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Worksheet Grade 10 Science

Redhawk Intake Channel: Flow, Velocity, and Continuity

A grade 10 hydraulics worksheet at the fictional Redhawk intake, applying Q = A x V to find flow and velocity and converting between cfs, gpm, and MGD.

Objectives

  1. Students will calculate flow rate from channel area and velocity using Q = A x V.
  2. Students will convert a flow rate among cubic feet per second, gallons per minute, and MGD.
  3. Students will rearrange the continuity equation to solve for an unknown 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, velocity, cross-sectional area, intake

Teacher overview

The Redhawk Intake is a fictional structure that draws river water into a treatment plant through an open channel and a round pipe. Operators use the continuity equation, Q = A x V, to relate the flow rate to the channel's cross-sectional area and the water's velocity. This worksheet has students compute flow, convert it among the units a plant reports, and rearrange the equation to solve for velocity. A common error is multiplying width by depth incorrectly or forgetting that a smaller area forces a higher velocity when the flow rate stays the same.

Redhawk Intake Channel: Flow, Velocity, and Continuity — Grade 10 Science
Name: ____________________________ Date: ______________

Student Worksheet

The Redhawk Intake (a fictional structure) has a rectangular channel that is 3 ft wide, with the water 2 ft deep. The water moves through the channel at 1.5 ft/s. Use the continuity equation Q = A x V, where A is the cross-sectional area in square feet and V is the velocity in feet per second, so Q comes out in cubic feet per second (cfs). Conversions: 1 cfs = 448.8 gpm, and 1 MGD = 1.547 cfs.

  1. Find the cross-sectional area A of the channel in square feet, using width 3 ft and depth 2 ft.
  2. Using your area from item 1 and the velocity of 1.5 ft/s, calculate the flow rate Q in cfs.
  3. Convert your flow rate from item 2 into gallons per minute, using 1 cfs = 448.8 gpm. Round to the nearest whole gpm.
  4. Convert your flow rate from item 2 into million gallons per day, using 1 MGD = 1.547 cfs. Round to the nearest hundredth of an MGD.
  5. In summer the plant needs 12 cfs through the same channel, so the area stays 6 sq ft. Rearrange Q = A x V to solve for the velocity needed, in feet per second.
  6. Explain what happens to the water's velocity if debris partly blocks the channel, cutting the area in half while the flow rate stays the same.
  7. A round intake pipe has a diameter of 2 ft. Its area is 0.785 x diameter squared. Calculate the pipe's cross-sectional area in square feet.
  8. To carry the original 9 cfs through that round pipe, what velocity is required? Use V = Q divided by the pipe area from item 7, and round to the nearest hundredth of a foot per second.
  9. Operators keep the velocity within a range rather than as slow as possible. Explain one problem caused by water moving too slowly through an intake.
  10. Compare the velocity in the open channel (1.5 ft/s, used in item 2) with the velocity in the round pipe (item 8). Explain why the pipe needs a higher velocity to carry the same 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: 6 sq ft. Area = width x depth = 3 ft x 2 ft = 6 square feet.
  2. Answer: 9 cfs. Q = A x V = 6 sq ft x 1.5 ft/s = 9 cubic feet per second.
  3. Answer: 4,039 gpm. 9 cfs x 448.8 gpm per cfs = 4,039.2, which rounds to 4,039 gpm.
  4. Answer: 5.82 MGD. 9 cfs divided by 1.547 cfs per MGD = 5.817, which rounds to 5.82 MGD.
  5. Answer: 2 ft/s. Rearranged, V = Q divided by A = 12 cfs divided by 6 sq ft = 2 feet per second.
  6. Answer: The velocity doubles. If the area is halved while Q stays the same, then V = Q divided by A must double to keep the same flow, so the water speeds up.
  7. Answer: 3.14 sq ft. Area = 0.785 x 2 squared = 0.785 x 4 = 3.14 square feet.
  8. Answer: 2.87 ft/s. V = Q divided by A = 9 cfs divided by 3.14 sq ft = 2.866, which rounds to 2.87 feet per second.
  9. Answer: When water moves too slowly, sediment and grit settle out inside the channel instead of staying suspended. That buildup narrows the channel and must be cleaned out, so operators avoid velocities that are too low.
  10. Answer: The pipe's area (3.14 sq ft) is smaller than the channel's area (6 sq ft). To push the same 9 cfs through the smaller area, the water must move faster, so the pipe velocity (about 2.87 ft/s) is higher than the channel velocity (1.5 ft/s).

Differentiation

Support

Give students a labeled diagram of the rectangular channel and the round pipe with the dimensions marked, and provide the rearranged forms V = Q divided by A and A = Q divided by V on a reference strip so they can focus on the arithmetic.

Extension

Ask students to find the diameter a round pipe would need so that the original 9 cfs flows at exactly 2 ft/s. They should rearrange the area formula, solve for the area first, then work back to the diameter, showing each step.

ELL supports

Pre-teach "flow rate," "velocity," "cross-sectional area," and "intake" with a diagram and a short definition for each. Provide a conversion reference card listing "1 cfs = 448.8 gpm" and "1 MGD = 1.547 cfs" so students do not need to recall these from memory.