Brown sugar is a mixture of white sugar and molasses. Unrefined brown sugar cont

Question

Brown sugar is a mixture of white sugar and molasses. Unrefined brown sugar contains a large portion of molasses. After one round of crystallization of this unrefined sugar, the brown sugar product contains 10.00 wt% molasses and the remainder white sugar. A second round of crystallization reduces the molasses content to 1.30 wt% molasses. A plant wishes to produce a brown sugar ideal for baking that contains 6.00 wt% molasses. In order to acheive this concentration of molasses, a plant feeds once- crystallized brown sugar into a recrystallizer. Some of this feed is diverted into a bypass stream that rejoins the sugar that leaves the crystallizer. Pure molasses leaves the crystallizer as a bypass stream. This process is described by the flow chart below m3 g/min 0.1000 g molasses/g 0.9000 g white sugar/g Crystallizer m2 g/min m1 g/min m5 g/min m6 g/min T 0.1000 g molasses/g 0.1000 g molass 0.0130 g molasses/g 0.0600 g molasses/g 0.9000 g white sugar/g 0.9000 g white sugar/g 0.9870 g white sugar/g 0.9400 g white sugar/g m4 g molasses/min

Explanation / Answer

Brown sugar flow rate required, m6= 150 g/min. It contains 0.06 molasses and rest of 0.94 is white sugar.

White sugar flow rate in the product= 150*0.94= 141 g/min

All this has to come from feed (m1) containing 0.9 fraction white sugar.

Flow rate of m1= 141/0.9= 156.7 g/min

Molasses in m6 =0.06*150= 9 g/min

Molasses in m1= 156.7*0.1= 15.67

Molasses (m4)= molasses entering through m1- molasses removed from m6= 15.67-9= 6.67 g/mmin

Writing overall balance after crystalizer, m3+m5= m6, m3+m5= 150, m5= 150-m3 (1)

Writng sugar balance, m3*0.9+m5*0.987= m6*0.94, m3*0.9+m5*0.987= 150*0.94= 141 (2)

Susbstituting Eq.1 in Eq.2 gives m3*0.9 +(150-m3)*0.987= 141

Hence m3*0.9+150*0.987-m3*0.987= 141

150-141= m3*(0.987-0.9), m3= 103.45 g/min

Hence m3/m1= 103.45/150= 0.689

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