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Sagot :
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 [tex]10^{5}[/tex] 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
[tex]W_{w}[/tex]/[tex]W_{w}[/tex] + [tex]W_{d}[/tex] x 100
[tex]W_{w}[/tex] + [tex]W_{d}[/tex] = 1000 kg when entering
it has 20% moisture content when entering
[tex]W_{w}[/tex] = 0.2 x 1000 = 200 kg
when leaving it has 3% moisture content then weight of dry material
[tex]W_{d}[/tex] = 1000 - 200 = 800 Kg
[tex]\frac{W_{w}^{'} }{W_{w}^{'} + W_{d}^{'} }[/tex] = 0.03
[tex]\frac{W_{w}^{'} }{W_{w}^{'} + 800 }[/tex] = 0.03
[tex]W_{w} ^{'}[/tex] = 0.03 x [tex]W_{w} ^{'}[/tex] + 0.03 x 800
[tex]W_{w} ^{'}[/tex] = 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 [tex]10^{5}[/tex] KJ of heat per hour.
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