Mans Part2

7/24/2019 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


8/28

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


25/28

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

Mans Part2

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