Quiz & test questions Grade 9 Engineering & Design

Sable Point Water System: Treatment and Hydraulics Test

A grade 9 engineering unit test worth 30 points, mixing multiple-choice, short-answer, and computation items on head, pressure, flow, detention, filters, and removal.

Grade: Grade 9
Subject: Engineering & Design
Material: Quiz & test questions
Time: 30 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: Computation items 3 through 7 require students to carry units through pressure, flow-conversion, detention-time, filter-loading, and removal-efficiency work, choosing consistent units at each step.

All materials for HSN-Q.A.1 →

HSN-Q.A.3 Common Core Math

Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.

How this material aligns: Items 5 and 6 ask students to report detention time and a loading rate at a level of accuracy that matches the plant's design ranges and supports a clear judgment.

All materials for HSN-Q.A.3 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-17.

Objectives

Students will apply head, pressure, and flow conversions to distribution and treatment problems.
Students will calculate detention time, filter loading rate, and removal efficiency and interpret the results.

Vocabulary

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: head, pressure, loading rate, removal efficiency

Teacher overview

This 30-minute unit test (30 points total) checks whether students can apply the core engineering math of the water unit. It uses one fictional system, Sable Point, and mixes two multiple-choice items, five computation items, and one short-answer item. The items cover head and pressure, flow conversions, detention time, filter loading, and removal efficiency.

Sable Point Water System: Treatment and Hydraulics Test — Grade 9 Engineering & Design

Name: ____________________________ Date: ______________

Assessment

The Sable Point regional water system (a fictional system used for this test) serves a small city. Use 1 foot = 0.433 psi and 1 MGD = 694 gpm where needed.

(2 points) One psi of pressure equals how many feet of water head?
- A) 0.433 feet
- B) 2.31 feet
- C) 7.48 feet
- D) 8.34 feet
(2 points) A distribution system usually operates in which pressure range?
- A) 10 to 15 psi
- B) 40 to 80 psi
- C) 100 to 150 psi
- D) 200 to 300 psi
(4 points) A tank's water surface sits 150 feet above a home. Find the water pressure at the home, using 0.433 psi per foot. Round to the nearest whole psi.
(4 points) A plant's design flow is 2.5 MGD. Convert this flow to gallons per minute (gpm).
(4 points) A sedimentation basin holds 150,000 gallons and the plant flow is 1.0 MGD. Find the detention time in hours. (Detention time in days = volume divided by daily flow; multiply by 24 for hours.)
(4 points) A filter has 200 square feet of surface area and receives 1,000 gpm. Find the filter loading rate in gpm per square foot.
(5 points) A plant lowers its influent BOD from 200 mg/L to an effluent of 20 mg/L. Find the removal efficiency as a percent. (Removal efficiency = (influent minus effluent) divided by influent, times 100.)
(5 points) Explain why a neighborhood on a hill above the water tank often needs a booster pump to keep good pressure. Use the idea of head in your answer.

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: B) 2.31 feet.
One psi equals 2.31 feet of head, the standard conversion between pressure and the height of a water column.
Answer: B) 40 to 80 psi.
Distribution systems typically run between 40 and 80 psi, with most states requiring at least 20 psi at every connection.
Answer: 65 psi.
Pressure = 150 feet x 0.433 psi per foot = 64.95, which rounds to 65 psi.
Answer: 1,735 gpm.
2.5 MGD x 694 gpm per MGD = 1,735 gpm.
Answer: 3.6 hours.
Detention in days = 150,000 gallons / 1,000,000 gallons per day = 0.15 day. In hours: 0.15 x 24 = 3.6 hours.
Answer: 5.0 gpm/sq ft.
Loading rate = 1,000 gpm / 200 square feet = 5.0 gpm per square foot, inside the usual 2 to 6 range.
Answer: 90%.
Removal efficiency = (200 - 20) / 200 x 100 = 180 / 200 x 100 = 90 percent.
Answer: The hill adds head the pump must overcome.
Homes on the hill sit above the tank's water surface, so gravity alone cannot give them enough pressure. A booster pump adds the extra head, about 0.433 psi for every foot of rise, needed to keep pressure in the normal range at the tap.

Differentiation

Support

Provide a conversion reference card with the conversion factors used in items 3 and 4, and set up the first computation item (150 x 0.433) so students complete only the final step. Allow a calculator for every computation item.

Extension

Ask students to find the pump run time to refill the 150,000-gallon basin at 1,000 gpm, and to explain how detention time and loading rate would both change if the plant's flow rose by 20 percent.

ELL supports

Provide a labeled diagram of a tank, a hill, and a home for item 8, and give a sentence frame such as "The hilltop home needs a booster pump because the tank sits ___ the home, so there is ___ head." Allow numeric answers to be shown as labeled steps.