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

a)

i)
`v'=(17)/(20)` ii)
`(m_v)/(m_f-m_v) =(17)/(83)`

b)
`m_r=752963.55\ kg`

Step-by-step explanation:

Given:

fraction of water in wet sugar of m kg by mass,
`m'_w=(1)/(5) * 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=(1)/(5) * m_f`

`m_w=(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' * m_w`

`m_v=(85)/(100) * 20`

`m_v=17\ kg`

i)

So, the fraction of of water leaving the evapourator:

`v'=(m_v)/(m_w)`

`v'=(17)/(20)`

ii)

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

`(m_v)/(m_f-m_v) =(17)/(100-17)`

`(m_v)/(m_f-m_v) =(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=(m_f)/(5)`

`m_w=(907185)/(5)`

`m_w=181437\ kg`

Mass of water vapourized after passing through the evaporator:

`m_v=(85)/(100)* m_w`

`m_v=(85)/(100)* 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.