A city with a population of 161.565 needs your assistance in evaluating its soli
ID: 298868 • Letter: A
Question
A city with a population of 161.565 needs your assistance in evaluating its solid waste collection system. Determine the mean time per collection stop plus the mean time to reach the next slop, the number of pickup locations per load, and the minimum number of trucks the city must own. The following information is given: Average truck capacity = 18.0 m^3 Average observed compaction ratio = 3.28 Crew size = 2 Number of pickups = l/wk (no rear-of-house service) Average number of cans per stop = 2.95/wk at 0.0911 m^3/can Average number of residents per stop = 2.5 Average uncompacted density = 122.0 kg/m^3 Average transport time to disposal site including delays and dumping = 1.50 h/trip Average number of trips to disposal site = 2/d Rest breaks = 2 at 15.0 min Average maintenance downtime = 36.0 min/d Average workday = 8.00 h Average percent of trucks out of service for major repairs = 15.0% V_T = V_p/rt_p [H/N_d - 2x/s - 2t_d - t_u - B/N_d] N_p = H/N_d - 2x/s - 2t_d - t_u - B/N_d/t_pExplanation / Answer
tp=tb+aCn/60
where tp=mean time per collection stop and the mean time required for reaching the next stop=??
tb=mean time between collection stop=1.50h/trip=90 minutes/stop
a= coefficient to regression fit to data point=1(assumed)
Cn=mean number of contain at each pickup location=2.95/week
Now tp=(90+1*2.95/60)
=(90+2.95)/60
=1.5
Now VT=Vp/rTp(H/Nd)(2x-s)-2(td)-(tu)-(B/Nd)
VT=truck capacity=18m3
Vp=volume of solid waste per stop=0.0911m3
r=compaction ratio=3.28
tp=mean time per collection stop+mean time to reach the next stop=1.5h
H=length of working day=8h
x=one way distance to disposal site=??
s=??
td=average delay time
tu=average unloading time
Nd=no of trips to disposal site per day=2/day
Td+tu=average transport time including delay and dumping=1.50 h/trip
B=off route time=36min/day=0.6/day
18=0.0911/3.28*1.5(8/2)-2x/s-2x-(2(td+tu)-0.6/2)
18=0.0911/4.92(4-2x/s-2(1.50)-0.3
18=0.018(4-2x/s-3-0.3)
18=0.018(0.7-2x/s)
18/0.018=0.7-2x/s
1000=0.7-2x/s
999.3=-2x/s
Now Np=(H/Nd-2x/s-2td-B/Nd)/tp
Tp=1.5h
Np=no of pickup locations per load=??
H=8
Nd=2/day
2x/s=999.3
B=0.6/day
Np=(8/2-999.3-(2(1.5)-0.6/2)/1.5
Np=4-999.3-3-0.3/1.5
=998.6/1.5
=665
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