Specific Energy

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# Specific Energy - PowerPoint PPT Presentation

Specific Energy . Channel transitions Steps, cross-section Slope change Local loss locations Gate Piers Hydraulic Jump. EGL. v 2 /2g. y. y c. D z. datum. 1. 2. If we assume no energy loss through the transition we can say:.

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Presentation Transcript
Specific Energy
• Channel transitions
• Steps, cross-section
• Slope change
• Local loss locations
• Gate
• Piers
• Hydraulic Jump

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EGL

v2/2g

y

yc

Dz

datum

1

2

• If we assume no energy loss through the transition we can say:

which says for a constant total energy E, there is a specific energy loss between 1-2 of Dz.

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### Open Channel Hydraulics

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Open Channel Hydraulics
• Three basic relationships
• Continuity equation
• Energy equation
• Momentum equation

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3a

Inflow

3

A

Change in Storage

3b

Outflow

1

2

A

Section AA

Continuity Equation

Inflow – Outflow = Change in Storage

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Area of the cross-section

(ft2) or (m2)

Avg. velocity of flow at a cross-section (ft/s) or (m/s)

Flow rate (cfs) or (m3/s)

General Flow Equation

Q = vA

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3

A

3

Section AA

1

2

A

Q1

Q2

Q1 – Q2 = Change in storage rate

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Energy loss between sections 1 and 2

Energy Equation (Bernoulli’s)

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Open Channel Energy Equation
• In open channel flow (as opposed to pipe flow) the free water surface is exposed to the atmosphere so that p/g is 0 (on the surface; or p/g is y at the invert), leaving:

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Q

Channel Bottom

z1

z2

Datum

2

1

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HGL

y1

y2

Channel Bottom

z1

z2

Datum

1

2

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EGL

HGL

y1

y2

Channel Bottom

z1

z2

Datum

2

1

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hL1-2

EGL

HGL

y1

y2

Channel Bottom

z1

z2

Datum

1

2

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A

E

y

Datum and channel bottom

A

Eqn. 2.2

Specific Energy

E is the specific energy.

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y

Section AA

q = vy

Where q is the flow per unit width.

(Eqn. 2.4)

Specific Energy (cont.)

Eqn. 2.2 becomes:

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y

y=E

y2

yc

y1

Ec

E

Eo

Specific Energy Diagrams

*Note q is constant.

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or

Critical Depth, yc
• The depth of flow corresponding to the minimum E is the critical depth, yc

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### Channel Transitions

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Channel transitions occur where there is a change in width, shape, slope, roughness, bottom elevation of the channel.
• For changes in slope and roughness we can use backwater curves to evaluate the effects of the changes.
• For other types of smooth transitions (transitions were energy loss is minimal), we can use energy relationships to evaluate the impact of the transition.

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EGL

v2/2g

y

yc

Dz

datum

1

2

• If we assume no energy loss through the transition we can say:

which says for a constant total energy E, there is a specific energy loss between 1-2 of Dz.

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y

y=E

y1

y2

Dz

yc

E2

E1

E

Specific Energy Representation of a Transition

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If the flow must pass through critical depth, the assumption of no energy loss may not be valid.
• This is especially true when going from supercritical to subcritical flow.
• A hydraulic jump accompanied by considerable energy loss occurs.

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Other Minimum Specific Energy Relations
• Froude Number: Minimum Specific Energy
• Emin:yc
• Wave speed of small disturbance

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DERIVATIONS
• E = y + v2/2g
• E = y + q2/2gy2
• dE/dy = 0 = 1- q2/gyc3
• yc = (q2/g) 0.333……(eqtn 2.6)……..SO…..!!!!!……
• Emin = yc + q2/2gyc2 = 3/2 yc
• Emin = 3/2 yc

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is known as the Froude Number, F

Froude Number, F
• Compare to (2.6) : q2/gyc3 = 1
• Then, vc2/gyc = 1 at critical conditions
• So, at critical conditions, the Froude number =1!

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……..Tie in with wave speed
• Speed of a small disturbance on water can be shown to be: vw = (gy)1/2
• So, at critical conditions…surface wave has same velocity as the river (vw = vc)

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Flow classification based on Froude number
• If F = 1, y = yc and flow is critical.
• If F < 1, y > yc and flow is subcritical.
• If F > 1, y < yc and flow is supercritical.
• F is independent of the slope of the channel, yc dependent only on Q.

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Critical Step Height
• Figure 2.6: Approach Froude number vs non-dimensional step height (Dz/y1)

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Example 2.1

For an approach flow in a rectangular channel with depth of 2.0 m (6.6 ft) and velocity of 2.2 m/s (7.2 ft/s), determine the depth of flow over a gradual rise in the channel bottom of Dz = 0.25 m (0.82 ft). Repeat the solution for Dz = 0.50 m (1.64 ft).

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EGL

v2/2g

V1 = 2.2

Y1= 2.0

y2

Dz = 0.25

datum

1

2

• If we assume no energy loss through the transition we can say:

which says for a constant total energy (TE), there is a specific energy loss between 1-2 of Dz (meters).

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Steps

• Determine U/S Condition (sub- or super-)
• Calculate critical depth, compare to u/s depth
• Alternately, calculate F1
• Solve specific energy equation for correct root for y2, v2
• Water surface elevation = y2 + Dz

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Repeat for a step of 0.50 m
• Check value of Emin
• Determine value of E2 = (E1-Dz)
• Find E2<Emin (CAN’T BE!!!!)
• Flow “backs-up” to allow passage of q
• Y1 increases; conditions at step are critical, with y2 = yc; E2 = Emin

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Contraction/Expansion in Width

width1

width2

Plan View

1

2

• If we assume no energy loss through the transition we can say:

V12/2g + y1 = V22/2g + y2 OR

q12/2gy12 + y1 = E1 = E2 = q22/2gy22 + y2

q2 and q1 related by continuity .. So two equations; two unknowns

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Continuity
• q1w1 = q2w2 = Q
• q2 = q1 w1/w2

Now, find y2 (3 roots, 2 real…find one that matches with upstream flow condition)

Ensure that flow isn’t “choked” by contraction condition

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q1

y

q2>q1

y1

y2

E

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Choking Condition

q1

y

q2

y1

E

E1< E2min

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Revised Upstream Condition

q1

y

q2

y1

y1

THIS IS THE CONTROL!

y2

E

E1 = E2 = Emin (q2)

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yc, General Form
• E = y + Q2/2gA2
• Differentiate with respect to y and set DE/dy = 0; find minimum point; ie condition where E = Emin; y = yc
• dE/dy = 0 =1 – Q2/gA3 (dA/dy)
• 1 = Q2/gA3 (dA/dy)
• (if you can determine dA/dy, and can write A(y), can find yc)

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General Channel Shape

B(y)

dy

dA=Bdy

y

A(y)

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dA/dy = B (topwidth, B(y))
• So, Q2Bc/gAc3 = 1 (2.18)
• Define hydraulic depth D = A/B
• V = Q/A
• F = V/(gD)1/2 or V/(gD/a)1/2
• Minimum specific energy, Emin = yc + Dc/2
• (From 2.16 and 2.18)

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Solve for yc in a non-rectangular channel:
• aQ2Bc/gAc3 = 1
• Find value of yc that satisfies this equality
• Geometric Elements in Table 2-1

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Example 2.2
• Find the critical depth in a trapezoidal channel with a 20 ft bottom width and 2:1 side slopes if Q = 1000 cfs. Use the bisection technique and compare the solution with that from Figure 2.13

1

2

20’

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F(y) = Q – (g1/2A3/2)/B1/2
• YOYC provides solution

OR

• Graphical technique (Figure 2.13)
• yc = 3.740 feet

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Overbank Flow

Flood Stage

Normal Stage

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Overbank Flow: calculate yc
• Can’t neglect a
• Must consider variation of a withy
• dE/dy = 1 – aQ2B/gA3 + Q2/2gA2 da/dy
• (2.16 differentiated with A(y) and a(y))

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Overbank Flow (Compound Channel) continued
• Fc=(aQ2B/gA3 – Q2/2gA2 da/dy)0.5
• Use Ycomp!

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