Non uniform flow_in_channels [compatibility mode]

11/11/2013

1

Non uniform flow in channels

By

Dr. Ajit Pratap Singh

Civil Engineering Department

Dynamic Equation of GVF� The basic differential equation for GVF

� When dy/dx = 0, So = Sf and the water surface profile is parallel to the channel bottom

� When dy/dx = +ve, water surface is rising and

� When dy/dx = -ve, water surface is falling

2

fo

Fr1

SS

dx

dy

−

−=

Dynamic Equation of GVF in Wide

Rectangular Channel� The differential equation for GVF in wide rectangular

channel

used isequation sChezy' if ,

y

y1

y

y1

Sdx

dy

used isequation sMannaing' if ,

y

y1

y

y1

Sdx

dy

3

c

3

n

o

3

c

10/3

n

o

−

−

×=

−

−

×=

Classification of Flow ProfilesClassification of Flow Profiles�� Bed slope SBed slope S00 is classified asis classified as

�� Steep : Steep : yynn < y< yc c or or ssoo>s>scc

�� Critical : Critical : yynn = y= yc c or or ssoo= s= scc

�� M ild : M ild : yynn > y> yc c oror ssoo< s< scc

�� Horizontal : SHorizontal : S00 = 0= 0

�� Adverse : SAdverse : S00 < 0< 0

�� Initial depth is given a zoneInitial depth is given a zone

�� Zone 1 : y > Zone 1 : y > yynn

•• The space above both critical and The space above both critical and normal depthnormal depth

�� Zone 2 : Zone 2 : yynn < y < y< y < ycc

•• The region lies between the normal The region lies between the normal and critical depthand critical depth

�� Zone 3 : y < yZone 3 : y < ycc

•• The lowest zone of space that lies The lowest zone of space that lies above the channel bed but below above the channel bed but below both critical and normal depth linesboth critical and normal depth lines

• 12 distinct configurations for surface profiles in GVF.

• It should be noted that a continuous flow profile usually occurs only in one zone

• Figure 16.4 shows various surface profiles which can be classified as backwater curves and drawdown curves depending on whether depth of flow increases or decreases in the direction of flow

• All the surface profiles with subscript 1 and 3 are backwater or rising curves while those with subscript 2 are drawdown or falling curves.

Non-uniform flow in Open Channel by Dr. Ajit Pratap Singh

MILD BACKWATER CURVES M1, M2 AND M3

M1: y1 >yn > yc

Again the case of constant bed slope S is considered. Recall that

M2: yn > y1 > yc

M3: yn > yc > y1

+

+=

−

−=

)(1

)(

12

1

1y

ySS

dx

dy fo

x Fr

Depth increases downstream,

decreases upstream

Depth increases downstream,

decreases upstream

Depth decreases downstream,

increases upstream

A bed slope is considered mild if yn > yc. This is the most common case in

alluvial rivers. There are three possible cases.

3

21/3

s

2rf3

2

ff3

2

2

gy

q

k

yα(y)Sor

gy

qC(y)S,

gy

q(y)Fr

−

−

===

+

−=

−

−=

)(1

)(

12

1

1y

ySS

dx

dy fo

x Fr

−

−=

−

−=

)(1

)(

12

1

1y

ySS

dx

dy fo

x Fr

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Non-uniform flow in Open Channel by Dr. Ajit

Pratap Singh

M1 CURVEM1 CURVEM1 CURVEM1 CURVE

� When y > yn > yc, what will be the limiting value of y??

side downstream on the y

side upstream on the yy n

∞→

→

o

n

Sdx

dy side, downstream on the y as and

0dx

dy side, upstream on the yy as and

vedx

dyget weHere

→∞→

→→

+→


Pratap Singh

Remember to note that

normally line ingcorrespond meet the curves specified that theindicatesit ,dx

dyWhen

normally line ingcorrespond meet the curves specified that theindicatesit ,dx

dyWhen

horizontal be to tendscurves specified that theindicatesit ,Sdx

dyWhen

ally asymptotic NDL meet the curves specified that theindicatesit 0,dx

dyWhen

o

−∞→

∞→

→

→

Thus M1 curves meet the NDL asymptotically on the u/s andit tends to be horizontal on the d/s. M1 curve becomeshorizontal as the depth becomes larger. An example of sucha flow is a river entering a lake or reservoir or Flow behindan overflow weir


Pratap Singh Non-uniform flow in Open Channel by Dr. Ajit Pratap Singh


� When yn >y> yc, what will be the limiting value of y??

side downstream on the yy

side upstream on the yy

c

n

→

→

−∞→→

→→

−→

dx

dy side, downstream on the yy as and

0dx


vedx

dyget weHere

c

n


Pratap Singh

Remember to note thatThus M2 curves meet the NDL asymptotically on the u/s andit meet the CDL normally on the d/s. An example of such aflow is over a free overfall. It is also obtained when a steepslope is preceded by a mild slope or a mild slope by a milderslope.


Pratap Singh


� When yn > yc >y, what will be the limiting value of y??

side downstream on the 0y

side upstream on the y c

→

→ y

undefined is dx

dy i.e.

dx

dy side, upstream on the 0y as and

dx


vedx

dyget weHere

c

∞→→

−∞→→

+→

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Pratap Singh

Remember to note thatThus M3 curves meet the CDL and channel bottom linenormally. An example of such a flow is d/s of a sluice gate. Itis also obtained when the bottom slope changes from steepto mild.


Pratap Singh


Pratap Singh


Pratap Singh

S1 CURVES1 CURVES1 CURVES1 CURVE

� When y>yc > yn, what will be the limiting value of y??


side upstream on the yy c

∞→

→

o

c

Sdx


dx


vedx

dyget weHere

→∞→

∞→→

+→


Pratap Singh


Thus S1 curves meet the CDL normally on the u/s and ittends to be horizontal on the d/s. Example Flow behind anoverflow weir or profile formed behind a dam constructed ona steep channel


Pratap Singh


� When yc >y> yn, what will be the limiting value of y??


side upstream on the yy

n

c

→

→

0dx


dx


vedx

dyget weHere

n

c

→→

∞→→

−→

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4

Remember to note thatThus S2 curves meet the CDL normally on the u/s and itmeet the NDL asymptotically on the d/s. An example of sucha flow is over a free overfall. It is also obtained when a steepslope is preceded by a mild slope or a mild slope by a milderslope.


Pratap Singh


� When yc > yn >y, what will be the limiting value of y??

side upstream on the 0y

side downstream on the y n

→

→ y

undefined is dx

dy i.e.

dx


0dx


vedx

dyget weHere

n

∞→→

→→

+→


Pratap Singh

Remember to note thatThus S3 curves meets the channel bed normally and it isasymptotic to the normal depth line NDL. An example ofsuch a flow is d/s of a sluice gate. It is also obtained whenthe bottom slope changes from steep to mild.


Pratap Singh

C1 CURVEC1 CURVEC1 CURVEC1 CURVE

� When y > yn = yc, what will be the limiting value of y??


side upstream on the yyy nc

∞→

=→

co

coc

SSdx


SSdx


vedx

dyget weHere

=→∞→

=→→

+→


Pratap Singh

Remember to note that horizontal be to tendscurves specified that theindicatesit ,S

dx

dyWhen

o→

Thus C1 curves will be more or less a horizontal line. For example flowbehind an overflow weir, flow behind a sluice gate.


Pratap Singh

C3 CURVEC3 CURVEC3 CURVEC3 CURVE

� When y < yc = yn, what will be the limiting value of y??


side downstream on the yyy nc

→

=→

co

coc

SSdx


SSdx


vedx

dyget weHere

=→→

=→→

+→

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5


Pratap Singh


horizontal be to tendscurves specified that theindicatesit ,Sdx

dyWhen

o→

Thus C3 curves will be more or less a horizontal line. Forexample back-water curve below a sluice gate provided in achannel with a critical slope is typical example of C3 profile.


Pratap Singh

H2 CURVEH2 CURVEH2 CURVEH2 CURVE

� When yn (=∞)>y> yc, what will be the limiting value of y??


side upstream on the y

c→

∞→

−∞→→

→∞→

−→

dx


0dx

dy side, upstream on the y as and

vedx

dyget weHere

c


Pratap Singh


Thus H2 curves meet the CDL normally at d/s end and at theu/s end it tends to approach horizontal line tangentially


Pratap Singh

H3 CURVEH3 CURVEH3 CURVEH3 CURVE

� When yn > yc >y, what will be the limiting value of y??


side downstream on the y c

→

→ y

undefined is dx

dy i.e.

dx


dx


vedx

dyget weHere

c

∞→→

−∞→→

+→


Pratap Singh


Thus H3 curves meet the CDL and channel bottom linenormally.


Pratap Singh

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Pratap Singh

� An M1 profile occurs behind a dam or a sluice gate located on a mildchannel. The dam or a sluice gate piles up water behind it such that theflow depth is greater than the normal depth. Far away from the dam orsluice gate on the upstream side, the flow would be occurring underuniform conditions and the flow depth would be normal.

� In a similar manner, S1 and C1 profiles occur on the upstream side of asluice gate located on a channel with steep and critical slopes,respectively.

� An M2 profile occurs on the upstream side of a free over fall at thedownstream end of a mild channel since a critical depth occurs in thevicinity of a free over fall.

� Similarly, a H2 profile occurs on the upstream side of a free over fall atthe downstream end of a horizontal channel.


Pratap Singh

� Critical flow conditions occur at the entrance to a steepchannel from a lake or a reservoir. However, flow shouldtend towards uniform flow conditions far away from theentrance if the channel is long. Therefore, a S2 profile occursin steep channels, on the downstream side of the entrance.

Gradually Varied FlowGradually Varied Flow Gradually Varied FlowGradually Varied Flow

�� Typical OC system Typical OC system involves several involves several sections of different sections of different slopes, with transitionsslopes, with transitions

�� Overall surface profile is Overall surface profile is made up of individual made up of individual profiles described on profiles described on previous slidesprevious slides


Pratap Singh

Steps to be followed to sketch the water

surface profiles

� Compute normal and critical depths for each reach of the channelsystem based on specified flow rate, roughness coefficient, slope ofthe reach,and the channel cross section.

� Plot the channel bed, the normal depth line (NDL) and the criticaldepth line (CDL) for each reach in the system.

� By comparing normal depth and critical depth, determinewhether the channel slope is mild, critical, steep, adverse orhorizontal

� Mark the control sections i.e., identify the sections where (i) theflow passes through a critical depth (ii) the flow is expected tooccur under uniform conditions, and (iii) there is a controlstructures such as a weir, a sluice gate, and a spillway. A controlsection is that at which for a given discharge the depth of flow isknown or it can be controlled to a required value


Pratap Singh

� It may be noted that uniform flow conditions occur in longprismatic channels, far away from control sections. Criticaldepth occurs at (i) the free overfall, and (ii) the entrance to asteep channel from a lake, when the water level in the lake isabove the elevation of the CDL at the entrance. Criticaldepth also occurs when channel bed slope changes from mildto steep.

� Knowing the normal depth and depth at the control section,determine the surface profile

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SURFACE WATER PROFILE� In practice, often the length of the surface profile of the gradually

varied flow needs to be known

� From all the equations the derivations show the changes of waterdepth (y) for some distances/lengths (x). By using the integrationmethod in those equations, we can know,

� Distances/lengths from one point to another point when bothdepth are known. For example, if a weir is constructed across ariver having a mild slope then it may be required to estimate thedistance on the u/s side up to which the effect of resulting M1profile exists.


Pratap Singh

� Starting from each control point, sketch the appropriate water surfaceprofile depending on the zone in which the depth at the controlsection falls and the nature of the slope.

� Qualitatively locate the hydraulic jumps wherever the flow changesfrom supercritical to sub critical.

� For example, if there is a sluice gate at the downstream end of a steepchannel, the flow is sub critical on the upstream side of the gate.However, if the channel is long, flow is supercritical far away from thegate on the upstream side. Therefore, a hydraulic jump occurs in such achannel. Also, on the downstream side of a sluice gate on a long mildchannel, the flow is supercritical immediately downstream of the gate.However, far away from the gate on the downstream side, flow issubcritical.Therefore,a hydraulic jump occurs in such a case also.


Pratap Singh

Real Life Cases of Water Surface Real Life Cases of Water Surface Real Life Cases of Water Surface Real Life Cases of Water Surface

ProfilesProfilesProfilesProfiles


Pratap Singh

� There are several method to obtain surface water profile.

� They are� Direct Integration

� Numerical Integration

� Direct Step Method

� Graphical Integration

� Numerical/Computer Methods

Direct Step Method

� In general, a step method is characterized by dividing the channel into short reaches and carrying the computation step by step from one end of the reach to the other. The direct step method is a simple step method applicable to prismatic channel.

� Equating the total head at the two end section 1 and 2, the following may be written;

2 2

1 1 2 21 1 2 2

2 2L

p V p Vz z h

g ga a

g g+ + = + + +

2 2

1 21 2

2 2o f

V Vy S dx y S dx

g g+ + = + +

Turbulent flow (α ≅ 1)z - measured from horizontal datum

2 2

1 21 2

2 2o f

V Vy S dx y S dx

g g+ + = + +

1 2 0z z S dx− =

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Short reach (dx) of the channel

2

1V

2g2

2V

2g

oS dx

dx

f fh S dx=

energy grade line

hydraulic grade line

velocity head

water surfacewater surface

1 2

Where

E = specific energy at one point = y + v²/2g

1 2o fE S dx E S dx+ = +

2 1

0 f

E Edx

S S

−=

−solving for dx

2 2

4 /3f

n VS

R=

2

2f

VS

C R=

Manning Chezy

� Limitation: channel must be _________ (so that velocity is a function of depth only and not a function of x)

� Method� Find yn and yc. Determine the type of slope.� Identify type of profile (determines whether ∆y is + or -)� choose ∆y and thus yn+1

� Compute the area of flow section A, wetted perimeter P, hydraulic radius R and velocity at known value of depths of flow yn and yn+1

� Compute the mean velocity, the velocity head, the specfic energy E and energy line slope Sf at the channel sections where the depth flow is known i.e. at yn and yn+1

� calculate average friction slope � calculate dx

prismaticprismaticDirect Step Method


Pratap Singh

� A rectangular channel conveys a discharge of 30 m3/s. It islaid at a slope of 0.0001. If at a section in this channel thedepth is 1.6 m, how far u/s or d/s from the section will thedepth be 2.0 m.Take Manning’s n = 0.015


Pratap Singh

� A very wide rectangular channel conveys a discharge of 3.50m3/s per metre width at a depth of 2.50 m. The bed slope is1 in 5000. Due to a weir placed across the channel the waterlevel is raised by 1.50 m just on the upstream of it. Find atwhat distance upstream of the weir the depth of water will be3 m. Take C = 51. Use step method and take two steps. Alsoclassify the type of water surface profile.

Example� A wide rectangular channel carries water at 10 m3/s with channel

width = 8m, bed slope = 0.001 and n= 0.025.

� Find the length of back water which is formed due to a dam and obtained the 2 m water depth at the dam’s back.

� Find yn and yc.

yn

1.093703

yc

0.542064

yn > yc Mild slope channel

PROFILE M1 TYPE

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1. y = water depth (m)

2. R = A/P = hydraulic radius or y for very wide rectangular

3. v = q/y = flow velocity

4. v²/2g = kinetic energy

5. y + v²/2g = E = specific energy

6. E2 - E1 = ∆E = energy loss

7. sf = slope energy grade line = n2v2 = v2

R4/3 C2R

8. (sf1 + sf2)/2 = EGL slope average

9. (so - sf ) = slope difference

10. dx = reach = ∆E / (so - sf )

11. L = length of surface water profile which is to be calculated from dam

1. Find yo and yc2. Fill in the table

A/P q/y

1 + 4 (y+v2/2g) E2 -E1

n2v2

/R4/3 6/9

1 2 3 4 5 6 7 8 9 10 11

y A R v v2/2g E ∆E Sf ∆x

- - - -

fS 0 fS S−

Solution for example

(See EXCEL FILE)

A/P q/y

1 + 5

(y+v2/2

g) E2 -E1

n2v2

/R4/3

1 2 3 4 5 6 7 8 9 10 11

y A R v v2/2g E ∆E Sf Sfbar S0-Sfbar dx

2 16 1.33333 0.625 0.01991 2.01991 0.00017

0.12714 0.00018 0.000815746 155.8520063

1.87 14.96 1.27428 0.66845 0.02277 1.89277 0.0002

0.12647 0.00023 0.000774196 163.3564655

1.74 13.92 1.21254 0.71839 0.0263 1.7663 0.00025

0.12558 0.00028 0.00071855 174.7695148

1.61 12.88 1.14795 0.7764 0.03072 1.64072 0.00031

0.12437 0.00036 0.000642175 193.6630303

1.48 11.84 1.08029 0.84459 0.03636 1.51636 0.0004

0.12266 0.00047 0.000534278 229.5819527

1.35 10.8 1.00935 0.92593 0.0437 1.3937 0.00053

0.12019 0.00062 0.000376502 319.231189

1.22 9.76 0.93487 1.02459 0.05351 1.27351 0.00072

Σ (Σ (Σ (Σ (dx)))) 1236.454159

� The calculation must be from the dam to upstream until the water surface is 1% higher than the normal depth.

52

Hydraulic Jump• A hydraulic jump occurs when flow changes from a supercritical flow

(unstable) to a sub-critical flow (stable).

• There is a sudden rise in water level at the point where the hydraulic

jump occurs.

• Rollers (eddies) of turbulent water form at this point. These rollers cause

dissipation of energy.

•A hydraulic jump occurs in practice at the toe of a dam or below a sluice gate

where the velocity is very high.

53

General Expression for HydraulicJump:

In the analysis of hydraulic jumps, the following assumptions are made:

(1) The length of hydraulic jump is small. Consequently, the loss of head

due to friction is negligible.

(2) The flow is uniform and pressure distribution is due to hydrostatic

before and after the jump.

(3) The slope of the bed of the channel is very small, so that the

component of the weight of the fluid in the direction of the flow is

neglected.

54

Location of hydraulic jump

Generally, a hydraulic jump occurs when the flow changes from

supercritical to subcritical flow.

The most typical cases for the location of hydraulic jump are:

1. Jump below a sluice gate.

2. Jump at the toe of a spillway.

3. Jump at a glacis.

(glacis is the name given to sloping floors provided in hydraulic structures.)

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Pratap Singh

MOMENTUM EQATION� Steady Flow

� Momentum is a vector quantity. The momentum equationcommonly used in most of the open channel flow problems is thelinear-momentum equation. This equation states that the algebraicsum of all external forces acting in a given direction on a fluidmass equals the time rate of change of linear-momentum of thefluid mass in that direction. In a steady flow the rate of change ofmomentum in a given direction will be equal to the net flux ofmomentum in that direction.


Pratap Singh

� Figure shows a control volume (a volume fixed in space) bounded by sections 1 and 2, the boundary and a surface lying above the free surface.


Pratap Singh

� The various forces acting on the control volume in the longitudinal direction are:

� (i) Pressure forces acting on the control surfaces, and

� (ii) Tangential force on the bed, ,

� (iii) Body force, i.e. the component of the weight of the fluid in the longitudinal direction, .

� By the linear-momentum equation in the longitudinal direction for a steady-flow discharge of ,

1F

2F

3F

4F

Q

124321MMFFFFF −=+−−=∑

� Momentum of the flow passing a channel section per unit time =

� Rate of change of momentum in the body of water flowing in a channel is equal to the resultant of all the forces that are acting on the body

� Where Ff total external force of frictional resistance acting in the direction opposite to the flow along the surface of contact between water and channel, W is the weight of water enclosed between the section, θ is angle of incination

g

QVwM =

( )f21

12 FWSinθPPg

VVQw−+−=

−

59

•The net force in the direction of flow = the rate of change of moment in that direction

1 2( )

w QV V

g= −

The net force in the direction of the flow, neglecting frictional resistance and the

component of weight of water in the direction of flow,

R = P1 - P2 .

Therefore, the impulse-moment yields

2 1 1 2( )

wQP P V V

g− = −

Where P1 and P2 are the pressure forces at section 1 and 2, respectively.

2 12 1 1 2( )wQ

wA z wA z V Vg

− = −

2

2 12 1

1 2

1 1( )

wQwA z wA z

g A A− = −

2 2

1 21 2

1 2

Q QA z A z

gA gA+ = +

z = the distance from the water surface to the centroid of the flow area60

Comments:

• This is the general equation governing the hydraulic jump for any shape

of channel.

• The sum of two terms is called specific force (F). So, the equation can

be written as:

M1 = M2

This equation shows that the specific force before the hydraulic jump is

equal to that after the jump.

2 2

1 21 2

1 2

Q QA z A z

gA gA+ = +

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61

Hydraulic Jump in Rectangular Channels

1 1A By= 1

1

2

yz =

2 2A By= 2

2

2

yz =

2 2

1 2

1 2

1 2

( )( ) ( )( )2 2

y yQ QBy By

gBy gBy+ = +

2 2

1 21 2

1 2

Q QA z A z

gA gA+ = +

Qq

B=

2 22

2 1 2 1

1 22

y y y yq

g y y

− −=

2

1 2 2 1

2( )

qy y y y

g= +

2

2 2

2 1 2 1

20

qy y y y

g+ − =

using

we get

2

1 2 1 2

2( )

qy y y y

g= +

62

2 2

1 12

1

2

2 2

y y qy

gy

= − + +

2 2

2 21

2

2

2 2

y y qy

g y

= − + +

This is a quadratic equation, the solution of which may be written as:

where y1 is the initial depth and y2 is called the conjugate depth. Both are called

conjugate depths.

These equations can be used to get the various characteristics of hydraulic jump.

2

2

3

1 1

811 1

2

qy

y g y

= − + +

2

1

3

2 2

811 1

2

qy

y gy

= − + +

63

2

3

c

qy

g=

3

2

1 1

11 1 8

2

cyy

y y

= − + +

3

1

2 2

11 1 8

2

cyy

y y

= − + +

( )221

1

11 1 8

2

yFr

y= − + +

( )212

2

11 1 8

2

yFr

y= − + +

1

1

1

VF

gy=

2

2

2

VF

gy=

But for rectangular channels, we have

Therefore,

These equations can also be written in terms of Froude’s number as:

64

1 2E E E∆ = −

2

22

s

qE y

gy= +

( )

2 22 2 2

2 1

1 2 2 122 3

1 2 1 2

( )2 2 2

y yq q qE y y y y

gy gy g y y

−∆ = + − + = − −

Due to the turbulent flow in hydraulic jump, a dissipation (loss) of energy occurs:

Where, E = specific energy

For rectangular channels:

hence,

Head Loss in a hydraulic jump (HL):

After simplifying, we obtain

3

2 1

1 2

( )

4

y yE

y y

−∆ =

65

2 1jh y y= −

6j jL h≅

Height of hydraulic jump (hj):

The difference of depths before and after the jump is known as the

height of the jump,

Length of hydraulic jump (Lj):

The distance between the front face of the jump to a point on the

downstream where the rollers (eddies) terminate and the flow becomes

uniform is known as the length of the hydraulic jump. The length of the

jump varies from 5 to 7 times its height. An average value is usually

taken:

Dam at Hiram Falls on the Saco River near Hiram, Maine, USA

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12

� A spillway is designed to discharge 5m3/sec/m length. After flowing over the spillway, water flows on to a horizontal concrete apron (Manning’s rugosity coefficient as 0.015). The velocity of flow at toe is observed to be 15 m/sec and tail water depth is limited to 3.0 m. Calculate the minimum length of apron to contain the jump on the apron and consequent energy lost.

Surges in Open Channel� A surge or surge wave is a moving wave front which brings

about an abrupt change in depth of flow

� It is also referred to as moving hydraulic jump and is caused by sudden increase or decrease of depth of flow, such as that caused by sudden opening or closing of a gate fixed in the channel.

� Positive surges: which results in an increase in depth of flow

� Negative surges: causes a decrease in depth of flow

� Positive Surge: Type A which is Advancing D/S and Vw = C+V1

� Positive Surge: Type B which is Advancing U/S and Vw = C-V1

� Negative Surge: Type C which is retreating D/S and Vw = C+V1

� Negative Surge: Type D which is retreating U/S and Vw = C-V1

� Where C is the celerity of the wave which is defined as the velocity of wave relative the velocity of flow and Vw is the absolute velocity of the wave

� Applying continuity equation to the control volume of fig., if ρ= density of water; A2 = flow area behind the wave and A1 = flow area ahead of the wave. Since ρ is a constant

� Applying momentum equation to the control volume of fig., if ρ= density of water; A2 = flow area behind the wave and A1 = flow area ahead of the wave. Since ρ is a constant

� The channel is prismatic, horizontal and frictionless. Therefore, the only force acting on the control volume is pressure force. Pressure force acts in the positive x - direction at the inlet section and in the negative x - direction at the outlet section. Above Equation can be written as

11/11/2013

13

� For Rectangular channels

Problem� A wide rectangular channel is carrying a flow of 3 m3/sec

per meter width of channel at a flow depth of 1.5m. What should be the increase in discharge at the upstream end to cause a surge of 0.5m? What is the corresponding surge velocity?

Date post:	15-Jul-2015
Category:	Engineering
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