A population of 125,000 has a per-capita waste generation rate of 1.6 kg/d. 28%
ID: 968830 • Letter: A
Question
A population of 125,000 has a per-capita waste generation rate of 1.6 kg/d. 28% of the waste is recovered or recycled and the remaining waste is landfilled. The landfilled waste is compacted to an in-place density of 850 kg/m^3 and the V_2/V_r ratio is 0.2. The operating life of the landfill is 30 years, what is the total volume required for the landfill over its operating life? the waste (waste only, not daily soil cover) has a moisture content of 32% and a chemical formula of C_40 H_35 O_37 N_. The unbalanced decomposition formula is: C_40 H_36O_37N rightarrow CH_4 + N_2 + CO_2 If natural gas (methane) can be sold for $3.22 per 1,000 ft^3, how much money could the landfill operators potentially make from its sale over the lifetime of the landfill? Assume 1 atm and 25 degree C conditions within the landfill.Explanation / Answer
Population (p)
p = 125.000 person
Waste per day (W)
W = 1.6 kg/day.person
Operating life of the landfill:
Op = 30 years = 10950 days (standard years)
In-place density of waste:
d = 850 kg/m3
Remaining waste (Rw) after recycling:
Rw = p.W.(100-28)/100
Rw = 125.000 person . 1.6 kg/day.person . 0.72
Rw = 125.000 person . 1.6 kg/day.person . 0.72
Rw = 144.000 kg/day
*rate of waste of the entire population
Total waste in 30 years (Tw):
Tw = Rw.Op
Tw = 144.000 kg/day . 10950 day
Tw = 1576800000 kg
Volume required for the lanfill (V):
V = Tw/d
V = 1576800000 kg/850 kg/m3
V = 1855059 m3
The waste has a moisture content of 32% and a chemical formula of C40H86O37N (M = 1173.0953 g/mol):
Dry waste (Dw)
Dw = Tw.(100-32)/100
Dw = 1576800000 kg . 0.68
Dw = 1072224000 kg
Mol of waste:
nw = 1072224000 kg/1.1730953 kg/mol
nw = 914012697.9 mol
Balanced decomposition equation:
2C40H86O37N --> 43CH4 + N2 + 37CO2
1 mol of gas in standard conditions is equivalent to 22.4 L (0.7910485 ft3)
Money could the landfill operators potentially make (M$):
M$ = 914012697.9 mol . (43 mol CH4/2 mol) . (0.7910485 ft3/1 mol CH4) . (3.22 $/1000 ft3)
M$ = 50055254.31 $
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