Gates and Orifices-Theory and Examples v2

8/10/2019 Gates and Orifices-Theory and Examples v2

http://slidepdf.com/reader/full/gates-and-orifices-theory-and-examples-v2 1/10

Orifice structures: Theory and Examples

1. Introduction: Hydraulic structures in channels.

From the Practical Hydraulics Textbook…

Examples of hydraulic structures: Orifices, gates, weirs, flumes, spillways, stilling basins, dropstructures, siphons, and culverts.

Functions of hydraulic structures in channels:

Measuring discharge

o

Weirs and flumes – better suited to measure discharges in rivers.

o

Orifices (not that useful) – upstream and downstream water levels are usually

required to determine the discharge, and constant attention for opening and

closing gates is needed, due to large variations in flow.

Discharge control (fair and equitable distribution of water, for example in large irrigation

networks; ensure minimum base flow on summer; and measure flows to avoid or control

flooding).

o

Orifices and gates.

o

Weirs and flumes (not that useful).

Controlling water levels (for example on irrigation schemes is used as discharge control).

o

Weirs and flumes

o

Spillways (dams)

o

Orifices and gates (not that useful)

Dissipating unwanted energy (hydraulic jumps).

o

Still basin

o

Drop structure

Some hydraulic structures only carry out one of the functions described above whilst others perform

all three functions at the same time, so a hydraulic structure may be used for discharge

measurement, and at the same time may be performing a water level control function and

dissipating unwanted energy.

From the point of view of measuring discharge and controlling water levels, there are only two types

of structure. Some structures allow water to flow through them and these are called orifice

structures (they include orifices and gates). Others allow water to flow over them and these are

called weirs or flumes. Hydraulically, they behave in quite different ways and so each has certain

applications for which they are best suited. The energy dissipating function can be attached to both

of these structure types.



2. Orifice structures.

a. Types of orifices and formula.

Figure 1. Left: round-edged orifices (Cc = 1). Right: Sharp-edged orifices (Cc < 1).

As reviewed in hydrodynamics and uses of the total energy equation:

= √ 2ℎ

Where:

is the discharge of the orifice

is the area of the orifice

ℎ is the water depth of the tank

is the gravity constant

is the discharge coefficient calculated as: =

is the coefficient of contraction: = ⁄

is the velocity coefficient and relates the real velocity discharged from an orifice with the

theoretic discharge calculated by Torricelli’s Law: √ 2ℎ.

All coefficients depend on the Reynolds Number, and for > 10, they are independent to the

Reynolds number and there values are:

= 0.99

= 0.605

= 0.60

For Reynolds Number that are less than 10, the following graph applies.



Figure 2. Variation of the velocity, contraction and discharge coefficients in comparison to the Reynolds

Number.

b. Methods to calculate the real velocity of an orifice.



EXAMPLE 1. An orifice on a vertical wall is located 2 m from the water surface. The orifice has a

diameter of 100 mm and a discharge of 29.5 lps.

A)

Applying the total energy equation and the trajectory method, calculate the coefficient of

contraction, coefficient of velocity and coefficient of discharge.

B)

Find the Reynolds number of the discharge of the orifice.

C)

Using the Reynolds number definition, identify the water depth at which the previous values

may no longer be used ( < 10).



c. Partial and total drowned flow

Total drowned flow: = √ 2∆

Where:

is the discharge of the orifice

is the area of the orifice

∆ is the difference between the

water depth of each side of the tank


is the discharge coefficient

calculated as: = , and Sotelo

recommends to use the same

discharge coefficient as free-discharge

flow.

Partial drowned flow:

= √ 2

= √ 2

Where:

Discharge: = +

Area of the orifice: = +

is the difference between the water

depth of each side of the tank

= − 2⁄


is the discharge coefficient of each

section. Where Schlag proposes the

values:

= 0.70 and

= 0.675.

IMPORTANT: The previous formula is for

orifices that are not located at the bottom of

the tank, as shown in the next images.

EXAMPLE 2. Calculate the discharge of the orifice shown in the picture. (Q = 0.0275 m3/s)



d. Free flow on gates.

Figure 3. Free flow on a vertical gate (Kay).

According to Munson et al . (1999) if…

< 0.2

Then… (Equation 1)

= √ 2

Where:

is the gate’s opening

is the water depth (upstream)

is the discharge of the channel

is the width of the channel

is the gravity constant, and

is the discharge coefficient that depends on the contraction coefficient: = ⁄ and

water depth: ⁄ (Henderson, 1966), but it also depends on the velocity coefficient

(Sotelo).

NOTE: varies from 0.55 – 0.6 with a vertical sluice gate and free flow. (See Figure 4.)

(NOTE: supercritical flow passing under the sluice gate.)

When…

a > 0.2

Then… (Equation 2)

= 2( − 3)1 − (3 ⁄ )

3



Figure 4. Discharge coefficient of the vertical gate (Henderson, 1966). Copia en Sotelo, pp. 216.

Submerged discharge

⁄

3 =



e. Drowned or submerged flow on gates.

Figure 5. Drowned outflow in a vertical sluice gate.

Equation 1 and 2 apply also to drowned flow, and as Figure 4 shows, if = 3, then = 0.

f. Types of gates.

Figure 6. Typical underflow gates (Henderson, 1966).



Figure 7. Radial or tainter gate.

Where:

is the water depth (upstream)

is the gate’s opening

is the height form the bed of the channel to the pin or bolt.

is the gate’s radius

3 is the water depth (downstream) after the hydraulic jump.

is the discharge coefficient that depends on the contraction coefficient: = ⁄ and

water depth: ⁄ (Henderson, 1966), but it also depends on the velocity coefficient

Figure 8. The discharge coefficient of the radial gate (Henderson, 1966). Copia en Sotelo, pp. 218.



Figure 9. Discharge coefficient of inclined gates with free discharge (Sotelo, pp. 215).

EXAMPLE 3. A vertical gate with a width of 1 m, discharges 2.6 m3/s, its upstream and downstream

water depth is 4.5 m and 3.45 m, respectively.

a)

Given a submerged discharge, calculate the opening of the gate assuming a Cd = 0.37.

b)

With the answer obtained in (a), calculate the discharge when water flows freely, and the

discharge coefficient.

Bibliography:

Henderson, F.M. 1966. Open Channel Flow. Macmillan, New York, USA.

http://www.lmnoeng.com/TankDischarge.php

Kay, M. 2008. Practical Hydraulics, 2nd Ed. Taylor & Francis, New York, USA.

Mott, R. L. 2006. Mecánica de Fluidos, 6ta Ed. Pearson Educación, México.

Munson, B. R., D. F. Young y T. H. Okiishi. 1999. Fundamentos de mecánica de fluidos. Limusa

Wiley, México D.F., México.

Sotelo, G. 1997. Hidráulica General Vol. 1. Limusa, México.

Streeter, V. L. 1962. Fluid Mechanics, 3d Ed. McGraw-Hill Book Company, Tokyo, Japan.




Documents

Gates and Orifices-Theory and Examples v2