1
Stoichiometry, Energy Balances, Heat Transfer, Chemical Equilibrium, and Adiabatic Flame
Temperatures
G f SilGeof Silcox
(801)581-8820
University of Utah
Chemical EngineeringChemical Engineering
Salt Lake City, Utah
May 2007
Stoichiometry
2
STOICHIOMETRY
• Outlinecombustion reactions– combustion reactions
– combustion with air
– equivalence ratio, stoichiometric ratio
– percent theoretical air, percent excess air
– percent excess oxygen in stack gases
– dry and wet basis for concentrationsdry and wet basis for concentrations
– actual and theoretical concentrations
– the effects of mixing and in leakage
STOICHIOMETRY
• Combustion reactions
2 2CH O CO H O In general4 2 2 2
2 2
2 2 2
2 2
1
21
2
CH O CO H O
CO O CO
H O H O
In general
2 2
2 24 2
nC nO nCO
m mmH O H O
or 2 2 24 2n m
m mC H n O nCO H O
3
STOICHIOMETRY
• Combustion with air
Assume air is 21% O2 and 79% N2, by volume.Then each mole O2 brings with it 3.76 mole N2.
2 23.764 4n m
m mC H n O n N
2 2 23.762 4
m mnCO H O n N
STOICHIOMETRY
• Equivalence ratio
• Stoichiometric ratio
mols fuelmols oxidant supplied by operator
mols fuelmols oxidant reqd for complete combustion
1SR
SR = 1/ = 1 means that conditions arestoichiometric or chemically balanced.
4
STOICHIOMETRY
• and SRF l i h ( d i ) 1 1SR d – Fuel rich (reducing):
– Fuel lean (oxidizing):
– Alternate forms w/ constant amount of fuel
1 1SR and 1 1SR and
mols oxidant reqd
l id t f d
mols oxidant fed
mols oxidant fedSR
mols oxidant reqd
STOICHIOMETRY
• Percent theoretical air• Percent theoretical air
• Percent excess air
( )100
%TA 100 SR
1 1%EA %TA 100 100 SR 1 100 1
5
STOICHIOMETRY
• Percent excess oxygen in stack gases– Defined as actual mole (volume) percent inDefined as actual mole (volume) percent in
stack. Not to be confused with %EA or %TA.
– Can be measured on a dry or wet basis.
• Dry or wet basis for stack measurement– Dry basis gives higher value because y g g
water vapor acts as a diluent and is 10 –15% of stack gases
STOICHIOMETRY• Actual and theoretical concentrations
0.18
0.2
sis effects of
imperfect mixing
0.06
0.08
0.1
0.12
0.14
0.16
s m
ole
fra
ctio
ns,
dry
ba
s
CO2 theoretical
O2 theoreticalCO theoretical
imperfect mixingeffects of airleakage
0
0.02
0.04
0.8 1 1.2 1.4 1.6 1.8 2
SR
Flu
e g
as
6
STOICHIOMETRY
• Controlling excess air
100 Th i i CO i t
50
100
ppm
CO
The rise in CO is steepas you approach SR = 1and it is insensitive toair leakage. Air flowcontrol by CO shouldlead to a close approachto SR = 1
1.0 1.1 1.2
Stoichiometric ratio
to SR = 1.
STOICHIOMETRY
• Spreadsheets for furnace material• Spreadsheets for furnace material balance calculations– FURN_MB.xls: balance for liquid and solid
fuels
– GAS_MB.xls: balance for gaseous fuels
• Author: Jost Wendt
7
ENERGY BALANCES
ENERGY BALANCES
• Outline– Conservation of mass and energyConservation of mass and energy– Example: drying salt with hot flue gas– Energy balances on systems that involve
reactions - the absolute enthalpy– Example: adiabatic combustion of H2 in air– Heats of combustion and higher and lower heating
values– Energy balance on a furnace – available heat and
the efficiency of combustion
8
ENERGY BALANCES
• Energy and mass are conserved
gz2
1ue 2 V
h = u + pv
Controlvolume
or system
Massentering (1)
Massleaving (2)
Q W
outinCV mm
dt
dm
2 2
, , 1 2
1 2
V V
2 2CV
in net in net
dEQ W m h gz m h gz
dt
ENERGY BALANCES
• Changes in kinetic and potential energy are usually small compared to changes in internal energy
, , 1 1 2 2CV
in net in net
dUQ W m h m h
dt Unsteady
, , 1 1 2 2 0in net in netQ W m h m h Steady
9
ENERGY BALANCES
• Drying salt with combustion gases
Flue gas at 1000F to be used to dry 40 tons/h saltthat is 3 wt. % water. The inlet temperature of thesalt is 40F. The dry salt exits the dryer at190-210F. What flow rate of flue gas do we need if we assume the flue gas leaves the dryer at 400F?
ENERGY BALANCES
• Drying salt with combustion gases
Dryer
1
2
3
4
5
6
1 ?m flue gas, 1000F
dry salt, 40F77,600 lb/h
H O(l) 40F
1 4m m flue gas, 400F
dry salt, 200F77,600 lb/h
H O(g) 400FH2O(l), 40 F2400 lb/h
H2O(g), 400 F2400 lb/h
1 4m m 3 6m m
2 5m m
10
ENERGY BALANCES
• Drying salt with combustion gases– energy balance over adiabatic dryer
Dryer
1
2
3
4
5
6
1 ?m flue gas, 1000F
dry salt, 40F77,600 lb/h
H2O(l), 40F
1 4m m flue gas, 400F
dry salt, 200F77,600 lb/h
H2O(g), 400F
1 1 4 2 2 5 3 3 6 0m h h m h h m h h
2 5 2 3 6 31
1 4
m h h m h hm
h h
2400 lb/h 2400 lb/h
ENERGY BALANCES
• Drying salt with combustion gasesenergy balance over adiabatic dryer– energy balance over adiabatic dryer
2 5 2 3 2 ( ) 5 3 2 ( ) 6 5
11 4
salt H O l fg pH O g
pgas
m c T T m c T T h c T Tm
c T T
978 0.254fg pgas
Btu Btuh c
lb lb F
2 ( )
2 1
0.21 0.47
1.0 36,500
fg pgas
salt pH O g
H O
lb lb FBtu Btu
c clb F lb FBtu lb
c mlb F h
11
ENERGY BALANCES
• Energy balances on reacting systems –the absolute enthalpypy
( ) ( )T
oi f i ref p ih T h T c dT
The reference state for systems in whichreactions are occurring is conveniently chosen to be the elements at Tref = 25 C (298.15 K) and 1 bar.
, ,( ) ( )ref
i f i ref p i
T
h T h T c dT Absoluteenthalpy
Enthalpyof formationat 1 bar, 298 K
Sensibleenthalpychange
ENERGY BALANCES
• Heat capacity as a function of T (K)
( ) ( )T
oh T h T dT, ,( ) ( )ref
oi f i ref p i
T
h T h T c dT 2 3 4
1 2 3 4 5
2 3 43 5 62 4
pca a T a T a T a T
Ra a aa ah
a T T T T
1 2 3 4 5a T T T T
RT T
where T is in Kelvin and R is the universal gasconstant, R = 8.3145 kJ/(kmol K).
Spreadsheet: ther_coef.xls
12
ENERGY BALANCES
• Adiabatic combustion of H2 with air, no dissociation. What is exit temperature?p
Controlvolume
Inlet (i)0.42 kmol/s H2
0.21 kmol/s O2
0.79 kmol/s N2
300 K, 1 bar
Exit (e)0.42 kmol/s H2O0.79 kmol/s N2
T = ?, 1 bar
0inQ 0inW
, , 0in net in net i i e ei e
Q W n h n h
Spreadsheet: h2_flame.xls
ENERGY BALANCES
• Adiabatic combustion of H2
0i i e ei e
n h n h
, ,( ) ( )ref
To
i f i ref p iT
h T h T c dT We have one equation in one unknown T This isWe have one equation in one unknown, T. This isconveniently solved in Excel with the Goal Seek toolto give an exit temperature of 2530K. If we allow fordissociation of the water (at equilibrium), Te = 2390K.
Spreadsheet: h2_flame.xls
13
ENERGY BALANCES
• Heats (enthalpies) of combustion and higher (gross) and lower (net) heating
l (HHV d LHV)values (HHV and LHV)
298rxn p fp r fr
prod react
h h h
4 2 2 22 2 ( )CH O CO H O liq
2984
298
74.600f
kJCH h
mol
4 2 2 2 ( )C O CO O q2982
2982
2982
2982
0
393.510
( ) 285.830
( ) 241.826
f
f
f
f
O h
CO h
H O l h
H O g h
4 2 2 22 2 ( )CH O CO H O gas
ENERGY BALANCES
• Heats of reaction and HHV and LHV
4 2 2 22 2 ( )CH O CO H O liq
298
4
( ) 393.51 2 285.83
74.6 0 890.57 /
rxnh l
kJ mol CH
4 2 2 22 2 ( )CH O CO H O gas
298
4
( )
890.57 /rxnHHV h l
kJ mol CH
Gross heating value
Net heating value4 2 2 2 ( )g
298
4
( ) 393.51 2 241.826
74.6 0 802.56 /
rxnh g
kJ mol CH
298
4
( )
802.56 /rxnLHV h g
kJ mol CH
Net heating value
14
ENERGY BALANCES
• Energy balance over a furnace
Controlvolume
Inlet (i) Exit (e)
inQ 0inW
Use absoluteenthalpies forenergy balancesinvolving reactions.
, 0in net i i e ei e
Q n h n h
, ,( ) ( )ref
To
i f i ref p i
T
h T h T c dT Tref = 298.15 K
ENERGY BALANCES
• Energy balance over a furnace –available heatavailable heat
, , , ,( ) ( )ei
ref ref
TTo o
avail i f i ref p i e f e ref p ei eT T
Q n h T c dT n h T c dT
Available heat = gross heat input Available heat = net heat inputg p– latent heat loss – unburned fuel loss – sensible flue gas heat loss
Available heat = net heat input – unburned fuel loss – sensible flue gas heat loss
15
ENERGY BALANCES
• Thermal efficiency of a combustion process– Definition in terms of HHV
– Definition in terms of LLV
avail
grossfuel
Q
m HHV
Definition in terms of LLV
avail
netfuel
Q
m LHV
HEAT TRANSFER
16
HEAT TRANSFER
• Outline• Outline– Conduction
– Convection
– Radiation
– Example: calculation of inside wall temperature
HEAT TRANSFER
• Review of basic heat transfer
T1
L
– Conduction (Fourier’s law)
1 2 ,K
kAQ T T W
L
M t i l k W/( K)
T1
T2
x
k = thermal conductivity, W/(m K) or W/(m C)
L Material k, W/(m K)
air 0.0263water 0.613copper 401
,K
dTQ kA W
dx
2, /dT
q k W mdx
17
HEAT TRANSFER
– Convection (Newton’s law of cooling)
Th h t t f ffi i t h i d fi d 2 ,C CQ h A T T W
Flow & fluid h, W/m2K
free conv, air 5-12f d i 10 300
T1
L
The heat transfer coefficient, hc, is definedby this equation. Newton’s “law” is not reallya law. It is a useful definition.
forced conv, air 10-300forced conv, water 300-12,000phase change 3,000-50,000(boiling)
T2
T
HEAT TRANSFER
– Radiation (Stefan-Boltzmann law) 4 4
2R surQ A T T
2 22 2 2
2
sur sur sur
R sur
A T T T T T T
h A T T
T1
T2
L
= 5.67x10-8 W/m2K4
T2
T
Tsur
Material , normal emissivity
polished Al 0.05water 0.95planed oak 0.90
18
HEAT TRANSFER
– Example. Given the outside wall temperature in a rotary kiln (T2), estimate the inside wall temperature (T1).the inside wall temperature (T1).
T1
C RK Q QQQq
A A A
At steady state
2 = 0.76h2C = 0.88 Btu/(h ft2 F)
T2
Steel shell, LS = 1“, k = 26 Btu/(h ft F)
Brick, LB = 6“, k = 0.98 Btu/(h ft F)
Deposit, LD = 0.262’, k = 0.9 Btu/(h ft F)
cL
ri = 7.75’T2 = 350 C = 1122 RT = Tsur = 15 C = 519 R
HEAT TRANSFER
– Solution to find T1
T2
T
qcond
qconv
Ti2Ti1T1
Ch
1
sur22
SBD
211TT
1TT
LLLTT
q
2qcond
qradTsur
i2i11
D
D
kL
S
S
kL
B
B
kL
Rh
1
r/rlnTT
length
Q baK
RCSBD hhkkk
where we have neglected the curvature of the wall.
Solving for T1 gives inside wall T = 2600 F = 1460 C.
k2r/rlnlength ab
why?
19
CHEMICAL EQUILIBRIUM
CHEMICAL EQUILIBRIUM
OutlineGibbs free energy and the criterion of gy
equilibriumGas-phase: water-gas shift reactionPresence of solid or liquid phase: steam-
carbon reactionEffect of inerts and pressure: steam-carbon
tireactionFuel-rich combustion of propane
20
CHEMICAL EQUILIBRIUM
Gibbs free energy and the criterion of equilibriumequilibriumG = U + pV – TS = H – TS
The criterion of equilibrium, for a closed system that is held at constant temperature and pressure, is that G has reached its minimum value:minimum value:
0 0dG
or dGdt
CHEMICAL EQUILIBRIUM
Gas-phase reaction: perfect gas mixtureConsider the general reactiong
The criterion of equilibrium means that the equilibrium constant, Kp, and the free energy change for the reaction, G, are related by
aA bB cC dD
y
0ln p TRT K G c dC D
p a bA B
p pK
p p and
21
CHEMICAL EQUILIBRIUM
Gas-phase reaction: water-gas shift
2 2 2CO H O CO H
T = 1000K, p = 1 bar. Initial CO and H2O, 1 mole each.
0 0 0 0 02 2 2
0
( ) ( ) ( ) ( )
395.865 0 ( 200.281 192.603) 2.981
T f f f f
T
G G CO G H G CO G H O
kJG
mol
0G = free energy of formation at 1 bar and 1000K
0 2981exp exp 1.431
8.314 1000
Tp
kJG kmol
KkJRT K
kmol K
fG = free energy of formation at 1 bar and 1000K.
CHEMICAL EQUILIBRIUM
Gas-phase reaction: water-gas shiftLet x moles of H2 form at equilibrium.
2 2
2
2 2 1.4311 1
tot totCO H
pCO H O
x xp pp p
Kx xp p p p
CO 1 – x
H2O 1 – x
CO
2 2 2CO H O CO H
2
2 2tot totp pCO2 x
H2 x
Total 2x = 0.5447
22
CHEMICAL EQUILIBRIUM
Immiscible solid or liquid phase: steam-carbon reaction
2 2( ) ( ) ( ) ( )C s H O g CO g H g
T = 1000K, p = 1 bar. Initially 1 mole H2O, excess C(s).
0 0 0 0 02 2
0
( ) ( ) ( ) ( )
200.281 0 (0 192.603) 7.678
T f f f f
T
G G CO G H G C G H O
kJG
0 7678exp exp 2.518
8.314 1000
Tp
kJG kmol
KkJRT K
kmol K
( )T mol
CHEMICAL EQUILIBRIUM
Immiscible solid or liquid phase: steam-carbon reactionLet x moles of H2 form at equilibrium
2 2( ) ( ) ( ) ( )C s H O g CO g H g
2
2
1 11
tot totCO H
pH O
x xp pp p x xK
xp p
CO x
H2O 1 – x2
2
2
1
2.5181
tot
tot
px
xp
x
H2 x
Total 1 + x
x = 0.8460Spreadsheet: minimizeG.xls
23
CHEMICAL EQUILIBRIUMDoes this value of x (0.8460) really
minimize G?Spreadsheet: minimizeG xls
-205000
-200000
-195000
-190000G
, kJ/
kmo
lSpreadsheet: minimizeG.xls
-215000
-210000
0 0.2 0.4 0.6 0.8 1
extent of reaction
G
CHEMICAL EQUILIBRIUM
Effect of pressure: steam-carbon reaction
2 2( ) ( ) ( ) ( )C s H O g CO g H g
1/ 2
1
/ 1x
K
At 1 bar and 1000 K, x = 0.8460. In general, for thisreaction,
/ 1tot pp K
At 10 bar and 1000 K, x = 0.4485.
Spreadsheet: minimizeG.xls
24
CHEMICAL EQUILIBRIUM
Effect of inert gases: steam-carbon reaction 2 2( ) ( ) ( ) ( )C s H O g CO g H g
Nitrogen will be present in gasifiers if air is used as an oxidizer. Assume that we start with 1 mole of steam, 1 mole of nitrogen, and
b S
CO x
H2O 1 – x
H2 xexcess carbon. Suppose we have x moles of H2 at equilibrium at 1000 K and 1 bar.
N2 1
Total 2 + x
Spreadsheet: minimizeG.xls
CHEMICAL EQUILIBRIUM
Effect of inert gases: steam-carbon reaction 2 2( ) ( ) ( ) ( )C s H O g CO g H g
2
2
2
2 212
( ) 2 0
tot totCO H
pH O
tot
x xp pp p x xK
xp px
p K x K x K
( ) 2 0
1 , 2.518, 0.8910
tot p p p
p
p K x K x K
p bar K x
Recall that with no N2, x = 0.8460
25
CHEMICAL EQUILIBRIUM
Fuel-rich combustion of propane in airReactants: O N C HReactants: O2, N2, C3H8
Products: N2, H2, CO, CO2, H2O, O2
1500K, 1 bar
ER = 1.20, SR = 0.833
1 kmol propane, 19.84 kmol air
Approach: minimize G subject to conservation of C, H, O. The Solver tool in Excel is convenient.
Spreadsheet: cxhy.xls
CHEMICAL EQUILIBRIUM
Fuel rich combustion of propane in airFuel-rich combustion of propane in airGram atoms C
Gram atoms H2x yC H CO COxn n n
2 22 2
x yC H H O Hyn n n Gram atoms O
2 2x y
2 2 22 * 0.21 2 2air CO CO H O On n n n n
Spreadsheet: cxhy.xls
26
CHEMICAL EQUILIBRIUM
Fuel-rich combustion of propane in airMinimization of GMinimization of G
0 ln
i i
i i i
G n
RT p
0 0fi ig = free energy of formation of i at 1 bar and T
ii tot
tot
np p
n
fi ig free energy of formation of i at 1 bar and T.
pi = partial pressure of ini = kmol i
Spreadsheet: cxhy.xls
CHEMICAL EQUILIBRIUM
Fuel-rich combustion Species mole frFuel-rich combustion of propane in air (1500K, 1 bar) –equilibrium concentrations (calculated using
Species mole fr.
N2 0.6913
H2 0.02909
CO 0.04444
CO2 0.08787( gSolver in Excel).
2
H2O 0.1473
O2 0.0000
Spreadsheet: cxhy.xls
27
CHEMICAL EQUILIBRIUM
SummaryAt equilibrium, for a system held at constant
T and p, the Gibb’s free energy will reach its minimum value.
The equilibrium constant is defined by
aA bB cC dD
0ln p TRT K G
aA bB cC dD c dC D
p a bA B
p pK
p p and
CHEMICAL EQUILIBRIUM
SummaryAt lo to moderate press res immiscibleAt low to moderate pressures, immiscible
liquids or solids do not appear in the equilibrium constant but must be included in the calculation of .
If c + d < a + b, an increase in pressure will increase the extent of reaction.
0TG
If c + d > a + b, an increase in pressure will decrease the extent of reaction.
aA bB cC dD
28
CHEMICAL EQUILIBRIUM
SummaryIf c + d > a + b, an increase in the partial pressure of
inert gases will increase the extent of reaction.
Complex equilibrium are conveniently solved with specialized software. One such program is GASEQ. It is available for download at http://www.gaseq.co.uk/.
The Solver in Excel is useful for simpler systems.
aA bB cC dD
ADIABATIC FLAME TEMPERATURE
29
ADIABATIC FLAME TEMPERATURE
OutlineGibbs free energy and the criterion ofGibbs free energy and the criterion of
equilibrium
Adiabatic flame temperature – no dissociation
Adiabatic flame temperature – dissociation
Calculation of g from heat capacity dataCalculation of gf from heat capacity data
ADIABATIC FLAME TEMPERATURE
Gibbs free energy and the criterion of equilibriumequilibriumG = U + pV – TS = H – TS
The criterion of equilibrium, for a closed system that is held at constant temperature and pressure, is that G has reached its minimum value:minimum value:
0 0dG
or dGdt
30
ADIABATIC FLAME TEMPERATURE
Combustion of H2 in air (no dissociation)
Controlvolume
0.42 kmol/s H2
0.21 kmol/s O2
0.79 kmol/s N2
300 K, 1 bar
0.42 kmol/s H2
0.79 kmol/s N2
T = ?, 1 bar
0inQ 0inW
, , 0in net in net i i e ei e
Q W n h n h
Spreadsheet: h2_flame.xls
ADIABATIC FLAME TEMPERATURE
Combustion of H2 in air (no dissociation)
0h h 0i i e ei e
n h n h
, ,( ) ( )ref
To
i f i ref p i
T
h T h T c dT Absoluteenthalpy
Enthalpyof formation
Sensibleenthalpy
We have one equation in one unknown, T. This isconveniently solved in Excel with the Goal Seek toolto give an exit temperature of 2530K.
enthalpy of formationat 1 bar, 298 K
enthalpychange
Spreadsheet: h2_flame.xls
31
ADIABATIC FLAME TEMPERATURE
Adiabatic combustion of H2 in air (allow dissociation of products)
Controlvolume
0.42 kmol/s H2
0.21 kmol/s O2
0.79 kmol/s N2
300 K, 1 bar
? kmol/s H2, H2O, OH,…0.79 kmol/s N2
T = ?, 1 bar
0inQ 0inW
, , 0in net in net i i e ei e
Q W n h n h
0i i e ei e
n h n h
ADIABATIC FLAME TEMPERATURE
Combustion of H2 in air (allow dissociation)dissociation)Reactants: O2, N2, H2
Products: N2, H2, H, OH, H2O, O2, O
ER = 1.0, SR = 1.0
Approach: minimize G subject to conservation of H, O while holding enthalpy constant. Enthalpy and p are specified rather than T and p.
32
ADIABATIC FLAME TEMPERATURE
Combustion of H2 in air (allow dissociation)element balances
Gram atoms H
Gram atoms O2 2 2
2 2 2H i H Oe H e OHe Hen n n n n
Gram atoms O
2 2 22 2O i H Oe O e OHe Oen n n n n
ADIABATIC FLAME TEMPERTURE
Combustion of H2 in air (allow dissociation)dissociation)Minimization of G
0 ln
i i
i i i
G n
RT p
0 0
ii tot
tot
np p
n
0 0fi ig = free energy of formation of i at 1 bar and T.
pi = partial pressure of ini = kmol i
33
ADIABATIC FLAME TEMPERATURE
Combustion of hydrogen in air
Species kmol
N 0 9hydrogen in air (enthalpy constant, p = 1 bar) – equilibrium composition calculated using GASEQ.
T d = 2390 K
N2 0.79
H2O 0.3964
O2 0.00687
H2 0.01794
OH 0 00913Tad 2390 K. OH 0.00913
H 0.00223
O 7.46x10-4
http://www.gaseq.co.uk/
ADIABATIC FLAME TEMPERATURE
Calculation of gf from ideal gas heat capacity datacapacity data
2 3 41 2 3 4 5
2 3 43 5 62 41 2 3 4 5
pca a T a T a T a T
Ra a aa ah
a T T T TRT T
3 5
2 3 43 541 2 7ln
2 3 4
a aasa T a T T T T a
R
Spreadsheet: therm_coef.xls
34
ADIABATIC FLAME TEMPERATURE
Calculation of gf from heat capacity data
Example: free energy of formation of CO
0
1
i iM
Example: free-energy of formation of CO.C(graphite) + 1/2O2 = CO
0 0 0
0 0 0
elements
fi i j jj
fi i j j
h h h
s s s
2
10 (1) ( 1)
2CO C O
0 0 0
elements
fi i j jj
fi fi fi
s s s
g h Ts
Spreadsheet: therm_coef.xls
ADIABATIC FLAME TEMPERATURE
Calculated values for t i hi t i
Fuel HHV LHV Tad (K)stoichiometric combustion with air; 300 K inlet, 1 bar, and allowing for dissociation. Enthalpy and pressure held
(kJ/kg) (kJ/kg)
H2 141,800 120,000 2390
CH4 55,528 50,016 2226
C3H8 50,368 46,357 2267and pressure held constant. n-
octane
C8H18
48,275 44,791 2275
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