Question

% water, the rest of the liquor stream is solids. Beet molasses contains 50 wt% sucrose, 1.0 wt% dextrose, and 18 wt% water, the rest of the molasses stream contains solid. Beet molasses is mixed with corn-steep liquor and water in a mixing tank to produce a dilute sugar mixture. The exit stream contains 2.0 wt%, dextrose and 12.6 wt% sucrose and is ready to be fed into a fermentation unit. See Figure 3.26. (Adapted from Doran PM, Bioprocess Engineering Principles, 1999.) (a) What is the basis in your solution to this problem? (b) What are the weight percents (wt%) of dextrose, sucrose, solids, and 3.4 Corn-steep liquor contains 2.5 wt% dextrose and 50 wt water in the exit stream? (c) What is the ratio of the mass flow rate of the water stream to the mass flow rate of the corn-steep liquor stream? System boundary Corn- 2.5 wt% Dextrose steep 50 wt% water- liquor Solids Mixer 2 wt% Dextrose 12.6 wt% Sucrose Solids Product 50 wt% Sucrose Beet molasses <) 1 wt% Dextrose Water Figure 3.26 Corn-steep liquor and beet molasses mixer. 18wt% Water Solids Water

Answer 1

Part a

Basis - 100 kg of corn steep liquor

Let mass of corn steep liquor = C kg

Mass of beet molasses = B kg

Mass of water inlet = W kg

Mass of product stream exit = P kg

Part b

Overall balance

C + B + W = P

100 + B + W = P

Dextrose balance

C*wcD + B*wbD + W*wwD = P*wpD

100*0.025 + B*0.01 + W*0 = P*0.02

2.5 + 0.01B = 0.02P............ eq1

Sucrose balance

C*wcSu + B*wbSu + W*wwSu = P*wpSu

100*0 + B*0.50 + W*0 = P*0.126

0.50B = 0.126P

B = 0.252 P

from eq1

2.5 + 0.01(0.252P) = 0.02P

2.5 + 0.00252P = 0.02P

2.5 = (0.02 - 0.00252)P

P = 143 kg

B = 0.252*143 = 36 kg

W = P - B - 100

= 143 - 36 - 100

= 7 kg

Solid balance

C*wcS + B*wbS + W*wwS = P*wpS

100*0.475 + 36*0.31 + 7*0 = 143*wpS

58.66 = 143*wpS

wpS = 0.41 = 41%

wpW = 44.4%

Dextrose % in product = wpD = 2%

Sucrose % in product = wpSu = 12.6%

Solid % in product = wpS = 41%

Water % in product = wpW = 44.4%

Now we can check our calculations

Water balance

100*0.50 + 36*0.18 + 7*1 = 143*wpW

63.48 = 143*wpW

wpW = 0.444 = 44.4%

It means we are correct in our math above.

Part c

Mass flow of water / mass flow of corn steep liquor

W/C = 7/100 = 0.07