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Notes from UMP
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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.