Water for Schools
Menu

Math problem set Grade 9 Science

Marlow Springs Aeration: Oxygen Demand and Blower Energy

A grade 9 problem set where students find the BOD load a fictional plant removes, convert it into oxygen demand, and compare blower energy, cost, and payback.

Objectives

  1. Students will use the pounds formula to convert a BOD concentration and a plant flow into a daily mass loading.
  2. Students will calculate the electricity two blower designs need to deliver the same oxygen demand.
  3. Students will compute annual savings and a simple payback period for an equipment upgrade.

Vocabulary

Biochemical oxygen demand (BOD)
Biochemical oxygen demand, or BOD, measures how much oxygen tiny microbes will use up as they eat the waste in water. Dirtier water has a higher BOD.
Aeration
Aeration means mixing air into water. At a wastewater plant, blowers bubble huge amounts of air through tanks so the helpful microbes have the oxygen they need to eat the waste.

Also: pounds formula, removal efficiency, payback period

Teacher overview

The bacteria that clean wastewater need oxygen, and at most plants the blowers that supply that oxygen are the largest single electricity user on site. How much oxygen the bacteria need depends on the mass of organic matter they break down each day, which is why operators convert a concentration in mg/L into pounds per day before sizing anything. This set follows the fictional town of Marlow Springs through that conversion and then compares its old blowers against a fine-bubble replacement. The misconception this set is built to correct is that oxygen demand follows the concentration alone; students who reason that way conclude that a plant with strong wastewater always needs more air than a plant with weak wastewater, when a large flow of weak wastewater can carry far more pounds of oxygen demand per day.

Marlow Springs Aeration: Oxygen Demand and Blower Energy — Grade 9 Science
Name: ____________________________ Date: ______________

Problems

The fictional Marlow Springs Water Reclamation Facility treats 2.4 MGD. Incoming wastewater carries 240 mg/L of biochemical oxygen demand, and treated water leaving the plant carries 12 mg/L. Aeration must supply 1.1 pounds of oxygen for every pound of oxygen demand removed. The plant's existing blowers deliver 1.4 pounds of oxygen per kWh, while proposed fine-bubble blowers would deliver 2.1 pounds per kWh. Electricity costs $0.095 per kWh. Use pounds per day equals concentration in mg/L times flow in MGD times 8.34. Round to the nearest whole unit unless a problem says otherwise.

  1. How many mg/L of biochemical oxygen demand does the plant remove?
  2. Using the pounds formula, find the pounds of oxygen demand the plant removes each day. Round to the nearest tenth of a pound.
  3. Find the plant's removal efficiency as a percent, using influent minus effluent divided by influent.
  4. Aeration must supply 1.1 pounds of oxygen for every pound removed. Using your answer from problem 2, find the pounds of oxygen the blowers must deliver each day.
  5. The existing blowers deliver 1.4 pounds of oxygen per kWh. Using your answer from problem 4, find the electricity they use each day, in kWh.
  6. The fine-bubble blowers deliver 2.1 pounds of oxygen per kWh. Using the same oxygen demand from problem 4, find their daily electricity use, in kWh.
  7. By what percent would daily blower electricity fall if the plant replaced the blowers? Round to the nearest tenth of a percent.
  8. Electricity costs $0.095 per kWh. Using the daily difference between problems 5 and 6, find the electricity cost the plant would avoid in a 365-day year. Round to the nearest dollar.
  9. The blower replacement would cost $520,000. Divide that cost by your answer from problem 8 to find the simple payback period in years. Round to the nearest tenth.
  10. Federal secondary treatment rules set a monthly average limit of 30 mg/L for biochemical oxygen demand, and this plant leaves 12 mg/L. An engineer suggests running less air to save electricity. Using your answers from problems 3, 7, and 8, give one reason to try it and one reason to be careful about it.

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: 228 mg/L. 240 mg/L minus 12 mg/L equals 228 mg/L removed.
  2. Answer: 4,563.6 pounds per day. 228 mg/L times 2.4 MGD equals 547.2, and 547.2 times 8.34 equals 4,563.6 pounds per day.
  3. Answer: 95 percent. 240 minus 12 equals 228, and 228 divided by 240 equals 0.95, or 95 percent.
  4. Answer: 5,020 pounds per day. 4,563.6 pounds per day times 1.1 equals 5,019.96, which rounds to 5,020 pounds of oxygen per day.
  5. Answer: 3,586 kWh per day. 5,020 pounds divided by 1.4 pounds per kWh equals 3,585.7, which rounds to 3,586 kWh per day.
  6. Answer: 2,390 kWh per day. 5,020 pounds divided by 2.1 pounds per kWh equals 2,390.5, which rounds to 2,390 kWh per day.
  7. Answer: 33.4 percent. 3,586 minus 2,390 equals 1,196 kWh per day saved, and 1,196 divided by 3,586 equals 0.3335, or 33.4 percent.
  8. Answer: $41,471. 1,196 kWh per day times $0.095 per kWh equals $113.62 per day, and $113.62 times 365 days equals $41,471.
  9. Answer: 12.5 years. $520,000 divided by $41,471 per year equals 12.54, which rounds to 12.5 years.
  10. Answer: Accept any answer that pairs a supported reason with a supported caution. A reason to try it: the plant removes 95 percent of the incoming oxygen demand and leaves 12 mg/L, far under the 30 mg/L limit, so there appears to be room to trim air, and blower electricity is large enough that even a partial cut is worth real money next to the $41,471 the new blowers would save. A reason to be careful: the bacteria that do the work need dissolved oxygen at all times, a storm or an industrial discharge can raise the incoming load without warning, and a plant that has cut its margin may exceed the monthly average before operators can respond.

Differentiation

Support

Provide a worked pounds-formula example using different numbers, showing each unit written out and cancelled, before students attempt problem 2. Give the oxygen demand from problem 4 to students who need it so they can still complete the energy and cost comparison in problems 5 through 9.

Extension

Have students find the electricity price at which the fictional plant's payback period would fall to 8 years, holding every other value constant, and then explain why a utility might approve the upgrade even at a longer payback.

ELL supports

Pre-teach influent, effluent, aeration, and payback with a labeled flow diagram of the fictional Marlow Springs plant showing where air enters the process. Provide a sentence frame for problem 10: "The plant could use less air because ___, but this would be risky because ___."