Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85% of the of the entering water is evaporated.

(a) Taking a basis of 100 kg feed, calculate (i) the mass fraction of water in the wet sugar leaving the evaporator, and (ii) the ratio (kg water vaporized/kg wet sugar) leaving the evaporator.

(b) If 1000 tons/day of wet sugar is fed to the evaporator, how much additional water must be removed from the outlet sugar to dry it completely.

a)

i)                                   ii)

b)

Explanation:

Given:

fraction of water in wet sugar of m kg by mass,

% of water evapourated from the total water after passing through the evapourator,

a)

amount of wet sugar fed to the evapourator,

Now the mass of water present in the fed amount of sugar:

Now the amount of water leaving from this total amount of water after passing through the evapourator:

i)

So, the fraction of of water leaving the evapourator:

ii)

Now the ratio of kg water vaporized/kg wet sugar leaving the evaporator.:

b)

amount of sugar fed per day,

Now the mass of water in the given amount of sugar per day:

Mass of water vapourized after passing through the evaporator:

Now the mass of water still remaining in the sugar:

is the mass of extra water to be evapourated.

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