Mass Balance

5.2ENERGY BALANCE 5.2.1Introduction Energy equilibrize in the crossroadion of fructosyltransferase the finale of high temperature being released and absorbed during the reactions. The inflame capacity equivalence at any constant pressure is; (J/mol.K) From that we can figure the heat content change that is given by, ?? = Then the molar predict rate is given by, Ni= No= Flowrateconcentrationmolecular weight The popular balance equation of the energy at steady evoke is Q = ?H + W where Q = heat absorbed or released ?H = enthalpy change W = work done 5.2.2Seed Fermenter ground on the chemic properties handbook, the Cp of water, ammonia and carbon dioxide are as survey: Table 3.2 Heat capacity, Cp for water, ammonia, and carbon dioxide Component|  | Cp= A+ BT + CT2 + DT3 + ET4 [T(K)]|  |  | A| B| C| D| E| body of wa ter| 92.053| -3.9953x10-2| -2.1103x10-4| 5.3467x10-7| -| ammonia| -182.157| 3.3618| 1.4398x10-2| 2.0371x10-5| -| Carbon dioxide| 27.437| 4.2315x10-2| -1.9555x10-5| 3.9968x10-9|  -2.

9872x10-13| Since the value of Cp for biomass, sucrose and fructosyltransferase is not procurable from the chemical properties handbook, therefore it should be calculated as quest for: From special Kopps rule, The specific heat capacity for individually touch are: C = 10.98 J/mol.K H = 7.56 J/mol.K O = 13.42 J/mol.K N = 18.74 J/mol.K Cp for sucrose is Cp = 1(10.98) + 1.83(7.56) + 0.916 (13.42) = 37.1075 J/mol.K Cp for product is Cp= 1(10.98) + 2.09(7.! 56) + 0.53(13.42) + 0.275(18.74) = 39.0465 J/mol.K Cp for biomass is Cp = 1(10.98) + 1.46(7.56) + 0.385(13.42) + 0.23(18.74) = 31.4945 J/mol.K To calculate heat capacity, ?H = T1T2Cp dTIf you indispensableness to get a full essay, order it on our website: OrderCustomPaper.com

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