Math problem set Grade 11 Math
Ashgrove Storage Tank: Modeling Chlorine Residual Decay
A grade 11 problem set contrasting linear and exponential decay of a fictional tank's chlorine residual, building a half-life model and comparing summer and winter.
Objectives
- Students will distinguish linear from exponential decay of a chlorine residual.
- Students will construct an exponential half-life model and evaluate it over time.
- Students will use the model to set a re-chlorination schedule against a residual minimum.
Vocabulary
- Chlorine residual
- Chlorine residual is the small amount of chlorine still left in the water after it leaves the treatment plant. It keeps protecting the water from germs all the way to your faucet.
Also: half-life, exponential decay, re-chlorination
Teacher overview
Chlorine residual does not hold steady once treated water sits in a storage tank; it fades as the chlorine reacts with organic matter and pipe walls. That fade is closer to exponential decay than to a steady loss, which changes how often an operator must add more chlorine. This set has grade 11 students build and read a half-life model for a tank at Ashgrove, a fictional town, and compare it against a simple linear model. A common student error is to treat the loss as a fixed amount per hour, a linear picture, when the residual actually loses a fixed percent per hour, so students overestimate how much chlorine remains after long holding times.
Problems
Ashgrove is a fictional town. Water enters its storage tank with a free chlorine residual of 2.0 mg/L. Use these facts:
- The operator wants the residual to stay at or above 0.5 mg/L in the tank.
- In a simple linear model, the residual would drop by a constant 0.2 mg/L every 12 hours.
- In the exponential model, the residual halves every 12 hours (a 12-hour half-life).
- Build the linear model first. Starting at 2.0 mg/L and dropping 0.2 mg/L every 12 hours, list the residual at 0, 12, 24, and 36 hours.
- Now build the exponential model, C(t) = 2.0 x (0.5) raised to the power (t divided by 12), where t is in hours. Find the residual at 12, 24, 36, and 48 hours.
- The operator's minimum is 0.5 mg/L. Using your exponential values, at what time does the residual reach 0.5 mg/L?
- Compare the linear and exponential models at 36 hours. Which predicts a lower residual, and by how much?
- The operator re-chlorinates the tank back to 2.0 mg/L each time it nears the minimum. Based on the exponential model, how often, in hours, must the tank be re-chlorinated to stay at or above 0.5 mg/L?
- Between 12 and 24 hours the exponential residual drops from 1.0 to 0.5 mg/L; between 24 and 36 hours it drops from 0.5 to 0.25 mg/L. Explain why the second 12-hour interval loses less chlorine than the first.
- In warmer summer water the residual halves every 8 hours instead of 12. Write the new model and find the residual at 24 hours. Compare it with the winter residual at 24 hours.
- The maximum residual disinfectant level is 4.0 mg/L. When the operator re-chlorinates to 2.0 mg/L, is that within the limit, and by what margin?
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: 2.0, 1.8, 1.6, 1.4 mg/L.
Starting at 2.0 and subtracting 0.2 each 12 hours gives 2.0 at 0 h, 1.8 at 12 h, 1.6 at 24 h, and 1.4 at 36 h. - Answer: 1.0, 0.5, 0.25, 0.125 mg/L.
C(12) = 2.0 x 0.5 = 1.0; C(24) = 2.0 x 0.25 = 0.5; C(36) = 2.0 x 0.125 = 0.25; C(48) = 2.0 x 0.0625 = 0.125. - Answer: 24 hr.
The exponential model reaches 0.5 mg/L at t = 24 hours, when the residual has halved twice. - Answer: exponential is lower by 1.15 mg/L.
At 36 hours the linear model gives 1.4 mg/L and the exponential model gives 0.25 mg/L; 1.4 minus 0.25 = 1.15 mg/L lower. - Answer: every 24 hr.
From 2.0 mg/L the exponential model reaches the 0.5 mg/L minimum at 24 hours, so the tank must be re-chlorinated about every 24 hours. - Answer: Each interval loses half of a smaller amount.
Exponential decay removes the same fraction (one half) each 12 hours, so the first interval halves 1.0 to give a 0.5 loss, while the second halves 0.5 to give only a 0.25 loss. - Answer: 0.25 mg/L at 24 hr.
The summer model is C(t) = 2.0 x (0.5) raised to the power (t divided by 8). At 24 hours, t divided by 8 = 3, so C = 2.0 x 0.125 = 0.25 mg/L, half the winter value of 0.5 mg/L at 24 hours. - Answer: yes, by 2.0 mg/L.
Re-chlorinating to 2.0 mg/L stays below the 4.0 mg/L maximum residual disinfectant level, with a 2.0 mg/L margin.
Differentiation
Support
Give students a blank two-row table for problems 1 and 2 with the time headers already filled in, and remind them that (0.5) raised to a whole-number power just means halving that many times. Work t = 12 and t = 24 together before students continue.
Extension
Ask students to estimate, using their tables, the time at which the winter exponential residual first drops below the 0.2 mg/L bottom of the usual distribution range, and to explain why a graph of the residual never actually reaches zero.
ELL supports
Pre-teach "residual," "half-life," and "decay" with a quick demonstration of repeatedly halving a strip of paper. Provide a sentence frame for problem 6: "The residual drops less in the second interval because ___."