Worksheet Grade 9 Science

Open-Channel Flow at Grantham: A Grade 9 Hydraulics Worksheet

A grade 9 hydraulics worksheet where students apply Q = A x V to an open channel, then convert flow between cubic feet per second, gpm, and MGD.

Grade: Grade 9
Subject: Science
Material: Worksheet
Time: 40 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: Items 1 through 4 chain an area calculation, the continuity formula, and two flow conversions into one multi-step problem, requiring the consistent unit tracking this standard specifies.

All materials for HSN-Q.A.1 →

HSA-CED.A.1 Common Core Math

Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

How this material aligns: Item 5 rearranges the continuity relationship Q = A x V to solve for the velocity the channel needs at a target flow, the one-variable equation work this standard describes.

All materials for HSA-CED.A.1 →

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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 calculate open-channel flow using the continuity equation Q = A x V.
Students will convert a flow rate between cubic feet per second, gallons per minute, and million gallons per day.
Students will rearrange Q = A x V to solve for a required velocity.

Vocabulary

Flow rate: Flow rate measures how much water moves past a point in a certain amount of time, like gallons per minute through a garden hose. Bigger pipes and stronger pushes give bigger flow rates.

Also: continuity equation, cross-sectional area, velocity, cubic feet per second

Teacher overview

Water entering a treatment plant often flows first through an open channel, a straight concrete trough with water running along the bottom. Operators track how fast that water moves and how much passes each second, because both numbers affect how well grit settles out and how the plant is sized. This worksheet uses the continuity equation, Q = A x V, at the fictional plant in Grantham to connect channel size, water speed, and flow. Watch for the common error of using the water depth alone as the area; the cross-sectional area is the channel width multiplied by the water depth.

Open-Channel Flow at Grantham: A Grade 9 Hydraulics Worksheet — Grade 9 Science

Name: ____________________________ Date: ______________

Student Worksheet

Grantham (a fictional town) runs incoming water through a straight, rectangular open channel at its plant. Use these numbers for all items:

The channel is 3 feet wide, and the water in it is 2 feet deep.
The water moves at a velocity of 1.5 feet per second.
Useful conversions: 1 cubic foot per second (cfs) = 448.8 gallons per minute (gpm), and 1 million gallons per day (MGD) = 694 gpm.
For a round pipe, cross-sectional area (sq ft) = 0.785 x diameter squared, with the diameter in feet.

Find the cross-sectional area of the water in the channel. (Area = width x depth.)
Use the continuity equation Q = A x V to find the flow in cubic feet per second.
Convert your flow from item 2 into gallons per minute.
Convert your gallons-per-minute answer from item 3 into million gallons per day. Round to the nearest hundredth.
During peak demand, operators need 12 cfs through the same channel, and the water depth stays 2 feet. What velocity, in feet per second, does the water need to reach? (Rearrange Q = A x V to V = Q divided by A.)
A round influent pipe at the plant has a diameter of 2 feet. Find its cross-sectional area using Area = 0.785 x diameter squared.
Water moves through that pipe at 3 feet per second. Find the flow through the pipe in cubic feet per second.
The channel (item 2) and the pipe (item 7) carry nearly the same flow, yet the pipe water moves twice as fast. Explain how this fits the equation Q = A x V.

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 sq ft.
Area = width x depth = 3 ft x 2 ft = 6 square feet.
Answer: 9 cfs.
Q = A x V = 6 sq ft x 1.5 ft/s = 9 cubic feet per second.
Answer: 4,039.2 gpm.
9 cfs x 448.8 gpm per cfs = 4,039.2 gallons per minute.
Answer: 5.82 MGD.
4,039.2 gpm divided by 694 gpm per MGD = 5.82 million gallons per day.
Answer: 2 ft/s.
Rearrange to V = Q divided by A = 12 cfs divided by 6 sq ft = 2 feet per second.
Answer: 3.14 sq ft.
Area = 0.785 x diameter squared = 0.785 x (2 ft x 2 ft) = 0.785 x 4 = 3.14 square feet.
Answer: 9.42 cfs.
Q = A x V = 3.14 sq ft x 3 ft/s = 9.42 cubic feet per second.
Answer: The pipe has a smaller area, so its water must move faster to carry the same flow.
In Q = A x V, the flow is the area times the velocity. The pipe's area (3.14 sq ft) is about half the channel's area (6 sq ft), so at nearly equal flow the pipe's velocity is about double the channel's. Area and velocity trade off to keep the flow the same.

Differentiation

Support

Give students a labeled diagram of the channel cross-section, with width and depth marked, so the area step is concrete. Provide the two conversion factors on a printed strip and work item 3 together before students try item 4 alone.

Extension

Ask students to find what flow, in cubic feet per second and in gpm, the channel would carry if the water depth rose to 2.5 feet while the velocity stayed at 1.5 feet per second, and to explain why deeper water raises the flow.

ELL supports

Pre-teach "area," "velocity," and "flow" with a simple sketch: a channel cross-section for area, an arrow for velocity, and the two multiplied for flow. Allow item 8 to be answered with a labeled diagram plus one sentence instead of a full paragraph.