---
title: "Stream discharge from current meter revolutions (V = 0.51N + 0.03)"
subject: "Hydro Power"
chapter: 2
key: "semester_6/hydro_power:stream-gauging-revolutions-0-51"
source: https://www.bctengineer.com/q/semester_6/hydro_power/stream-gauging-revolutions-0-51
site: BCT Engineering
---

# Stream discharge from current meter revolutions (V = 0.51N + 0.03)

Hydro Power · Chapter 2: Introduction to Hydrology

- Asked in: 2080 Asoj, 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 Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/stream-gauging-revolutions-0-51
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-2

## Question

The data pertaining to a stream-gauging operation at a gauging site are given below. The rating equation of the current meter is V = 0.51 N + 0.03 m/s where, N = revolutions per second. Calculate the discharge in the stream.

| Distance from left water edge (m) | Depth (m) | Revolutions at 0.6 depth | Duration of observations (s) |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 1 | 1.1 | 39 | 100 |
| 3 | 2.0 | 58 | 100 |
| 5 | 2.5 | 112 | 150 |
| 7 | 2.0 | 90 | 150 |
| 9 | 1.7 | 45 | 100 |
| 11 | 1.0 | 30 | 100 |
| 12 | 0 | 0 | 0 |

## Short answer

Using mid-section method: compute rev/s N, velocity V=0.51N+0.03, segment widths from half-distances to neighbors. Summing w_i y_i V_i gives discharge ≈ 6.24 m³/s.

## Answer

The **mid-section method** is used: each vertical represents a segment extending halfway to the neighbouring verticals, so $\Delta Q_i = w_i\,y_i\,V_i$ with $w_i = (x_{i+1} - x_{i-1})/2$. Velocity is from the rating equation $V = 0.51N + 0.03$, with $N$ = revolutions/second at 0.6 depth (gives mean velocity).

#### Computation
| x (m) | Width $w_i$ (m) | Depth (m) | N (rev/s) | V (m/s) | Area (m²) | ΔQ (m³/s) |
|---|---|---|---|---|---|---|
| 1 | 1.5 | 1.1 | 39/100 = 0.390 | 0.2289 | 1.65 | 0.3777 |
| 3 | 2.0 | 2.0 | 58/100 = 0.580 | 0.3258 | 4.00 | 1.3032 |
| 5 | 2.0 | 2.5 | 112/150 = 0.747 | 0.4108 | 5.00 | 2.0540 |
| 7 | 2.0 | 2.0 | 90/150 = 0.600 | 0.3360 | 4.00 | 1.3440 |
| 9 | 2.0 | 1.7 | 45/100 = 0.450 | 0.2595 | 3.40 | 0.8823 |
| 11 | 1.5 | 1.0 | 30/100 = 0.300 | 0.1830 | 1.50 | 0.2745 |
| **Total** | | | | | **19.55** | **6.2357** |

(At the edges, x = 0 and 12 m, depth and velocity are zero, so they add nothing.)

Sample: at x = 5 m, $N = 112/150 = 0.7467$ rev/s, $V = 0.51 \times 0.7467 + 0.03 = 0.4108$ m/s, $\Delta Q = 2.0 \times 2.5 \times 0.4108 = 2.054$ m³/s.

$$
Q = \sum \Delta Q_i = 6.24\ \text{m}^3/\text{s}
$$

**Answer:** Discharge ≈ 6.24 m³/s.

(Subramanya's textbook version gives the end verticals a width $\bar{W}_1 = (W_1 + W_2/2)^2/(2W_1) = 2.0$ m instead of 1.5 m, which gives Q ≈ 6.45 m³/s. Either method is accepted if stated.)

## Check yourself

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

1. In the mid-section method, how is the width of the strip for vertical i found?
   A. w = (x(i+1) - x(i-1))/2
   B. w = x(i+1) - x(i)
   C. w = x(i) / 2
   D. w = depth x 2
   Answer: A. Each vertical represents a strip extending halfway to the neighbouring verticals. (Key point: Mid-section width)

2. At x = 5 m there were 112 revolutions in 150 s. What are N and V by V = 0.51N + 0.03?
   A. N = 0.747 rev/s, V = 0.256 m/s
   B. N = 1.34 rev/s, V = 0.714 m/s
   C. N = 0.747 rev/s, V = 0.411 m/s
   D. N = 112 rev/s, V = 57.2 m/s
   Answer: C. N = 112/150 = 0.747 rev/s and V = 0.51(0.747) + 0.03 = 0.411 m/s. (Key point: Revolutions per second)

3. What is the strip width for the vertical at x = 1 m (neighbours at 0 and 3 m)?
   A. 1.0 m
   B. 3.0 m
   C. 2.0 m
   D. 1.5 m
   Answer: D. w = (3 - 0)/2 = 1.5 m. (Key point: End vertical width)

4. What is the total discharge from the sum of w*y*V for all verticals?
   A. About 19.55 m3/s
   B. About 6.24 m3/s
   C. About 12.5 m3/s
   D. About 0.62 m3/s
   Answer: B. Summing the segment discharges gives 6.2357, about 6.24 m3/s (19.55 m2 is the total area). (Key point: Total discharge)

## Close variants set in other exams

- [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×

## Related in Hydro Power

- [Stream gauging with calibration equation](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Astream-gauging-calibration-no-data) — Hydro Power, chapter 2
- [Gauging site selection and current meter use](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Astream-gauging-site-current-meter) — Hydro Power, chapter 2

## Related in other subjects

- [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
- [Current meter discharge, rating equation](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Acurrent-meter-0-53n) — 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
- [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