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Mars2501 [29]
3 years ago
12

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

vaporated.
(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.
Physics
1 answer:
Natalka [10]3 years ago
8 0

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

<u>Now the mass of water in the given amount of sugar per day:</u>

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.

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