Mans Part2
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Transcript of Mans Part2
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Prof. M.N. SABRY 1
ThermodynamicsPart 2: Applications
Psychrometry Equilibrium Thermochemistry
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Prof. M.N. SABRY 2
Ch. 1: Special mixture: Air Water vapor
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Prof. M.N. SABRY 3
Psychrometry
Problem considered
Problem considered
A perfect mixture of perfect gases: Each component : Perfect Gas
A small quantity of matter may change phase
(Gas Liquid ou Gas Solid)
Simplified Model
Simplified Model
gaseous phase is homogeneous and composed of 2 perfect gases
vapor is at low pressure
Treated is a perfect gas, even when close to saturation
Liquid and solid phases are void of dissolved gases
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Prof. M.N. SABRY 4
State of vap.In mixtureT
s
Pvap_sat
Pvap
T s Diagramme forCondensable material
Definitions
Absolute (Specific) HumidityAbsolute (Specific) Humidity
= mvap/ ma mvap mass of vapour
mamass of other gases
= (Pvap V / Rvap T)/ (PaV / RaT) = (vap/a) (Pvap/ Pa)
(For air and water:= 0.622 Pvap/ Pair )
Relative HumidityRelative Humidity
= Pvap/ Pvap_sat
( For air and water: = Pair/ (0.622 Pvap_sat (T))
Pvap_sat(T ) Saturation pressureAt mixture temperature
Max Quantity of vapor when: = = = =100% 13
Dew
point2
Add vapor at Tconst. Until state 3 (reaches 100%)
Starting from state 1, we can : Cool at constPuntil state 2 (reaches 100%)
(beyond points 2 or 3: condensation)
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Prof. M.N. SABRY 5
Adiabatic Saturation
f
1 2Air entering
< 1Air leaving
= 1
1st Law1st Law
T1 : Temperature of state 1 (dry bulb)
T2 : Temperature of state 2 (wet bulb)
maha1+ mv1hv1 + mwhf2= maha2+ mv2hvsat2
mw = mv2- mv1
ha1+ 1 (hv1 - hf2) = ha2+ 2hfg2
But
T wet bulb T dry bulb
Evaporationof water
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Prof. M.N. SABRY 6
Psychometric Chart
T
= 1
= Const < 1(obtained by dividingOrdinates proportionally)
h Tot. = Constwet bulb = Const
Obtained from: = 1 = Pair/ (0.622 Pvap_sat (T))
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Prof. M.N. SABRY 7
Air conditionning processes
1
22
=(2-1)
(2- 1)
Humidification
Drying
Cooling Heating
Cooling and Dehumidification
Dew point
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Prof. M.N. SABRY 8
ThermodynamicsCh2 : Thermodynamic Equilibrium
For Single Component
Equilibrium and thermodynamic potentials
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Prof. M.N. SABRY 9
Equilibrium of an Isolated System
Since isolated: dS/dt = dSirrev/dt 0
At equilibrium: dS/dt = 0 S = max
dS = dSa+ dSb= (dUa+PadVa)/Ta+ (dUb+PbdVb)/Tb
But: dUa= - dUb; dVa= - dVb
dS = 0 (1/Ta - 1/Tb) dUa+ (Pa/Ta- Pb/Tb) dVb=0
Ta= Tb; Pa= Pb
Q = m = 0:
At equilibrium:
Consider an isolated systemComposed of 2 sub-systems (a, b)
exchanging Heat & WorkSub
System
a
SubSystem
b
An isolated
system
. .In general for a control mass : TdS/dt = Q + TdSirrev/dt
.dSirrev/dt 0
Exchange of:-Heat- Work
Single component
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Prof. M.N. SABRY 10
Equilibrium depends on constraints
Tamb = 300 KPamb = 0.1 M PaP = 2 M Pa
T = 300 K T = 500 K
P = 0.1 M Pa
(exchanging only heat ) (exchange only work)
SystemA SystmB
Ambient
Both systems are in equilibrium with ambient,But final equilibrium depends on constraints
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Prof. M.N. SABRY 11
Thermodynamic Potentials
Entropie variation has allowed the study of equilibrium of an isolated system :
T dS = dU + PdV
(isolated: dU total= 0; dV total= 0 dS = 0at equilibrium)
For other constraints, we need other potentials
Gibbs Formula
dU = T dS PdV
dH = dU + d(VP)
Define:
Gibbs Free Energy : G = H TS
dG = -SdT + VdP
Helmholtz Free Energy : A = U TS
dA = -SdT PdV
dH = TdS + VdP
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Prof. M.N. SABRY 12
Variation of potentials
P V
T S
A G U H
PropertiesRelated to work
PropertiesRelated to Heat
To be used ifT, Vconst.
To be used ifT, Pconst. etc.
For a control masse atP, Tconstant :
TdS/dt = Q + TdSirrev/dt.
Tconst.
Pconst.
d(TS)/dt dH/dt
dG/dt = TdSirrev/dt 0
At equilibrium : dG/dt = 0
G = min. (if P,T const.)
Also, if V, Tconst. :Q TdS/dt = dA/dt = TdSirrev/dt 0
.
At equilibrium : dA/dt = 0
A = min. (if V,T const.)
Also,at equilibrium:
U = min. (if S,V const.)
H = min. (ifS,P const.)
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Prof. M.N. SABRY 13
Phase Equilibrium (single component)
Let us study the transition : Liquid vapor atP = const. T = const.
During the transition : Uvap Uliq = Q + W
If transition in equilibrium :
Uvap Uliq = T(Svap Sliq) P(Vvap Vliq)
Q=T(Svap Sliq)W = P (V
vap
Vliq
)
Gvap = Gliq
Another approch (standard) for the study of equilibrium:
Equilibrium at T&Pconst. implies:
dGmix = 0
dGmix = Gliq dmliq + Gvap dmvap
Molecules will cross the interface in both directions.At equilibrium: same number crosses in each direction.
dmliq = - dmvapBut Gvap = Gliq0 = (Gliq Gvap) dmliq
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Prof. M.N. SABRY 14
Clausius-Clapeyron Relation
If the pressure changes by dP , and temperature by dTto restore equilibrium :
dGvap = dGliq
d(Gvap Gliq) = 0 = (Vvap Vliq) dP (Svap Sliq) dT
dP / dT = (Svap Sliq)/ (Vvap Vliq)
dP / dT = (Hvap Hliq)/ T (Vvap Vliq) Clausius Clapeyron Relation
Variation ofP wrt Talong saturation line Latent Heathfg
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Prof. M.N. SABRY 15
Maxwell Relations
Since G, A, H and U are state variables
Hence, dG, dA, dH and dU are total differentials
dU = T dS PdVdG = SdT + VdP dA = SdT PdV dH = TdS + VdP
STGP =
PT TVPS =
PT TVPS =
VPGT =
STAV
=
VT TPVS =
VT TPVS =
PVAT =
TSHP=
PS SVPT =
PS SVPT =
VPHS =
TSUV =
VS SPVT =
VS SPVT =
PVUS =
Maxw1 Maxw2 Maxw3 Maxw4
Maxwell Relations
Starting frommesurable properties (P,V,T)
Get Non - mesurableproperties (S,H,G)
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Prof. M.N. SABRY 16
Enthalpy of a real gas
v+=TT
PsTPh
PT TPs = v
( ) ( ) ( ) += P
PdPTTThTPh
0*, vv( ) ( ) ( ) +=
P
PdPTTThTPh
0*, vv
T
PWe can easily measure:
vvvv=V/m = f(P,T)
h* (T) = h|P0(T) By calorimetry :h* =
vvvv =
h (P,T)= ?
h* (T)=
m h*(T1
).
m h*(T2
).
Q.
To geth(P,T) :
h (P,T)= [h(P,T) h*(T)] + h*(T)
P
TdPPh0
But, the potentielh : dh = dH/m = Tds + vvvv dP
=0 for an ideal gas
But (Maxw1) :
Gibbs Formula
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Prof. M.N. SABRY 17
Entropy of a real gas
T
P
h* =
vvvv, h =
s (P,T)= ?
s* (T)=
s* = Gibbs Formula
To get entropys* @P 0:
Tds = dh vvvv dP
= T
T
Tdhs 0
)(**
PT TPs = v
( ) ( ) = P
PdPTTsTPs
0*, v( ) ( ) =
P
PdPTTsTPs
0*, v
To gets(P,T) :
s (P,T)= [s(P,T) s*(T)] + s*(T)
P
TdPPs0
N.B.: knowing vvvv, h andsas a function ofP and Twe can get :
But (Maxw1) :
u = h Pvvvv = g = h Ts= a = u Ts=
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Prof. M.N. SABRY 18
Constructing thermodynamic tables
Starting from:
vvvv = f(P,T)
h* (T) = h|P=c(T)
P = f(T)
For each phase
For each phase transition
We can calculate for each phase:
( ) ( ) ( ) += P
c PdPTTThTPh vv*,( ) ( ) ( ) +=
P
c PdPTTThTPh vv*, =
TTref T
Tdhs )(** = TTref T
Tdhs )(** ( ) ( ) = P
c PdPTTsTPs v*,( ) ( ) =
Pc P
dPTTsTPs v*,
u = h Pv = g = h Ts= a = u Ts=
We can calculate for each phase transition:
h = T vvvv dP / dT Clausius Clapeyron Relation
s = h / T u =h Pvvvv g =h Ts =0 a = Pvvvv
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Prof. M.N. SABRY 19
ThermodynamicsCh. 3: Thermochemistry
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Prof. M.N. SABRY 20
Thermochemistry
Sources of energy : 38%
23%
24%
7%7%1%
Petrol
Gaz Naturel
Charbon
Hydraulique
Nucleaire
Nouvelle
Types of fuel:
Combustion offossil fuel: 85%
Solid
Liquid
Gaseous
Essentially coal C, rural & urban wastes
Essentially petrol, mixtures of hydrocarbons CnHm (octane=C8H18)
Essentially natural gas, methane CH4 , hydrogen H2
Water electrolysisbiomass
Apply to all chemical reactions,Concentrate on combustion
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Prof. M.N. SABRY 21
Stoichiometric Mixture
If air quantity is just sufficient for COMPLETE combustion
CnHm + aO2 + bN2 cCO2 + dH2O + bN2
air
Reactants Products
We neglect rare gases in air (Argon 1%, )
We neglect formation of nitrogen oxides (except at very high temperature)
Nitrogen exists because it is heated
Rich mixture: air < stoichiometric: incomplete combustion but fast, high T
Poor mixture: air > stoichiometric: complete combustion, T moderate
Stoichiometric mixture
N.B.Composition by volume: 79% N
2, 21% O
2
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Prof. M.N. SABRY 22
Mass balance
If stoichiometric:
CnHm + aO2 + bN2 cCO2 + dH2O + bN2
air
Reactants Products
Equality of C atoms: n = c
Equality of H atoms: m = d/2Equality of O atoms: 2a = 2c + d
Composition of air: b = 79/21 a = 3.76 a
Example :
C8H18+ 12.5O2 + 12.5*3.76N2 8CO2 +9H2O + 12.5*3.76N2
(Air to Fuel Ratio AFR):Mass of air / kg fuel
Mass of fuel: 8*12 + 18*1 = 114Mass of air : 12.5 (16*2 + 3.76*14*2)= 1382
Soichiometric AFR for C8H18= 1382 / 114 = 12.123
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Prof. M.N. SABRY 23
Heat (enthalpy) of formation
For a chemical reaction :
1 kmol of C
1 kmol of O2
1 kmol of CO2Entering at NPT
Leaving at NPT
NPT:T = 25 + 273 K
P = 0.1 MPaReaction
C + O2 CO2
Q (produced) = 393.522 MJ
It is important to fix the reference (i.e. level 0) of uandh.To define u, h
Rules:
by convention, enthalpy of an element (C, O2, N2, ) is 0 at TPN enthalpy of a chemical compound at NPT is:
Enthalpy (or heat) of formation : hfo(in J/kmol) hf
o(in J/kg)
Heat received during formation reaction
hfoofCO2 = -393.522 MJ/kmol hf
oofCO2 = -32.794 MJ/kg
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Prof. M.N. SABRY 24
Applying 1st Law
( ) ( ) ( )[ ]ssinininin PEKEumdt
dPEKEhmPEKEhmWQ ++=++++++ &&&& ( ) ( ) ( )[ ]ssinininin PEKEum
dt
dPEKEhmPEKEhmWQ ++=++++++ &&&&
+=Ri iTPNiinifiinin hhhmhm ,,
0
,&&
+=Ri iTPNiinifiinin hhhmhm ,,
0
,&&
where
+= Pj jTPNjoutjfioutout hhhmhm ,,0
,&& += Pj jTPNjoutjfioutout hhhmhm ,,
0
,&&
R =ReactantsP =Products
Definitions
Adiabatic Flame Temperature
Temperature of products of adiabatic combustion at P const, reactants at NPT
Higher calorific value
Heat produced if reactants et products at NPT, with liquid H2O in products
Lower calorific value
Heat produced if reactants et products at NPT, with vapor H2O in products
Reference Delta
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Prof. M.N. SABRY 25
Third Law of thermodynamics
To fix the reference ofs
Entropy of a perfect crystal at T = 0is 0
3rd Law:3rd Law:
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Prof. M.N. SABRY 26
Equilibrium Many components
Let us mix reactantsA & Band products C & D.at arbitrary proportions
At equilibrium, the number of molespresent for all components:
chemical reaction isreversible& stoichiometric
AA + BB C C + DD
The degree of advance of a reaction: Stoichiomtricnumber of moles
nA, nB, nC, nD = ?nA, nB, nC, nD = ?
dnA = A d
dnB= Bd
dnC = C d
dnD= Dd
0000 = (= (= (= ( A gA B gB+C gC + D gD) d
dG|T,P= gA dnA + gBdnB+ gC dnC + gDdnD= 0
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Prof. M.N. SABRY 27
Chemical Potential
U = U (S, V, n1, n2, ) ni: Numbre of moles of componant i
dU = (U/S|V,ni) dS + (U/V|S,ni) dV + i(U/ni|S,V,nj) dnidU = (U/S|V,ni) dS + (U/V|S,ni) dV + i(U/ni|S,V,nj) dni
If dni=0 T -Pi: Chemical Potential
1 componant Work of changing volume only
2 Degrees of Freedom.dU = TdS PdV (Gibbs Formula)
i.e. U=f(S,V)
For many componants :
For a system with:
In general dU = TdS PdV + iidni
Also: dH = d(U+PV) = TdS + VdP + iidni
dA = d(U-TS) = -SdT - PdV + iidni
dG = d(H-TS) = -SdT + VdP + iidniijij
ijij
nPTinVTi
nPSinVSii
n
G
n
A
n
H
n
U
=
=
=
=
,,,,
,,,,
ijij
ijij
nPTinVTi
nPSinVSii
n
G
n
A
n
H
n
U
=
=
=
=
,,,,
,,,,
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Prof. M.N. SABRY 28
Chemical Equilibrium
d gA|T= RT dln PA
gA= gA|Po+ RT ln (aA)
0000 = (= (= (= ( A GA BGB+C GC + D GD) d
0000 = (= (= (= ( AgA0BgB
0+CgC0+ DgD
0)
++++R (((( A lnaA BlnaB+C lnaC + D lnaD)
Put: G0
Put: ln K = G0/RT
For an ideal gas: aA = yAP/P0
BADC
BA
DC
P
P
yy
yyK
BA
DC
+
=
0
BADC
BA
DC
P
P
yy
yyK
BA
DC
+
=
0
For constant T, ideal gas:
d gA|T= dhA - Tds A