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Math problem set Grade 9 Engineering & Design

Ostermill Orthophosphate Feed: Corrosion Control Dose Math

A grade 9 set on a fictional district adding orthophosphate after a high lead result, using the pounds formula to size the feed, the day tank, and the yearly cost.

Objectives

  1. Students will apply the pounds formula to size a chemical feed at average and peak plant flows.
  2. Students will convert a chemical feed rate in pounds per day into gallons of delivered product and a yearly cost.
  3. Students will write and solve a one-variable equation for the dose that produces a required feed rate.

Vocabulary

Orthophosphate
A chemical that utilities add to water in very small amounts. It coats the inside of pipes so lead cannot easily get into the water.
Action level
A trigger number in lead and copper testing. If too many homes test above this level, the utility must take steps like improving treatment and telling the public.

Also: corrosion inhibitor, pounds formula, day tank

Teacher overview

The Ostermill Water District is a fictional utility that draws from one well field and serves a small town. Its most recent round of tap sampling came back high for lead, so the district decided to feed orthophosphate, a corrosion inhibitor that builds a thin protective film on the inside of older service lines and fixtures. This set has students size that chemical feed with the pounds formula, convert the result into gallons of delivered product, and measure the sampling result against the action level that takes effect in 2027. The most common error in this math is entering the plant flow in gallons per day instead of million gallons per day, which inflates every answer by a factor of one million.

Ostermill Orthophosphate Feed: Corrosion Control Dose Math — Grade 9 Engineering & Design
Name: ____________________________ Date: ______________

Problems

The town of Ostermill and its water district are fictional. Use only the figures given below.

The Ostermill plant treats an average of 2.4 MGD. Its peak day flow is 3.6 MGD. The target orthophosphate dose is 3.0 mg/L. The district buys a liquid product that holds 2.5 pounds of orthophosphate in every gallon of solution, and it pays $2.50 for each pound of orthophosphate. One gallon of water weighs 8.34 pounds. One MGD equals 694 gpm.

  1. The pounds formula is pounds per day = dose in mg/L x flow in MGD x 8.34. Find the pounds of orthophosphate fed each day at the average flow of 2.4 MGD and a dose of 3.0 mg/L. Round to the nearest tenth.

  2. Now find the pounds fed on a peak day. The flow is 3.6 MGD and the dose is still 3.0 mg/L. Round to the nearest tenth.

  3. Use 60.0 pounds per day as the average feed rate. The district pays $2.50 for each pound. What does one 365-day year of orthophosphate cost?

  4. The liquid product holds 2.5 pounds of orthophosphate in each gallon. How many gallons of product does the plant feed on an average day at 60.0 pounds per day?

  5. The day tank in the feed room must hold a three-day supply at the peak rate of 90.1 pounds per day. How many gallons of product must the tank hold?

  6. Convert the plant's average flow of 2.4 MGD to gallons per minute. Use the conversion 1 MGD = 694 gpm.

  7. Ostermill's most recent ninetieth percentile lead result was 18 ppb. The action level drops to 10 ppb in 2027. How far above that 2027 level is the district's result?

  8. Use the same two numbers from problem 7. By what percent must Ostermill lower its ninetieth percentile result to reach 10 ppb? Round to the nearest tenth of a percent.

  9. Let D stand for the dose in mg/L. Write an equation for the dose that would feed exactly 75.0 pounds per day at the average flow of 2.4 MGD, then solve it. Round to the nearest hundredth.

  10. A new operator suggests doubling the dose to 6.0 mg/L so the film forms faster. Use your answer to problem 3. State what doubling the dose does to the yearly chemical cost, then give one reason a higher dose is not automatically the better choice.

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: 60.0 pounds per day.
    Multiply the dose, the flow, and the constant: 3.0 mg/L x 2.4 MGD = 7.2. Then 7.2 x 8.34 = 60.048 pounds per day, which rounds to 60.0 pounds per day. Students who write 2,400,000 for the flow will get an answer near 60 million and should be sent back to the units.

  2. Answer: 90.1 pounds per day.
    3.0 mg/L x 3.6 MGD = 10.8. Then 10.8 x 8.34 = 90.072 pounds per day, which rounds to 90.1 pounds per day. The dose did not change, so the feed rate rose in the same ratio as the flow.

  3. Answer: $54,750 per year.
    60.0 pounds per day x 365 days = 21,900 pounds in a year. Then 21,900 pounds x $2.50 per pound = $54,750.

  4. Answer: 24.0 gallons per day.
    60.0 pounds per day divided by 2.5 pounds per gallon = 24.0 gallons of product per day.

  5. Answer: 108.1 gallons.
    90.1 pounds per day divided by 2.5 pounds per gallon = 36.04 gallons per day at the peak rate. Then 36.04 gallons per day x 3 days = 108.12 gallons, which rounds to 108.1 gallons. A tank sold in standard sizes would be the next size up.

  6. Answer: 1,665.6 gpm.
    2.4 MGD x 694 gpm per MGD = 1,665.6 gpm.

  7. Answer: 8 ppb.
    18 ppb minus 10 ppb = 8 ppb above the level that takes effect in 2027.

  8. Answer: 44.4 percent.
    The required drop is 8 ppb out of a starting value of 18 ppb. 8 divided by 18 = 0.4444, which is 44.4 percent to the nearest tenth.

  9. Answer: 3.75 mg/L.
    The equation is D x 2.4 x 8.34 = 75.0, which simplifies to 20.016D = 75.0. Divide both sides by 20.016 to get D = 3.747, which rounds to 3.75 mg/L. Checking the rounded dose: 3.75 x 20.016 = 75.06 pounds per day, close enough for a feed setting.

  10. Answer: The cost doubles to $109,500 per year.
    Feed rate is directly proportional to dose, so doubling 3.0 mg/L to 6.0 mg/L doubles both the pounds fed and the bill: $54,750 x 2 = $109,500. Accept any sound reason a higher dose is not automatically better. Strong answers note that phosphate leaving the plant ends up in the wastewater stream and can feed algae downstream, that too much inhibitor does not build film any faster once the pipe surface is coated, and that the day tank and feed pump were sized for the lower rate.

Differentiation

Support

Give these students a formula card that prints pounds per day = dose x flow x 8.34 with each variable labeled by its unit, and work one example with different numbers than the set uses: a different, smaller fictional plant feeding 1.5 mg/L at 1.0 MGD needs 12.5 pounds per day. Supply problem 9 as a partly completed equation with only D missing, and allow a calculator throughout.

Extension

Have students find the dose in mg/L that would keep the yearly cost at or below a stated budget of $40,000 for the same fictional district, then explain in writing whether that dose is workable. Ask them to state which of the two figures, cost or protection, is the criterion and which is the constraint, and to defend the choice.

ELL supports

Pre-teach dose, feed rate, and day tank with a labeled sketch showing the storage drum, the feed pump, and the pipe carrying treated water. Provide the sentence frame "The plant feeds ___ pounds each day because the dose is ___ and the flow is ___" for problems 1 and 2. Allow problem 10 to be answered with a short list rather than a paragraph.

Extensions

Have students build a small table for the fictional town of Ostermill showing pounds per day at doses of 2.0, 2.5, 3.0, and 3.5 mg/L, all at the average flow of 2.4 MGD. Ask them to graph dose on the horizontal axis and pounds per day on the vertical axis, describe the shape of the graph, and explain what the slope of the line represents in plain language. Students can then mark the point on their graph that matches the 75.0 pounds per day feed rate from problem 9.