Question

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

Answer 1

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.