Question

Question 2: Suppose it takes 0.35 hours to drive from the garage to the beginning of the route, 0.3 h to drive between the route and disposal site, and 0.35 hours to return from the disposal site to the garage. It takes 0.25 hours to offload a truck at the disposal site. The crew is given two 20-minute break per day and another 20 minutes are allowed for unexpected delays. If three runs are made to the disposal site each day a. how much time is left in ap 8-hour shift for actual refuse collection? R c b. If it takes 88 sec to collect the waste at each stop, How many stops could be made by one truck each run? c. If the volume collected from each stop is 20 fthow large should the truck be if it has a compaction ratio of 2.5:1?

Answer 1

Solution:

First we need to add all the time intervals to derive the total time of operation

garage to route = 0.35

route to disposal = 0.3

disposal to garage = 0.35

offloading = 0.25 hrs

breaks + delay = 20*2 + 20 mins = 60 mins = 1 hr

Thus total operation time for 1 run = 0.35 + 0.3 + 0.35 + 0.25 = 1.25 hrs

3 runs are made so total time = 3.75 hrs

Now we have to add the breaks and the delay = 3.75 + 1 = 4.75 hrs

so time remaining = 8 - 4.75 = 3.25 hrs = 3 hr and 15 mins

although here it is not specified whether the vehicle returns to the garage every time or not.

b. 88 sec to collect waste.

route to disposal site = 0.3 hrs = 0.3 * 60 * 60 = 1080 secs

thus Number of stops = 1080 / 88 = 12.27

thus, 12 complete stops can be made

c. volume collected from each stop = 20 ft3

total volume collected = 20 * 12 = 240 ft3

compaction ratio = 2.5 :1

compaction ratio = compacted volume / initial volume

but compacted volume is always smaller than the initial.

so the ratio would be used in the following form:

(1/ 2.5) = compacted volume / 240

or compacted volume = 240/ 2.5 = 96 ft3

Thus the minimum volume the truck should carry is 96 ft3