Worksheet Grade 11 Careers & CTE
Shalecombe Creek Stream Gauging: A Field Technician Worksheet
A grade 11 careers worksheet in which a watershed technician gauges a creek by the mid-section method, totals discharge, and converts it to gpm, MGD, and a daily volume.
Objectives
- Students will compute subsection areas and discharges from measured widths, depths, and velocities.
- Students will total a stream cross-section's discharge and convert it to gallons per minute and million gallons per day.
- Students will explain why a technician measures velocity at several points across a channel rather than at one.
Vocabulary
Key words: discharge, mid-section method, cross-sectional area, mean velocity, stream gauging
Teacher overview
Measuring how much water a stream carries is a field job with a tape, a wading rod, and a current meter, and the arithmetic that follows is what turns a morning of measurements into a number a watershed council can act on. This worksheet uses the fictional Shalecombe Creek, where a technician gauges the channel before and after a restoration project that widened it. Students compute each subsection's area and discharge, total them, and then convert the result into the units a utility, a regulator, and a landowner each expect. The misconception this worksheet corrects is the belief that a wider or deeper channel automatically carries more water; discharge is area times velocity, so a channel that widens and slows can carry exactly what it carried before.
Student Worksheet
A technician for the fictional Shalecombe Creek watershed council gauges the creek by the mid-section method. The channel is divided into five subsections, each 2.0 feet wide. Depth and velocity are measured at the middle of each. Each subsection's area is its width times its depth, and each subsection's discharge is its area times its velocity. Use these conversions: one cubic foot per second equals 448.8 gallons per minute, and one million gallons per day equals 1.547 cubic feet per second.
| Subsection | Width (ft) | Depth (ft) | Velocity (ft/s) |
|---|---|---|---|
| S1 | 2.0 | 0.6 | 0.8 |
| S2 | 2.0 | 1.1 | 1.5 |
| S3 | 2.0 | 1.4 | 1.9 |
| S4 | 2.0 | 1.0 | 1.4 |
| S5 | 2.0 | 0.5 | 0.7 |
- Find the cross-sectional area of subsection S3, in square feet.
- Find the discharge of subsection S3, in cubic feet per second.
- Find the discharge of subsection S1, in cubic feet per second.
- Find the total cross-sectional area of the channel by adding all five subsection areas, in square feet.
- Find the creek's total discharge by adding all five subsection discharges, in cubic feet per second.
- Find the mean velocity across the whole cross-section by dividing your answer from item 5 by your answer from item 4. Round to the nearest hundredth of a foot per second.
- Convert the total discharge from item 5 to gallons per minute. Round to the nearest tenth.
- Convert the total discharge from item 5 to million gallons per day. Round to the nearest hundredth.
- Using your answer from item 7, find the volume the creek carries in one full day, in gallons. Round to the nearest gallon.
- Two years after a restoration project widened the channel, the technician gauges the same reach at a similar dry-weather base flow and measures a total cross-sectional area of 12.46 square feet and a mean velocity of 1.05 feet per second. Find the new discharge, in cubic feet per second, rounded to the nearest hundredth.
- Compare your answers from items 5 and 10, and explain in two or three sentences how the channel can carry the same discharge at a lower velocity.
- Explain in two or three sentences why the technician measures velocity at the middle of every subsection rather than once in the middle of the creek.
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: 2.8 square feet.
Area equals width times depth: 2.0 feet times 1.4 feet equals 2.8 square feet. - Answer: 5.32 cfs.
Discharge equals area times velocity: 2.8 square feet times 1.9 feet per second equals 5.32 cubic feet per second. - Answer: 0.96 cfs.
S1's area is 2.0 times 0.6, or 1.2 square feet, and 1.2 times 0.8 feet per second equals 0.96 cubic feet per second. - Answer: 9.2 square feet.
The five areas are 1.2, 2.2, 2.8, 2.0, and 1.0 square feet. Their sum is 9.2 square feet. - Answer: 13.08 cfs.
The five discharges are 0.96, 3.30, 5.32, 2.80, and 0.70 cubic feet per second. Their sum is 13.08 cubic feet per second. - Answer: 1.42 ft/s.
13.08 cubic feet per second divided by 9.2 square feet equals 1.42174, which rounds to 1.42 feet per second. This is lower than S3's measured 1.9 feet per second because the shallow edges pull the average down. - Answer: 5,870.3 gpm.
13.08 cubic feet per second times 448.8 gallons per minute per cubic foot per second equals 5,870.304, which rounds to 5,870.3 gallons per minute. - Answer: 8.46 MGD.
13.08 cubic feet per second divided by 1.547 cubic feet per second per million gallons per day equals 8.4551, which rounds to 8.46 million gallons per day. - Answer: 8,453,238 gallons.
5,870.304 gallons per minute times 1,440 minutes per day equals 8,453,237.76 gallons, which rounds to 8,453,238 gallons. This agrees with the 8.46 million gallons per day found in item 8. - Answer: 13.08 cfs.
Discharge equals area times mean velocity: 12.46 square feet times 1.05 feet per second equals 13.083, which rounds to 13.08 cubic feet per second. - Answer: The discharge is the same because the channel gained about as much area as it lost in velocity.
Discharge is the product of area and velocity, so the two can trade against each other. The restored reach carries 12.46 square feet of water instead of 9.2, an increase of about 35 percent, while mean velocity fell from 1.42 to 1.05 feet per second, a decrease of about 26 percent. Multiplying the larger area by the smaller velocity returns the same 13.08 cubic feet per second, which is what the technician expected when gauging at a similar base flow, since a restoration project reshapes a channel rather than adding water to the watershed. - Answer: Velocity is not uniform across a channel, so one reading in the middle would overstate the whole.
Friction against the banks and the streambed slows the water at the edges and near the bottom, which is why S1 and S5 read 0.8 and 0.7 feet per second while S3 reads 1.9. Using the middle reading for the whole cross-section would have given 9.2 times 1.9, or 17.48 cubic feet per second, which overstates the true discharge by about a third. Measuring each subsection separately keeps each velocity paired with the area it actually applies to.
Differentiation
Support
Provide the five subsection areas for the fictional creek already computed so students can begin the discharge column in items 2 and 3 without stalling, and give a printed conversion card listing the two flow conversions used on the worksheet.
Extension
Ask students to find what mean velocity the restored reach of the fictional creek would need in order to carry twice the discharge measured in item 5 at the same 12.46 square feet of area, then judge whether that velocity is plausible for a restored channel with a gravel bed.
ELL supports
Pre-teach "discharge," "cross-section," "subsection," and "mean velocity" with a labeled front-view sketch of the fictional creek showing the five vertical strips, the wading rod, and the current meter. Provide the sentence frame "Discharge stayed the same because the area went ___ while the velocity went ___" for item 11.
Extensions
Have students sketch the fictional creek's cross-section to scale on graph paper from the width and depth table, shade the wetted area, and compare their drawn area to the 9.2 square feet computed in item 4, explaining any difference.