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Unformatted text preview: 5.6. SOME CONSERVATION AND EVOLUTION EQUATIONS 197 Solving, we get d dt = 1 T dT dt N i =1 i r N i =1 i N i =1 i . (5.449) One takes dT dt from Eq. (5.442) to get d dt = 1 T r P N i =1 h i i c P N i =1 i r N i =1 i N i =1 i . (5.450) Now recall that = /M and c P = c P M , so c P = c P . Then Equation (5.450) can be reduced slightly: d dt = r P N i =1 h i i c P T =1 bracehtipdownleft bracehtipuprightbracehtipupleft bracehtipdownright N summationdisplay i =1 i N i =1 i N i =1 i , (5.451) = r N i =1 h i i c P T N i =1 i N i =1 i , (5.452) = r N i =1 i parenleftBig h i c P T 1 parenrightBig P RT , (5.453) = r RT P N summationdisplay i =1 i parenleftbigg h i c P T 1 parenrightbigg , (5.454) = rM N summationdisplay i =1 i parenleftbigg h i c P T 1 parenrightbigg , (5.455) where M is the mean molecular mass. Note for exothermic reactionis the mean molecular mass....
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.
- Fall '11