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 15.00 wt% molasses and the remainder white sugar. A second round of crystallization reduces the molasses content to 1.10 wt% molasses. A plant wishes to produce a brown sugar ideal for baking that contains 6.50 wt% molasses. In order to achieve 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 0.1500 g molasses/g 0.8500 g white sugar/g Crystallizer m1 g/min 0.1500 g molasses/g 0.8500 g white sugarlg Im2 g/min 0.1500 g molasses 0.8500 g white sugarlg ms g/min 0.0110 g molassesig 0.0650 g molasses/g 0.9890 g white sugarg 0.9350 g white sugar/g ms g/min m4 g molasses/min a) If the plant wishes to produce 350.0 g/min of brown sugar today, what is the feed rate, b) What is the flow rate of pure molasses from the crystallizer, m4? What is the fraction of the feed that bypasses the crystallizer? c)

Explanation / Answer

Production of brown sugar, m6 = 350 g/min

Feed rate, m1 =?

Overall Material balance of white sugar

m1 x 0.85 = m6 x 0.9350

m1 = 350*0.9350/0.85 = 385 g/min

Overall Balance of the molasses

m1 x 0.15 = m4 + m6 x 0.065

385 x 0.15 = m4 + 350 x 0.065

m4 = 35 g/min

Molasses balance around the mixing point

0.011 x m5 + 0.15 x m3 = 0.065 x m6

0.011 x m5 + 0.15 x m3 = 0.065 x 350 = 22.75

m5 + 13.636 m3 = 2068.181 ...... Eq1

White sugar balance around the crystallizer

m2 = m5 x 0.989/0.85 = 1.163 m5

Molasses balance around the split point

m1 x 0.15 = m2 x 0.15 + m3 x 0.15

385 = 1.163 x m5 + m3

m5 + 0.8598m3 = 331.04 ...... Eq2

m5 + 13.636 m3 = 2068.181 ....... Eq1

Solve eq1 and Eq2 simultaneously for m3

8.267 m3 = 1269.367

m3 = 135.966 g/min

Part C

Fraction of feed bypass the crystallizer

m3/m1 = 135.966/385

= 0.353

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