---
title: "Stream gauging with calibration equation"
subject: "Hydro Power"
chapter: 2
key: "semester_6/hydro_power:stream-gauging-calibration-no-data"
source: https://www.bctengineer.com/q/semester_6/hydro_power/stream-gauging-calibration-no-data
site: BCT Engineering
---

# Stream gauging with calibration equation

Hydro Power · Chapter 2: Introduction to Hydrology

- Asked in: 2075 Bhadra, 7 marks
- Video: [Discharge (Stream flow) Measurement Methods: [ Mid-Section  and Mean-Section Methods ]](https://www.youtube.com/watch?v=QoeX_fyuDCE) — Hydraulics and Hydrology Tutorials
- 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-calibration-no-data
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-2

## Question

The following are the data obtained in a stream-gauging operation. A current meter with a calibration equation V = (0.55N+0.04) m/s, where N = revolutions per second was used to measure the velocity at 0.6 depth. Calculate the discharge in the stream. (The data table is not printed in the paper.)

## Short answer

Use mid-section method: for each vertical, compute N=revolutions/duration, V=0.55N+0.04, segment width = (x_{i+1} - x_{i-1})/2, then ΔQ = w y V. Sum all ΔQ to get total discharge.

## Answer

The data table was not printed in the paper, so the full method is shown; the student puts in the given values.

#### Principle
Discharge is found by the **velocity–area method**: the cross-section is divided into vertical segments, the velocity in each is measured by a current meter, and
$$
Q = \sum \Delta Q_i = \sum w_i\,y_i\,\bar{V}_i
$$

#### Steps
1. **Revolutions per second** at each vertical: $N_i = \dfrac{\text{revolutions}}{\text{time (s)}}$.
2. **Velocity** at 0.6 depth from the calibration equation:
$$
V_i = 0.55N_i + 0.04\ \text{m/s}
$$
   The velocity at 0.6y below the surface is taken as the mean velocity of that vertical.
3. **Width of each segment (mid-section method):**
$$
w_i = \frac{x_{i+1} - x_{i-1}}{2}
$$
   where $x$ is the distance of the vertical from the bank.
4. **Segment discharge:** $\Delta Q_i = w_i\,y_i\,V_i$.
5. **Total discharge:** $Q = \sum \Delta Q_i$. The edge verticals (depth 0) give zero flow.

#### Table layout
| x (m) | $w_i$ (m) | $y_i$ (m) | N (rev/s) | $V_i$ (m/s) | $w_iy_i$ (m²) | $\Delta Q_i$ (m³/s) |
|---|---|---|---|---|---|---|
| ... | ... | ... | ... | ... | ... | ... |

#### Illustration
For a vertical at 4 m (neighbours at 1 m and 7 m) with depth 2.5 m and 60 revolutions in 100 s:
$N = 0.6$ rev/s, $V = 0.55 \times 0.6 + 0.04 = 0.37$ m/s, $w = (7 - 1)/2 = 3$ m, $\Delta Q = 3 \times 2.5 \times 0.37 = 2.775$ m³/s.

Repeating for every vertical and adding gives the stream discharge. (For the similar question with verticals at 1, 4, 7 and 10 m from the left edge, this method gives Q ≈ 10.99 m³/s.)

## Check yourself

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

1. What is the principle of discharge measurement used here?
   A. Sum of w*y*V over vertical segments of the cross-section
   B. Manning's equation using water surface slope
   C. Rainfall multiplied by catchment area
   D. Weir crest height only
   Answer: A. The velocity-area method divides the section into segments and sums the segment discharges Q = sum(wi yi Vi). (Key point: Velocity-area principle)

2. How is the velocity at each vertical obtained?
   A. Revolutions are used directly as velocity
   B. N = revolutions/time, then V = 0.55N + 0.04
   C. V = revolutions x depth
   D. V = 0.55 x time + 0.04
   Answer: B. Convert to revolutions per second, then apply the calibration equation. (Key point: Revolutions and rating)

3. For the vertical at 4 m (neighbours 1 m and 7 m), depth 2.5 m, 60 revolutions in 100 s, what is the segment discharge?
   A. 1.11 m3/s
   B. 2.775 m3/s
   C. 5.55 m3/s
   D. 0.925 m3/s
   Answer: B. V = 0.55(0.6) + 0.04 = 0.37 m/s, w = 3 m, so dQ = 3 x 2.5 x 0.37 = 2.775 m3/s. (Key point: Illustration check)

## Related in Hydro Power

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