---
title: "Discharge from depth and mean velocity data"
subject: "Hydro Power"
chapter: 2
key: "semester_6/hydro_power:discharge-depth-mean-velocity"
source: https://www.bctengineer.com/q/semester_6/hydro_power/discharge-depth-mean-velocity
site: BCT Engineering
---

# Discharge from depth and mean velocity data

Hydro Power · Chapter 2: Introduction to Hydrology

- Asked in: 2073 Bhadra, 8 marks
- Video: [Streamflow Measurement | Discharge Measurement | Area Velocity Method](https://www.youtube.com/watch?v=URVjuEDUUgk) — IntuitiveLearn
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/discharge-depth-mean-velocity
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-2

## Question

At known distances from an initial point on the stream bank, the measured depth and velocity of a river are shown in table. Calculate the corresponding discharge at this location.

| Distance from initial point (m) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 99 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Depth (m) | 0 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 3.0 | 2.0 | 1.5 | 0 |
| Mean velocity (m/s) | 0 | 0.6 | 0.8 | 1.0 | 1.2 | 1.4 | 1.6 | 1.3 | 1.0 | 0.5 | 0 |

## Short answer

Mid-section method with strip widths determined from adjacent distances. Multiplying each strip width by depth and given mean velocity, then summing gives total discharge ≈ 232 m³/s.

## Answer

Mean velocities are given directly, so the **mid-section method** is applied: each vertical represents a strip from halfway to the previous vertical to halfway to the next, $w_i = (x_{i+1} - x_{i-1})/2$, and $\Delta Q_i = w_i d_i V_i$.

#### Computation
| x (m) | Width (m) | Depth (m) | V (m/s) | Area (m²) | ΔQ (m³/s) |
|---|---|---|---|---|---|
| 10 | 10 | 1.0 | 0.6 | 10.00 | 6.000 |
| 20 | 10 | 1.5 | 0.8 | 15.00 | 12.000 |
| 30 | 10 | 2.0 | 1.0 | 20.00 | 20.000 |
| 40 | 10 | 2.5 | 1.2 | 25.00 | 30.000 |
| 50 | 10 | 3.0 | 1.4 | 30.00 | 42.000 |
| 60 | 10 | 3.5 | 1.6 | 35.00 | 56.000 |
| 70 | 10 | 3.0 | 1.3 | 30.00 | 39.000 |
| 80 | 10 | 2.0 | 1.0 | 20.00 | 20.000 |
| 90 | (99 − 80)/2 = 9.5 | 1.5 | 0.5 | 14.25 | 7.125 |
| **Total** | | | | **199.25** | **232.125** |

$$
Q = \sum w_i d_i V_i = 232.1\ \text{m}^3/\text{s}
$$

**Answer:** Discharge ≈ 232 m³/s.

(By the mean-section method, averaging depth and velocity of adjacent verticals for each 10 m panel, $Q = 225.9$ m³/s; the mid-section result is normally reported.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Since mean velocities are given, what is done with them?
   A. Averaged at 0.2d and 0.8d again
   B. Converted using a rating equation
   C. Used directly as V in w*d*V
   D. Multiplied by 0.6
   Answer: C. The velocities are already mean values, so dQ = w d V is applied directly. (Key point: Mean velocity given)

2. What width is used for the vertical at 90 m (neighbours at 80 m and 99 m)?
   A. 9.5 m
   B. 10 m
   C. 9 m
   D. 19 m
   Answer: A. w = (99 - 80)/2 = 9.5 m. (Key point: Last vertical width)

3. What is dQ at 60 m (width 10 m, depth 3.5 m, V = 1.6 m/s)?
   A. 35 m3/s
   B. 56 m3/s
   C. 5.6 m3/s
   D. 28 m3/s
   Answer: B. dQ = 10 x 3.5 x 1.6 = 56 m3/s. (Key point: Segment discharge)

4. What is the discharge by the mid-section method?
   A. 199 m3/s
   B. 226 m3/s
   C. 232 m3/s
   D. 250 m3/s
   Answer: C. The sum of all segment discharges is 232.125, about 232 m3/s (199.25 is the area). (Key point: Total discharge)

## Related in Hydro Power

- [Discharge from 0.2d and 0.8d velocities (right bank)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Astream-gauging-right-bank) — Hydro Power, chapter 2
- [Discharge from 0.2d and 0.8d velocities (left bank)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Astream-gauging-left-bank) — Hydro Power, chapter 2
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