Brays Bayou

Brays Bayou

Concrete Channel

CH 7 - Open Channel Flow

1. Uniform & Steady

2. Non-uniform and Steady

3. Non-Uniform and Unsteady

Open Channel Flow

1. Uniform & Steady - Manning’s Eqn in prismatic channel - Q, v, y, A, P, B, S and roughness are all constant

2. Critical flow - Specific Energy Eqn (Froude No.)

3. Non-uniform flow - gradually varied flow (steady flow) - determination of floodplains

4. Unsteady and Non-uniform flow - flood waves

Uniform Open Channel Flow

Manning’s Eqn for velocity or flow

€

v =

1

n

R

2 / 3

S S.I. units

v =

1 . 49

n

R

2 / 3

S English units

where n = Manning’s roughness

coefficient R = hydraulic radius = A/PS = channel slope

Q = flow rate (cfs) = v A

Normal depth is function of flow rate, and

geometry and slope. One usually solves for normal

depth or width given flow rate and slope information

B

b

Normal depth implies that flow rate, velocity, depth,

bottom slope, area, top width, and roughness remain

constant within a prismatic channel as shown below

A

Q = ConstV = Cy = CS0 = CA = CB = Cn = C

UNIFORM FLOW

y

So

B

V

Common Geometric Properties Cot = z/1

z

1

1 + Z2

Optimal Channels - Max R and Min P

Uniform FlowEnergy slope = Bed slope or dH/dx = dz/dx

Water surface slope = Bed slope = dy/dz = dz/dx Velocity and depth remain constant with x

H = z + y + v2/2g = Total Energy

E = y + v2/2g = Specific Energy

often near 1.0 for most channels

H

= vi2 Qi

V2 QT

Energy Coeff.

My son Eric

Critical Depth and Flow

Critical depth is used to characterize channel flows -- based on addressing specific energy E = y + v2/2g :

E = y + Q2/2gA2 where Q/A = q/y and q = Q/b

Take dE/dy = (1 – q2/gy3) and set = 0. q = const

E = y + q2/2gy2

y

E

Min E Condition, q = C

Solving dE/dy = (1 – q2/gy3) and set = 0.

For a rectangular channel bottom width b,

1. Emin = 3/2Yc for critical depth y = yc

2. yc/2 = Vc2/2g

3. yc = (Q2/gb2)1/3

Froude No. = v/(gy)1/2

We use the Froude No. to characterize critical flows

Y vs E E = y + q2/2gy2

q = const

In general for any channel shape, B = top width

(Q2/g) = (A3/B) at y = yc

Finally Fr = v/(gy)1/2 = Froude No.

Fr = 1 for critical flowFr < 1 for subcritical flowFr > 1 for supercritical flow

Critical Flow in Open Channels

Non-Uniform Open Channel Flow

With natural or man-made channels, the shape, size, and slope may vary along the stream length, x. In addition, velocity and flow rate may also vary with x. Non-uniform flow can be best approximated using a numerical method called the Standard Step Method.

Non-Uniform Computations

Typically start at downstream end with known water level - yo.

Proceed upstream with calculations using new water levels as they are computed.

The limits of calculation range between normal and critical depths. In the case of mild slopes, calculations start downstream.In the case of steep slopes, calculations start upstream.

Q

Calc.

Mild Slope

Non-Uniform Open Channel Flow

Let’s evaluate H, total energy, as a function of x.

H = z + y + v

2

/ 2 g

( )

dH

dx

=

dz

dx

+

dy

dx

+

2 g

dv

2

dx

⎛

⎝

⎜

⎞

⎠

⎟

Where H = total energy headz = elevation head,

v2/2g = velocity head

Take derivative,

Replace terms for various values of S and So. Let v = q/y = flow/unit width - solve for dy/dx, the slope of the water surface

– S = − S

o

+

dy

dx

1 −

q

2

gy

3

⎡

⎣

⎢

⎤

⎦

⎥

since v = q / y

1

2 g

d

dx

v

2

[ ]=

1

2 g

d

dx

q

2

y

2

⎡

⎣

⎢

⎤

⎦

⎥

= −

q

2

g

1

y

3

⎡

⎣

⎢

⎤

⎦

⎥

dy

dx

Given the Froude number, we can simplify and solve for dy/dx as a fcn of measurable parameters

Fr

2

= v

2

/ gy

( )

dy

dx

=

S

o

− S

1 − v

2

/ gy

=

S

o

− S

1 − Fr

2

where S = total energy slopeSo = bed slope, dy/dx = water surface slope

*Note that the eqn blows up when Fr = 1 and goes to   zero if So = S, the case of uniform OCF.

Mild Slopes where - Yn > Yc

Uniform Depth

Yn > Yc

Now apply Energy Eqn. for a reach of length L

y

1

+

v

1

2

2 g

⎡

⎣

⎢

⎤

⎦

⎥

= y

2

+

v

2

2

2 g

⎡

⎣

⎢

⎤

⎦

⎥

+ S − S

o

( )L

L =

y

1

+

v

1

2

2 g

⎡

⎣

⎢

⎤

⎦

⎥

− y

2

+

v

2

2

2 g

⎡

⎣

⎢

⎤

⎦

⎥

S − S

0

This Eqn is the basis for the Standard Step Method Solve for L = x to compute water surface profiles as function of y1 and y2, v1 and v2, and S and S0

Backwater Profiles - Mild Slope Cases

x

Backwater Profiles - Compute Numerically

Computey3 y2 y1

Routine Backwater Calculations1. Select Y1 (starting depth)

2. Calculate A1 (cross sectional area)

3. Calculate P1 (wetted perimeter)

4. Calculate R1 = A1/P1

5. Calculate V1 = Q1/A1

6. Select Y2 (ending depth)

7. Calculate A2

8. Calculate P2

9. Calculate R2 = A2/P2

10. Calculate V2 = Q2/A2

Backwater Calculations (cont’d)

1. Prepare a table of values

2. Calculate Vm = (V1 + V2) / 2

3. Calculate Rm = (R1 + R2) / 2

4. Calculate Manning’s

5. Calculate L = ∆X from first equation

6. X = ∑∆Xi for each stream reach (SEE SPREADSHEETS)

€

S =nVm

1.49Rm2

3

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2

€

L =

y1 + v12

2g

⎛

⎝ ⎜

⎞

⎠ ⎟−y2 + v2

2

2g

⎛

⎝ ⎜

⎞

⎠ ⎟

S − S0

Energy Slope Approx.

100 Year Floodplain

Main Stream

Tributary

Cross Sections

Cross Sections

A

B

C

D

QA

QD

QC

QB

Bridge Section

Bridge

Floodplain

The Floodplain

Top Width

Floodplain Determination

The Woodlands planners wanted to design the community to withstand a 100-year storm.

In doing this, they would attempt to minimize any changes to the existing, undeveloped floodplain as development proceeded through time.

The Woodlands

HEC RAS or (HEC-2)is a computer model designed fornatural cross sections in natural rivers. It solves the governing equations for the standard step method, generally in a downstream to upstream direction. It can Also handle the presence of bridges, culverts, and variable roughness, flow rate, depth, and velocity.

HEC RAS (River Analysis System, 1995)

HEC - 2

Orientation - looking downstream

Multiple Cross Sections

River

HEC RAS (River Analysis System, 1995)

HEC RAS Bridge CS

HEC RAS Input Window

HEC RAS Profile Plots

3-D Floodplain

HEC RAS Cross Section Output Table

Brays Bayou-Typical Urban System• Bridges cause unique problems in hydraulics

Piers, low chords, and top of road is considered

Expansion/contraction can cause hydraulic losses

Several cross sections are needed for a bridge

288 Bridge causes a 2 ft

Backup at TMC and is being replaced by TXDOT

288 Crossing