SHWPCE
Homework 4
4.1.Football field dimensions ~ 120 (50) = 6000 yds^2
Landfill volume 6000 (20/3) = 40 000 yd^3
Landfill volume occupied by daily soil cover = 20%
Effective landfill volume available for refuse = 40 000 x 0.8 = 32 000 yd^3
Waste density (in landfill) ~ 1200 lb/yd^3
Landill capacity = 1200 x 32 000 = 38 400 000 lb
Waste Generation ~ 4 lb/capita day
Waste Diversion ~ 50%  Recycled refuse = 2 lb/cap day ; Refuse to landfill = 2 lb/cap day
Population (Hartford) ~ 125 000  Waste generated =125 000 (2) = 250 000 lb/day
Time it would take to fill landfill = 38 400 000/250 000 = 153 days ~ 5 months
4.2.- Same procedure as above.
Volume of landfill required (w/diversion & incl soil cover ) ~ 15 000 yd^3/yr
Volume of landfill (220x220 yd w/side slopes) ~ 180x180x20 + 4*20*20/2 = 648 800 =
Time it would take to fill landfill = 648 800/15 000 = 43 yrs
4.3.Assume a calculation basis of 100 lb of refuse, then the numbers in table are lb of constitutents
Density of uncompacted refuse
= 100/[(10/23) + (20/4.45) + (40/3.8) + (10/1.87) + (10/18.45) + (10/6.36)] =
= 100/[0.43 + 4.5 + 10.5 + 5.34 + 0.54] = 4.69 lb/ft^3 =126 lb/yd^3 (urealistically low!)
It is likely that the table has a misprint and the fractions are VOLUME fractions; If so
Density of uncompacted refuse
= 0.1x23 + 0.2x4.45 + 0.4x3.8 + 0.1x1.87 + 0.1x18.45 + 0.1x 6.36 = 7.4 lb/ft^3 = 200 lb/yd^3
Compaction factor F =1/5  Density of compacted refuse ~ 7.4 x F ~ 37 lb/ft^3 = 590 kg/m^3
Density of uncompacted & compacted refuse (aft recovery) ~ 7.6 4 lb/ft^3 and 38 4 lb/ft^3
Diversion ~ 35%  Life of landfill (w/diversion) = 10/0.65 = 15.5 yrs
4.5.Methane generation in landfill ~ 100 m^3/ton SW
Gas emission constant = 0.0307 /yr
Year 1 (Eqn p. 120)
Q = 2 x 0.0307 x 100 x 1000 x 365 x exp(-0.0307 x 1) ~ 2 200 000 m^3
Year 2
2 x 0.0307 x 100 x 1000 x 365 x exp(-0.0307 x 2) +
+ 2 x 0.0307 x 100 x 1000x1.1 x 365 x exp(-0.0307 x 1) ~ 4 200 000 m^3
4.8.- Expected landfill life ~ 10-20 yrs , depending on assumptions
4.9.25 yr-24 hr design storm rainfall (CT) ~ 6 in  0.00017 cm/s
Using Eqn on p. 128 P ~ 2020 cm = 67 ft
4.10.-
Continuity eqn. v = (750/60)/0.0884 = 141.4 ft/s = 43 m/s
M_landfill gas = f_CH4 M_CH4 + f_CO2 M_CO2 = 0.5x16 + 0.5x44 = 30 g/mol ;
P = 1958 lb/ft^2 = 93 750 Pa ;
R = 8.314 J/K mol ; T = 115 oF = 46.11 oC ~ 319 K
Gas Density = M P/(RT) = ( 0.03 x 93750 )/(8.314 x 319) ~ 1 kg/m^3 = 0.06 lb/ft^3
Gas Viscosity ~ 1e-5 Pa s
Re = rho x v x D/mu ~ (1)(43)(4 x 0.0254)/1e-5 ~ 430 000  f =0.014 (Moody diagram)
dP = (0.0096) (0.0618) (0.016) (800) (141.4)^2 /(4/12)x(32.2) = 14 ft of water