3.1 Specific energy

3.2 Determination of normal depth using various method

3.3 Control Section

3.4 Rapidly Varied Flow (RVF)

3.5 Gradually Varied Flow (GVF)

CHAPTER 3 NON UNIFORM FLOW

IN OPEN CHANNEL

Introduction

The energy grade line, water surface and channel bottom are not parallel;

that is,

Sf ≠ Sw ≠ So

where;

Sf = slope energy grade line

Sw= slope of the water surface

So= slope of the channel bed

Where;

z is the elevation of the channel’s centerline

y the water depth

Specific Energy

The concept of specific energy as introduced by Bakhmeteff (1932) has

proven to be useful in the analysis of open channel flow.

It arises quite naturally from a consideration of steady flow through a

transition defined by a gradual rise in the channel bottom elevation. (shown

in the figure1).

For a given approach flow conditions of velocity and depth, the unknown

depth y2, after a channel bottom rise of height ∆z is of interest.

v²/2g

y1 Q

Q y2 = ?

∆z

EGL

Figure 1 :Transition with bottom step

Cont’ : Specific Energy

The specific energy decreases in the flow direction, but it would be equally

possible for the specific energy to increase in the flow direction by

dropping rather than raising the channel bottom.

If for the moment we neglect the energy loss, the energy equation

combined with continuity can be written as

E1 = E2

y1 + Q²/2gA1² = y2 + Q²/2gA2² + ∆z (4.1)

Where is; (From Equation 4.1)

y= depth

Q= discharge

A= Cross-sectional area of flow

∆z= z2 –z1 = change in bottom elevation from cross section 1 to 2

If the specific energy is defined as the sum of depth and velocity head, it follows that the possible solutions of the problem for the depth depend on the variation of specific energy with depth.

A more formal definition of specific energy is the height of the energy grade line above the channel bottom.

In uniform flow, for example, the energy grade line by definition is parallel to the channel bottom, so that the specific energy is constant in the flow direction. (The component of the gravity force in the flow direction is just balanced by the resistance boundary friction.)

Specific Energy Equation in Open Channel

(Specific energy = energy of water per unit weight (water) at cross section which is measured from channel bed)

Total of Energy = height + depth of flow + energy of velocity

H = z + y + av²/2g (4.2)

if bed of channel given as datum, so Equation 4.2 may be written as;-

E = y + av²/2g (4.3)

note; a : Coriolly coefficient

(non uniform velocity correction coefficient)

Normally in an open channel, velocity head is av²/2g

where as 'a' = 1.0 --> 1.36

From 'v = Q/A', therefore;

E = y + aQ²/2gA² (4.4)

av²/2g

y

E

H

z

Ө

EGL

Datum

=

Figure 2 :Non Uniform Flow schematic

From Equation (4.2);-

H = z + y + av²/2g (total of Energy)

Z = potential head (distance from the datum to channel

bed)

Y = Depth of flow

av²/2g = velocity head (energy of velocity)

Considered square channel (prismatic and straight)from Eq 4.4;

E = y + q²/2gy² (4.5)

where, q = flowrate per unit width (m³/s/m)

q = Q/b , a = 1.0 and A = by

Equation 4.5 (energy,E, depth of flow,y and flowrate,q) may be written/defined in 2 conditions as below;-

E and y if q is constant

q and y if E is constant

i) E and y if q is constant

From Eq (4.5)

E = y + q²/2gy² y³ – Ey² + q²/2g = 0 (4.6)

It is apparent from Equation (4.5) that there indeed is a unique functional variation between y and E for a constant value of q, and it is sketched as the specific energy diagram.

Specific Energy Diagrams (E-y)

y

o E = y + q²/2gy²

D B

A

C yc

Emin

y

y v²/2g (head of

velocity)

y > yc

y < yc

45º

*Note q is constant.

Emin

y

yc

E Eo

y1

y2

Specific Energy Diagrams

At C, specific energy is minimum and normal depth at this point is 'critical depth', yc

If

y > yc ; v < vc ==> Subcritical flow (steady)

y < yc ; v > vc ==>Supercritical flow (turbulant)

Differentiation of Equation (4.4)

E = y + aQ²/2gA²

dE/dy = 1 - a (Q²/2g)(2/A³)dA/dy

dE/dy = 1 – a(Q²/gA³).T

= 1 – (av²/gA).T

= 1 – av²/gD

At critical point, E is minimum i.e. dE/dy = 0

therefore;-

v²/gD = 1 ; (froude,Fr = 1)

v²/2g = D/2 or v/√(gD) = 1

T

dy

dA = T.dy

dA/dy = T

For a rectangular channel,

Hydraulic Depth, D = A /T = by/b = y

Therefore, at critical condition ==>>

Fr = 1

y = yc

v = vc

vc /√(g yc )= 1 (a)

vc²/2g = yc/2 (b)

From the schematic diagram;- (E = min, y = yc )

E = y + q²/2gy

dE/dy = 1 – q²/gyc³ = 0

q² = gyc³

yc = ³√(q²/g) (c)

q² = gyc³ but q = vy = vcyc

vc ²yc² = gyc³

vc ² = gyc

vc =√(gyc) or vc ²/2g = ½ yc (d)

Emin = yc + q²/2gyc²

= yc +(g yc³) / (gyc²)

= yc + yc√2

Emin = 1.5yc or yc =(2/3)Emin (e)

The point of minimum E is found by setting dE/dy equal to zero, and solving for y. The result is yc = 2E/3, which is called the critical depth yc. The corresponding velocity V is called the critical velocity Vc. The critical depth divides the energy curve into two branches. On the upper branch, y increases with E, while on the lower branch y decreases with E.

ii) q and y if E is constant

From Eq (4.5)

E = y + q²/2gy²

q² = 2gy² (E - y) (4.7)

E constant

y

y1

yc

y2

q qmax q

E

yc=2/3E

q – y curve

At critical point, dq/dy = 0

Differentiation of Equation (4.7):

q² = 2gy²

(E - y) = 2g(Ey² – y³)

2qdq/dy = 2g(E 2yc – 3yc² = 0

2ycE = 3yc²

E = 1.5yc

yc = 2/3E (f)

qmax² = 2gy² (E – y) (from Eq. 4.7)

= 2gy² (1.5yc – yc)

= gyc³

qmax = √(gyc³) (g)

Note;

Subcritical and supercritical flow normally depend on the channel slope, S. Therefore,

for the supercritical flow, value of S is high.

*Critical Slope = slope at critical depth.

Critical flow criteria (square/rectangular channel)

Fr = 1.0

'E' is minimum for 'q' constant

Emin = 1.5yc

yc = ³√(q²/g)

'q' is maximum at E constant

yc = 2/3Emin

qmax = √(gyc³)

Velocity head (vc²/2g) is one-half of critical depth, yc

vc²/2g = yc /2

Critical velocity (vc);

vc = √(gyc)

In general, critical flow will occur when Fr = 1.0, it will expressed as below;-

dE/dy = 1 – (Q²/gA³). T = 0

Q² T/gA³ = 1 (for all channels) (4.8)

cgyv

is known as the Froude Number, F cgyv

Froude Number, Fr and Flow classification

q2/gyc3 = 1

Then, vc2/gyc = 1 at critical conditions

So, at critical conditions, the Froude number =1

If F = 1, y = yc and flow is critical.

If F < 1, y > yc and flow is subcritical.

If F > 1, y < yc and flow is supercritical.

F is independent of the slope of the channel, yc dependent only on Q.

Flow characteristics of flow in rectangular channels

Critical Depth in non-rectangular channels

Critical conditions for channels of various shape

Cross Section Factor for Critical Flow (Z)

From Equation (4.8) :

Q² /g = A³/ T (4.9)

or

Z = Q /√g = A√D

where; D = A/T (4.10)

Equation (4.10) may be define as Cross Section Factor (z) for critical flow. (Where as A√D is cross section factor,z)

If Fr = a, therefore

z = Q/√(ga) = A√D (4.11)

Equation (4.11) is use generally in critical flow analysis...

Q = z√g (4.12)

or

Q = z√(ga) (4.13)

z = Q/√(ga) ===>> cross section factor for Non uniform Flow

Example

Example 1

Water flows in a rectangular channel with 5 m width and 8m³/s flowrate.

Depth of channel is 1m. Determine the specific energy for this channel.

Example 2

From example 4.1, if the channel is a trapezoidal channel with side slope is

1.5:1 and width of channel is 2m. Determine the specific energy for this

channel.

Example:

a) A wide and straight river was flows with 3.5m³/s/m flow rate. What is the

value of critical depth? If normal depth is 4.6m, calculate the Froude number

for this flow rate. (Type of flow: sub critical or supercritical). Calculate the

critical slope if Manning’s Coeffcient is 0.035.

b) Refer to question (a), calculate the depth (y2) for the same specific energy.

What is Froude number for this condition?

*For (b), there are 2 solutions; trial and error and graphical

Solution:

q = 3.5m³/s/m

yc = ³√(q²/g) = ³√[(3.5²)/9.81] = 1.08m (answer)

At normal depth,y = 4.6m,

Flow Velocity, v = q/y = 3.5/4.6 = 0.76m/s

Froude Number at y = 4.6 is;-

Fr = v/√(gy)

= 0.76/√(9.81)(4.6)

= 0.113 (answer)

Note: Fr < 1.0, therefore , flow in this river is subcritical flow

From Manning Formula:

Q = AR(2/3)√S / n

Note:-

for a rectangular channel, q= Q/b; for a very wide channel, R = y

Therefore;

q = y(5/3)S(1/2)/n

At critical flow in Non-Uniform flow;-

q = yc(5/3)Sc(1/2)/n

Sc = (qn/yc (5/3))²

= [3.5 x 0.035 / (1.08)(5/3)]²

= 0.012 or 1/86 (answer)

Solution (3 b) :

Specific Energy for y1 = 4.6 is :

E = 4.6 + (3.5)²/19.62 (4.6)² = 4.63m

but

E = y2 + q²/(2gy2²)

Where as y2 = depth at the same specific energy

THERE ARE 2 METHODS:-

Trial & Error Method

y2 should be in supercritical flow, therefore, the value of y2 is smaller than yc.

If y2 >>>> ;E <<<<

E – y Curved Method

Graph 'y' vs 'E = y+ (3.5)²/19.62 (y)²'

E = y + 0.624/y²

y = 0 – 5 meter

y = 0.383 m

Example 4

A rectangular channel with 3m width flows water at 12m³/s flow rate when

Froude number is 0.8. Determine the depths (y1 and y2) for the same flow

rate and specific energy.

Solution 4

Q = 12m³/s

v= Q/A = 12/3y1 = 4/y1

If Fr = 0.8 (subcritical flow)

v/√(gy1) = 0.8

(4/y1)/√(9.81y1) = 0.8

y1 = 1.366m (depth for subcritical flow,y1)

Specific Energy;-

E1 = y1 + q²/2gy1²

= 1.366 + (4)²/2(9.81)(1.366)²

= 1.803m

Calculation for y2 at the same flow rate and specific energy

E1 = E2 = y2 + q²/2gy2²

1.803 = y2 + 0.815/y2²

Critical depth, yc= (q²/g)(1/3)

= (4²/9.81)(1/3)

= 1.177m ( as a reference for trial & error method)

E = 1.803

1 1.821.01 1.81

1.02 1.8

Y2

Trial & Error Method:-

y2 = 1.02 m (depth at supercritical flow)

Exercise

1. A trapezoidal channel designed with 6 m width and side slope 1:2, calculate the critical depth when the flow rate is 17 m3/s using ;

-Trial and Error

-Graph

-Design Chart

2. A trapezoidal channel with side slope of 2 horizontal to 1 vertical is to carry a flow for 16.7 m3/s. For the bottom width of 3.6 m, calculate

a. critical depth

b. critical velocity

using :

-Trial and Error

-Design Chart

3.3 : Control Section

Control Section may be define as :

A section where a certain relationship can be established between flowrate

and water level, Q and h

It also controls the flow so that it can prevent the changes of flow types

from happening (critical flow, subcritical & supercritical)

Gauge station – to get flow rating curve which represents the 'flowrate' vs

'depth' relationship for the channel.

Control point

Point where depth of steady flow can be determined

due to grade change, dam, weir, etc.

Examples;

The change of slope from mild to steep

Free drop

Entrance point from reservoir to steep channel

Outlet point from steep channel to reservoir

Flow over weir

Presence of Broad Crested Weir

A rectangular channel with width b (constant along the channel) flows with

q m³/s/m. Assume this channel's slope is 0 degree (flat) and no roughness

coefficient (subcritical flow).

E1

y1

q²/2gy2²

EGL

q²/2gy1²

y2

E2

∆z

q

Broad Crested Weir

A

B

Curve of Depth; y (Channel Bed) versus Specific

Energy, E for the Presence of Broad Crested Weir

45º

y

E E2

A

A'

E1

C

B

yc

y2

y1

∆Z

q (constant)

E = y + q²/2gy²

E1 = E2 + ∆Z

note:- ∆Z = height of Broad crested weir

y1 + v1²/2g = y2 + v2²/2g+ ∆Z

or

E2 = E1 - ∆Z

From that figure, depth of water flow become lesser from point A to point B,

Specific Energy at point A, E1 > E2 (at point B)

If y2 = yc ; E2 = Emin ;

therefore

∆z = ∆zc (critical flow and this broad crested weir represent as control point)

If the weir increase more than before, specific energy will be decreased and

water depth, y2 become lower until one point (point C). Specific energy, E2

become minimum and y2 turn to yc. At this point, ∆z = ∆zc, flow is critical and

weir known as control point.

If the height of weir increase greater than

(∆z > ∆zc), E-y curve for same q can not be used because minimum

point for this curve is achieved and E2 < Emin. Therefore, for E2 and water

depth above weir is constant, yc so ∆z move to right side from point C. At this

point, E1 not enough for same q.

E1’ = Emin + ∆z

In this condition, total flow at point A cannot flow over weir but maintain at the

back of the weir. This condition called ‘choke’ and water depth at the upstream

is increase. The depth at the upstream called backwater situation. At the

downstream of the weir supercritical flow will be happened.

Example 5

Water flows uniformly at 15m³/s in a rectangular channel with 3 m width and

2.5 m depth. If broad crested weir is constructed, calculate the minimum height

of this weir which can cause critical flow above the weir (critical depth).

Solution:-

E1 = Emin + ∆Zc

= 1.5 yc + ∆Zc

Example 6

A rectangular channel with 5 m width, constructed on a mild slope conveying 8

m³/s and the normal depth is 1.25 m.

a) Determine the critical depth

b) How the broad crested weir height effect the normal depth at the

upstream and downstream of channel (assume no energy loss)

c) Shows that, if broad crested weir in critical condition, it can be use as

a flow gauge (use depth at upstream)

Examples of Weirs

Change of channel’s width (Narrowing channel)

The similar concept like weir will be applied in this topic but the relationship

between q-y is used because the width of channel will be changed and q also.

For same Q :

Q = q1b1 = q2b2

q1 = Q/b1

q2 = Q/b2

b2 < b1

therefore q2 > q1

Plan

Side plan

At critical point, b2 become minimum and q is maximum at same specific

energy.

When the channel’s become smaller, E1 do not enough to support q so E1 need

to increase for achieve suitable specific energy, E’ for critical depth yc’.

Therefore, the depth at upstream, y1 increase for E’.

y1’ + q1²/2gy1²’ = E’

y1’+ Q²/2gb1²’y12’ = E’

where E’ =1.5yc

and

yc ‘ = ³√(q’²/g) = ³√(Q²/b’2g)

Therefore,

if E’ > E1, b’ < bmin and specific energy at upstream E’ = 1.5y’c

If control situation happened, the structure will be controlled flow at upstream

and this structure called venturi flume.

Example of flume

Example 7

A rectangular channel flows at 3 m3/s with 2.0m width. The normal depth is

0.8m. The width will be decreased at downstream.

a. Determine the maximum width for critical

flow obtained at this part (downstream)

b. Calculate the depth at upstream (before

throat) if the throat is 1.2 m

Example 8

The water flows uniformly at 16.5 m3/s in a rectangular channel with 3.0 m

width and 1.8 m depth. If one part of the channel was narrowing, calculate the

maximum width of narrowing that can obtain critical water depth.

Example 9

A rectangular channel with 3.0 m width and water depth 3.0 m at velocity 3.0

m/s. If the channel bed increase at 0.61 m, how much the width will be

increased for maintain the same flow at the upstream?

3.4 : RAPIDLY VARIED FLOW

3.4.1 Hydraulic Jump Types and Uses

3.4.2 Momentum Principle, Conjugate Depths, Dissipated

Energy and Power

3.4.3 Length and Location of hydraulic jump

Rapidly Varied Flow

Developed mainly at hydraulic structures and most of the related problems can

be solved by using the continuity equations and energy principles provided

that the energy losses are known

However, if losses are unknown, the momentum principle must be used;

Net force= rate of change of momentum

For RVF, the momentum equation will be introduced in the context of the

HYDRAULIC JUMP (an important phenomenon in open channel flow and an

example of RVF – stationary surge wave)

Time

Space

Flow in Open Channel

Steady Flow

Uniform Flow Non Uniform Flow

Rapidly varied Flow Gradually Varied Flow

Unsteady Flow

The following classification is made according to

the change in flow depth with respect to time and space.

The primary criteria of classification is the variation of the depth of flow y in time, t and space, x.

Time

a flow can be classified as being:

Steady - which implies that the depth does not change with time ( y/ t = 0)

Unsteady - which implies that the depth does change with time ( y/ t ≠ 0)

Space

a flow can be classified as being:

Uniform – if the depth of flow does not vary with distance ( y/ x = 0)

Non uniform (varied flow) - if the depth varies with distance ( y/ x ≠ 0)

Rapidly varied – the depth of flow changes rapidly over a relatively short distance such as is the case with hydraulic jump

Gradually varied – the depth of flow changes rather slow with distance such as is the case of a reservoir behind the dam

Next

3.4.1 Hydraulic Jump: Types and Uses

Hydraulic jump analysis is the most common application of the momentum

equation in open channel flow.

The hydraulic jump, an abrupt change in depth from supercritical to

subcritical flow, always is accompanied by a significant energy loss.

A hydraulic jump primarily serves as an energy dissipator to dissipate the

excess energy of flowing water downstream of hydraulic structures such as

spillway and sluice gate.

Hydraulic Jump

A hydraulic jump occurs in the transition from supercritical to subcritical flow.

The intense turbulence in the jump cause mixing and energy dissipator.

The hydraulic jump is often used down stream of spillways and drop structures

to dissipate energy and prevent erosion in the downstream channel

supercritical subcritical

hydraulic jump

one two

three four

3.4.1 (a)Types of Jump

Hydraulic jumps on horizontal floor are of several distinct types. These types

can be conveniently classified according to the Froude number, Fr.

For Fr = 1, ; the flow is critical, and hence no jump can form.

For Fr = 1 to 1.7, ; the water surface shoes undulations, and the jump is

called undular jump.

For Fr = 1.7 to 2.5, ; a series of small rollers develop on the surface of

the jump, but the downstream water surface remains smooth. The velocity

throughout is fairly uniform, and the energy lost is low. This jump may be

called weak jump.

For Fr = 2.5 to 4.5, ; there is an oscillating jet entering the jump bottom to

surface and back again with no periodicity. Each oscillation produces a

large wave of irregular period which, very commanly in canals. This jump

may called an oscillating jump.

For Fr = 4.5 to 9.0, ; the downstream extremity of the surface roller and the point at which the high velocity jet tends to leave the flow occur at practically the same vertical section. The action and position of this jump are least sensitive to variation in tailwater depth. The jump is well-balanced and the performance is at its best. The energy dissipation ranges from 45 to 70%. This jump may be called a steady jump.

For Fr = 9.0 and larger, ; the high-velocity jets grabs intermittent slugs of water rolling down the front face of the jump, generating waves downstream, and a rough surface can prevail. The jump action is rough but effective since the energy dissipation may reach 85%. This jump may be called a strong jump.

3.4.1 (b) The use

Practical Applications of the hydraulic jump are many, it is used;-

To dissipate energy in water flowing over dams, weirs, and other hydraulic

structures and thus prevent scouring downstream from the structures

To recover head or raise the water level on the downstream side of a

measuring flume and thus maintain high water level in the channel for

irrigation or other water-irrigation or other water-distribution purposes

To increase weight on an apron and thus reduce uplift pressure under a

masonry structure by raising the water depth on the apron

To increase the discharge of a sluice by holding back tailwater, since the

effective head will be reduced if the tailwater is allowed to drown the

jump

Cont'd...

To indicate special flow conditions, such as the existence of supercritical

flow or the presence of a control section so that a gauging station may be

located

To mix chemicals used for water purification

To aerate water for city water supplies

To remove air pockets from water supply lines and thus prevent air locking.

3.4.2 Momentum Principles, Conjugate depth,

Dissipated energy, Power

Relationship between hydraulic jump equation and momentum equation.

There are several assumptions;

Flat channel bed

Uniform channel cross section

Uniform velocity and water depth

Ignore the stress at channel surface

Frictionless

Hydraulic jump occurs at short distance

Momentum for water flows in the channel section per unit time (N)

From Newton Second Law, the changing of momentum per time equal to

combination of external forces

∑F = wQ(v2 – v1)/g

M1 = M2

Hydraulic Jump Height/Depth (the different of height before & after hydraulic jump) ,

yj =y2 – y1 (in m)

Conjugate Depth (determine the depth before & after hydraulic jump) y1/y2 = ´ (√1 + 8 Fr22 – 1) (in m)

y2/y1 = ´ (√1 + 8 Fr12 – 1) (in m)

Energy Loss from jump (into heat),

∆E = E1 – E2 (in m)

∆E = (y2 – y1)3 /4y1y2 (in m)

Power dissipated or obtained from jump,

P = gQ ∆E (in W)

Example 1

At the bottom of spillway, a rectangular channel with 30 m width, the

velocity of flow, 28.2 m/s and depth before jump is 0.96 m. The hydraulic

jump is immediately (abruptly) occurred. Calculate the height of hydraulic

jump and the power dissipated.

Example 2

In a rectangular channel with 0.6 m width, hydraulic jump occurs when

Froude Number is 3. Normal depth before jump is 0.6 m. Determine the

energy loss and power dissipated in this situation.

Example 3

A hydraulic jump is formed in a rectangular channel conveying water.

If the velocity and the depth before the jump are 8.0 m/s and 0.3 m

respectively, calculate the depth after the jump. What is the head loss due to

the jump?

Example 4

Water flows at a rate 20 m3/s through a rectangular channel 4 m

wide from a ‘steep channel’ to a ‘mild channel’, creating a hydraulic jump. The

upstream depth of flow is 1.2 m. Determine;

The downstream depth of flow

The energy (head) loss in the jump

The upstream and downstream velocities

Power dissipated

Example 5

Water discharging into a 10m wide rectangular horizontal channel

from a sluice gate is observed to have undergone a hydraulic jump. The flow

depth and velocity before the jump are 0.8m and 7m/s, respectively.

Determine;

the flow depth and the Froude number after the jump

the head loss and the dissipation ratio

the wasted power production potential due to the hydraulic jump

3.4.3 Hydraulic Jump Positions

Length and location of Hydraulic Jump

Hydraulic jump will be occurred when

y0S < yc < y0M or y1 < yc < y2

Example

A rectangular channel 3m width carries water at 12m3/s. At one point, the

slope changes abruptly from 0.015 to 0.0016. The Manning’s coefficient,

n=0.013. Determine;

Is the hydraulic jump occurs?

The position of the jump (if its occurred)

Power dissipated

3.5 : Gradually varied flow (GVF)

Computations of depths in a GVF

3.5.1 Computation of GVF by numerical integration

3.5.2 Computation of GVF by direct step method

84

Cont’ : Gradually Varied Flow

We will discuss the gradually varied flow, which is the steady flow whose

depth varies gradually along the length of the channel

EFFECT OF BED SLOPE AND CHANNEL FRICTION

For a RVF, the influence of bed slope and channel friction was not mentioned

and assumed that frictional effects may be ignored. (as changes take place

over a very short distance)

However, in GVF, bed slope and channel friction are very important and they

actually determine the flow regime.

For a given specific energy or discharge, there are two possible flow depths at any section of a channel. Solution to the manning equation results in only one possible flow depth (normal depth)

For a given channel and discharge,

normal depth;

manning’s equation

Critical depth;

For a given channel chape and roughness, only one value of slope will produce the critical depth and known as critical slope (Sc)

86

31

2

g

qyc

Classification of Open-Channel Flows

Obstructions cause the flow depth to vary.

Rapidly varied flow (RVF) occurs over a short distance near the obstacle.

Gradually varied flow (GVF) occurs over larger distances and usually connects UF and RVF.

In GVF, y and V vary slowly, and

the free surface is stable

In contrast to uniform flow,Sf S0.

Now, flow depth reflects the

dynamic balance between gravity,

shear force, and inertial effects

To derive how the depth varies

with x, consider the total head

Let’s evaluate H, total energy, as a function of x.

H z y v 2 / 2g

dH

dx

dz

dx

dy

dx 2 g

dv2

dx

Where H = total energy head

z = elevation head,

v2/2g = velocity head

Take derivative,

Slope dH/dx of the energy line is equal to negative of the

friction slope

Bed slope has been defined

Inserting both S0 and Sf gives

Characteristic of Flow Profile

The dynamic equation of GVF developed expresses the longitudinal surface

slope of the flow with respect to the channel bottom.

It can be used to describe the characteristics of various flow profile or profile

of the water surface of flow.

Assume the channel is prismatic

The flow profile represent the surface curve of the flow.

Backwater curve if the depth of flow increases in the direction of flow

(dy/dx = +ve)

Drawdown curve if the depth of flow decreases in the direction of flow

(dy/dx = - ve)

Numerical Analysis of Water Surface Profile

There are several method to obtain surface water profile.

Direct Integration

Numerical Integration

Direct Step Method

Graphical Integration

Numerical/Computer Methods

Those method can identified;

Depth (y) at some distances/lengths (L/x)

Distances/lengths from one point to one point when both depth are known

3.5.1 : Numerical Integration

For this method, all equation before can rewrite in finite diference.

For any prismatic channel

For rectangular channel

32

2

1

1

gATQ

KKS

x

y

dx

dy oo

2

32

1

1

KK

gATQ

S

yx

oo

3

2

1

1

yy

KKS

x

y

dx

dy

c

oo

2

3

1

1

KK

yy

S

yx

o

c

o

For very wide rectangular channel

Chezy

Manning

If the channel length (L) divide by several small distances /lengths, it can call reach (∆x), therefore

∆x = length for each reache

= L / no of reaches

∆y = the change of water depth in each reach

(yi+1 – yi)

y = average for water depth in each reach (yi+1 – yi)/2

3

3

1

1

yy

yy

o

c

s

yx

310

3

1

1

yy

yy

o

c

s

yx

Example

The very wide rectangular channel carry the water at 2.5 m3/s/m with

channel bed slope, 0.001 and n= 0.025.

Find the length of back water which is happened from one dam and obtained

the 2 m water depth at the dam’s back.

The calculation must from the dam to upstream until the water surface is 1%

higher than normal depth.

Show your calculation until level 4 only.

3.5.2 : 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.

Exercise

Exercise

Exercise

A rectangular channel was designed;

Width = 3 m, Channel bed slope = 1:1500

Manning’s coefficient = 0.013

Determine the flow rate of this channel if the water depth is 1.2 m.

A weir constructed at the downstream and the M1 profiles was obtained at the

back of the structure. Sketch the M1 profiles clearly.

Using the Numerical Integration Method (3 Level), calculate the distance which

is the water depth change from 1.5m to 1.26m for this channel.

The calculate the distance which the water depth change from 1.5m to 1.26m

for this channel using Direct Step Method.