Uniform Flow in Open Channel - University of Memphis Flow in Open Channel.… · 2 Flow in Open...

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Civil Engineering Hydraulics

Open Channel Flow Uniform Flow

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Flow in Open Channel – Flow Conditions

¢ While we used the Froude Number to calculate the critical flow conditions, the Reynolds number is still used to characterize the flow as laminar, transitional, or turbulent

Uniform Flow 2 Friday, November 2, 2012

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¢ The calculation of the Reynolds Number for Open Channel flow is a bit different than for closed conduits.



¢ For our purposes, we will assume that transition occurs at a Reynolds number of 1000.

¢ Most common open-channel flows are turbulent.


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Friday, November 2, 2012 Uniform Flow 5

Open Channel ¢  If we consider flow in an open channel

Some of the notation in this section may be different from that we have used earlier. This material is taken from a different text. I will try and point out where the differences are.


Open Channel ¢  If we consider flow in an open channel

In this case, the z is taken to the bottom of the channel from some reference datum and y is the depth of flow from the bottom of the channel.

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Open Channel ¢ The bottom of the channel is sloped at

some angle α in the direction of flow


Open Channel ¢  It is very common to have the slope of the

channel expressed as a fall of so many feet per 100 or 1000 feet

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Open Channel ¢  If we choose two points in the flow (1) and

(2) we can write a continuity expression at the two points.


Open Channel ¢ At point 1, we have a velocity v1 and a

depth of flow (relative to the channel bottom) of y1

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Open Channel ¢ So we can write the specific energy as

y1 +

v12

2g


Open Channel ¢ There is also a potential energy term that is

the elevation of the channel bottom above the datum

¢ Adding this we have

y1 +

v12

2g+ z1

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Open Channel ¢ We can follow the same steps at point 2 to

determine the energy there

y1 +v1

2

2g+ z1

y2 +v2

2

2g+ z2


Open Channel ¢ The energy line in the drawing represents

the sum of the specific and potential energy with respect to the datum.

y1 +v1

2

2g+ z1

y2 +v2

2

2g+ z2

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Open Channel ¢  If we consider friction losses between the

two points ¢ Remember hL is negative

y1 +v1

2

2g+ z1 + hL

y2 +v2

2

2g+ z2


Open Channel ¢ For continuity, we can equate the energy

values at the two points.

y1 +v1

2

2g+ z1 + hL =

y2 +v2

2

2g+ z2

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Open Channel ¢  z2 can be expressed in terms of z1, the

length of the channel between points 1 and 2, and the slope of the channel.

y1 +v1

2

2g+ z1 + hL = y2 +

v22

2g+ z2

z2 = z1 − Lsinα


Open Channel ¢ Substituting for z2 in the energy equation

y1 +v1

2

2g+ z1 + hL = y2 +

v22

2g+ z1 − Lsinα

z2 = z1 − Lsinα

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Open Channel ¢ Manipulating the expression we then have

y1 +

v12

2g+ hL = y2 +

v22

2g− Lsinα


Open Channel ¢  If we constrain α to very small values, we

can use the approximation

y1 +v1

2

2g+ hL = y2 +

v22

2g− Lsinα

sinα ≅ tanα

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Open Channel ¢ So we can move the L sin α to the other

side of the expression and make the substitution of tan for sin

y1 +

v12

2g+ hL + L tanα = y2 +

v22

2g


Open Channel ¢ The slope, and therefore the tangent of α, is

given as S0 so the expression becomes

y1 +

v12

2g+ hL +S0L = y2 +

v22

2g

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Open Channel ¢  If the rate of head loss due to friction is

constant, we can define a slope for the energy line, Sf

y1 +

v12

2g+ hL +S0L = y2 +

v22

2g


Open Channel ¢ Assuming a constant rate of head loss with

distance the slope of the energy line would be

Sf =hL

L

y1 +v1

2

2g+SfL +S0L = y2 +

v22

2g

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Open Channel ¢ Now if we consider the flow to be uniform

Sf =hL

L

y1 +v1

2

2g+SfL +S0L = y2 +

v22

2g


Open Channel ¢ Uniform flow exists if the depth is uniform

(i.e., unchanging) at every section of the channel.

¢ Uniform flow occurs when volume flow rate does not change during the interval of interest.

Sf =hL

L

y1 +v1

2

2g+SfL +S0L = y2 +

v22

2g

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Forces in Uniform Flow

¢  If we consider the forces acting on an element of a uniform flow. We will have diagram similar to what is shown below.



¢ The element goes from the free surface to the contact with the channel

¢  It has a cross-sectional area in the zy plane of A


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¢ There are forces developed by the pressure acting on the upstream and downstream faces of the element.

¢ The partial differential represents the change in pressure in the downstream direction.



¢ There is no shear force acting at the top of the element as it is only in contact with a free surface (air)

¢ The shear force on the bottom is from the contact with the channel bottom and sides.


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¢ The final forces on the diagram represent the weight on the element and how it is broken into components along the z and x axis.



¢ Since uniform flow is considered as a steady condition, there is no acceleration to the element.

¢ Also there is no change in the pressure in the x direction.


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¢ Therefore


∂p∂x

= 0

Fx∑ = 0


¢ So


−τwdAc + ρgAdxsin θ( ) = 0

Ac = Area of contact with the channel

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¢ The area of contact is the wetted perimeter times the distance along the channel.


−τwdAc + ρgAdxsin θ( ) = 0

Ac = Px

−τwPdxc + ρgAdxsin θ( ) = 0


¢  Isolating the shear


τw =ρgAdxsin θ( )

Pdx

τw =ρgAsin θ( )

P

τw = ρgsin θ( ) AP

τw = ρgsin θ( )Rh

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¢ The shear can be expressed in terms of the velocity of the flow, the density, and the friction at the contact surface


τw = ρgsin θ( )Rh

τw = f4ρv 2

2f4ρv 2

2= ρgsin θ( )Rh


¢  Isolating the velocity term


f4ρv 2

2= ρgsin θ( )Rh

v 2 =8ρgsin θ( )Rh

f ρ

v =8gsin θ( )Rh

f

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Forces in Uniform Flow ¢ The θ in this expression

is the same as the α used earlier.

¢ So the sin(θ) is equal to the channel slope.


f4ρv 2

2= ρgsin θ( )Rh


f ρ

v =8gsin θ( )Rh

f

Forces in Uniform Flow ¢  In pipe flow, the f is

dependent only on the material of the pipe.

¢ Open channels are more complex.


f4ρv 2

2= ρgsin θ( )Rh


f ρ

v =8gsin θ( )Rh

f

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Forces in Uniform Flow ¢ The f can depend of the variable geometry of the

channel, the roughness of the surface, and changes in alignment of the channel.


v =

8gsin θ( )Rh

f

Forces in Uniform Flow ¢ To reflect this, a different term was developed

for the roughness ¢  It is known as the Manning roughness coefficient

and is related to f by


8gf

= 1n

Rh

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Uniform Flow ¢ Returning to the channel with uniform flow

both the depth of flow and the velocity are the same at points 1 and 2

y1 +v1

2

2g+SfL +S0L = y2 +

v22

2gy1 = y2

v1 = v2


Uniform Flow ¢  In this case, the right and left hand sides of

the expression we developed reduces to

SfL +S0L = 0SfL = −S0L

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Normal Flow ¢ The slope of the head loss due to friction is

exactly equal to the slope of the channel ¢ Remember this is for normal/uniform flow

SfL +S0L = 0SfL = −S0LSf = −S0


Normal Flow ¢ We calculated the head loss over a length

of conveyance in the work we did on pipes ¢ That hasn’t changed

hL = f L

Dh

v 2

2g

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Normal Flow ¢ Dh is the hydraulic diameter and was

defined as 4 times the wetted area divided by the wetted perimeter

S0L = SfLS0 = Sf

hL = f LDh

v 2

2g


Normal Flow ¢ Dh is the same as 4 times the hydraulic

radius, Rh

Sf = −SO

hL = f LDh

v 2

2g= f L

4Rh

v 2

2g

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Normal Flow ¢ Sf was the ratio of hL to L, so we have

Sf = −SO

hL = f LDh

v 2

2g= f L

4Rh

v 2

2g

hL

L=

f L4Rh

v 2

2gL


Normal Flow ¢ So the slope can be equated to

Sf = −SO

SO = −hL

L= −

f L4Rh

v 2

2gL

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Normal Flow ¢ Combining the two expression we derived

S0 = −f v 2

8gRh

8gf

= 1n

Rh

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8gf

= 1n2 Rh

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v 2 =S08gRh

f

v 2 =S08gRh

8g1n2 Rh

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v 2 = S0

1n2 Rh

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v = 1n

Rh

23S0

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Normal Flow ¢ The final expression is known as the

Manning Equation

v = 1

nRh

23S0

12

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Normal Flow ¢ R is given in meters, and v is in meters/sec ¢ To achieve dimensional consistency, the

units of 1/n are m1/3/s ¢ You text gives units of n as L1/6 but the

units of 1/n above are consistent

v = 1

nRh

23S0

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Normal Flow ¢ The Manning equation in USCS units is

v = 1.486

nRh

23S0

12

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Normal Flow ¢ The units of R are feet and v are in feet/

second ¢ The units of 1.486/n are ft1/3/s

v = 1.486

nRh

23S0

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Normal Flow ¢ Values for n are found in tables such as

Table 7.2 in your text.

v = 1.486n

Rh

23S0

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v = 1n

Rh

23S0

12

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13–54 A trapezoidal channel with a bottom width of 5 m, free surface width of 10 m, and flow depth of 2.2 m discharges water at a rate of 120 m3/s. If the surfaces of the channel are lined with asphalt (n=0.016), determine the elevation drop of the channel per km.

We know the geometry of the channel, the flow depth, and the rate of flow along with the Manning coefficient for the wetted surfaces.

We can start by writing the Manning equation and see is we can isolate the slope, S0.



Since the units are in SI, we will use the SI version of the Manning equation.

v = 1

nRh

23S0

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Isolating S0 we have.

nv

Rh

23

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

2

= S0



We can calculate the velocity from the flow rate if we first calculate the cross sectional area.

⎛ ⎞⎜ ⎟

=⎜ ⎟⎜ ⎟⎝ ⎠

2

023h

QnA SR

widthtop 10m:= widthbottom 5m:= yn 2.2m:=

Q 120m3

sec:= n 0.016:=

Areawidthtop widthbottom+

2yn⋅:= Area 16.5 m2=

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13–54 A trapezoidal channel with a bottom width of 5 m, free surface width of 10 m, and flow depth of 2.2 m discharges water at a rate of 120 m3/s. If the surfaces of the channel are lined with asphalt (n=0.016), determine the elevation drop of the channel per km. Since we know the flow

area, all we need is the wetted perimeter to calculate the hydraulic radius.

⎛ ⎞⎜ ⎟

=⎜ ⎟⎜ ⎟⎝ ⎠

2

023h

QnA SR

Assume that the channel slopes are the same on both sides of the channel.

Lengthsidewidthtop widthbottom−

2

⎛⎜⎝

⎞⎟⎠

2

yn2+:= Lengthside 3.33 m=



⎛ ⎞⎜ ⎟

=⎜ ⎟⎜ ⎟⎝ ⎠

2

023h

QnA SR

WP 2 Lengthside⋅ widthbottom+:= WP 11.66 m=

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⎛ ⎞⎜ ⎟

=⎜ ⎟⎜ ⎟⎝ ⎠

2

023h

QnA SRS0

nQArea⋅

AreaWP

⎛⎜⎝

⎞⎟⎠

⎛⎜⎝

⎞⎟⎠

23 m

13

sec⋅

⎡⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎦

2

:= S0 8.523 10 3−×=



This is the elevation drop per meter of length, the question asked for the drop per kilometer.

⎛ ⎞⎜ ⎟

=⎜ ⎟⎜ ⎟⎝ ⎠

2

023h

QnA SR

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Homework 23-1

¢  Cooling water used in condensers of power plants is sometimes conveyed to cooling ponds before being returned to the body of water from which it came. One such pond is fed by a channel that is rectangular in cross section, 3 m wide, and formed by dredging. The channel is well maintained, uniform, and weathered. If the water depth is 2 m and the channel slope is 0.09, calculate the flow rate in the channel.


Homework 23-2

¢  A trapezoidal channel carries water at a flow of 200 ft3/s. Its width is 8 ft, and m = 1.0. The channel is made of riveted steel. l  Calculate the slope necessary to maintain a uniform depth of

3 ft. l  Calculate the critical depth of the flow.

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Homework 23-3

¢  Water is flowing in a 6 m diameter culvert at a depth of 4.7 m.The culvert is uniform, made of concrete, and laid on a slope of 1°. Determine the volume flow rate through the culvert. HINT(The expressions from Wednesday’s class might be useful here).

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