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: