Worksheet Grade 10 Engineering & Design
Pennyford Sanitary Sewer: Inflow and Infiltration Math
A grade 10 worksheet measuring storm water leaking into a fictional town's sanitary sewer, then pricing the extra treatment and testing what a lining project would fix.
Objectives
- Students will separate storm-driven inflow and infiltration from a collection system's dry-weather flow.
- Students will normalize a leakage rate per inch-diameter-mile and price the extra treatment it causes each year.
- Students will test whether a sewer lining project brings a wet-weather peak back within a plant's rated capacity.
Vocabulary
- Infiltration
- Infiltration is when rain or melted snow soaks down into the soil instead of flowing away across the top of it. Water that infiltrates can feed plant roots or keep sinking to become groundwater.
Also: inflow, dry-weather flow, peaking factor, inch-diameter-mile, rated capacity
Teacher overview
Rainwater enters a sanitary sewer through cracked pipe, leaking manhole covers, and roof drains that were never supposed to be connected, and the treatment plant has to handle every gallon of it. This worksheet follows one storm through the collection system of the fictional town of Pennyford, measuring how much extra water arrives, pricing what it costs to treat, and testing whether a lining project would keep the plant within its rating. The misconception this worksheet targets is that a wet-weather overflow means the plant is undersized; the plant here is sized correctly for the town it serves, and the excess flow is rainwater that should never have entered the sewer at all.
Student Worksheet
Pennyford is a fictional town created for this worksheet. Its collection system and plant data are below. Round money to the nearest cent and percentages to one decimal place.
| System data | Value |
|---|---|
| Length of sanitary sewer | 34 miles |
| Average pipe diameter | 10 inches |
| Dry-weather average daily flow | 1.20 MGD |
| Wet-weather peak daily flow | 3.45 MGD |
| Days the peak lasts | 3 |
| Treatment cost | $1,450 per million gallons |
| Storms of this size in a typical year | 12 |
| Plant rated capacity | 3.00 MGD |
| Share of inflow and infiltration a lining project would remove | 40 percent |
One million gallons equals 1,000,000 gallons. An inch-diameter-mile is one mile of pipe one inch across, so it is found by multiplying the diameter in inches by the length in miles.
Find the inflow and infiltration during the storm. Subtract the dry-weather flow from the wet-weather peak. Give the answer in MGD.
Find the peaking factor. Divide the wet-weather peak by the dry-weather flow. Round to two decimal places.
Find the inch-diameter-miles of sewer. Multiply the 10-inch average diameter by the 34 miles.
Find the inflow and infiltration per inch-diameter-mile. First write your answer from item 1 in gallons per day. Then divide by the inch-diameter-miles from item 3.
The peak lasts 3 days. Find the extra volume the plant treats, in million gallons.
Treatment costs $1,450 per million gallons. Find the cost of treating that extra volume.
The town sees 12 storms like this in a typical year. Find the yearly cost of treating the extra water.
The plant is rated for 3.00 MGD. By how much does the 3.45 MGD peak exceed that rating?
What percent above the rated capacity is the peak? Divide your answer from item 8 by 3.00 and multiply by 100.
Lining would remove 40 percent of the inflow and infiltration. Find how much would remain, in MGD.
Find the new wet-weather peak after lining. Add your answer from item 10 to the dry-weather flow.
How far below the 3.00 MGD rating would the new peak sit?
Find the yearly treatment saving after lining. Take 40 percent of your answer from item 7.
Explain why a utility spends money to keep clean rainwater out of a sanitary sewer.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
Answer: 2.25 MGD.
Subtracting the dry-weather flow from the peak: 3.45 minus 1.20 equals 2.25 MGD of inflow and infiltration.Answer: 2.88.
Dividing 3.45 by 1.20 gives 2.875, which rounds to a peaking factor of 2.88. Anything much above two signals a leaky system.Answer: 340 inch-diameter-miles.
Multiplying the 10-inch average diameter by 34 miles gives 340 inch-diameter-miles of sewer.Answer: about 6,618 gallons per day per inch-diameter-mile.
The inflow and infiltration of 2.25 MGD is 2,250,000 gallons per day, and dividing by 340 inch-diameter-miles gives 6,617.6, which rounds to about 6,618 gallons per day per inch-diameter-mile.Answer: 6.75 million gallons.
Multiplying 2.25 MGD by the 3 days the peak lasts gives 6.75 million gallons of extra water.Answer: $9,787.50.
Multiplying 6.75 million gallons by 1,450 dollars per million gallons gives 9,787.50 dollars for the storm.Answer: $117,450.
Multiplying 9,787.50 dollars by 12 storms gives 117,450 dollars per year in treatment the town would not otherwise buy.Answer: 0.45 MGD.
Subtracting the rating from the peak: 3.45 minus 3.00 equals 0.45 MGD above what the plant is built to handle.Answer: 15.0 percent.
Dividing 0.45 by 3.00 gives 0.15, and multiplying by 100 gives 15.0 percent above the rated capacity.Answer: 1.35 MGD.
Removing 40 percent leaves 60 percent, so 2.25 multiplied by 0.60 equals 1.35 MGD of inflow and infiltration remaining.Answer: 2.55 MGD.
Adding the remaining inflow and infiltration to the dry-weather flow: 1.20 plus 1.35 equals a new peak of 2.55 MGD.Answer: 0.45 MGD.
Subtracting the new peak from the rating: 3.00 minus 2.55 equals 0.45 MGD of margin. The project turns a peak 0.45 MGD above the rating into one 0.45 MGD below it.Answer: $46,980.
Taking 40 percent of the yearly cost: 117,450 multiplied by 0.40 equals 46,980 dollars saved each year in treatment.Answer: Clean rainwater adds treatment cost and can overload the plant.
Accept any answer explaining that clean stormwater costs the same per gallon to pump and treat as wastewater, and that it can push flow past the plant's rating and cause overflows of untreated sewage. Strong answers add that the plant would otherwise be correctly sized, so repairing pipe is cheaper than expanding a plant to handle water that should never have entered it.
Differentiation
Support
Give these students items 1 and 2 worked as examples with each step labeled, and provide a conversion reminder that one MGD equals 1,000,000 gallons per day. Let them skip items 3 and 4, which carry the least familiar unit, and mark the remaining items in two blocks, with items 5 through 7 as the cost block and items 8 through 12 as the capacity block.
Extension
Ask these students to find the share of inflow and infiltration a lining project would have to remove for the fictional Pennyford peak to land exactly at the plant's rating, and to compare that share with the 40 percent the project promises. Have them also explain why removing a roof drain connection is usually cheaper per gallon than lining a mile of pipe.
ELL supports
Provide a labeled diagram showing three ways water enters a sanitary sewer, which are a cracked pipe joint, a leaking manhole cover, and a roof downspout connected to the sewer. Post a term strip pairing inflow with rain entering directly, infiltration with groundwater seeping in, dry-weather flow with normal sewage, and rated capacity with what the plant is built to treat.
Extensions
Have students find whether their own community has a sewer system evaluation survey or a consent decree addressing wet-weather overflows, and summarize which repairs it requires. Students can also design a simple smoke-testing or dye-testing plan for one block of the fictional Pennyford system and explain what result would identify an illegal roof drain connection.