---
title: "Design of orifice type intake"
subject: "Hydropower Engineering"
chapter: 5
key: "semester_7/hydropower_engineering:intake-orifice-design-5cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/intake-orifice-design-5cumec
site: BCT Engineering
---

# Design of orifice type intake

Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants

- Asked in: 2082 Baishakh, 7 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/intake-orifice-design-5cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5

## Question

Design an orifice type intake for a ROR plant having design discharge of 5 m³/s. The difference of elevations between the weir crest level and the river bed level is 2.5 m. The difference of elevations between the weir crest level and the designed canal water surface level is 0.3 m. The intake velocity is limited to 0.8 m/s. Take coefficient of discharge as 0.6, coefficient of contraction as 0.03, and coefficient of trash rack bar as 1.83.

## Short answer

Net area 6.25 m2, gross 6.44 m2; adopt opening 5.4 m by 1.2 m in two bays, sill 1.0 m above bed. Required head is 0.084 m, less than 0.3 m available; trash rack loss about 0.0067 m.

## Answer

**Given**: $Q=5$ m³/s; weir crest is 2.5 m above the river bed; canal water level is 0.3 m below the crest; intake velocity $\le0.8$ m/s; $C_d=0.6$; bar-contraction coefficient 0.03; trash-rack bar shape factor $\beta=1.83$.

#### Step 1: Net opening area from the velocity limit
$$
A_{net}=\frac{Q}{V}=\frac{5}{0.8}=6.25\ \text{m}^2
$$

#### Step 2: Gross area allowing for the bars
The bars reduce the opening by the contraction coefficient 0.03:
$$
A_{gross}=\frac{A_{net}}{1-0.03}=\frac{6.25}{0.97}=6.44\ \text{m}^2
$$

#### Step 3: Dimensions of the opening
Sill level 1.0 m above the river bed (to keep bed load out). Take the height of the orifice $h=1.2$ m so that its top is at 2.2 m above the bed, i.e. at the canal water level (0.3 m below the crest), so the opening stays submerged.
$$
b=\frac{A_{gross}}{h}=\frac{6.44}{1.2}=5.37\ \text{m}\ \Rightarrow\ \text{adopt }5.4\ \text{m (2 bays of 2.7 m)}
$$
Adopted opening area $A=1.2\times5.4=6.48$ m².

#### Step 4: Head required across the orifice (check)
$$
Q=C_dA\sqrt{2g\,\Delta h}\ \Rightarrow\ \Delta h=\frac{1}{2g}\left(\frac{Q}{C_dA}\right)^2=\frac{1}{19.62}\left(\frac{5}{0.6\times6.48}\right)^2=0.084\ \text{m}
$$
Available head difference $=0.3$ m $>0.084$ m, so the opening can pass the design flow; the extra head is absorbed by throttling the gate.

For comparison, the area that would pass 5 m³/s under the full 0.3 m head is $A=\dfrac{5}{0.6\sqrt{2g(0.3)}}=3.43$ m², so the velocity limit (6.25 m²) controls the size.

#### Step 5: Trash rack loss (Kirschmer)
Assume bars 10 mm thick at 50 mm clear spacing, inclined at $75^\circ$:
$$
h_t=\beta\left(\frac{s}{b}\right)^{4/3}\frac{V^2}{2g}\sin\theta=1.83\left(\frac{10}{50}\right)^{4/3}\frac{0.8^2}{19.62}\sin75^\circ=0.0067\ \text{m}
$$

```text
 SECTION                         PLAN
 crest 2.5 ~~~~~~~~~~~~          river ==>
        |   |gate|  2.2         |rack|gate|--> canal
  rack  |## |    |~~~~ canal    | 5.4 m opening |
 sill   |___|____|              2 bays of 2.7 m
 1.0 m above bed
```

**Answer**: orifice opening $5.4\ \text{m}\times1.2$ m with sill 1.0 m above the bed (2 bays of 2.7 m); required head $0.084$ m < 0.3 m available; rack loss $\approx0.0067$ m.

## Check yourself

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

1. For the orifice intake, how is the net opening area found from the intake velocity limit?
   A. A_net = Q/V = 5/0.8 = 6.25 m^2
   B. A_net = Q x V = 5 x 0.8 = 4.0 m^2
   C. A_net = Q/Cd = 5/0.6 = 8.33 m^2
   D. A_net = V/Q = 0.8/5 = 0.16 m^2
   Answer: A. Continuity gives area = discharge / velocity, so limiting velocity to 0.8 m/s needs 6.25 m^2 of net opening. (Key point: Net area from velocity limit)

2. How is the gross area obtained from the net area when the bar contraction coefficient is 0.03?
   A. A_gross = A_net x (1 - 0.03) = 6.06 m^2
   B. A_gross = A_net/(1 - 0.03) = 6.44 m^2
   C. A_gross = A_net/Cd = 10.4 m^2
   D. A_gross = A_net = 6.25 m^2, bars do not matter
   Answer: B. The bars block 3% of the opening, so the gross area must be larger than the net area: divide by 0.97. (Key point: Gross area allowing for bars)

3. Why is the sill placed 1.0 m above the river bed and the orifice top kept at the canal water level (2.2 m above the bed)?
   A. The high sill reduces the trash rack loss, and the top at canal level increases the intake velocity
   B. The high sill raises the discharge coefficient, and the top at canal level lets floating debris enter
   C. The high sill keeps bed load out, and the top at canal level keeps the opening submerged
   D. The sill is only for fixing the gate, and the top level is chosen arbitrarily
   Answer: C. Bed load travels near the bed so a raised sill excludes it; a top at the canal level means the orifice works submerged as an orifice. (Key point: Sill level and submergence)

4. Area needed to pass 5 m^3/s under the full 0.3 m head is 3.43 m^2, while the velocity limit needs 6.25 m^2. What does this show?
   A. The head controls the size, so the opening is 3.43 m^2
   B. The opening must be 3.43 + 6.25 = 9.68 m^2
   C. The design fails because the head is too small
   D. The velocity limit controls the size, and the extra head is absorbed by throttling the gate
   Answer: D. The larger area (6.25 m^2) is adopted; since this opening needs only 0.084 m of head, the surplus of the 0.3 m available is throttled at the gate. (Key point: Which criterion controls the size)

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