Water for Schools
Menu

Lesson plan Grade 11 Math

Harnsworth's Carbon Contactor: Reading a PFAS Breakthrough Curve

A 55-minute grade 11 lesson where students read key features of a granular activated carbon breakthrough curve and compute average rates of change across its intervals.

Objectives

  1. Students will interpret the intercept, the steepest interval, and the end behavior of a breakthrough curve.
  2. Students will calculate average rates of change over stated intervals of a table.
  3. Students will use linear interpolation to estimate the point at which a curve crosses a regulatory limit.

Vocabulary

Granular activated carbon
Small black grains of specially treated carbon that grab contaminants as water flows through them. Utilities use big tanks of it to remove PFAS, bad tastes, and odors.
Influent
Influent is the incoming dirty water that arrives at a treatment plant, ready to be cleaned. Operators test it to see how much work the plant has to do.
Effluent
Effluent is the cleaned water that leaves a treatment plant at the end of the process. It is tested carefully before being released to a river, lake, or the ocean.

Also: breakthrough curve, bed volume

Teacher overview

A granular activated carbon vessel does not fail all at once. It holds a contaminant almost completely at first, then lets a trace through, then more, and finally passes water nearly as contaminated as what entered, which is why operators track effluent against the number of bed volumes treated rather than against the calendar. This lesson works from a fictional breakthrough table for the Harnsworth Regional Water District's Ockley contactor, where the influent holds a fictional 32 ppt of PFOA and the maximum contaminant level is 4.0 ppt. The misconception it targets is that a curve which is nearly flat at both ends must be flat in the middle too, so students who read only the first and last rows conclude the carbon is either fine or spent and miss the narrow interval where the decision actually has to be made.

Procedure

Launch (6 min): Show only the first two and the last two rows of the fictional Harnsworth table and ask students to sketch what happens in between. Collect two or three sketches on the board, which typically include a straight line and a sudden jump.

Reading key features (12 min): Reveal the full table and have the class plot it. Name the features aloud as students mark them: the value at zero bed volumes, the interval where the curve rises fastest, and the level the curve approaches at the right. Ask why, in this simple single-contaminant model, the curve cannot rise above the influent concentration, and establish that the influent value is a horizontal asymptote rather than a coincidence.

Average rate of change (12 min): Model the average rate of change over the interval from 20 to 40 thousand bed volumes, keeping the units as parts per trillion per thousand bed volumes throughout. Have students compute the rate over 40 to 60 and compare. Draw the two secant lines on the plotted curve so the numerical comparison has a visual counterpart.

Independent practice (20 min): Students complete questions 1 through 11 on the handout, including the two interpolation estimates in questions 5 and 9.

Closure and formative check (5 min): Students answer handout question 11 on a slip of paper. Collect the slips and sort them for any answer claiming quarterly sampling is adequate because the curve is flat at the start, which shows the student has not connected the steep middle interval to sampling frequency, and address those first next session.

Harnsworth's Carbon Contactor: Reading a PFAS Breakthrough Curve — Grade 11 Math
Name: ____________________________ Date: ______________

Student Handout

The fictional Harnsworth Regional Water District runs a granular activated carbon contactor at its Ockley wellfield. The water entering the contactor holds a fictional 32 ppt of PFOA. The table gives the concentration leaving the contactor after each stated number of bed volumes treated.

Bed volumes treated (thousands) Effluent PFOA (ppt)
0 0.0
10 0.2
20 0.8
30 3.0
40 9.6
50 19.0
60 26.5
70 30.4
  1. Find the average rate of change of effluent concentration from 20 to 40 thousand bed volumes, in parts per trillion per thousand bed volumes.
  2. Find the average rate of change from 40 to 60 thousand bed volumes.
  3. Find the average rate of change from 60 to 70 thousand bed volumes.
  4. Name the twenty-thousand bed-volume interval in the table over which the effluent rises fastest.
  5. The maximum contaminant level for PFOA is 4.0 ppt. Using linear interpolation between the 30 and 40 thousand rows, estimate the number of bed volumes at which the effluent reaches that level.
  6. The contactor treats 90 thousand bed volumes each year. Using your answer to question 5, find the time to that level in months, to one decimal place.
  7. State the value the effluent concentration approaches as the bed volumes keep increasing, and give a one-sentence reason.
  8. Find the percent of PFOA removed at 50 thousand bed volumes, using removal efficiency = (influent - effluent) divided by influent, times 100.
  9. Suppose the district decides to change the carbon at 2.0 ppt instead. Using linear interpolation between the 20 and 30 thousand rows, estimate the number of bed volumes at that point.
  10. Using your rounded answers from questions 5 and 9, find the percent reduction in the carbon's run length caused by moving the changeout from 4.0 ppt to 2.0 ppt, to one decimal place.
  11. Explain in one or two sentences what the shape of this curve means for a district that samples the contactor only once each quarter.

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: 0.44 ppt per thousand bed volumes.
    The change is 9.6 - 0.8 = 8.8 ppt across 40 - 20 = 20 thousand bed volumes, and 8.8 divided by 20 = 0.44.

  2. Answer: 0.845 ppt per thousand bed volumes.
    The change is 26.5 - 9.6 = 16.9 ppt across 20 thousand bed volumes, and 16.9 divided by 20 = 0.845.

  3. Answer: 0.39 ppt per thousand bed volumes.
    The change is 30.4 - 26.5 = 3.9 ppt across 10 thousand bed volumes, and 3.9 divided by 10 = 0.39.

  4. Answer: From 40 to 60 thousand bed volumes.
    That interval has the largest average rate of change of the three computed, 0.845 against 0.44 and 0.39, so the curve is steepest there.

  5. Answer: 31.5 thousand bed volumes.
    Between the two rows the effluent climbs 9.6 - 3.0 = 6.6 ppt. Reaching 4.0 ppt requires 4.0 - 3.0 = 1.0 of that climb, and 1.0 divided by 6.6 = 0.152, so the estimate is 30 + 0.152(10) = 31.5 thousand bed volumes.

  6. Answer: 4.2 months.
    Dividing gives 31.5 divided by 90 = 0.35 year, and 0.35 x 12 = 4.2 months.

  7. Answer: 32 ppt.
    The effluent approaches the influent concentration, because once the carbon holds no more PFOA the water leaves the vessel carrying what it carried on the way in, and in this simple single-contaminant model it cannot leave with more than it entered with.

  8. Answer: 40.6%.
    Substituting gives (32 - 19.0) divided by 32, times 100, which is 13.0 divided by 32 = 0.40625, or 40.6 percent removed.

  9. Answer: 25.5 thousand bed volumes.
    Between the two rows the effluent climbs 3.0 - 0.8 = 2.2 ppt. Reaching 2.0 ppt requires 2.0 - 0.8 = 1.2 of that climb, and 1.2 divided by 2.2 = 0.545, so the estimate is 20 + 0.545(10) = 25.5 thousand bed volumes.

  10. Answer: 19.0%.
    The run length falls from 31.5 to 25.5 thousand bed volumes, a drop of 6.0, and 6.0 divided by 31.5 = 0.190, or 19.0 percent shorter.

  11. Answer: Quarterly sampling can miss the steep interval entirely.
    The contactor passes about 22.5 thousand bed volumes each quarter, which is wider than the whole interval over which the effluent climbs from below the limit to well above it, so a district sampling four times a year could record a clean result and an exceedance in consecutive samples with nothing in between.

Differentiation

Support

Give these students the fictional Harnsworth table already plotted on labeled axes, so the first task is reading the graph rather than building it. Provide the average-rate-of-change formula with the subtraction set up and only the numbers left blank, and let them answer question 4 by comparing the steepness of three drawn secant lines before computing anything.

Extension

Ask these students to estimate the instantaneous rate of change at 50 thousand bed volumes for the fictional Harnsworth contactor by shrinking the interval around that point, and to describe what happens to their estimates as the interval narrows. Then have them explain why a plant would run two carbon vessels in series rather than one larger vessel.

ELL supports

Pre-teach "influent," "effluent," and "bed volume" with a labeled drawing of a vessel showing water entering at the top and leaving at the bottom, with the carbon layer shaded. Provide sentence frames for questions 7 and 11, such as "The effluent gets close to ___ because ___," and "Sampling once a quarter is a problem because ___."

Extensions

Students can research how a utility reports a contactor changeout in its annual water quality report and describe what a customer would and would not see there. A second task is to sketch, without new numbers, what the fictional Harnsworth curve would look like for a contaminant the carbon holds poorly, and to say which key feature changes most.