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Gradually-varied Flow
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Gradually-varied flow (GVF): A steady non-uniform flow in a
prismatic channel with gradual changes in its water-surface
elevation.Examples of GVF:(a) Drawdown produced by sudden drop in
the channel(b) Backwater produced by increased in bed elevation
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Two basic assumptions are involved in the analysis of GVF:1. The
pressure distribution at any section is hydrostatic.2. The
resistance to flow at any depth can be assumed using uniform-flow
equation, such as the Manning's equation, with the condition that
the slope term to be used in the equation is the energy slope and
not the bed slope. Thus, if in a GVF the depth of flow at any
section is y, the energy slope Sf is:where R is the hydraulic
radius of the section at depth y.Differential Equation of GVFThe
total energy H of a gradually-varied flow in a channel of small
slope is:
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Since the water surface varies in the longitudinal x-direction,
the depth of the flow and the total energy are functions of
x.Differentiating Eq. (4.1) with respect to x,
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Eq. (4.2) can now be rewritten as(3.21)
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(a) If K = conveyance at any depth y and Ko = conveyance
corresponding to the normal depth yo, thenOther forms of Eq.
(4.3)Similarly, if Z = section factor at depth y and Zc = section
factor at the critical depth yc,
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Using Eqs. (4.4) and (4.5), Eq. (4.3) can be written asThis
equation is useful in developing direct integration techniques.(b)
If Qn represents the normal discharge at a depth y and Qc denotes
the critical discharge at the same depth y,
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(c) Another form of Eq. (4.2) is This equation is called the
differential-energy equation of GVF to distinguish it from the GVF
differential equations of Eqns. 4.3, 4.6 and 4.7. This energy
equation is very useful in developing numerical techniques for the
GVF profile computation.Classification of Flow ProfilesIn a given
channel, yo and yc are two fixed depths if Q, n and So are fixed.
Also, there are three possible relations between yo and yc as:(i)
yo > yc(ii) yo < yc(iii) yo = ycFurther, there are two cases
where yo does not exist, i.e. whenthe channel bed is horizontalwhen
the channel has an adverse slope So is -ve.
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Based on the above, the channels are classified into five
categories as indicated in Table 4.1
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Figure 4.3. Various GVF profilesAll curves in region 1 have
positive slopes - backwater curvesAll curves in region 2 have
negative slopes - drawdown curves
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Example 4.1A rectangular channel with a bottom width of 4 m and
bottom slope of 0.0008 has a discharge of 1.5 m3/s. In a
gradually-varied flow in this channel, the depth at a certain
location is found to be 0.30 m. Assuming n = 0.016, determine the
type of GVF profile.Solution:Find the normal depth yo and the
critical depth ycThe normal depth yoManning equationThe critical
depth ycThe channel is a mild slope channelThe profile is of the M2
type
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Examples of Occurrence of Type M Flow ProfilesM1 profile is a
subcritical-flow condition. Obstructions to flow such as weirs,
dams, control structures and natural features such as bends,
produce M1 backwater curves. These extends to several kilometres
upstream before merging with the normal depth.M2 profiles occur at
a sudden drop in the bed of the channel, at constriction type of
transitions and at the canal outlet of pools.M3 profiles occur when
there is a supercritical flow enters a mild-slope channel. Examples
are the flow leading from a spillway or a sluice gate to a mild
slope. M3 normally followed by a small stretch of rapidly-varied
flow and the downstream is generally terminated by a hydraulic
jump. Compared to M1 and M2 profiles, M3 profiles are relatively
short length.
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Examples of Occurrence of Type S Flow ProfilesS1 profile is
produced when the flow from a steep channel is terminated by a deep
pool created by an obstruction, such as a weir or dam. At the
beginning of the curve, the flow changes from the normal depth
(supercritical flow) to subcritical flow through a hydraulic jump.
The profiles extend downstream with a positive water-surface slope
to reach a horizontal asymptote at the pool elevation.S2 profiles
occur at the entrance region of a steep channel leading from a
reservoir and at a break of grade from mild slope to steep slope.
Generally S2 profile are of short length.
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S3 profile is produced when the flow exited from a sluice gate
with a steep slope on its downstream.S3 curve is also produced when
a flow exists from a steeper slope to a less steep slope.
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Examples of Occurrence of Type C Flow ProfilesExamples of
Occurrence of Type H Flow ProfilesC1 and C3 profiles are very rare
and are highly unstable.A horizontal channel can be considered as
the lower limit reached by a mild slope as its bed slope becomes
flatter. It is obvious that there is no region 1 for a horizontal
channel as yo = . The H2 and H3 profiles are similar to M2 and M3
profiles respectively. However, the H2 curve has a horizontal
asymptote.Adverse slopes are rather rare and A2 and A3 curves are
similar to H2 and H3 curves respectively. These profiles are of
very short length.
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Example 4.2A triangular channel has side slope 1 horizontal : 1
vertical and bed slope is 0.001. Estimate and categorize this
channel whether mild, steep or critical when the flow rate is given
as much as 0.2 m3/s through this channel. Assume that Mannings
roughness coefficient n = 0.015. Give various of depths that
categorize the flow profile in type 1, 2 and 3.Solution:The normal
depth yoManning equationThe critical depth ycThe channel is a mild
slope channelCurve M1 y > 0.536 mCurve M2 0.536 m > y >
0.382 mCurve M3 y < 0.382 m
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Control sectionA control section is defined as a section in
which a fixed relationship exists between the discharge and depth
of flow. Weirs, spillways, sluice gates are some typical examples
of structures which give rise to control sections. The critical
depth is also a control point. Any GVF profile will have at least
one control section.
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Profile of gradually varied flow can be determined using dynamic
equation of gradually varied flow.
Equations below are differential equations for gradually varied
flow and they represent depth of water y at a certain distance
x.
Calculations are carried out to:i. determine the length L if the
depth y1 and y2 known. ii. determine either one of water depth (y1
or y2), if the length of L and either one of depth are
known.Calculations of Flow ProfileMethods to determine flow
profile:1. Direct Integration*2. Numerical Integration*3. Multiple
Intergration a. Direct Step Method*b. Standard Step Method4.
Graphical Integration5. Numerical/computer methods* Methods that
will applied for this subject
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Direct IntegrationReplacing u = y/yodu /dy = 1 / yody = yo
du
Substituting in the above equation:From previous
equationIntergrate above equation, hence it become
:(3.27)Substituting = F(u,N)Replacing with
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Then, Applying the above equation between two section (x1, y1)
and (x2, y2) in one channel: Values of F(u,N) can be obtained from
table provided in appendix I. If (u,N) is replaced with (v,J), then
the table can also be used to find the value of F(v, J).
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A very wide river have depth of 3.0 m and slopes of 0.0005.
Estimate length of flow profile produced by a weir that caused
water surface increased as much as 1.50 m at the upstream of weir
(n = 0.035).Example 4.3Solution:1. Determine yo and yc. Given yo =
3.0 m, y = 4.5 m Critical depth , yc = (q2/g )1/3 = (3.987/
9.81)1/3 = 1.175 m
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2. Determine value of N & M For a very wide rectangular
channel,N = 10/3 dan M = 33. Calculate J,J = N / (N M+1)= 10/3= 2.5
(10/3 3 + 1)4. Calculate u1, u2 and v1, v2 u1 = y1/yo = 4.5 / 3 =
1.5u2 = y2/yo = 3.03 / 3 = 1.01v1 = u1N/J = (1.5)3.33/2.5 = 1.72v2
= u2N/J = (1.01)3.33/2.5 = 1.01
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5.Obtain F (u, N) and F ( v, J) from table F(u1, N) = F (1.5 ,
3.33) = 0.189 F(u2, N) = F (1.01 , 3.33) = 1.220 F(v1, J) = F (1.72
, 2.50) = 0.333 F(v2, J) = F (1.01 , 2.50) = 1.867
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5.Obtain F (u, N) and F ( v, J) from table F(u1, N) = F (1.5 ,
3.33) = 0.189 F(u2, N) = F (1.01 , 3.33) = 1.220 F(v1, J) = F (1.72
, 2.50) = 0.333 F(v2, J) = F (1.01 , 2.50) = 1.867
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5.Obtain F (u, N) and F ( v, J) from table F(u1, N) = F (1.5 ,
3.33) = 0.189 F(u2, N) = F (1.01 , 3.33) = 1.220 F(v1, J) = F (1.72
, 2.50) = 0.333 F(v2, J) = F (1.01 , 2.50) = 1.867
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5. Obtain F (u, N) and F ( v, J) from table F(u1, N) = F (1.5 ,
3.33) = 0.189 F(u2, N) = F (1.01 , 3.33) = 1.220 F(v1, J) = F (1.72
, 2.50) = 0.261 F(v2, J) = F (1.01 , 2.50) = 1.867
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L = - 8711 m ; L = 8711m from back of weir. 6. Finding length L,
from equation.
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Numerical Integration ii) For rectangular channelEquations used
are:i) For any shape of channel
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iii) For wide rentangular channel equation (using Chezy, C)
(using Manning)
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A very wide channel (rectangular) have depth 3.0 m and slope
0.0005. Determine type of flow profile and estimate length of
gradually-varied flow profile produced by a weir that has elevated
the upstream flow as much as 1.50 m (assumed n = 0.035). Take N = 4
steps/ section.Example 4.4 Calculate yo and yc Given yo = 3.0 m, y
= 4.5 m Critical depth, yc = (q2/g )1/3= (3.9872/ 9.81)1/3= 1.175 m
y > yo > yc; GVF profile is M1Solution:
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Equation used for wide rectangular channel is: Table 1:
Calculation for flow profileStop calculation at, y = 3.00 myc =
1.175 m; yo = 3.0 m; So = 0.0005
y (m)dy (m)(m)1 (yc/yave)31 (yo/yave)10/3dx (m)4.500
4.1250.3754.31250.97980.70171047.234.125
3.7500.3753.93750.97350.59601224.903.750
3.3750.3753.56250.96420.43611658.263.375
3.0000.3753.18750.95000.18303893.92L = x =7824.31
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Length of flow profile M1 is 7824.31 m from back of weir.7824.31
m1047.231224.901658.263893.92
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A river of width 6 m conveys water at normal depth 2.0 m along a
slope of 0.0002 and n = 0.035, upstream of a water fall. The water
falls to a depth of 1.9 m. Taking N = 3 steps/ section,
Determine
type of flow profile, length of gradually-varied flow profile
produced, andprofile sketch.Assignment Q1 Calculate yo and yc Given
yo = 2.0 m, y = 1.9 mSolution:
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Critical depth, yc = (q2/g )1/3 = (0.91262/ 9.81)1/3= 0.4395 myo
= 2 m
y = 1.9 m
yc = 0.4395 m
yo > y > yc GVF profile is M2
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Equation used for rectangular channel is: Table 1: Calculation
for flow profileStart calculation from y = 1.9 m
Stop calculation at y = 2.00 myc = 0.4395 m; yo = 2.0 m; y = 1.9
m, So = 0.0002,
n = 0.035, N = 3 step/section
y (m)dy (m)yave (m)Aave (m2)Pave (m)Rave (m)KaveKo1 (yc/yave)31
(Ko/Kave)2dx (m)1.900
1.9330.0331.91711.59.8331.169364.721387.1690.98790.12691297.7
(1284.50)1.933
1.9670.0331.95011.79.9001.182373.667387.1690.98860.07362239.39
(2222.33)1.967
2.0000.0331.98311.99.9671.194382.658387.1690.98910.02376950.19
(6886.14)L = x =10487.3 (10392.97)
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1.9 m1.933 m1.967 m519 m896 m2780 mM22.0 m
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Direct Step MethodUsing concept of energy continuity Length AC =
BD,
Energy equation , E1 + z1 = E2 + z2 + hL Gradually varied flow
with respect to length of dx
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If Hence, equation used for calculating gradually varied flow
is: Remember! 1. End calculation at y = ( 1 0.01 ) yo 2. Accuraccy
of calculation, depending on the number of numerical, N
choosen.
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A 100 m wide and 3.0 m deep channel has 0.0005 slope. Determine
type of flow profile and estimate length of gradually varied low
profile produced by a weir that elevated the upstream flow as much
as 1.50 m.(assumed n = 0.035).
Example 4.5SolutionAssumed the river as a very wide channel,
flow/ width are: Critical depth, = 1.175 mq = 3.987 m3/s/m
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y > yo > yc; GVF profile is M1
Take N = 4 steps / section
Stop calculation at y = 3.01 m R = y, q = 3.987 m3/s/m , n =
0.035 , So = 0.0005Table 2:Calculation of flow profile by using
direct step method
1y (m)2R (m)3v (m/s)4v2/2g(m)5E (m)6E (m)7S x 10-48Savg x
10-49(So - Savg) x10-410x(m)11L
(m)4.54.50.8860.04004.540_1.294___04.134.130.9650.04754.1780.3621.7221.5083.4921036.710363.763.761.0600.05733.8170.3612.3542.0382.9621218.822553.393.391.1760.07043.4600.3573.3272.8412.1591653.539083.033.031.3160.08833.1180.3424.8394.0830.9173729.67638
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Hence the length of flow profile is 7525 m from weir.
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A river of width 6 m conveys water at normal depth 2.0 m along a
slope of 0.0002 and n = 0.035, upstream of a water fall. The water
falls to a depth of 1.9 m. Taking N = 5 steps/ section, Using
direct step method, determine
type of flow profile, length of gradually-varied flow profile
produced, andprofile sketch.Assignment Q2
