Combustion. Adiabatic Flame Temperature - Keenan and Kaye Example
Combustion. Adiabatic Flame Temperature - Keenan and Kaye Example
Combustion. Adiabatic Flame Temperature - Keenan and Kaye Example
Equation for a combustion with excess air. Combustion air is "r" times the theorical air.
C8H18(g) + r * ( 12.5 O2 + 47 N2 )
C8H18(g) + r * 12.5 O2 + r 47
C8H18(g) + r 12.5 O2 + r 47
C8H18(g) + 25 O2 + 94 N2
C8H18(g) + 25 O2 + 94 N2
Reactants Products
NC8H18 = 1 kmolf NH2O = 9
NO2 = 25 kmol/kmolf NCO2 = 8
NN2 = 94 kmolkmolf NO2 = 12.5
NN2 = 94
Na = 119 kmola/kmolf Np = 123.5
1
h P=
NP
{
mf⋅[ h f −h'f −h'rp ] +ma⋅[ ha −h'a ] +h'P } Eq . d
Enthalpy of products
hP = (1/NP) * ( Mf * ( (hf,tR - hf,tref) - hrp ) + ma * ( ha,ta - ha,tref ) ) + hP,tref
hP = #VALUE! [kJ/kmolP]
Enthalpy difference
Dh = hP,bal - hP,comb
Dh = #VALUE!
#VALUE!
Specific enthalpy of octane liquid Constant pressure heat capacity of
[10], page 220 octane liquid
hf = 0.5*TR -287 Btu/lb-liq VBA function
TR: [R] Cpliq = Gas_OctaneSpecificHeatC
1 Btu/lb = 2.326 kJ/kg t= 40 ºC
287 Btu/lb = 668 kJ/kg T= 313.15 kJ/(kg*K)
Cpliq = #VALUE! kJ/(kg*K)
hf = 0.5*TK*1.8 -667.56 Validity range
hf = 0.9 * TK - 667.56 kJ/kg 20.0 ºC <= t <= 45 ºC
The enthalpy of the reactants may be considered to be the sum of the enthalpies
of liquid octane and of air.
Com
C8H18(g)
C 8 H 18 + a O2 +
N: 3.76 * a
3.76 * a
3.76 * a
C8H18(g) + 12.5 O2 + 47 N2
C8H18(g) + 12.5 O2 + 47 N2
Equilibrium equation for the combustion of octane with 200% of theorical air
C8H18(g)
C 8 H 18 + 2 ( 12.5
C 8 H 18 + 25
C8H18(g) + 25 O2 + 94 N2 --->
Reactants Products
NC8H18 = 1 Kmolf NH2O = 9
NO2 = 25 Kmol/Kmolf NCO2 = 8
NN2 = 94 Kmol/Kmolf NO2 = 12.5
NN2 = 94
Reactants Products
NC8H18 = 1 Kmolf NH2O = 9 Kmol/Kmolf
NO2 = 25 Kmol/Kmolf NCO2 = 8 Kmol/Kmolf
NN2 = 94 Kmol/Kmolf NO2 = 12.5 Kmol/Kmolf
NN2 = 94 Kmol/Kmolf
Na = 119 kmola/kmolf Np = 123.5 kmolp/kmolf
Na = 119 kmola/kmolf
Ma = #VALUE! kga/kmola Air #VALUE!
ma = #VALUE! kga/kmolf
Notes
- Valid when reactants and products are at the
same pressure and temperature (Avogadro)
- Water in products is vapor, if the emperature
is high enough to prevent water condensation
iquid octane Air: Products
emperature Ait supply temperature Products reference temperature
ºC ta = 260 ºC tref = 25
at a reference temperature
ta_ref = 25 ºC
9 H2O + 8 CO2 + 47 N2
N2 ---> 9 H2O + 8 +
CO2 (r-1) +1 47
N2 ---> 9 H2O + 8 +
CO2 r * 47
reference temperature
h Nx h
kJ/kmol kmol/kmolF kJ/kmolF
#VALUE! 9 #VALUE! 9898.7428 #VALUE!
#VALUE! 8 #VALUE! 9368.6385 #VALUE!
#VALUE! 12.5 #VALUE! 8669.4126 #VALUE!
#VALUE! 94 #VALUE! 8659.5353 #VALUE!
hP,ref = #VALUE! kJ/kmolF
Np = 123.5 kmolP/kmolF
hP,ref = #VALUE! kJ/kmolP
Combustion
* a = b + 2 * c
* a = 9 + 2 * 8
* a = 9 + 16
* a = 25
a= = 12.5
= 3.76 * a
= 3.76 * 12.5
= 47
Exc = 2 -
O2 + 47 N2 ) ---> 9 H2O +
O2 + 94 N2 ---> 9 H2O +
+ 12.5 O2
Kmol/Kmolf
Kmol/Kmolf
Kmol/Kmolf
Kmol/Kmolf
5 O2 + 47 N)
eference temperature
ºC
page 2 of 9
N2 + (r - 1) * 12.5 O2
47 N2
N2 + (r- 1) * 12.5 O2
N2 + (r - 1) * 12.5 O2
N2 + 1* 12.5 O2
N2 + 12.5 O2
page 3 of 9
page 4 of 9
Enthalpy of combustion of
liquid n-octane
[10], Table 10, page 95, at a
reference temperature
tf,ref = 25 ºC
tf,ref = 77 ºF
Tf,ref = 536.7 R
hRP = -19,100 Btu/lb
hRP = -44,427 kJ/kgf-liq
page 5 of 9
kJ/kmolF
kmolP/kmolF
kJ/kmolP
R
K
ºC
page 6 of 9
page 7 of 9
CO2 + 3.76 *a N2
Solution
a= 12.5 KmolO2/ Kmolf
b= 9 KmolH2O/ Kmolf
c= 8 KmolCO2/ Kmolf
3.76 * a = 47 KmolN2/ Kmolf
N2
N2
page 8 of 9
8 CO2 + 47 N2
8 CO2 + 47 N2
+ 47 N2
+ 94 N2
page 9 of 9
9 * (hH2O(g) - hH2O(g),ref)
8 * (hCO2 - hCO2,ref)
Products
12.5 * (hO2 - hO2,ref)
94 * (hN2 - hN2,ref)
[10], Example 10, page 220 Fuel: Liquid octane
Fuel supply temperature
Flame temperature tF = 15.6
Adiabatic combustion at constant pressure.
Liquid octane, originaly at tF, is burned at Molar mass of fuel
constant pressure in steady flow with air at t a. (sheet Gas data)
The combustion is with Ex % of theorical air. fuel: C8H18
Mf = #VALUE!
From equation
kJ Btu
[ H R−H 'R] =[ H P -H 'P ] +H 'rp kmolf
or
lbmol f
Eq . a
' ' '
HR −H R−H rp=H P -H P
[ H P -H'P ]=[ H R−H 'R ]−H 'rp
Introducing specific enthalpies,
N P⋅[ hP -h 'P ] =[ H R−H 'R ]−H 'rp Eq . b
and disaggregating the enthalpy of the reactants
in the enthalpies of the fuel and of the air
[ H R−H 'R] −H RP =[ H f −H 'f ] + [ H a−H 'a ] −H 'rp
one gets
[ H R−H 'R] −H RP =mf⋅[ h f −h'f ]+ma⋅[ ha−h 'a ] −mf⋅h'rp
[ H R−H 'R] −H RP =mf⋅[ h f −h'f −h'rp ] +ma⋅[ ha−h'a ] Eq . c
mf
[ ]
kg f
kmol f
h f , h'f , and h 'rp
[ ]
kJ
kgf
h P=
1
NP
{
mf⋅[ h
ma
[ ]
kga
kmol f
h a and h'a
[ ]
kJ
kga Microsoft Equation
3.0
Equation for a combustion with excess air. Combustion air is N times the theorical air.
C8H18(g) + r * ( 12.5 O2 + 47 N2 )
C8H18(g) + r * 12.5 O2 + r 47
C8H18(g) + r 12.5 O2 + r 47
C8H18(g) + 25 O2 + 94 N2
C8H18(g) + 25 O2 + 94 N2
Reactants Products
NC8H18 = 1 kmolf NH2O = 9
NO2 = 25 kmol/kmolf NCO2 = 8
NN2 = 94 kmolkmolf NO2 = 12.5
NN2 = 94
Na = 119 kmola/kmolf Np = 123.5
1
h P=
NP
{
m f⋅[ h f −h 'f −h'rp ] +m a⋅[ h a−h'a ] +h'P } Eq . d
Enthalpy of products
hP = (1/NP) * ( Mf * ( (hf,tR - hf,tref) - hrp ) + ma * ( ha,ta - ha,tref ) ) + hP,tref
hP = ### [kJ/kmolP]
Enthalpy difference
Dh = hP,bal - hP,comb
Dh = #VALUE!
#VALUE!
hf = 0.5*TK*1.8 -667.56
hf = 0.9 * TK - 667.56 kJ/kg
The enthalpy of the reactants may be considered to be the sum of the enthalpies
of liquid octane and of air.
Com
C8H18(g)
C 8 H 18 + a O2 +
N: 3.76 * a
3.76 * a
3.76 * a
C8H18(g) + 12.5 O2 + 47 N2
C8H18(g) + 12.5 O2 + 47 N2
Equilibrium equation for the combustion of octane with 200% of theorical air
C8H18(g)
C 8 H 18 + 2 ( 12.5
C 8 H 18 + 25
C8H18(g) + 25 O2 + 94 N2 --->
Reactants Products
NC8H18 = 1 Kmolf NH2O = 9
NO2 = 25 Kmol/Kmolf NCO2 = 8
NN2 = 94 Kmol/Kmolf NO2 = 12.5
NN2 = 94
Reactants Products
NC8H18 = 1 Kmolf NH2O = 9 Kmol/Kmolf
NO2 = 25 Kmol/Kmolf NCO2 = 8 Kmol/Kmolf
NN2 = 94 Kmol/Kmolf NO2 = 12.5 Kmol/Kmolf
NN2 = 94 Kmol/Kmolf
Na = 119 kmola/kmolf Np = 123.5 kmolp/kmolf
Na = 119 kmola/kmolf
Ma = #VALUE! kga/kmola
ma = #VALUE! kga/kmolf
Notes
- Valid when reactants and products are at the
same pressure and temperature (Avogadro)
- Water in products is vapor, if the emperature
is high enough to prevent water condensation
Liquid octane Air: Products
emperature Ait supply temperature Products reference temperat
ºC (Note 1) ta = 260 ºC (Note 1) tref =
at a reference temperature
ta_ref = 25 ºC
kJ/kg (Note 1)
temperature
ºC
es at this reference The enthalpy of the reactants The enthalpy of the reactants may be
may be considered to be the
rimed symbols, considered to be the sum of the enthalpies
e sum of the enthalpies of
liquid octane and of air .
of liquid octane and of air .
The enthalpy of liquid octane The enthalpy of octane can be given as
H 'P -H 'P -H'R h f =0 . 9⋅TK−667
can be given as
H 'P ] + ( H 'P -H'R ) h f =0 .5⋅T −287 kJ
=H RP is the Btu h f : octane enthalpy
h f : octane enthalpy kgf-liq
stion . Thus lbf-liq
TK:Absolute temperture K
H 'P ] +H RP Eq . a T:Absolute temperture R
The state of zero enthalpy
The state of zero enthalpy is 0 K
rakets are
base state selected . is 0 R .
Microsoft Equation
Microsoft Equation 3.0
3.0
Microsoft Equation
Microsoft Equation 3.0
3.0
Replacing equation
superscripts
[ H R−H 'R] −H RP =mf⋅[ h f −h'f −h'rp ] +ma⋅[ ha−h'a ] Eq . c ' : refers to a temperatu
into equation subscripts:
N P⋅[ hP -h 'P ] =[ H R−H 'R ]−H RP Eq . b R:reactants
P:products
N P⋅[ hP -h 'P ] =m f⋅[ h f −h'f −h'rp ] +ma⋅[ h a−h'a ] f:fuel
1
h P -h 'P=
NP {
mf⋅[ hf −h 'f −h 'rp ] +ma⋅[ h a −h'a ] } a:air
H : enthalpy [Btu ]
1 h: specific enthalpy [ B
h P=
NP
{ }
mf⋅[ h f −h 'f −h'rp ] +m a⋅[ h a−h'a ] +h'P Eq . d
h̄: specific enthalpy [ B
N : specific molar mass
Microsoft Equation
3.0
9 H2O + 8 CO2 + 47 N2
+
N2 ---> 9 H2O + 8 +
CO2 (r-1) +1 47
N2 ---> 9 H2O + 8 +
CO2 r * 47
kmol/kmolf
kmol/kmolf
kmol/kmof
kmol/kmolF
kmolp/kmolf
reference temperature
h Nx h
kJ/kmol kmol/kmolF kJkmolF
#VALUE! 9 #VALUE!
#VALUE! 8 #VALUE!
#VALUE! 12.5 #VALUE!
#VALUE! 94 #VALUE!
hP,ref = #VALUE! kJ/kmolF
Np = 123.5 kmolP/kmolF
hP,ref = #VALUE! kJ/kmolP
Specific enthalpy of products Average fuel temperature
at flame temperature Enthalpy change as required in
hP = [kJ/kmolP] equation (d)
Eq . d tf,ave = (tf + tf,ref)/2
Specific enthalpy of products tf = 15.6 ºC
at reference temperature tf,ref = 25 ºC
From page 5 tf,ave = 20.28 ºC
hP,ref = #VALUE! [kJ/kmolP] Constant pressure heat capacity
of octane liquid at average
Mols of products temperature.
From page 12 VBA function with validity range
Np = 123.5 kmolp/kmolF 20 ºC <= t <= 45 ºC
Cpliq_ave = Gas_OctaneSpecificHeatConstantP
Molar mass of fuel tf,ave = 20.3 ºC
fuel: C8H18 T,ave = 293.4 K
Mf = #VALUE! kgF/kmolF Cpliq_ave = #VALUE! kJ/(kg*K)
Although name is Gas_Octane, it corresponds to liquid
Mass of air per mol fuel Fuel enthalpy change
From page 12 hf,t - hf,tref = Cpf,ave * (tf - tf,ref)
ma = #VALUE! kga/kmolf Cpliq_ave = #VALUE! kJ/(kg*K)
tf = 15.6 ºC
tf,ref = 25.0 ºC
hf,t - hf,tref = #VALUE! kJ/kg
Gas_OctaneSpecificHeatConstantPressureLiquid_tK
Enthalpy of products
hP = (1/NP) * ( Mf * ( (hf,tR - hf,tref) - hrp ) + ma * ( ha,ta - ha,tref ) ) + hP,tref
hP = (1/NP) * ( Mf * ( ( (0.9*Tf,TK -668) - (0.9*Tf,Ref -668) ) - Hrp) + ma * ( ha,ta - ha,tref ) ) + hP,tref
hP = (1/NP) * ( Mf * ( 0.9*(Tf,TK -Tf,Ref ) - hrp) + ma * ( ha,ta - ha,tref ) ) + hP,tref
hP = (1/NP) * ( Mf * ( 0.9*(tf,ºC -tf,Ref-ºC ) - hrp) + ma * ( ha,ta - ha,tref ) ) + hP,tref
hP = #VALUE! kJ/kmolP
Return
* a = b + 2 * c
* a = 9 + 2 * 8
* a = 9 + 16
* a = 25
a= = 12.5
= 3.76 * a
= 3.76 * 12.5
= 47
Exc = 2 -
O2 + 47 N2 ) ---> 9 H2O +
O2 + 94 N2 ---> 9 H2O +
+ 12.5 O2
Kmol/Kmolf
Kmol/Kmolf
Kmol/Kmolf
Kmol/Kmolf
5 O2 + 47 N)
Kelv = 273.15 K
Rank = 459.67 R
eactants may be
um of the enthalpies
of air .
ne can be given as
kJ
kgf-liq
ure K
alpy is 0 K
Microsoft Equation
3.0
3
superscripts
' : refers to a temperature of 25 ºC
subscripts:
R:reactants
P:products
f:fuel
a:air
H : enthalpy [Btu ]
h: specific enthalpy [ Btu/lb ]
h̄: specific enthalpy [ Btu/lbmol ]
N : specific molar mass [ Btu/lbmol f ]
m: specific mass [ lb/ lbmol f ]
M f : molar mass of fuel [lb f / lbmol f ]
Microsoft Equation
3.0
N2 + (r - 1) * 12.5 O2
47 N2
*
N2 + (r- 1) * 12.5 O2
N2 + (r - 1) * 12.5 O2
N2 + 1* 12.5 O2
N2 + 12.5 O2
6
Enthalpy of combustion of
liquid n-octane
[10], Table 10, page 95, at a
reference temperature
tf,ref = 25 ºC
tf,ref = 77 ºF
Tf,ref = 536.67 R
hRP = -19,100 Btu/lb
hRP = -44,427 kJ/kgf-liq
eatConstantPressureLiquid_tK(tK)
Microsoft Equation
3.0
ma * ( ha,ta - ha,tref ) ) + hP,tref
kJ/kmolF
kmolP/kmolF
kJ/kmolP
CO2 + 3.76 *a N2
Solution
a= 12.5 KmolO2/ Kmolf
b= 9 KmolH2O/ Kmolf
c= 8 KmolCO2/ Kmolf
3.76 * a = 47 KmolN2/ Kmolf
N2
N2
11
8 CO2 + 47 N2
8 CO2 + 47 N2
+ 47 N2
+ 94 N2
12
9 * (hH2O(g) - hH2O(g),ref)
8 * (hCO2 - hCO2,ref)
Products
12.5 * (hO2 - hO2,ref)
94 * (hN2 - hN2,ref)
Adiabatic flame temperature Denoting the values at this referenc
temperature with primed symbols,
Application of the first law indicates it is posible to write
' '
H R −H R= H P -H R
equality of the enthalpy of reactants
H R and products H P [ H R−H 'R] =H P+H 'P -H 'P -H'R
Thus [ H R−H 'R] =[ H P -H 'P ] +( H 'P -H'R )
H R =H P where ( H 'P -H 'R )=H RP is the
For the evaluation of the enthalpy enthalpy of combustion . Thus
it is considered its value related [ H R−H 'R] =[ H P -H 'P ] +H RP Eq
to the value it has at a reference The quantities in brakets are
temperature t ref=25ºC . independent of the base state selec
Microsoft Equation
3.0
From equation
kJ Btu
[ H R−H 'R] =[ H P -H 'P ] +H 'rp kmolf
or
lbmol f
Eq . a
mf
[ ]
kg f
kmol f
h f , h'f , and h 'rp
[ ]
kJ
kgf
ma
[ ]
kga
kmol f
h a and h'a
[ ]
kJ
kga
Replacing equation
Eq . a [ H R−H 'R] −H RP =mf⋅[ h f −h'f −h'rp ] +ma⋅[ ha−h'a ] Eq . c
into equation
N P⋅[ hP -h 'P ] =[ H R−H 'R ]−H RP Eq . b
N P⋅[ hP -h 'P ] =m f⋅[ h f −h'f −h'rp ] +ma⋅[ h a−h'a ]
1
Eq . b h P -h 'P=
NP {
mf⋅[ hf −h 'f −h 'rp ] +ma⋅[ h a −h'a ] }
N P⋅[ hP -h 'P ] =[ H R−H 'R ]−H RP Eq . b
N P⋅[ hP -h 'P ] =m f⋅[ h f −h'f −h'rp ] +ma⋅[ h a−h'a ]
1
Eq . b h P -h 'P=
NP { }
mf⋅[ hf −h 'f −h 'rp ] +ma⋅[ h a −h'a ]
1
h P=
NP
{ }
mf⋅[ h f −h 'f −h'rp ] +ma⋅[ h a−h'a ] +h'P Eq . d
Microsoft Equation
3.0
1
.c h P=
NP
{
mf⋅[ h f −h'f −h'rp ] +ma⋅[ ha −h'a ] + h'P } Eq . d
Microsoft Equation
3.0
P
Const=
R g⋅T
Microsoft Equation
3.0
ρ=Const⋅MM
1
superscripts
Eq . c ' : refers to a temperature of 25 ºC
subscripts:
Eq . b R:reactants
P:products
f:fuel
a:air
H : enthalpy [Btu ]
Eq . b R:reactants
P:products
f:fuel
a:air
H : enthalpy [Btu ]
.d h: specific enthalpy [ Btu/lb ]
h̄: specific enthalpy [ Btu/lbmol ]
N : specific molar mass [ Btu/lbmol f ]
m: specific mass [ lb/ lbmol f ]
Microsoft Equation
3.0
Gas density
P⋅v= R⋅T
P 1 ρ=Const⋅MM
= =ρ Const=
P
R⋅T v R g⋅T
P
ρ=
R⋅T
with
Rg
R=
MM
P
ρ=
Rg
⋅T
MM
P
ρ= ⋅MM
R g⋅T
P
Const=
R g⋅T
ρ=Const⋅MM
Microsoft Equation
3.0
Octane properties [11] and [13]
101.3 kPa <= P <= 2486 kPa 398.8 K <= T <= 568.8 K
Octane properties
'
1.- Saturation pressure of n-octane
2.- Saturation temperature of n-octane
3.- Specific enthalpy of saturated Liquid octane
4.- Specific enthalpy of saturated Gas octane
5.- Enthalpy of evaporation of octane
6.- Specific heat at constant pressure of saturated Liquid
7.- Specific heat at constant pressure of saturated Gas
8.- Specific Heat at Constant Pressure of Octane Gas
9.- Specific Heat at Constant Pressure of Octane Liquid
- Enthalpy of saturated gas 7.-Specific heat at constant pressure of saturated gas
398.8 K <= T <= 568.8 K 200 [K] <= T <= 1500 [K]
Specific heat at constant pressure of saturated liquid 9.- Constant pressure heat capacity of octane liquid
398.8 <= tK <= 555 293.7 [K] <= T <= 318.15 [K]
Although name is Gas_Octane, it corresponds to liquid 20.55 [ºC] <= t <= 45 [ºC]
Although name is Gas_Octane, it corresponds to liquid
Gas_OctanePsat_tK(tK)
Gas_OctaneTsat_PkPa(PkPa)
Gas_OctaneEnthalpySaturatedLiquid_tK(tK)
Gas_OctaneEnthalpySaturatedGas_tK(tK)
Gas_OctaneEnthalpyEvaporation_tK(tK)
Gas_OctaneSpecificHeatConstantPressureSaturatedLiquid_tK(tK)
Gas_OctaneSpecificHeatConstantPressureSaturatedGas_tK(tK)
Gas_OctaneSpecificHeatConstantPressureGas_tK(tK)
Gas_OctaneSpecificHeatConstantPressureLiquid_tK(tK)
rev.cjc.24.01.2018
tantPressureSaturatedGas_tK
eatConstantPressureGas_tK
eatConstantPressureLiquid_tK
e, it corresponds to liquid
TP = 3032.5 R
TP = 1684.7 K
tP = 1411.6 ºC
1
2
3
4
Gas data
Molecular mass
VBA function
GasMolarMass Gas M HHV Ref
M kg/kmol MJ/kg
kg/kmol (Note 1)
#VALUE! Methane CH4 16.04246 #VALUE! [14]
#VALUE! Ethane C2H6 30.06904 #VALUE! [14]
#VALUE! Propane C3H8 44.09562 #VALUE! [14]
#VALUE! Butane C4H10 58.12220 #VALUE! [14]
#VALUE! Pentane C5H12 72.14878 #VALUE! [14]
#VALUE! Hexane C6H14 86.17536 #VALUE! [14]
#VALUE! Heptane C7H16 100.20194 #VALUE! [14]
#VALUE! Octane C8H18 114.22852 #VALUE! [14]
#VALUE! Nonane C9H20 128.25510 #VALUE! [14]
#VALUE! Decane C10H22 142.28168 #VALUE! [14]
#VALUE! Undecane C11H24 156.30826 #VALUE! [14]
#VALUE! Dodecane C12H26 170.33484 #VALUE! [14]
#VALUE! Hydrogen H2 2.01588 141.8 [14]
#VALUE! Carbon monoxide CO 28.01010 10.112 [14]
#VALUE! Carbon dioxide CO2 44.00950 0
#VALUE! Sulfur (solid) S 32.06500 9.163
#VALUE! Hidrogen sulfide SH2 34.08088 #VALUE! Note 3
#VALUE! Sulfur dioxide SO2 64.06380 0
#VALUE! Water H2O 18.01528 0
#VALUE! Nirogen N2 28.01340 0
#VALUE! Argon Ar 39.94800 0
#VALUE! Oxygen O2 31.99880 0
#VALUE! Carbon (graphite) C 12.01070 32.808 [14]
Neon Ne 20.1797 0
Helium He 4.002602 0
Xenon Xe 131.293 0
0
SH2
M= #VALUE! kg/kmol
Dhreaction = Dhreaction / M
Dhreaction = -614700 kJ/kmol
M= #VALUE! kg/kmol
Dhreaction = #VALUE! kJ/kg
SH2
M= #VALUE! kg/kmol
Dhreaction = Dhreaction / M
Dhreaction = -658700 kJ/kmol
M= #VALUE! kg/kmol
Dhreaction = #VALUE! kJ/kg
1 CO2(g) + 1 H2O(g)
1 -393.5 + 1 -241.8
-393.5 + -241.8
-635.3
1 CO2(g) + 1 H2O(liq)
1 -393.5 + 1 -285.8
-393.5 + -285.8
-679.3
10.04.2018
3] , page 1-7
[ HHV-LHV ] =N H 2O
[ ] [ ]
molw
mol F
⋅( 44 )
kJ
mol w [ ]
kJ
mol
Equation valid for an oxidation process where the Microsoft Equation
3.0
N H 2O
[ kmolw
]
[ ] [ ]
kJ kmol F kJ
( HHV-LHV ) = ⋅( 44000 )
[ ]
kg kg fuel kmol w
MF
kmol F
N H 2O
[ kmolw
]
[ ] [ ]
kJ kmol F kJ
( HHV-LHV ) = ⋅( 44000 )
[ ]
kg kg fuel kmol w
MF
kmol F
1 CH4 + 2 O2 ---> 1
1 C2H6 + 3.5 O2 ---> 2
1 C3H8 + 5 O2 ---> 3
1 C4H10 + 6.5 O2 ---> 4
1 C5H12 + 8 O2 ---> 5
1 C6H14 + 9.5 O2 ---> 6
1 C7H16 + 11 O2 ---> 7
1 C8H18 + 12.5 O2 ---> 8
https://en.wikipedia.org/wiki/Heat_of_combustion
HHV-LHV|F =
HHV-LHV|F =
Difference between the heating values
[ kJ
molfuel ] For a specific fuel with combustion products with N
mols of water per mol of fuel and with a molar mass
the difference between the heating values is
N H2 O
[ kmol w
]
[ ] [
id state kJ kmol F kJ
( HHV-LHV ) = ⋅( 44000 )
[ ]
or state kg kg fuel km
MF
kmol F
J
ol ] DHV [kJ/kg] = 44000 * NH2O (kmolH2O/kmolF) / MF (kgF/kmolF
[ kJ
kmolw ]
[ kJ
kmolw ]
ce in the heating values of some hydrocarbons DHV = HHV - LHV
DHV =
(NH2O / MF) * 44000
ls (complete combustion) Faires (1)
NH2O MF DHV DHV HHV LHV
molw/molF kg/kmol kJ/kg kcal/kg kcal/kg kcal/kg
CO2 + 2 H2O #VALUE! #VALUE! #VALUE! 13,256 11,945
CO2 + 3 H2O #VALUE! #VALUE! #VALUE! 12,391 11,342
CO2 + 4 H2O #VALUE! #VALUE! #VALUE! 12,025 11,072
CO2 + 5 H2O #VALUE! #VALUE! #VALUE!
CO2 + 6 H2O #VALUE! #VALUE! #VALUE!
CO2 + 7 H2O #VALUE! #VALUE! #VALUE!
CO2 + 8 H2O #VALUE! #VALUE! #VALUE!
CO2 + 9 H2O #VALUE! #VALUE! #VALUE!
[ ]
N H2O [ molw ] J
HV-LHV|F =Hv⋅ [ J
mol w ]
⋅
N F [ mol F ] mol F
HHV-LHV|F⋅ kmol [ kJ
]
HV-LHV|F =
F
[ ]
kJ
MF
[ kg
kmol F ] kgF Microsoft Equation
3.0
1
ues
mol w
]
[ ]
mol F kJ
⋅( 44000 )
fuel
mol F ] kmolw
kmolF) / MF (kgF/kmolF)
2
DHV
kcal/kg
1,311
1,049
953
Microsoft Equation
3.0
Higher and lower heating values
[14] LHV
Lower heating value for some organic compounds (at 15.4°C)
HHV LHV
Higher (HHV) and Lower (LHV) Heating values
of some common fuels (NIST Chemistry)
Fuel HHV MJ/kg HHV BTU/lb HHV kJ/mol LHV MJ/kg
Hydrogen 141.8 61,000 286 121
Methane 55.5 23,900 889 50
Ethane 51.9 22,400 1,560 47.8
Propane 50.35 21,700 2,220 46.35
Butane 49.5 20,900 2,877 45.75
Pentane 45.35
Gasoline 47.3 20,400 44.4
Paraffin 46 19,900 41.5
Kerosene 46.2 43
Diesel 44.8 19,300
Coal
27 14,000
(Anthracite)
Coal (Lignite) 15 8,000
Wood 15 6,500
Peat (damp) 6 2,500
Peat (dry) 15 6,500
Adapted from [14]
Isoparaffins
Isobutane 45.613 — — —
Isopentane 45.241 — — —
2-Methylpentane 44.682 — — —
2,3-Dimethylbutane 44.659 — — —
2,3-Dimethylpentane 44.496 — — —
2,2,4-Trimethylpentane 44.31 - — —
Naphthenes
Cyclopentane 44.636 — — —
Methylcyclopentane 44.636 — — —
Cyclohexane 43.45 — — —
Methylcyclohexane 43.38 — — —
Monoolefins
Ethylene 47.195 — — —
Propylene 45.799 — — —
1-Butene 45.334 — — —
cis-2-Butene 45.194 — — —
trans-2-Butene 45.124 — — —
Isobutene 45.055 — — —
1-Pentene 45.031 — — —
2-Methyl-1-pentene 44.799 — — —
1-Hexene 44.426 — — —
Diolefins
1,3-Butadiene 44.613 — — —
Isoprene 44.078 - — —
Nitrous derivated
Nitromethane 10.513 — — —
Nitropropane 20.693 — — —
Acetylenes
Acetylene 48.241 — — —
Methylacetylene 46.194 — — —
1-Butyne 45.59 — — —
1-Pentyne 45.217 — — —
Aromatics
Benzene 40.17 — — —
Toluene 40.589 — — —
o-Xylene 40.961 — — —
m-Xylene 40.961 — — —
p-Xylene 40.798 — — —
Ethylbenzene 40.938 — — —
1,2,4-Trimethylbenzene 40.984 — — —
Propylbenzene 41.193 — — —
Cumene 41.217 — — —
Alcohols
Methanol — — —
Ethanol 28.865 — — —
n-propanol 30.68 — — —
Isopropanol 30.447 — — —
n-Butanol 33.075 — — —
Isobutanol 32.959 — — —
Tertiobutanol 32.587 — — —
n-Pentanol 34.727 — — —
Ethers
Methoxymethane 28.703 — — —
Ethoxyethane 33.867 — — —
Propoxypropane 36.355 — — —
Butoxybutane 37.798 — — —
Aldehydes and ketones
Methanal 17.259 — — —
Ethanal 24.156 — — —
Propionaldehyde 28.889 — — —
Butyraldehyde 31.61 — — —
Acetone 28.548 — — —
Other species
Carbon (graphite) 32.808 — — —
Hydrogen 120.971 — — —
Carbon monoxide 10.112 — — —
Ammonia 18.646 — — —
Sulfur (solid) 9.163 — — —
Note that there is no difference between the lower and higher
heating values for the combustion of carbon, carbon monoxide and
sulfur since no water is formed in combusting those substances
(1)
Mols of water in products / mol fuel
(2)
DeltaHV = 44000 * NH2O (kmolH2O/kmolF) / MF (kgF/kmolF)
which consideres the evaporation enthalpy of the water in the products
See sheet Diff HHV to LHV
Air composition according CRC
Published by Anne Marie Helmenstine
Argon -- Ar -- 0.934%
Ozone -- O3 -- 0.000007%
Iodine -- I2 -- 0.000001%
Carbon Monoxide -- CO -
[2] Termodinamica
Virgil Moring Faires
[3] http://acmg.seas.harvard.edu/people/faculty/djj/book/bookchap1.html
CHAPTER 1. MEASURES OF ATMOSPHERIC COMPOSITION
[5] http://acmg.seas.harvard.edu/people/faculty/djj/book/bookchap1.html
CHAPTER 1. MEASURES OF ATMOSPHERIC COMPOSITION
[7] [7]
Heat ans mass transfer
Yunus A. Cengel & Afshin
[8]
[9]
[13] Termopedia
http://www.thermopedia.com/es/content/996/
[14] Thermal fluids Central
http://www.thermalfluidscentral.org/encyclopedia/index.php/Heat_of_Combustion#References
Heat of combustion
[17] Fandom
Data from NASA, normalized
http://renewableenergy.wikia.com/wiki/Molecular_weight_of_dry_air
[18]
https://www.thoughtco.com/chemical-composition-of-air-604288
Microsoft Equation
3.0
Microsoft Equation
3.0