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Worksheet Grade 10 Science

Aldermere Plant UV Disinfection: Dose and Log Inactivation

A grade 10 science worksheet where students calculate ultraviolet dose from intensity and contact time, then test it against a fictional plant's inactivation targets.

Objectives

  1. Students will calculate ultraviolet dose as intensity multiplied by contact time.
  2. Students will predict how flow rate and water clarity change the delivered UV dose.
  3. Students will compare a calculated dose to a stated inactivation target and state the margin.

Vocabulary

Key words: ultraviolet dose, intensity, contact time, log inactivation, transmittance

Teacher overview

Many drinking water plants disinfect with ultraviolet (UV) light instead of, or alongside, chlorine. A UV reactor works by exposing flowing water to lamps, and the "dose" a microbe receives is the lamp intensity multiplied by how long the water stays in the light. This worksheet uses the fictional Aldermere Regional Water Plant to practice that calculation and to reason about what raises or lowers the dose. The dose numbers here are invented for this scenario, not measured values from a real plant. The inactivation targets are also simplified for practice. Federal rules require much higher UV doses for virus credit, such as 186 mJ/cm2 for 4-log virus inactivation.

Aldermere Plant UV Disinfection: Dose and Log Inactivation — Grade 10 Science
Name: ____________________________ Date: ______________

Student Worksheet

Suppose the fictional Aldermere Regional Water Plant runs one UV reactor. Use these scenario values for every item. UV dose is reported in millijoules per square centimeter (mJ/cm2), and it equals the lamp intensity in milliwatts per square centimeter (mW/cm2) multiplied by the contact time in seconds.

  • Lamp intensity at the design condition: 10 mW/cm2
  • Contact time at the design flow: 6 seconds
  • Target dose for 4-log virus inactivation: 40 mJ/cm2
  • Target dose for 3-log Cryptosporidium inactivation: 12 mJ/cm2
  1. Calculate the UV dose the reactor delivers at the design condition (intensity 10 mW/cm2, contact time 6 seconds).
  2. Does the dose from item 1 meet the 4-log virus target of 40 mJ/cm2? By how many mJ/cm2 does it clear or miss the target?
  3. When the plant's flow doubles, the contact time is cut in half, from 6 seconds to 3 seconds. Calculate the new dose at the same 10 mW/cm2 intensity.
  4. At the doubled flow (your dose from item 3), does the reactor still meet the 4-log virus target of 40 mJ/cm2? By how many mJ/cm2 does it clear or miss it?
  5. Explain in one or two sentences why a higher flow rate through the same reactor lowers the delivered dose.
  6. To restore the 40 mJ/cm2 target at the doubled flow (contact time 3 seconds), what lamp intensity in mW/cm2 would the reactor need? Round to the nearest tenth.
  7. A storm makes the incoming water cloudier, which lowers how much UV passes through it, so the intensity drops to 7 mW/cm2 while the contact time stays at 6 seconds. Calculate the dose and state whether it still meets the 40 mJ/cm2 target.
  8. Explain why cloudy water with more suspended particles makes UV disinfection less effective, even when the lamps are working normally.
  9. The reactor's sensor reads intensity only to the nearest whole mW/cm2. Explain why logging a dose as "60.000 mJ/cm2" claims more precision than that sensor can support.
  10. The plant adds a second identical reactor in series, and each delivers 60 mJ/cm2 at the design condition. What total dose does the water receive after passing through both?

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 mJ/cm2. Dose = intensity x contact time = 10 mW/cm2 x 6 s = 60 mJ/cm2, because one milliwatt applied for one second delivers one millijoule.
  2. Answer: Yes, by 20 mJ/cm2. The delivered 60 mJ/cm2 is above the 40 mJ/cm2 target: 60 - 40 = 20 mJ/cm2 of margin.
  3. Answer: 30 mJ/cm2. Halving the contact time to 3 seconds gives 10 mW/cm2 x 3 s = 30 mJ/cm2.
  4. Answer: No, short by 10 mJ/cm2. The 30 mJ/cm2 dose is below the 40 mJ/cm2 target: 40 - 30 = 10 mJ/cm2 short.
  5. Answer: Higher flow moves water through the reactor faster, so each parcel of water spends less time in the light. Because dose depends on contact time, less time means a smaller dose.
  6. Answer: 13.3 mW/cm2. Intensity = dose / time = 40 mJ/cm2 / 3 s = 13.33, which rounds to 13.3 mW/cm2.
  7. Answer: 42 mJ/cm2, which still meets the target. Dose = 7 mW/cm2 x 6 s = 42 mJ/cm2, and 42 is above 40 by 2 mJ/cm2.
  8. Answer: Suspended particles absorb and scatter UV light and can shield microbes hiding behind or inside them, so less UV energy reaches each organism and the effective dose falls.
  9. Answer: The sensor rounds intensity to a whole number, so the dose is only known to about that same coarseness. Writing three decimal places implies the plant measured to a thousandth of a unit, which the sensor cannot do.
  10. Answer: 120 mJ/cm2. Two reactors in series add their doses: 60 mJ/cm2 + 60 mJ/cm2 = 120 mJ/cm2.

Differentiation

Support

Give students a filled-in first example (intensity 5 mW/cm2 for 4 seconds equals 20 mJ/cm2) and a labeled unit strip showing that milliwatts times seconds equals millijoules. Allow a calculator for every item.

Extension

Ask students to build a small table showing the dose at flows that give contact times of 6, 4, 3, and 2 seconds at a fixed 10 mW/cm2, then describe in words the relationship between contact time and dose.

ELL supports

Pre-teach "intensity," "contact time," and "dose" with a simple diagram of water passing lamps, labeled in English and, where possible, students' home languages. Provide the sentence frame "The dose is ___ because the intensity is ___ and the time is ___" for the explanation items.