---
title: "Stream flow from weir and float measurements"
subject: "Energy Resources"
chapter: 5
key: "semester_6/energy_resources:stream-flow-weir-float-numerical"
source: https://www.bctengineer.com/q/semester_6/energy_resources/stream-flow-weir-float-numerical
site: BCT Engineering
---

# Stream flow from weir and float measurements

Energy Resources · Chapter 5: Micro and small hydro power systems

- Asked in: year not given, 6 marks
- Video: [13 Discharge Measurement-Q-Method: Weir](https://www.youtube.com/watch?v=sH54xFValFY) — USGS Trainings
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/energy_resources/stream-flow-weir-float-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/energy_resources/chapter-5

## Question

(a) A rectangular sharp-crested weir of crest length 0.6 m with two end contractions has a measured head of 0.18 m. Find the flow using Francis' formula $Q = 1.84\,(L - 0.1nH)H^{3/2}$ (SI units). (b) In a rectangular channel of width 1.2 m and mean water depth 0.40 m, a float takes 25 s to travel 20 m. Taking the mean velocity as 0.85 times the surface velocity, find the flow.

## Short answer

(a) Effective length 0.6−0.1×2×0.18=0.564 m, H^1.5=0.0764, Q=1.84×0.564×0.0764=0.0793 m³/s (79 L/s). (b) Surface velocity 20/25=0.8 m/s, mean 0.85×0.8=0.68 m/s, area 1.2×0.4=0.48 m², Q=0.48×0.68=0.326 m³/s (326 L/s).

## Answer

#### (a) Weir
Here $L = 0.6$ m, $n = 2$, $H = 0.18$ m.

Effective length: $L - 0.1nH = 0.6 - 0.1(2)(0.18) = 0.564$ m

$H^{3/2} = 0.18^{1.5} = 0.0764$

$$
Q = 1.84 \times 0.564 \times 0.0764 = 0.0793\ \text{m}^3/\text{s}
$$

That is 79 litres per second.

#### (b) Float
Surface velocity:

$$
v_s = \frac{20}{25} = 0.80\ \text{m/s}
$$

Mean velocity:

$$
v = 0.85 \times 0.80 = 0.68\ \text{m/s}
$$

Area: $A = 1.2 \times 0.40 = 0.48$ m².

$$
Q = A v = 0.48 \times 0.68 = 0.326\ \text{m}^3/\text{s}
$$

**Answer:** (a) $Q = 0.0793$ m³/s (79 L/s); (b) $Q = 0.326$ m³/s (326 L/s).

## Check yourself

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

1. In Francis' formula with two end contractions, what is the effective crest length for L = 0.6 m and H = 0.18 m?
   A. 0.6 + 0.036 = 0.636 m
   B. 0.6 m with no correction
   C. 0.564 m (L - 0.1nH)
   D. 0.582 m (L - 0.1H)
   Answer: C. Each end contraction reduces the length by 0.1H, so 0.6 - 0.1 x 2 x 0.18 = 0.564 m. (Key point: Francis effective length)

2. What is the weir discharge?
   A. About 0.0793 m3/s (79 L/s)
   B. About 0.326 m3/s
   C. About 0.79 m3/s
   D. About 0.0076 m3/s
   Answer: A. Q = 1.84 x 0.564 x 0.0764 = 0.0793 m3/s. (Key point: Weir discharge)

3. In the float method, how is the mean velocity obtained?
   A. Surface velocity 0.80 m/s times 0.85 = 0.68 m/s
   B. Surface velocity divided by 0.85
   C. Distance 20 m divided by the depth
   D. Equal to the surface velocity
   Answer: A. A float moves at surface speed (20/25 = 0.8 m/s); the mean velocity is smaller, 0.85 times that. (Key point: Float method velocity)

4. Which gives the flow in the channel?
   A. Q = 0.80 x 0.48
   B. Q = A x v = 0.48 x 0.68 = 0.326 m3/s
   C. Q = 1.2 x 0.68
   D. Q = 0.40 x 0.85
   Answer: B. Flow is cross-section area (1.2 x 0.40) times mean velocity. (Key point: Float method flow)

## Related in Energy Resources

- [Methods of stream flow measurement](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Aflow-measurement-methods-micro-hydro) — Energy Resources, chapter 5
- [Micro hydro net head, power and energy](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Amicro-hydro-power-output-numerical) — Energy Resources, chapter 5

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