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 10.50 wt% molasses and the remainder white sugar. A second round of crystalization reduces the molasses content to 1.20 wt% molasses. A plant wishes to produce a brown sugar ideal for baking that contains 3.50 wt% molasses. In order to acheive this concentration of molasses, a plant feeds once-crystallized brown sugar into a recrystalizer. Some of this feed is diverted into a bypass stream that rejoins the sugar that leaves the crystalizer. Pure molasses leaves the crystalizer as a bypass stream. This process is described by the flow chart below.

Explanation / Answer

Applying the conservation of mass principle for each intersection, we can easily see that,

m1 = m2 + m3 (since the composition of m1. m2, m3 is the same) ..........(1)

m2 is the flowrate actually entering the crystalliser

thus rate of entry of molasses = 0.1050*m2 g/min

Rate of entry of white sugar = 0.8950*m2 g/min

Applying conservation of mass at the second junction after exit from crystallizer,

0.1050*m3 + 0.0120*m5 = 0.0350*m6 (conservation of mass for molasses) .............(2)

0.8950*m3 + 0.9880*m5 = 0.9650*m6 (conservation of mass for white sugar) ...............(3)

Applying conservation of mass at the crystalliser,

for molasses, 0.1050*m2 = 0.0120*m5 + m4 ................(4)

for white sugar, 0.8950*m2 = 0.9880*m5 .....................(5)

Equations 1 to 5 are the material balance equations.

Thus we see that there are 6 unknowns, and five equations,

given that, m6 = 300

this reduces the no. of variables to 5. With 5 unknowns and 5 equations, we can solve these equations using equation solver (http://www.idomaths.com/simeq.php)

we get,

m1 = 323.464

m2 = 249.27

m3 = 74.194

m4 = 23.464

m5 = 225.806

Feed rate m1 = 323.464 g/min

flow rate of pure molasses from the crystallizer, m4 = 23.464 g/min

Fraction of the feed that bypasses the crystallizer = (feed that bypasses / Total feed) = m3/m1 = 74.194/323.464 = 0.23 or 23 %

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