Lecture 4
Flow in channel with equivalent roughness, Equivalent roughness estimation.
Flow in close conduit with open channel flow
Flow in channel with equivalent roughness
In some channels different parts of the channel perimeter may have different roughness.
Canals in which only the sides are lined, laboratory flumes with glass side walls and rough bed, natural rivers with sandy bed and sides with vegetation, flood plains with different land uses are some typical examples of such case.
For such channels it is necessary to determine an equivalent roughness coefficient that can be applied to the entire cross- sectional perimeter for use in Manning formula.
This equivalent roughness, also called composite roughness, represents a weighted average value for roughness coefficient.
In applying the Manning€s formula to such channels, it is sometimes necessary to compute an equivalent n value for the entire perimeter and use this equivalent value for the computation of the flow in the whole section.
Figure 2.4: A typical open channel section
Consider channel having its perimeter composed of N types of roughness P1,P2,P3...Pnare the lengths of these N parts and n1,n2....nN are the respective roughness coefficients. Let each part Piassociated with a partial area Aisuch that
∑ 1 = A1+A2+...+AN= A = total area
It is assumed that the mean velocity in each partial area is the mean velocity V for the entire area of flow
V1= V2= V3= ...= VN= V By Manning€s formula
= … … =
Where n = equivalent roughness
P1,n1
P2,n2 P3,n3 Pi,ni Pn,nn
From above equation we can get
=
⎣⎢
⎢⎢
⎡∑
∑
⎦⎥
⎥⎥
⎤ /
This equation commonly known as Horton's equation affords a means of estimating the equivalent roughness of a channel having multiple roughness types in its perimeter.
However if the total force resisting to flow is made equal to the sum of forces resisting to flow over the different segments of the perimeter, we get
= (∑ )
∑
/
This formula is attributed to Einstein and Banks (Chow 1959)
In case of rivers thee are flood plain adjacent to channel which are wide and shallow compared to the main channel. Channels of such kinds are known as compound channel.
Figure 2.5 A compound channel
When the depth of flow is within the deep channel, i.e. y < h,discharge can be calculated using Manning's formula.
When the depth of flow is greater than h then flow spills into flood plains. Therefore, the problem of discharge calculation is complicated, which gives a smaller hydraulic radius for the system; therefore, underestimates the discharge.
The underestimation of the discharge will happen in only a small range i.e. for ym<y < h.
When y > ym the computation of discharge considering whole section will be adequate.
The ym is known as maximum value of y beyond which the underestimation of discharge does not occur.
y 2
1 3
h
For ym<y<h. the channel is considered to be made up of sub-area and discharge in each sub-area can be computed separately. We can know = A R where i is the ith subsection. Total discharge Q is the sum of discharge in each subsection.
Q = ∑ K S /
Example: - Compute discharge for the composite channel section shown in figure below
Take Manning n as = 0.015, = 0.012, = 0.016, = 0.022 and channel bottom slope is given by , = 0.0002.
Solution:- From the channel geometry we can write = 5 , = 3 ,
= √2 +2 = 2.82 , = 3 The equivalent roughness can be evaluated as
Using Horton formula we get = ∑∑
/
0162 . ) 0
3 82 . 2 3 5 (
) 022 . 0 .(
3 ) 016 . 0 .(
82 . 2 ) 012 . 0 .(
3 ) 015 . 0 .(
5 1.5 1.5 1.5 1.52/3
neq
5m
3m
3m
2m n1
P1
n3 n2 P3
P2
n4 P4
2m
A1
A2
Using Einstein and Banks formula:-
= (∑ )
∑
/
0164 . ) 0
3 82 . 2 3 5 (
) 022 . 0 .(
3 ) 016 . 0 .(
82 . 2 ) 012 . 0 .(
3 ) 015 . 0 .(
5 2 2 2 2 0.5
neq
It can be seen that quite similar values of equivalent n is obtained by using both the formulas. The higher value of Manning's n is taken for discharge computations.
The flow area will be the sum of rectangular area A1 and the trapezoidal area A2 A = A1 + A2
= 5 ∗ 3 = 15
=5 ∗ 3
2 ∗ 2 = 8
= 15 + 8 = 23
P = P1 + P2
= 5 + 3 = 8
= 2.82 + 3 = 5.82
= 8 + 5.82 = 13.82
= = . = 1.664m
Q =1
nAR S /
Q = . (23). (1.664) (0.0002 . )= 27.88 cumec
Flow in close conduit with open channel flow
As discussed earlier, the channels with closing top width can be classified as channels of second kind for example, circular section. Here K may not be a single valued function of depth. Therefore, in such channels for a particular discharge there can be two depths possible. In case of channels of second kind it is important for us to find out at which depth this channel carries maximum discharge.
Which can be written mathematically as = 0. When n andS are constant then we have
/
/ = 0, when channel geometry is known this equation can be used to get depth for maximum discharge.
The depth for maximum discharge can also be found simply by equating to zero the first derivative of AR2/3with respect to y, since the discharge computed by Manning formula is proportional to AR2/3for constant n and S.
Similarly, as the velocity by Manning's formula is proportional to R2/3, the depth for maximum velocity can be obtained by equating the first derivative of R2/3to zero.
Here it is assumed that roughness coefficient remains constant as depth changes.
The exact depths for maximum discharge and velocity however will depend on the shape and roughness variation of specific conduit section.
For example, in case of circular section the maximum discharge and velocity occurs at about0.938dand0.81dwheredis the diameter in case of circular channel.
However, in practice for sewer pipe or drain tile does not remain constant as the depth changes; therefore, in practice it is assumed that the maximum discharge and depth occurs as0.97dand0.94drespectively.
Since the maximum discharge and velocity of a closed conduit of gradually closing top do not occur at the full depth, this means that the conduit will not flow full at the maximum capacity as long as it maintains an open channel flow on a uniform grade free from obstructions.
For practical purpose however it may sometimes be assumed that the maximum discharge of a circular conduit with gradually closing top occurs at the full depth because the depth for maximum discharge is so close to the top that there is always a possibility of slight backwater to increase this depth closer to and eventually equal to the full depth.
Example:- The rate of flow of water through a circular channel of diameter 0.5m is 100lit/sec. Find the slope of bed of the channel for maximum velocity. Assume C=50.
Solution:Q=100 lit/sec or 0.1cumecs; D= diameter of the channel=0.5m; C=50;
= 0.81 = 0.405
D y 1 2 cos 1 θ
= 2.247 (128 45′)
= = 0.15m
= 2 = 1.1235
= 0.15 = 0.168
=
0.1 = (0.168 ) × 50√0.15
On solving the above equation we get
= 1 1058.43
Objective questions:-
1. In most economical rectangular section of a channel, depth is kept equal to a. One-fourth of the width
b. Three times the hydraulic radius c. Hydraulic mean depth
d. Half the width
2. In a hydraulically most efficient trapezoidal channel section the hydraulic radius R=
a. y/2 b. y/4 c. 2y d. 4y
3. Most efficient channel section, is a. Trapezoidal
b. Rectangular c. Circular
d. Half hexagon in the form of trapezoid
4. Most economical section of a triangular channel is a. Equilateral triangle
b. Right angled triangle
c. Isosceles triangle having vertex angle equal to 450 d. Right angled triangle with equal sides
5. In a hydraulically most efficient trapezoidal channel section the ratio of the bed width to depth is
a. 1.155 b. 0.50
c. 0.707 d. 1.10
6. Most economical section of a circular channel for maximum velocity, is if a. Depth of water = 0.81 diameter
b. Hydraulic mean depth = 0.304 diameter c. Wetted perimeter = 2.245 diameter d. All the above.
Answers:-
1(d) 2(a) 3(d) 4(d) 5(a) 6(d)
Subjective questions:-
1 What do you mean by uniform flow? What are the characteristics of uniform flow?
2 Enlist the factors affecting the Manning's roughness coefficient.
3 Explain what you mean by the •most economical€ section of a channel. Develop the criterion for the most economical trapezoidal section of a channel if the side slopes are fixed.
4 Show that top width is equal to the length of sloping side for the most economical trapezoidal section.
5 Develop the criterion for the most economical trapezoidal section of a channel if the side slopes are fixed.
6 A circular channel is proposed to lay on a slope of 1 in 2000 and is required to carry 1.5cumec. What size of circular channel should be used if it has to flow half-full take n=0.015. (Ans. 2.1m)
7 A circular sewer of diameter 1m carries storm water to a depth of 0.75m. Compute the hydraulic radius, hydraulic depth and section factor. (Ans. 0.198m, 0.479 and 0.287)
8 A triangular channel with an apex angle of 90° carries a flow of 0.7cumec at a depth of 0.6m. If bed slope is 1 in 100. Find the Manning's roughness of channel. (Ans. 0.0147) 9 A trapezoidal channel is 15m wide and has a side slope 1H: 1V. Bottom slope is 1 in 2500.
The channel is lined with smooth concrete having n=0.015. Compute discharge and velocity for the depth of 3.6m. (Ans. 171.42 cumec)
10 In the above example, find the bed slope, if it carries 76 cumec of flow at 3.6m. (Ans. 1 in 12709)
11 The rate of flow of water through a circular channel of diameter 0.5m is 100 lit/sec. find the slope of bed of the channel for maximum velocity. Assume C = 50. (Ans. 1 1058.43) 12 Determine the dimensions for the economical section of a trapezoidal channel having
discharge area 7m2. Take the side slope of the channel as 2 vertical to 1 horizontal. (Ans.
2.48 )
13 Determine the maximum discharge of water through a circular channel of diameter 1.2m, when the bed slope of the channel is 1 in 1500. Assume C=50. (Ans. 0.839cumec)
14 A trapezoidal channel of most efficient section has side slopes of 1:1. It is required to carry
a discharge of 15cumec with a slope of 1in 1500. Design the section if n=0.0135. (Ans. B=
1.7m, H=2.05m)
15 A circular channel conveys 3.25 cumecs of water when ¾ of vertical diameter is immersed.
The slope of the channel is 0.2m per kilometre. Determine the diameter of channel, using Manning€s formula. Take C= 87.5. (Ans. D =1.9 )
16 Find the bed slope of trapezoidal channel of bed width 2.5m, depth of water 1.2m and side slope of 2H to 3V, when the discharge through the channel is 10cumecs.assume Manning€s n=0.03. (Ans.1 90.91)
References:-
1. Chaudhry, M.H., 1994, Open-Channel Flow, Prentice Hall of India Pvt. Ltd. New Delhi.
2. Chow, V.T., 1959. Open-channel hydraulics, McGraw Hill, New York.
3. Nagaratnam,S., 1976. Fluid mechanics, Khanna publishers, Delhi.
4. Ojha, C.S.P., Berndtsson, R., and Chandramouli, P.N., 2010. Fluid Mechanics and Machinery, Oxford University Press, India.
5. Subramanya, K., 1991. Flow in open channels, Tata McGraw-Hill New Delhi.
