OPEN CHANNEL FLOW Open channel flow is a flow of liquid, basically water in a conduit with a free surface. The open channel flows are driven by gravity alone, and the pressure gradient at the atmospheric interface is negligible. Closed duct flows are full of fluid, have no free surface within, and are driven by a pressure gradient along the duct axis. That is a surface on which pressure is equal to local atmospheric pressure. Open channel flows are characterized by the presence of a liquid-gas interface called the free surface. Some are natural flows such as rivers, creeks and floods, some are human made systems such as fresh water aquaducts, irrigation, sewers and drainage ditches.

An open channel always has two sides and a bottom, where the flow satisfies the no-slip condition. Therefore even a straight channel has a three-dimensional velocity distribution. Some measurements of straight channel velocity contours are shown below. The profiles are quite complex, with maximum velocity typically occurring in the midplane about 20% below the surface.

COMPARISON OF OPEN CHANNEL FLOW AND PIPE FLOW OPEN CHANNEL FLOW PIPE FLOW Open channel flow must have a free surface. No free surface in pipe flow. A free surface is subject to atmospheric pressure.

No direct atmospheric pressure, hydraulic pressure only.

Flow area is determined by the geometry of the channel plus the level of free surface, which is likely to change along the flow direction and with as well as time.

Flow area is fixed by the pipe dimensions. The cross section of a pipe is usually circular.

The cross section may be of any from circular to irregular form of natural streams, which may change along the flow direction and as well as with time.

The cross section of a pipe is usually circular.

The depth of flow, discharge and the slopes of channel bottom and of the free surface are interdependent.

No such dependence.

TYPES OF OPEN CHANNEL FLOWS The most common method of classifying open channel flow is by the rate of change of the free surface depth. The simplest and most widely analyzed case in uniform flow. Type of flow Description Steady flow When discharge (Q) does not change with time. Uniform flow When depth of fluid does not change for a selected length or section of

the channel. Uniform steady flow When discharge does not change with time and depth remains constant

for a selected section. Cross section should remain unchanged, referred to as a prismatic channel.

Varied steady flow When depth changes but discharge remains the same. Varied unsteady flow When both depth and discharge change along a channel length of

interest. Rapidly varying flow Depth change is rapid. Gradually varying flow

Depth change is gradual.

1. Rapidly varying flow 2. Gradually varying flow 3. Hydraulic jump 4. Weir and waterfall 5. Gradually varying 6. Hydraulic drop due to change in channel slope

PARAMETER USED IN OPEN CHANNEL FLOW ANALYSIS Hydraulic radius, R of open channel flow R is a ratio of flow cross sectional area, A and wetted perimeter (WP)

WPAR =

R : Hydraulic radius A : Flow cross sectional area WP : Wetted perimeter

Reynolds number for open channel flow.

υμρ VRVR

==Re

Re < 500 – laminar flow Re > 2000 – turbulent flow Reynolds number for pipe flow

υμρ VDVD

==Re

Re < 2000 – laminar flow Re > 4000 – turbulent flow

Froude number, Fr The Froude number (Fr) is a dimensionless number defined as the ratio of channel velocity to the speed of propagation of a small disturbance wave in the channel. For a rectangular or very wide constant depth channel, Froude number can be defined as :

hgyVFr ==

speed waveSurface velocityFlow

where, V = Velocity g = gravity yh = Hydraulic depth

TAyh =

A = Area T = Top width of the channel Fr < 1.0 → Sub-critical flow Fr = 1.0 or when hgyV = → Critical flow Fr > 1.0 → Super-critical flow A combination of both numbers is used to describe channel flow conditions.

THE CHEZY FORMULA Uniform flow can occur in long straight runs of constant slope and constant channel cross section. The water depth is constant at nyy = , and the velocity is constant at 0VV = . The slope be

θtan0 =S , where θ is the angle the bottom makes with the horizontal, considered positive for downhill flow. From Bernoulli equation, the head loss becomes:

LSzzh f 021 =−= where L is the horizontal distance between section 1 and 2. Head loss from Darcy-Weisbach is:

gV

DLfh

hf 2

2

=

radius Hydraulicdiameter Hydraulic

44

==

×==

h

h

hh

RD

WPARD

21

0212

1

08 SRfgV h ⋅⋅⎟⎟⎠

⎞⎜⎜⎝

⎛=

For a given shape and bottom roughness, the quantity 21

8⎟⎟⎠

⎞⎜⎜⎝

⎛fg is constant and can be denoted by C.

21

8⎟⎟⎠

⎞⎜⎜⎝

⎛=

fgC

Finally, the velocity V0 can be expressed as :

( )21

021

0212

1

08 SRCSRfgV hh ⋅=⋅⋅⎟⎟⎠

⎞⎜⎜⎝

⎛=

( )2

1

0SRCAQ h⋅= These are called Chezy formulas, first developed by the French engineer Antoine Chezy in conjuction with his experiments on the Saine River and the Courpalet Canal in 1769. The quantity C, called the Chezy coefficient, varies about 60 ft½/s for small rough channels to 160 ft½/s for large smooth channels (30 m½/s to 90 m½/s in SI units).

EXAMPLE A straight rectangular channel is 6 ft wide and 3 ft deep and laid on a slope of 2°. The friction factor is 0.022. SOLUTION

sft 108022.0

2.3288 21

=×

==fgC

2ft 1836 =×=A ft 5.1

36318

=++

==WP

ARh °= 2tan0S

( ) ( )( ) ( )( ) /sft 4502tan5.118108 3

21

21

0 =°×=⋅= SRCAQ h For SI units, it can be used directly.

Uniform steady flow and Manning’s equation When discharge remain the same and depth does not change, then we have uniform steady flow.

In this condition, the surface of water is parallel to the bed of the channel. To make sure water (liquid) flow inside the channel, it must have certain angle of inclination, or the channel’s slope. The slope of the channel (S) can be expressed as :-

• An angle = 1 degree • As percent = 1% • As fraction = 0.01 or 1 in 100

Manning’s equation is used to estimate the velocity of flow in a channel. The SI units form of Manning’s equation:-

21

320.1 SR

nV ⋅⋅=

V = Velocity of flow in a channel (m/s) n = Channel surface roughness. Values developed through experimentation. R = Hydraulic radius (A/WP) in meter S = Slope of the channel The English units form of Manning’s equation:-

21

3249.1 SR

nV ⋅⋅=

V = Velocity of flow in a channel (ft/s) R = Hydraulic radius (A/WP) in feet

Example of the n values:-

The flowrate of a channel could be determined by :-

21

320.1 SAR

nAVQ ⋅⋅==

where Q is in m3/s. For uniform flow, Q is referred to as normal discharge. The above equation can also be re-arranged such that :-

21

32

S

nQAR =

The left hand side equation is based on channel geometry.

Example #01 Determine normal discharge for a 200 mm inside diameter common clay drainage tile running half-full if the slope drops 1m over 1000m.

21

320.1 SAR

nQ ⋅⋅=

013.0=n

22

m 0157.042

1=×=

DA π

m 3142.021 perimeter, Wetted =×= DWP π

m 05.03142.00157.0

===WP

AR

001.0=S /sm 1018.50.1 332

132

−×=⋅⋅= SARn

Q

Example #02 Calculate slope of channel if normal discharge is 50 ft3/s. Channel is formed, unfinished concrete.

English unit !!!

21

3249.1 SAR

nQ ⋅⋅= →

32

21

49.1 AR

QnS =

017.0

ft 24.1WPAR

ft 66.9ft 12 2

=

==

==

n

WPA

00169.0=S

(Channel should drop 1.69 feet for every 1000 feet length.

Example #03 Design a rectangular channel in formed, unfinished concrete with below mention specifications. Normal flowrate = 5.75 m3/s S = 1.2% Normal depth = half of the width of the channel. Since we have to design the channel, use this equation.

21

32

S

nQAR =

Y=B/2

B

( )892.0

012.0

75.5017.0

21

21

32

=×

==S

nQAR

22

2BBBBYA =×==

BYBWP 22 =+=

4B

WPAR ==

892.042

32

232

=⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

BBAR

m 88.0m 76.1

==

YB

Example #04 In a rectangular channel as mention in Example #03, the final width was set at 2m and the maximum discharge is 12m3/s. Find the normal depth for this maximum discharge.

21

32

S

nQAR =

Y=B/2

2

( )862257.1

012.0

12017.0

21

21

32

=×

==S

nQAR

YA 2= YWP 22+=

YY

WPAR

222 +

==

( ) 862257.122

2232

32

=⎟⎠⎞

⎜⎝⎛

+=

YYYAR

Cannot solve directly. Use trial and error method. Can use MS Excel datasheet. From Excel, The normal depth must be 1.348 meter.

From MS Excel worksheet:

CONVEYANCE AND MOST EFFICIENT CHANNEL SHAPES

21

320.1 SAR

nQ ⋅⋅=

Other than the S term, all other terms are related to channel cross section and its features. These terms together are referred to as the Conveyance (K) of the channel.

320.1 AR

nK ⋅=

Then,

21

SKQ ⋅=

WPAR =

K is maximum when wetted perimeter (WP) is the least for a give area. This is also the most efficience cross section for conveying flow. For circular section, half full flow is the most efficient.

COMPOUND SECTIONS It is occurs when channel shape changes with flow depth. It is a typical idea in natural stream sections during flooding.

At normal condition, water flows in the main channel. During floods, water spills over the flood plain. We need to know the flowrate, Q at various depths or vice-versa. So that we could design channels or determine channel safety for various flood magnitudes.

Example #05 Channel type: Natural channel with levees. Slope: 0.00015 Determine the normal discharge, Q for depth 3 ft and 6 ft.

COMPOUND SECTION This is more realistic situation, where the channel roughness (value of n) may be different for floodplain than the main channel. In this case, we need to determine velocity for each sub-section, and then sum up the discharges for the sections. Example #06 Slope: 0.5% n for bank = 0.06 n for main channel = 0.03 Calculate discharge for depth of 8 feet?

5005100340085023204801

=×==×==×=

AAA

105510035735042

844801

=+==++=

=+=

PPP

∑=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

n

iii

i

i

ii

VAQ

SPA

nV

1

213

2

49.1

( )( ) ( ) ( ) ( )

/sft 9010

06.0500105/500

03.040057/400

06.032084/320005.049.1

2

32

32

32

21

=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ ×+

×+

×=

Q

Q

ENERGY PRINCIPLES FOR OPEN CHANNEL FLOW Energy at particular point in the channel is potential energy and kinetic energy. Specific energy:

2

22

22 gAQy

gVyE +=+=

area flow sectional CrossDischargeVelocity

flow ofDepth

====

AQVy

Total energy:

2

22

22 gAQzy

gVzyE ++=++=

datum thefrom bottom channel theofHeight =z

Example: Rectangular channel width = 2 m Depth = 1 m Q = 4.0 m3/s 2 m

1 m

2 m

Height above datum = 2 m Determine the specific energy and total energy. Specific energy:

( )( )( )m 20.1

281.9241

2 2

22

=+=+=g

VyE

Datum Total energy: E = datum height + specific energy = 2.0 + 1.2 = 3.2 m

Specific energy diagram The specific energy can be plotted graphically as a function of depth of flow.

k

s

EgAQ

EygAQy

gVyE

energy, Kinetic2

energy) (potential energy, Static22

2

2

2

22

=

=

+=+=

Relationship between y and static energy, Es

Relationship between y and kinetic energy, Ek

2

2

2gAQEk =

For a rectangular channel; Substitute Q with specific discharge (discharge per unit width),

BQq =

Substitute area, A with, yBA ⋅=

2

2

2

2

22 gyq

gAQEk ==

EXAMPLE A rectangular channel, width is 4 m, flowrate is 12 m3/s and depth of flow is 2.5 m. Draw specific energy diagram Find critical and alternate depth?

2

2

2

2

2

2

gyqyE

gAQyE

+=

+=

/sm 34

12 2===BQq

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 1 2 3 4 5 6Energy

y

static E = ykinetic Etotal E

Meaning from the graph:

1. The diagram applies for a given cross section and discharge. 2. As the depth of flow increases, the static energy increases, and the kinetic energy

decreases 3. The total energy curve approaches the static energy curve for high depths and the

kinetic energy curve for small depths 4. The specific energy is minimum (Emin) for a particular depth – this depth happens to

be the critical depth – Depth for which the Froude’s number = 1.0. velocity = Vc. 5. Emin – only energy value with a singular depth! 6. Depths less than the critical depths – supercritical flow. Froude Number > 1.0. V > Vc. 7. Depths greater than the critical depths – subcritical flow. Froude Number < 1.0. V < Vc. 8. For all other energy values – there are two depth associated – one greater than the

critical depth and one less than the critical depth. 9. The two depths associated with the same energy values are referred to as – Alternate

depths 10. As discharge increases, the specific energy curves move to the upper right portion of

the chart.

2

2

2

2

2

2

gyqyE

gAQyE

+=

+=

There is a minimum value of E at a certain value of y called the critical depth.

3

2

2210gyq

dydE

−==

It shows that Emin occurs at yc

31

2

231

2

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

gbQ

gqyc

The associated minimum energy is ;

( ) cc yyEE23

min ==

The depth, yc corresponds to channel velocity equal to the shallow-water wave propagation speed, C0.

gyVFr =

( ) 22232ccccc yVygygyq ===

By comparison it follows that the critical channel velocity is:

( ) 021

CgyV cc ==

1=Fr For , no solution exists. minEE <

For , two solutions are possible: minEE >

(1) Large depth with

cVV < , called sub-critical. (2) Small depth with , called super-critical. cVV >

In sub-critical flow, disturbances can propagate upstream because wave speed . VC >0

In super-critical flow, waves are swept downstream: Upstream is a zone of silence, and a small obstruction in the flow will create a wedge-shape wave exactly analogous to the Mach waves. The angle of these waves must be:

( )V

gyVC 2

1

101 sinsin −− ==μ

The wave angle and the depth can thus be used as a simple measurement of super-critical flow velocity.

Critical depth, yc

m 971.081.9

3 31

231

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

gqyc

Minimum specific energy, Emin

( ) m 457.1971.023

23

min === cyE

Since given depth was 2.5 m > 0.971 m, the given depth is sub-critical and the other depth should be super-critical. Now, determining alternate depths (Energy at 2.5 m)

m 57.25.281.92

35.22 2

2

2

2

=××

+=+=gyqyE

This energy value is the same for the other alternate (super-critical) depth, so;

2

2

81.92357.2

yy

××+=

Determine value of y by trial and error method. (use Excel ok)

super-critical alternate depth, y = 0.467 m

HYDRAULIC JUMP In open channel flow a supercritical flow can change quickly back to a subcritical flow by passing through a hydraulic jump The upstream flow is fast and shallow. The downstream flow is slow and deep. The hydraulic jump is quite thick, ranging in length from 4 to 6 time the downstream depth. It is very important that such jumps be located on specially designed aprons; otherwise the channel bottom will be badly scoured by the agitation. Jumps also mixed fluids very effectively and have application to sewage and water treatment designs.

EXAMPLE #01

Data:

1. Triangular channel with side slopes having ratio of 1:1.5 2. Flowrate is 0.68 ft3/s 3. Channel type – clean, excavated earth

Calculate:

1. Critical depth 2. Minimum specific energy 3. Plot specific energy curve 4. Determine energy for 0.45 ft and alternate depth 5. Velocity of flow and Froude number 6. Calculate require slope for given flow

SOLUTION FOR EXAMPLE #01 Used Excel. z = 1.5 Q = 0.68 ft3/s n = 0.022 y = independence number Area, ( )( ) 22

21 zyyzyA ==

Flow velocity, AQV =

Free surface width, zyT 2=

Hydraulic depth,

TAyh =

We cannot solve this question directly because the dimension of the triangular weir is not given. Trial and error method need to be applied. First step, create excel worksheet and calculate all the important parameter.

(1) Critical depth Critical depth occur at minimum energy, or at Froude number, Fr = 1.0 Find in the excel worksheet. Critical depth is 0.53 ft

(2) Minimum specific energy, Emin Minimum specific energy occur at critical depth. Emin = 0.66275 ft

(c) Plot the graph y-axis is E (energy) and x-axis is depth (y)

Specific energy

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 0.5 1 1.5 2

Energy (ft)

Dep

th (f

t)

(d) depth 0.45 ft, Energy = 0.705438 ft Alternate depth = 0.645 ft (which has same specific energy)

(e) Velocity for y = 0.450 ft, V = 2.24 ft/m, Fr = 1.51, super-critical flow for y = 0.645 ft, V = 1.09 ft/m, Fr = 0.61, sub-critical flow (f) slope ? Take y = 0.45 ft For triangle, Cross section area, A = 1

22zy( ) y( ) = zy2 = 0.30375

Wetted perimeter, A = 2y 1+ z2 =1.6225

Hydraulic radius, R = A

WP= 0.30375

1.6225= 0.1872

Flowrate, Q = 1.49

nAR

23S

12

So, slope, S = Qn

1.49( )AR23

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

2

= 0.01

Or 1 foot drop for every 100 feets.

EXAMPLE #02 (10.2) Water at 20°C flows in a 30 cm wide rectangular channel at a depth of 10 cm and a flowrate of 80,000 cm3/s. Estimate (a) the Froude number and (b) the Reynolds number. EXAMPLE #03 (10.8) An earthquake near the Kenai peninsula, Alaska, creates a single “tidal” wave called tsunami that propagates southward across the Pacific Ocean. If the average ocean depth is 4 km and seawater density is 1025 kg/m3, estimate the time of arrival of this tsunami in Hilo, Hawaii. EXAMPLE #04 (10.11) A rectangular channel is 2 m wide and contains water 3 m deep. If the slope is 0.85° and the lining is corrugated metal, estimate the discharge for uniform flow.

EXAMPLE #05 (10.17-10.18-10.19) The trapezoidal channel as shown below is made of brickwork and slopes at 1:500.

1. Determine the flowrate if the normal depth is 80 cm. 2. Determine the normal depth for which the flowrate will be 8 m3/s. 3. Let the surface be clean earth, which erodes if V exceeds 1.5 m/s. What is the maximum

depth to avoid erosion?

EXAMPLE #06 (10.84) Consider the flow under the sluice gate. If y1=10 ft and all losses are neglected except the dissipation in the jump, calculate y2 and y3 and the percentage of dissipation, and sketch the flow to scale with the EGL included. The channel is horizontal and wide.

EXAMPLE #07 (10.89) Water 30cm deep is in uniform flow down a 1° unfinished concrete slope when a hydraulic jump occurs. If the channel is very wide, estimate the water depth y2 downstream of the jump.

EXAMPLE #08 (10.93) Water in horizontal channel accelerates smoothly over a bump and then undergoes a hydraulic jump. If y1 = 1m and y3 = 40cm, estimate (a) V1, (b) V3, (c) y4 and (d) the bump height, h.

EXAMPLE #09 (10.95) A 10 cm high bump in a wide horizontal water channel creates a hydraulic jump just upstream. Neglecting losses except in the jump, for case y3 = 30cm, estimate (a) V4, (b) y4, (c) V1 and (d) y1.

EXAMPLE #10 (C10.6) Figure shows a horizontal flow of water through a sluice gate, a hydraulic jump, and over 6 ft sharp crested weir. Channel, gate, jump and weir are all 8 ft wide unfinished concrete. Determine (a) the flow rate, (b) the normal depth, (c) y2, (d) y3 and (e) y4.