University of Babylon /College Of Engineering Electrochemical Engineering Dept. Second Stage /Thermodynamics Adiabatic reaction This reaction mean that there is no heat absorbed or ejected , if the reaction exothermic the products temperature raised while for endothermic the products temperature decreased , a good example for this kind of reaction is fuel burned with oxygen. Adiabatic flame temperature , the highest temperature reached for products during oxidation reaction for the fuel, in reality these kind of reaction described that is an adiabatic because they are occurred too fast and in this case there is no time for heat exchange between system and surroundings Ex-4 :What is the maximum temperature that can be reached by combustion of CH4 with 20% air excess ? both methane and air enter the burner at 25 ○C CH4 + 2O2 → CO2 + 2H2O ( g ) DH o298 = - 393509 + ( 2 )( - 241818 ) - ( - 74520 ) = - 802625 J Assumptions · Assume complete adiabatic combustion ( Q = 0) · Kinetic and potential energy change are negligible For overall energy balance result ∆H = 0 From chemical equation reaction for 1 mole methane 2 Mole of O2 required ,mole of O2 excess = (0.2 ) ( 2 ) = 0.4 ,and mole of N2 entering = ( 2.4 ) ( 79 / 21) = 9.03 Gas leaving the burner contain 1 mole CO2 ,2 moles H2O ,0.4 mole O2 and 9.03 N2 Since enthalpy change must be same regardless of path University of Babylon /College Of Engineering Electrochemical Engineering Dept. Second Stage /Thermodynamics DH o298 + DH oP = DH = 0 DH o298 = -802625 J DH oP = å ( ni C Pmh ,i )(T2 - 298 .15) · Because mean heat capacity depend on T2 ,and it is sought · Assume T2 ( because of T1 = 298 , thus the value of assumed T2 is ≥ 298) å ni Di ] · Use equation å ( ni C Pmh ,i ) = R[ å ni Ai + ( å ni Bi )Tam + T1T2 DH o298 + 298 .15 · Calculate T2 by use T2 = å n I C pmh ,i · Repeat the above steps until calculated T2 converges with assumed T2 ån A i = 1 ´ 5.457 + 2 ´ 3.470 + 0.4 ´ 3.639 + 9.03 ´ 3.280 = 43 .489 i ån B i i ån D i i = 9.504 ´ 10 -3 = -0.645 ´ 10 5 \ å ni C Pmh ,i = 8.314 ( 43 .489 + 9.504 ´ 10 Tam -3 By continue try and error , T2 = 2066 ○K 0.645 ´ 10 5 T1T2