Fuels and energy
heat
mechanical
work
heat
fuel
mechanical
work
heat
mechanical
work
light
electric energy
light
sound
Heatgenerators
2
Heatgenerators
3
Fuel composition
1
Fuel, air and flue gases
fuel
solid,liquid:
C,H,S,O,N,H2O,ash
gaseous:
H2,CxHy,CO2,N2,...
+
flue gases
N 2,O2,H2O,CO2,SO2
+
air
N 2,O2,(H2O,CO2,Ar,...)
heat
Q g =mg H i
Heatgenerators
6
Heatgenerators
7
Combustion
2
Combustion - air
C+O2 → CO 2
1kmol C+ 1kmol O2 → 1 kmol CO2 + 407353 kJ
12,011 kg C + 31,9988 kg O2 → 44,0098 kg CO2 + 407353 kJ
1kg C+ 2,6641 kg O2 → 3,6641 kg CO2 + 33915 kJ
2H 2 +O2 → 2H 2O
1kgH2 + 7,9370 kg O2 → 8,9370 kg H2O + 141800 kJ
S+O2 → SO 2
1kgS+ 0,9981 kgO2 → 1,9981 kg SO2 + 10467 kJ
requiredquantityofoxygen/air(kg/kgfuel)
mO,min = 2,6641 wC + 7,9370 w H + 0,9981 wS − wO
mO,min
mA,min =
0 ,23
actualquantityofair(kg/kgfuel)
mA = λ mA,min
λ – airratio
Heatgenerators
10
Combustion - air
C+O2 → CO 2
1kmol C+ 1kmol O2 → 1 kmol CO2 + 407353 kJ
12,011 kg C + 31,9988 kg O2 → 44,0098 kg CO2 + 407353 kJ
1kg C+ 2,6641 kg O2 → 3,6641 kg CO2 + 33915 kJ
2H 2 +O2 → 2H 2O
1kgH2 + 7,9370 kg O2 → 8,9370 kg H2O + 141800 kJ
S+O2 → SO 2
1kgS+ 0,9981 kgO2 → 1,9981 kg SO2 + 10467 kJ
requiredmassflowrateofoxygen/air(kg/kgfuel)
m
& O,min = m
& f mO,min
m
& A,min = m
& f m A,min
actualmassflowrateofair(kg/kgfuel)
m
&A =m
& f mA = m
& f λ mA,min
Heatgenerators
11
3
Combustion - flue gases
C+O2 → CO 2
1kmol C+ 1kmol O2 → 1 kmol CO2 + 407353 kJ
12,011 kg C + 31,9988 kg O2 → 44,0098 kg CO2 + 407353 kJ
1kg C+ 2,6641 kg O2 → 3,6641 kg CO2 + 33915 kJ
2H 2 +O2 → 2H 2O
1kgH2 + 7,9370 kg O2 → 8,9370 kg H2O + 141800 kJ
S+O2 → SO 2
1kgS+ 0,9981 kgO2 → 1,9981 kg SO2 + 10467 kJ
quantityofproducedfluegas(kg/kgfuel)
mCO2 =3,6641wC
carbondioxide
mH2O =8,9370wH +wH2O
water
mSO2 =1,9981wS
sulphurdioxide
mN2 =wN +0,77mA,min
nitrogen
mair =(λ – 1)mA,min
air=nitrogen+oxygen
Heatgenerators
13
Combustion - flue gases
C+O2 → CO 2
1kmol C+ 1kmol O2 → 1 kmol CO2 + 407353 kJ
12,011 kg C + 31,9988 kg O2 → 44,0098 kg CO2 + 407353 kJ
1kg C+ 2,6641 kg O2 → 3,6641 kg CO2 + 33915 kJ
2H 2 +O2 → 2H 2O
1kgH2 + 7,9370 kg O2 → 8,9370 kg H2O + 141800 kJ
S+O2 → SO 2
1kgS+ 0,9981 kgO2 → 1,9981 kg SO2 + 10467 kJ
quantityofproducedfluegas(kg/kgfuel)
mfg =mCO2 +mH2O +mSO2 +mN2 +mair
massflowrateofproducedfluegases(kg/s)
m
& fg = m
& f m fg
Heatgenerators
14
4
Combustion - flue gases
CO2
H2 O
SO2
N2
m CO2
m H2O
m SO2
m N2th
air
m air
=
CO2
H2 O
SO2
m CO2
m H2O
m SO2
N2
m N2
O2
m O2
Heatgenerators
15
Combustion - example
Fuel:biodiesel
wC =77%
wH =12%
wO =11%
mO ,min = 2,6641·0,77 + 7 ,9370·0 ,12 + 0 ,9981·0 − 0 ,11 = 2,89 kg/kg
m A,min =
2,89
= 12,59kg/kg
0 ,23
m A = 1 ,1·12,59 = 13,84kg/kg
m
& A = 0,2·13,84 = 2,77kg/s
Airratio:
λ =1,1
Fuelmassflowrate:
mg =0,2kg/s
m CO2 = 3,6641·0,77 = 2,82 kg/kg
m H2O = 8,9370·0,12 + 0 = 1,07 kg/kg
m SO2 = 1,9981·0 = 0 kg/kg
m N2 = 0 + 0,77·12,59 = 9,69 kg/kg
m air = (1,1 – 1)·12,59 = 1,26 kg/kg
m fg = 2,82 + 1,07 + 9,69 + 1,26 = 14,84 kg/kg
m
& fg = 0,2·14 ,84 = 2,97kg/s
Heatgenerators
19
5
Combustion - heat
C+O2 → CO 2
1kmol C+ 1kmol O2 → 1 kmol CO2 + 407353 kJ
12,011 kg C + 31,9988 kg O2 → 44,0098 kg CO2 + 407353 kJ
1kg C+ 2,6641 kg O2 → 3,6641 kg CO2 + 33915 kJ
2H 2 +O2 → 2H 2O
1kgH2 + 7,9370 kg O2 → 8,9370 kg H2O + 141800 kJ
S+O2 → SO 2
1kgS+ 0,9981 kgO2 → 1,9981 kg SO2 + 10467 kJ
releasedenergy- lowerandhigherheatingvalue(MJ/kgfuel)
wO 

H i = 33,9 ⋅ wC + 121,4 ⋅  w H −
 + 10,5 ⋅ wS − 2,5 ⋅ w H2O
8 

H s = H i + 2,5 ⋅ mH2O
releasedheatflowrate(MW)
Q& f = m
& f ⋅ Hi
Heatgenerators
21
Carbon dioxide
VCO2
Vd ,0 , s
maximumCO2 volumefractioninfluegases
φ(CO2 ) =
volumeofCO2 (Nm3/kgfuel)
VCO2 =
theor.volumeofdryfluegases(Nm3/kgfuel)
V fg ,0 ,d = VCO 2 + VH2O + VSO2
massofCO2 perunitreleasedheat(kg/MJ)
mCO2
ρ CO2 ,0
ˆ CO2 =
m
= 1,8563·wC
mCO2
Hi
Normalcubicmeter(Nm3)representsthequantity(mass)ofagasthatfillsavolume
of1m3 atnormalconditions(T0 =0°C,p0 =1,01325bar).Atotherconditionsvolume
mightchange,butthemassremainsconstant.
Heatgenerators
24
6
Enthalpy of flue gases
specificenthalpy(kJ/kgfg)
h − h0 = c p T ⋅ (T − T0 )
T
0
h=cp·T
h fg = T ⋅ ∑ (c p ⋅ w fg )i = ∑ (h ⋅ w fg )i
heatflowrate(kW)
Q& = m
& fg ⋅ (h fg ,1 − h fg ,2 )
Q& = m
& fuel ⋅ m fg ⋅ (h fg ,1 − h fg ,2 ) =
=m
& fuel ⋅ (m fg ⋅ h fg ,1 − m fg ⋅ h fg ,2 )
Q& = m
& fuel ⋅ (H fg ,1 − H fg ,2 )
enthalpyoffluegases(kJ/kgfuel) H fg = ∑ (h ⋅ m )i
Heatgenerators
27
Enthalpy of flue gases, Hfg - T diagram
temperatura Td / °C
0
400
800
1200
1 600
2000
2 400
30
28
7
26
24
6
22
20
5
entalpija H d / (MJ/kgg )
18
16
4
14
12
3
8
2
6
4
1
entalpija H d / (MJ/kgg)
10
2
0
0
-200
-100
0
100
200
300
400
te mpe ratura Td / °C
Heatgenerators
28
7
Theoretical temperature of combustion
energybalanceinfurnace:
m
& f Hi + m
& f hf + m
& a ha = m
& fg h fg
heatflowrateoffluegases
heatflowrateofcombustionair
sensibleheatflowrateoffuel(negligible)
heatflowratereleasedfromfuel
Hfg
m
& f Hi + m
& f λ mA ,min ha = m
& f m fg ,m h fg
H i + mA ha = H fg ,th
Hfg,th
Tfg,th
T
Heatgenerators
31
Energy of flue gases
composition
CO 2
heat
Q=md cp (T– T0)
Q =mCO2 cp,CO2 (T– T0)+
steam
water
SO 2
+ mw cp,w (T– T0)+ m s cp,s (T– T0)+m s (h” – h’)+
+mSO2 cp,SO2 (T– T0)+
+mN2 cp,N2 (T– T0)+
N2
+mO2 cp,O2 (T– T0)
O2
Heatgenerators
32
8
Enthalpy of flue gases
H s = H i + 2500 mH2O
theoretical point after
adiabatic combustion
enthalpy decrease during
coolingof flue gases
enthalpy
enthalpy in the temperature
interval of condensation
virtual enthalpy
curve
parallel move to the
origin of the diagram
virtual point after
adiabatic combustion
Hs
Hi
temperature
Heatgenerators
33
9
Univerza v Ljubljani
Fakulteta za strojništvo
Laboratorij za termoenergetiko
HEAT GENERATORS / FUELS AND COMBUSTION
4
Fuel properties
1)
For a given fuel calculate the following stoichiometric parameters:
a) lower and higher heating value
b) minimal required quantity of oxygen and combustion air
c) minimal quantity of dry and wet flue gases
d) maximal fraction of CO2 in flue gases
e) quantity of CO2 produced per unit released heat
2) Analyse the influence of excess air ratio on flue gas composition. Show the results in a
diagram for excess air ratios from 1 to 5.
3) Draw the Hfg-T chart for flue gases for temperature range of 0 to 2000 °C and for
various excess air ratios (from 1 to 2).
4) Find theoretical temperature of combustion when combustion air temperature is
0 °C. Graphically show the relation of theoretical temperature and excess air ratio
(from 1 to 2).
5) Find fractions of combustible and incombustible mass of fuel and their ratio as well as
ratio between fractions of carbon and hydrogen in the fuel.
6) Compare compositions of various fuel, show and comment noticable characteristics of
the observed fuel compared to others.
The report should include the purpose of the work, properties of fuel, calculation
procedure and results. All diagrams should have appropriate notation on axes, legends etc.
Results should have appropriate comments.
Data
Costa
Mota
Salema
Simoes
fuel composition
SouthAfrica
Indonesia
Czech republic
Slovenia
wC = 65,9 %
wH = 3,6 %
wS = 0,6 %
wO = 7,3 %
wN = 1,6 %
wp = 13,6 %
wH2O = 7,4 %
wC = 54,79 %
wH = 3,91 %
wS = 0,1 %
wO = 13,9 %
wN = 0,65 %
wp = 0,7 %
wH2O = 25,95 %
wC = 44,5 %
wH = 3,43 %
wS = 1,1 %
wO = 9,84 %
wN = 0,66 %
wp = 9,56 %
wH2O = 30,94 %
wC = 30,8 %
wH = 2,6 %
wS = 1,5 %
wO = 12,5 %
wN = 0,6 %
wp = 32 %
wH2O = 20 %