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

Answer:

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.

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

$\frac{W_{w}^{'} }{W_{w}^{'} + W_{d}^{'} }$ = 0.03

$\frac{W_{w}^{'} }{W_{w}^{'} + 800 }$ = 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.