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 2r0 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 ) 74.2933T2 450 K 1.40 (63076 5.67 x10 8 T24 ) 104T2 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 hr0 (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 12