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

CJM

slide2

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.

CJM

open channel hydraulics5
Open Channel Hydraulics
  • Three basic relationships
    • Continuity equation
    • Energy equation
    • Momentum equation

CJM

continuity equation

3a

Inflow

3

A

Change in Storage

3b

Outflow

1

2

A

Section AA

Continuity Equation

Inflow – Outflow = Change in Storage

CJM

general flow equation

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

CJM

slide8

3

A

3

Section AA

1

2

A

Q1

Q2

Q1 – Q2 = Change in storage rate

CJM

energy equation bernoulli s

elevation head

velocity head

Energy loss between sections 1 and 2

pressure head

Energy Equation (Bernoulli’s)

CJM

open channel energy equation
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:

CJM

slide11

Q

Channel Bottom

z1

z2

Datum

2

1

CJM

slide12

HGL

y1

y2

Channel Bottom

z1

z2

Datum

1

2

CJM

slide13

EGL

HGL

y1

y2

Channel Bottom

z1

z2

Datum

2

1

CJM

slide14

hL1-2

EGL

HGL

y1

y2

Channel Bottom

z1

z2

Datum

1

2

CJM

specific energy15

A

E

y

Datum and channel bottom

A

Eqn. 2.2

Specific Energy

E is the specific energy.

CJM

specific energy cont

y

Section AA

q = vy

Where q is the flow per unit width.

(Eqn. 2.4)

Specific Energy (cont.)

Eqn. 2.2 becomes:

CJM

specific energy diagrams

y

y=E

y2

yc

y1

Ec

E

Eo

Specific Energy Diagrams

*Note q is constant.

CJM

critical depth y c

or

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

CJM

slide20
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.

CJM

slide21

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.

CJM

slide22

y

y=E

y1

y2

Dz

yc

E2

E1

E

Specific Energy Representation of a Transition

CJM

slide23
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.

CJM

other minimum specific energy relations
Other Minimum Specific Energy Relations
  • Froude Number: Minimum Specific Energy
  • Emin:yc
  • Wave speed of small disturbance

CJM

slide25
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

CJM

froude number f

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!

CJM

tie in with wave speed
……..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)

CJM

flow classification based on froude number
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.

CJM

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

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example 2 1
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).

CJM

slide31

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).

CJM

slide32

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

CJM

repeat for a step of 0 50 m
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

CJM

slide35

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
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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slide37

q1

y

q2>q1

y1

y2

E

CJM

choking condition
Choking Condition

q1

y

q2

y1

E

E1< E2min

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revised upstream condition
Revised Upstream Condition

q1

y

q2

y1

y1

THIS IS THE CONTROL!

y2

E

E1 = E2 = Emin (q2)

CJM

y c general form
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)

CJM

general channel shape
General Channel Shape

B(y)

dy

dA=Bdy

y

A(y)

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slide42
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)

CJM

slide43
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

CJM

example 2 2
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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slide45
F(y) = Q – (g1/2A3/2)/B1/2
  • YOYC provides solution

OR

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

CJM

overbank flow
Overbank Flow

Flood Stage

Normal Stage

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overbank flow calculate y c
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))

CJM

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

CJM