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Math problem set Grade 10 Science

Glenmoor Clearwell: Chlorine Contact Time and CT Values

A grade 10 problem set on disinfection where students size a clearwell, find contact time with a baffling factor, and compute the CT a fictional plant achieves.

Objectives

  1. Students will calculate a contact basin's volume and detention time from its dimensions and flow.
  2. Students will apply a baffling factor and compute the CT value a disinfection step achieves.

Vocabulary

Clearwell
A clearwell is a large storage tank at a treatment plant that holds finished water right after it comes out of the filters. It gives the disinfectant enough time to work before the water is pumped out to homes and businesses.
Detention time
Detention time is how long water stays inside a tank or basin before flowing out. Water must stay long enough for each treatment step, like settling or disinfecting, to finish its job.
CT value
CT value is a way operators check that water has been disinfected long enough. It is found by multiplying how much disinfectant is in the water by how many minutes the water has been in contact with it.

Also: baffling factor, contact time

Teacher overview

After a treatment plant filters its water, it holds the water in a clearwell so chlorine has time to inactivate pathogens. Operators measure that protection with the CT value, the chlorine residual multiplied by the effective contact time. This set has students size a fictional plant's clearwell, find how long water actually stays in it, and calculate the CT it reaches. A common misconception in this math is using the full detention time as the contact time, which ignores short-circuiting; the baffling factor corrects for water that races through faster than average.

Glenmoor Clearwell: Chlorine Contact Time and CT Values — Grade 10 Science
Name: ____________________________ Date: ______________

Problems

The Glenmoor plant clearwell is a fictional basin, and every value below is given for it. Use these numbers and formulas:

  • Clearwell dimensions: 80 ft long, 30 ft wide, 12 ft of water depth.
  • Flow through the clearwell: 1,500 gpm.
  • Baffling factor: 0.5.
  • Chlorine residual leaving the clearwell: 1.2 mg/L.
  • Required CT for the target pathogen inactivation: 36 mg/L-min.
  • Rectangular volume (cu ft) = length times width times depth; gallons = cubic feet times 7.48.
  • Detention time (min) = volume in gallons divided by flow in gpm.
  • Effective contact time (min) = detention time times the baffling factor.
  • CT value = chlorine residual times effective contact time.
  • Cylinder volume (cu ft) = 0.785 times diameter squared times depth.
  1. Calculate the clearwell's volume in cubic feet.
  2. Convert that volume to gallons.
  3. Calculate the detention time in minutes at the 1,500 gpm flow.
  4. Apply the 0.5 baffling factor to find the effective contact time in minutes.
  5. Calculate the CT value the clearwell achieves, using the 1.2 mg/L residual.
  6. Does the plant meet the required CT of 36 mg/L-min? By how much does it clear or miss the target?
  7. During peak summer demand the flow rises to 2,500 gpm. Recalculate the effective contact time and the CT value, and state whether the plant still meets the 36 mg/L-min requirement.
  8. Glenmoor considers replacing the rectangular clearwell with a cylindrical tank of the same 12 ft depth that must hold at least the original volume in cubic feet. Find the minimum diameter, rounded up to the next whole foot.

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: 28,800 cu ft. Volume = 80 ft times 30 ft times 12 ft = 28,800 cubic feet.
  2. Answer: 215,424 gallons. 28,800 cubic feet times 7.48 gallons per cubic foot = 215,424 gallons.
  3. Answer: about 143.6 min. Detention time = 215,424 gallons divided by 1,500 gpm = 143.6 minutes.
  4. Answer: about 71.8 min. Effective contact time = 143.6 minutes times 0.5 = 71.8 minutes.
  5. Answer: about 86.2 mg/L-min. CT = 1.2 mg/L times 71.8 minutes = 86.2 mg/L-min.
  6. Answer: Yes, by about 50.2 mg/L-min. The achieved CT of 86.2 is above the 36 required, a margin of 86.2 minus 36 = 50.2 mg/L-min.
  7. Answer: about 51.7 mg/L-min. At 2,500 gpm the detention time is 215,424 divided by 2,500 = 86.2 minutes; times 0.5 gives 43.1 minutes; times 1.2 mg/L gives 51.7 mg/L-min, which still clears the 36 requirement.
  8. Answer: 56 ft. Set 0.785 times D squared times 12 equal to 28,800, so D squared = 28,800 divided by 9.42 = 3,057; D = 55.3 ft, which rounds up to 56 ft to hold at least the original volume.

Differentiation

Support

Provide the volume and conversion steps for problems 1 and 2 already completed so students begin at the detention-time calculation, and give the formula chain on a labeled reference card in the order the problems use it.

Extension

Ask students to find the highest flow, in gpm, at which the clearwell would still exactly meet the required CT with the given residual, and then explain how raising the chlorine residual would let the plant tolerate a higher flow.

ELL supports

Pair "clearwell," "detention time," and "baffling factor" with a diagram of a long basin showing average water taking a winding path while some water shortcuts straight across. Provide the formula chain as labeled arrows and allow answers to be shown as numbers with units rather than sentences.