Imbibition.docx

Assume Double Simple Imbibition

IMBIBITION WATER (HOT) Temperature = 80˚C (176˚F) w = total weight of imbibition water to be distributed, per unit cane = 0.35 x = 0.58-0.51 (optimum) xw = proportion to be applied before the last mill = (0.54)(0.35) = 0.189 (1-x) w = proportion to be applied before the penultimate mill = (1-0.54)( 0.35) = 0.161 F = fiber in bagasse Energy Balance Solving for the mass flow rate of steam in the first heater, assume that the effective steam usage is ɛ=0.95

Qgained =Qlost Q water =Qsteam •

•

m water Cp water ΔT=ε m steam λ •

•

m steam =

m water Cp water  T2 -T`  ελ

The saturation temperature of the steam at 116 °C or 240.8 °F, the value of λ is

λ=hg-hf BTU BTU -209.3012 lb lb BTU λ=951.48 lb

λ=1160.78

Substituting the values •

•

m steam =

m Juice Cp Juice  T2 -T`  ελ

'

lb   BTU    648159.008   0.916  130F-104F  h  lb.F   m steam = BTU   0.95  951.48  lb   • lb m steam = 17077.61087 h •

Solving for the Log Mean Temperature Difference,

ΔTlm =

ΔTlm =

ΔT2 -ΔT1  ΔT  ln  2   ΔT1 

 240.8-130  -  240.8-104    240.8-130   ln     240.8-104  

ΔTlm = 123.345F

For a condensing liquid in the shell-side, correction factor, F=1

Q=UAΔTlm F •

msteam  λ  =UAΔTlm F Solving for the heat transfer coefficient,

U = 0.035  T -3216.4 + V  Where U= heat transfer coefficient

T= temperature of the exhaust steam V=juice velocity in heater tubes

U=0.035  240.8-32 16.4+6.5 U = 167.3532 Solving for the heat transfer area:

Q=UAΔTlm F •

msteam1  λ  =UAΔTlm F •

A1 =

msteam1  λ  UTlm F

lb  BTU   951.48  h lb  A1 = BTU   167.3532  123.345F 1 hr.ft 2 F   A1 = 787.17484 ft 2 17077.61087

BTU ft 2 °F