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Worksheet Grade 12 Math

Hollow Creek Desalination Facility: RO Recovery and Rejection Rates

A grade 12 worksheet using a fictional reverse osmosis plant to calculate percent recovery, salt rejection, and concentrate flow from feed and permeate data.

Objectives

  1. Students will calculate percent recovery and salt rejection for a reverse osmosis system from feed, permeate, and concentrate data.
  2. Students will apply a mass-balance equation to find an unknown concentration in a reverse osmosis system.

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: percent recovery, rejection rate, mass balance, concentrate

Teacher overview

Hollow Creek (a fictional desalination facility used for this worksheet) treats seawater with a reverse osmosis (RO) system, a membrane process that separates most dissolved solids from the feed water, producing two output streams: a low-salt permeate and a concentrated reject stream. Two quantities describe how well the system performs: recovery, the percent of the feed water that becomes usable permeate, and rejection, the percent of dissolved solids the membrane keeps out of that permeate. This worksheet uses one feed stream and two performance targets to work through both quantities and the mass-balance relationship that connects them.

A common misconception is assuming that a high rejection rate and a high recovery rate are the same thing, or that improving one automatically improves the other. Rejection describes how clean the permeate is; recovery describes how much of the feed water is converted into that clean permeate. In fact, as problems 7 and 8 show, pushing recovery higher without changing the membrane's rejection performance concentrates the same total amount of dissolved solids into a smaller reject stream, which raises the concentrate's concentration rather than lowering it.

Hollow Creek Desalination Facility: RO Recovery and Rejection Rates — Grade 12 Math
Name: ____________________________ Date: ______________

Student Worksheet

This fictional scenario follows the Hollow Creek desalination facility. The feed seawater enters the system at a flow of 5.0 MGD with a total dissolved solids (TDS) concentration of 35,000 mg/L. The facility's current recovery target is 45 percent, and the permeate leaves the system at a TDS concentration of 250 mg/L. Use these formulas as needed: Recovery (%) = permeate flow ÷ feed flow × 100; Rejection (%) = (feed concentration − permeate concentration) ÷ feed concentration × 100; and the mass-balance equation, feed flow × feed concentration = permeate flow × permeate concentration + concentrate flow × concentrate concentration.

  1. Using the recovery formula and the stated 45 percent recovery target, calculate the permeate flow in MGD.

  2. Calculate the concentrate (reject) flow in MGD, using the fact that concentrate flow equals feed flow minus permeate flow.

  3. Convert your permeate flow from problem 1 to gallons per minute (gpm), using 1 MGD = 694 gpm. Round to the nearest whole gpm.

  4. Calculate the salt rejection percentage using the feed and permeate TDS concentrations given above. Round to the nearest tenth of a percent.

  5. Using the mass-balance equation and your flow values from problems 1 and 2, solve for the concentrate concentration, in mg/L. Round to the nearest whole mg/L.

  6. Explain, in terms of where the dissolved solids that do not pass through the membrane must end up, why the concentrate concentration you found in problem 5 is so much higher than the feed concentration of 35,000 mg/L.

  7. Hollow Creek's engineers propose raising recovery from 45 percent to 50 percent without changing the 5.0 MGD feed flow. Calculate the new permeate flow and new concentrate flow, in MGD, at 50 percent recovery.

  8. Without recalculating the exact concentrate concentration at 50 percent recovery, predict whether it will be higher, lower, or about the same as the 45 percent recovery answer from problem 5. Justify your prediction using the mass-balance idea: the same dissolved solids load is now carried by a smaller concentrate volume.

  9. Hollow Creek's planners are also studying a second feed source that would raise the total feed flow to 6.0 MGD while keeping the 45 percent recovery target. Calculate the new permeate flow, in MGD, at this larger feed flow.

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.
    Permeate flow = 0.45 × 5.0 MGD = 2.25 MGD.

  2. Answer: 2.75 MGD.
    Concentrate flow = 5.0 MGD − 2.25 MGD = 2.75 MGD.

  3. Answer: 1,562 gpm.
    2.25 MGD × 694 gpm/MGD = 1,561.5 gpm, which rounds to 1,562 gpm.

  4. Answer: 99.3 percent.
    Rejection = (35,000 − 250) ÷ 35,000 × 100 = 34,750 ÷ 35,000 × 100 = 99.2857 percent, which rounds to 99.3 percent.

  5. Answer: 63,432 mg/L.
    5.0 × 35,000 = 175,000 (total feed solids in flow-concentration units). 2.25 × 250 = 562.5 (permeate solids). Concentrate solids = 175,000 − 562.5 = 174,437.5. Concentrate concentration = 174,437.5 ÷ 2.75 = 63,431.8 mg/L, which rounds to 63,432 mg/L.

  6. Answer: The membrane blocks almost all of the dissolved solids from passing into the permeate, so nearly all of the solids that entered with the 5.0 MGD feed flow end up carried by the much smaller 2.75 MGD concentrate stream, concentrating them to a level far above the original feed concentration.

  7. Answer: 2.5 MGD permeate and 2.5 MGD concentrate.
    Permeate = 0.50 × 5.0 MGD = 2.5 MGD. Concentrate = 5.0 MGD − 2.5 MGD = 2.5 MGD.

  8. Answer: Higher.
    Raising recovery from 45 percent to 50 percent shrinks the concentrate flow from 2.75 MGD to 2.5 MGD while the total dissolved solids removed from the feed stays about the same, so that similar solids load is packed into a smaller volume, which raises the concentrate concentration above the 63,432 mg/L found in problem 5.

  9. Answer: 2.7 MGD.
    Permeate = 0.45 × 6.0 MGD = 2.7 MGD.

Differentiation

Support (below grade level)

Provide a labeled flow diagram showing the feed stream splitting into the permeate and concentrate streams, with arrows for flow and concentration values, before students begin problem 1. Walk through problem 1 and problem 2 as a class, and give students the mass-balance equation from problem 5 already rearranged to solve for concentrate concentration.

Extension (above grade level)

Ask students to write a single equation for concentrate concentration purely in terms of feed flow, recovery (as a decimal), feed concentration, and permeate concentration, without plugging in Hollow Creek's specific numbers first, then verify that equation reproduces the answer to problem 5 when the Hollow Creek values are substituted in.

ELL supports

Pre-teach "recovery," "rejection," "permeate," and "concentrate" using a simple filter diagram with the four terms labeled directly on the arrows they describe. Allow students to write their mass-balance work as a labeled diagram with numbers plugged into each box before writing the final equation in words.