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Worksheet Grade 12 Science

Flood Frequency and Culvert Sizing on Tullimore Creek

A grade 12 science worksheet using 20 years of peak-flow records to rank floods, find return periods and exceedance risk, and size a replacement culvert.

Objectives

  1. Students will rank an annual peak-flow record and compute return periods using a plotting-position formula.
  2. Students will convert an annual exceedance probability into the risk of at least one exceedance over ten years.
  3. Students will size a replacement culvert from a design flow and a maximum allowable velocity.

Vocabulary

Key words: annual peak flow, return period, exceedance probability, plotting position, design flow, continuity equation

Teacher overview

Winford is a fictional town whose main road crosses Tullimore Creek on a culvert installed decades ago. The road floods every few winters, and the town has twenty years of annual peak-flow records for the creek. This worksheet has students rank that record, convert ranks into return periods and exceedance probabilities, and then size a replacement culvert from a stated design flow and a velocity limit.

The recurring misreading here is treating a return period as a schedule. A twenty-one-year flow is not a flow that arrives once every twenty-one years in an orderly rotation; it is a flow with about a 4.8 percent chance of being exceeded in any single year, which is why item 5 asks students to convert the return period into a probability over the life of a structure.

Flood Frequency and Culvert Sizing on Tullimore Creek — Grade 12 Science
Name: ____________________________ Date: ______________

Student Worksheet

The table lists the largest instantaneous flow recorded each year in Tullimore Creek, at the gauge just above the fictional Winford road crossing. All values are in cubic feet per second.

Year Peak Year Peak
2006 103 2016 176
2007 61 2017 99
2008 204 2018 51
2009 131 2019 140
2010 48 2020 220
2011 159 2021 83
2012 89 2022 154
2013 255 2023 113
2014 118 2024 72
2015 67 2025 185

Use these formulas and stated design values.

  • Plotting position: return period T = (n + 1) / m, where n is the number of years of record and m is the rank of a peak when the peaks are sorted from largest to smallest
  • Annual exceedance probability: p = 1 / T
  • Risk of at least one exceedance in N years: R = 1 - (1 - p) raised to the power N
  • Continuity: Q = A x V
  • Circular area: A = 0.785 x D squared
  • The fictional Winford design manual sets the design flow for a road culvert at 290 cfs and caps the velocity through the culvert at 12 ft/s
  • Culverts are stocked in whole-foot diameters only

  1. Sort the twenty peaks from largest to smallest. What rank m does the 2008 peak of 204 cfs receive?

  2. Calculate the return period T for the largest peak in the record, using n = 20. Round to the nearest whole year.

  3. Calculate the annual exceedance probability for that largest peak. Express it as a percent rounded to one decimal place.

  4. Calculate the return period for the 2019 peak of 140 cfs. Round to two decimal places.

  5. Using the unrounded probability from item 3 (p = 1/21), calculate the risk that the largest recorded peak is equaled or exceeded at least once during the next ten years. Express it as a percent rounded to one decimal place.

  6. Using the design flow of 290 cfs and the 12 ft/s velocity cap, calculate the minimum cross-sectional area the new culvert needs, in square feet. Round to two decimal places.

  7. Calculate the minimum diameter, in feet, that gives the area from item 6. Round to two decimal places.

  8. Culverts come in whole-foot diameters. State the diameter the town must order, then calculate the actual velocity through that culvert flowing full at the 290 cfs design flow. Round the velocity to two decimal places.

  9. Convert the 290 cfs design flow to million gallons per day, using 1 MGD = 1.547 cfs. Round to one decimal place.

  10. The existing culvert is 4.0 feet in diameter. Calculate the flow it carries when it is full and the velocity in it reaches the 12 ft/s cap. Round to one decimal place.

  11. Count how many of the twenty recorded peaks exceed the existing culvert's capacity from item 10.

  12. Divide the twenty years of record by your count from item 11 to estimate the average return period, in years, of a flow that overtops the existing crossing. Round to one decimal place.

  13. Explain why twenty years of record cannot establish the flow with a 1 percent annual chance of being exceeded, and state what the plotting-position formula can and cannot tell the town.

  14. The replacement culvert costs far more than a repair to the existing one. Using your answers to items 5 and 12, explain to the fictional town board why the larger culvert is worth the money.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: Rank 3.
    Sorted from largest to smallest, the record runs 255, 220, 204, 185, 176, 159, 154, 140, 131, 118, 113, 103, 99, 89, 83, 72, 67, 61, 51, 48. The 204 cfs peak from 2008 sits third.

  2. Answer: 21 years.
    The largest peak is 255 cfs at rank m = 1. T = (20 + 1) / 1 = 21 years.

  3. Answer: 4.8 percent. p = 1 / 21 = 0.0476, which is 4.8 percent when rounded to one decimal place.

  4. Answer: 2.63 years.
    The 140 cfs peak sits at rank m = 8. T = 21 / 8 = 2.625, which rounds to 2.63 years.

  5. Answer: 38.6 percent.
    R = 1 - (1 - 0.0476) raised to the tenth power. (0.9524) squared is 0.90707, raised to the fourth is 0.82278, raised to the eighth is 0.67697, and multiplying the eighth power by the second gives the tenth power, 0.61405. R = 1 - 0.61405 = 0.38595, or 38.6 percent.

  6. Answer: 24.17 square feet.
    A = Q / V = 290 cfs / 12 ft/s = 24.1667 square feet, which rounds to 24.17 square feet.

  7. Answer: 5.55 feet.
    D squared = 24.1667 / 0.785 = 30.786. The square root of 30.786 is 5.548, which rounds to 5.55 feet.

  8. Answer: 6 feet, giving 10.26 ft/s.
    The next whole-foot size above 5.55 feet is 6 feet. Its area is 0.785 x 6 x 6 = 28.26 square feet, so V = 290 / 28.26 = 10.262 ft/s, which rounds to 10.26 ft/s. That is under the 12 ft/s cap.

  9. Answer: 187.5 MGD.
    290 cfs / 1.547 cfs per MGD = 187.46 MGD, which rounds to 187.5 MGD. The figure is large because a flood peak is many times a town's daily water use.

  10. Answer: 150.7 cfs.
    A = 0.785 x 4.0 x 4.0 = 12.56 square feet. Q = 12.56 x 12 = 150.72 cfs, which rounds to 150.7 cfs.

  11. Answer: 7 peaks.
    The peaks above 150.7 cfs are 255, 220, 204, 185, 176, 159, and 154, which is seven of the twenty years.

  12. Answer: 2.9 years.
    20 years of record / 7 exceedances = 2.857 years per exceedance, which rounds to 2.9 years. On that record the crossing is overtopped roughly every three years.

  13. Answer: Twenty years of record supports return periods no longer than 21 years, so a 1 percent flow has to be extrapolated.
    The plotting-position formula assigns the longest return period to the largest observed peak, and with n = 20 that ceiling is (20 + 1) / 1, or 21 years. A flow with a 1 percent annual chance has a 100-year return period, far outside the record. The formula can rank and space the observed floods and give a defensible estimate of frequent events, yet estimating a 100-year flow requires fitting a statistical distribution to the record or borrowing data from similar gauged watersheds, and the result carries wide uncertainty.

  14. Answer: The existing crossing floods about every three years, while the replacement handles a flow the record has never reached.
    Item 12 shows the current culvert is overtopped on average every 2.9 years, so the road closures and repair costs recur every few years. Item 5 shows that even the largest peak in the record has a 38.6 percent chance of being matched or beaten within ten years, and the 6-foot culvert passes the 290 cfs design flow with velocity to spare. Over the life of the crossing, one larger investment can cost less than repeated repairs and road closures.

Differentiation

Support

Give students the fictional Tullimore Creek peaks already sorted from largest to smallest with a blank rank column, so ranking is not the barrier to the return-period work. Work item 2 together, then have students complete items 3 and 4 before continuing. For items 6 through 8, supply a worked example using a different design flow so the sequence of area, then diameter, then rounded size, then recomputed velocity is modeled before students try it.

Extension

Have students plot the twenty peaks against their return periods on semi-log paper, fit a straight line by eye, and read off an estimate for the 50-year flow in the fictional Winford record. Ask them to compare that estimate with the 290 cfs design flow and to write two sentences on how much confidence the extrapolation deserves.

ELL supports

Provide a term card that separates the three ideas students most often merge: rank, return period, and probability, each with one example drawn from the table. Give a sentence frame for items 13 and 14, such as "The record supports return periods up to ___ years because ___." Allow the item 14 response as a short list of reasons with numbers attached rather than a paragraph.

Extensions

Students can pull the annual peak-flow record for a real gauged stream near the school from the United States Geological Survey and repeat items 1 through 5 on that record, then compare the length of the real record with the twenty years available in the fictional Winford case. A second option is to have students calculate the risk of at least one exceedance across a 50-year culvert design life rather than ten years, and explain why designers of long-lived structures use the longer window.