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Many cereals are made with high moisture content so that the cereal can be formed into various shapes before it is dried. A cereal product containing 57% by mass is produced at the rate of 1000. kg/h. How much water must be evaporated per hour if the final product contains only 21% water

User Carl K
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1 Answer

4 votes

Answer:

278.482 kilograms of water must be evaporated each hour.

Explanation:

Initial mass of cereal is equal to the product of cereal mass-to-mass ratio and total mass flow:


\dot m_(in,c) = r_(in,c)\cdot \dot m_(in)

Where:


\dot m_(in,c) - Inlet cereal mass flow, measured in kilograms per hour.


r_(in,c) - Inlet cereal mass ration, dimensionless.


\dot m_(in) - Cereal product mass flow, measured in kilograms per hour.

If
r_(in,c) = 0.57 and
\dot m_(in) = 1000\,(kg)/(h), then:


\dot m_(in,c) = (0.57)\cdot \left(1000\,(kg)/(h) \right)


\dot m_(in,c) = 570\,(kg)/(h)

The initial mass flow of water is:


\dot m_(in,w) = \dot m_(in) - \dot m_(in, c)


\dot m_(in,w) = 1000\,(kg)/(h)-570\,(kg)/(h)


\dot m_(in,w) = 430\,(kg)/(h)

Given that final total mass flow contains only 21 % water, the final water mass flow, measured in kilograms per hour, is:


\dot m_(out) = (\dot m_(out, c))/(r_(out, c))

(
\dot m _(out, c) = 570\,(kg)/(h),
r_(out) = 0.79)


\dot m_(out) = (570\,(kg)/(h) )/(0.79)


\dot m_(out) =721.518\,(kg)/(h)


\dot m_(out,w) = \dot m_(out) - \dot m_(out, c)


\dot m_(out,w) = 721.518\,(kg)/(h)-570\,(kg)/(h)


\dot m_(out,w) = 151.518\,(kg)/(h)

Finally, the amount of water that must be evaporated per hour is:


\dot m_(evap) = \dot m_(in,w)-\dot m_(out,w)


\dot m_(evap) = 430\,(kg)/(h) - 151.518\,(kg)/(h)


\dot m_(evap) = 278.482\,(kg)/(h)

278.482 kilograms of water must be evaporated each hour.

User Aswan
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