Energy Concept

2g

21

V

1y

1z

1H ++=

EGL

HGL

2g

V 21

2g

V 22

xS o ∆

2y

1y

x∆

xSf∆

� Components of the energy equation

• z is the elevation head

• y is the pressure head-potential head

• V2/2g is the dynamic head-kinetic head

SPECIFIC ENERGY CONCEPT

� Specific Energy, E, is the energy referred as the channel bed as datum

� Taking the datum z=0 as the bottom of the channel, the specific energy E is the sum of the depth of flow and the velocity head.

y

A=A(y)

( ) 2

22

][22 yAg

Qy

g

VyE +=+=

SPECIFIC ENERGY CONCEPT

x)SS(2g

22

Vy

2g

21

Vy of21 ∆−++=+

x)SS(EE of21 ∆−+=

2g

22

Vy

2g

21

Vy

22

11

+=

+=

E

E

If channel bottom is horizontal and no head loss 21 EE =

For uniform flow Sf=So then

Specific-Energy Curve

� For a given Q, E is only a function of y, i.e: E=E(y)

( ) ( ) constant2

*2

2][ ==−g

QyAyE( ) 2

22

][22 yAg

Qy

g

VyE +=+=

45o

Q=cons.

E=E

1

A

BE

y

E=y

Specific-Energy Curve

� For a given Q, E is only a function of y, i.e: E=E(y)

The plot of E vs y is called specific-energy curve. Above equation has two asymptotes:

(E-y)=0 and y=0

in fact one section of the curve falls within the 45o angle between these two asymptotes in the first quadrant.

There is another section of the curve shown as broken line, but this is of no practical interest as it yield negative values for y.

( ) 2

22

][22 yAg

Qy

g

VyE +=+= ( ) ( ) constant

2*

22][ ==−

g

QyAyE

g2V 21

45o

Q=cons.

E=E1

y1

A

B

E

y

g2V 22

y2

y1

y2E=y

( ) ( ) constant2

22][ ==−

g

QyAyE

� If we regard this curve as a means of solving Eq.:

� for y, given E and Q, the three solutions of cubic are clearly shown by drawing a vertical line coresponding to the given value of E.

� Only two of them are physically real, so for given values of E and Q, there are two possible depths of flow, unless the vertical line referred misses the curve altogether.

� These two possible flow depths, for a given E and Q, are referred as alternate depths.

� One may say that the curve represents two possible regimes of flow-slow and deep on the upper limb,fast and shallow on the lower limb-meeting at the crest of the curve, C.

� Other curves might be drawn for other values of Q; since, for a given value of y, E increases with Q, curves having higher values of Q will occur inside and to the right of those having lower values of Q.

( ) 2

22

][22 yAg

Qy

g

VyE +=+=

E

g2V 21

45o

Q1

C

E=E1

y1

yc

y

g2V 22

y2

y1

y2E=y

Q 2>Q 1

Q 3>Q

2>Q

1

Specific Energy for rectangular channels

� For rectangular channels, one can define unit discharge, q, as:

� Specific Energy, E, for rectangular channels is defined in terms of unit discharge q as:

� and specific- energy curve is drawn for a given unit discharge.

2

22

gy2

q+y=

g2

V+y=E

Vyb

byV

b

Qq ===

)(

y

q=V

E

g2V 21

45o

q 1

C

E=E1

y1

yc

y

g2V 22

y2

y1

y2E=y

q 2>q 1

q 3>q

2>q

1

Specific Energy� Plot E vs y for constant Q

– Easy to see breakdown of E into pressure (y) & dynamic (V2/2g) heads

– E → ∞ as y → 0

– E → y for large y

– E reaches a minimum

0

1

2

3

4

0 1 2 3 4

y

E45°

yV2/(2g)

y

What is the physical interpretation of local

minimum?

Minimum Specific Energy

=== �ËgD

AQ

gD

VFr

Recall Froude Number

3

222

gA

TQ

AT

Ag

QFr ==

2

2222

gDA

Q

gD

AQFr ==

3

22

gA

TQFr =

Minimum Specific Energy

dy

dA

AdA

d

g

Q

dy

dE

+=

2

2 1

2 1

2

2

gA2

QyE +=

0dy

dE=For a given Q, when specific energy is minimum?

Fr=1, i.e: flow is critical

Tdy

dA=

dA=T*dy

T

dy

yA=f(y)

sinceor 13

2

gA

TQ

dy

dE−=

dy

dA

Ag

Q3

2 2

21 −=

01dy

dE 2=−= rF3

22

gA

TQFr =

g2V 21

45o

Q=cons.

C

E=E1

y1

yc

A

B

E

y

g2V 22

y2

y1

y2E=y

F r<1

Fr>1

Fr=1

( ) ( ) constant2

][2

2==−

g

QyAyE

SPECIFIC ENERGY CONCEPT

�For a given Q, E = E(y) only. The plot of E vs. y gives Specific Energy Curve.

�As Q increase, the curves will move towards right

2

22

22 gA

Qy

g

VyE +=+=

Characteristics Of The Specific Energy Curve

� Curve has 2 asymptotes:E = y & y = 0 lines

� Curve has 2 limbs AC and BC

� Limb AC approaches the horizontal axis as y → 0

� Limb BC approaches to E = y line as y → ∞

� On this curve Q is constant

� At any point P on this curve, the ordinate represents the depth and the abscissa represents the specific energy.

� For a given specific energy E, there are 2 possible flow depths: y1 and y2

Emin

yc

Critical depth

Characteristics Of The Specific Energy Curve

Emin

yc

3

22

gA

TQFr

gD

VFr ==⇒=

y1

y2

E

At point C, the specific energy is minimum

�Minimum specific energy corresponds to critical state of flow, i.e. Fr = 1.

�At the critical state, the two alternate depths become one, which is known as critical depth, yc.

�If y <<<< yc , V >>>> Vc →→→→ Fr >>>> 1

�the limb AC corresponds to supercritical flow and y2 is supercritical depth

�If y >>>> yc , V <<<< Vc →→→→ Fr <<<< 1

�the limb BC corresponds to subcritical flow and y1 is subcritical depth

� The depths y1 and y2 are called alternate depths.

Specific Energy For rectangular Channel

VbyAVQ )(== dischargeunitqb

Q=

g2

VyE

2

+=

2

2

gy2

qyE +=

Ay

b

Critical Flow:Rectangular channel

yc

Tc

Ac

3c

c2

gA

TQ1 =

qb=Q by=A cc

3

2

33

32

1cc gy

q

bgy

bq==

3/12

cg

qy

=

3cgyq =

Only for rectangular channels!

b=Tc

Given the depth one can find the discharge

cccc yV=q gy=V

Critical Flow Relationships:Rectangular Channels

3/12

cg

qy

= cc yVq =

=

g

yVy

2c

2c3

c

g

Vy

2

cc =

1gy

V

c

c = Froude number for critical flow

velocity head = 0.5 * depth

since

g2

V

2

y 2

cc =

2

yyE c

cc +=

cc E3

2y =

g2

VyE

2

+=

Critical DepthMinimum energy for q

kinetic = potential!

Fr=1

Fr>1 = Supercritical

Fr<1 = Subcritical

0dy

dE=

0

1

2

3

4

0 1 2 3 4

E

y

g2

V

2

y 2

cc =

Characteristics of Critical Flow

Arbitrary Cross Section Rectangular Cross Section

• c

c

T

A

g

QFr

32

1 ====→→→→==== • 3

231

g

q ygyq Fr

cc2

============

• 222

2

ccc

ccD

yED

g

V++++====→→→→==== •

cccc yE

y

g

V

2

3

22

2

====→→→→====

• For a given Q, E = Emin • For a given q, E = Emin

• For a given specific energy, • For a given specific energy

Eo, Q = Qmax Eo, q = qmax

Discussion

21 HH = zEE 21 ∆+=

zEE 21 ∆+=

A long rectangular channel carries water with a flow depth of y1 on a horizontal channel. If there is a rise on the channel bed:a)What is the relation of total head between section 1 and 2, if head loss assumed to be negligible?

b)What is the relation of specific energy between section 1 and 2?

c)How does the water surface profile change?

zEE 21 ∆+=

21 HH =

Side view

CHANNEL TRANSITIONfor rectangular channels

� Change on the bottom elevation of channel

� Change on the width of the channel

� Change on the bottom elevation and width of the channel

Upward Step-Constant width

21

2211 )()(

qq

byVbyVVAQ

=

===21 HH =

zEE 21 ∆+=

1) Subcritical flow3/1

2

cg

qy

=

Upward Step-Constant width

21

21

HH

qq

=

=

zEE

g

qyc

∆+=

=

21

3/12

Downward Step-Constant width

21 HH =

zEE 12 ∆+=

1) Subcritical flow

3/12

cg

qy

=

21

2211 )()(

qq

byVbyVVAQ

=

===

zEE

g

qy c

∆+=

=

12

3/12

Downward Step-Constant width

21

21

HH

qq

=

=

Upward & Downward Step-Constant width

zEE 21 ∆+= zEE 23 ∆+=

zEE 21 ∆+=

Subcritical flow

(3)

y3

Channel Expansion (constant bed elevation)

21

21

3/12

c

21

EE

HH

g

qy

qq

=

=

=

≠

21

21

3/12

c

21

EE

HH

g

qy

qq

=

=

=

≠Channel Contraction (constant bed elevation)

Figure E10.7

Specific Energy: Step UpAdditional

Consideration

∆z0

1

2

3

4

0 1 2 3 4

E

y

ZEE

zEE

HH

g

qy

qq

12

21

21

3/12

c

21

∆−=

∆+=

=

=

=

E1=3.3 m

CHOKING

CHOKING

Example 3Water is flowing in a rectangular channel. Find the change in depth and in absolute water level produced by a smooth downward step of 0.30 m if the upstream velocity and depth are given as.

a) V1=3 m/s and y1=3 m.

b) V1=5 m/s and y1=0.60 m.

V1=3m/s

∆z=30 cm

y1=3 m

∆yabs=?

y

(1) (2)

Datum

b

Example 3 Solution part a

m76.330.046.3Em46.362.19

33

g2

VyE 2

221

11 =+=⇒=+=+=

s/m933yVq,yg2

qyE 2

1122

2

22 =×==+=

2

2

22

2

2

2y

128.4y

y26.19

9y76.3 +=+=

.flowlsubcritica1553.0381.9

3

gy

VF

1

11 <=

×==

Energy Eq’n between (1) and (2) :E1+Dz=E2 +hloss

There are 2 possible solutions. To determine which one occurs compute upstream Froude number.

Therefore, y2 will correspond to subcritical flow.

≅0

Example 3 Solution part a

y2 must be greater than 3 m. The root of greater than 3 m can found by trial and error as;

y2 = 3.40 m

∆yabs = y2 - ( ∆z + y1 )

= 3.40 - ( 3.0 + 0.30 )

∆yabs = 0.10 m.

q=9 m2/s

E1=3.46 m

y2=3.40 m

y1=3.00 m

E2=3.76 m

∆z=0.30 m

E

yc

V1=3m/s

∆z=30 cm

y1=3 m

∆yabs=?

y

(1) (2)

Datum

b

Example 3 Solution part b

V1=5m/s

∆z=30 cm

y1=60cm ∆yabs=? y

b

lhEEz 11 +=+∆ m874.1

62.19

560.0

g2

VyE

22

111

=+=+=

2

2

2

22gy2

qym174.2874.130.0E +==+= , s/m36.05yVq 2

11 =×==

22

222

2

2y

4587.0ym174.2

y62.19

3ym174.2 +=⇒+=

≅0

flowcalsupercriti,106.260.081.9

5

gy

VFr

1

11 >=

×==

Energy Eq’n between (1) and (2) :E1+Dz=E2 +hl

There are 2 possible solutions. To determine which one occurs compute upstream Froude number.

Example 3 Solution part by2 must be smaller than 0.60 m

can be solved by trial and error to obtain y2=0.528 m.

Or y2 = 0.53 m.

∆yabs = ( ∆z + y1 ) - y2= ( 0.30 + 0.60 ) - 0.53

∆yabs = 0.37 m.

q=3m2/s

E1=1.874m

y1=0.60m

y2=0.53m

E2=2.174m

∆z=0.30m

yc

V1=5m/s

∆z=30 cm

y1=60cm ∆yabs=? y

b

Solution of Specific Energy Equation

2

2

2gy

qyE +=

22

2

y

CyE

gy2

qyE +=⇒+=

2

*

y

CEy −=

Subcritical Root

C

yEy* −

=

Supercritical root

Koch Parabola

� For a given E,

� Q = Q(y) [q=q(y) for rectangular channels]

� The plot of Q vs. y [q vs y] gives Koch Parabola.

Q = Qmax

E0

c

Q

Fr > 1

Fr < 1

Fr = 1

y

yc

q = qmax

E0

c

q

Fr > 1

Fr < 1

Fr = 1

y

yc

2/12 )]yE(gA2[Q −=

Variation of specific energy and (unit) discharge with depth: (a) E versus y for constant q; (b) Q versus y for constant E.

y2

v2/2g

EGL

∆z = 0.25datum

Assuming no energy loss through the transition :

zyg2

vy

g2

v2

22

1

21 ∆++=+

1 2

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

Y1= 2.0V1 = 2.2

Example 4 For an approach flow in a rectangular channel with depth of 2.0 m and velocity of 2.2 m/s, determine the depth of flow over a gradual rise in the channel bottom of ∆z = 0.25 m. Repeat the solution for ∆z = 0.50 m.

� Determine Condition (sub- or super-)

� Calculate critical depth, compare to sub – super depth

Yc = [(2* 2.2)2 / 9.81]1/3Yc = 1.254 m ……Thus, the flow is subcritical

� Solve specific energy equation for correct root for y2, V2

or V2 = (2.2*2)/y2

2+ [(2.2)2 / (2*9.81)] = y2 + {[(2.2*2) / y2 ]2 }/2*9.81+ 0.25

� Solve for y2

� Water surface elevation = y2 + ∆z

Steps

3/12

cg

qy

=

)()( 2211 byVbyVVAQ ===

Repeat for a step of 0.50 m

� Check value of Emin

� Determine value of E2 = (E1-∆z)

� 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