Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 5 (Q.No. 48)
48.
An evaporator while concentrating an aqueous solution from 10 to 40% solids evaporates 30000 kg of water. The amount of solids handled by the system in kg is
Discussion:
1 comments Page 1 of 1.
Shivaprasad said:
1 decade ago
Let x will be the total mixture.
Therefore, amount of solids initially = 0.1*x.
Amount of solids after evaporation = 0.4(x-30000).
Equating mass of solids.
So, 0.1*x = 0.4*(x-30000) => x = 40000.
Therefore amount of solids handled is = 0.1*40000 = 4000.
Therefore, amount of solids initially = 0.1*x.
Amount of solids after evaporation = 0.4(x-30000).
Equating mass of solids.
So, 0.1*x = 0.4*(x-30000) => x = 40000.
Therefore amount of solids handled is = 0.1*40000 = 4000.
(3)
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