Quiz & test questions Grade 11 Math
Coalport Plant Process Math: A Grade 11 Treatment Unit Test
A 45-minute, 30-point grade 11 unit test mixing multiple-choice, short-answer, and computation items on treatment process math for a fictional plant.
Objectives
- Students will apply the pounds formula, volume, and detention-time formulas to treatment scenarios.
- Students will compute removal efficiency and filter loading rate and judge them against ranges.
- Students will solve the pounds formula for an unknown dose.
Vocabulary
- 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.
Also: pounds formula, removal efficiency, filter loading
Teacher overview
This 45-minute unit test (30 points total) checks whether grade 11 students can carry out the core process math a treatment-plant operator uses every shift. It is built around the Coalport Water Treatment Plant, a fictional plant, and mixes two multiple-choice items, two short-answer items, and five computation items covering the pounds formula, tank volume, detention time, removal efficiency, and filter loading. The items move from recall of a formula to solving that formula for an unknown, so the test samples more than one level of understanding.
Assessment
The Coalport Water Treatment Plant is a fictional plant. Show your work on computation items.
- (2 points) The pounds formula for chemical feed is:
- A) pounds per day = mg/L x MGD x 8.34
- B) pounds per day = mg/L divided by MGD x 8.34
- C) pounds per day = mg/L x MGD divided by 8.34
- D) pounds per day = mg/L x gpm x 8.34
- (4 points) Coalport doses 2.0 mg/L of chlorine at 3.0 MGD. How many pounds of chlorine per day does that require?
- (4 points) A rectangular clearwell is 60 ft long, 30 ft wide, and 16 ft deep. Find its volume in gallons, using 1 cubic foot = 7.48 gallons.
- (4 points) The clearwell holding 215,424 gallons passes 3.0 MGD. Find the detention time in hours.
- (3 points) A basin lowers turbidity from 12 NTU (influent) to 3 NTU (effluent). Its removal efficiency is:
- A) 25%
- B) 40%
- C) 75%
- D) 300%
- (4 points) A filter passes 900 gpm across 300 square feet of surface. Find the filter loading rate in gpm per square foot, and state whether it falls inside the typical 2-to-6 range.
- (3 points) A plant must feed 45 pounds of chlorine per day and runs at 3.0 MGD. Solve the pounds formula for the dose in mg/L this requires.
- (3 points) The chlorine maximum residual disinfectant level is 4.0 mg/L. Explain whether a 1.8 mg/L dose complies, and why an operator still keeps the residual well below that limit.
- (3 points) Convert 3.0 MGD to gallons per minute, using 1 MGD = 694 gpm.
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: A.
The pounds formula multiplies concentration in mg/L by flow in MGD by 8.34 pounds per gallon; the other options misplace a term. - Answer: 50.0 lb/day.
2.0 mg/L x 3.0 MGD x 8.34 = 50.04, which rounds to 50.0 pounds per day. - Answer: 215,424 gal.
60 ft x 30 ft x 16 ft = 28,800 cubic feet; 28,800 x 7.48 = 215,424 gallons. - Answer: 1.7 hr.
215,424 gallons divided by 3,000,000 gallons per day = 0.0718 days; 0.0718 x 24 = 1.72 hours. - Answer: C.
Removal efficiency = (12 minus 3) divided by 12 x 100 = 75%, so C is correct; 300% would require the effluent to be negative. - Answer: 3.0 gpm per sq ft.
900 gpm divided by 300 square feet = 3.0 gpm per square foot, which falls inside the typical 2-to-6 range. - Answer: 1.8 mg/L.
45 = dose x 3.0 x 8.34, so dose = 45 divided by 25.02 = 1.80 mg/L. - Answer: yes, it complies.
The residual left after chlorine demand cannot exceed the 1.8 mg/L applied, so it stays below the 4.0 mg/L maximum residual disinfectant level, which limits the residual in distribution rather than the dose. Operators keep the residual well below the limit to hold down disinfection byproducts and taste-and-odor complaints while still disinfecting. - Answer: 2,082 gpm.
3.0 MGD x 694 gpm per MGD = 2,082 gallons per minute.
Differentiation
Support
Provide a formula reference strip listing the pounds formula, the rectangular-volume formula, and the detention-time formula, along with the conversions 1 cubic foot = 7.48 gallons and 1 MGD = 694 gpm. Allow partial credit for correct setup even when the final arithmetic slips.
Extension
Ask students to write and solve one additional item of their own: choose a flow and a target pounds-per-day feed, then solve for the dose, and explain why the residual that reaches customers, not the dose, is what must stay below the 4.0 mg/L maximum residual disinfectant level.
ELL supports
Provide a labeled diagram of the treatment train (basin, filter, clearwell) with each item's key term marked, and allow short-answer items 6 and 8 to be answered with a labeled calculation plus a short phrase rather than full sentences.