---
title: "Orifice plate flow in hydropower canal"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_3/fluid_mechanics:orifice-plate-canal-flow"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/orifice-plate-canal-flow
site: BCT Engineering
---

# Orifice plate flow in hydropower canal

Fluid Mechanics · Chapter 6: Flow measurement

- Asked in: 2079 Baisakh, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/orifice-plate-canal-flow
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6

## Question

An orifice plate is used to measure the flow in hydropower canal 500 m wide with a water depth of 300 mm. A rectangular orifice size of 300 mm wide and 100 mm high is placed 5 cm above the canal bed. If downstream water depth in canal is 225 mm, what is the flow in the canal? Take coefficient of discharge of an orifice plate 0.63.

## Short answer

The orifice is fully submerged; discharge Q = Cd a √(2g(H1-H2)) = 0.63 × 0.03 × √(2×9.81×0.075) ≈ 0.0229 m³/s (22.9 L/s).

## Answer

**Given:** orifice $b = 0.30$ m wide, $d = 0.10$ m high, bottom edge 5 cm above the bed, so the top edge is 0.15 m above the bed. Upstream depth $H_1 = 0.300$ m, downstream depth $H_2 = 0.225$ m, $C_d = 0.63$.

Both depths are above the top edge of the orifice (0.15 m), so the orifice is **fully submerged** (drowned): the discharge is governed by the difference of the water levels, not by the depth below the surface.

```text
 upstream 0.300 m  |  |  downstream 0.225 m
 ~~~~~~~~~~~~~~~~~~|  |~~~~~~~~
                   |__|  orifice  (0.15 m top)
                   |  |  (0.05 m bottom)
 ------------------------ bed
```

Head difference:

$$
H = H_1 - H_2 = 0.300 - 0.225 = 0.075\ \text{m}
$$

Area of orifice $a = 0.30\times 0.10 = 0.03\ \text{m}^2$.

$$
\begin{aligned}
Q &= C_d\,a\sqrt{2gH}\\
&= 0.63\times 0.03\times\sqrt{2\times 9.81\times 0.075}\\
&= 0.63\times 0.03\times 1.2130\\
&= 0.02293\ \text{m}^3/\text{s}
\end{aligned}
$$

The velocity of approach is small (it is $0.153$ m/s if the canal is 0.5 m wide), so it is neglected. The canal width does not enter the calculation.

**Answer: $Q \approx 0.0229$ m³/s (about 22.9 L/s).**

## Check yourself

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

1. Why a submerged orifice?
   A. Canal is wide
   B. Cd is low
   C. Both depths exceed the top edge 0.15 m
   D. Depth is 300 mm
   Answer: C. Both sides above the orifice. (Key point: Submerged)

2. Head used?
   A. 0.3 m
   B. 0.225 m
   C. 0.2 m
   D. 0.075 m
   Answer: D. Level difference. (Key point: Head)

3. Orifice area?
   A. 0.03 m^2
   B. 0.3 m^2
   C. 0.15 m^2
   D. 0.015 m^2
   Answer: A. 0.30 x 0.10. (Key point: Area)

4. Q =
   A. 0.229 m^3/s
   B. 0.0229 m^3/s
   C. 0.0036 m^3/s
   D. 1.2 m^3/s
   Answer: B. 0.63*0.03*sqrt(2*9.81*0.075). (Key point: Discharge)

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