Question

Sugar crystals enter a dryer at the rate of 1000 kg h-1 and at 20% w.b. moisture content. They leave the dryer at 3% w.b. moisture content. If the drying process requires 3000 kJ kg-1 of water removed, estimate the amount of heat required per hour and the rate of dry crystals out of the dryer.

Answer 1

The dryer evaporate 200 - 24.74 kg of water per hour

To remove 1kg of water it need 3000 K J

So to remove, 175.26 Kg

it need 5.257x
`10^(5)` KJ of heat per hour.

Step-by-step explanation:

In one hour, the amount of sugar entering = 1000 kg

w.b moisture content is defined as,

weight of water / weight of water + weight of dry

`W_(w)`/
`W_(w)` +
`W_(d)` x 100

`W_(w)` +
`W_(d)` = 1000 kg when entering

it has 20% moisture content when entering

`W_(w)` = 0.2 x 1000 = 200 kg

when leaving it has 3% moisture content then weight of dry material

`W_(d)` = 1000 - 200 = 800 Kg

`(W_(w)^(') )/(W_(w)^(') + W_(d)^(') )` = 0.03

`(W_(w)^(') )/(W_(w)^(') + 800 )` = 0.03

`W_(w) ^(')` = 0.03 x
`W_(w) ^(')` + 0.03 x 800

`W_(w) ^(')` = 24.74 kg

When leaving the dryer, the crystals has total weight of the water = 24.74 kg per hour.

The dryer evaporate 200 - 24.74 kg of water per hour

To remove 1kg of water it need 3000 K J

So to remove, 175.26 Kg

it need 5.257x
`10^(5)` KJ of heat per hour.