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 70wt% water and 30wt% solids is passed through a drier , and resulting product contains 25wt% water , how much water is evaporated per ton of sludge sent to the drier ?
I feel this requires slight engineering knowledge .
Since it has 70% of water at the inlet , and25% of water left in the outlet ,
I took 70-25 = 45%
So 45% of the water has been evaporated in 30% of sludge .
Am I right to say now that the Amt of water evaporated per ton of sludge is 45/30 ?
Let us say $x$% of the sludge evaporates, then we should have $70-x$ of water and $30$ of solids left, so the percentage of water in the resultant is $(70-x)/(70-x+30)*100$%. Equate that to $25$% and solve for $x$, you're done.