---
title: "River discharge from two-point current meter data"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:current-meter-0-02ns"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/current-meter-0-02ns
site: BCT Engineering
---

# River discharge from two-point current meter data

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2070 Asar, 14 marks
- Video: [Stream Discharge Measurement](https://www.youtube.com/watch?v=9MpyTopzon8) — Boat of Knowledge Ohio University
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/current-meter-0-02ns
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

Calculate the discharge of river section as given:

| Distance (m) | 0 | 1 | 2 | 3 | 4 | 6 | 8 | 12 | 16 | 17 | 18 | 19 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Depth (m) | 0 | 1 | 4.3 | 7.2 | 8.5 | 7.4 | 5.6 | 4.7 | 3.5 | 2.1 | 1.4 | 0 |
| Revolution/s at 0.2d | 0 | 1.4 | 1.0 | 2.6 | 2.9 | 2.7 | 2.5 | 2.3 | 2.1 | 1.8 | 1.5 | 0 |
| Revolution/s at 0.8d | 0 | 0.7 | 1.2 | 1.8 | 2.0 | 1.9 | 1.7 | 1.5 | 1.3 | 1.1 | 1.0 | 0 |

The current meter formula is $v = 0.02 N_s - 0.02$, v = velocity (m/s) and $N_s$ = revolution per minute.

## Short answer

Mid-section method with v = 0.02×(60×rev/s)−0.02 and mean velocity from 0.2d and 0.8d points gives total discharge 211.3 m³/s. Q ≈ 211 m³/s.

## Answer

**Method:** mid-section method with the two-point velocity: $v_m=(v_{0.2}+v_{0.8})/2$.

The meter equation is $v = 0.02N_s - 0.02$ with $N_s$ in revolutions per minute. The readings are in rev/s, so they are converted: $N_s = 60\times$(rev/s). Example at 4 m, 0.2 d: $N_s = 2.9\times60 = 174$ rpm, $v = 0.02\times174-0.02 = 3.46$ m/s.

Strip width $w_i=(x_{i+1}-x_{i-1})/2$; the bank verticals at 0 m and 19 m have zero depth.

| x (m) | d (m) | $v_{0.2}$ | $v_{0.8}$ | $v_m$ | $w$ (m) | Area (m$^2$) | $q$ (m$^3$/s) |
|---|---|---|---|---|---|---|---|
| 1 | 1 | 1.66 | 0.82 | 1.240 | 1 | 1.00 | 1.24 |
| 2 | 4.3 | 1.18 | 1.42 | 1.300 | 1 | 4.30 | 5.59 |
| 3 | 7.2 | 3.10 | 2.14 | 2.620 | 1 | 7.20 | 18.86 |
| 4 | 8.5 | 3.46 | 2.38 | 2.920 | 1.5 | 12.75 | 37.23 |
| 6 | 7.4 | 3.22 | 2.26 | 2.740 | 2 | 14.80 | 40.55 |
| 8 | 5.6 | 2.98 | 2.02 | 2.500 | 3 | 16.80 | 42.00 |
| 12 | 4.7 | 2.74 | 1.78 | 2.260 | 4 | 18.80 | 42.49 |
| 16 | 3.5 | 2.50 | 1.54 | 2.020 | 2.5 | 8.75 | 17.68 |
| 17 | 2.1 | 2.14 | 1.30 | 1.720 | 1 | 2.10 | 3.61 |
| 18 | 1.4 | 1.78 | 1.18 | 1.480 | 1 | 1.40 | 2.07 |
| **Total** | | | | | | **87.90** | **211.3** |

Mean velocity $= 2.40$ m/s.

**Answer: Discharge $Q \approx 211\ \text{m}^3/\text{s}$.**

## Check yourself

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

1. The meter equation uses Ns in revolutions per minute, but data are rev/s. What is done?
   A. Ns = 60 x (rev/s)
   B. Ns = (rev/s)/60
   C. No change needed
   D. Ns = 100 x (rev/s)
   Answer: A. Example: 2.9 rev/s gives 174 rpm. (Key point: Unit conversion)

2. For 2.9 rev/s at 0.2 d, v = 0.02 Ns - 0.02 gives
   A. 2.9 m/s
   B. 3.46 m/s
   C. 0.038 m/s
   D. 58 m/s
   Answer: B. Ns = 174, v = 0.02 x 174 - 0.02 = 3.46 m/s. (Key point: Velocity at 4 m 0.2d)

3. Mean velocity of a vertical here is
   A. v0.2 - v0.8
   B. v0.8
   C. (v0.2 + v0.8)/2
   D. v0.2
   Answer: C. Two-point method using readings at 0.2 d and 0.8 d. (Key point: Two-point mean)

4. The strip width for the vertical at 8 m (neighbours 6 and 12 m) is
   A. 2 m
   B. 6 m
   C. 4 m
   D. 3 m
   Answer: D. (12 - 6)/2 = 3 m. (Key point: Strip width)

## Related in Engineering Hydrology

- [Discharge from multi-point current meter data](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Acurrent-meter-two-point-0-3n) — Engineering Hydrology, chapter 4
- [Mid-section method discharge from current meter data](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Amidsection-0-3n-0-004) — Engineering Hydrology, chapter 4
- [Discharge of Nakhu Khola from current meter data](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Acurrent-meter-nakhu-khola) — Engineering Hydrology, chapter 4
- [River discharge with current meter constants a and b](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Acurrent-meter-a0-51-b0-03) — Engineering Hydrology, chapter 4
- [Discharge from current meter at 0.6d, stream gauging data](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Acurrent-meter-0-32n-0-032-2) — Engineering Hydrology, chapter 4
- [Discharge from current meter at 0.6 depth](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Acurrent-meter-0-32n-0-032) — Engineering Hydrology, chapter 4

## Related in other subjects

- [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
- [Discharge from depth and mean velocity data](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Adischarge-depth-mean-velocity) — Hydro Power, chapter 2
- [Stream discharge from current meter revolutions (V = 0.51N + 0.03)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Astream-gauging-revolutions-0-51) — Hydro Power, chapter 2, asked 2×