CHAPTER 3 NON UNIFORM FLOW IN OPEN CHANNEL 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)
Introduction The energy grade line, line, water water surface and channel bottom b ottom are not parallel; that is,
Sf ≠ Sw ≠ So where; Sf = slope energy grade grade line Sw= slope of the water surface So= slope of the channel bed
Introduction The energy grade line, line, water water surface and channel bottom b ottom are not parallel; that is,
Sf ≠ Sw ≠ So where; Sf = slope energy grade 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.
Cont’ : Specific Energy EGL
v²/2g Q
y2 = ?
y1
Q
∆z
Figure 1 :Transition with bottom step
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 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
(4.3)
(4.4)
av²/2g
EGL =
E y
H
z Datum
Ө
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 D y
B
v²/2g (head of velocity)
y > yc
y yc
o
C A
45º
Emin
y < yc
E = y + q²/2gy²
Specific Energy Diagrams
y
y2
yc y1
Emin
*Note q is constant.
Eo
E
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 vc²/2g = yc/2 From the schematic diagram;(E = min, y = yc ) E = y + q²/2gy dE/dy = 1 – q²/gyc³ = 0 q² = gyc³ yc = ³√(q²/g)
(a) (b)
(c)
q²
= gyc³
but
q = vy
vc ²yc²
= gyc³
vc ²
= gyc
vc
=√(gyc) or vc ²/2g
Emin
= yc + q²/2gyc²
= vcyc
= ½ yc
(d)
= yc +(g yc³) / (gyc²)
= yc + yc√2 Emin (e)
= 1.5yc or
yc
=(2/3)Emin
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.
y E constant y1 yc E yc=2/3E y2 q – y curve
q
qmax
q
ii) q and y if E is constant From Eq (4.5) E = y + q²/2gy² q² = 2gy²
(E - y)
(4.7)
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
qmax²
= 2gy²
(E – y)
= 2gy²
(1.5yc – yc)
(f) (from Eq. 4.7)
= 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 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
Q² T/gA³ = 1
(for all channels)
=0 (4.8)
Froude Number, Fr and Flow classification q2/gyc3 = 1 Then, vc2/gyc = 1 v
at critical conditions
gyc
So, at critical conditions, the Froude number =1 v
gyc
is known as the Froude Number, F
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 or
(4.9)
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) z = Q/√(ga) ===>> cross section factor for Non uniform Flow
(4.13)
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
If Fr
= 0.8 (subcritical flow)
v/√(gy1) = 0.8 (4/y1)/√(9.81y1)
= 4/y1
= 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)
Trial & Error Method:-
y2
Y2
E = 1.803
1
1.82
1.01
1.81
1.02
1.8
= 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). EGL q²/2gy1²
q²/2gy2²
E2
E1 y1
y2
q
B ∆z
A Broad Crested Weir
Curve of Depth; y (Channel Bed) versus Specific Energy, E for the Presence of Broad Crested Weir
y
q (constant)
y1
A
y2
E = y + q²/2gy²
B
yc
C A' 45º E2
∆Z
E1
E
E1 = E2 + ∆Z note:y1 +
∆Z = height of Broad crested weir 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)
Flow in Open Channel
Steady Flow
Unsteady Flow
Time Uniform Flow
Non Uniform Flow
Space Rapidly varied Flow
Gradually Varied 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
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 c hange 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) (in m) y2/y1 = ´ (√1 + 8 Fr12 – 1) Energy Loss from jump (into heat),
∆E = E1 – E2 ∆E = (y2 – y1)3
(in m) /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
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 c hannel EFFECT OF BED SLOPE AND CHANNEL FRICTION For a RVF, RVF, the influence of bed slope and channel c hannel friction was not mentioned and assumed that frictional effects may be ignored. (as c hanges take place over a very short distance) However However,, in GVF, GVF, bed slope and c hannel friction are very important and they 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 c hannel and discharge disc harge,, normal depth;
manning’s equation Critical depth; 1
yc
q
2
3
g
For a given channel chape c hape and roughness, roughness, only one value of slope will produce the critical depth and known as critical slope (Sc)
86
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 Take derivative,
dH dx
z y dz dx
v dy dx
Where H = total energy head z = elevation head, v2 /2g = velocity head
2
/ 2 g
dv 2 g dx
2
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 dy
y
dx
x
x
S o
y S o
1
K o K
2
1
Q 2 T gA3
1
Q 2 T gA3
1
For rectangular channel
K o K
2
dy
y
dx
x
S o
1 1
K o K yc y 3
x
y
1
y c y
S o
1
K o K
2
3
2
For very wide rectangular channel
x
x
y s y s
1
yc y
1
yo y
1
yc y
1
yo y
3
3
Chezy 3
10 3
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
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
