FLOW IN CHANNELS

What is a channel and channel flow?

• A passage through which fluid flows with a free surface.

• To make fluids flow in channels, these are constructed with downward slope in
the direction of flow i.e. flow in a channel is a gravitational flow.

❖ Head causing flow is due to the slope of channel.

❖ Flow in a pipe is a pressure flow.

TYPES OF CHANNELS
❖ Various types of channels are
• Natural Channels
• Artificial Channels
• Open Channels
• Closed Channels
• Prismatic Channels
• Non-prismatic Channels
• Rigid Boundary Channels (lined channels)
• Mobile Boundary Channels (unlined channels)

Natural Channels

• have irregular sections of varying shapes developed in natural ways.

Examples: Rivers, streams etc.

Artificial Channels

• built artificially for carrying liquid for various purposes

• have cross-sections with regular geometric shapes such as

➢ triangular, rectangular, trapezoidal, circular or parabolic etc.

Examples: Laboratory flumes, irrigation canals, sewer pipes etc.

Open Channels

• have no cover at the top.

Examples: Laboratory flumes, irrigation canals etc.

Closed Channels

• have cover at the top.

Examples: Pipes running partly full, underground drains and tunnels running partly full etc.

1
Prismatic Channels

• having constant cross-section and longitudinal slope

Non-prismatic Channels

• if either slope or cross-section varies

Rigid Boundary Channels (lined channels)

• channels with immovable bed and sides

Examples: Concrete, brick channels etc.

Mobile Boundary Channels (unlined channels)

• channels with movable bed and sides

Example: Alluvial channel

TYPES OF FLOW IN CHANNELS

❖ various types of flow in channels can be:

• Steady and unsteady flow
• Uniform and non-uniform flow
• Laminar and turbulent flow
• Subcritical, critical and supercritical flow
• Steady flow- flow in channel (discharge) does not change with time

• Unsteady - flow changes with time

• Uniform - depth of flow remains constant over a given length of the channel

• Non-uniform - depth of flow varies over a given reach of the channel

• Laminar flow - various fluid particles move in layers or laminae or well defined
paths, with one layer of fluid smoothly sliding over an adjacent layer

• Turbulent flow- various fluid particles move in zigzag or irregular manner

❖ To determine the nature of flow, Reynolds number of flow (Re) is calculated

• Re is defined as the ratio inertia force to viscous force

VL VL
• Given by, Re = =
 
➢ L is the characteristic dimension; V is the mean velocity of flow,  is the dynamic
viscosity and  is the kinematic viscosity

2
 • For flow through pipes, L = D, D is the diameter of pipe

• For flow in channels, L = Rh

❖ Rh is known as the hydraulic radius or hydraulic mean depth

➢ Rhis the ratio of cross-sectional area of flow to the wetted perimeter of channel (Rh = A/P)

Type of System Laminar flow Turbulent flow

Pipes Re < 2000 Re > 4000
Channels Re < 500 Re > 2000
between these two values, flow is in transitional state

Subcritical, critical or supercritical flow (defined for flow in channels only)

• criticality of flow is defined in terms of Froude number of flow (Fr)

❖ defined as the ratio of square root of inertia force to the gravity force of fluid

V
• Fr is given by: Fr =
gL
❖ For channels, L = Dh and is called hydraulic depth

• defined as the ratio of cross sectional area of flow to the top width of flow of
channel i.e. Dh = A/ T

Type of flow Froude number

Subcritical <1
Critical =1
Supercritical >1

3
CONTINUITY EQUATION FOR FLOW IN CHANNELS

• Consider L-section of flow in a channel having X-sectional area of flow A.

• Let 1-1 and 2-2 are the two sections (control volume), distance dx apart as
shown in Figure. To start with, consider unsteady incompressible flow through
the control volume)

• Inflow through section 1-1 = Q

 Q 
• Outflow through section 2 -2 =  Q + dx 
 x 
Q
Net volume of flow accumulated per unit time = (Inflow – Outflow) = − dx
x
• Applying the principle of conservation of mass (for flow in channels, its volume
per unit time), any accumulation of volume per unit time is equal to the rate of
increase of volume in the control volume.
• Volume of liquid between sections 1-1 and 2-2 is equal to (A  dx) and its rate of

increase is equal to (Adx )
t
Equating the two expressions, to get

(Adx ) = − Q dx 
 A Q
+ =0 (1)
t x t x
• Eq. (1) is the general expression for continuity equation for flow in channels.
Observations:
A Q
(i) For steady flow, =0  =0 or Q = Constant
t x
(ii) Continuity eq. for flow in rectangular channels
• Consider a rectangular channel of bed b and depth of flow y.
❖ For rectangular channels, discharge Q is generally expressed in terms of
discharge per unit width q, also known as discharge intensity i.e. q = Q/b.
• A=by

From Eq. (1), we get

4
   y q
(b  y) + (q  b) = 0  + =0
t x t x
y 
 + (V  y ) = 0 (q = Q/b = AV/b = V  y)
 t x
y y V
+V +y =0
t x x
• Required expression.
UNIFORM FLOW EQUATIONS

• Consider uniform flow of a liquid in an open channel having inclination  to the horizontal.

• Let 1-1 and 2-2 are two the vertical sections, horizontal distance L apart.

➢ At section 1-1, y1is the depth of flow (vertical distance measured from the channel bed
upto the free surface), V1 is the mean velocity of flow and Z1 is the datum head.

❖ various forces acting between sections 1-1 and 2-2 are:

(i) Weight of liquid, W = Specific weight  Volume of liquid = g  AL,

➢ A is cross-sectional area of flow

❖ component of weight in the direction of flow is Wsin

(ii) Frictional resistance, FR offered to the flowing liquid

➢ Given by: FR = f  PLV , f’ is a constant

2

(iii) Hydrostatic forces P1 and P2 are acting at the ends of two sections
❖ For steady-uniform flow, there is no acceleration in the direction of flow
(velocity of flow is constant)
Sum of components of all the external forces in direction of flow is equal to zero i.e.
P1 + W sin  − P2 − FR = 0
5
 • As depths of flow at the two sections are same and therefore P1= P2.

g A g
g  AL sin  − f PLV 2 = 0 V=  sin  = R h sin 
f P f
• For small values of , sin tan = Sb, Sb is the bed slope of channel

 V = C RhSb
 g 
• C is called Chezy’s constant or Chezy’s resistance coefficient  = 
 f  

❖ As Q = AV,
 Q = AC RhS
• Eq. is known as Chezy’s eq. for calculating discharge of uniform flow in channels.

To prove that in uniform flows Sf = Sb = Sw

• Applying Bernoulli’s eq. between 1-1 and 2-2, to write

V12 V22
+ y1 + z 1 = + y2 + z 2 + hf  hf = (z1 − z 2 ) [ V1 = V2 , y1 = y2 ]
2g 2g
hf  z 1 − z 2 
 =   S f = Sb
L  L 

• Also, Sb = Sw(y1 = y2)

Sf = Sb = Sw

➢ Sf, Sb and Sw are the energy line, bed and water surface slopes, respectively

Observations: (i) Chezy’s Eq. is dimensionally non-homogeneous.

• Variables should be used in same system of units as that of C.
• Generally value of C is given.
• Alternatively, the following Eqs. in SI units can be used to find the value of C.

Bazin’s Equation Kutter’s Eq. Manning Eq.

157.6 23 +
0.00155
+
1 1 1/ 6
C= or Sb K2
C= Rh
K C= N
1.81 + 1 
Rh 0.00155  K 2
1 +  23 + 
 Sb  Rh
87
K1
1 + 0.55
Rh

6
 • K1 is called Bazin’s constant, K2 - Kutter’s constant and N - Manning’s coefficient
or rugosity coefficient
❖ Value of these constants depends on the type of channel surface (i.e. smooth
cement surface to rivers)

➢ K1 varies from 0.11 to 3 for smooth cement surface to rivers

➢ K2 varies from 0.01 to 0.03

➢ N varies from 0.01 to 0.045

1 1/ 6 1 2 / 3 1/ 2
(ii)Substituting C = R h in Chezy’s Eq., to get V= R h Sb
N N
1
As Q = AV Q = AR h2 / 3 S 1b/ 2
N
• Eq. is known as Manning’s Equation.

❖ Both Chezy’s and Manning’s eqs. are used to find the discharge of liquid in any
channel cross-section of any shape.

Let us calculate A and P of different shapes of channels

Rectangular channel

• Area of flow cross-section, A = b  y y

• Wetted perimeter, P = (b +2y)

b
Trapezoidal channel

A D

y 1
n

B C
b

A = (b + ny)y P = (b + 2y 1 + n2 )

Triangular channel section

7
 1
A= Ty
2
T = (AO + OC) = 2AO = 2y tan

 A = y2 tan 
P = (AB + BC) = 2 AB [AB = BC]

 P = 2  y sec 

Circular channel section

O
r

2
A B
C

• Let 2 be the angle (in radians) subtended at the centre by the free surface of liquid

• Area of flow section, A = Area of minor segment ABDA = (Area of minor sector OADBO-
Area of AOB)
2
➢ Area of minor sector OADBO = r 2 = r 2
2
➢ Area of AOB = Area of AOC + Area of COB = 2(Area of AOC)
1 
✓ Area of AOC =   r sin   r cos  
2 
2
r
Area of AOB = r sin  cos  = sin 2
2
2
• Wetted perimeter, P = length of minor arc ABD = = 2r 2 = 2r
2

8
MOST ECONOMICAL CROSS-SECTIONS OF CHANNELS
A A3
• As Q = AV = AC R hSb = AC Sb = C Sb
P P
For a given cross-sectional area of flow A, bed slope Sb and Chezy’s constant C,
discharge Q is maximum, when wetted perimeter P is minimum.

❖ Minimum perimeter reduces the cost of lining the channel and thus resulting in
economy of the project.
• Most economical section is also called most efficient section(has max. discharge
carrying capacity but min. perimeter)

❖ Minimum perimeter condition is used to determine dimensions of most

economical sections of various shapes of channels viz. rectangular, triangular
and trapezoidal.

Most Economical Rectangular Channel Section

• Consider a rectangular channel section of bed width, b and depth of flow, y as
shown in Figure.

b
• Area of flow cross-section, A = b  y (1)
• Wetted perimeter, P = (b +2y) (2)

A 
 P =  + 2y  [Using Eq. (1)] (3)
 y 
• For a given A (i.e. keeping A constant), rectangular section will be most
economical, when P is minimum i.e.
• Differentiating Eq. (3) with r. t. to y and put the derivative equal to zero i.e.
dP A
=- 2 +2=0  A = 2y2  by = 2 y 2  b = 2 y
dy y
A by 2y  y y
• Hydraulic mean depth, R h = = =  Rh =
P b + 2y 2y + 2y 2
❖ Rectangular channel section will be most economical when
➢ Bed width is two times the depth of flow and
➢ Hydraulic mean depth is half the depth of flow
❖ Observation:

• Same results are obtained when P is kept constant and A is minimized i.e. (dA/dy = 0).

9
Most Economical Trapezoidal Channel Section

• Consider a trapezoidal channel section of bed width b and depth of flow y and
sides slope n : 1 (n horizontal to 1 vertical)

A O D

y
y 1
F n:1 n
 B C
E
ny b

( AD + BC) (b + 2ny) + b
• Area of flow cross-section, A = y= y
2 2
A 
 A = (b + ny)y  b =  − ny 
y 
• Wetted perimeter, P = (AB + BC + CD) = (2AB + BC); [AB = CD]
A 
 P = (2  y 1 + n2 + b)  P =  − ny  + 2y 1 + n2
y 
❖ For most economical section, for a given A, either depth of flow y or side slope
n can be varied i.e.

(i) Keep A & n constant and vary y (Generally done)

(ii) Keep A & y constant and vary n

Case (i): For given A and n (i.e. keeping A and n constant), trapezoidal section will be
most economical, when P is minimum i.e.

• Differentiating P with respect to y and put the derivative equal to zero, to get

dP A A
 = - 2 - n + 2 1 + n2 = 0  2
+ n = 2 1 + n2
dy y y

(b + ny)y (b + 2ny)
 2
+ n = 2 1 + n2  = y 1 + n2
y 2
❖  Half of the top width of flow = One of the sloping side

10
 A (b + ny)y (b + ny)y y y
Rh = = = =  Rh =
P b + 2  y 1 + n2 b + (b + 2ny) 2 2

❖  Hydraulic mean depth is equal to half the depth of flow

Observation:

To prove that most economical trapezoidal section is a half hexagon with one side horizontal

• Let O be the centre of AD (top width of flow)

• Draw OF perpendicular to AB. (To prove OF = y)

• If  is the angle made by AB with horizontal, then from right angled OAF, OF =
OA sin 

To find OA and sinθ

(b + 2ny)
• OA = Half of the top width = = y 1 + n2
2
AE y 1
• From ABE, sin  = = =
AB y 1 + n2 1 + n2
1
 OF = y 1 + n2   OF = y
1+ n 2

• If a semicircle is drawn with top surface as centre and radius equal to the
depth of flow, then three sides of the most economical trapezoidal section viz.
bottom and the two sloping sides will be tangential to the semicircle.
•
❖ In other words, the most efficient trapezoidal channel section is a half hexagon
with one side horizontal

 Three conditions for most economical trapezoidal channel section are:

➢ Half of the top width = One of the sloping side

➢ Hydraulic mean depth = Half the depth of flow

➢ Half hexagon with one side horizontal

❖ Any one of these conditions must be satisfied for a trapezoidal channel section to
be the most economical.

❖ These conditions are applied only when A and n (side slope) are fixed.

Case (ii): Keeping A & y fixed and vary side slope n

11
 • If side slope varies, then the best side slope of the most economical trapezoidal
section can be determined as:
A 
P =  − ny  + 2 y 1 + n2
 y 
• For a given A and y, (i.e. keeping A and y constant), side slope will be best when
P is minimum i.e. differentiating P with respect to n and put the derivative equal to
zero and solve, to get
1
n=
3
1
• Also, tan  = = 3 = tan 60o   = 60o
n
Best side is at 60o to the horizontal.

To find the perimeter of best side slope section

 2 
b + y
(b + 2ny)  3  1 2y
= y 1 + n2  = y 1+  b =
2 2 3 3
• (
Substitute in P = b + 2y 1 + n
2
) and solve, to get
P = 3b (n = 1/3)

❖ For best sides slope, length of each sloping side is equal to the bed width of
trapezoidal section

Observation:

(i) For economical trapezoidal section of any shape, Rh = y/2.

(ii) (ii) For trapezoidal cross section to be minimum (Amin), section will also
have the dimensions of efficient section.

Most Economical Triangular Channel Section

• Consider a triangular channel section of vertex angle 2, depth of flow y and top
width of flow T as shown in Figure:

T
O
A C

y
2

B
12
 1
• Area of flow cross-section, A = Ty
2
➢ T = (AO + OC) = 2AO = 2y tan(AO = y tan)

 A = y2 tan 
➢ Wetted perimeter, P = (AB + BC) = 2 AB [AB = BC]

OB
From ABO, cos  =  AB = OB sec  = y sec   P = 2  y sec 
AB
A
 P = 2 sec 
tan 
➢ For a given A, triangular section will be most economical, when P is minimum i.e.

dP d  A  d  sec  
= 2 sec  =0   =0
d d   tan      
 d tan
➢ Solve , to get; 2 tan  = sec   d (sec ) = sec  tan , d (tan ) = sec2  
2 2

 d d 
1
 sin  =   = 45o
2
Triangular section will be most economical when each of its sloping side makes
an angle of 45o with the vertical or each side slope is 1:1.

❖ Most efficient triangular channel is a half square with one diagonal horizontal or
a right angled triangle with vertex at base.

A y2 tan  y2 tan 45o y2 y

❖ Rh = = = =  Rh =
P 2  y sec  2  y sec 45 o
2y  2 2 2

Most Economical Circular Channel Section

• For circular channel sections, as depth of flow changes, flow area also changes
(due to convergence of the boundary towards top).
❖ Condition of keeping area of flow section constant cannot be applied for circular
channel sections.
❖ Two separate conditions are derived, one for maximum discharge and the other
for maximum velocity of flow
❖ Consider a circular channel of radius r and depth of flow y (y < r) .
• Let 2 be the angle (in radians) subtended at the centre by the free surface of liquid

13
 O

r
2
A B
C

2 sin 2 
• Area of flow section, A = r   − 
 2 
• Wetted perimeter, P = 2r

Condition for maximum discharge:

A3
Q=C Sb
P
• For a given C and Sb (i.e. keeping C and Sb constant), discharge will be maximum,
when (A3/P) is maximum.

 Differentiating (A3/P) with respect to  and put the derivative equal to zero i.e.

d  A3 
 =0
d  P 
• On solving, to get
 dA   dP 
3P  − A =0
 d   d 

dP dA
• = 2r and = r 2 (1 − cos 2)
d d
 sin 2 
 3  2r  r 2 (1 − cos 2 ) − r 2   −   2r = 0
 2 

 sin 2 
  2 − 3 cos 2 + =0
 2 

• Solving Eq. by trial and error method, to get

• 2 = 308o

14
 • Depth of flow, y= (OD - OC) = r (1 - cos)

 Depth of flow at max. discharge, y = r (1 − cos 154 ) = 1.9r = 0.95 d

o

➢ d is the diameter of circular channel

❖ Max. discharge is obtained when depth of flow is 0.95 times the diameter of channel

sin 2  r 2  2.686 − sin 308 

o
2
r  −   
Rh = = 
2 
=   = 0.572r = 0.29d
A 2
P 2r 2  r  2.686
❖ For maximum discharge, hydraulic mean depth is equal to 0.29 times the
diameter of channel.
Note:
• Above conditions for maximum discharge are derived by using Chezy’s equation.

• If Manning’s equation is used, then 2 ~ 302.33oand depth of flow = 0.938d.

Condition for maximum velocity

A
V = C R h Sb V=C Sb
P
• For given C and Sb, velocity will be maximum, when (A/P) is maximum.

Differentiating (A/P) with respect to  and put the derivative equal to zero i.e.
d  A dA dP
 =0 P −A =0
d  P  d d
• Put the values of dA/d and dP/d, to get

 sin 2 
• 2r  r 2 (1 − cos 2 ) − r 2   −   2r = 0
 2 
sin 2
 =  cos 2  tan 2 = 2
2
• Solve Eq. by trial and error method, to get

❖ 2 = 257 .5 o   = 128 .75 o or  = 2.247 radians

• Depth of flow at max. velocity, y = r (1 − cos 128 .75 o ) = 1.62r = 0.81d

❖ Max. velocity is obtained when the depth of flow is 0.81 times the dia. of channel.

 sin 257.5 o 
r 2  2.247 − 
Rh = =   = 0.6086r = 0.3d
A 2
P 2  r  2.247

15
 ❖ For maximum velocity, hydraulic mean depth is equal to 0.3 times the diameter
of circular channel.

Problems:

• A rectangular channel carries water at the rate of 400 lps. The bed slope of channel is 1
in 2000. Find the most economical dimensions of channel. Given, Chezy’s constant = 50
in SI units.

(1.15 m, 0.58 m)

• A rectangular channel has: width = 4 m, depth of water = 1.5 m, bed slope = 1 in

1000 and Chezy’s constant = 55. It is desired to increase the discharge to a
maximum by changing the dimensions of section but keeping the same cross-
sectional area of flow, bed slope and roughness of the channel. Find the new
dimensions of channel and increase in discharge.

(3.46 m, 1.73 m, 0.05 m3/s)

• A trapezoidal channel is required to carry 7 cumec of water at a velocity of 1.75

m/s. The sides’ slope of channel is 2H: 1V. Design the most economical cross-
section of channel. For conveying the same discharge, if this trapezoidal channel
is replaced by a rectangular channel 3.6 m wide and having depth of flow 1.25 m,
find the saving in head per km length of the channel. Take C = 60 in SI units.

(0.6 m, 1.27 m, 1 in 746, 0.427 m)

16