Estimation of Flame Temperature
By
Cesar Olmedo
6/12/09
ME-598
For
MFDC Combustion lab
Dr .Guillaume
Professor
Sara Esparza
Johanna Lopez
Purpose:
To able to calculate the flame temperature of the combustion chamber, using a surface
temperature reading on the outside of the chamber.
Background:
The MFDC lab had fabricated a combustion chamber to study the emission of different
fuels and efficiency of combustion. The have introduced the concept of recalculating air
to increase combustion efficiency. They also have burned different fuels to study how the
combust and release emissions. The most common fuel used in the chamber is Kerosene;
one of the so many challenges that the MFDC lab has encountered is actual flame
temperature.
They have introduced a window to view the flame color. With the use of an infrared
laser temperature gauge we took a temperature measurement on the surface of the
stainless steel chamber. Using this temperature reading as a base line; a heat transfer
analysis was developed to estimate the actual flame temperature, the analysis was based
on a few assumption and a surface temperature was estimated. The estimated temperature
value was compared with the actual reading to analysis how much error the theoretical
calculation produce.
The assumptions made in the heat transfer calculations were developed to simplify the
approach to the problem. These assumptions can be eliminated and the introduction of
actual parameter can be done to obtain a better results.
Theory:
The math model for estimating the flame temperature of the combustion chamber
was developed using Heat Transfer theories and Thermodynamic theories. The first step
in the calculation was to calculate the adiabatic flame temperature. The assumption here
is that the chamber is adiabatic so it can neither gain or loss heat. This isolates the overall
heat generated by reaction to heat of combustion. The following is the general equations
used in the calculations of the adiabatic flame temperature.
C12 H 26 l    18.5(O2  3.76 N 2 )  12CO2  13H 2 O  69.56 N 2
 N product (h f0  h  h 0 ) product   N react h f0 , react  ( Nh f0 ) C12H 26
2
Kerosene is made up of many different hydrocarbons depending on the suppliers, but
most kerosene fuels are made up of dodecane regardless of the supplier. That is why
dodecane was used in the calculation of the adiabatic flame temperature. Lambda
represents the amount of theoretical air in the combustion it is the stoichmetric ratio.
(h f0  h  h 0 )
Equals the enthalpy of a chemical component at a specified state; the enthalpy is the
sum of the enthalpy of the component at 25 C° and 1 atm ( h f0 ), and the sensible enthalpy
of the component relative to 25 C° and 1 atm ( h  h 0 ).
The assumptions made in these calculations were
1. Steady state flow combustion.
2. The combustion chamber is adiabatic.
3. There are no work interactions.
4. All gasses involved in are ideal.
5. Kinetic and potential energies are negligible.
The first calculation of the adiabatic flame temperature was calculated with lambda equal
to one. The result was theoretically too high for the flame temperature the lambda then
pushed up to 4 the result was more reasonable.
Once the adiabatic flame temperature was calculated with in a reasonable result, a
radiation exchange between two surfaces enclosure was developed. They were assumed
to both be gray and the general formula for this type of calculation is
 (T

 T24 )
q12  q1  q2 
1  1
1  2
1


 1 A1 A1 F12  2 A2
4
1
3
Air
P=1 Atm
Tm= 450K

kg
m  .1
s
T2 ,  2  .8
r0  .0762m
T1  abidiatic flame temp
1  .8
The assumption of the open flame being a gray body makes this problem more feasible, it
translate the open flame into a type heater coil. The interesting thing about a circular
surface is that it may be cut in half to produce a semicircular surface.
The assumptions made for this part of the problem are the following
1. Open flame converted into a heating coil.
2. Steady state conditions
3. Diffuse , gray surfaces
4. Negligible tube end effects and axial variation in gas temperature.
5. Fully developed flow
Once the surface temperature of the combustion chamber was calculated ( T2 ) a simple
conduction equation was used to determine the hea tlost between the thickness of the
stainless steel material.
qx 
qx K
 (Ts1  Ts 2 )
A L
Where K is the thermal conductivity, L is the thickness of the material and (Ts1  Ts 2 ) is
the change in temperature.
4
Ts1
q "x
Ts 2
Apparatus:
The equipment being used in the calculations include
1. One amazing combustion chamber designed by MFDC lab students, the
combustion chamber was operated with kerosene for the temperature
measurements.
2. MT pro Micro Temp infrared laser temperature gauge.
Sample Calculations
C12 H 26 l    18.5(O2  3.76 N 2 )  12CO2  13H 2O  69.56 N 2
 N product (h f0  h  h 0 ) product   N react h f0 ,react  ( Nh f0 )C12H 26
Since all the reactants are at standard reference state
h f0 =0 for O2 and N 2
Substance
h f0 ( KJ/Kmol)
C12 H 26
-290720
O2
N2
H 2O
0
0
-241820
h298K (KJ/Kmol)
8682
8669
9904
5
CO2
-393520
9364
Substituting into the equation we obtain
12CO2 Kmol(393520  hCO 2  9364)
 69.56 Kmol(0  hN 2  8669)
KJ
KJ
 13H 2 OKmol (241820  hH 2O  9904)
KmolCO2
KmolH 2O
KJ
KJ
 C 2 H 12 Kmol(290720)
KmolN 2
KmolC2 H 12
Simplied
12hCO 2  13hH 2O  69.56hN 2  8419315.64 KJ
It appears that there is three unknown but since the assumption of the ideal gas was
introduced. Then the T product  h
Therefore we can obtain the temperature of the flame by interpolation and an educated
guess. First divide the sum of the equation by the total number of moles
8419315
KJ
 89063.8
12  13  69.56
Kmol
This value would correspond to CO2 , H 2O or N 2 noting that the majority of the moles
belong to N 2 . It can be determined that T product  h will be close to 2700K
The starting point will be 2600K
12(137449)  13(114273)  69.56(86650)  9162311KJ Higher than the 8419315.94 KJ
value, the next guess is at 2400 K
12(125152)  13(103508)  69.56(79320)  8364927 KJ
The value is very close to 8419315.94 KJ by interpolation the actual temperature is
2413.6 K this result is much too high, if this was true most of the chamber would have
melted. In reality there is excess air in the reaction this fact changes the lambda to
estimated 4 instead of one. This makes sense too since the flame will achieve the highest
temperature with a lambda of 1.
With a lambda the results change.
6
12CO2 Kmol(393520  hCO 2  9364)
 278.24 N 2 Kmol(0  hN 2  8669)
KJ
KJ
 13H 2OKmol (241820  hH 2 O  9904)
KmolCO2
KmolH 2O
KJ
KJ
KJ
 37O2 Kmol(0  hO2  8682)
 C2 H12 Kmol(290720)
KmolN 2
KmolO2
KmolC2 H12
Simplied
12hCO 2  13hH 2 O  278.24hN 2  37hO2  10839969 KJ
Dividing
10839969
KJ
 31859.8
12  13  278.24  37
Kmol
This value would correspond to CO2 , H 2O or N 2 noting that the majority of the moles
belong to N 2 . It can be determined that T product  h will be close to 1040K
The starting point will be 1020K
12(43859)  13(36709)  278.24(30784)  37(32088)  10756121.16 KJ is lower than the
calculated value of 10839969 KJ therefore 1020k is the lower limit.
At 1060 K
12(38380)  13(46051)  278.24(32101)  37(33490)  11230135KJ is higher than the
calculated value of 10839969 KJ.
By interpolation
The adiabatic flame temperature is 1027 K this more reasonable result. Once we have
obtained a flame temperature. A radiation and convection analysis was done in the
chamber. The following assumption were made
1. Adiabatic flame was assumed to be a gray body.
2. Steady state conditions
3. Negligible tube effects and axial variations in the gas temperature
4. Fully developed flow
5. Assume average temperature in flow to be 450 K
6. Assume .1 Kg/s flow rates
7. Chamber is insulated
8. Using a slightly lower Pr value
Air properties at 1 atm and 450K
7
 w 
K  .0373

 mK 
Pr  .686
J
Cp  1021
Kg
Air
P=1 Atm
Tm= 450K

kg
m  .1
s
  250 x10 7
kg
ms
T2 ,  2  .8
r0  .0762m
T1  abidiatic flame temp
1  .8
Insulation
Since the chamber is insulated there is no external heat addition therefore the surface
energy balance yields
q 2 radition  q 2
Since it is assumed that the chamber is a two surface enclosure the following equation
can be used.
8
 (T

 T24 )
q12  q1   q2 
 hA2 (T2  Tm )
1  1
1
1 2


1 A1 A1F12  2 A2
4
1
Where
the
viewing
factor
F12  1
A1  2r0
A2  r0
r0  .0762m
The Reynolds number is calculated first

m Dh
Re D  2
r0

2
4A
2r0 2 (.0762m)
Dh  c 

 .093241m
P
 2
 2

m Dh
(.1kg / s)(. 093241m)
Re D  2 
 40784

r0
2
7
 ( * (.0762m) (250 x10 kg / ms)
2
2
The Reynolds number fall under the turbulent section, to calculate the Nusselt number the
use of Dittus- Boelter equation is used since it is valid when Reynolds numbers are above
10000 , Pr is very close to between .7 and 160 and(L/D)is greater than 10.
4
NuD  .023 Re D 5 Pr n
n  .4
for
Ts  Tm heating
4
4
NuD  .023 Re D 5 Pr.4  (.023)( 40834) 5 (.686)  96.622
h
.0373W / mK
96.22  47.2966
.0762m
Dividing both side of the equation by A1
The equation becomes
9
q12  q1  q2 
 (T

 T24 )
A
 h 2 (T2  Tm )
1  1 1 1   2
A1


A
1
F12  2
2
A1
4
1


W
1027 4  T24 K 4
2
4
 
m K
 47.296 *  T2  450 K 
1  .8
1  .8  2 
2
1
 
.8
.8   
(5.67 x10 8
(63076  5.67 x10 8 T24 )
 74.2933T2  450 K 
1.40
(63076  5.67 x10 8 T24 )
 104T2  450 K 
(63076  5.67 x10 4 T24 )  104T2  46804.8
5.67 x10 8 T24  104T2  109881  0
Solving for T
T2  815.4 K
Calculating the heat flux for the surface of the surface of the combustion chamber
q2  hr0 (T2  Tm )
q2  47.2966 *  * .0762m(815.4  450)
4132.64
W
m
The final step in the calculation was the conduction of the heat through the stainless steel
material.
10
qx 
qx K
 (Ts1  Ts 2 )
A L
K  19.76(W / mK )
L  .005m
Ts1  815.4 K
4132.64(W / m) 
19.76
815.4  Ts 2 
.005
Ts 2  814.0 K
Ts 2  1005F 0
The actual temperature measurement is 650F the percent error is
1005  650 *100  54.6%
error % 
650
Discussion
The value obtained by the calculation was high when compared to the actual
value measured. This error can be attributed to some assumptions that were made, in the
radiation and convection heat transfer analysis. Increasing the flowrate would increase
the Reynolds number which would result in a higher convection coefficient h. This would
reduce the temperature of T2 this new temperature would give a better result. The Tm of
air may have been to low this I believe might be the hardest part to determine in the
experiment since currently don’t have a method to measure this temperature. The DittusBoelter equation may not have been the best equation to use since the Pr value is slightly
below the required range. The assumption of making an open flame into a resistant heater
grey body may have been too much of an assumption that they can not be comparable; so
there will be a need to determined the 1 for an open flame, the  2 for the chamber I think
is good one since the inside is cover with carbon.
The adiabatic flame temperature also could be adjusted since only two values were
calculated for Lambda 1 and 4. It was notices that an increase of air would reduce the
flame temperature so this needs to be determined more accurately.
I believe that the steady state assumption is a good one since the chamber can be run for
a few minutes.
Conclusion
In the end the calculation did prove to be of value; it enable us to get an idea of how
to obtain a flame temperature without giving outrageous results. Most of the assumption
made in the problem statements can be eliminated with more information being obtained
from the chamber. It also has lead to the development of a MATlab program (that is still
under construction) that can calculate the adiabatic flame temperature for different fuels
11
and lambda values. Another idea that is being investigated is the determination of the
mass flowrate for the air in the chamber and the Tm temperature; along with being able to
model an open flame in a closed gray body that is experiencing radiation from the open
flame.
In the development of the calculations a book that can offer useful information is
Perry Chemical Engineering Handbook. The book has empirical formulas that simulate
combustion in oven and how to determine flame temperature in order to find the oven
efficiency. The development of a program that can get close to actual temperature
measured for the combustion chamber using the following heat transfer equations would
be a great goal to achieve.
References
1. Introduction to Heat Transfer 4th edition, Frank P. Incropera and David P. Dewitt,
John Wiley and Sons.2002
2. Thermodynamics an engineering approach 4th edition, Yunus A Cengel and
Michael A. Boles, Mc Graw Hill, 2002
3. Perry’s Chemical Engineering Hand Book 7th edition , Robert H . Perry and Don
W. Green, Mc Graw Hill, 1997
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