Sludge is wet solids that result from the processing in municipal sewage systems. The sludge has to be dried before it can be composted or otherwise handled. If a sludge containing 70% water and 30% solids is passed through a dryer, and the resulting produce contains 25% water, how much water is evaporated per ton of sludge sent to the dryer?
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Given:
The percentage of water = 70 %
The percentage of solids = 30 %
The percentage of water after drying = 25 %
To Find:
The amount of water evaporated per ton of sludge.
Calculation:
- Let the mass of sludge be 1 ton, i.e., 1000 kg.
-Then according to the question:
The mass of water = 700 kg
The mass of solids = 300 kg
- After drying, the percentage of water = 25 %
⇒ The percentage of solids = 75 %
⇒ 75 % of the final sludge = 300 kg
⇒ 25 % of the final sludge = (300/75) × 25 = 100 kg
⇒ The mass of water in the final sludge = 100 kg
⇒ The mass of water evaporated = 700 - 100 = 600 kg
- So, 600 kg of water is evaporated per ton of sludge.
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