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OPEN CHANNEL FLOW

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Question – What is the most obvious difference between pipe flow and open channel flow????????????? (in terms of flow conditions and energy situation)

Typical open channel shapes – Figure 14.1

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Types of open channel flows – 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 (how can this happen?) 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

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Figure 14.3

Section 1 – rapidly varying flow Section 2 – gradually varying flow Section 3 – hydraulic jump Section 4 – weir and waterfall Section 5 – gradually varying Section 6 – hydraulic drop due to change in channel slope

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Hydraulic radius of open channel flow A parameter that is used often Ratio of flow cross sectional area (A) and wetted perimeter (WP)

R = A/ WP Hydraulic radius R for various channel shapes – Figure 14.1

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Kinds (types) of open channel flow Reynolds number for pipe flow –

υvDN R =

Reynolds number for channel flow –

υvRNR =

For pipe flow – NR < 2000 – laminar NR > 4000 – turbulent For channel flow – NR < 500 – laminar NR > 2000 – turbulent

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Another “number” for channel flow! Froude Number [NF] (gravity versus inertial forces)

hF gy

vN =

Where yh is referred to as the hydraulic depth and given as –

yh = A/T where A is the area and T is the top width of the channel NF = 1.0 or when v = (gy)1/2 - critical flow NF < 1.0 – subcritical flow NF > 1.0 – super critical flow A combination of both the numbers is used to describe channel flow conditions.

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Uniform steady flow and Manning’s Equation When discharge remains 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

Or S = Sw

Where S is the slope of the channel The slope of the channel can be expressed as –

- An angle = 1 degrees - As percent = 1% - Or as fraction = 0.01 or 1 in 100

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Velocity of flow (v) in a channel can be computed numerous empirical equations – One of them is Mannings equation –

2/13/20.1 SRn

v =

This the SI units form of the equation with v (meters/sec) and R (meters). Where n is the Manning’s coefficient (dimensionless) – values developed through experimentation Possible n values for various channel surfaces – Table 14.1

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In English units the Manning’s equation form is –

2/13/249.1 SRn

v =

Where v is in feet/sec and the R value is in feet. If velocity is known, the discharge (Q) can then be computed as –

Q = A*v

2/13/20.1 SARn

Q =

Where Q is in m3/s For uniform flow, Q is referred to as Normal discharge

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The above equation can also be re-arranged such that –

2/13/2

SnQAR =

The left hand term is simply based on channel geometry.

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Problem 14.2 Determine normal discharge for a 200 mm inside diameter common clay drainage tile running half-full if the slope drops 1 m over 1000 m.

S = 1/1000 = 0.001 A = (1/2) * (π D2/4) = 0.5*π*(0.2)2/4 = 0.0157 m2 WP = (1/2) * (π D) = 0.5*π*0.2 = 0.3141 m R = 0.05 m From Table 14.1 n for clay tile = 0.013 Substitute these values in the equation –

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2/13/20.1 SARn

Q = And we get

2/13/2 )001.0()05.0(*0157.0*013.00.1

=Q Q = 5.18 x 10-3 m3/s

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Problem 14.3 Calculate slope of channel below If Q = 50 ft3/s Formed unfinished concrete channel

Equation that you will use

2/13/20.1 SARn

Q =

Or

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3/22/1

49.1 ARQnS =

Compute A = 12 ft2 WP = 9.66 ft R = A/WP = 12/9.66 = 1.24 ft Manning’s n for concrete channel = 0.017 Substitute And S = 0.00169 Drop 1.69 ft for every 1000 ft.

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Problem 14.4 Design rectangular channel in formed unfinished concrete Q = 5.75 m3/s S = 1.2% Normal depth = ½ of the width of the channel Since we have to design the channel – the equation that should be used –

2/13/2

SnQAR =

RHS is known. RHS = 0.017*5.75/(0.012)1/2 = 0.0892 Now we know that y = b/2 Express Area and the hydraulic radius in terms of b. A = by = b2/2

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WP = b+ 2y = 2b R = A/WP = b/4 Therefore, LHS = AR2/3 = b2/2 * (b/4) 2/3 = RHS = 0.892 B = 1.76 m y = 1.76/2 m

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Problem 14.5 In the problem above the final width was set at = 2m and the maximum Q = 12 m3/s; find the normal depth for this maximum discharge. OK again,

2/13/2

SnQAR =

RHS = 0.017*12/(0.012)1/2 = 1.86 B = 2m A = 2y WP = 2+2y R = 2y/(2+2y) Therefore LHS =

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86.122

223/2

=⎟⎟⎠

⎞⎜⎜⎝

⎛+ y

yy

Cannot solve this directly, will have to do trial and error. Set up a Table and compare y (m) A (m2) WP

(m) R (m) R2/3 AR2/3 Required

change in y

2.0 4.0 6.0 0.667 0.763 3.05 Make y lower

1.5 3.0 5.0 0.600 0.711 2.13 Make y lower

1.35 2.7 4.7 0.574 0.691 1.86 OK

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Conveyance and most efficient channel shapes Look at the RHS of the equation

2/13/20.1 SARn

Q =

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

3/20.1 ARn

K =

OR

2/1SKQ =

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K is maximum when WP is the least for a give area

this is also the most efficient cross section for conveying flow

For circular section – half full flow is the most efficient For other shapes – see Table 14.3 from the text.

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Compound Sections When channel shape changes with flow depth – typical in natural stream sections during flooding During floods – water spills over the flood plain You need to know Q at various depths or vice-versa – so that you can design channels or determine channel safety for various flood magnitudes

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Cazenovia Creek in Buffalo during “normal” flow conditions

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Cazenovia Creek during flood!

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Problem 14.21E Figure 14.21 – natural channel with levees

Channel – earth with grass cover, n = 0.04 S = 0.00015 Determine normal Q for depth = 3 and 6 ft.

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Assignment # 9

- 14.3E - 14.9M - 14.10M - 14.14M

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Compound section – More realistic situation – channel roughness n may be different for floodplain than the main channel Why?????

In that case

- determine velocity for each sub section - and then sum up the discharges for the sections

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2/13/2

49.1 SPA

nv

i

i

ii ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

i

n

ii AVQ ∑

=

=1

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Example Problem –

Compute discharge for depth of 8 feet S = 0.5% n for bank = 0.06 n for main channel = 0.03 A1 = 80*4 = 320 A2 = 50*8 = 400 A3 = 100*5 = 500 P1 = 80+4 = 84 P2 = 4+50+3 = 57 P3 = 100+5 = 105

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2/13/2

49.1 SPA

nv

i

i

ii ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

i

n

ii AVQ ∑

=

=1

⎥⎦

⎤⎢⎣

⎡++=

06.0500)105/500(

03.0400)57/400(

06.0320)84/320()005.0(49.1

3/23/23/22/1Q

Q = 9010 cfs.

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Energy Principles for Open Channel flow Energy at a particular point in the channel = PE + KE

gvyE2

2

+=

Where y is the depth of flow and v is the velocity Note – no pressure term! This is energy with respect to the channel bottom – Specific Energy When energy is measured with respect to another fixed datum – Total energy

gvzyE2

2

++=

Where z is the height of the channel bottom from the datum In terms of Q the specific energy can be expressed as –

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2

2

2gAQyE +=

Where Q is the discharge and A is the cross-sectional flow area

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Example Problem Channel width (rectangular) = 2m Depth = 1m Q = 4.0 m3/s Height above datum = 2m Compute specific and total energy A = by = 2.0*1.0 = 2 m2 Specific energy =

2

2

2gAQyE +=

2

2

2*81.9*241 +=E

E = 1.20 m Total energy = = Datum height + specific energy = 2.0 + 1.20 = 3.20 m

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

2

2

2gAQyE +=

E = Es + Ek Es = y (static energy) Ek = Q2/2gA2 (kinetic energy) Relationship between y and Es & Ek

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Combining the two relationships – specific energy diagram

Key points from the diagram –

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

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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.

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Additional Equations Specific energy equation –

2

2

2gAQyE +=

Taking a derivative and equating it to zero (critical flow conditions when energy is minimum) We get – Condition at critical flow

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2

=gA

BQ

Solving these further, for a rectangular channel (A = By), we get –

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2

=cgy

q

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Or

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2

gqyc =

Critical depth can be determined explicitly Also, for rectangular channel -

cyE23

min = Explicit equations that can quickly give you the critical depth and minimum specific energy for a rectangular channel – no need to interpolate from graph

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Example Problem: • Rectangular channel • Width = 4 m • Q = 12.0 m3/s • Depth of flow = 2.5 m

• Draw specific energy diagram • Find critical and alternate depth

2

2

2gAQyE +=

2

2

2gyqyE +=

Specific discharge – discharge per unit width = q = Q/B The advantage of using specific discharge is that we avoid using B and relate q directly to y q = 12/4 = 3m2/s

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Set up a table and compute the specific energy for every 0.2m depth increment.

2

2

2gyqyE +=

y KE total E

0.20 11.47 11.67 0.40 2.87 3.27 0.60 1.27 1.87 0.80 0.72 1.52 1.00 0.46 1.46 1.20 0.32 1.52 1.40 0.23 1.63 1.60 0.18 1.78 1.80 0.14 1.94 2.00 0.11 2.11 2.20 0.09 2.29 2.40 0.08 2.48 2.60 0.07 2.67 2.80 0.06 2.86 3.00 0.05 3.05 3.20 0.04 3.24 3.40 0.04 3.44 3.60 0.04 3.64 3.80 0.03 3.83 4.00 0.03 4.03

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00.5

11.5

22.5

33.5

44.5

5

0 1 2 3 4 5

E

y

00.5

11.5

22.5

33.5

44.5

5

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

E

y

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Explicit computation –

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2

=cgy

q

Or

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2

gqyc =

971.081.9

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2

==cy

cyE23

min =

457.1971.0*23

min ==E

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Since given depth 2.5 m > 0.971 – the given depth is subcritical and the other depth should be supercritical Now determining alternate depths – Energy at 2.5 m =

2

2

2gyqyE +=

57.25.2*81.9*2

35.2 2

2

=+=E

This energy value is the same for the other alternate (supercritical) depth, so –

2

2

*81.9*2357.2

yy +=

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Solve equation by trail and error y E

0.40 3.27 0.41 3.14 0.42 3.02 0.43 2.91 0.44 2.81 0.45 2.72 0.46 2.63 0.47 2.55 0.48 2.47 0.49 2.40 0.50 2.33 0.51 2.27 0.52 2.22 0.53 2.16 0.54 2.11

y = 0.467 m – supercritical alternate depth.

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Problem 14.41E from text GIVEN – • Triangular channel with side slopes having ratio of 1:1.5

• Q = 0.68 ft3/s

• Channel – clean, excavated earth

CALCULATE –

a. critical depth b. Emin c. Plot specific energy curve d. Determine energy for 0.25 ft and alternate depth e. Velocity of flow and Froude number f. Calculate required slopes if depths from d are to be normal

depths for given flow

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Solution –

y, A = zy2, v = Q/A, T =2zy, yh=A/T ,

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y, A = zy2, v = Q/A, T =2zy, yh=A/T ,

hF gy

vN =

2

2

2gAQyE +=

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Assignment # 10 - Specific Energy of Channel Flow • 14.39M • 14.42E