3.3 Reynolds num ber at dextrose rest of wt% sucrose, 1.0 wt% Corn-steep liquor

Question

3.3 Reynolds num ber at dextrose rest of wt% sucrose, 1.0 wt% Corn-steep liquor contains 2.5 wt% and 50 wt% water; the solids. the liquor stream is solids. Beet contains 50 contains to dextrose, and 18 wt% water; the rest of the molasses stream mixing tank Beet molasses is mixed with corn-steep liquor and water in a wt% dextrose produce a dilute sugar mixture. The exit stream contains 2.0 unit. See and 12.6 wt% sucrose, and is ready to be fed into a fermentation Figure 3.24. (Adapted from Doran PM, Bioprocess Engineering Principles, 1999.) solution to this problem? solids, and (a) What is the basis in your sucrose, (b) What are the weight percents (wt%) of dextrose, water in the exit stream? to the mass (c) What is the ratio of the mass flow rate of the water stream flow rate of the corn-steep liquor stream? System boundary Corn- 2.5 wt% Dextrose 2 wt% Dextrose steep 50 wt% Water Product 12.6 wt% sucrose liquor solids Mixer 50 wt% Sucrose Water Beet 1 wt% Dextrose Figure 3 molasses 18 wt% Water Corn-steep liquor and beet Solids molasses mixer. Water

Explanation / Answer

(a) The basis in the solution to this problem that whatever enters the system, leaves the system. There is no increase or decrease in the mass of the system.

(b) 100 g of Input Corn steep liqour contains

      2.5 g Dextrose, 50 g water and 47.5 g solids

      100 g of Input beet molasses contains

      1 g Dextrose, 18 g water, 50 g Sucrose and 31 g solids

      Inputs = Output , Therefore the exit stream contains 3.5 g Dextrose, 68 g water,50 g Sucrose and 78.5g solids

      Output weight is 200 g

      wt % Dextrose   = 1.75 %

       wt % Water = 34 %

       wt % sucrose = 25 %

       wt % solids = 34.25 %

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