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Worksheet Grade 12 Math

Sunmere Brackish RO Plant: Recovery, Rejection, and Energy Use

A grade 12 worksheet on a fictional brackish desalination plant, chaining flow, salt-balance, and energy calculations for a reverse osmosis system.

Objectives

  1. Students will compute product and concentrate flows from a stated recovery ratio.
  2. Students will use a flow-weighted salt balance to find concentrate concentration.
  3. Students will convert a specific-energy figure into a plant's daily energy demand.

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.

Also: recovery, permeate, concentrate, salt rejection, specific energy

Teacher overview

The fictional Sunmere Brackish Water Plant pushes salty groundwater through reverse osmosis membranes, splitting the feed into clean product water (permeate) and a saltier reject stream (concentrate). This worksheet walks students through the flow split, the salt balance that sets the concentrate strength, and the plant's daily energy demand. A common mistake at this grade is treating recovery as if it applied to the concentrate rather than the product, or forgetting that the salt removed from the permeate does not vanish but concentrates in the reject stream.

Sunmere Brackish RO Plant: Recovery, Rejection, and Energy Use — Grade 12 Math
Name: ____________________________ Date: ______________

Student Worksheet

All figures below describe the fictional Sunmere plant. The membrane system runs a feed flow of 3.0 MGD at a recovery of 75 percent, meaning 75 percent of the feed leaves as product. The feed water carries 2,400 mg/L of dissolved salts (TDS); the product carries 120 mg/L. The plant uses 2.6 kWh of electricity per 1,000 gallons of product, and buys power at $0.11 per kWh. Use the conversion that 1 MGD equals 694 gpm.

  1. Find the product (permeate) flow in MGD. Recovery times feed flow gives product flow.

  2. Find the concentrate (reject) flow in MGD, using the fact that feed equals product plus concentrate.

  3. Convert the product flow from item 1 into gallons per minute, using 1 MGD = 694 gpm. Round to the nearest gpm.

  4. Find the plant's salt rejection as a percent, where rejection (%) = (1 − product TDS ÷ feed TDS) × 100.

  5. The product is 2.25 million gallons per day, or 2,250,000 gallons. Using the plant's 2.6 kWh per 1,000 gallons, find the daily energy used to make the product, in kWh.

  6. Find the plant's daily electricity cost for that product, using the $0.11 per kWh price.

  7. The salt that leaves the product must show up in the concentrate. Using a flow-weighted balance, concentrate TDS = (feed flow × feed TDS − product flow × product TDS) ÷ concentrate flow. Find the concentrate TDS in mg/L.

  8. The operator considers raising recovery from 75 percent to 80 percent at the same 3.0 MGD feed. Find the new product and concentrate flows, and explain in one sentence the tradeoff of running at higher recovery.

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: 2.25 MGD.
    Product = recovery × feed = 0.75 × 3.0 = 2.25 MGD.

  2. Answer: 0.75 MGD.
    Concentrate = feed − product = 3.0 − 2.25 = 0.75 MGD.

  3. Answer: 1562 gpm.
    2.25 MGD × 694 gpm/MGD = 1561.5 gpm, which rounds to 1562 gpm.

  4. Answer: 95 percent.
    Rejection = (1 − 120 ÷ 2400) × 100 = (1 − 0.05) × 100 = 95 percent. The membranes hold back 95 percent of the incoming salt.

  5. Answer: 5,850 kWh.
    2,250,000 gallons ÷ 1,000 = 2,250 thousand-gallon units. 2,250 × 2.6 kWh = 5,850 kWh per day.

  6. Answer: $643.50.
    5,850 kWh × $0.11/kWh = $643.50 per day.

  7. Answer: 9,240 mg/L.
    Concentrate TDS = (3.0 × 2,400 − 2.25 × 120) ÷ 0.75 = (7,200 − 270) ÷ 0.75 = 6,930 ÷ 0.75 = 9,240 mg/L. The reject stream is nearly four times as salty as the feed, which is why brine disposal is a real cost. The pounds formula (mg/L × MGD × 8.34) gives the same answer because the 8.34 factor cancels on both sides of the balance.

  8. Answer: 2.4 MGD product and 0.6 MGD concentrate.
    Product = 0.80 × 3.0 = 2.4 MGD; concentrate = 3.0 − 2.4 = 0.6 MGD. Higher recovery makes more product from the same feed, but it squeezes the leftover salt into a smaller, saltier concentrate stream, which raises the risk of mineral scaling on the membranes and can make the brine harder to discharge. Accept answers that cite scaling, higher pressure or energy, or brine disposal.

Differentiation

Support

Give students a labeled block diagram of the membrane system with feed, permeate, and concentrate arrows before they start, and pre-fill item 1 so the product flow of 2.25 million gallons per day is available for the later steps. Provide the conversion 1 MGD = 694 gpm on a reference card.

Extension

Ask students to compute the specific energy per 1,000 gallons of feed water rather than product water, and to explain why the two figures differ. Then have them estimate how the daily energy cost would change if recovery rose to 80 percent while product energy intensity stayed the same.

ELL supports

Pre-teach "feed," "permeate," "concentrate," "recovery," and "rejection" with the block diagram and a home-language gloss where possible. Provide a worked unit-conversion strip showing million gallons per day converting to gallons per minute, so the arithmetic in item 3 is a matter of following the arrows.