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1CHAPTER 3
NON - UNIFORM FLOW IN OPEN CHANNEL
(PART 1)
3.1 Use of Specific energy3.2 Determination of critical depth by various method3.3 Control Sections
Non Uniform FlowThe energy grade line, water surface and channel bottom are not parallel;
Sf Sw So
Sf = slope energy grade line , Sw= slope of the water surface , So= slope of the channel bed
2Total Energy
3.1 : USE OF SPECIFIC ENERGY
The concept of specific energy introduced by Bakhmeteff (1932)
Specific energy is defined as the sum of depth and velocity head.
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.
3The 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.
v2/2gv1/2g
y1Q
Qy2 = ?
z
EGL
Figure 1 :Transition with bottom step
If for the moment we neglect the energy loss, the energy equation combined with continuity can be written as
E1 = E2 + zy1 + Q/2gA1 = y2 + Q/2gA2 + z (4.1)
Where isy = depthQ = dischargeA = Cross-sectional area of flowz = z2 z1
= change in bottom elevation from cross section 1 to 2
4Considered square channel (prismatic and straight)E = y + q/2gy (4.2)
where, q = flowrate per unit width (m/s/m)q = Q/b , and A = by
Equation 4.2 (energy, E, depth of flow, y and flowrate, q) may be written/defined in 2 conditions as below;-
i. E and y if q is constantii. q and y if E is constant
Specific Energy in Open Channel
From Eq (4.2)
E = y + q/2gy0 = y Ey + q/2g (4.3)
It is apparent from Equation (4.3) 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.
i) E and y if q is constant
5Specific Energy Diagrams (E-y)
y
oE = y + q/2gy
DB
A
Cyc
Emin
y
yv/2g (head of velocity)
y > yc
y < yc45
*Note q is constant.
Emin
y
yc
EEo
y1
y2
Specific Energy Diagrams
6At C, specific energy is minimum and normal depth at this point is 'critical depth', ycIf
y > yc ; v < vc ==> Subcritical flow (steady)y < yc ; v > vc ==>Supercritical flow (turbulant)
Differentiation of Equation (4.2)E = y + Q/2gAdE/dy = 1 - (Q/2g)(2/A)dA/dydE/dy = 1 (Q/gA).T
= 1 (v/gA).T= 1 v/gD
At critical point, E is minimum i.e. dE/dy = 0therefore;-
v/gD = 1 ; (froude,Fr = 1)v/2g = D/2 or v/(gD) = 1
Tdy
dA = T.dydA/dy = T
For a rectangular channel,
Hydraulic Depth, D = A /T = by/b = yTherefore, at critical condition ==>>
Fr = 1 (y = yc, v = vc)vc /(g yc ) = 1 (a)vc/2g = yc/2 (b)
From the schematic diagram;- (E = min, y = yc )E = y + q/2gydE/dy = 1 q/gyc = 0q = gyc yc = (q/g) (c)
7q = gyc but q = vy = vcycvc yc = gycvc = gycvc =(gyc) or vc /2g = yc (d)
Emin = yc + q/2gyc= yc +(g yc) / (gyc)= yc + yc2
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 constantE = y + q/2gyq = 2gy(E - y) (4.4)
At critical point, dq/dy = 0
Differentiation of Equation (4.4): q = 2gy(E - y) = 2g(Ey y)2qdq/dy = 2g(E 2yc 3yc = 02ycE = 3ycE = 1.5ycyc = 2/3E (f)
qmax = 2gy (E y) (from Eq. 4.7)= 2gy (1.5yc yc)= gyc
qmax = (gyc) (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.
8E constanty
y1
yc
y2
q qmax q
E
yc=2/3E
q y curve
Specific Energy Diagrams (q-y)
Critical flow criteria (square/rectangular channel)Fr = 1.0
'E' is minimum for 'q' constantEmin = 1.5yc yc = (q/g)
'q' is maximum at E constantyc = 2/3Eminqmax = (gyc)
Velocity head (vc/2g) is one-half of critical depth, ycvc/2g = yc /2
Critical velocity (vc);vc = (gyc)
9is known as the Froude Number, Fcgyv
Froude Number, Fr and Flow classification
q2/gyc3 = 1Then,
vc2/gyc = 1 at critical conditionsSo,
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
10
Critical Depth in non-rectangular channels
Critical conditions for channels of various shape
11
Cross Section Factor for Critical Flow (Z)Q /g = A/ T(4.9)
or
Z = Q /g = ADwhere; D = A/T
Equation (4.10) may be define as Cross Section Factor (z) for critical flow. (Where as AD is cross section factor,z)
If Fr = a, thereforez = Q/(ga) = AD
Equation (4.11) is use generally in critical flow analysis.
Q = zgor
Q = z(ga)z = Q/(ga) ===>> cross section factor for
Non uniform Flow [email protected]
12
Example 1Water flows in a rectangular channel with 5 m width and 8m/s flowrate. Depth of channel is 1m. Determine the specific energy for this channel.
Example 2From 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 3
a) A wide and straight river was flows with 3.5m/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 Mannings Coefficient 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
13
Solution 3(a): q = 3.5m/s/myc = (q/g) = [(3.5)/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,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 = yTherefore;
q = y(5/3)S(1/2)/n
At critical flow in Non-Uniform flow;-q = yc(5/3)Sc(1/2)/nSc = (qn/yc (5/3))
= [3.5 x 0.035 / (1.08)(5/3)]= 0.012 or 1/86 (answer)
14
Specific Energy for y1 = 4.6 E = 4.6 + (3.5)/19.62 (4.6) = 4.63m
butE = y2 + q/(2gy2)
Where as y2 = depth at the same specific energy
THERE ARE 2 METHODS:-Trial & Error Methody2 should be in supercritical flow, therefore, the value of y2 is smaller than yc.
If y2 >>>> ;E
15
Solution 4:Q = 12m/sV = Q/A = 12/3y1 = 4/y1If 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 + q/2gy1
= 1.366 + (4)/2(9.81)(1.366)= 1.803m
Solution Calculation for y2 at the same flow rate and specific energy
E1 = E2 = y2 + q/2gy21.803 = y2 + 0.815/y2
Critical depth, yc = (q/g)(1/3) = (4/9.81)(1/3) = 1.177m (as a reference for trial & error method)
Trial & Error Method:-
y2 = 1.02 m (depth at supercritical flow)
16
Exercise1. 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 velocityusing :
-Trial and Error-Design Chart
