Part II: Non-uniform Flow
Rapidly varied flow
Changing of conditions takes place over a short distance
Gradually varied flow
Changing conditions takes place over a long distance
Reference books 1. Hydraulics in Civil and Environmental Envi ronmental Engineering, Andrew Chadwick and John Morfett, nd 2 Ed, E & FN Spon, 1995. 2. Introduction to Fluid Mechanics, Robert W. Fox, Alan T. McDonald, P. J. Pritchard, 6th 1 Ed, John Willey & Sons, 2004.
1. Rapidly varied flow: •Occurs when there is a sudden sud den change in geometry, Flow over a sharp-crested weirs, Flow
─
Flow through regions of rapidly varied cross-section, Flow
─
e.g. venturi flumes and broad-crested weirs.
•A sudden change in flow regime Normally associated with hydraulic jump phenomena, i.e. Normally
─
flow with high speed and low depth is rapidly changed to low speed and high depth. In open channel channel flow, flow, flow regime regime is defined defined by Froude numb number: er:
Fr = V / gL ( L : characteristic length)
(6.1)
2
1. Rapidly varied flow: •Occurs when there is a sudden sud den change in geometry, Flow over a sharp-crested weirs, Flow
─
Flow through regions of rapidly varied cross-section, Flow
─
e.g. venturi flumes and broad-crested weirs.
•A sudden change in flow regime Normally associated with hydraulic jump phenomena, i.e. Normally
─
flow with high speed and low depth is rapidly changed to low speed and high depth. In open channel channel flow, flow, flow regime regime is defined defined by Froude numb number: er:
Fr = V / gL ( L : characteristic length)
(6.1)
2
1.1 Characteristics of rapidly varied flows: •Water surface profile changes suddenly and has curvature, •Pressure distribution departs from hydrostatic hydro static distribution, •The assumption of parallel streamline does not apply.
1.2 Methods of analysis for rapidly varied flows: •The use of energy and momentum principles •Main task: to determine of water depth •Bernoulli equation and continuity equation ─
Specific energy E
─
Specific discharge q
─
Alternate depths
3
1.3 Energy considerations Example: In a rectangular channel, fluid flows over a section in which the bed gradually rises by ∆z. The upstream depth y1 and discharge Q are known. What is the depth of the flow at position 2? (1)
(2) p2 / γ= 0
p1 / γ =0
Figure 6.1 datum
Assuming no energy loss between (1) and (2), apply the energy Eq: 2
y1 +
V 1
2g
2
=
y2
+
V 2
2g
+ ∆ z
(6.2)
4
Apply the continuity equation between points (1) and (2):
q
=
V1 y1
=
V2 y2
(6.3)
where q (m3 /s/m) is the specific discharge (discharge per unit width). Substitute (6.3) into (6.2):
y1 +
3
q
2 2 1
2 gy
=
y2
+
q
2 2 2
2 gy
2
2 gy2 + y2 (2 g∆ z − 2 gy1 −
q
+ ∆ z
(6.4)
2 2
= 0 2 ) + q
y1
Cubic equation
3 solutions for y2. Which one(s) is(are) correct?
5
To answer the above question, we use the concept of specific energy 2
V / 2 g
The energy at a location is given by: H = z+ y +
where E
=
V
2
2g y+
=
V
z + y +
Q
2
2g
=
y +
Q
Water surface
2
2 gA
2
=
z + E
y Channel bed
2
2 gA
EL
2
(6.5)
datum
z
E: specific energy=energy between energy line (EL) and channel bed. (Please note the definition of E: E = local water depth + velocity head) For a rectangular channel: Q/A = bq/by = q/y E
=
y+
q
2
2 gy
(6.6)
2
Eq. (6.6) gives two asymptotes: y → 0 ⇒ E →
∞
y→
∞
⇒
E
→
y
6
Using Eq (6.6), Eq. (6.4) can be written as
E1
=
E2
+ ∆
z
Figure 6.2
(6.7)
Eq. (6.6)
alternate depths
• Point A is defined by E1 corresponds to location 1, • Location 2 in the channel have E2 and is on the specific energy curve. Possible B1 and B2 whose depths are known as alternate depths • To arrive at B2 from A requires that E2 < E1- ∆z at some intermediate points. This means energy loss in a frictionless system – impossible. Therefore the flow depth at (2) must be B1 on the specific energy curve. 7
Example 6.1: The discharge in a rectangular channel of width 5m is 8 m 3 /s. The normal depth is 1.25 m. Determine the depth of flow above a section in which the bed gradually rises by 0.2 m. Use graphical methods to find the solution.
Solution: Q = 8 m3 /s; b = 5 m; q = Q/b = 8/5 = 1.6 m2 /s 2
2
E = y + q /( 2 gy ) = y + 0.13 / y 2 1
E 1 = y1 + 0.13 / y = 1.33m
From energy Eq., E 1 = E 2 + ∆ z
E 2 = E 1 − ∆ z = 1.13m From the figure we get: y2=1m
Specific Energy
2
∆z
1.8 1.6 1.4 ) 1.2 m 1 ( y 0.8 0.6 0.4 0.2 0
E2=1.13
E1=1.33
y2
0
0.2 0.4 0.6 0.8
1
1.2 1.4 1.6 1.8
E (m)
8
2
Example 6.2 Discharge in a rectangular channel of width 5m is 8 m 3 /s. The normal depth is 1.25 m. Determine the flow depth where the section contracts by 1.0 m. Use graphical and numerical methods to find the solution.
V 1 = Q / A1 = 8 /(5 ×1.25) = 1.28m / s
2.0
E 1 = y1 + V 12 / 2 g = 1.33m
1.5
V 2 = Q / A2 = 8 /(4 × y2 ) = 2 / y2 2
y
2
E 2 = y2 + V 2 / 2 g = y2 + 0.204 / y2
From energy Eq. E 1 = E 2 we get y2=1.18m
y2=1.18m
) m 1.0 (
0.5 E1=1.33m=E2
0
0
0.5
1.0 E (m)
1.5
2.0 9
Example 6.3 The discharge in a rectangular channel of width 5m is 8 m 3 /s. The normal depth is 1.25 m. Determine the depth of flow where the section contracts by 1.0 m and the bed gradually rises by 0.2 m. Use graphical methods to find the solution.
y2
y1
∆z =
Elevation
Plan view
V 1 = Q / 1 = 8 /(5 × 1.25) = 1.28m / s V 2 = Q / A2 = 8 /( 4 × y2 ) = 2 / y2
E 1 = y1 + V 12 / 2 g = 1.33m 2
2
E 2 = y2 + V 2 / 2 g = y2 + 0.204 / y2
From energy equation, we have: E 1 = E 2 + ∆ z or E 2 = E 1 − ∆z Plot the specific energy curve for E2, the water depth corresponding 10 to E2 is y2.
(1)
2.0 Eq (1)
1.5
∆z=0.2
) m 1.0 ( y
y2=0.83 m
0.5 E2
0
0
0.5
1.0 E (m)
E1=1.33m
1.5
2.0 11
1.4 Subcritical, critical and supercritical flow From Eq. (6.6), E
=
y+
q
2
2 gy
(q is constant)
2
(6.6)
The specific volume flow rate q can be expressed as a function of y, q=
2 gy 2 ( E − y )
(E is constant)
(6.8)
Eq (6.6) is plotted in (a) and Eq(6.8) is plotted in (b). (b)
(a) Figure 6.3
Eq(6.6)
Critical point subcritical
supercritical
Eq(6.8)
12
12
Derivation of the critical depth for a rectangular channel The critical depth yc can be evaluated from Eq(6.6) by setting the derivative of E with respect to y equal to zero: From E = y +
q2
2 gy
2
dE / dy = 1 −
, to let E have an extremum value, q
2
gy
3
=0
(6.9) 2
Since q=Vy
1−
V
gy
= 1 − Fr 2 = 0
Fr = 1
Thus, for critical depth, Fr = 1. For a rectangular channel, we have
Fr =
V
=
q 3
(6.10)
Solving for depth in Eq (6.9) we have: y = q 2 / g 1 / 3 = y c
(6.11);
gy
(
Vc =
gyc (6.11a)
and
E c = 3 yc / 2
gy
)
(6.12)
13
A generalized derivation of the critical depth For a generalized cross-section, the specific energy can be written in terms of the total discharge Q and the cross-sectional area A: E = y +
V 2
2g
= y +
Q2
2 gA
2
(6.13)
The minimum-energy is obtained by differentiating (6.13) with respect to y, i.e. dE dy
= 1−
Q
2
dA
3
gA dy
=0
(6.14)
For incremental changes in depth. The change in area is
dA = Bdy
(6.15) 14
Substituting (6.15) to (6.14), we get: 2 3 1− Q B/gA = 0
(6.16)
Therefore, at the critical depth, we have: 2
3
Qmax Bc / gAc = 1
(6.17)
At the critical depth, Q is maximum and E is minimum. The second term in Eq(6.16) is the Froude number squared, Fr = Q 2 B / g A3 =
Q / A gA / B
=
V gA / B
(6.18)
The ratio A/B is the hydraulic depth (B is the surface width). For a rectangular channel, A/B=y, Eq (6.18) is simplified to (6.10). 15
1.5 Some more discussion to Figure (6.3): 1. For a given constant discharge (Figure 6.3a) •
The specific energy curve has a minimum value E
•
For any other E, there are two possible depths ca alternate depths, one is called subcritical and the supercritical.
For supercritical flow, y < y c (Fr > 1)
For subcritical flow, y > y c (Fr < 1)
2. For a given constant specific energy (Figure 6.3b) •
The discharge is maximum at depth yc;
Summary of critical, subcritical and supercritica • Critical Flow (requires Fr = V / gy = 1 ) V: water velocity in channel gy : celerity of long wave when there is a disturbance in
– requires Eminimum or qmaximum; – For For re rect ctan angu gula larr chan channe nels ls:: yc =
3
2 q / g, E m i n = 3 y c / 2 ,
• Subcritical Fl Flow: Fr
< 1,
Vc =
gyc
• Supercritical Flow: Fr
> 1,
Water
velocity < Wave velocity
Disturbances Upstream
travel downstream
water levels are unaffected by downstream control
• Slope classification: mild, steep and critical slopes A mild slope: on which uniform flow is subcritical: y 0 > y c ( o r F r < 1 o r S 0 < S c )
A steep slop: on which uniform flow is supercritical: y 0 < y c ( o r F r > 1 o r S 0 > S c )
A critical slope: on which uniform flow is critical: y 0 = y c ( o r F r = 1 o r S 0 = S c )
where y0 is the uniform depth.
18
Example 6.4 The discharge in a rectangular channel of width 5m is 8 m 3 /s. The normal depth is 1.25 m. Determine the depth of flow where the section contracts by 1.0 m and the bed gradually rises by 0.3 m. m. Use graphical methods to find the solution. y2
y1
∆.. z = ∆z=0.3m
q1 = Q / b1 = 8 / 5 = 1.6
E 1 = y1 + q12 /( 2 gy12 ) = y1 + 0.1305 / y12 = 1.33m (plot curve) q2 = Q / b2 = 2
2
2
2
E 2 = y2 + q2 /( 2 gy2 ) = y2 + 0.204 / y2 (plot curve) (1)
From energy equation, we have: E 1 = E 2 + ∆ z or E 2 = E 1 − ∆z = 1.33 - 0.3 = 1.03m
19
Corresponding to E2=1.03m, there are no crossing points on the specific energy curve. Therefore, there are no solutions. This result suggests that the depth y2 should be the critical depth. From the specific energy curve, we get: ycritical = 0.741m and E2=Emin = 1.112m. (refer to Figure a) From E1=E2+∆z
E1new=1.412m (refer to Figure b)
From the specific energy curve at location (1), we get the new water depth at location (1): y1new=1.34m Therefore, at the contraction, the water depth is y2 = 0.74m And the original water depth is increased to y 1new = 1.34m. 20
2.0 (a) No crossing points with the energy curve No solution. Therefore, y2 should
1.5 ) m 1.0 ( y
E2=1.03
E1~y1
∆z=0.3
be at critical depth
0.5
E2~y2
E1=1.33m
0
0
0.5
1.0 E (m)
1.5
2.0 21
2.0 (b) Move by ∆h
1.5
∆z=0.3
y1=1.34m (new depth) y1=1.2m (original depth)
) m 1.0 ( y
∆z=0.3
E1=1.33m E1=1.412m
ycritical=0.735 m
2
q2=2m /s
Emin=1.112m
0.5
Move by ∆h
E2=1.03m
at location (2) 2
q1=1.6m /s
Emin=1.112m
0
at location (1) 0
0.5
1.0 E (m)
1.5
2.0 22
Example 6.5 The discharge in a trapezoidal channel is 30 m3 /s. The channel has a base width of 4 m and side slopes of 1:2. Determine the critical water depth corresponding to the flow rate in the channel, n=0.022. Solution:
B
(1) At critical water depth, Q 2 B
Fr =
3
gA
yc
1 2
=1
4m
3
With Q = 30m / s, B = 4 + 4 yc , A = ( 4 + B ) yc / 2 = ( 4 + 2 yc ) yc
30 2 (4 + 4 yc ) 3
9.81× (4 + 2 yc ) yc
3
=1
yc = 1.4m 23
(2) Critical slope: Using Eq. (5.6),
Q=
A n
Rh
2/3
1/ 2
Sc
(1)
At critical flow, y = yc
30 =
Ac n
Rhc
2 / 3
Sc
1 / 2
Using n = 0.022, Ac = ( 4 + 2 yc ) yc , Rhc =
(2) ( 4 + 2 yc ) yc 4 + 2 5 yc
Substituting into (2) with yc = 1.4m
S c = ....... 24
Example 6.6 The discharge in a rectangular channel of width 5m is 8 m 3 /s. The normal depth is 1.25 m. Determine the depth of flow immediately prior to, above and after a section in which the bed gradually rises by 0.5 m as shown.
(3) y0=
y3
y1
y2
∆z=
Solution: We define: y0 = original depth y1 = immediately prior to the hump y2 = depth after the hump y3 = depth above the hump
25
Write the energy Eq between (0) and (3):
E 0 = E 3 + ∆ z
(1)
2
For q = Q / B = 8 / 5 = 1 . 6 m / s , y 0 = 1 . 25 m , ∆ z = 0 . 5 m 2 2 E0 = y0 + q /(2 gy0 ) = 1.33 m
E 3 = 1.33 − 0.5 = 0.83m
Plot E = y + q 2 /( 2 gy 2 ) = y + 0.1305 / y 2 (plot curve)
(2) (3)
From the figure, we can see that for hump higher that 0.37m, there would be no crossing points (no solution) between E3 and specific energy curve, i.e. the fluid does not have enough energy to pass over a hump higher than 0.37m. To pass over a hump higher than 0.37m, water dams up (known as damming action or backwater) before the hump until it acquires sufficient energy to pass over, after which the hump acts as a control. 26 The minimum energy required is the E min (or Ec) at the critical point.
The depth of water at a control for a known flow rate is given by (6.11):
yc =
(
2
1/ 3
)
q/ g
3
= 1.62 / 9.81 = 0.64 m= y3
(4)
The minimum energy for the flow to pass the hump is:
E min = 3 yc / 2 = 3 × 0.64 / 2 = 0.96m
(5)
Since y0 > yc , the flow prior to the hump is subcritical. The E1 prior to the hump is obtained by using
E 1 = E min + ∆ z = 0.96 + 0.5 = 1.46m
(6)
From the figure, corresponding to E1=1.46m, we have y1 = 1.39m or 0.34m. However, because the flow prior to the hump is subcritical, It can only increase its specific energy by increasing its depth since there is no control point prior to the hump. Thus y 1 = 1.39m. (Note: y1 can also be obtained using E 1 = y1 + q 2 /(2 gy12 ) = 1.46 )27
Assuming no energy loss from (1) to (2), i.e. E1 = E2, 1.46 = y2 +
q
2
2 2 gy2
= E 2
(7)
From the figure, we get y2 = 1.39m or 0.34m. Since a control usually causes the flow to change from subcritical to supercritical or vice versa, thus the depth y2 = 0.34m is expected downstream of the hump as shown in the figure.
Damming action
28
2.0 (a)
E3=0.83m
Emin=0.96m ∆z
=0.37m
max
2
y1=1.39m
1.5
q=1.6m / s
y0=1.25m ) m 1.0 ( y
No crossing points No solution. E1=1.46m = E2
yc=0.64m 0.5 0.34m ∆z=0.5m
0
0
0.5
1.0 E (m)
E0=1.33m 1.5
2.0 29
1.6 Flow characteristics over humps and contractions y1
y2
y2
y
y2
y2
E
30
Example 6.7
The discharge in a 6 m wide rectangular concrete channel (n = 0.013) is 50 m3 /s. Determine: (a) The critical water depth and critical slope of the channel (b) Types of flow if the flow at different sections of the channel has water depths of 3.72, 1.92 and 1.55 m. (c) The Froude Number of the flow corresponding to flow depths of 3.72, 1.92 and 1.55 m (d) The normal water depths and the type of slopes if the channel slopes are 0.001 and 0.04 respectively
Solution: Given : Q = 50m 3 / s, b = 6m, n = 0.013 (a) yc =
3
2 2 q / g = 3 (50 / 6) / 9.8 a= 1.92 m, From Eq. (5.6),
A 2 / 3 1 / 2 Q = Rh Sc , A = yc b, Rh = yc b /(b + 2 yc ) n
S c = 0.0258 31
(b) At section of: y=3.73m, since y > yc, y=1.92m,
⇒ subcritical flow since y = yc, ⇒ critical flow
y=1.55m,
since y < yc,
⇒ supercritical flow
(c) The Froude number: Fr = V / gy
, V=Q/(yb) =50/(yb)
At y=3.73m, V=2.24m/s,
⇒ Fr = V/(gy)1/2 = 0.37
At y= 1.92m, V=4.34m/s,
⇒ Fr = V/(gy)1/2 = 1
At y=1.55m, V=5.76m/s,
⇒ Fr = V/(gy)1/2 = 1.38 2 / 3
1 / 2
(Eq 5.6) (d) The normal water depths: Since Q = Rh S0 , n 3 With A=by, Rh = by /(b + 2 y ) , Q = 50m /s, S0 = 0.001 and 0.004, ⇒ S0
Normal depth y
Slope type
0.001
2.71m > yc
Mild slope
< Sc = 0.0258
0.04
0.76m < yc
Steep slope
> Sc = 0.0258
32
2. Hydraulic Jumps A hydraulic jump is a steplike increase in fluid depth in an open channel flow. Commonly seen below weirs and sluice gates.
Hydraulic jump length
Supercritical flow (upstream)
Hydraulic jump Subcritical flow 33
Consideration of specific energy: •Downstream: Subcitical flow, y2 > yc Specific energy: E2,
•Upstream: Supercritical flow, y1 < yc, Specific energy: E1,
•Energy difference: ∆Es = E1 − E2 > 0 •So after a hydraulic jump, there is energy loss
y
E2
y2
y2 V1
y1
V2
yc
E1
y1 Emin
∆Es
E
34
2.1 Hydraulic jump-formation and characteristics Formation:
− Formed when transit from supercritical flow to subcritical flow Characteristics:
− Transition from supercritical to subcritical flows is rapid, − Length of hydraulic jump: about 7 times downstream depth, − The transition involves large energy loss due to turbulence, − Energy loss can be found in terms of y1, y2 and upstream Fr, −Hydraulic jump can be used to dissipate water energy, such as in a spillway of a dam to reduce damage to riverbed. 35
2.2 Downstream or sequent depth for a rectangular channel The basic equations involved are: •Continuity equation •Momentum equation
Assumptions: • Ignore boundary friction (due to a short length of hydraulic jump), • The channel bed slope is very small, • Pressure distributions at (1) and (2): hydrostatic
36
•Momentum equation: F1 − F2
=
F1 + M1
=
∑ F x
M2
−
M1
F2
+
M 2 =
= ∆ M
(6.19) Μ
(6.20)
where M, the pressure force plus momentum flux, is constant for any given flow rate q. y (2)
(1)
p1
y2
p2 Ignore friction
y1
Sequent depth
Initial depth
F + M
37
Assuming hydrostatic pressure at (1) and (2), we have: 2 F1 = ρ gy1 / 2, F2
= ρ gy 2
M2
= ρ qV 2
1
q
= ρ qV 1 ,
2
2
+
gy1
y1
=
2
q
2
gy2
2
/ 2
2
+
y2
(6.21)
2
Momentum function Μ =
q2 gy
+
y 2 2
(for rectangular channel)
1 2 = 1 + 8 Fr 1 y1 2
y 2 y1 y2
(
=
1 2
[(1
+
2 2
8 Fr
1/ 2
)
1 / 2
)
−
−1 ,
]
1
(6.22)
(6.23)
(6.24) 38
2.3 Energy loss h L
= ∆E =
Since Fr 2
E1 − E2
=
V
2
gy
h L
=
=
q
( y1 +
q
2 2
2 gy 1
) − ( y2 +
q
2 2
2 gy 2
)
(6.25)
2
gy
3
= ∆ E =
And use Eqs. (6.23) and (6.24):
( y2 − y1 )
3
(6.26)
4 y1 y2
• Energy loss increases dramatically as the relative height of the jump increases.
39
Example 6.8 Determine the depth y4 immediately upstream of the hydraulic jump that will form in the situation shown below if q = 1.6 m2 /s. From Example 6.6, we have: y1 = 1.39m, y2 = 0.34m, yc = y3 = 0.64m
Damming action
y0 = 1.25m
y3=yc
y4
y5=y0
40
Assuming y5 = y0 = 1.25m,
Fr 2 =
V 5 gy5
=
q / y5 gy2
=
1 = 1 + 8 Fr 22 2 y 5
y 4
(
As y5 = 1.25m,
Energy loss:
h L =
1.6 / 1.25 9.81× 1.25 1/ 2
)
= 0.37
− 1 = ........
y4 can be calculated
( y5 − y4 ) 3 4 y4 y5
= ....... 41
2.4 Classification of hydraulic jumps: The actual structure of a hydraulic jump depends on Fr
42
2.5 Some further discussion on hydraulic jumps •The relations for hydraulic jumps developed in the present unit are based on a rectangular and horizontal channel, •Hydraulic jumps in other channel configurations are similar to those for rectangular channels, •The expression of sequent depth and energy loss in other channel configurations are somewhat different from jumps in rectangular channels.
43
3. Analysis of Gradually Varied Flows •Characteristics of gradually varied flow ─
Water depth and velocity change gradually,
─
Flow is non-uniform,
─
Water surface changes smoothly and continuously,
─
Friction loss along the channel is not negligible.
• Tasks – Deduce the trend of water surface change (classification of surface profiles) – Calculate water levels and velocity along the course of the channel (quantitative evaluation) • Analysis method – Energy equation – Continuity (mass conservation) equation
44
3.1 The equations for gradually varied flow Assuming the change in the total energy is due to frictional losses over some distance ∆ x, the energy grade line (EGL) and hydraulic grade line (HGL) can be drawn as shown below. Friction slope S f
datum
45
Conservation of the energy gives: 2
S0 ∆ x + y1 +
V 1
2g
2
=
S f ∆ x + y2
S0 ∆ x
=
+
V 2
(7.1)
2g
S f ∆ x + y2
2
−
y1
+
V 2
2g
2
−
V 1
(7.2)
2g
The differential form by choosing ∆x small → d x: S0dx
=
S f dx + dy + d ( 2
Since
d V ( ) dy 2 g
=
d
Q
V 2 2g
)
2
( 2 ) dy 2 gA
= −
(7.3) Q
2
dA
3
gA dy
2
= −
Q B gA
3
,
(7.4)
A: the cross-sectional area; Q: discharge (m 3 /s). Substitute (7.4) into (7.3) and rearrange: dy dx
=
S0 − S f
1 − Q 2 B / gA3
(7.5)46
Eq. (7.5) is valid for any cross section shapes For a rectangular channel, since Froude number is defined as Fr =
V gy
Eq. (7.5) can then be rearranged as
dy dx
=
S0 − S f 2
1 − V / gy
=
S0 − S f 2 r
1 − F
(7.6)
• Eq. (7.5) allows the water surface profiles of gradually varied flow to be deduced, • y is measured vertically from the channel bottom, • the slope of the water surface dy/dx is relative to channel bottom. 47
3.1.1 A note on the friction slope Strictly speaking, the slope term in Manning’s Eq (5.6) is the friction slope Sf , since for steady uniform flow, S0 = S f . Rearranging Manning’s equation to find S f gives: 2
S f
=
2
n Q P
4 / 3
(7.8)
10 / 3
A
Eq. (7.8) implies that S f decreases as the water depth y increases, which in combination with (7.6) means that
y < y0
⇒
S f
>
S0
y > y0
⇒
S f
<
S0
S f > S 0
(7.9) S f < S 0
48
3.2 Calculation of Gradually Varied Flow Profile 3.2.1 Direct step method: Task: calculate the distance between two cross sections • Resources: channel slope S0, channel properties, water depth y 1 at section 1 and water depth y2 at section 2 are known • How? 2
From Eq. (7.1) S0 ∆ x + E1
=
S0 ∆ x + y1 +
2
V 1
=
2g
S f ∆ x + E2
S f ∆ x + y2
∆ x =
E1 − E 2 S f
−
S0
=
+
V 2
2g
∆ E
(7.11)
∆S
S f can be calculated using Manning’s equation: 2
S f
=
2
n Q P 10 / 3
A
4 / 3
(7.12)
49
Example 7.1 The water depth of a gradually varying flow ( q = 1.6 m2 /s) in a wide rectangular channel (n = 0.015) is 0.34 m right behind a hump and 0.38 m at a location further downstream. What would be the distance between the two locations?
(7.11) Sf : Slope of water surface (Eq. (7.12) S0: Slope of channel bed (Eq. (5.6) 50
q2
51
52
53
3.2.2 Numerical methods:
y0 y1
•Task: calculate yi (i=1 to n) • Resources: channel properties;
∆ x
y2
...
yi
yi+1
∆ x
• Method: An iterative procedure From Eq.(7.5), we can get y ' =
dy dx
=
S 0 − S f 2 r
1 − F
Using Taylor’s expansion:
y i + 1 = y i +
dy dx
Qn S 0 − 2 / 3 AR h = 2 1−
2
(7.13)
Q b gA 3
∆ x = y i + y ' ∆ x
(7.14)
2 2 Qn Q b 1 / 1 − where y ' = ( y ' i + y ' i +1 ) and y ' i = S 0 − 3 2 / 3 gA i 2 Ai R hi 54
The Iterative Procedure: 1. Use Eq. (7.13) to calculate y'i where yi is known either as initial point or from a previous cycle of this calculation 2. Set y 'i+1 = y ' i as a first estimate 3. Use current values of y ' i and y ' i+1 to calculate yi+1 from Eq. (7.14) for a selected ∆x 4. Find a revised estimate of y ' i+1 from Eq. (7.13) using the yi+1 value from step 3 5. If the new y ' i+1 value is not close enough to the previously calculated value, repeat step 3, 4 and 5 using the latest estimate of y’i+1 found in step 4 6. Once the iteration procedure has yielded successive estimates of y ' i+1 and yi+1 within acceptable limits of accuracy, proceed to the next section of channel and repeat the process 7. Terminate the process when the desired reach has been covered 55
3.3 Water surface profiles and their classifications Take a channel with mild slope as an example. The surface profile may occupy the three regions shown below and the sign of dy / dx can be found for each region:
y0 y c
56
(a) Region 1 for the mild slope: • Since y>y0 > yc, we have Sf < S0 from Eq. (7.8) • Since y > yc, we have Fr2 < 1 from Eq. (7.10)
⇒ From Eq. (3.2.6) we get dy/dx is positive. The asymptotic behaviour of the free surface M1: As y → ∞, S f and Fr → 0 ⇒ dy/dx → S 0
⇒ hence the water surface is asymptotic to a horizontal line. As y → y0 , S f → S 0 ⇒ dy/dx → 0
⇒ hence the water surface is asymptotic to the line y = y n. 57
(b) Region 2 on a mild slope: y→ y0
⇒
S f
→
S0
y > yc ⇒ F < 1 2 r
dy / dx → 0
For the M2 profile, as y → yc , dy / dx → ∞ This is physically impossible. This is because as the fluid enters a region of rapidly varying flow and Eq. (7.6) is no longer valid.
(c) Region 3 on a mild slope: 2 y 0 > yc > y, S f > S0 and Fr > 1 ⇒ dy / dx is positive.
For the M3 profile, as y → yc , dy / dx → ∞ This is physically impossible. In practice, an hydraulic jump will form before y = yc. Note: above discussion is for mild slope. For steep, critical, horizontal and adverse slopes, the surface profiles are given in next page. 58
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