Problem: [ChE Boards April 2014] From sugar cane, ethanol is commercially produced at a rate of 100000 L/day (99.
5% v/v).
From here, fermentation occurs at a rate of 0.511 kg ethanol per kilogram of fermentable sugar. If the syrup is 53% fermentable sugar
and cane juice is 25% syrup with distillation and fermentation being 99% and 90% efficient, respectively, calculate the following:
1. mass flowrate of the ethanol solution produced
a. 176.32 MT/day c. 322.38 MT/day
b. 1302.54 MT/day d. 78.58 MT/day
2. mass flowrate of the syrup entering the fermenter
a. 176.32 MT/day c. 322.38 MT/day
b. 1302.54 MT/day d. 78.58 MT/day
3. mass flowrate of the cane juice feed
a. 176.32 MT/day c. 322.38 MT/day
b. 1302.54 MT/day d. 78.58 MT/day
Solution:
Density of H2O = 997.042 kg/m3 (Perry’s 8e, Table 2-30, p. 2-96)
Density of ethanol = 784.718 kg/m3 (Perry’s 8e, Table 2-32, p. 2-98)
Density of ethanol soln. = 0.995(784.718) + 0.005(997.042) = 785.780 kg/m3
1.
100000 L ethanol soln. 1 m3 785.780 kg 1 MT
( )( )( 3
)( ) = 78.58 MT/day
1 day 1000 L 1m 1000 kg
2.
78.58 MT ethanol soln. 1 kg fermentable sugar 100 kg theo. prodn. 100 kg syrup
( )( )( )( ) = 322.38 MT/day
1 day 0.511 kg ethanol soln. 90 kg actual prodn. 53 kg fermentable sugar
3.
322.38 MT syrup 100 kg theo. prodn. 100 kg cane juice
( )( )( ) = 1302.54 MT/day
1 day 99 kg actual prodn. 25 kg syrup