16 FLOW NOPEN

CHAPTER CHANNELS

16.1 INTRODUCTION

IElow in open channels is defined as the flow of a liquid with a free surface. A free surface is a
surface having constant pressure such as atmospheric pressure. Thus a liquid flowing at
atmospheric
nressure through a passage is known as flow in open channels. In most of cases, the liquid is taken
as water. Hence flow of water through a passage under atmospheric pressure is called flow in
open
channels. The flow of water through pipes at atmospheric pressure or when the level of water in the
pipe is below the top of the pipe, is also classified as open channel flow)
In case of open channel flow, as the pressure is atmospheric, the flow takes place under the force
of gravity which means the flow takes place due to the slope of the bed of the channel only. The
hydraulic gradient line coincides with the free surface of water.

º 16.2 CLASSIFICATION OF FLOW IN CHANNELS

The flow in open channel is classified intothe following types :

1. Steady flow and unsteady flow, 2. Uniform flow and non-uniform flow,
3. Laminar flow and turbulent flow, and 4. Sub-critical, critical and super critical flow.

16.2.1 Steady Flow and Unsteady Flow. If the flow characteristics such as depth of flow,
velocity offlow, rate of flow at any point in open channel flow do not change with respect to time, the
tlow is said to be steady flow. Mathematically, steady flow is expressed as

=0, =0 dy =0 ..(16.1)
Where V= velocity, Ì = rate of flow and y= depth of flow.
at any point in open channel flow, the velocity of flow, depth of flow or rate of flow changes
u Tespect to time, the flow is said to be unsteady flow. Mathematically, unsteady flow means
#0 0r dy #0 r +0.

16.2.2 Uniform Flow and Non-uniform Flow. If for a given length of the channel, the
velocity of flow, depth of flow, slope of the channel and cross-section remain constant, the flow is

737
738 Fluid Mechanics
said to be unifornm. On the other hand, if for a given length of the channel, the velocity of
of flow etc.,, donot remain constant, the flow is said to be non-uniform flow. Mathematically low dep
and non-unifom flow are written as :
dy = 0, = 0for uniform flow

and dy +0, 0for non-uniform flow.

classified in
Non-uniform flow in open channels is also called varied flow, which is the fol owing
(wo types as :
(i) Rapidly Varied Flow(R.V.F.), and
(i) Gradually Varied Flow (G.V.F.).
Rapidly Varied Flow (R.V.F.). Rapidly varied flow is defined as that flow in which denik
thee i
tlow changes abruptly over a small length of the channel. As shown in Fig. 16.1 when
obstruction in the path of flow of water, the level of water rises above the obstruction and then fi
and again rises overa small length of channel. Thus the depth of flow changes rapidly over asho
length of the channel( For this short length of the channel the flow is called rapidly varied fo
(R.VE)) GRADUALLY VARIED RAPIDLY
FLOW VARIED FLOW

UNIFORMNON-UNIFORMFLOW
FLOW

Fig. 16.1 Uniform and non-uniform flow.

kín) Gradually Varied Flow (G.V.F). If the depth of flow in a channelchanges gradually over:
long length of the channel, the flow is said to be gradually varied flow and is denoted by G.V.F.
16.2.3 Laminar Flow and Turbulent Flow. The flow in open channel is said to be laminar i
the Reynold number (R) is less than 500 or 600. Reynold number in case of open channels is defined as
pVR ...(16.2)
R, =
where V= Mean velocity of flow of water
R= Hydraulic radius or Hydraulic mean depth
Cross-section area of flow normal to the direction of flow
Wetted perimeter
p and u = Density and viscosity of wate.
If the Reynold number is more than 2000, the flow is said to be turbulent in open channel flow. I
R. lies between 500 to 2000, the flow is considered to be in transition state.
A6.2.4 Sub-critical, Critical and Super Critical Flow. The flow in open channel is aa
be sub-critical if the Froude number (F,) is less than 1.0. The Froude number is defined as
V ...(16.31
F,=
gD
 Flow inOpen Channels 739
flow
where V= Mean velocity of
D= Hydraulic depth of channel and is equal to the ratio of wetted area to the top width of chaninei
A
where T = Top width of channel.
Cah.critical flow is also called tranquil or streaming flow. For sub-critical flow. F.<10.
The flow is called critical if F, = 1.0. And if F, > 1.0, the flow is called super critical or shooting
)
or rapid or torrential.
16.3 DISCHARGE THROUGH OPEN CHANNEL BY CHEZY'S FORMULA
Consider uniform flow of water in a channel as shown in Fig. 16.2. As the flow is uniform. it
means the velocity, depth of flow and area of flow will be constant for a given length of the channel.
Consider sections 1-1 and 2-2.

2
W sin i

DATUM LINE 2
Fig. 16.2 Uniform flow in open channel.
Let L= Length of channel,
A = Area of flow of water,
i= Slope of the bed,
V= Mean velocity of flow of water,
P=Wetted perimeter of the cross-section,
f= Frictional resistance per unit velocity per unit area.
The weight of water between sections 1-1 and 2-2.
W= Specific weight of water x volume of water
= WXA XL
Component of Walong direction of flow = W xsin i= WAL sini ....i)
Frictional resistance against motion of water =fx surface area x (velocity Y
The value of n is found experimentally equal to 2 and surface area = P x L
Fictional resistance against motion =fx PxL x V ...(iiy
Ine forces acting on the water between sections 1-1 and 2-2 are:
.Component of weight of water along the direction of flow,
. Fnction resistance against flow of water,
3. Pressure force at section
l-1,
*. Pressure force at section 2-2.
As the depths of water at the sections 1-1 and 2-2 are the same, the pressure forces on these two
sections are same and acting in the opposite direction. Hence they cancel each other. In case of
uniform flow, the velocity of flow is constant for the given length of the channel. Hence there is no
accel
Zero. eration acting on the water. Hence the resultant force acting in the direction of flow must be
740 Fluid Mechanics
Resolving all forces inthe direction of flow, we get
WAL sin i-fx PxL× V= 0
WAL sin i=fx PxLx V
WAL Sini wAX X sin i
fx PxL f P

or V= X sin i

A
But = m
P
= hydraulic mean depth or hydraulic radius,
W
=C= Chezy's constant
f
A w
Substituting the values of and in equation (ii), V= CJm sin i
P

For small values of i. sin i tani i . V=Cymi ...(164

Discharge, Q= Area x Velocity =A x V
=Ax Cmi ..(165
Veroblem 16.1 Find the velocity of flowand rate offlow of water through a rectangular chan:
of6 mwide and 3 mdeep, when it is running full. The channel is having bed slope as lin 2000. Tai:
Chezy's constant C = 55.
Solution. Given :
Width of rectangular channel, b=6 m
Depth of channel, d=3m
Area, A =6%3 = 18 m
Bed slope, i=l in 2000 =
2000
Chezy's constant, C= 55
Perimeter, P=b+ 2d =6 +2 x3 = 12 m
A 18
Hydraulic mean depth, = 1.5 m
P 12
Velocity of flow is given by equation (16.4) as,
1
V= Cymi = = 1.506 m/s. Ans.
2000
Rate of flow, Q= Vx Area = VxA = 1.506 × 18 = 27.108 m'/s. Ans.
Problem 16.2 Find the slope of the bed of arectangular channel of width 5 mwhen depth
water is 2m and rate of flow is given as 20 m'/s. Take Chezy's constant, C= 50.
Solution. Given :
Width of channel, b=5 m
Depth of water, d= 2 m
Rate of flow, Q= 20 m'/s
 Flow in Open Channels 741
C= 50
Chezy'sconstant
slope
Letthebed ACAmi
Using
cquation (16.5), we have Q=
A= Area=bxd= 5 x 2 = 10m?
where 10 10
A 10 10
pb+2d 5+2x2 5+ 4
10 20.0 2
20.0 = 10 x 50 x Vo xiorV9 i=
500 50
10 4
have
Squaring both sides, we 9 2500
4 9 36 1 1
i= Ans.
2500 10 25000 25000 694.44
36
Bed slope is l in 694.44.
Problem 16.3 Aflow of water of 100 litres per second flows down in a rectangular flume of width
600mm and having adustable bottom slope. If Chezy's constant C is 56, find the bottom slope
necessary for uniform flow with a depth of flow of 300 mm. Also find the conveyance Kof the flume.
Solution. Given :
100
Discharge. Q= 100 litres/s = = 0.10 m/s
1000
Width of channel, b= 600 mm = 0.60m
Depth of flow, d= 300 mm = 0.30 m
.:. Area of flow. A=bxd= 0.6x 0.3 = 0.18 m'
Chezy's constant, C= 56
Let the slope of bed =i
A 0.18 0.18 0.18
Hydraulic mean depth, m =
P b+ 2d 0.6+ 2 >x0.30
= 0.15 m
1.2

Using equation (16.5), we have Q=ACAmi

0.10 = 0.18 x 56 x J0.15 xi or 0.10
/O.15i = 0.18 × 56
2
0.10
Squaring both sides, we have 0.15 i= = ,000098418
0.18 x 56

.000098418 1 1
i=
0.15
=,0006512 = 1

.0006512
- 1524
. Ans.

Slope of the bed is 1l in 1524.

Veonveyance Kof the channel
Equation (16.5) is given as Q= ACmi
742 FIuid Mechanics

which can be written as

section.
where K = ACNm and Kis called conveyance of the channel
=3.9039 ms. Ans
K= ACm = 0.18 x 56 x /o.15
channel of width 8 m and tide
Problenm 16.4 Find the discharge through a trapezoidal mand value of Chezy'sconstan:
2.4
Ihorizontal to 3 vertical. The depth of flow of water is
The slope of the bed of the channel is given 1 in
4000.
Solution. Given :
Width, b=8 m
Side slope = 1hor. to 3 vertical 224m
d= 2.4 m
Depth.
Chezy's constant, C= 50 A

1 8m
i= 0.8
Bed slope, 4000
Fig. 16.3
From Fig. 16.3 when depth, CE =2.4,
1
the horizontal distance BE = 2.4 x = 0.8 m
3

Top width of the channel, 0.8 = 9.6 m

CD = AB + 2 x BE = 8.0 + 2 x
Area of trapezoidal channel, ABCD is given as, 2.4 = 17.6 x 1.2 =21:
A= (AB + CD )×
CE = (8 + 9.6) x
2 2
(: BC=
P= AB+ BC+ AD = AB + 2BC
Wetted perimeter,
BC = JBE + CE= /(0.8) + (2.4) 2.529 m
But
P=8.0 +2 × 2.529 = 13.058 m
A 12.12
Hydraulic mean depth, m= =1.617 m
P 13.058

The discharge, is given by equation (16.5) as

1
m's. Ans.
Q= ACNmi =21.12x 50, 1.617 x- 4000 = 21.23
Problem 16.5 Find the bed slope of trapezoidal channel of bed width 6 m, depth of water
discharge through the channel is 30 ms
side slope of 3horizontal to 4 vertical, when the
Chezy's constant, C= 70.
Solution. Given: D
Bed width, b= 6.0 m
Depth of flow, d= 3.0 m 3 m
=3 horizontal to 4vertical
Side slope
Discharge, Q= 30 m'ls
A
Chezy's constant, C= 70 -6 m
From Fig. 16.4, for depth of flow
=3m= CE Fig. 16.4
3
Distance, BE = 3 x = 2.25 m
4 4
744 Fluid Mechanics
Depth of flow, d= 4.0 m
Angle made by each side with vertical,
i.e., ZABD =2CBD= 30°
From Fig. 16.6, we have
Area, A = Area of ABC 30

=2x Area ABD =

2x ADxBD B
= AD x BD
2
Fig. 16.6
AD
= BD tan 30° x BD tan 30°=
BD AD= BD tan 3
= 4 tan 30° x 4 =9.2376 m²
Wetted perimeter, P= AB + BC= 2AB
(:: AB= Br
=2 BD² +AD² =24.0 +(4 tan 30°)*
= 2/16.0 +5.333 =9.2375 m.
.. Hydraulic mean depth, m= A 9.2376
= 1.0 mn
9.2375
Using equation (16.5) for discharge,
1
Q= ACmi = 9.2376 x 55 x = 16.066 m'/s. Ans.
1000

º I6.4 EMPIRICAL FORMULAE FOR THE VALUE OF CHEZy's

CONSTANT
Equation (16.4) is known Chezy's formula after the
who developed this formula in 1975. In this equation C isname of a French Engineer, Antoine Che
dimensionless co-efficient. The dimension of C is known as Chezy's constant, which is no

=
LIT LIT L
T/i T
{iis dimensionles
Hence the value of Cdepends upon the system of units. The following are the empirical formula:
after the name of their inventors, used to determine the value of C:
1. Bazin formula ( In MKS units): C=: 157.6 ...(16.6
K
1.81+
Vm whose
where K = Bazin's constant and depends upon the
roughness of the surface of channel,
values are given in Table 16.1.
m = Hydraulic mean depth or hydraulic radius.
 Flow in Open Channels 745
Ganguillet-Kutter Formula. The value of Cis given in MKS unit as
2.
0.00155
23+
C= N
0.00155 N ...(16.7)
1+| 23+
Vm
where N= Roughness co-efficient which is known as Kutter's constant, whose value for different
surfaces are given in Table l6.2
i= Slope of the bed
m = Hydraulic mean depth.
Table 16.I Values of K in the Bazin's Formula
S. No. Nature of Channel inside surface Value of K
Smooth cemented or planned wood 0.11
2. Brick or concrete or unplanned wood 0.21
3. Rubble masonary or Ashlar or poor brick work 0.83
4. Earthen channel of very good surface 1.54
5. Earthen channel of ordinary surface 2.36
6. Earthen channel of rough surface 3.17

Mable 16.2 Value of Nin the Ganguillet-Kutter Formula

S. No. Nature of Channel inside surface Value ofN
1 Very smooth surface of glass, plastic or brass 0.010
2. Smooth surface of concrete 0.012
3. Rubble masonary or poor brick work 0.017
4. Earthen channels neatly excavated 0.018
5. Earthen channels of ordinary surface 0.027
6. Earthen channels of rough surface 0.030
7. Natural streams, clean and straight 0.030
8. Natural streams with weeds, duppools etc. 0.075 to .15

3. Manning's Formula. The value of Caccording to this formula is given as

mll6
...(16.8)
N
where m= Hydraulic mean depth
N= Manning's constant which is having same value as Kutter' s constant for the normal
range of slope and hydraulic mean depth. The values of Nare given in Table 16.2.
Problem 16.8 Find the discharge through a rectangular channel 2.5 m wide, having depth of
warer 1.5mand bed slope as l in 2000. Take the value of k = 2.36 in Bazin's formula.
Solution. Given :
Width of channel, b= 2.5 m
Depth of flow, d= 1.5 m
.:. Area, A= bx d= 2.5 x 1.5 = 3.75 m
Wetted Perimeter, P=d+b+ d= 1.5 + 2.5 + 1.5 = 5.5 m
 746 Fluid Mechanics
A 3.75 = 0.682
Hydraulic mean depth, m=
P
f=
5.50
1
Bed slope, i=
2000
Bazin's constant, K= 2.36
Using Bazin's formula given by equation (16.6), as
157.6 157.6
C= 2.36 = 33.76
K
1.81+ 1.81+
/0.682
Discharge, Qis given by equation (16.5), as
Q=A¢Vmi
1
=3.75 x 33.76 x j0.682 X = 2.337 m's. Ans,
2000
Problem 16.9 Find the discharge through a rectangular channel 14 mwide, having d¡nt.
water 3 m and bed slope Iin1500. Take the value of N=0.03 in the Kutter's formul.
Solution. Given :
Width of channel, b=4 m
Depth of water, d=3 nm
1
Bed slope, i= = 0.000667
1500
Kutter's constant, N=0.03
Area of flow, A =bxd= 4x3= 12 m
Wetted perimeter, P=d+b +d=3+4+3= 10 m
12
Hydraulic mean depth, m = = 1.2 m
P 10
Using Kutter's formula given by equation (16.7), as
.00155 1 .00155
23+ +
23+
C= N .000667 .03
.00155) N .03
.00155 x
1+ 23+ 1+23 +.000667, V1.20
i m
23 + 2.3238 +33.33 58.633 = 32.01
1+(23 +2.3238)x.03286 1.832
Discharge, Qis given by equation (16.5), as
m%Ar
Q= AC mi = 12 x 32.01 x12x000667 = 10.867havinga k
Sroslem 16.10 Find the discharge through a rectangular channel of width 2 m,
slope of4 in 8000. The depth of flow is 1.5 mand take the value of Nin Manning sformula
Solution. Given :
Width of the channel, b=2m
Depth of the flow, d= 1.5 m
Area of flow, A = bx d=2 x 1.5 = 3.0 m²
 Flow in Open Channels 747
Wetted perimeter, P=b+ d+d= 2+1.5 + 1.5 = 5.0 m
A 3.0
Hydraulicmean depth, m = P S.0
= 0.6

4 1
Bed slope, i=4in 8000 =
8000 2000
Value ofN = 0.012
IIsing Manning's formula, given by equation (16.8), as
C= mll6 x 0.6 = 76.54
N 0.012
Discharge, is given by equation (16.5), as
Q= ACymi
= 3.0x 76.54 /0.6x 2000 m²/s =3.977 m's. Ans.
Problem 16.11 Find the bed slope of trapezoidal channel of bed width 4 m, depth of water 3 m
and side slope of 2 horizontal to 3 vertical, when the discharge through the channel is 20 m's.
Take Manning's N= 0.03 in Manning's formula C = N
Solution. Given :
Bed width, b=4 cm 3 m.

Depth of flow, d= 3 m
Side slope =2 hor. to 3 vert. 4m 2m

Discharge, Q= 20.0 m's Fig. 16.7

Manning's, N=0.03
From Fig. 16.7, we have

Distance, BE =dx 3 3x 3 =2m

.:. Top width, CD = AB + 2BE

= 4+2x 2= 8.0 m
Area of flow, A = Area of trapezoidal section ABCD

(AB+ CD) Xd=

(4+ 8) x3 = 18 m?
2
Wetted perimeter, P= AD + AB + BC = AB + 2BC (:: AD = BC)
=4.0+ 2/BE + EC? =4.0 +2/2' +3² =4.0+2x /13 = 11.21 m
18
Hydraulic mean depth, m= = 1.6057
11.21
1
Using Manning's formula, C=l m 1/6
0.03
x(1.6057) = 36.07
N
748 Fluid Mechanics

Using equation (16.5) for discharge,

0= ACymi = 18 x 36.07 x /1.6057 xi or 20.0 = 8)),s
20.0 1
= 0.0005909 Ans.
822.71) 1692
Problem 16.12 Find the diameter of acircular sewer pipe which is laid at a slope of lin
carries a discharge of 800 litres/s when flowing half ful. Take the vaue of. Manning'sN =
Solution. Given : 0.020
Slope of pipe, i=
8000
Discharge, Q= 800 litres/s = 0.8 m'ls
Manning's, N= 0.020
Let the dia. of sewer pipe, =D
d

Depth of flow,
2
Fig. 16.8
Area of flow, A=D²y D
4 2 8
TD
Wetted perimeter, P=
2

D
A 8 D
Hydraulic mean depth, m=
P D12 4

Using Manning's formula given by equation (16.8), C = 1

N
The discharge, Q through pipe is given by
equation (16.6), as
Q= ACymi
D? 1
X -mo.mi
8
or 1
0.80=D x xmxi
.020

-ml6+ 1/2) 1 1 Xm23x00118

8 .020 8000 8 .020
D)/3
0.2195 x D' ×
4 m=
.2195
42/3 x D' x D - 0.0871 D3
D83 0.80
.0871
=9.1848
D= (9.1848) = (9.1848)0 =
2.296 m. Ans.
 Flow in Open Channels 749
16.5 MOST ECONOMICAL SECTION OF CHANNELS

(Asection of achhannel is said to be most economical when the cost of construction of the channel
minimum But the cost of construction of a channel depends upon the excavation and the lining. To
iskeep the cost down or minimum,the wetted perimeter, for agiven discharge, should be minimum.
This conditionis utilized for determining the dimensions of ?a economical sections of different form of
channels.
Mosteconomical section is also called the best section or most efficient section as the discharge,
nassing through a most economical sectionof channel for a given cross-sectional area (A), slope of the
hed () and a resistance co-efficient, is maximum. But the discharge. Qis given by equation (16.5) as
|Axi
Q
=AC\mi = AC P
For a given A, i and resistance co-efficient C, the above equation is written as
1
Q= K, where K=AC\Ai =constant
Hence the discharge, will be maximum, when the wetted perimeter Pis minimum. This condition
willbe used for determining the best section of a channel i.e., best dimensions of a channel for a
given area.
The conditions to be most economical for the following shapes of the channels willbe considered :
1. Rectangular Channel, 2. Trapezoidal Channel, and 3. Circular Channel.
Most Economical Rectangular Channel.< The condition for most economical sec
tion, is that for a given area, the perimeter should be minimum. Consider a rectangular channel as
shown in Fig. 16.9
Let b= width of channel,
d= depth of the flow,
Area of flow, A = bxd ...(i)
Wetted perimeter, P=d+b+ d=b+2d ...(ii)
A
From equation (1), b=
d
Substituting the value of b in (i), Fig. 16.9 Rectangular channel.
A
P=b+2d = + 2d ...(iüi)
d
For most economical section, P should be minimum for a given area.
dP
=0
d(a)
Differentiating the equation (iii) with respect to d and equating the same to zero, we get
A
0 + 2 =0 or A = 2d
d(d) Ld ...(16.9)
But A=bxd, .. bx d= 2d or b=2d
A bxd (: A bd, P =b+ 2d)
Now hydraulic mean depth, m=
P b+ 2d
 750 Fluid Mechanics
2d xd
2d + 2d

2d' d
(:b)
4d 2
From equations (16.9) and (16.10), it is clear that rectangular channel will be most economical
depth of flow.
(() Either b = 2d means width is two times
(in) Or m= means hydraulic depth is half the depth of flow.>
VProblem 16.13 Arectangular channel of width, 4 mis having a bed slope of 1 in 1500. Find t,
maximum discharge through the channel. Take value of C = 50.
Solution. Given :
Width of channel, b=4 m
1
Bed slope. i=
1500
Chezy's constant, C= 50
Discharge will be maximum, when the channel is most economical. The conditions for n.
economical rectangular channel are :
(1) b= 2d or d b4 = 2.0mn
2 2
(ü) d 2
m= = = 1.0 m
2 2
Area of most economical rectangular channel, A= bxd =
4.0 × 2.0 = 8 m
Using equation (16.5) for discharge as

Q=ACmi = 8.0 >x 50x 1

Problem 16.14 A rectangular channel carries

in 2000. Find the most
Solution. Given : economical
10X1500
dimensions the channel if C = 50.
of
= 10.328 m'/s. Ans.
water at the rate of 400litres/s when bed slope i.

Discharge, Q= 400 litres/s = 0.4 m'ls

Bed slope, i=
2000
Chezy's constant, C= 50
For the rectangular
() Width, channel to be most economical,
b= 2d
(ii) Hydraulic mean depth, m=
d
2
Area of flow, A =bxd= 2d xd=
Using equation (16.5) for 2d
discharge,
Q= AC\mi
 752 Fluid Mechanics
(New hydraulic mean depth, m' corresponds to the condition of maximum discharge. And
also equal to
1.732
0.866 m)
2 2

Max. discharge, Q', is given by Q= AC\m'i = 6.0 x 55 x0.866 x 1000 =9.71 m''s
Increase in discharge = -Q=9.71 - 9.66 = 0.05 m's. Ans.
<16.5.2 Most Economical Trapezoidal Channel. The trapezoidal section of a chan.
be most economical, when its wetted perimeter is minimum. Consider atrapezoidal section of
as shown in Fig. 16.10.
b= width of channel at bottom,
chranmr.
Let
d= depth of flow,
0=angle made by the sides with horizontal,
D

t-nd
Fig. 16.10 Trapezoidal section.
(i) The side slope is given as 1 vertical to n horizontal.

Area of flow, A= (BC + AD) Xd= b+(b+2nd) Xd (: AD =b+ 2nd

2
2b + 2nd
xd=(b +nd) xd
A
=b+ nd
d
A
b= - nd
.d
Now wetted perimeter, P= AB + BC + CD = BC+ 2CD (:" AB = CD
=b+ 2CE +DE = b+ 2/n'a² +d =b+2dyn +1 ..0
Substituting the value of b from equation (ii), we get
A
P=-nd+
d 2dyn² +1
For most economical section, P should be minimum or dP
=0
d(d)
Differentiating equation (iii) with respect to d and equating it equal to zero, we ge
d [A
d(d)Ld - nd +2d yn'+1|
=0

A
2n+2yn' +1 = o (:" n is
constao!
 Flow in Open Channels 753
A
+n = 2yn' +1
or

Substitutingthe value of Afrom equation (i) in the above equation,

(b+ nd)d b+ nd
+n =2,n' +1
d +n= 2/n' +1
b+ nd+ nd b+ 2nd
b+ 2nd
or d d -=2/n' +1 or = dyn'+1 ...(16.11)
2
b+ 2nd
But from Fig. 16.10, 2
= Half of top width

and dln + l = CD = one of the sloping side

Fauation (16.11) is the required condition for a trapezoidal section to be
can be expressed as half of the top width must be equal to one of the sloping most economical, which
ó Hydraulic mean depth sides of the channel.
A
Hydraulic mean depth, m=
P
Value of A from (i), A = (b + nd) xd
Value ofP from (ia), P=b+ 2dn² +1 =b+ (b +2nd) (: From equation (16.11)
b+ 2nd =2d/n² +1)
= 2b + 2nd = 2(b+ nd)
A
(b+ nd)d d
Hydraulic mean depth, m=
...(16.12)
P 2(b + nd)
Hence for a trapezoidal section to be most economical hydraulic mean depth must be
the depth of flow,
equal to half
(iii) The three sides of the trapezoidal section of most economical section are
tangential to the
semi-circle described on the water line. This is proved as :
Let Fig. 16.11 shows the trapezoidal channel of most economical section.
Let
e= angle made by the sloping side with horizontal, and
O= the centre of the top width, AD.
Draw OF perpendicular to the sloping side AB.
AOAF is a right-angled triangle and angle OAF =
OF
sin = OF = A0 sin 9 ..(iv)
OA

In AAEB, AE d
sin = A
AB Ja +n'd?
d 1
d
dy1 +n² G
B

Substituting sin =
J1+n?
in equation (iv), we get
Fig. 16.11
754 Fluid Mechanics
1
OF =AOx

AO= half of top width

But
(16.11)
=dln' +1 from equation
b+ 2nd

Substituting this value of AO in equation (v),

dyn'+1
OF = = d depth of flow
Jn' +1 of flow d.
semi-circle is drawn with O as centre and radius equal to the depth the
Thus, if a be tangential to the semi-circle.
section will
sides of most economical trapezoidal economical trapezoidal section are:
Hence the conditions for the most
b+ 2nd
= dyn' +1
2
d
m =
2
flow will touch the three sides
. A semi-circle drawn from O with radius equal to depth of
channel. vertical and the sicr: -
Problem 16.16 A trapezoidal channel has side slopes of lhorizontal to 2 if it is
the bed is 1in 1500. The area of the section is 40m. Find the dimensions of the section
economical. Determine the discharge of the most economical section if C = 50.
Solution. Given :
Horizontal
Side slope, n=
Vertical 2
1
Bed slope, i=
1500
Area of section, A = 40 m²
Chezy's constant, C= 50 Fig. 16.12
For the most economical section, using equation (16.11)
b+ 2nd b+2x} xd
2
=dyn' +1 Or
2 -a+ +1

b+d 1
+1 =1.118 d
Or b=2x 1.118d-- d= 1.236 d
But area of trapezoidal section, A = b+ (b+2nd) Xd= (b + nd) d
2
and n:
=(1.236 d+B ) d (:" b= 1236 d
= 1.736 d
But A= 40 m
40 = 1.736 d²