4. [25 points] A waste has the following composition: I sw abyd ) Component % by
ID: 1866782 • Letter: 4
Question
4. [25 points] A waste has the following composition: I sw abyd ) Component % by weight Food waste 10 Plastic Yard waste Paper Glass Metals MC (96) 20 20 30 10 10 110 170 150 270 270 60 Determine the overall specific weight of the waste a. Determine the total moisture content (lb) of ONLY the compostable wastes (food waste, paper, and yard waste) b. If the waste is generated at 4.4 lbs/person per day from a town with a population of 15,000 people, determine the pounds per day of sludge (at 95% moisture) necessary to add to the compostables to achieve the desired 45% moisture (assuming 100% of those materials are composted) c.Explanation / Answer
Answer (a)
Let overall specific weight be S lb/yd3
Thus for 1 yd3 of mixture
Volume of food waste = 0.1S/500 yd3
Similarly volume of all individual wastes can be found and summation of them be equated to 1
Thus
S( 0.1/500 + 0.2/110 + 0.2/170 + 0.3/150 + 0.1/270 + 0.1/270) =1
Thus S = 168.48 lb/yd3
Answer (b) calculating moisture contents in 1 yd3 of mix
Moisture content of food waste = 0.7 *0.1*168.48 lb = 11.7936 lb
Moisture content of paper = 0.06*0.3*168.48 = 3.03264 lb
Moisture content of yard waste = 0.6*0.2*168.48 = 20.2176 lb
Thus total moisture content = 35.04384 lb
Answer (c)
Total weight of waste per day = 4.4*15000 lbs= 66000 lbs
Thus volume of waste per day = 66000/168.48 = 391.74 yd3
% Moisture content of waste = 36.39*100/(168.48-36.39) = 27.549
Let volume of sludge added per day = V yd3
Thus total volume = 391.74 + V
Thus 0.95V + 0.27549*391.74 = 0.45(391.74 +V )
Thus V = 136.725 yd3
Now moisture content of sludge = 0.95 thus weight of liquid = 1685.55 * 136.725* 0.95 lb
weight of solids = 1685.55*136.725*0.95 /0.95 = 230456.824 lb
Thus total weight of sludge = 449390.807 lb
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