Chapter 2. Spatially Varied Flow (SVF) PDF
Chapter 2. Spatially Varied Flow (SVF) PDF
Chapter 2. Spatially Varied Flow (SVF) PDF
2.1. Introduction
A steady spatially varied flow represents a gradually-varied flow with non-uniform
discharge. The discharge varies along the length of the channel due to lateral addition or
withdrawal.
For examples: side channel spillways, side weirs, channel with permeable boundaries,
gutters for conveying storm water, drop structures in bottom of channels.
Roof gutter
1
2.2. SVF with increasing discharge
SVF with increasing discharge finds considerable practical applications. Flow in side
channel spillway, wash-water troughs in filter plants, roof gutter, highway gutters are some
typical instances. The lateral flow enters the channel normal to the channel flow direction
causing considerable turbulence. It is difficult to assess the net energy imparted to the flow
and as such the energy equation is not of much use in developing the equation of motion.
2.2.1 Differential Equation of SVF with Increasing Discharges
In applying the momentum equation, the following assumptions are made:
1. The pressure distribution is assumed to be hydrostatic. This amounts to assuming the
water surface curvatures to be moderate.
2. The one-dimensional method of analysis is adopted. The momentum correction factor
is used to adequately represent the effect of non-uniformity of velocity distribution.
3. The friction losses in SVF are assumed to be adequately represented by a uniform
flow resistance equation, such as Manning’s formula.
4. The effect of air entrainment on forces involved in the momentum equation is
neglected.
5. It is assumed that the lateral flow does not contribute any momentum in the
longitudinal direction.
6. The flow is considered to be steady
7. The channel is prismatic and is of small slope.
2
M2 – M1 = P1 – P2 + Wsin - Ff (1-1)
or M = - P + Wsin - Ff (1-1a)
in which M = momentum flux = Q2/A
P = pressure force = Ah
where h = depth of the center of gravity of the flow cross-section from the water
surface; Wsin = component of the weight of the CV in x direction and Ff = frictional force =
ASfx.
dividing eq.(1.1a) by x and taking limits as x 0,
dM dP
AS0 AS f (2-2)
dx dx
In this:
dM 2Q dQ Q 2 dA
(i) 2
dx A dx A dx
2Q Q 2T dh
q* 2
A A dx
By taking moments of the areas about the new water surface after a small change dh in
depth
A h dh dA
dh
2
A dA h d h
dAdh
Adh hdA Ad h dAd h
2
By neglecting second-order small quantities:
Ad h hdA Adh
dP dh
Thus A
dx dx
Hence, eq. (2.2) simplifies to
2Qq* Q 2T dh
dh
S0 S f
gA2 gA3 dx dx
3
2Qq*
S 0 S f
dh gA2
or (2-3)
dx Q 2T
1
gA3
equation (2-3) is the basic differential equation governing the motion in SVF with
increasing discharge.
In general, q* is function of x. However, in lateral spillway channel q* is constant.
In view of the high non-uniform velocity distribution in the channel cross-section, it is
necessary to use proper value of momentum correction factor . In lateral spillway channel,
value of as high as 1.60 are not uncommon. It may be note that if = 1.0 and q* = 0, eq.(2-
3) will be the same that of the differential equation of GVF.
Eq. (2-3) is a non-linear equation and is more complex than GVF equation. A numerical
solution of the equation is feasible.
Starting from a section where the flow properties are known (such as control section),
the water surface profile can be computed.
gA3
Qc = critical discharge modified by .
T
Equation (2-4) reduces to:
4
K 2Q 2 Q 2 2K 2 q*
1 2
dh K 2Qn
2
Qn gA2Q
S0 2
dx Q
1
Qc
2
Q 2K 2 q*
1 1
dh Qn gA2Q
S0 2
(2-5)
dx Q
1
Qc
Redefining Qn1 = modified normal discharge
Qn
Qn1
2K 2 q*
1
gA2 Q
2K 2 q* Qn2T
or 3 1 (2-6)
gA2 Q Ag
q* 1 S 0T gA2
(2-7)
Q 2 A K 2
Which is the equation of the transitional profile for SVF with increasing discharge.
In general SVF with increasing discharge,
5
x
Q Qi q*dx
0
Transitional profile and the CDL for given q* are plotted and control point is determined
by the intersection of these two lines as:
q* = 3.0 m3/s.m hc = 4.0 m xc = 56.0 m
q* = 2.0 m3/s.m hc = 2.65 m xc = 40.0 m
6
h A T Qc P R K 1/xt xt xc = Qc/q*
gA3
Qc
(m) (m 2 ) (m) T (m) (m) m 3 /s eq.(2.8) (m) q * = 3.0 q * = 2.0
1,0 6,00 7,0 15,56 7,828 0,7664 335 0,05707 17,52 5,19 7,78
1,5 9,75 8,0 30,15 9,243 1,0549 674 0,04020 24,87 10,05 15,08
2,0 14,00 9,0 48,92 10,657 1,3137 1120 0,03153 31,72 16,31 24,46
2,5 18,75 10,0 71,93 12,071 1,5533 1677 0,02618 38,20 23,98 35,96
3,0 24,00 11,0 99,31 13,485 1,7797 2350 0,02251 44,43 33,10 49,66
3,5 29,75 12,0 131,23 14,899 1,9967 3145 0,01982 50,46 43,74 65,61
4,0 36,00 13,0 167,83 16,314 2,2067 4068 0,01775 56,34 55,94 83,91
4,5 42,75 14,0 209,28 17,728 2,4115 5125 0,01610 62,11 69,76 104,64
5,0 50,00 15,0 255,73 19,142 2,6120 6322 0,01475 67,78 85,24 127,87
5,5 57,75 16,0 307,36 20,556 2,8094 7665 0,01363 73,37 102,45 153,68
6,0 66,00 17,0 364,31 21,971 3,0040 9161 0,01268 78,89 121,44 182,15
2,65 20,2725 10,3 79,674959 12,4953319 1,62240589 1866,046 0,02494075 40,095 26,5583197 39,8374795
7
dH dZ
In which: S f and S0
dx dx
d V 2 d Q 2 2Q dQ 2Q 2 dA dh
3
dx 2 g dx 2 gA2 2 g A2 dx A dh dx
dA dQ
T and q*
dh dx
Equation (2.10) simplifies to
Qq*
S0 S f
dh gA2
(2.11)
dx Q 2T
1
gA3
Equation (2.11) is the basic differential equation governing the motion of SVF with
decreasing discharges.
Note the difference between eq. 2.11 and eq. 2.3:
- When q* = 0, eq. 2.11 will be the same differential equation of GVF; unlikely the SVF
with increasing discharges, in this case q* is not externally controlled but will be implicitly
governed by the flow conditions.
c. Computations
- The determination of the critical-flow control point in the SVF with decreasing
discharge is difficult as q* is not explicitly known. Normally, SVF with lateral out flow occurs
in a relatively small portion of length of canals and the upstream or downstream depth,
depending upon the flow, is known through the characteristics of the outflow structure and
main channel, this forms, the starting point for the SVF computations.
- It is necessary to establish a relationship for q* as a function of the relevant flow
conditions. The SVF profile is then computed by using a numerical procedure. The method if
approach depends upon the understanding of the particular flow phenomenon.
- A few specific examples of flow situation where SVF with lateral outflow occurs are
described below.
2.4. Side weir
2.4.1 General concepts
8
Definition of Sketch of side weir flow
- The above figure is a definition of the flow over a side weir. Side weirs are usually
short structures with L/B 3.
- The longitudinal water surface should increase in the downstream direction when the
main channel flow is subcritical throughout.
- The water profile would be a decreasing curve for supercritical flow in the channel.
- The possible flow profiles can be broadly classified into the following three
categories:
Type 1: the channel is on mild slope and the weir heights s > hc1 where hc1 is the critical depth
corresponding to the incoming discharge Q1 at section 1 (fig. a). At the downstream end the
normal depth corresponding to discharge Q2 will prevail. Thus h2 = ht, the tail depth. At the
section 1, the depth h1 will be such that hc1 < h1 < h0, where h0 = normal depth for Q0 = Q1.
Along the weir the depth increases from h1 to h2. Upstream of section 1 will be an M2 curve
from h0 to h1. The control for the SVF will be the downstream depth h2 = ht.
Type 2: The channel is on mild slope (h0 > hc1) and with s < hc1 (fig.b). If the weir is long,
flows below critical depth are possible. At the upstream end of weir, the depth h1 can be
considered to be equal to hc1. At the downstream end the depth h2 will rise to the tailwater
depth ht through a jump. Depending upon the tailwater depth, the jump can also advance into
the weir portion. The control point for this type 2 profile is at section 1.
9
Type 3: The channel is on steep slope (h0 < hc1) and with s < hc1 (fig.c). The upstream depth
h1 = h0 decreasing depth water profile will start from Section 1. At section 2 the depth reaches
a minimum value and in the downstream channel the water surface rises through an S 2 curve
to meet the tailwater depth ht. The control for this profile is h1 = h0 at section 1.
dQ
Qh
i.e.
dh
dx
(2.12)
dx gh3 B 2 Q 2
The outflow rate = discharge over the side weir per unit length
dQ 2
= CM 2 g h s
3/ 2
(2.13)
dx 3
10
In which CM = a discharge coefficient known as the De Marchi coefficient. Also, since
the specific energy E is assumed to be constant, the discharge in the channel at any cross-
section is given by:
Q Bh 2 g ( E h) (2.14)
From eq. 2.12, 2.13 and 2.14:
dh 4 CM E hE h3
(2.15)
dx 3 B 3h 2 E
Assuming that CM is independent of x, on integration,
M h, E , s const
3B
x (2.16)
2CM
In which
2 E 3s Eh Eh
M h, E, s 3 sin 1
Es hs Es
Equation 2.16 is known as the De Marchi equation and the function M h, E, s is known
as the De Marchi varied flow function. Applying eq. 2.16 to section 1 and 2:
x2 x1 L
3B
M 2 M 1 (2.17)
2CM
Knowing L, s and (Q and h) at either 2 or 1, the discharge over side weir Qs can be
computed by eq. 2.17 and by the continuity equation
Qs = Q1 – Q2 (2.18)
De Marchi coefficient CM
- In subcritical approaching flow: the major flow parameter affecting CM is the
Froude number of the approaching flow:
3Fr12
CM 0,611 1
Fr12 2 (2.19)
V1
Where Fr1 and equation 2.19 can be simplified as:
gh1
1 Fr12
CM 0,864 (2.20)
2 Fr12
11
condition of constancy of specific energy (E = constant). It is apparent that iterative
procedures have to be adopted in the calculation of Qs or L.
Example 1
A rectangular channel, b = 2.0 m, n = 0.014, is laid on as slope S0 = 0.001. A side weir
is required at a section such that it comes into operation when the discharge is 0.6 m 3/s and
divers 0.15 m3/s when the canal discharge us 0.9 m3/s. Design the elements of the side weir.
Solution
The normal depths at the two discharges are found by referring to the following table:
Applying the Chezy’s equation:
2/3
bh0
Q A.V bh0 R 2 / 3 S 01/ 2 bh0
1 1
S 01/ 2
n n b 2h0
2/3 2/3
Qn bh0 2h0
bh0 2h0
b 2h0 2 2h0
1/ 2
S
By using trial and error method, the normal depth can be obtained for each case:
For Q = 0.6 m3/s => h0 = 0.33 m
For Q = 0.9 m3/s => h0 = 0.44 m
- The height of weir crest: s = 0.33 m
3
- For discharge of 0.9 m /s:
1/ 3 1/ 3
Q2 0.9 2
The critical depth hc1 2 2
0.274 m
b g 2 9.81
Because s > hc1 and h0 = 0.45 m > hc1 => the flow is of type 1
By using the De Marchi equation: h1 = h0 = 0.45 m
Q1 0.9 V1 1
V1 1,023m / s Fr1 0.4924
bh1 2 0.45 gh1 9.81 0.45
V12 1.0232
Specific energy E1 h1 0.44 0.4933m E2
2g 2 9.81
Discharge over the side weir Qs = 0.15 m3/s => Discharge at the end of the weir:
Q2 = 0.90 – 0.15 = 0.75 m3/s
V22 Q22
At section 2, E2 h2 h2 E1
2 g bh2
2
2g
0.752
h2 0.4933
2 9.812 h2
2
By using trial and error method, finding the depth at the section 2: h2 = 0.46 m
Q2 0.75 V2 0.815
V2 0.815m / s Fr2 0.384
bh2 2 0.46 gh2 9.81 0.46
De Marchi’s function:
12
2 E 3s Eh Eh
M h, E , s 3 sin 1
Es hs Es
2 E 3s 2 0.4933 3 0.33
0.02082
Es 0.4933 0.33
0.4933 0.44 0.4933 0.44
M 1 0.02082 3 sin 1 1.840
0.44 0.33 0.4933 0.33
The discharge over the side weir Qs = Q1 - Q2 = 0.900 – 0.623 = 0.277 m3/s
PROBLEMS
2.1. Show that the following equation is applicable to a control section where critical depth
occurs in a frictionless lateral spillway channel:
S 02 gAcTc
1
4q*2
2.2. A side spillway channel is 100 m long and is rectangular in cross-section with b = 5 m, n
= 0.020, = 1.30 and S0 = 0.15. If the lateral inflow rate is 1.75 m3/s.m, find the critical depth
and its location.
Answer: hc = 3.14 m xc = 43.7 m
13
2.3. A lateral spillway channel is trapezoidal in cross-section with b = 10.0 m, side slope m =
0.5, and Manning’s roughness n = 0.018. The bed slope is 0.08. If the lateral inflow rate is 2.5
m3/s.m length, find the critical depth and its location. Assume = 1.20.
Answer: hc = 3.30 m xc = 74.75 m
2.4. A 3.0 m wide rectangular channel can carry a discharge of 3.6 m3/s at a normal depth of
1.20 m. Design a side weir so that it will pass all the flow in the canal when the discharge is
2.0 m3/s and will divert 0.6 m3/s when canal discharge is 3.6 m3/s.
Answer: L = 1.353 m s = 0.79 m
2.5. A rectangular canal of 2.0 m width carries a flow with a velocity of 8.75 m/s and depth of
1.25 m. A side weir of height 0.75 m and length 1.20 m is provided in one of it walls. Find the
quantity of flow diverted by the side weir.
Answer: Qs = 0.206 m3/s
2.6. A rectangular channel is 2.0 m wide and carries a flow of 3.0 m3/s at a Froude number of
0.3. A uniformly discharging side weir having contouring on the sides only is set at a height
of 0.4 m above the bed with it crest horizontal. If the length of the side weir is 1.80 m,
estimate the total flow diverted by side weir.
Answer: Qs = 0.206 m3/s
14