Question 4 A 600 metric tonne/day resource-recovery receives municipal solid was

Question

Question 4 A 600 metric tonne/day resource-recovery receives municipal solid waste in 14-m3compactor vehicle at 250 kg/m3. Collection occurs 5 days/week in one shift per day, whereas the facility processes waste 24 hours per day, 7 days per week. Actual work is done only 6 hours per shift as 2 hours are spent on lunch, breaks and washing up. (i) Determine the nominal processing rate in tonne per day (ii) Calculate the vehicle capacity in tonnes per vehicle (iii) Determine the truckloads per hour, arriving to unload. (iv) Calculate the actual receiving rate in tonnes per hour and truckloads per hour (v) Calculate the peak receiving rate for the facility (in tonnes per hour and in truckloads per hour) taking a peaking factor of 150% and a seasonal peaking factor of 125% {1}

Explanation / Answer

Given waste is collected by 14m3 compactor vehicle at 250kg/m3.

Hence waste collected by one vehicle = 250kg/m3*14m3 = 3500 kg = 3.5 MT

Total amount of waste collected = 600 MT/day

Hence no. of vehicles required for collecting the entire waste = 600/3.5 = 171.43 (nearly 172 vehicles)

i) Nominal processing rate in tonne per day = 600 MT/day as the facility processes waste 24 hours per day.

ii) Vehicle capacity = 250*14 = 3.5 tonnes/vehicle.

iii) Truckloads required in a day for unloading the waste = 172

But only 6 hours are spent in a day for collection.

Hence no. of truckloads arriving per hour to unload = 172/6 = 28.66 (nearly 29 truckloads per hour)

iv) Actual receiving rate in tonnes/hr = 600 MT/24 hrs as the facility operates for 24 hours in a day.

= 25 tonnes/hr

  Actual receiving rate in truckloads/hr = 172/24 = 7.166 (nearly 8 truckloads/hr)

v) Given peaking factor of 150% and seasonal peaking factor of 125%

Peak receiving rate in tonnes/hr = 25*1.5*1.25 = 46.875 tonnes/hr.

Peak receiving rate in truckloads/hr = 8*1.5*1.25 = 15 truckloads/hr

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