For a given q:

• any y can exist, and each y is linked to a specific V and E.

• any V can exist, and each V is linked to a specific y and E.

• E must be greater than some minimum value, and if it is, each value can exist at two possible y’s and corresponding V’s. (Note: in uniform flow, y is constant, so E is constant. However, constancy of E does not assure uniform flow.

Specific Energy, E

Wat

er D

epth

, y

q4

q1q2q3

Review – Equations used thus far:

2

2VE yg

Specific Energy, E

Wat

er D

epth

, y

Q VA Vqb

• No mention of slope, roughness, hydraulic radius, etc.; i.e., nothing about channel itself.

• To incorporate information about the channel, need to consider energy equation and its relation to E.

Energy equation between two locations separated by distance L in an open channel with steady flow:

The change in specific energy between two locations equals the decline in elevation of the channel bottom minus the headloss. Thus, in uniform flow, E is constant because the increase in E due to elevation drop is exactly balanced by the decrease in E due to friction.

2 1 o 1 2 channelf Lbottom

E E S S L z z h

2 2

1 21 2 o2 2 fV Vy y S S Lg g

2 2

1 22 o 1 2 22 2f

V Vz S L y S L z yg g

What happens as water moves under conditions of non-uniform flow?

• Specific energy can increase or decrease, depending on relative magnitudes of Dz and hL.

• In a system with steady flow, the flow regime remains on the same E vs. y curve (constant qavg); a decrease in E leads to lower y and higher V for sub-critical flow, and the reverse for super-critical flow.

• If E declines below Emin for the given q, the flow rate leaving that location drops below q, and water backs up (‘damming’).

Specific Energy, E

Wat

er D

epth

, y

q4

q1q2q3

Example. What happens when water flowing sub-critically in an open channel encounters a step change in the channel bottom elevation?

2 21 1 2 2

1 22 2 Lp V p Vz z h

g g

Specific Energy, EW

ater

Dep

th, y

2 1 1 2 0channel Lbottom

E E z z h

Example. What happens if the step change gets bigger?

Specific Energy, E

Wat

er D

epth

, y

(3,4)

Specific Energy, E

Wat

er D

epth

, y

Specific Energy, E

Wat

er D

epth

, y

(1) (2) (3) (4)

Upstream Over Hump Downstream

(0,1,2,3)

(1)

(0)

(4)

(2)

(0,1,2)

(3) (4)

Modeling Hydraulic Jumps

What happens to y, V, Fr, and E across a hydraulic jump?

• y _Increases________

• V _Decreases________

• Fr _Decreases from >1 to <1

• E _Decreases________

How do we analyze acceleration/ deceleration in fluids?

Assuming that wall friction and gravitational forces are negligible compared to pressure forces over the distance of interest:

1 1 2 2 2 1xF h A h A Q V V 1 1 2 2xF h A h A 1 1 2 2 2 1 2 1xF h A h A Q V V Q V Vg

1 1 2 2 2 11h A h A Q V Vg

For rectangular channels, substitute 1 21 2

2 1

12 2y y q qy y q

g y y

2, , , :h y A by Q bq V q y

Typically: 25jumpL y

1 21 2

2 1

12 2y y q qy y q

g y y

2 22 21 2

1 22 2y yq q

gy gy

21

2 31

1 8 12y qy

gy

21Fr 2

2Fr2

21 3

2

1 8 12y qy

gy

For a given y1, this eqn is cubic in y2. But, one root is negative, and the constraint that y2 >yc allows us to choose between the other two and solve the equation as a quadratic.

For rectangular channels: 3

2 1

1 24L

y yh

y y

2 21 2

1 2 1 22 2LV Vh E E y yg g

Example. A hydraulic jump occurs in a 5-m-wide rectangular channel carrying 8 m3/s. The depth after the jump is 1.75 m.

(a) What is the depth before the jump?

(b) What are the losses of head and power across the jump?

a) 3 28 m /s m1.6

5 m sq

22

1 32

1 8 1 0.156 m2y qy

gy

3

3

kN m9.81 8 3.70 m 290 kWm sL LP Qh

1.59 myD

b) 3

2 1

1 2

3.70 m4L

y yh

y y

• Consider uniform flow with a CV in the shape of a rectangular box.

• Pressure forces upstream and downstream are equal and opposite, so only forces on the water are gravity and friction. Since water is not accelerating, these forces must also be equal and opposite.

• Pressure head and velocity head are constant over l, so loss of total head (hL) equals loss of gravitational head (Dz). Correspondingly, So= Sw= Sf.

osinw flow wettedA l P l

Gravitational force in direction of flow is Wsin; opposing frictional force is oPwettedl, where o is shear stress on the bottom, and Pwetted is the wetted perimeter. So:

o sin sinw floww h

wetted

A lR

P l

o is important for predicting scour of sediments, settled solids (e.g., in runoff or sanitary sewers), fish eggs, etc., from bottoms of open channels.

2 2

o * 2 8L

h

h f V f VSL D g R g

By analogy with pipe flow, assume that hL/l in open channels can be expressed as (f/D*)(V2/2g), where D* is a representative length; choose D* to be 4Rh. Then, for uniform flow:

Above equation is known as the Chezy equation (1768), and C as the Chezy coefficient. Equation is analogous to D-W eqn for pipe flow, with prediction of C the key to its usefulness.

o o8

h hgV R S C R Sf

Based on empirical results, Manning reported that C varies with Rh1/6, so:

2/3 1/2o

1hV R S

n1/61

hC Rn

Equation for V is the Manning Eqn (1890) , and n is the Manning coefficient. Increase in n corresponds to increased friction and hL. Values tabulated for a wide range of channel characteristics. Formally applicable only to uniform flow (where V and Rh are constant), but often applied to gradually varying flow to relate average values of Sf , V, and Rh in a reach.

n has units of s/m1/3, but is always reported as dimensionless. Using that convention, if Rh is in ft instead of m, coefficient becomes 1.49.

For a given channel shape (i.e., cross-section), Rh depends only on y, so a given y leads to one and only one value of V (or Q). This value of y is known as the uniform depth or normal depth for the specified V or Q.