(N)CEWB222 Chapter 4 - Specific Energy

7/24/2019 (N)CEWB222 Chapter 4 - Specific Energy

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CEWB222 HYDRAULIC

ENGINEERING

Chapter 4 - Specific Energy

(Alternate Depths)

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The Energy Principles

ENERGY EQUATION

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Lh Z g

Where:Z1 = elevation of streamline above the datum at cross-Section 1.

P1 = Pressure

= Fluid specific weight

P/ = Pressure head

= streamline velocityV2/2g = velocity head

g = local acceleration of gravity

hL= hydraulic losses

= correction factors

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consider the special case where the channel bottom

is horizontal (S0 =0); HL= 0 and 1 = 2 = 1.0

ENERGY EQUATION

and Y z

X S Y z

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Apply Continuity Equation:

ENERGY EQUATION

For rectangular channel or wide:

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Channel Bottom

z E z y gA

V Head

22Specific

Energy

ENERGY IN OPEN CHANNELS

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Specific Energy, EE = energy (head) measured with respect to

the channel bottom

Multiply through by y2 and arrange to find

y3 – E y2 + q2/2g = 0

What kind of equation is this? How

many roots? Significance of roots?

SPECIFIC ENERGY

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SPECIFIC ENERGY CURVEFor a given specific energy, there are 2

possible depths, low stage Y1 and high stage

. The low stage is the alternate depth of the

high stage and vice versa.

-The specific energy is a minimum at point C.

-The critical depth of flow , Yc corresponds to

this condition of minimum specific energy.

When the depth of flow is lesser than the

critical depth, Y1 <Yc and the velocity of flow isalso greater than the critical velocity

V1 >Vc

Therefore the flow is supercritical flow

When the depth of flow is greater than the

critical depth,

Y2 > Yc and the velocity of flow is

also less than the critical velocity

V2 <VC

Therefore the flow is subcritical flow

The position of the specific energy curve

changes when discharge changes.

Specific Energy Curve

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Specific Energy Curve

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Specific Energy Diagram

0.0 2.0 4.0 6.0 8.0 10.0

Specific Energy (ft)

Depth(f

Rectangular

channel:

B = 20 ft

Q = 600 ft3/s.Fr = 1

Fr = 2

Fr = V/(gy)

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Critical Flow & Depth (Yc)

Critical flow occurs when the specific energy is minimum for a given discharge.The depth for this condition may be determined if we solve for dy/dE

E = y + Q2 /2gA2 and

dE/dy = 0 (set dE/dy equal to zero)

Where, dA=Tdy

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Critical Flow & Depth (Yc)

General

Geometry

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Critical Flow for Rectangular Channel

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Specific Energy Equations

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EXAMPLE 1Determine the critical depth for water flowing at 10 m3 /s

in a trapezoidal channel with bottom width 3 m and sideslopes of 2:1 (H:V).

T=3+4y

A=3y+2y2

Solution:

Under critical flow conditions

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EXAMPLE 1

3)2cy2cy3(

And since Q=10m3

/s and g=9.81m/s2

, then under criticalflow conditions

Solving for yc by trial and error yields the critical depth

yc=0.855m

EXAMPLE 2

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EXAMPLE 2

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Thank You

(N)CEWB222 Chapter 4 - Specific Energy

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