Worksheet Grade 10 Math

Quillfeather RO Plant: Percent Recovery and Flow Conversions

A grade 10 worksheet in which students calculate reverse osmosis percent recovery, reject flow, and gallon-to-MGD conversions for a fictional groundwater plant.

Objectives

  1. Students will calculate percent recovery for a reverse osmosis system from feed and permeate flow rates.
  2. Students will convert flow rates between gallons per minute and million gallons per day.
  3. Students will solve a one-variable equation to find a flow rate that meets a target percent recovery.

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: permeate, reject stream, percent recovery, feed flow

Teacher overview

Quillfeather Regional Water Authority (a fictional utility used for this worksheet) treats brackish groundwater with a reverse osmosis system before sending it into the distribution system. Every RO system splits its incoming feed water into two streams: clean permeate that becomes drinking water, and a concentrated reject stream that carries away the salts the membrane filters out. This worksheet uses one plant's feed and permeate readings to practice percent recovery, unit conversion, and solving for an unknown flow rate.

Answer Key

  1. Answer: 70 percent.
    630 ÷ 900 × 100 = 70%.

  2. Answer: 270 gpm.
    900 gpm − 630 gpm = 270 gpm.

  3. Answer: 1.30 MGD.
    900 gpm ÷ 694 gpm/MGD = 1.2968, which rounds to 1.30 MGD.

  4. Answer: 907,200 gal.
    630 gpm × 1,440 minutes/day = 907,200 gallons per day.

  5. Answer: 675 gpm.
    Let p = the new permeate flow. 0.75 = p ÷ 900, so p = 0.75 × 900 = 675 gpm.

  6. Answer: 45 gpm.
    New reject = 900 − 675 = 225 gpm. Original reject (item 2) was 270 gpm, so the plant sends 270 − 225 = 45 fewer gpm to reject.

  7. Answer: The salts the membrane holds back must leave the system in some of the water.
    Some water must always carry the rejected salts away as concentrate, so a fraction of the feed flow is permanently unavailable as permeate, which is why recovery stays below 100 percent.

  8. Answer: $9,720.
    Reject flow of 270 gpm × 60 min/hr × 20 hr/day = 324,000 gallons per day. 324,000 × $0.03 = $9,720 per day.

  9. Answer: 756 gpm.
    New feed = 900 × 1.20 = 1,080 gpm. New permeate = 1,080 × 0.70 = 756 gpm.

  10. Answer: Reject flow decreases as recovery increases.
    Comparing items 2 and 6 shows that raising recovery from 70% to 75% at the same feed flow lowered the reject flow from 270 gpm to 225 gpm, so a higher percent recovery leaves less water in the reject stream.

Differentiation

Support (below grade level)

Provide the percent recovery formula and the MGD conversion factor (694 gpm per MGD) as a printed reference card for this fictional scenario, and complete item 1 together as a class before students work items 2 through 10 independently.

Extension (above grade level)

Ask students to find the feed flow rate the fictional Quillfeather plant would need, at 70 percent recovery, to produce a permeate flow of 900 gpm, and to explain what that result implies about scaling up the plant.

ELL supports

Provide a labeled diagram of an RO train showing the feed stream splitting into permeate and reject arrows, with the home-language terms for "feed," "permeate," and "reject" alongside the English vocabulary list.

Extensions

Have students use the plant's given flow numbers to show how the daily disposal cost in item 8 would change if the fictional utility negotiated a lower disposal rate, recomputing the cost at a rate of $0.02 per gallon instead of $0.03.