Math problem set Grade 11 Science
Cheswick's UV Reactor: Dose, Transmittance, and Log Inactivation
A grade 11 problem set on ultraviolet disinfection, moving from exposure time and lamp aging to absorbance from transmittance and the survivors left after inactivation.
Objectives
- Students will calculate ultraviolet exposure time from a reactor volume and a plant flow rate.
- Students will compute delivered dose after lamp aging and sleeve fouling and compare it with a required dose.
- Students will convert ultraviolet transmittance to absorbance and apply it to intensity at depth.
Vocabulary
Key words: ultraviolet dose, transmittance, absorbance, log inactivation, fouling factor, end of lamp life
Teacher overview
Ultraviolet light disinfects by damaging the genetic material of organisms as they pass a lamp, and what matters is the dose delivered, which is intensity multiplied by the time each organism spends in the beam. This set follows one reactor at the fictional Cheswick Water Treatment Plant from a clean new lamp to an aged, fouled one, then asks what flow the plant can still treat. The misconception to correct early is the belief that dose is a property of the lamp. Students often assume a brighter lamp guarantees disinfection, when a faster flow shortens exposure time and cloudy water absorbs the light before it reaches the far side of the chamber, and either effect can leave the dose short. A second point worth stating: ultraviolet light leaves no residual, so a chlorine residual is still needed in the distribution system.
Problems
Every problem describes the Cheswick Water Treatment Plant, a fictional facility whose state permit requires a delivered dose of at least 40 mJ/cm2. Use 694 gallons per minute per million gallons per day, and treat dose in mJ/cm2 as intensity in mW/cm2 multiplied by exposure time in seconds.
| Design value | Reading |
|---|---|
| Plant flow through the reactor | 2.0 MGD |
| Volume inside the ultraviolet chamber | 65 gallons |
| Lamp intensity when new | 12.0 mW/cm2 |
| End-of-lamp-life output factor | 0.75 |
| Sleeve fouling factor | 0.90 |
| Ultraviolet transmittance of the source water | 85.0 percent per centimeter |
| Required dose under the permit | 40 mJ/cm2 |
- Cheswick sends 2.0 MGD through the reactor. Using 694 gallons per minute per million gallons per day, calculate the flow in gallons per minute.
- The chamber holds 65 gallons. Using your item 1 answer, calculate the average exposure time in seconds, rounded to the nearest hundredth. Convert the flow to gallons per second first.
- A new lamp delivers an average intensity of 12.0 mW/cm2. Using your item 2 exposure time, calculate the delivered dose in mJ/cm2, rounded to the nearest tenth.
- At the end of lamp life the output factor is 0.75, and sleeve fouling multiplies intensity by a further 0.90. Calculate the delivered intensity in mW/cm2 for an aged, fouled lamp that started at 12.0 mW/cm2.
- Using your item 4 intensity and the item 2 exposure time, calculate the delivered dose in mJ/cm2 at the end of lamp life, rounded to the nearest tenth. State whether it meets the permit requirement of 40 mJ/cm2.
- The source water has an ultraviolet transmittance of 85.0 percent per centimeter. Absorbance equals the negative base-ten logarithm of the transmittance expressed as a decimal. Calculate the absorbance per centimeter, rounded to four decimal places.
- Intensity at a distance d centimeters from the lamp sleeve equals the starting intensity multiplied by 10 raised to the power of negative absorbance times d. Using a starting intensity of 12.0 mW/cm2, your item 6 absorbance, and d equal to 5 centimeters, calculate the intensity at the far wall of the chamber, rounded to the nearest hundredth.
- Using the end-of-life intensity from item 4, calculate the exposure time in seconds that would be needed to deliver the required 40 mJ/cm2, rounded to the nearest hundredth.
- The chamber still holds 65 gallons. Using your item 8 exposure time, calculate the maximum flow in gallons per minute that Cheswick could treat with aged lamps and still meet the permit dose, rounded to the nearest gallon per minute.
- This reactor's validated performance delivers one log of inactivation for every 10 mJ/cm2 of dose, so the item 3 dose of 33.7 mJ/cm2 gives 3.37 logs. If the water entering the reactor carries 22,000 organisms per liter, calculate the number surviving per liter, using survivors equal to the entering count multiplied by 10 raised to the power of negative 3.37. Round to the nearest tenth.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
- Answer: 1,388 gallons per minute.
2.0 MGD x 694 gallons per minute per MGD = 1,388 gallons per minute. - Answer: 2.81 seconds.
1,388 gallons per minute / 60 seconds per minute = 23.13 gallons per second. 65 gallons / 23.13 gallons per second = 2.8098, which rounds to 2.81 seconds in the beam. - Answer: 33.7 mJ/cm2.
12.0 mW/cm2 x 2.81 seconds = 33.72, which rounds to 33.7 mJ/cm2. A milliwatt-second per square centimeter is a millijoule per square centimeter, so the units resolve directly to dose. - Answer: 8.10 mW/cm2.
12.0 x 0.75 = 9.00 mW/cm2 after lamp aging, and 9.00 x 0.90 = 8.10 mW/cm2 after fouling. The two factors multiply because each removes a fraction of what the previous step left. - Answer: 22.8 mJ/cm2.
8.10 mW/cm2 x 2.81 seconds = 22.76, which rounds to 22.8 mJ/cm2. That is well below the 40 mJ/cm2 the permit requires, so an aged, fouled reactor cannot hold the permit dose at this flow. - Answer: 0.0706 per centimeter.
Transmittance of 85.0 percent is 0.850 as a decimal. Absorbance = -log10(0.850) = 0.070581, which rounds to 0.0706 per centimeter. - Answer: 5.32 mW/cm2.
The exponent is -0.0706 x 5 = -0.353, and 10 raised to -0.353 = 0.4436. Then 12.0 x 0.4436 = 5.323, or about 5.32 mW/cm2, so water at the far wall of the chamber receives well under half the intensity found at the sleeve. - Answer: 4.94 seconds.
40 mJ/cm2 / 8.10 mW/cm2 = 4.938 seconds, which rounds to 4.94 seconds of exposure. - Answer: 790 gallons per minute.
65 gallons / 4.938 seconds = 13.1625 gallons per second, and 13.1625 x 60 = 789.75, which rounds to 790 gallons per minute. That is roughly 1.14 MGD, so aged lamps would cut the plant's rated output nearly in half unless the lamps are replaced or another reactor is brought online. - Answer: 9.4 organisms per liter.
10 raised to the power of -3.37 = 0.00042658. 22,000 x 0.00042658 = 9.38, which rounds to 9.4 organisms per liter. Each additional log of inactivation divides the survivors by another factor of ten.
Differentiation
Support
Give students the item 1 and item 2 conversions worked out so they can begin at the dose calculation, and provide a printed reminder that a milliwatt-second per square centimeter equals a millijoule per square centimeter. For items 6 and 7, supply a short table of transmittance values with their absorbances already computed so students practice applying the exponent rather than the logarithm.
Extension
Have students find the transmittance at which the intensity 5 centimeters from the sleeve would fall to half its starting value, then explain what a source-water change, such as a spring runoff event carrying more dissolved organic matter, would do to the plant's ability to hold its dose. Ask them to state whether they would recommend holding the flow to the problem 9 rate or replacing the lamps, and what each choice costs the utility.
ELL supports
Pre-teach dose, intensity, exposure time, transmittance, and fouling with a labeled cross-section of the reactor showing the lamp, the quartz sleeve, and the water path, with arrows for light weakening as it travels. Provide the frame "Dose equals ___ multiplied by ___" and let students state each problem's target quantity aloud before computing it.