Math problem set Grade 10 Math
Nyefield Blending Plan: A Two-Source System of Equations
A grade 10 problem set where students solve a two-variable system to blend purified recycled water with hard groundwater for a fictional district in drought.
Objectives
- Students will write a volume equation and a mass balance equation for a two-source blend.
- Students will solve a two-variable linear system to find each source's required flow.
- Students will recompute a blend when a supply cap or a source quality changes.
Vocabulary
- Potable reuse
- Cleaning used water so thoroughly that it becomes safe drinking water again. Special plants use several layers of treatment and constant testing to make sure it is safe.
Also: blend, mass balance, hardness, system of equations
Teacher overview
The Nyefield Water Authority is fictional, and its flows, hardness values, and targets are set for this problem set. During a drought the authority stretches its supply by blending purified recycled water, which is very soft, with groundwater, which is hard, so the mixed water meets a hardness target while still delivering the full summer demand. The misconception this set is built to correct is averaging the two hardness values instead of weighting each one by its flow; the simple average is correct only in the rare case when both sources contribute the same flow, as problem 5 shows. Students also tend to forget that the mass balance equation multiplies each concentration by that source's flow, not by the total flow.
Problems
The Nyefield Water Authority is a fictional utility serving a town of 40,000 residents. In summer the system must deliver 6.0 million gallons per day (MGD). Purified recycled water leaves the reuse plant with a hardness of 20 mg/L. Groundwater from the district's wells has a hardness of 260 mg/L. The blended water entering the distribution system must have a hardness of 120 mg/L or less. Let r stand for the recycled flow in MGD and g for the groundwater flow in MGD. Use 1 MGD = 694 gallons per minute where it is needed.
Write two equations in r and g: one for the total flow of 6.0 MGD, and one hardness balance stating that 20r + 260g equals the hardness target times the total flow.
Solve your system for g, the groundwater flow in MGD.
Find r, the recycled flow in MGD.
Convert the recycled flow from problem 3 to gallons per minute, using 694 gallons per minute per MGD.
The reuse plant can produce at most 3.0 MGD. Suppose the authority runs it at 3.0 MGD and takes the other 3.0 MGD from the wells. Find the hardness of that blend in mg/L.
Keeping the reuse plant at its 3.0 MGD cap, find the groundwater flow g, in MGD, that would hold the blend at exactly 120 mg/L. Round to the nearest hundredth.
Find the total blended flow, in MGD, for the problem 6 answer. Round to the nearest hundredth.
The system still needs 6.0 MGD. Find the shortfall from problem 7 in gallons per day.
The authority asks its 40,000 residents to close the problem 8 shortfall through conservation. How many gallons per person per day is that? Round to the nearest tenth.
Suppose the reuse plant's hardness rises from 20 mg/L to 40 mg/L. With the plant uncapped again and the full 6.0 MGD target, find the new groundwater flow g in MGD. Round to the nearest hundredth.
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: r + g = 6.0 and 20r + 260g = 720. The hardness target times the total flow gives 120 × 6.0 = 720, so the mass balance reads 20r + 260g = 720.
Answer: 2.5 MGD. Substitute r = 6.0 − g into the balance: 20(6.0 − g) + 260g = 720, so 120 − 20g + 260g = 720, then 240g = 600 and g = 2.5 MGD.
Answer: 3.5 MGD. From r = 6.0 − g, r = 6.0 − 2.5 = 3.5 MGD. Check: 20(3.5) + 260(2.5) = 70 + 650 = 720, and 720 ÷ 6.0 = 120 mg/L.
Answer: 2,429 gpm. 3.5 MGD × 694 gallons per minute per MGD = 2,429 gallons per minute.
Answer: 140 mg/L. The hardness load is 20(3.0) + 260(3.0) = 60 + 780 = 840, and 840 ÷ 6.0 = 140 mg/L. This blend misses the 120 mg/L target. The simple average of 20 and 260 also gives 140 here, and only because the two flows happen to be equal.
Answer: 2.14 MGD. Set (20 × 3.0 + 260g) ÷ (3.0 + g) = 120, so 60 + 260g = 360 + 120g. Then 140g = 300 and g = 2.142857, which rounds to 2.14 MGD.
Answer: 5.14 MGD. The blend is 3.0 + 2.14 = 5.14 MGD.
Answer: 860,000 gallons per day. The shortfall is 6.0 − 5.14 = 0.86 MGD, and 0.86 × 1,000,000 = 860,000 gallons per day.
Answer: 21.5 gallons per person per day. 860,000 ÷ 40,000 residents = 21.5 gallons per person each day.
Answer: 2.18 MGD. The balance becomes 40r + 260g = 720 with r = 6.0 − g, so 240 − 40g + 260g = 720. Then 220g = 480 and g = 2.1818, which rounds to 2.18 MGD. Softer recycled water would have allowed more groundwater, so the rise in reuse-plant hardness cuts the groundwater share.
Differentiation
Support
Provide the two equations from problem 1 already written, and walk students through the substitution step for problem 2 on the board before they work independently. Give a filled-in blending diagram showing two arrows entering a mixing box with flow above each arrow and hardness below it, so students can see which numbers multiply.
Extension
Ask students to graph both equations from problem 1 on the same axes with r on the horizontal axis, label the intersection point, and explain what the coordinates of that point mean for the fictional Nyefield district. A second challenge is to find the largest hardness the reuse plant could have and still allow the district to meet both the 6.0 MGD demand and the 120 mg/L target.
ELL supports
Give students a sentence frame for problem 1 that reads "flow one plus flow two equals total flow" above the algebraic version, so the equation and the words line up term by term. Provide a card defining hardness, blend, and shortfall with one short sentence each, and allow answers to problems 5 and 8 to be shown as labeled arithmetic rather than sentences.
Extensions
Have students find their own water utility's most recent consumer confidence report, locate the hardness value if the report lists one, and compare it with the 120 mg/L target used for the fictional Nyefield blend. Students can then state, in a short paragraph, whether their own supply would need blending under Nyefield's rule.