Math problem set Grade 11 Math
Torvale Coast RO Blending: Recovery and a TDS Mass Balance
A grade 11 problem set on a fictional brackish RO plant: find recovery, solve a blended-TDS mass balance for the bypass fraction, and estimate daily energy.
Objectives
- Students will calculate reverse-osmosis recovery and product flows from a stated feed flow.
- Students will solve a flow-weighted mass balance for the bypass fraction that meets a TDS target.
- Students will estimate daily treatment energy from a per-thousand-gallon rate.
Vocabulary
- Reverse osmosis
- Reverse osmosis cleans water by pushing it through a membrane with holes so tiny that salt and most other dissolved things cannot fit through. It can even turn seawater into fresh water.
- Total dissolved solids (TDS)
- Total dissolved solids is a measure of everything dissolved in water that you cannot see or filter out, like minerals and salts. Water with very high TDS can taste salty or bitter, even though it might look perfectly clear.
Also: permeate, recovery, mass balance
Teacher overview
Brackish-water reverse-osmosis plants rarely treat every gallon through the membranes. Instead they blend clean, low-salt permeate with a small stream of untreated water, which saves energy while keeping the delivered water below a total-dissolved-solids (TDS) target. This set has grade 11 students work through recovery, a flow-weighted blending mass balance, and an energy estimate for the Torvale Coast Water District, a fictional district. A common student error is to average the two TDS values directly instead of weighting each stream by its flow; the blend depends on how much of each stream is used, not on a plain average of their concentrations.
Problems
The Torvale Coast Water District is a fictional district. Its brackish RO plant uses these values:
- RO feed water TDS: 2,100 mg/L. RO permeate TDS: 100 mg/L.
- District target for delivered water: at most 500 mg/L TDS.
- RO unit feed flow: 2.0 MGD. RO recovery: 75%.
- Energy: this brackish RO plant uses about 3 to 4 kWh per 1,000 gallons.
- The RO unit takes 2.0 MGD of feed water at 75% recovery. How much permeate (MGD) does it make, and how much concentrate?
- Recovery is permeate flow divided by feed flow, written as a percent. Confirm that 1.5 MGD of permeate from 2.0 MGD of feed gives 75% recovery.
- The district blends permeate (100 mg/L) with untreated feed water (2,100 mg/L). For a blend that is fraction f untreated water and (1 minus f) permeate, write the blended-TDS expression, then solve for the fraction f that gives exactly 500 mg/L.
- Using f = 0.20, the plant delivers all 1.5 MGD of permeate as the 80% permeate portion. What total blended flow (MGD) results, and how much untreated bypass water (MGD) is added?
- Verify the blend from problem 4 by a mass balance: add the TDS load from each stream and divide by the total flow. Does it equal 500 mg/L?
- The plant's RO units use about 3 to 4 kWh per 1,000 gallons. Using 3.5 kWh per 1,000 gallons, estimate the daily RO energy for treating the 2.0 MGD feed.
- After a dry spell the feed water TDS rises to 2,600 mg/L while permeate stays at 100 mg/L. Recompute the bypass fraction f that still meets the 500 mg/L target.
- Using your problem 3 and problem 7 answers, explain why saltier feed water forces the plant to bypass less untreated water and lean harder on the RO units.
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: 1.5 MGD permeate and 0.5 MGD concentrate.
Permeate = 0.75 x 2.0 = 1.5 MGD. Concentrate = 2.0 minus 1.5 = 0.5 MGD. - Answer: 75%.
1.5 MGD divided by 2.0 MGD = 0.75, or 75% recovery, which matches the stated value. - Answer: 0.20.
The blended TDS is 100(1 minus f) + 2,100 f. Setting it equal to 500 gives 100 + 2,000 f = 500, so 2,000 f = 400 and f = 0.20, meaning 20% untreated water. - Answer: 1.875 MGD total, with 0.375 MGD bypass.
The 1.5 MGD of permeate is the 80% portion, so total = 1.5 divided by 0.80 = 1.875 MGD. Bypass = 1.875 minus 1.5 = 0.375 MGD. - Answer: 500 mg/L.
Permeate load = 100 x 1.5 = 150; bypass load = 2,100 x 0.375 = 787.5; total load = 937.5. Divide by 1.875 MGD to get 500 mg/L, which matches the target. - Answer: 7,000 kWh/day.
2.0 MGD = 2,000,000 gallons per day; divided by 1,000 gives 2,000 units; 2,000 x 3.5 kWh = 7,000 kWh per day. - Answer: 0.16.
With feed at 2,600 mg/L: 100(1 minus f) + 2,600 f = 500 gives 100 + 2,500 f = 500, so 2,500 f = 400 and f = 0.16, meaning only 16% bypass. - Answer: Saltier feed allows less bypass.
As the untreated water gets saltier, each unit of bypass adds more salt to the blend, so a smaller bypass fraction is needed to stay at 500 mg/L. The plant must then route more water through the membranes, which raises energy use.
Differentiation
Support
Provide the blended-TDS equation already set up for problem 3, with only the final algebra left to complete. Give students a two-stream diagram showing permeate and bypass merging into the delivered flow so the mass balance is concrete.
Extension
Ask students to find the bypass fraction and total delivered flow if the fictional district lowered its target to 350 mg/L at the original feed strength, then explain in one sentence how a stricter target changes the plant's energy use.
ELL supports
Pre-teach "permeate," "bypass," "blend," and "recovery" with a labeled sketch of two pipes joining into one. Provide a sentence frame for problem 8: "When the feed water is saltier, the plant can bypass ___ water because ___."