced
TTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPaTTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPa
Phase Change of Water- Pressure ChangePhase Change of Water- Pressure ChangePhase Change of Water- Pressure ChangePhase Change of Water- Pressure Change
Compressed liquidCompressed liquid: Good : Good estimation for properties estimation for properties by taking yby taking y = y = yf@T f@T where where
y can be either y can be either , u, h or , u, h or s.s.
2 =
f@
30 °C
4.246
3
2
5
4 =
g@
30 °C
, m3/kg
1
100
P, kPa
30 C 3 = [f + x f g]@ 30 °C
4 = g@ 30 °C
1 = f@ 30 °C
2 = f@ 30 °C
5= @2kPa, 30 °C
TTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPaTTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPa
Phase Change of WaterPhase Change of WaterPhase Change of WaterPhase Change of Water
1,553.8
200 C
1.2276 10 C
g@100 C
P, C
, m3/kg
101.35 100 C
f@
100 C
Phase Change of WaterPhase Change of WaterPhase Change of WaterPhase Change of Water
P, C
, m3/kg
101.35
g@
100 C
1,553.8
1.2276
200 C
10 C
100 C
f@
100 C
22,090
P- diagram with respect to the saturation
lines
P- diagram with respect to the saturation
lines
T, C
, m3/kg
T – v diagram - Example
70
=f@70 C = 0.001023
50 kPa
P, kPa T, C
50 70
Phase, Y?
Compressed Liquid,
T < Tsat or P > Psat
, m3/kg
f@70 C
Tsat, C
81.33
Psat, kPa
31.19
81.3
3.2400.001030
P, kPa
, m3/kg
P – v diagram - Example
P, kPa T, C
50 70
Phase, Y?
Compressed Liquid,
P > Psat or T < Tsat
, m3/kg
f@70 C
70 C
31.19
5.0420.001023
Tsat, C
81.33
Psat, kPa
31.19
=f@70 C = 0.001023
50
400 C
P – v diagram - Example
P, kPa T, C
200 400
P- diagram with respect to the saturation
lines
P- diagram with respect to the saturation
lines
Phase, Why?
Sup. Vap., T >TSup. Vap., T >Tsatsat
Psat, kPa Tsat, C
NA 120.2120.2
P, kPa
, m3/kg
22,090.0
= 1.5493
, m3/kg
1.5493
200
f@200 kPa
= 0.001061
g@200 kPa
= 0.8857
120.2 C
T – v diagram - Example
T, C
, m3/kg
1,00
0 kP
a
P, kPa u, kJ/kg
1,000 2,000
T- diagram with respect to the saturation
lines
T- diagram with respect to the saturation
lines
Phase, Why?
Wet Mix., uWet Mix., uf f < u < u< u < ugg
Psat, kPa Tsat, C
179.9179.9
374.1
f@1,000 kPa
= 0.001127
179.9
g@1,000 kPa
= 0.19444
T, C
179.9179.9
= [f + x f g]@1,000 kPa
Property TableSaturated water – Temperature table
Temp
Tsat, C
10
50
100
200
300
374.14
Specific internal energy, kJ/kg
uf, kJ/kg ufg, kJ/kg ug, kJ/kg
42.00 2347.2 2389.2
209.32 2234.2 2443.5
418.94 2087.6 2506.5
850.65 1744.7 2593.3
1332.0 1231.0 2563.0
2029.6 0 2029.6
Specific volume, m3/kg
f, m3/kg g, m3/kg
0.001000 106.38
0.001012 12.03
0.001044 1.6729
0.001157 0.13736
0.001404 0.02167
0.003155 0.003155
Sat. P.
P, kPa
1.2276
12.34912.349
P, MPaP, MPa
0.102350.10235
1.55381.5538
8.5818.581
22.0922.09
Saturated Liquid-Vapor Mixture
H2O:Sat. Liq.
Sat. VaporSat. Vapor
gf VVV t
g
m
mx
Given the pressure, P, then T = Tsat, yf < y <yg
Given the pressure, P, then T = Tsat, yf < y <yg
Vapor Phase:, VVapor Phase:, Vgg, m, mgg, , gg, u, ugg, h, hgg
Liquid Phase:, VLiquid Phase:, Vff, m, mff, , ff, u, uff, h, hff
Mixture:, V, m, Mixture:, V, m, , u, h, x, u, h, x
gft mmm
ggffavgt mmm Specific volume of Specific volume of mixture?? Since mixture?? Since
V=m V=m
Specific volume of Specific volume of mixture?? Since mixture?? Since
V=m V=m
Mixture’s qualityMixture’s qualityMixture’s qualityMixture’s quality
More vapor, higher More vapor, higher qualityquality
x = 0 for saturated x = 0 for saturated liquidliquid
x = 1 for saturated x = 1 for saturated vaporvapor
More vapor, higher More vapor, higher qualityquality
x = 0 for saturated x = 0 for saturated liquidliquid
x = 1 for saturated x = 1 for saturated vaporvapor
Saturated Liquid-Vapor Mixture
H2O:Sat. Liq.
Sat. VaporSat. Vapor
t
g
m
mx
Vapor Phase:, VVapor Phase:, Vgg, m, mgg, , gg, u, ugg, h, hgg
Liquid Phase:, VLiquid Phase:, Vff, m, mff, , ff, u, uff, h, hff
Mixture:, V, m, Mixture:, V, m, , u, h, x, u, h, xMixture’s qualityMixture’s qualityMixture’s qualityMixture’s quality
Divide by Divide by total mass, total mass,
mmtt
Divide by Divide by total mass, total mass,
mmtt
gft
gavg x
m
m
1
ggffavgt mmm
fgfg wherewherewherewhere
fgf xx fgf x
ggfgt mmm
Given the pressure, P, then T = Tsat, yf < y <yg
Given the pressure, P, then T = Tsat, yf < y <yg
Saturated Liquid-Vapor Mixture
H2O:Sat. Liq.
Sat. VaporSat. Vapor
t
g
m
mx
Vapor Phase:, VVapor Phase:, Vgg, m, mgg, , gg, u, ugg, h, hgg
Liquid Phase:, VLiquid Phase:, Vff, m, mff, , ff, u, uff, h, hff
Mixture:, V, m, Mixture:, V, m, , u, h, x, u, h, xMixture’s qualityMixture’s qualityMixture’s qualityMixture’s quality
Given the pressure, P, then T = Tsat, yf < y <yg
Given the pressure, P, then T = Tsat, yf < y <yg
fgfg wherewherewherewherefgf x
fgfg yyy wherewherewherewherefgf xyyy
If x is known or has been determined, use If x is known or has been determined, use above relations to find other properties. If above relations to find other properties. If
either either , , u, h are known, use it to find quality, x. h are known, use it to find quality, x.
If x is known or has been determined, use If x is known or has been determined, use above relations to find other properties. If above relations to find other properties. If
either either , , u, h are known, use it to find quality, x. h are known, use it to find quality, x.
y can be y can be , u, h, u, hy can be y can be , u, h, u, h
Interpolation: Example – Refrigerant-134a
T, C
, m3/kg
TH
L
TL
H
T = ??
m1
m2
21 mslopemslope
LH
LH
L
L TTTT
LH
LLHL
TTTT
P, kPa , m3/kg
200 0.10600
Phase, Why?
Sup. Vap., Sup. Vap., > > gg
Psat, kPa Tsat, C
- -10.09-10.09
T, C
??
Assume properties are linearly dependent. Perform interpolation in superheated vapor phase.
Assume properties are linearly dependent. Perform interpolation in superheated vapor phase.
Interpolation: Example – Refrigerant-134a
P, kPa , m3/kg
200 0.10600
Assume properties are linearly dependent.
Perform interpolatio
n in superheate
d vapor phase.
Assume properties are linearly dependent.
Perform interpolatio
n in superheate
d vapor phase.
Phase, Why?
Sup. Vap., Sup. Vap., > > gg
Psat, kPa Tsat, C
- -10.09-10.09
T, C
??
T, C , m3/kg u, kJ/kg
TL= 0 L = 0.10438 uL = 229.23
0 < T0 < TLL < T < THH = 0.10600 0 < u0 < uLL < u < uHH
TH= 10 H = 0.10922 uH = 237.05
LH
LLHL
TTTT
LH
LLHL
uuuu
Interpolation: Example – Refrigerant-134a
P, kPa , m3/kg
200 0.10600
P- diagram with respect to the saturation
lines
P- diagram with respect to the saturation
lines
Phase, Why?
Sup. Vap., Sup. Vap., > > gg
Psat, kPa Tsat, C
- -10.09-10.09
P, kPa
, m3/kg
22,090.0
T, C
3.353.35
f@200 kPa
= 0.0007532
g@200 kPa
= 0.0993
200
-10.09 C
u, kJ/kg
231.85231.85
= 0.10600
T = 3.35 CPsat ??
Properties of Pure Substances- Ideal Gases Equation of State
Ideal Gases
Ideal Gases
• Equation of StateEquation of State– An equation relating pressure, temperature and An equation relating pressure, temperature and
specific volume of a substance.specific volume of a substance.
– Predicts P- Predicts P- -T behaviour quite accurately-T behaviour quite accurately
– Any properties relating to other propertiesAny properties relating to other properties
– Simplest EQOS of substance in gas phase is Simplest EQOS of substance in gas phase is ideal-gasideal-gas ((imaginary gasimaginary gas) equation of state) equation of state
Ideal Gases
• Equation of State for ideal gasEquation of State for ideal gas– Boyle’s Law: Pressure of gas is inversely Boyle’s Law: Pressure of gas is inversely
proportional to its specific volume P proportional to its specific volume P VP /1
• Equation of State for ideal gasEquation of State for ideal gas– Charles’s Law: At low pressure, volume is Charles’s Law: At low pressure, volume is
proportional to temperatureproportional to temperature TV
Ideal Gases
• Equation of State for ideal gasEquation of State for ideal gas– Combining Boyles and Charles laws:Combining Boyles and Charles laws: kgkJRTP / ,
M
RR u
and where and where RRuu is universal gas constant is universal gas constant RRuu = 8.314 kJ/kmol.K = 8.314 kJ/kmol.Kand where and where RRuu is universal gas constant is universal gas constant RRuu = 8.314 kJ/kmol.K = 8.314 kJ/kmol.K
and where M is molar massand where M is molar massand where M is molar massand where M is molar mass
RTm
VP So,So,So,So, kJmRTPV ,
TM
RMNPV uEQOSEQOS: : since the masssince the mass m = MNm = MN
where N iswhere N is number of moles:number of moles:EQOSEQOS: : since the masssince the mass m = MNm = MN where N iswhere N is number of moles:number of moles:
So,So,So,So, kJTNRPV u ,
where gas constant R iswhere gas constant R iswhere gas constant R iswhere gas constant R is
EQOSEQOS: : Since the total volume Since the total volume is V = mis V = m, so : , so : = V/m = V/m
EQOSEQOS: : Since the total volume Since the total volume is V = mis V = m, so : , so : = V/m = V/m
Ideal Gases
• Equation of State for ideal gasEquation of State for ideal gas– Real gases with low densities behaves like an ideal gasReal gases with low densities behaves like an ideal gas
kgkJRTP / ,
M
RR u
P << PP << Pcrcr, T >> T, T >> TcrcrP << PP << Pcrcr, T >> T, T >> Tcrcr
where,where,where,where,
kJmRTPV , kJTNRPV u ,
Hence real gases satisfying conditionsHence real gases satisfying conditionsHence real gases satisfying conditionsHence real gases satisfying conditions
RRuu = 8.314 kJ/kmol.K, = 8.314 kJ/kmol.K,RRuu = 8.314 kJ/kmol.K, = 8.314 kJ/kmol.K, V = m andV = m andm = MNm = MNm = MNm = MN
Obeys EQOSObeys EQOSObeys EQOSObeys EQOS
Ideal Gases
Gas Mixtures – Ideal GasesGas Mixtures – Ideal Gases
Low density (mass in 1 m3) gases
Molecules are further apart
Real gases satisfying condition
PPgasgas << P << Pcritcrit; T; Tgasgas >> T >> Tcrit crit , have low density and can be treated as ideal gases
High density
Low densityMolecules far apart
Ideal Gases
Gas Mixtures – Ideal GasesGas Mixtures – Ideal Gases
Equation of StateEquation of State - P--T behaviour
PP=RT=RT (energy contained by 1 kg mass) where is the specific volume in m3/kg, RR is gas constant, kJ/kgK, TT is absolute temp in Kelvin.
High density
Low densityMolecules far apart
Ideal Gases
Gas Mixtures – Ideal GasesGas Mixtures – Ideal Gases
Equation of StateEquation of State - P--T behaviour
PP=RT=RT , since = V/m then,P(V/m)=RT. So,
PV =mRTPV =mRT, in kPam3=kJ.Total energy of a system.
Low density
High density
Ideal Gases
Gas Mixtures – Ideal GasesGas Mixtures – Ideal Gases
Equation of StateEquation of State - P--T behaviour
PV =mRTPV =mRT = NMRT = N(MR)THence, can also write PV = PV = NRNRuuTT whereNN is no of kilomoles, kmol,MM is molar mass in kg/kmole and
RRuu is universal gas constant; RRuu=MR=MR.
RRuu = 8.314 kJ/kmol = 8.314 kJ/kmolKK
Low density
High density
T – v diagram - Example
T, C
, m3/kg
1,00
0 kP
a
P, kPa u, kJ/kg
1,000 2,000
T- diagram with respect to the saturation
lines
T- diagram with respect to the saturation
lines
Phase, Why?
Wet Mix., uWet Mix., uf f < u < u< u < ugg
Psat, kPa Tsat, C
179.9179.9
374.1
f@1,000 kPa
= 0.001127
179.9
g@1,000 kPa
= 0.19444
T, C
179.9179.9
= [f + x f g]@1,000 kPa