---
title: "Current meter discharge, rating equation"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:current-meter-0-53n"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/current-meter-0-53n
site: BCT Engineering
---

# Current meter discharge, rating equation

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2079 Baisakh, 8 marks
- Video: [Discharge measurement using Currentmeter by Raj KC](https://www.youtube.com/watch?v=TWr5feW-SFQ) — Raj Kumar KC
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/current-meter-0-53n
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

A current meter (rating equation: $V = (0.53N + 0.05)$ m/s, where N = revolutions per second) was used to measure the velocity at 0.6 depth. If current meter readings at various locations at a cross section are as in the following table, calculate the discharge in the stream.

| Distance from right bank (m) | 0 | 1 | 3 | 5 | 7 | 9 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|
| Depth (m) | 0 | 1.2 | 2.1 | 2.6 | 2.0 | 1.7 | 1.1 | 0.0 |
| No. of revolutions | 0 | 39 | 58 | 112 | 90 | 45 | 30 | 0 |
| Time (seconds) | 0 | 100 | 100 | 150 | 100 | 100 | 100 | 0 |

## Short answer

Mid-section method: velocities from V = 0.53N+0.05 m/s, strip widths from mid-points, summed to total discharge = 7.71 m³/s.

## Answer

The **mid-section method** is used. Each vertical represents a strip from the mid-point of the previous vertical to the mid-point of the next vertical:
$$
q_i = v_i\,d_i\,\frac{b_{i+1}-b_{i-1}}{2},\qquad Q=\sum q_i
$$
The rating equation $v = 0.53N + 0.05$ gives the velocity at 0.6 depth, which is taken as the mean velocity of the vertical ($N=$ revolutions / time). The end points (banks) have zero depth.

| Distance (m) | Depth d (m) | N = rev/time (rps) | v (m/s) | Width w (m) | Area $w\,d$ (m$^2$) | $q = v\,w\,d$ (m$^3$/s) |
|---|---|---|---|---|---|---|
| 1 | 1.2 | 39/100 = 0.390 | 0.257 | 1.5 | 1.800 | 0.462 |
| 3 | 2.1 | 58/100 = 0.580 | 0.357 | 2 | 4.200 | 1.501 |
| 5 | 2.6 | 112/150 = 0.747 | 0.446 | 2 | 5.200 | 2.318 |
| 7 | 2 | 90/100 = 0.900 | 0.527 | 2 | 4.000 | 2.108 |
| 9 | 1.7 | 45/100 = 0.450 | 0.289 | 2 | 3.400 | 0.981 |
| 11 | 1.1 | 30/100 = 0.300 | 0.209 | 1.5 | 1.650 | 0.345 |
| **Total** | | | | | **20.250** | **7.715** |

Sample calculation at 5 m: $N = 112/150 = 0.747$ rps; $v = 0.53\times0.747+0.05 = 0.446$ m/s; width $=(7-3)/2 = 2$ m; area $=2\times2.6 = 5.2$ m$^2$; $q = 0.446\times5.2 = 2.318$ m$^3$/s.

Total area $= 20.25$ m$^2$ and mean velocity $= 0.381$ m/s.

**Answer: Discharge $Q \approx 7.71\ \text{m}^3/\text{s}$.**

## Check yourself

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

1. In the mid-section method, what is the width of the strip assigned to vertical i?
   A. Half the distance between the verticals on either side, (b(i+1) - b(i-1))/2
   B. The full distance between the two banks divided by the number of verticals
   C. The distance b(i) - b(i-1)
   D. The distance b(i+1) - b(i)
   Answer: A. Each vertical represents a strip from the mid-point of the previous interval to the mid-point of the next, so w = (b(i+1) - b(i-1))/2. (Key point: Strip width in mid-section method)

2. How is N obtained before putting it in V = 0.53N + 0.05?
   A. N = revolutions in minutes
   B. N = revolutions / time in seconds (rps)
   C. N = revolutions x time
   D. N = time / revolutions
   Answer: B. The rating equation takes N in revolutions per second, so N = revolutions divided by the time in seconds. (Key point: Velocity from rating equation)

3. Why can the velocity at 0.6 depth be used directly as the vertical's velocity here?
   A. It is the surface velocity
   B. It is the bed velocity
   C. It is taken as the mean velocity of the vertical
   D. It is the maximum velocity of the vertical
   Answer: C. The velocity at 0.6 depth approximates the mean velocity of the vertical. (Key point: Meaning of 0.6 depth velocity)

4. At the 5 m vertical (d = 2.6 m, v = 0.446 m/s, neighbours at 3 m and 7 m), what is q?
   A. 0.446 x 2.6 x 1 = 1.16 m3/s
   B. 0.446 x 2 = 0.89 m3/s
   C. 0.446 x 2.6 x 4 = 4.64 m3/s
   D. 0.446 x 2.6 x 2 = 2.32 m3/s
   Answer: D. Width = (7 - 3)/2 = 2 m, so q = 0.446 x 2.6 x 2 = 2.318 m3/s. (Key point: Partial discharge at 5 m)

## Related in Engineering Hydrology

- [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
- [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 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, V=0.05+0.3N](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Acurrent-meter-0-05-0-3n) — Engineering Hydrology, chapter 4
- [Stream flow from current meter data](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Acurrent-meter-0-035-0-74n) — Engineering Hydrology, chapter 4, asked 2×
- [Stream flow from current meter, V=0.045+0.76N](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Acurrent-meter-0-045-0-76n) — Engineering Hydrology, chapter 4, asked 2×

## Related in other subjects

- [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×
- [Stream discharge from current meter revolutions (v = 0.55N + 0.04)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Astream-gauging-revolutions-0-55) — Hydro Power, chapter 2, asked 2×
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- [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