Math problem set Grade 10 Math
Pallenford Main Canal: Trapezoid Area and Flow Rate Problems
A grade 10 problem set where students model a fictional irrigation canal as a trapezoid, find its flow with Q = A × V, and convert cubic feet per second to acre-feet.
Objectives
- Students will calculate the cross-sectional area of a trapezoidal channel from its widths and depth.
- Students will apply Q = A × V and convert the result through gallons per minute to acre-feet per day.
- Students will explain why flow falls faster than depth when a trapezoidal canal runs lower.
Vocabulary
- Acre-foot
- The amount of water it takes to cover one acre of land, about the size of a football field, one foot deep. Western states use acre-feet to measure large amounts of water.
Also: trapezoid, cross-sectional area, velocity, cubic feet per second
Teacher overview
The Pallenford Irrigation District and its main canal are fictional, and every dimension below is set for this problem set. Students find the area of the water's trapezoidal cross-section, apply the continuity relationship Q = A × V, and carry the result through the units an irrigation district actually reports. The misconception this set targets is treating the channel as a rectangle by multiplying the top water width by the depth, which gives 56 square feet instead of 40 and overstates the flow by 40 percent. A second frequent error is assuming that lowering the water depth by one quarter lowers the flow by one quarter, which problems 8 through 10 disprove.
Problems
The Pallenford Irrigation District is fictional, and the canal below is used only for this set. The main canal is an earthen channel whose water cross-section is a trapezoid: the bottom is 6 feet wide, the water surface is 14 feet wide, and the water is 4 feet deep. The banks rise 1 foot vertically for each 1 foot horizontally. Water moves at an average velocity of 2.5 feet per second. Use these conversions: area of a trapezoid = (base one + base two) ÷ 2 × height; Q = A × V; 1 cubic foot per second = 448.8 gallons per minute; 1 day = 1,440 minutes; 1 acre-foot = 325,851 gallons.
Find the cross-sectional area of the water in the canal, in square feet.
Using Q = A × V with a velocity of 2.5 feet per second, find the canal flow in cubic feet per second.
Convert the problem 2 flow to gallons per minute.
Convert the problem 3 flow to gallons per day.
Convert the problem 4 volume to acre-feet per day. Round to the nearest tenth.
Seepage through the earthen banks loses 12 percent of the water before it reaches the fields. Find the acre-feet per day actually delivered. Round to the nearest tenth.
Each irrigated acre needs 0.25 inches of water per day, which is 1/48 of an acre-foot per acre. Using the problem 6 delivery, how many acres can the canal serve? Round to the nearest whole acre.
In late summer the district lowers the canal to a 3-foot depth, and the water surface narrows to 12 feet. Find the new cross-sectional area in square feet.
Find the canal flow at the 3-foot depth, in cubic feet per second, with the velocity still at 2.5 feet per second.
By what percent does the flow fall when the depth drops from 4 feet to 3 feet?
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: 40 square feet. Area = (6 + 14) ÷ 2 × 4 = 10 × 4 = 40 square feet. Using the 14-foot top width alone would give 56 square feet, which counts the sloped corners as if they held water.
Answer: 100 cfs. Q = A × V = 40 square feet × 2.5 feet per second = 100 cubic feet per second.
Answer: 44,880 gpm. 100 × 448.8 = 44,880 gallons per minute.
Answer: 64,627,200 gallons per day. 44,880 × 1,440 = 64,627,200 gallons in a day.
Answer: 198.3 acre-feet per day. 64,627,200 ÷ 325,851 = 198.33, which rounds to 198.3 acre-feet per day.
Answer: 174.5 acre-feet per day. Seepage takes 12 percent, so 88 percent arrives: 198.3 × 0.88 = 174.50 acre-feet per day.
Answer: 8,376 acres. Each acre needs 1/48 of an acre-foot per day, so the canal serves 174.5 × 48 = 8,376 acres.
Answer: 27 square feet. Area = (6 + 12) ÷ 2 × 3 = 9 × 3 = 27 square feet.
Answer: 67.5 cfs. Q = 27 × 2.5 = 67.5 cubic feet per second.
Answer: 32.5 percent. The drop is 100 − 67.5 = 32.5 cfs, and 32.5 ÷ 100 = 0.325, or 32.5 percent. The depth fell by 25 percent, yet the flow fell by more, because the sloped banks make the channel narrower as the water drops.
Differentiation
Support
Give students a drawing of the trapezoidal cross-section with the two widths and the depth labeled, and the formula written beneath it with blanks for each value. Work problem 1 together, then let students complete problems 2 through 5 with a printed conversion ladder that lists each multiplier in order.
Extension
Ask students to find the velocity, in feet per second, that the 3-foot-deep canal would need to carry the same flow as the full canal, then say whether that faster water would be likely to erode an earthen bank. A second challenge is to find the depth at which the flow would be exactly half of the full-canal value, using the 1-to-1 bank slope to write the top width in terms of depth.
ELL supports
Provide a card showing the trapezoid with the words bottom width, top width, and depth placed directly on the figure, along with a second card matching each unit to what it measures: square feet for area, feet per second for velocity, and cubic feet per second for flow. Allow students to show problems 3 through 5 as a labeled chain of multiplications rather than sentences.
Extensions
Have students measure the top width, bottom width, and depth of a drainage ditch or a small channel near the school, then repeat problems 1 and 2 using a velocity measured by timing a floating object over a known distance. Do this only with adult supervision, measure from the bank or in a dry or ankle-deep channel, stay out of moving water, skip the activity during or after heavy rain, and have students wash their hands afterward. Comparing their result with the fictional Pallenford canal's 100 cfs shows how much larger an irrigation main canal is than a local ditch.