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

Answer:

278.482 kilograms of water must be evaporated each hour.

Step-by-step 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 $\dot m_{in} = 1000\,\frac{kg}{h}$, then:

$\dot m_{in,c} = (0.57)\cdot \left(1000\,\frac{kg}{h} \right)$

$\dot m_{in,c} = 570\,\frac{kg}{h}$

The initial mass flow of water is:

$\dot m_{in,w} = \dot m_{in} - \dot m_{in, c}$

$\dot m_{in,w} = 1000\,\frac{kg}{h}-570\,\frac{kg}{h}$

$\dot m_{in,w} = 430\,\frac{kg}{h}$

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

$\dot m_{out} = \frac{\dot m_{out, c}}{r_{out, c}}$

($\dot m _{out, c} = 570\,\frac{kg}{h}$, $r_{out} = 0.79$)

$\dot m_{out} = \frac{570\,\frac{kg}{h} }{0.79}$

$\dot m_{out} =721.518\,\frac{kg}{h}$

$\dot m_{out,w} = \dot m_{out} - \dot m_{out, c}$

$\dot m_{out,w} = 721.518\,\frac{kg}{h}-570\,\frac{kg}{h}$

$\dot m_{out,w} = 151.518\,\frac{kg}{h}$

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

$\dot m_{evap} = \dot m_{in,w}-\dot m_{out,w}$

$\dot m_{evap} = 430\,\frac{kg}{h} - 151.518\,\frac{kg}{h}$

$\dot m_{evap} = 278.482\,\frac{kg}{h}$

278.482 kilograms of water must be evaporated each hour.