Question

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.

Answer 1

Answer:

a)

i) $v'=\frac{17}{20}$                                   ii) $\frac{m_v}{m_f-m_v} =\frac{17}{83}$

b) $m_r=752963.55\ kg$

Explanation:

Given:

fraction of water in wet sugar of m kg by mass, $m'_w=\frac{1}{5} \times m$

% of water evapourated from the total water after passing through the evapourator, $m'_v=85\%$

a)

amount of wet sugar fed to the evapourator, $m_f=100\ kg$

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

$m_w=\frac{1}{5} \times m_f$

$m_w=\frac{100}{5}$

$m_w=20\ kg$

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

$m_v=m_v' \times m_w$

$m_v=\frac{85}{100} \times 20$

$m_v=17\ kg$

i)

So, the fraction of of water leaving the evapourator:

$v'=\frac{m_v}{m_w}$

$v'=\frac{17}{20}$

ii)

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

$\frac{m_v}{m_f-m_v} =\frac{17}{100-17}$

$\frac{m_v}{m_f-m_v} =\frac{17}{83}$

b)

amount of sugar fed per day, $m_f=907185\ kg$

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

$m_w=\frac{m_f}{5}$

$m_w=\frac{907185}{5}$

$m_w=181437\ kg$

Mass of water vapourized after passing through the evaporator:

$m_v=\frac{85}{100}\times m_w$

$m_v=\frac{85}{100}\times 181437$

$m_v=154221.45\ kg$

Now the mass of water still remaining in the sugar:

$m_r=m_w-m_v$

$m_r=907185-154221.45$

$m_r=752963.55\ kg$

is the mass of extra water to be evapourated.