Specific Energy ConceptIn

Open Channel Flow

Applying Bernoulli Equation at any location along the channelgives the sum of the vertical distance measured from ahorizontal datum ‘z’, the depth of flow ‘y’, and the kineticenergy ‘vav

2/ 2g’. That sum defines the energy grade line and istermed the total energy, H.

Accepting that, the pressure distribution in open channel flow ishydrostatic (i.e. p = γ.y) and using the channel bottom as thedatum (i.e., z = 0), then total head above the channel bottom canbe defined as Specific Energy, E.

where Ei : (m),point at that availableEnergy Specific

yi: (m),point at that depth flowdeepest the

vav i: (m/s),point at that velocity flow averege

g: gravitational acceleration (m2/s). g: gravitational acceleration (m/s2).

Using continuity equation (Q = vav . A), the specific energy can be expressed in terms of the discharge as

Knowing that area of the cross-section ‘A’ is a function of depth ‘y’,it is obvious that, the energy ‘E’ and the flow depth ‘y’ is relatedwith a power of 3. Taking the rectangular cross-section as anexample, this relationship can be observed very easily.

or

3rd degree polynomial hence, having 3 roots.

In fact one of the roots is imaginary but the remaining two roots are real.

These real roots are called alternate depths in

literature. Implying that any flow can create the same energy attwo different depths:

one at sub-critical regime and the other at super critical regime.

Among these two real roots only one value is correct for anyspecific case and the proper one is determined with the helpof Froude number.

For a fixed flow rate (discharge), this equation is represented by the types of curves shown:

Note that for every specific energy ‘E’ and flow rate ‘Q’, there are at most 2associated +ve flow depths ‘yupper’ and ‘ylower’.These two depths are the ALTERNATE depths (stages).• The upper depth ‘yupper’ occurs if the flow is at sub-critical ‘tranquil’ flow regime;• The lower depth ‘ylower’ occurs if the flow is at super critical ‘rapid’ flow regime.

Depth ‘d’ [m]

Energy ‘E’ [m]

Discgarge ‘Q’

[m3/s]

E

dupper

dlower Discharge

y

y

‘y’[m]

45°

Sub-critical regime

Super critical regime

Figure illustrates the alternate depths y1 and y2 for which E1=E2 or

of super critical flow along with the location of the critical flowstates.

, the range of subcritical flow and the range

Specific Energy Curve

Upper asymptote

Lower asymptote

45°

2

Schematic representation of alternate and critical depths locationsand the flow regimes on the specific energy curve.

Sub-critical regime

Super critical regime

Critical regime

45°

Ecr≡Emin

Specific Energy curve showing subcritical and super criticalflow ranges and the variation of the curve as discharge varies[Q3 > Q2 > Q1]

45°

At a constant discharge, minimum specific energy occurs at the critical flow condition Fr =1.0 (i.e., dE/dy = 0), so that:

Knowing that the top width is defined as T= dA/dy so the above equation can be expressed as:

or

to denote the critical conditions one can use the subscript “cr” to define T, A, vav and y as Tcr, Acr, vav cr and ycr. so

Rearranging the above equation yields,

The Hydraulic depth is defined as D=A/T so the final form of theequation becomes

or

This is basically the Froude Number Fr, which is 1 at critical flow.

For a rectangular channel Dcr=Acr/Tcr=ycr , so the Froude number for critical flow becomes:

Energy minimum ‘Emin’for any cross-section isdetermined by insertingthe critical depth ycr valuefor defining area A:

𝐄𝐦𝐢𝐧 = 𝐲𝐜𝐫 +𝐐𝟐

𝟐𝐠𝐀𝐜𝐫𝟐

For RECTANGULAR cross-section, by inserting therelevant value of ycr (and also Tcr and Acr where B =Tcr) through Froude number (Fr=1.0):

Fr2 = 𝟏 =𝐐𝟐𝑻

𝐠𝐀𝐜𝐫𝟑 =

𝐐𝟐𝑻

𝐠𝐀𝐜𝐫𝟐 .𝐀𝐜𝐫

=𝐐𝟐𝑻

𝐠𝐀𝐜𝐫𝟐 .𝑩.𝒚𝐜𝐫

=𝐐𝟐

𝐠𝐀𝐜𝐫𝟐 .𝒚𝐜𝐫

So 𝒚𝐜𝐫=𝐐𝟐

𝐠𝐀𝐜𝐫𝟐

therefore the minimum Energy relationship forRectangular channel cross-section is determinedas:

This means that, at the critical stage of flow(critical depth), the velocity head equals halfthe critical hydraulic depth (for rectangularcross-section only).

So when y = ycr and Fr = 1, the critical stage of flow represents the boundary between super critical and sub-critical flow regimes.

Question 9.1A rectangular channel carries a discharge of Q = 12.0 m3/s of

Manning’s roughness n= 0.013. Channel bed slope is s0= 1.2 % andthe channel bottom is B = 2.0 m. Determine:a- ycr ,b- Emin ,c- draw the Specific Energy ‘E-y’ curve for different y values of this Q ,d- determine the normal and alternative depths that satisfies the

same energy,e- show the energy and both occuring and alternative depths of this

flow on the same curve.Solution

31

2

2

crgb

Qy

m 542.1

81.9x2

12y

31

2

2

cr

a-

b- For rectangular channels Emin =3/2 ycr

Emin = 2.314 m

c- Select depths below and above ycr to get the specific energy curve data

y

(m)

Area

(m2)

Q2/(2gA2)

(m)

E

(m)

0.6 1.2 5.10 5.70

1.0 2.0 1.83 2.83

1.2 2.4 1.27 2.47

1.4 2.8 0.94 2.34

1.6 3.2 0.72 2.32

1.8 3.6 0.57 2.37

2.0 4.0 0.46 2.46

2.2 4.4 0.38 2.58

3.0 6.0 0.20 3.20

4.0 8.0 0.11 4.11

d- to obtain the available energy the normal flow depth fromMannings should be used.

By trial and error y = 1.0 m E = 10.617 ≠ 12 m3/s increase y

y = 1.2 m E = 13.501 ≠ 12 m3/s decrease y

y = 1.10 m E = 12.047 ≈ 12 m3/s OK

so

s0

2.616 x1.12

The roots (depths) that satisfy this enery value are:yn= + 1.10 m occuringyn= + 2.26 m alternativeyn= - 0.74 m - value so eliminate.

2.616 19.62 x 22x 1.12

[m]

y

g2

vyE

2

[m]

Q= 12 m3/s

E= 2

.61

6normal 1.10

critic 1.542E m

in=2

.31

4supercritical

45°

alternative2.26

Question 9.2A rectangular channel whose bottom width is 4.0 m carries

a discharge of Q = 3 m3/sec, having Manning roughness coefficient of n = 0.02 and the channel bottom slope is s0= 0.004.a) compute the critical depth, b) determine the normal depth y,c) calculate the Froude number of the given cross-sectiond) classify the flow regime i- using Froude Number

ii- using flow depthiii- using channel bed slope

e) obtain the specific energy of this reach,f) calculate Emin

g) obtain the alternate depth h) draw the longitudinal reach andi) draw the specific energy curve.

Solutiona)Froude Number for critical flows is equivalent to 1 should be used.

b) The normal depth is obtained by Manning’s equation

either by trial and error or by using programmable calculator

= 0.459 m.

= 0.459 m > flow regime is sub-critical.

c)

Fr = 0.770 < Frcr= 1.0 flow regime is subcritical

d) i- Using Froude number

ii- Using flow depth

iii- Using channel bed slope

Obtain critical slope using critical depth using Manning equation

scr = 0. 00685

sb = 0.004 < 0.00685 hence MILD slope thereforeflow regime is subcritical

e)

Emin = 1.5 ycr Emin = 0.578 mf)

g)

The solution gives 3 roots:y1 = 0.459 m subcritical regime (given)

y2 = 0.327 m super critical regime (alternate)

y3 = - 0.191 not possible

h)

i)

(GIVEN)

(ALTERNATE)

Question 9.3 For a rectangular channel of 20.0 m width, construct a

family of specific energy curves for Q = 0, Q = 1.36, Q = 2.7 and Q = 8.1 m3/sec.

SolutionComputing critical water depth ycr and the minimum energy Emin

for each discharge values ‘Q’

0gb

Qy 0Q

31

2

2

cr

Emin = 0 m

m 078.020x81.9

36.1y 36.1Q

31

2

2

cr

Emin = 1.5x0.078 = 0.117 m

m 123.020x81.9

7.2y 7.2Q

31

2

2

cr

Emin = 1.5x0.123 = 0.184 m

m 256.020x81.9

1.8y 1.8Q

31

2

2

cr

Emin = 1.5x0.256 = 0.384 m

The specific energy is computed using the following equation for different discharges and are listed below:

y [m]

E [m]

ESTIMATION OF GRADUALLY VARIED FLOW LENGTHSESTIMATION OF GRADUALLY VARIED FLOW LENGTHS

The length of the gradually varied flow (M1, M2, M3, S1, S2, S3, …) can be roughly estimated with the

help of specific energies and the average slope of the formed water profile slopes.

Eend : specific energy at the end of the curve (m)

Ebegin : specific energy at the beginning of the curve (m)

sb : bed slope of the channel below that formed curve

sfbegin : water surface profile slope at the beginning of the curve

sfend : water surface profile slope at the end of the curve

Δx : estimated length of the curve (m)

2

sfsfs

EExΔ

endbegin

b

beginend

from the Manning’s equation inserting ‘ybeginning ’from the Manning’s equaton inserting ‘yend’.

Example 9.4:Estimate the length for the given surface profile between the given flow depths,

occurring within the rectangular channel of width B= 2.85 m, neq = 0.0157 that carries thedischarge of Q = 6.780 m3/s.

s0

1.765 1.765

1.765

m 2.659

2

000286.0000615.00013.0

876.1436.2xΔ

Question 9.5:A rectangular channel of bottom width B = 7.55 m carries an

average discharge of Q = 8.965 m3/s. There are two succesive long reaches. The first reach has s01=0.0018 and n1=0.018; and the second reach has s02=0.0013 and n2=0.021. a) Classify the slopes,b) Draw the longitudunal flow profile by showing the flow depths c) Name the occurance of non-uniform portion (G.V.F.)d) Determine the length of the occuring G.V.F. surface profilee) Draw the specific energy curve and show all the important points.

7.55 m

0.0018

s02

s01 0.0018

ANSWER:a- s01: Mild s02 : more mild scr1=0.004684 scr2=0.00637b- y1: 0.710 m, y2: 0.870 m , ycr:0.524 mc- E1= 0.853 m, E2= 0.965 m, Ecr(min)= 0.786 md- M1 occurse- ΔxM1 = 448 m

Example 9.6:A rectangular channel of bottom width B = 6.55 m carries

an average discharge of Q = 8.965 m3/s. There are two succesivelong reaches where first reach s01=0.0032 and n1=0.018; and thesecond reach s02=0.013 and n2=0.021.a) Classify the slopes,b) Draw the longitudunal flow profile by showing the flow depths c) Name the occurance of non-uniform portion(s) (G.V.F.)d) Determine the length of the occuring G.V.F. surface profile(s)e) Draw the specific energy curve and show all the important points.

ANSWER:a- s01: Mild s02 : Steepscr1=0.0047 scr2=0.00645b- y1: 0.653 m, y2: 0.462 m , ycr:0.576 mc- M2 - S2d- ΔxM2 = 17.33 m, d- ΔxS2 = 14.046 me- E1= 0.877 m, E2= 0.910 m, Ecr= 0.864 m

Question 9.7:A rectangular channel of bottom width B = 6.55 m carries an

average discharge of Q = 8.965 m3/s. There are two succesive long reaches. The first reach has s01=0.0032 and n1=0.018; and the second reach has s02=0.0013 and n2=0.021. a) Classify the slopes,b) Draw the longitudunal flow profile by showing the flow depths,c) Name the occurance of non-uniform portion,d) Draw a proper Specific Energy (E) – Water depth (y) graph and

show the details around the break point,e) Estimate the length of the established GVF.

s01=0.0032n1= 0.018

s02=0.0013 n2= 0.021

Old Exam Question

neqi=0.0171, A=4.127 m2, P=5.407 mQ=12.499 m3/s, Fr<1.0 Sub-criticalE=2.118 m, ycr=1.54 m, Emin=2.105 m,yalt=1.45 m, neqi-cr= 0.0172 scr=0.00503

Question 9.8:Below given open channel is the cross-section of a simple but compound channel with

main and floodway parts. Its longitudinal bottom slope sb = 0.00065 and the Manning’s

roughness coefficient for all over the channel is n = 0.017. The water depth at the floodway is

y = 2.855 m. Use vertical interface excluded (VIE) to calculate:

a) The discharge ‘Q’ passing within this channel,

b) The regime of this flow,

c) The existing specific energy ‘E’ for this given flow,

d) The critical depth ‘ycr’ of this cross-section,

e) The minimum energy ‘Emin’ of this flow,

f) The alternate depth ‘yalt’ of this flow.

g) Show all the above calculated values on specific energy – depth ‘E-y’ curve.

3.525 m

a) AA= 8.852 m2, PA= 6.38 m, AB= 25.095 m2, PB=11.645 m, QA= 16.514 m3/s QB= 62.790 m3/s QTOTAL= 79.304 m3/s

a) Fr = 0.435 < 1.00 Sub-critical regime

b) E = 3.803 m

c) ycr = 2.276 m

d) Emin = 3.119 m

e) yalt = 1.643 m

3.525 m

A B

For the given composite and compound channel cross-section with sides of different Manning

roughness coefficients having a uniform flow depth yn = 2.25 m which is flowing over the

longitudinal channel bottom slope s0 = 0.00165.

a) determine the discharge ‘Q’ passing within this cross-section?

b) determine the regime of the flow?

c) calculate the critical depth ‘ycr’ of this given cross-section?

d) find the specific energy ‘E’ of this cross-section?

e) calculate the minimum specific energy ‘Emin’ of this cross-section?

f) obtain the alternate depth ‘yalt’ of this cross-section?

g) show them on longitudunal channel profile and on ‘E-y’ graph.

Old Exam Question

Q=23.451 m3/s, Sub-criticalycr= 1.952 m

Old Exam Question:

Below given is a composite and compound channel cross-section with main

and floodway parts. Its longitudinal bottom slope is sb = 0.0085. The normal water depth within

the main part is y = 2.135 m and within the floodway is y = 83 cm. Calculate:

a) the amount of discharge ‘Q’ passing within this channel, (30)

b) the regime of this flow, (5)

c) the existing specific energy ‘E’ within this channel, (5)

d) the critical depth ‘ycr’ of this cross-section, (20)

e) the minimum energy ‘Emin’ of this cross-section, (5)

f) the alternate depth ‘yalt’ of this cross-section, (15)

g) the critical slope ‘scr’ of this reach, (5)

h) show all of these above relevant calculated information on a specific energy – depth (E-y)

curve. (15)

nequi MAIN= 0.0233 nequi FLOOD= 0.0147

Q MAIN= 58.0235 m3/s Q FLOOD= 57.0089 m3/s Q TOTAL= 115.1224 m3/s

Fr2=2.335 super critical regime E = 3.417 m y cr= 2.505 m Emin= 3.213 m

y alternate= 3.06 m s cr= 0.00352