Math problem set Grade 9 Engineering & Design

Brindlecote Intake Screen: Sizing for Through-Screen Velocity

A grade 9 set where students convert a fictional town's river intake flow from MGD to cfs, then size a cylindrical wedge-wire screen to hold through-screen velocity down.

Grade: Grade 9
Subject: Engineering & Design
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

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 1 through 3 chain MGD to cfs to gpm and then divide a flow by a velocity to produce an area, with units checked at every step.

All materials for HSN-Q.A.1 →

HSG-MG.A.1 Common Core Math

Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder).

How this material aligns: Problem 5 models the intake screen as a cylinder and uses the curved surface formula to convert a required face area into a screen length.

All materials for HSG-MG.A.1 →

HS-ETS1-3 NGSS

Evaluate a solution to a complex real-world problem based on prioritized criteria and trade-offs that account for a range of constraints, including cost, safety, reliability, and aesthetics as well as possible social, cultural, and environmental impacts.

How this material aligns: Problem 6 tests the leaf-fouled screen against the fisheries permit's 0.5 foot per second limit, problems 7 and 8 evaluate the proposed 7.0 MGD expansion against that same limit, and problem 9 measures the withdrawal against the river's seven-day low flow.

All materials for HS-ETS1-3 →

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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 convert a design flow among MGD, cfs, and gpm using stated conversion factors.
Students will apply the continuity relationship Q equals A times V to size a screen face for a velocity limit.
Students will model an intake screen as a cylinder and solve for the length that supplies a required area.

Vocabulary

Key words: through-screen velocity, open area, continuity, intake screen, low flow

Teacher overview

Brindlecote is a fictional town that pulls its raw water from the Vachell River through a screened intake sitting in the channel. The screen keeps debris and small fish out of the pump station, and the speed of the water passing through the screen openings, called through-screen velocity, is what decides whether a small fish drawn against the screen can swim away from it. This set has students convert the design flow into cubic feet per second, use the continuity relationship to find the open screen area that keeps through-screen velocity at the permitted limit, and then model the screen as a cylinder to find its length. The error to watch for is treating the whole screen face as open water when only part of the face is open, since the wire itself blocks the rest.

Brindlecote Intake Screen: Sizing for Through-Screen Velocity — Grade 9 Engineering & Design

Name: ____________________________ Date: ______________

Problems

The town of Brindlecote, the Vachell River, and every figure below are fictional. Use only what is given here.

Brindlecote's intake is designed for 4.5 MGD. The fisheries permit sets a maximum through-screen velocity (the speed of water passing through the screen openings) of 0.5 feet per second. The wedge-wire screen is 40 percent open area; the wire blocks the rest. The screen is a cylinder 3.0 feet in diameter, and water enters only through its curved surface. Use 1 MGD = 1.547 cfs, 1 MGD = 694 gpm, and 3.14 for pi.

Convert the design flow of 4.5 MGD to cubic feet per second. Round to the nearest hundredth.
Convert the same design flow of 4.5 MGD to gallons per minute.
The continuity relationship is Q = A x V. Using your flow in cubic feet per second from problem 1 and the velocity limit of 0.5 feet per second, find the open screen area in square feet. Round to the nearest hundredth.
Only 40 percent of the screen face is open. Using your answer from problem 3, find the total face area the screen must have. Round to the nearest hundredth.
The screen is a cylinder 3.0 feet in diameter, and water enters only through the curved surface. That surface area equals 3.14 x diameter x length. Using your answer from problem 4, find the screen length in feet. Round to the nearest hundredth.
Leaves pack the screen and block half of its open area, leaving 6.96 square feet open. The intake is still drawing 6.96 cubic feet per second. What is the through-screen velocity now, in feet per second?
Brindlecote may expand the intake to 7.0 MGD later. At the same 0.5 feet per second limit, what open screen area would that flow require? Round to the nearest hundredth.
Compare the open area for 7.0 MGD from problem 7 with the open area for 4.5 MGD from problem 3. By what percent does the required open area increase? Round to the nearest tenth of a percent.
In a dry September the Vachell River drops to a seven-day low flow of 42 cubic feet per second. At the original design flow from problem 1, what percent of the river's low flow does the intake withdraw? Round to the nearest tenth of a percent.
A classmate says that doubling the screen diameter would double its open area if the length stays the same. Using the curved surface formula from problem 5, state whether that claim is correct and explain why in one or two sentences.

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: 6.96 cfs.
4.5 MGD x 1.547 cfs per MGD = 6.9615 cfs, which rounds to 6.96 cfs.
Answer: 3,123 gpm.
4.5 MGD x 694 gpm per MGD = 3,123 gpm.
Answer: 13.92 square feet.
Rearranging Q = A x V gives A = Q divided by V. A = 6.9615 cfs divided by 0.5 feet per second = 13.923 square feet, which rounds to 13.92 square feet. Note that cubic feet per second divided by feet per second leaves square feet.
Answer: 34.81 square feet.
The open area is 40 percent of the face, so the face is the open area divided by 0.40. 13.923 divided by 0.40 = 34.8075 square feet, which rounds to 34.81 square feet. Accept 34.80 from a student who carried the rounded 13.92 forward.
Answer: 3.70 feet.
The curved surface area is 3.14 x 3.0 feet x length = 9.42 x length. Setting 9.42 x length = 34.8075 gives length = 3.695 feet, which rounds to 3.70 feet. Accept 3.69 from a student who carried 34.80 forward.
Answer: 1.0 feet per second.
V = Q divided by A = 6.96 cubic feet per second divided by 6.96 square feet = 1.0 feet per second. That is twice the permitted limit, which is why intake screens carry brushes or air bursts to clear debris.
Answer: 21.66 square feet.
7.0 MGD x 1.547 = 10.829 cfs. Then 10.829 divided by 0.5 = 21.658 square feet, which rounds to 21.66 square feet.
Answer: 55.6 percent.
The increase is 21.658 minus 13.923 = 7.735 square feet. Then 7.735 divided by 13.923 = 0.5555, or 55.6 percent. The flow also rose 55.6 percent, because at a fixed velocity the area scales directly with the flow.
Answer: 16.6 percent.
6.9615 cfs divided by 42 cfs = 0.1658, which is 16.6 percent of the river's seven-day low flow.
Answer: Yes, the open area doubles.
The curved surface area is 3.14 x diameter x length, so with the length held fixed the area is directly proportional to the diameter. Doubling the diameter doubles the area. Accept answers that make this proportionality clear. Students who answer that the area quadruples are applying the rule for a circle's area, where the radius is squared, to a surface where it is not.

Differentiation

Support

Give these students a conversion strip listing 1 MGD = 1.547 cfs and 1 MGD = 694 gpm, plus a labeled sketch of the fictional Brindlecote intake showing the cylinder's diameter, its length, and an arrow marked through-screen velocity. Provide problem 3 with the rearranged formula A = Q divided by V already written, and allow them to stop after problem 6.

Extension

Ask students to size the screen for a site where the fisheries permit is stricter, allowing only 0.33 feet per second through the screen, and to report the new length for the same 3.0-foot cylinder in this fictional scenario. Then have them explain why a utility might install two shorter screens instead of one long one, naming at least one maintenance reason and one hydraulic reason.

ELL supports

Pre-teach through-screen velocity, open area, and low flow using a drawing of a screen face with the wire shaded and the open slots left white. Provide the sentence frame "The open area is ___ square feet because the flow is ___ and the velocity limit is ___" for problem 3. Allow problem 10 to be answered with a labeled sketch and one sentence.

Extensions

Have students plot required open area against design flow for the fictional Brindlecote intake at 2.0, 4.5, 7.0, and 9.0 MGD, holding the velocity limit at 0.5 feet per second. Ask them to describe the graph's shape and to explain what the slope means in units of square feet per MGD. Students can then extend the line beyond 9.0 MGD and mark the flow at which the intake would need more open area than a single 3.0-foot cylinder ten feet long could provide.