Math problem set Grade 10 Science

Fallowbrook Reclamation: Solids Loading and Digester Detention

Ten linked grade 10 problems that follow solids from a primary clarifier through an anaerobic digester to dewatered cake, using the pounds formula and tank volumes.

Grade: Grade 10
Subject: Science
Material: Math problem set
Time: 50 min
Prep: No prep
Materials: Calculator

Standards

HS-ESS3-4 NGSS

Evaluate or refine a technological solution that reduces impacts of human activities on natural systems.

How this material aligns: Problem 10 has students quantify how much less residual mass leaves the plant after digestion and dewatering, then judge that technology by the hauling it avoids.

All materials for HS-ESS3-4 →

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Problems 2 through 9 chain milligrams per liter, pounds per day, cubic feet, gallons, days, and tons, so a wrong unit choice shows up immediately as an impossible answer.

All materials for HSN-Q.A.1 →

Download Word file

This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will calculate solids removal and solids loading using the pounds formula.
Students will convert a sludge mass at a known percent solids into a daily volume in gallons.
Students will calculate digester volume and detention time and interpret the result.

Vocabulary

Total suspended solids (TSS): Total suspended solids is the amount of solid material floating or suspended in water, measured by filtering a sample and weighing what gets left behind. High TSS makes water look cloudy and can smother fish habitat and clog treatment equipment.
Detention time: Detention time is how long water stays inside a tank or basin before flowing out. Water must stay long enough for each treatment step, like settling or disinfecting, to finish its job.

Also: percent solids, volatile solids, anaerobic digestion, dewatering

Teacher overview

Much of what a wastewater plant removes from the water leaves the site as solids, and tracking that mass through each unit process is the work of solids handling. This set follows one day of solids at the fictional Fallowbrook Water Reclamation Facility, from the suspended solids captured in the primary clarifier through the anaerobic digester and out as dewatered cake. The error to watch for is confusing the pounds of dry solids with the pounds of sludge that carries them: at four percent solids, ninety-six percent of what leaves the clarifier is water, so students who skip that step underestimate the sludge volume by a factor of twenty-five. Percent solids is always a mass ratio, never a volume ratio.

Fallowbrook Reclamation: Solids Loading and Digester Detention — Grade 10 Science

Name: ____________________________ Date: ______________

Problems

Suppose the fictional Fallowbrook Water Reclamation Facility treats 4.2 MGD for a town of 32,000 people. Use these constants throughout: one gallon of water weighs 8.34 pounds, one cubic foot holds 7.48 gallons, a cylindrical tank holds 0.785 times diameter squared times depth in cubic feet, and pounds per day equals concentration in mg/L times flow in MGD times 8.34.

The influent to the primary clarifier carries 240 mg/L of total suspended solids, and the primary effluent carries 96 mg/L. Calculate the percent of the suspended solids the clarifier removes.
The clarifier removes the difference between 240 mg/L and 96 mg/L at a flow of 4.2 MGD. Using the pounds formula, calculate the pounds of dry solids removed per day.
The sludge leaving the clarifier is 4.0 percent solids by mass, and it carries 5,044 pounds of dry solids per day. Calculate the total pounds of sludge that must be pumped each day.
Sludge weighs about the same as water, at 8.34 pounds per gallon. Using a sludge mass of 126,100 pounds per day, calculate the daily sludge volume in gallons.
The anaerobic digester is a cylinder 60 feet in diameter with a liquid depth of 24 feet. Using volume equals 0.785 times diameter squared times depth, calculate the digester volume in cubic feet.
Convert a digester volume of 67,824 cubic feet to gallons, using 7.48 gallons per cubic foot.
The digester holds 507,324 gallons and receives 15,120 gallons of sludge per day. Calculate the detention time in days.
Of the 5,044 pounds per day of dry solids fed to the digester, 75 percent is volatile solids, and the digester destroys 55 percent of those volatile solids. Calculate the pounds of volatile solids destroyed per day.
After digestion, 2,963 pounds per day of dry solids remain. The dewatering press produces a cake that is 22 percent solids by mass. Calculate the tons of wet cake hauled off site each day, using 2,000 pounds per ton.
Compare hauling 126,100 pounds per day of raw sludge with hauling 13,468 pounds per day of dewatered cake. Calculate how many times more mass the raw option would move, then explain in two sentences what digestion and dewatering accomplish for the plant and for the places that receive the residuals.

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: 60 percent.
Use the removal efficiency form: influent minus effluent, divided by influent, times 100. That is 240 minus 96 = 144 mg/L removed, and 144 / 240 = 0.60, so the clarifier removes 60 percent.
Answer: 5,044 lb/day.
Apply the pounds formula to the concentration actually removed: 144 mg/L x 4.2 MGD x 8.34 = 5,044.0 pounds per day. Multiplying the influent concentration instead of the difference gives about 8,407 pounds per day and counts solids that stayed in the water.
Answer: 126,100 lb/day.
Percent solids is dry solids divided by total sludge mass, so total mass equals dry solids divided by the decimal fraction: 5,044 / 0.04 = 126,100 pounds per day. The other 121,056 pounds is water.
Answer: 15,120 gal/day.
Divide mass by the weight of a gallon: 126,100 / 8.34 = 15,119.9, or about 15,120 gallons per day. Students who divide the dry solids instead get 605 gallons, which is far too small to represent a day of primary sludge from a 4.2 MGD plant.
Answer: 67,824 cu ft.
Square the diameter first: 60 x 60 = 3,600. Then 0.785 x 3,600 = 2,826 square feet of cross-section, and 2,826 x 24 = 67,824 cubic feet. Using the radius in place of the diameter is the classic error and gives one quarter of the correct volume.
Answer: 507,324 gallons.
Multiply cubic feet by the conversion: 67,824 x 7.48 = 507,323.5, or about 507,324 gallons.
Answer: 33.6 days.
Divide volume by daily flow: 507,324 gallons / 15,120 gallons per day = 33.55 days, or about 33.6 days. That sits in the usual range for an anaerobic digester, which needs several weeks for the bacteria to work through the volatile solids.
Answer: 2,081 lb/day.
Find the volatile portion first: 5,044 x 0.75 = 3,783 pounds per day of volatile solids. Then apply the destruction fraction: 3,783 x 0.55 = 2,080.65, or about 2,081 pounds per day destroyed. That mass leaves mostly as methane and carbon dioxide.
Answer: 6.7 tons/day.
Divide the remaining dry solids by the cake fraction: 2,963 / 0.22 = 13,468 pounds per day of wet cake. Convert to tons: 13,468 / 2,000 = 6.73, or about 6.7 tons per day.
Answer: 9.4 times.
Divide the two masses: 126,100 / 13,468 = 9.36, so hauling raw sludge would move about 9.4 times the mass. Digestion converts roughly two thousand pounds a day of volatile solids into gas rather than truckloads, and dewatering removes most of the remaining water, so the plant sends out a stable, far smaller load. Fewer truck trips means less fuel and less road wear, and the digested material is more stable when it reaches a farm field or a landfill than raw sludge would be.

Differentiation

Support

Give students a flow diagram of the fictional Fallowbrook plant with boxes for the clarifier, the digester, and the press, and let them write each answer into the box it belongs to so the arc stays visible. Provide a worked twin for problems 3 and 4 using a different plant, such as 2,000 pounds per day of dry solids at 5.0 percent solids, so students practice the division by the decimal fraction before the real numbers arrive.

Extension

Have students calculate what the detention time would become at the fictional Fallowbrook plant if flow rose to 5.6 MGD with the same influent and effluent concentrations, and determine whether the digester would still hold the sludge for at least 20 days. Ask them to state one operating change the plant could make if the answer falls short.

ELL supports

Post a two-column card that pairs each term with its plain-language meaning: dry solids as the material with no water, sludge as the material plus its water, and cake as the material after most water is squeezed out. Provide the formulas as a labeled strip in the order the problems use them, and let students record their answers on the flow diagram before writing full sentences for problem 10.

Extensions

Ask students to prepare a short briefing for the fictional Fallowbrook town council that reports the plant's daily solids numbers and recommends whether to invest in a second dewatering press. The briefing should state the current cake tonnage, the detention time, and what would happen to both if the town grew enough to raise plant flow to 5.6 MGD.