You have been asked to determine the optimum number of manholes for use in your subdivision. The sewer run is 2000 feet long. The pipe has a 3 foot/100 ft drop from end to end. Pumps cost $1,200 each. Manholes cost $600 per foot^1.2. Thus a 17 foot deep manhole would cost you $600*17^1.2 = $17,975.87. The pipe costs $20/foot. Digging and embedding the pipe depends on the average depth of the pipe segment at $0.60 per foot^1.4. Thus if the pipe segment were 400 foot long it would be embedded to an average depth of 11 feet and cost $6889.08, for a total job embedment cost of $34445.38. Minimum cover is 5 feet.
Determine the optimum number of manholes to use across the property.
EES solution:
“A sewer line layout requires a slope of 3ft/100ft drop, over a 2000 foot run. Manhole pumps cost $1,200 each.
Digging and installing a manhole costs $600/foot * manholedepth^1.2
Trenching and laying the sewer line costs $0.60/foot * averagepipedepth^1.4 per foot of pipe
Minimum cover = 5 feet
Pipe costs $20/foot
Cost of laying sewer lines: k:\my documents\classes\homework\422\EES\Sewer line.ees”
slope = 3/100
totallength = 2000
cover = 5
numberofmanholes = 5
piperun = totallength/numberofmanholes
pipedrop= piperun*slope
manholedepth = piperun*slope + cover
averagepipedepth = cover + pipedrop/2
Pumpcost = 1200 * numberofmanholes
Manholecost =600*(manholedepth^1.2)*numberofmanholes
costofpipe = 20 * totallength “pipe costs 20/foot”
costtoburypipe = (0.6 * (averagepipedepth^1.4)) * totallength
Totalcost = pumpcost + manholecost + costofpipe + costtoburypipe
Solution:
| numberofmanholes=5 | costofpipe=40000 | cover=5 | Pumpcost=6000 | totallength=2000 | slope=0.03 |
| piperun=400 | pipedrop=12 | averagepipedepth=11 | manholedepth=17 | ||
| costtoburypipe=34445 | Manholecost=89879 | Totalcost=170325 |
Table solution:
