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1 CTC 261 Review Hydraulic Devices Orifices Weirs Sluice Gates Siphons Outlets for Detention...

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1 CTC 261 Review Hydraulic Devices Orifices Weirs Sluice Gates Siphons Outlets for Detention Structures
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Page 1: 1 CTC 261 Review Hydraulic Devices Orifices Weirs Sluice Gates Siphons Outlets for Detention Structures.

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CTC 261 Review

Hydraulic Devices Orifices Weirs Sluice Gates Siphons Outlets for Detention Structures

Page 2: 1 CTC 261 Review Hydraulic Devices Orifices Weirs Sluice Gates Siphons Outlets for Detention Structures.

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Subjects

Open Channel Flow Uniform Flow (Manning’s Equation) Varied Flow

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Objectives

Know how to use Manning’s equation for uniform flow calculations

Know how to calculate Normal Depth

Know how to calculate Critical Depth

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Open Channel Flow Open to the atmosphere

Creek/ditch/gutter/pipe flow Uniform flow-EGL/HGL/Channel

Slope are parallel velocity/depth constant

Varied flow-EGL/HGL/Channel Slope not parallel velocity/depth not constant

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Uniform Flow in Open Channels

Water depth, flow area, Q and V distribution at all sections throughout the entire channel reach remains unchanged

The EGL, HGL and channel bottom lines are parallel to each other

No acceleration or deceleration

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Manning’s Equation Irish Engineer “On the Flow of Water in Open Channels

and Pipes” (1891) Empirical equation See more:

http://manning.sdsu.edu/ http://el.erdc.usace.army.mil/elpubs/pdf/

sr10.pdf#search=%22manning%20irish%20engineer%22

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Manning’s Equation-EnglishSolve for Flow

Q=AV=(1.486/n)(A)(Rh)2/3S1/2

Where:Q=flow rate (cfs)A=wetted cross-sectional area (ft2)Rh=Hydraulic Radius=A/WP (ft)

WP=Wetted Perimeter (ft)S=slope (ft/ft)n=friction coefficient (dimensionless)

Page 8: 1 CTC 261 Review Hydraulic Devices Orifices Weirs Sluice Gates Siphons Outlets for Detention Structures.

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Manning’s Equation-MetricSolve for Flow

Q=AV=(1/n)(A)(Rh)2/3S1/2

Where:Q=flow rate (cms)A=wetted cross-sectional area (m2)Rh=Hydraulic Radius=A/WP (m)

WP=Wetted Perimeter (m)S=slope (m/m)n=friction coefficient (dimensionless)

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Manning’s Equation-EnglishSolve for Velocity

V=(1.486/n)(Rh)2/3S1/2

Where:V=velocity (ft/sec)A=wetted cross-sectional area (ft2)Rh=Hydraulic Radius=A/WP (ft)

WP=Wetted Perimeter (ft)S=slope (ft/ft)n=friction coefficient (dimensionless)

Page 10: 1 CTC 261 Review Hydraulic Devices Orifices Weirs Sluice Gates Siphons Outlets for Detention Structures.

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Manning’s Equation-MetricSolve for Velcocity

V=(1/n)(Rh)2/3S1/2

Where:V=flow rate (meters/sec)A=wetted cross-sectional area (m2)Rh=Hydraulic Radius=A/WP (m)

WP=Wetted Perimeter (m)S=slope (m/m)n=friction coefficient (dimensionless)

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Manning’s Friction Coefficient

See Appendix A-1 of your book http://www.lmnoeng.com/

manningn.htm Typical values:

Concrete pipe: n=.013 CMP pipe: n=.024

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Triangular/Trapezoidal Channels

Must use trigonometry to determine area and wetted perimeters

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Pipe Flow

Hydraulic radii and wetted perimeters are easy to calculate if the pipe is flowing full or half-full

If pipe flow is at some other depth, then tables/figure are usually used

See Fig 7-3, pg 119 of your book

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Example-Find Q

Find the discharge of a rectangular channel 5’ wide w/ a 5% grade, flowing 1’ deep. The channel has a stone and weed bank (n=.035).

A=5 sf; WP=7’; Rh=0.714 ft

S=.05Q=38 cfs

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Example-Find S

A 3-m wide rectangular irrigation channel carries a discharge of 25.3 cms @ a uniform depth of 1.2m. Determine the slope of the channel if Manning’s n=.022

A=3.6 sm; WP=5.4m; Rh=0.667m

S=.041=4.1%

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Friction loss

How would you use Manning’s equation to estimate friction loss?

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Using Manning’s equation to estimate pipe size Size pipe for Q=39 cfs Assume full flow Assume concrete pipe on a 2%

grade Put Rh and A in terms of Dia. Solve for D=2.15 ft = 25.8” Choose a 27” or 30” RCP Also see Appendix A of your book

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Normal Depth

Given Q, the depth at which the water flows uniformly

Use Manning’s equation Must solve by trial/error (depth is in

area term and in hydraulic radius term)

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Normal Depth Example 7-3

Find normal depth in a 10.0-ft wide concrete rectangular channel having a slope of 0.015 ft/ft and carrying a flow of 400 cfs.

Assume: n=0.013

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Normal Depth Example 7-3

Assumed D (ft)

Area (sqft)

Peri. (ft)

Rh (ft)

Rh^.66 Q (cfs)

2.00 20 14 1.43 1.27 356

3.00 30 16 1.88 1.52 640

2.15 21.5 14.3 1.50 1.31 396

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Stream Rating Curve

Plot of Q versus depth (or WSE) Also called stage-discharge curve

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Specific Energy

Energy above channel bottom Depth of stream Velocity head

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Depth as a function of Specific Energy

Rectangular channel Width is 6’ Constant flow of 20 cfs

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Specific Energy D+v^2/2gStart Depth 0.2 ftDepth Increment 0.2 ftFlow 20 cfsRect Channel Width 6 ftg 32.2 ft/sec^2Critical Depth 0.70 ft

Depth Area Velocity Specific Energy0.20 1.20 16.67 4.510.40 2.40 8.33 1.480.60 3.60 5.56 1.080.80 4.80 4.17 1.071.00 6.00 3.33 1.171.20 7.20 2.78 1.321.40 8.40 2.38 1.491.60 9.60 2.08 1.671.80 10.80 1.85 1.852.00 12.00 1.67 2.042.20 13.20 1.52 2.242.40 14.40 1.39 2.432.60 15.60 1.28 2.632.80 16.80 1.19 2.823.00 18.00 1.11 3.02

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Specific Energy Curve

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

0.0 1.0 2.0 3.0 4.0 5.0

Specific Energy (ft)

Ch

ann

el D

epth

(ft

)

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Critical Depth

Depth at which specific energy is at a minimum

Other than critical depth, specific energy can occur at 2 different depths Subcritical (tranquil) flow d > dc

Supercritical (rapid) flow d < dc

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Critical Velocity

Velocity at critical depth

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Critical Slope

Slope that causes normal depth to coincide w/ critical depth

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Calculating Critical Depth

a3/T=Q2/g A=cross-sectional area (sq ft or sq m) T=top width of channel (ft/m) Q=flow rate (cfs or cms) g=gravitational constant (32.2/9.81)

Rectangular Channel—Solve Directly Other Channel Shape-Solve via trial & error

Page 31: 1 CTC 261 Review Hydraulic Devices Orifices Weirs Sluice Gates Siphons Outlets for Detention Structures.

Critical Depth (Rectangular Channel) Width of channel does not vary with

depth; therefore, critical depth (dc) can be solved for directly:

dc=(Q2/(g*w2))1/3

For all other channel shapes the top width varies with depth and the critical depth must be solved via trial and error (or via software like flowmaster)

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Froude Number F=Vel/(g*D).5

F=Froude # V=Velocity (fps or m/sec) D=hydraulic depth=a/T (ft or m) g=gravitational constant

F=1 (critical flow) F<1 (subcritical; tranquil flow) F>1 (supercritical; rapid flow)

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Varied Flow Rapidly Varied – depth and velocity

change rapidly over a short distance; can neglect friction hydraulic jump

Gradually varied – depth and velocity change over a long distance; must account for friction backwater curves

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Hydraulic Jump

Occurs when water goes from supercritical to subcritical flow

Abrupt rise in the surface water Increase in depth is always from

below the critical depth to above the critical depth

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Hydraulic Jump

Velocity and depth before jump (v1,y1) Velocity and depth after jump (v2,y2) Although not in your book, there are

various equations that relate these variables. Can also calculate the specific energy lost in the jump

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Hydraulic Jump http://www.engineering.usu.edu/classes/cee/3500/openchannel.htm

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Varied FlowSlope Categories

M-mild slope S-steep slope C-critical slope H-horizontal slope A-adverse slope

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Varied FlowZone Categories Zone 1

Actual depth is greater than normal and critical depth

Zone 2 Actual depth is between normal and critical

depth Zone 3

Actual depth is less than normal and critical depth

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Water-Surface ProfileClassifications

H2, H3 (no H1) M1, M2, M3 C1, C3 (no C2) S1, S2, S3 A2, A3 (no A1)

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Water Surface Profileshttp://www.fhwa.dot.gov/engineering/hydraulics/pubs/08090/04.cfm

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Water Surface Profiles-Change in Slopehttp://www.fhwa.dot.gov/engineering/hydraulics/pubs/08090/04.cfm

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Backwater Profiles Usually by computer methods

HEC-RAS Direct Step Method

Depth/Velocity known at some section (control section)

Assume small change in depth Standard Step Method

Depth and velocity known at control section Assume a small change in channel length


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