---
title: "Discharge from current meter at 0.6d, stream gauging data"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:current-meter-0-32n-0-032-2"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/current-meter-0-32n-0-032-2
site: BCT Engineering
---

# Discharge from current meter at 0.6d, stream gauging data

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2081 Bhadra, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/current-meter-0-32n-0-032-2
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

The data pertaining to a stream gauging operation at a gauging site are given below. The rating equation of the current meter is $v = (0.32N + 0.032)$ m/s where N is revolution per second. Calculate the discharge in the stream.

| Distance from right bank (m) | Depth (m) | Revolution of CM at 0.6d | Time (s) |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 2 | 0.5 | 80 | 180 |
| 4 | 1.1 | 83 | 120 |
| 6 | 1.95 | 131 | 120 |
| 9 | 2.25 | 139 | 120 |
| 12 | 1.85 | 212 | 120 |
| 15 | 1.75 | 114 | 120 |
| 18 | 1.65 | 109 | 120 |
| 20 | 1.5 | 92 | 120 |
| 22 | 1.25 | 85 | 120 |
| 23 | 0.75 | 70 | 150 |
| 24 | 0 | 0 | 0 |

## Short answer

Using the mid-section method with velocity v = 0.32N + 0.032 at 0.6d, discharge is computed as the sum of v × depth × segment width for each vertical. Total Q ≈ 13.17 m³/s.

## Answer

The mid-section method is used with $v = 0.32N + 0.032$, where $N$ = revolutions / time (rps), at 0.6 d.
$$
q_i = v_i\,d_i\,\frac{b_{i+1}-b_{i-1}}{2}
$$

| 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) |
|---|---|---|---|---|---|---|
| 2 | 0.5 | 80/180 = 0.444 | 0.174 | 2 | 1.000 | 0.174 |
| 4 | 1.1 | 83/120 = 0.692 | 0.253 | 2 | 2.200 | 0.557 |
| 6 | 1.95 | 131/120 = 1.092 | 0.381 | 2.5 | 4.875 | 1.859 |
| 9 | 2.25 | 139/120 = 1.158 | 0.403 | 3 | 6.750 | 2.718 |
| 12 | 1.85 | 212/120 = 1.767 | 0.597 | 3 | 5.550 | 3.315 |
| 15 | 1.75 | 114/120 = 0.950 | 0.336 | 3 | 5.250 | 1.764 |
| 18 | 1.65 | 109/120 = 0.908 | 0.323 | 2.5 | 4.125 | 1.331 |
| 20 | 1.5 | 92/120 = 0.767 | 0.277 | 2 | 3.000 | 0.832 |
| 22 | 1.25 | 85/120 = 0.708 | 0.259 | 1.5 | 1.875 | 0.485 |
| 23 | 0.75 | 70/150 = 0.467 | 0.181 | 1 | 0.750 | 0.136 |
| **Total** | | | | | **35.375** | **13.172** |

Total area $= 35.38$ m$^2$; mean velocity $= 0.372$ m/s.

**Answer: Discharge $Q \approx 13.17\ \text{m}^3/\text{s}$.**

Note: the reading of 212 revolutions at 12 m is much higher than its neighbours (139 and 114). It has been used as given. If it is a misprint for 121, the discharge would be 11.82 m$^3$/s.

## Check yourself

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

1. What is notable about the 212 revolutions at 12 m?
   A. It is the lowest value
   B. It was discarded
   C. It is much higher than neighbours and was used as given
   D. It was halved
   Answer: C. The answer used it as given but noted a possible misprint for 121. (Key point: Suspect data)

2. What velocity results at 12 m?
   A. N = 212 rps
   B. v = 0.403 m/s
   C. v = 0.355 m/s
   D. N = 212/120 = 1.767 rps, v = 0.597 m/s
   Answer: D. v = 0.32 x 1.767 + 0.032 = 0.597 m/s. (Key point: Velocity at 12 m)

3. What is the discharge with 212 and with 121?
   A. 13.17 m^3/s with 212; 11.82 m^3/s if 121
   B. 11.82 with 212; 13.17 with 121
   C. Both 35.4 m^3/s
   D. Both 13.17 m^3/s
   Answer: A. The vertical's partial discharge is 3.315 vs 1.968 m^3/s, a difference of 1.35 m^3/s. (Key point: Effect on discharge)

## Related in Engineering Hydrology

- [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
- [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×
- [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 Jhiku khola from stream gauging data](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Acurrent-meter-jhiku-khola) — Engineering Hydrology, chapter 4

## Related in other subjects

- [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×
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