Split_00063 - 9.53 a. Energy balance: U = 0 n lb m fuel...

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9- 63 9.53 a. b. Energy balance : Δ Δ U nU mC T v =⇒ + − ° = 07 7 0 lb fuel burned (Btu) lb F mc o m out bg ± ⇒+ ° ° ° = 0 00215 4 62 0 900 87 06 77 00 0 . ± .. a f b g a f Δ U c o mm lb Btu lb F F F ⇒= Δ ± U c o m Btu lb 19500 The reaction for which we determined Δ ± U c o is 1 lb oil + O g) CO g)+ H O(v) (1) m 22 2 abc (( The higher heating value is Δ ± H r for the reaction CO g)+ H O(l) (2) m 2 Eq. (9.1-5) on p. 441 ΔΔ ±± () HU R T b c a c1 o c1 o =++ Eq. (9.6-1) on p. 462 −= + Δ ± ( )) HH c H HHV LHV c2 o ( c1 o ( v2 HO , 77F ) D To calculate the higher heating value, we therefore need a b c = = = lb - moles of O that react with 1 lb fuel oil lb - moles of CO formed when 1 lb fuel oil is burned lb - moles of H O formed when 1 lb fuel oil is burned 2m 9.54 a. CH OH v + O g CO g 2H O l 32 2 2 3 2 →+ . r o c o CH OH v 3 kJ mol == ej 764 0 Basis : 1 mol CH OH fed and burned 3 1 mol CH OH( ) 25°C, 1.1 atm n 0 3 l (mol O ) 2 vaporizer 1 mol CH OH( ) 3 v 100°C 1 atm reactor Q 1 (kJ) 3.76 n 0 (mol N ) 2 100°C Effluent at 300°C, 1 atm n p (mol dry gas) 0.048 mol CO /mol D.G. 2 0.143 mol O /mol D.G.
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This note was uploaded on 01/07/2012 for the course CH E 210H at Penn State.

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