A channel cross-section shows the shape of a river’s bed and banks at one point. You can use it to find the cross-sectional area, then combine it with velocity to find discharge, the volume of water passing that point each second.
This lesson follows erosion, transport and deposition, because channel shape is the result of those processes. It belongs to rivers and drainage.
How do you read a cross-section?
Look first at the labels. The horizontal axis gives width across the channel, and the vertical gives depth. Depths are measured from the water surface down to the bed at set intervals.
Then work in steps.
- Find the width and the interval between depth readings.
- Find the area by adding up the strips. For equal intervals, add every depth and count the two end readings as half each. Then multiply by the interval.
- Find discharge with Q = A × v.
- Judge the efficiency by comparing area with the wetted perimeter. The ratio of area to wetted perimeter is called the hydraulic radius. A higher value means less friction.
Worked example
All figures are invented for practice. A channel is 6 m wide. Depths were measured every 1 m from the left bank:
0, 0.8, 1.2, 1.4, 1.0, 0.6, 0 (metres).
Step 1, area of each strip (average of two depths, times 1 m):
- 0 to 1 m: (0 + 0.8) ÷ 2 = 0.4 m²
- 1 to 2 m: (0.8 + 1.2) ÷ 2 = 1.0 m²
- 2 to 3 m: (1.2 + 1.4) ÷ 2 = 1.3 m²
- 3 to 4 m: (1.4 + 1.0) ÷ 2 = 1.2 m²
- 4 to 5 m: (1.0 + 0.6) ÷ 2 = 0.8 m²
- 5 to 6 m: (0.6 + 0) ÷ 2 = 0.3 m²
Total area = 0.4 + 1.0 + 1.3 + 1.2 + 0.8 + 0.3 = 5.0 m².
Check by the shortcut: the middle depths add to 0.8 + 1.2 + 1.4 + 1.0 + 0.6 = 5.0, and the end readings are both 0. So the area is 5.0 × 1 = 5.0 m². Both methods agree.
Step 2, discharge: the mean velocity was measured as 0.8 m/s, so Q = 5.0 × 0.8 = 4.0 m³/s.
Step 3, efficiency: the wetted perimeter, measured along the bed from bank to bank, is about 6.7 m. The hydraulic radius is 5.0 ÷ 6.7 ≈ 0.75 m.
Now compare it with a second channel with the same area, 5.0 m², but a wide, shallow shape and a wetted perimeter of 9.0 m. Its hydraulic radius is 5.0 ÷ 9.0 ≈ 0.56 m. The first channel has the higher value, so it loses less energy to friction and is the more efficient shape.
The mistake to watch for
Mistaken working: area = width × deepest depth = 6 × 1.4 = 8.4 m².
The student treated the channel as a rectangle. The bed slopes up at both banks, so the real area is 5.0 m², and 8.4 m² overestimates it by 3.4 m². That mistake carries into discharge: 8.4 × 0.8 = 6.72 m³/s, far above the correct 4.0 m³/s.
The correction is to add strips between readings. A quick sense-check helps: the mean depth is 5.0 ÷ 6 ≈ 0.83 m, which is clearly below the deepest reading of 1.4 m.
Check yourself
All data are invented.
1. A channel is 4 m wide with a mean depth of 0.5 m. Velocity is 0.6 m/s. Find discharge.
Show answer
Area = 4 × 0.5 = 2.0 m². Q = 2.0 × 0.6 = 1.2 m³/s.
2. A river carries 6 m³/s through a section with an area of 3 m². What is the mean velocity?
Show answer
Rearrange Q = A × v to v = Q ÷ A = 6 ÷ 3 = 2 m/s.
3. Channel X has an area of 4.0 m² and a wetted perimeter of 5.0 m. Channel Y has an area of 4.0 m² and a wetted perimeter of 8.0 m. Which is more efficient, and why?
Show answer
X: 4.0 ÷ 5.0 = 0.8 m. Y: 4.0 ÷ 8.0 = 0.5 m. Channel X is more efficient because its hydraulic radius is higher, so less water touches the bed and banks and there is less friction.
Where this leads next
Next, trace a landform sequence to see how channel shape changes from source to mouth. The mixed practice set includes cross-section questions, and the statistics and distribution explorer helps with means.
Calculation questions are easier to trust when someone checks your method with you, which is what a teacher in online one-to-one Geography tuition can do.