### ENGR 691-73 _ Lecture 03 _ Spring 11

```ENGR 691 – 73: Introduction to Free-Surface Hydraulics in Open Channels
Lecture 03: Conservation Laws
Energy Equation and Critical Depth
Uniform Flow and Normal Depth
Yan Ding, Ph.D.
Research Assistant Professor, National Center for Computational
Hydroscience and Engineering (NCCHE), The University of Mississippi,
Old Chemistry 335, University, MS 38677
Phone: 915-8969; Email: [email protected]
Course Notes by: Mustafa S. Altinakar and Yan Ding
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
1
Outline
• Review of Reynolds Transport Theorem, Control Volume, and Conservation
Laws
• Concept of Energy in Open Channel Flow
• Energy equation for Open Channel Flow
• Specific Energy Curve and Specific Discharge Curve
• Critical Depth and its Computation
• Uniform Flow and Normal Depth
• Computation of Uniform Flow
• Friction Coefficient
• Chezy and Manning Coefficients
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
2
System vis-a-vis Control Volume
System: A particular collection of matter, which is identified and viewed as being
separated from everything external to the system by an imagined or real
closed boundary.
Control Volume: A volume in space through whose boundary matter, mass,
momentum, energy, and the like can flow. Its boundary is called a control
surface. The control volume may be of any useful size (finite and
infinitesimal) and shape; the control surface is a closed boundary.
Inertial Reference Frame: a frame of reference that describes time homogeneously
and space homogeneously, isotropically, and in a time independent manner.
All inertial frames are in a state of constant, rectilinear motion with respect
to one another; they are not accelerating.
Noninertial Reference Frame: a frame of reference that is under acceleration
Intensive Property: does not depend on the system size or the amount of material in
the system, e.g., B  M
Extensive Property: directly proportional to the system size or the amount of
M
B
b


1
material in the system, e.g.,
M
M
sys
sys
sys
Lecture 03.
sys
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
3
Conservation of Mass: The 1-D Continuity Equation (without free surface)
A Fluid System: All the matter (fluid) within control volume (I+R) at time t,
but within the control volume (R+O) at time t+dt
Control Volume: a volume fixed in space (between Section 1 and 2)
From the conservation of system mass:
( m I  m R ) t  ( m R  m O ) t  dt
( m I ) t  ( m O ) t  dt
( m I ) t   1 A1 ds1
( m O ) t  dt   2 A2 ds 2
 1 A1
Then,
ds1
dt
  2 A2
ds 2
dt
Mass flowrate =  1 A1V1   2 A2V 2
Volume flowrate
Q  AV
Q  A1V1  A2V 2 If density variation is negligible
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
4
The 1-D Continuity Equation for Open Channel Flows
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
5
Review: Derivation of Reynolds Transport Theorem
D B out

D B in
Time t
System
Control surface
System
Control surface
m
Quantity of property B present in the fluid system at time t
m
Quantity of property B present in the fluid system at time t+Dt
Bt
B t  Dt

Time t+Dt
Quantity of property B present in the control volume at time t
Bt

B t  Dt
Quantity of property B present in the control volume at time t+Dt

Quantity of property B which entered the control volume through the control surface during the time interval Dt
D B out

Quantity of property B which left the control volume through the control surface during the time interval Dt
DB
m
Total change in the quantity of property B in the fluid system during the time interval Dt
D B in
DB

Lecture 03.
Total change in the quantity of property B in the control volume during the time interval Dt
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
6
Review: Derivation of Reynolds Transport Theorem
System
Control surface
D B out


D B in
Bt  Bt
m
Time t
Time t+Dt

At time t, the control volume and the fluid system coincide
Bt  Bt
At time t+Dt, the quantity of B present in the system is
B t  D t  B t  D t  D B out  D B in
m

m


Total change in the quantity of property B in the fluid system during the time interval Dt
DB
Total change in the quantity of property B in the control volume during the time interval Dt
D B  B t  Dt  B t
Divide both sides of equation by Dt
At the limit Dt  0
DB
m
Dt
DB
B

m
Dt


t  Dt
B

 Bt
Dt

t

DB
  D B

out
 DB

out

 D B in
Dt
  DB
Dt

 B t  Dt  B t

DB
Combining the first three relationships the change in B in the fluid system is written as
m
m
D B

m
m


 B t  D t  D B out  D B in   B t


out


 D B in



Dt

in
Dt
The material derivative operator “D/Dt” underlines the fact that the derivative applies to a fluid system moving in the coordinate
system (derivative contains both local and convective changes).
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
7
Review: Derivation of Reynolds Transport Theorem
System
Control surface
D B out


D B in
Bt  Bt
m
Time t
The interpretation of different terms :
Recalling the definition
of extensive property:
B

Time t+Dt
DB

Dt
 rate of change 


 of B in the  
 fluid system 


b dm 

B
m

D B out  D B in

t
Dt
 rate of change 
 net efflux of



 of B in the    B through the
 control volume 
 control surface



b  d
B
m


b  d
B
and
s

M










b  d
 cv
 DB   DB 
 
out
in
b  d 
  b  d  
Using this definition we can write:



Dt 
t 
Dt
s
 cv

The integral on the right hand side is carried out over the control volume which is invariant in time. Therefore, one can bring the
derivative sign inside the integral
D
 
b

d


  b d

Dt 
t  
s
 cv
D
Lecture 03.
 DB   DB 
out
in



Dt


 cv

t
b   d  


D B out  D B in
Dt
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
8
Review: Derivation of Reynolds Transport Theorem
System
Control surface
D B out
V

B
m
t
B
V

t
D B in
n
Time t
n
Time t+Dt
(V  n ) C S . o u t  0
Now let us take a look at the last term on the right hand side more
closely. We have already shown that the net efflux of property B
through the control surface can be expressed as:





D B out  D t   b  V  dA   D t   b  V  n dA 
 cs . out

 cs . out



Considering that d A  n dA

where n is the vector normal to the surface element dA
(V  n ) C S .in  0





D B in  D t    b  V  dA   D t    b  V  n dA 
 cs .in

 cs .in

We have, therefore


D B out  D B in
Dt
Finally

D
Dt
Lecture 03.



b  V  n dA  ( 
C S . out
 b  d  
s



b  V  n dA ) 
C S . in
 cv

C S . out

b   d    b 
t
CS
V  n  dA


b  V  n dA 



b  V  n dA 
C S . in
 b  V  n  dA
CS
Reynolds transport theorem in its
general form
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
9
Review: Conservation of Mass
Recall :
D
Reynolds Transport Theorem
Dt
b 
d 
sv

  t  b   d    b  V  n  dA
cv
CS
B  M sys
Extensive property : System Mass
Reynolds Transport Theorem for mass
Intensive property :
D
Dt

 d 
SV

CV

t
sv = System Volume
cv = Control Volume
cs = Control Surface
B
b
M sys

M sys
1
M sys
  V  n  dA
 d 
CS
 M sys  Const .
0
Equation for Conservation of Mass  Continuity Equation

0
CV
Equation for Conservation of Mass  Continuity Equation


CV
If the flow is steady or of uniform density (i.e. time
derivative of density is equal to zero)
Lecture 03.
0

t

t
   V  n  dA
 d 
 d 
CS
   V  n  dA
CS
  V  n  dA
CS
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
10
Review: Conservation of (Linear) Momentum
Recall :
Reynolds Transport Theorem
D
Dt
Extensive property : System Momentum
b
SV

CV
B  M sys
Reynolds Transport Theorem for Momentum
d 

t

V
D
CS
 d 
sv

  t V   d    V
cv

D M sys
Dt


Equation for Conservation of Momentum  Momentum Equation

M sys

V
M sys
  V  n dA
Momentum of the fluid system



dM
V

dV
sys
F  M sys a  M sys

dt
dt

F
demonstration
F


CV
If the flow is steady (i.e. time derivatives are equal to zero)
B

M sys V
CS

 M sys
Recall from physics: The rate change of change of
Momentum for a system is equal to the net sum of
the external forces acting on the system.
  V  n  dA
Intensive property : b 
V
Dt
b   d    b
sv = System Volume
cv = Control Volume
cs = Control Surface
F


t
V
 V   d    V  V  n  dA
CS
  V  n  dA
CS
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
11
Review: Conservation of Energy
Recall :
D
Reynolds Transport Theorem
Dt
b 
B
b
M sys
  t  b   d    b  V  n  dA
d 
sv
cv
B  E  E u
Extensive property : System Total Energy
Intensive property :


E

M sys
Eu
Ep

M sys
D
Reynolds Transport Theorem for Energy
Dt
M sys
e 
CS
sv = System Volume
cv = Control Volume
cs = Control Surface
 2

V

 E p  E k    M sys u  M sys g z  M sys
2


 2
V
Ek

 e u  e p  e k  e  u  gz 
M sys
2
d

sv






  t  e   d    e  V  n  dA
cv
CS
 E sys
Recall from physics: The 1st principle of
thermodynamics.
DE
sys
Dt

DQ
Dt

DW
Dt
Rate of change of Total Work
accomplished by the system
Rate of change of Total Energy of the System
Lecture 03.
Rate of change of Net Heat Efflux (heat
entering and/or leaving the system)
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
12
Review: Conservation of Energy
Using the 1st equation of thermodynamics we can now write:
dQ
Equation for Conservation of Energy  Energy Equation
dQ

dt
dQ
dt
dW

dt

dW
dt

 t  e
u
 ep  ep   d 
cv


cv
 e
u

dt

dW
dt


  t  e   d    e  V  n  dA
cv
CS

 e p  e p   V  n dA
CS
2

V
 
u  gz 
 t 
2


  d 


2

V

  u  gz  2
CS


  V  n dA




The terms on the left side of the energy equation are written in a very general context. Let us now analyze these two
terms in more detail:
1.Rate of change of work done on the fluid contained in the system, dW/dt, and
2.Rate of change of heat energy in the fluid system, dQ/dt.
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
13
Review: Conservation of Energy
Rate of change of work, dW/dt
In a fluid system the work can be done in two ways:
Work due to a mechanical device (shaft)
W

Positive when
fluid does
work on
machine and
negative when
work is done
on the fluid
by a machine

 Wp
dt
Lecture 03.

dW s
dt



p V  dA 
CS
 Wt
Pump is a mechanical
device which does
work on a fluid system
to increase its energy
(negative sign)
dW
Wf
Work done by the flow of fluid, i.e. pressure forces
Wf 
Ws

Ws
p (V  n ) dA
CS
Turbine is a mechanical
device on which the
fluid system does work
and looses some of its
energy (positive sign)
dW f
dW
dt
dt


dW p
dt
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels

dW t
dt

 p V  n  dA
CS
14
Review: Conservation of Energy
Rate of change of heat energy, dQ/dt
dQ
If heat is added to the system:
0
dt
If heat is extracted from the system:
dQ
0
dt
Let us investigate the physical interpretation of the change of variation of the heat energy of the system. We
will consider two cases:
Case of Ideal Fluid:
Ideal Fluid is defined as a nonviscous fluid. In reality
all fluids are viscous. Ideal fluid is a simplification of
the reality.
Case of Real Fluid:
Real Fluid is defined as a viscous fluid. In reality all
fluids are viscous.
In case of an ideal fluid, if the flow process is
adiabatic (i.e. no energy is transferred in or out of fluid
system) the internal energy of the fluid system remains
constant.
In case of a real fluid, even if the flow process is
adiabatic (i.e. no energy is transferred in or out of fluid
system) the internal energy of the fluid system
decreases.
Ideal fluid does not experience any internal energy
loss, since there is no friction.
The reason for this is the loss of a portion of the
mechanical energy by conversion into heat (due to
internal friction and friction with the surroundings).
The lost mechanical energy cannot be recovered by the
flow, and it is forever lost.
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
15
Review: Conservation of Energy
dQ

dW p
dt
dt
dQ
dW p

dt
dt

dW t
dt

dW t
dt
2
2


V 
V
 



  t  u  gz  2   d     u  gz  2
CS
CV
CS



2
2


V 
V 
 


  V  n dA
u  gz 
 d    u  gz 



t 
2 
2 
CS 




 dW p
dW t
 


dt
dt
dt

By combining all
together :
dQ

 p  V  n  dA  
CS

CV



CV


p V  n dA  

2

V
 
u  gz 
 t 
2


  d 


2

V

  u  gz  2
CS


  V  n dA 




Equation for Conservation of
Energy  Energy Equation
If the flow is steady (i.e. time
derivatives are equal to zero)

dW p
dt
dt
dQ
dW p
dt

dt

dW t
dt


cv
2

V
 
u  gz 
 t 
2



 

dt
CS 

dW t

 p V  n  dA
CS


  d 



p

  V  n  dA
2

V
p

  u    gz  2
CS


  V  n dA





  V  n dA
2 

2
u

p

 gz 
V

Specific enthalpy
(enthalpy per unit
mass)
Lecture 03.


CS
dQ

  V  n dA


Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels

h u 
p

16
Concept of Energy in Open Channel Flow
We will get back to these notions later.
Let us now start discussing the concept of energy for open channel flow.
We will first introduce the definition of energy
Then we will look into difference energy between two cross sections. This will be used
to derive an equation of energy for open channel flow.
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
17
Equation of Energy for Open Channel Flow
Let us consider the open channel flow on the left.
The total energy for the fluid element at point P,
located at elevation zP, where local velocity is u,
can be written as:
u
2

2g
u
p
zP
u

pt

 Const
Velocity head, i.e. energy per unit
weight of fluid
Pressure head
Elevation of point P, i.e. potential energy
2

2g
p

 zP 
2
2g

p
p

 zP 
 zP 
Lecture 03.
p*

pt

 H
Total mechanical energy head or simply “total head”
Piezometric head
Important note: If the pressure distribution over the depth h
is hydrostatic, the piezometric head is constant along the
direction normal to the bed.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
18
Equation of Energy for Open Channel Flow
Note that the pressure head at the bottom of the
channel, i.e. zP = z, can be written as:
 p
   h cos 
  b
with
For
dz
tan  
dx
 6
  Sf
or
we have cos   1
If we consider an ideal fluid, inviscid fluid with no friction losses,
the velocity is constant over the depth, u(z) = U, we have then
U
If we consider a real fluid, viscous fluid with friction losses, the
velocity has a distribution over the depth u(z) = U+f(z); we have then
U
Lecture 03.
thus
 p
   h
  b
2
 h z
pt
 h z 
pt
2g
e
2
2g


 H  Const
 H  Const
3
1
 u 
dA



where  e 

A AU 
QU
1
S f  0 .1
2
u
A
3
dA
is the “kinetic energy correction coefficient”, which
accounts for the non-uniform velocity distribution.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
19
Energy Correction Coefficient
Refer to Open Channel Flow (M.H. Chaudhry)
on Page 12
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
20
Equation of Energy for Open Channel Flow
Let us now write the equation of energy between two cross sections:
Consider the flow of a real fluid in an open channel as shown in
the figure. The conservation of energy between cross sections 
and  can be written as:
e
U1
2
2g
 h1  z1   e
Total energy at 
U2
2
2g
 h2  z 2 
Total energy at 
h

dx
Total head loss
between  and 
Referring to the figure let us write the above equation in a more explicit form:
2
 U2

U 
1 U
1  o dP
   h  dh    z  dz  
e
 h  z   e
 d   e
dx

dx

2g
2
g
2
g
g

t
g

dA



U
2
where
hr 
1  o dP
g  dA
1 U
g t
Lecture 03.
dx
dx
Head loss (or energy loss per unit weight) due to friction (this is also
called linear head loss or regular head loss)
Head loss (or energy loss per unit weight) due to acceleration in the
flow in x-direction
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
21
Equation of Energy for Open Channel Flow
Simplifying the previous equation, we have
2


U
1 U
1  o dP
d   e
 h  z   
dx 
dx
2g
g t
g  dA


or
2


U
1 U
d   e
 h  z   
dx  h r
2
g
g

t


Head difference
between  and 
Total head loss
between  and 

h

dx
The above equations are the energy equations for unsteady non-uniform flow. They express the conservation of
energy between two cross sections. Note that the energy slope Se = hr/dx and the bed slope Sf = -dz/dx., it can also be
written as follows:
1 U
U U
h


 S f  Se
g t
g x
x
So far we have not proposed any method to calculate energy loss due to friction. This point will be developed later in
detail and various methods will be discussed.
The energy equation for unsteady non uniform flow developed above can be manipulated to obtain the equation of
conservation of (linear) momentum for unsteady non uniform flow, which is also called dynamic equation of open
channel flow.
This is what we propose to do in the next slide.
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
22
Concept of Specific Energy
Let us consider the energy equation for a steady flow:
hf
2
V1 / 2 g
2
V2 / 2 g
Q
z1  h1 
U1
2
2g
U2
 z 2  h2 
2
2g
 hr
h1
Total Energy
h2
z1
H  z h
z2
U
2
2g
ref. line
Specific Energy: H s  h 
L
U
2
2g
Specific energy is the total mechanic energy with respect to the local invert elevation of the channel.
Note that since
U Q/A
we can also write
Hs  e
For a given cross section, the flow area, A, is a function of h;
therefore, the specific energy is a function of Q and h.
We can thus study the variation of
Lecture 03.
Q / A 2
 h  e
2g
Hs 
Q
Q
2
2 gA
2
h
Q
2
2 gA
2
h
2
2 gA
2
 h
h as a function of Hs
for Q = constant
Specific Energy Curve
h as a function of Q
for Hs = constant
Specific Discharge Curve
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
23
Specific Energy Curve
We wish to plot the Specific Energy Curve (i.e. h as a function of Hs for constant Q) :
One immediately sees that the
curve has two asymptotes:
Hs  h 
Q
2
2 gA
2
for
h 0
we have
A 0
therefore
Hs  
for
h 
we have
A 
therefore
Hs  h
In addition, for a given Q, the curve has a minimum value, Hsc. We will see about this minimum later in detail.
Some observations impose:
•
•
•
•
For a given Hs, there are always (except when Hs = Hsc) two depths h1 and h2. They are called alternate depths.
The depth corresponding to the minimum specific energy, Hsc, is called critical depth, hc.
Minimum specific energy, Hsc, increases with increasing discharge, Q.
There are three possible flow regimes: subcritical (h > hc), critical (h = hc), and supercritical (h < hc).
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
24
Critical Depth and Specific Energy
The critical depth hc, can be investigated by taking the derivative of specific energy, Hs, with respect
to the depth h, and then equating it to zero (point of minimum); i.e. dHs/dh = 0.
Hs  h 
Specific Energy
U
2
h
2g
Q
2
2 gA
2
B
dH
s
1
dh
2
Q
gA
dA
3
0
dh
B
Therefore :
dH s
dh
1
Q
gA
Let us work on this equation
to see what it means:
because
2
3
dA  B dh
Q
B0
2
gA
1 Q
2
g A
2
B
A
1
U
g
2
3
B 1
1
1
Dh
Fr 
U
1
gD h
This shows that the critical flow condition (h = hc and Hs is minimum)
is reached when Froude number is equal to one.
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
25
How to plot the specific energy curve for a cross section
h
It is important
to note that for
some h = hc, the
specific energy
curve is at its
minimum value.
Curve plotted for a constant Q
Hs  h 
U
2
2g
h1
hc
h1  hc
Alternate depths
2
U c / 2g
hc
To plot the specific energy curve:
1. Select a discharge Q
2. Assume an h value
3. Calculate A knowing h
4. Calculate U = Q/A
5. Calculate U2/2g
6. Calculate Hs = h + U2/2g
7. Repeat steps 2 to 6 by assuming
other h values.
2
h2
U 2 / 2g
Supercritical flow
h 2  hc
Es  H s
The specific energy, Hs, which is always measured with respect to the
channel bed, is composed of pressure energy (h) and kinetic energy
(V2/2g).
Specific Energy
Hs  h 
Lecture 03.
U
2
2g
h
E s or H s
Q
2
2 gA
2
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
26
Specific Discharge Curve for a Cross Section
h
Specific Energy
Curve plotted for
E s  Const . or H s  Const .
Hs  h 
Hs  h 
q  q max
H s
U
2
2g
2
Q
h
2g
2
2 gA
 h  2 gA  Q
2
2
2
2 g H s  h 
Q  A
hc
U
For a rectangular section
A h B
2
2
 h  2 gB
2
Q
H s
Instead of plotting h vs Hs for a constant discharge Q
(or q), i.e. specific energy curve, one can also plot h vs
Q (or q) for a constant Hs. This will be called specific
discharge curve.
Lecture 03.
H s
 h 2g h 
qh
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
2
2
h Q
2
Q
2
B
2
2
q
2
2 g H s  h 
27
Critical Depth and Specific Discharge Curve
We can easily see that
for
h0
we have
Q 0
for
h  Hs
we have
Q 0
Discharge curve has a maximum, Qmax, for
dQ
d

dh
dh
dA
Since
We can write
dQ

gB 2  H s  h   D h 
dh
For a rectangular channel
For a triangular channel
For a parabolic channel
Lecture 03.
2 g  H s
Dh  h
Dh 
h
Dh 
2h
2
3
 h 
2
A

2 g H s  h  
 B
and
dh
0
The expression
is zero if
2 g  H s  h   dA / dh   Ag
2 g  H s
Dh 
hc
2  H sc  h c  
2 hc
0
A
B
2  H sc  hc   D hc  0
2  H sc  hc   hc  0
2  H sc  h c  
 h 
2
0
2
0
3
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
hc 
2
hc 
4
hc 
3
5
3
4
H sc
H sc
H sc
28
Critical Depth and its Importance
Critical depth, hc, in a channel is the flow depth at which:
• The specific energy is minimal, Hsc, for a given discharge, Q,
• The discharge is maximal, Qmax, for a given specific energy Hsc.
2  H sc  hc   D hc
For critical flow in a channel:
Recall that:
Q max  Ac
2 g  H sc  hc 
Q max  Ac
Uc 
The average velocity corresponding to the critical depth is :
g D hc
g D hc
or
Uc
2
2g

D hc
2
For critical flow in a channel, the velocity head is equal to half of the hydraulic depth
One can also write:
Uc 
Critical velocity is given by
Flow regimes can be
classified according to Fr
Lecture 03.
Uc
g D hc
Uc 
1
g D hc
g D hc  c
Fr  1
Subcritical flow
Fr  1
Critical flow
Fr  1
Supercritical flow
Fr c  1
Propagation velocity of small
perturbations in still water of depth h
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
29
Critical Depth for the Special Case of Rectangular Channel
Dh  h
In a rectangular channel, we have
Recall that the critical depth, hc, in a rectangular channel is given by: h c 
Using the definition of unit discharge:
One obtains:
hc
2

q
2 g hc
3
H sc
 H sc
or
 hc  
hc

2
Uc
2
2g
q  Uh
2
2
2
hc 
q
2
3
This is valid for a rectangular channel
g
The maximum unit discharge, q, which may exist in a rectangular channel is:
q
3
g hc 
2

g  H sc 
3

3
Critical flow is unstable and, generally, it cannot be maintained over a long distance. Critical flow is rather a local
phenomenon.
For a given cross section shape, the critical depth depends only on discharge. This property is exploited to design flow
measuring methods and devices in open channels.
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
30
Example: Plotting a Specific Energy Curve
A trapezoidal channel has a bottom width of b = 3.0m and side slopes of m = 1.5. Calculate and plot
the specific energy curve for a discharge of Q = 2.0m3/s.
Specific Energy Curve for a Trapezoidal Channel
Specific energy is
defined as:
B
m
b
b  mhh
h
A
m
0.100
0.147
0.194
0.218
0.242
0.265
0.289
0.301
0.317
0.331
0.336
0.349
0.389
0.436
0.469
0.535
0.602
0.668
0.734
0.867
1.000
m2
0.315
0.474
0.640
0.726
0.813
0.901
0.992
1.038
1.103
1.159
1.178
1.231
1.395
1.592
1.737
2.036
2.348
2.674
3.012
3.730
4.500
Lecture 03.
V = Q/A
m/s
6.349
4.218
3.125
2.757
2.461
2.219
2.016
1.927
1.814
1.726
1.698
1.624
1.434
1.256
1.152
0.982
0.852
0.748
0.664
0.536
0.444
V2/2g
m
2.055
0.907
0.498
0.387
0.309
0.251
0.207
0.189
0.168
0.152
0.147
0.134
0.105
0.080
0.068
0.049
0.037
0.029
0.022
0.015
0.010
Hs  h 
2
h
2g
Q
2
2 gA
2
h
1
b  2mh
Hs
B
m
2.155
1.054
0.692
0.605
0.550
0.516
0.496
0.490
0.485
0.483
0.483
0.484
0.494
0.516
0.537
0.584
0.639
0.697
0.757
0.882
1.010
m
3.300
3.442
3.583
3.654
3.725
3.796
3.867
3.902
3.952
3.994
4.008
4.048
4.168
4.307
4.407
4.606
4.805
5.004
5.203
5.602
6.000
V
gDh
Fr 
Dh = A/B
m
0.095
0.138
0.179
0.199
0.218
0.237
0.257
0.266
0.279
0.290
0.294
0.304
0.335
0.370
0.394
0.442
0.489
0.534
0.579
0.666
0.750
Fr
(-)
6.56
3.63
2.36
1.98
1.68
1.45
1.27
1.19
1.10
1.02
1.00
0.94
0.79
0.66
0.59
0.47
0.39
0.33
0.28
0.21
0.16
The calculation of Hs for different h was
carried out on an MS Excel spreadsheet as
shown on the left. The calculated values are
plotted below:
1.200
1.000
0.800
h (m)
Channel and flow data
b=
3m
m=
1.5 (-)
3
Q=
2 m /s
V
0.600
0.400
0.200
0.000
0.000
0.500
1.000
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
1.500
2.000
2.500
Hs (m)
31
Example: Plotting a Specific Discharge Curve
Solved Problem 14.2
A trapezoidal channel has a bottom width of b = 3.0m and side slopes of m = 1.5. Calculate and plot
the specific discharge curve for a specific energy of Hs = 0.6m.
Channel and flow data
b=
3.000 m
m=
1.500 (-)
Hs =
0.600 m
hc =
0.421 m
Dh =
Fr =
B
h
1
m
b  2mh
m
0.100
0.164
0.228
0.260
0.292
0.325
0.357
0.373
0.395
0.414
0.421
0.424
0.435
0.448
0.457
0.475
0.492
0.510
0.528
0.564
0.600
B
m
3.300
3.492
3.685
3.781
3.877
3.974
4.070
4.118
4.185
4.243
4.262
4.273
4.305
4.343
4.370
4.424
4.477
4.531
4.585
4.692
4.800
Lecture 03.
Hs  h 
b
0.358 m
1.00 (-)
h
Specific energy is
defined as:
b  mhh
A
2
m
0.315
0.533
0.763
0.883
1.006
1.132
1.261
1.326
1.419
1.501
1.528
1.543
1.589
1.644
1.683
1.761
1.841
1.922
2.004
2.170
2.340
V
gDh
Fr 
Q
3
m /s
0.987
1.558
2.061
2.279
2.470
2.631
2.755
2.801
2.846
2.864
2.865
2.865
2.859
2.842
2.822
2.764
2.674
2.548
2.376
1.820
0.000
V = Q/A
m/s
3.132
2.924
2.700
2.581
2.456
2.325
2.185
2.112
2.005
1.909
1.875
1.856
1.799
1.729
1.677
1.569
1.453
1.326
1.186
0.839
0.000
V2/2g
Hs
m
0.500
0.436
0.372
0.340
0.308
0.275
0.243
0.227
0.205
0.186
0.179
0.176
0.165
0.152
0.143
0.125
0.108
0.090
0.072
0.036
0.000
m
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
0.600
Dh = A/B
m
0.095
0.153
0.207
0.233
0.259
0.285
0.310
0.322
0.339
0.354
0.358
0.361
0.369
0.378
0.385
0.398
0.411
0.424
0.437
0.462
0.488
Fr
(-)
3.24
2.39
1.89
1.71
1.54
1.39
1.25
1.19
1.10
1.02
1.00
0.99
0.95
0.90
0.86
0.79
0.72
0.65
0.57
0.39
0.00
Q
Hs  h 
2
2 gA
Q  A
2
V
2
h
2g
Q
2
2 gA
2
2 g H s  h 
The calculation of Q for different h was carried
out on an MS Excel spreadsheet as shown on the
left. The calculated values are plotted below:
0.7
0.6
0.5
h (m)
Specific Discharge Curve for a Trapezoidal Channel
0.4
0.3
0.2
0.1
0.0
0.0
1.0
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
2.0
Q (m3/s)
3.0
4.0
32
Hydraulic Jump
• Refer to Open Channel Flow (M.H. Chaudhry)
on Page 43
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
33
Homework
Open-Channel Flow, 2nd Edition, by M.H.
Chaurdhry
• Problems 2.11, 2.12, 2.19, and 2.24
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
34
Critical Flows in Different Types of Channels
• Refer to Chapter 3, Open Channel Flow (M.H.
Chaudhry) on Pages 55-85
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
35
Concept of Uniform Flow
Now we will introduce an important concept:
The Uniform Flow
In relation with uniform flow, we will also define:
Normal Depth or Uniform Flow Depth
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
36
Pioneers who Introduced the Concept of Uniform Flow
Willi Hager (2003) : “Hydraulicians in Europe, 1800-2000”; IAHR Monograph, IAHR, Delft, Netherlands
Antoine de Chézy
born at Chalon-sur-Marne,
France, on September 1, 1718,
died on October 4, 1798
Robert Manning
born on Oct 22, 1816 in
Normandy, died on Dec 9, 1897
in Dublin
Strickler
born on July 27, 1887 in
Wädensville, died on Feb 1, 1963
in Küsnacht
Chézy was given the task to determine the
cross section and the related discharge for a
proposed canal on the river Yvette, which is
close to Paris, but at a higher elevation. Since
1769, he was collecting experimental data
from the canal of Courpalet and from the river
Seine. His studies and conclusions are
contained in a report to Mr. Perronet dated
October 21, 1775. The original document,
written in French, is titled "Thesis on the
velocity of the flow in a given ditch," and it is
signed by Mr. Chézy, General Inspector of des
Ponts et Chaussées
At the age of 30, Robert Manning entered the
service of the commissioners of public works
to work on the projects of arterial drainage. In
1855 he started his own business and was
involved in harbor works in Dundrum. In 1869,
he returned to the public service and was
promoted chief engineer in 1874. In 1880 he
was in charge of the improvement of river
Shannon and later he worked on fishery piers.
Manning retired in 1881.
Obtained his diploma of mechanical
engineering at ETH Zurich in 1916. He
submitted a Ph.D. thesis related to turbine
design. He was appointed head of section in
Federal Water Resources Office, where he was
involved with low head power plants. In 1928
he was elected the director of the Swiss
Power Transmission Society in Bern. Later he
founded a private company and worked on
projects in eastern Switzerland.
He developed the formula that bears his name
from Ganguillet-Kutter formula based on the
data by Henry Basin.
He is well known for his uniform flow formula
that he established using his own data and
data from literature.
http://chezy.sdsu.edu/
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
37
Concept of Uniform Flow
Consider a channel defined by the following characteristics :
Cross section shape
Bed slope (S = tg)
The roughness of the bed (ks)
i.e. the relationships: A = f(h), B = f(h), and P = f(h)
Assume also that the channel is sufficiently long.

h=?
The question is: what will be the flow depth in the channel for a given discharge ?
To answer this question we must consider the equilibrium between the forces driving the
flow (gravitational force) and forces resisting the flow (friction due to viscous forces).
The flow depth will become constant when an equilibrium is reached between driving
and resisting forces (i.e. no net force is acting on the flow).
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
38
Concept of Uniform Flow
Consider a prismatic open channel (the section does not change along the flow direction) with S o
Consider a free-surface flow of constant depth in this channel
i.e. the free surface is parallel to the bed
Equilibrium of all forces in the flow direction
(no acceleration)
L
W sin   F f
W sin 
Ff   o (P L)
with
W   ( A L)
 gh
W
with
Lecture 03.
sin   tan 
 gh
Ff
tan   S o
 ( AL ) tan    o ( P L )

small
and
 A
 tan 
P
o   
A
P
 Rh
 o   Rh S o
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
39
Concept of Uniform Flow
with
remember also

u* 
2
u*  g Rh S o

u 
 *
8 V 
f
by rearranging terms, we get
2
f
2
u* 
So  f
1
8
V
V
2
f
8
V
2
 g Rh S o
2
4 Rh 2 g
which is Darcy-Weissbach eqn
It tells us that in case of uniform flow the slope of the energy gradient line (right hand side
of Darcy-Weissbach eqn) is also parallel to the bed slope, So.
In uniform flow in an open channel, the water surface, the bed and the energy
line are all parallel to each other.
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
40
Methods for Computing Uniform Flow
Several methods are available for calculating the uniform flow in an open channel:
• Using Darcy-Weissbach equation and friction factor,
• Using Chezy equation, and
• Using Manning-Strickler equation.
We will now study these three methods is detail.
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
41
Uniform Flow Calculation Using Darcy-Weisbach Equation
So  f
The Darcy-Weissbach equation for open channel flow is given as:
This equation can be rewritten as:
U
2

1
f
since
Q  UA
1
Q  UA 
2
4 Rh 2 g
S o 4 R h  2 g 
We can also write
U
U 
8g
f
8g
f
Rh S o
Rh S o A
In this equation both hydraulic radius and flow area are functions of depth h: A = f(h) and Rh = f(h)
The friction coefficient can be computed either using Moody-Stanton diagram or Colebrook and White equation.
Colebrook and White equation for friction coefficient in pipes (for all flow regimes) was adapted for open channel
flows (valid for all regimes) by Silberman et al. (1963) as follows:
 k / R 

bf
h
  2 . 0 log  s


f
a
Re
f

f

1
Reynolds number is computed as:
Lecture 03.
with
Re 
12  a f  15
and
0  bf  6
U 4 Rh

Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
42
Uniform Flow Calculation Using Darcy-Weisbach Equation
Uniform flow problems can be solved by
solving the following two equations
simultaneously:
Q  UA 
8g
f
Rh S o A
START
Read Q, cross section
data, ks, and So
Estimate hn
and
Calculate A, P, Rh, and ks/Rh
 k / R 

bf
1
h
  2 . 0 log  s


f
Re
f 
 a f
U = Q/A and Re = 4URh/
Calculate discharge with
Estimate f
Note that if it is required to solve for the flow
depth for a given discharge and cross section
geometry, a trial and error procedure, such as
the one shown on the right, must be used.
The trial and error procedure has two loops.
The outer loop iterates the value of h until we
reach the normal depth hn. The criteria for
stopping the iteration is that the computed
discharge is equal to the given discharge. The
inner loop finds the value of f iteratively. The
criteria for stopping the iteration is that the
computed friction factor is equal to the
estimated friction factor.
Lecture 03.
Qc 
Use Colebrook and White
equation to calculate f’
f
Rh S o A
is
Q = Qc
?
Use Colebrook and White
equation to calculate f’
is
f = f’
?
8g
no
yes
yes
Output the results:
hn, A, P, Rh, ks/Rh
Re, f, Q, U
no
take f = f’
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
END
43
Uniform Flow Calculation Using Darcy-Weisbach Equation
In rough channels of large width, Rh = h, the friction coefficient, f , can be obtained making in situ measurements of
two point velocities and assuming a logarithmic velocity distribution:
u
u*

 z 
  8 .5
ln 
  k s 
u
1
u*
 z
 5 . 75 log 
u*
 ks
u

 z 
  8 .5
ln( 10 ) log 


k
 s
1

  8 .5


 30 z
 5 . 75 log 
u*
 ks
u
u

2 . 3026
u*

 z 
  8 .5
log 

k
 s




It is customary to measure and use point velocities at 0.2h and 0.8h.
z  0 .2 h
u ( z  0 . 8 )  u 0 .2
 ( 30 ) ( 0 . 2 h ) 
 6h 

u 0 .2  5 . 75 u * log 
  5 . 75 u * log 

k
k
s


 s 
z  0 .8 h
u ( z  0 . 2 )  u 0 .8
 ( 30 ) ( 0 . 8 h ) 
 24 h
u 0 .8  5 . 75 u * log 
  5 . 75 u * log 
ks


 ks
Eliminating u* from these two equations, one obtains:
The expression for the average velocity for
turbulent rough flow in a wide channel (Rh ≈ h) is:
The expression for the frcition coefficient for
turbulent rough flow in a wide channel (Rh ≈ h) is:
Lecture 03.
 h  0 . 78   1 . 38

log 

k
1
 s
 h 
  6 . 25
 5 . 75 ln 

u*
k
 s
U
 h 
  2 .2
 2 . 03 log 

f
k
 s
1
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels




 
with
u 0 .8
u 0 .2
U

1 . 78   0 . 95 
 1
u*
1
f

1 . 78   0 . 95 
8
 1
44
Uniform Flow Calculation Using Darcy-Weisbach Equation
The table given below is not exhaustive.
Consult other references for a more detailed table.
Types of Wall
ks (mm)
Glass, copper, brass
< 0.001
Lead
0.025
Steel pipe, old
0.03 to 0.1
Steel pipe, new
0.4
Wrought iron, new
0.25
Wrought iron, old
1.0 to 1.5
Wrought iron, coated
0.1
Concrete, smooth
0.3 to 0.8
Concrete, rough
<3.0
Wood
0.9 to 9
Stone, worked rough
8 to 15
Lecture 03.
Since open channel cross sections are not circular in general, a
correction factor must be used to multiply the hydraulic radius.
This correction factor takes into account the influence of the
shape of the channel.
Rectangular cross section (B = 2h)
  0 . 95
Large trapezoidal cross section
  0 . 80
Triangular (equilateral) cross section
  1 . 25
Using these corrections, in the formulate replace Rh by Rh.
1.0 to 2.5
Riveted Steel
Rock
The values given in the table are for circular industrial pipes.
However, they are generally assumed to be valid also for openc
channel flows.
90 to 600
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
45
Uniform Flow Calculation Using Darcy-Weisbach Equation
We begin by dividing the flow into several regions:
• The viscous sublayer (~0.00 ≤ z’/h ≤ ~0.05) is where the viscous forces are dominant. The velocity profile varies
linearly with the distance from the bed.
• The inner region (~0.05 ≤ z’/h ≤ ~0.2) is where the turbulence production is important. The length and velocity
scales are /u* and u*, respectively,
• The outer region (z’/h ≥ ~0.6) is where the free surface properties are important. The length and velocity scales are
flow depth h and maximum flow velocity Uc, respectively,
• The intermediate region (~0.2 ≤ z’/h ≤ ~0.6) is where turbulent energy production and dissipation are
approximately equal.
Outer region
Outer region
Intermediate
region
Inner region
Inner region
Viscous sublayer
From here on, however, we will assume that there are only two layers:
• The inner region will be assumed to include also the viscous sublayer. The inner region, therefore, is defined as:
~0.00 ≤ z’/h ≤ ~0.20 ,
• The outer region will be assumed to include also the intermediate region. The outer region is, therefore, defined as:
~0.2 ≤ z’/h ≤ 1.00 .
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
46
Uniform Flow Calculation Using Darcy-Weisbach Equation
Without going into details, the derivation of the velocity profile for the inner region leads to:
u
u*

1

ln z '  C
This equation is called law of the wall, or logarithmic velocity profile. It is only valid in the inner region. It is
important to remember that it has been derived by assuming that the longitudinal pressure distribution is negligible and
the shear stress is constant and equal to the wall shear stress over the entire inner region. The integration constant C
needs to be determined experimentally.
To summarize, inn the inner region the velocity has the following functional relationship:
u  f  o ,  ,  , k s , z ' 
In the above equation, ks represents Nikuradze’s equivalent sand roughness, which can be interpreted as the
characteristics length scale corresponding to the height of the roughness elements.
Note: Although it is not correct, for simplicity, sometimes the logarithmic velocity profile is assumed to apply
over the entire flow depth.
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
47
Flow Regimes and Friction Coefficient
The conclusion of Nikuradse’s experiments was that there is no unique relationship between friction factor, f , Reynolds
number, VD/n, and the relative pipe roughness, ks /D. Different relationships must be used for different flow types.
The classification of flow types is done using “Reynolds number” and “shear Reynolds number” as criteria.
u (r)
Reynolds number
Re 
UD
u* 
Shear
Reynolds
Number
Re * 
roughness
o

D
U

(r )
ks
Re < 2000
Shear velocity
Laminar flow
2000 < Re < 3000
Transition flow
Re > 3000
Turbulent flow
1st level of classification
o
Hydraulic smooth
Hydraulic transition
Hydraulic rough
u* k s

5
u* k s


5
u* k s

u* k s
 70
 70
2nd level of classification
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
48
Friction Coefficient Formulae for Different Flow Regimes
Laminar Flow
f 
64
Re
Turbulent Smooth Flow
1

 2 log Re
f
f
  0 .8
Turbulent Rough Flow
Colebrook-White Formula
 k / D 
2 .5 
s
  2 . 0 log 


f
3
.
7
Re
f


1
Swamee and Jain Formula
f 
0 . 25

 log

Lecture 03.
5 . 74  



10
0 .9  
Re
 3 .7 D

ks
2
hf  f
L V
2
Darcy-Weisbach equation for head loss
D 2g
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
49
Developing a Diagram for Friction Coefficient
The Darcy-Weisbach equation was not made universally useful until the development of the Moody diagram (Moody,
1944) based on the work of Hunter Rouse. Rouse always felt that Moody was given too much credit for what he himself
and others did (http://biosystems.okstate.edu/darcy/DarcyWeisbach/Darcy-WeisbachHistory.htm)
Sir Thomas Ernest Stanton
December 12, 1865, Atherstone, GB
August 30, 1931, Eastbourne, GB
Received his BSc from Owen’s College , Manchester, in 1891 and worked as
assistant to Osborne Reynolds. In 1896 took a position of lecturer in engineering at
the University College, Liverpool, together with Henry S. Hele-Shaw. Submitted
his Ph.D. Thesis in 1898 and became professor of civil and mechanical
engineering at Bristol University in 1899. In 1901, he was appointed
superintendent of the newly inaugurated National Physical Laboratory,
Teddington, where he stayed until his retirment in 1930. He did research on
sterngth of materials, lubrication, heat transmission, and hydrodynamics. His main
contribution is his 1914 paper with J.R. Pannell: “Similarity relations of motion in
relation to the surface friction of fluids, Philosophical Transactions 214: 199-224”.
Stantaon received numerous prizes. He became a fellow of the Royal Society in
1914. He was knighted in 1928. He drowned in the sea near Pevensey.
Willi H. Hager (2003) :”Hydraulicians in Europe, 1800-2000”, IAHR Monograph.
Published by IAHR, Delft, The Netherlands
Lecture 03.
Lewis F. Moody
Professor of Hydraulic Engineering,
Princeton University
The current form of the Moody-Stanton diagram (or
chart) was proposed by Moody in his paper:
“Moody, L. F., 1944. Friction factors for pipe flow.
Transactions of the ASME, Vol. 66”.
* http://biosystems.okstate.edu/darcy/DarcyWeisbach/Darcy-WeisbachHistory.htm
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
50
Moody-Stanton Diagram for Friction Coefficient in Pipes (and in open channel)
Moody-Stanton Diagram for Industrial Pipes
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
51
Colebrook and White Equation for Friction Coefficient
We know that:
u* k s

5
Turbulent smooth flow
5
In between
u* k s

 70
u* k s

 70
Turbulent transition flow
Turbulent rough flow
Colebrook and White equation for friction coefficient in pipes (for all flow regimes) was adapted for open channel
flows (valid for all regimes) by Silberman et al. (1963) as follows:
 k / R 

bf
h
  2 . 0 log  s


f
Re
f 
 a f
1
Reynolds number is computed as:
with
Re 
and
0  bf  6
a f  12
and
b f  3 .4
U 4 Rh

For wide channels it is recommended to take:
ks  0
Consider the following limiting cases:
Re  
Lecture 03.
12  a f  15
 bf

  2 . 0 log 

f
f 
 Re
Turbulent
smooth flow
 k / R  
h
  2 . 0 log  s

f
 a f

Turbulent rough
flow
1
1
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
52
Chezy and Manning-Strickler Equations for Computing Uniform Flow
Chezy Equation
start with Darcy-Weissbach eqn
rearrange
Manning-Strickler Equation
So  f

1
S o R h   f
 8g
1
V
2
4 Rh 2 g

 V

 8g
C  
 f
define the Chezy coefficient
to obtain the Chezy equation
V C
1
n
2



using
S o Rh
Rh
V  K Rh
Strickler defined
which means
2
V 
Manning defined
K 
2/3
2/3
So
So
1/ 2
1/ 2
1
n
Q  VA
Q 
Manning eqn for discharge
A
n
Rh
2/3
So
1/ 2
This is the average velocity for uniform flow in a channel
using
Q  VA
we get discharge
Q  K A Rh
Strickler eqn for discharge
Q  VA  CA
S o Rh
Manning eqn for
traditional unit system
V 
1 . 49
n
Rh
2/3
2/3
So
So
1/ 2
1/ 2
Relationship between the two friction coefficients
C 
Rh
1/ 6
n
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
53
Chezy and Manning Coefficients
ATTENTION ! DIMENSIONAL COEFFICIENTS
Chezy Equation
C
m
1/ 2
s
1
Manning-Strickler Equation


n m
Tables are available for various surfaces
Lecture 03.
1 / 3
s

Tables are available for various surfaces
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
54
Chezy and Manning Coefficients
Chezy Equation
Chezy equation is valid only for turbulent rough flows. It should not be used for laminar flows or turbulent smooth
flows.
Note that one can write the following relationship between the Chezy Coefficient, C, and
the friction factor, f:
C 
8g
1
f
Therefore one can use two-point velocity measurements to calculate the Chezy coefficient (Graf & Altinakar 1998,
Pp77-78):
C 
8g
1
f
 1 . 78
g

 0 . 95 
 1
Tables of Chezy coefficients for different types of channel materials are given in various textbooks and reference
books:
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
55
Manning Coefficients for Various Types of Channels
n (m-1/3s)
n (m-1/3s)
Taken from  http://harris.centreconnect.org/Table%20A-1.htm
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
56
Manning Coefficient for Natural Channels
Hydraulic computations in natural stream require an evaluation of the roughness characteristics of the channel. In the absence of a
satisfactory quantitative procedure this evaluation remains chiefly an art. The ability to evaluate roughness coefficients must be developed
through experience. One means of gaining this experience is by examining and becoming acquainted with the appearance of some typical
channels whose roughness coefficients are known.
The USGS web site http://wwwrcamnl.wr.usgs.gov/sws/fieldmethods/Indirects/nvalues/index.htm displays photos, characteristics and
Manning-Strickler coefficients for a wide range of channel conditions. It would be a good idea to study these photos. Familiarity with the
appearance, geometry, and roughness characteristics of these channels will improve your ability to select roughness coefficients for
channels that you will encounter in your professional life.
n
(m-1/3s)
Stream
Photo 1
0.024
Columbia River at Vernita, Washington
0.028
Clark Fork at St. Regis, Montana
0.030
Clark Fork above Missoula, Montana
0.032
Salt River below Stewart Mountain Dam, Arizona
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
Photo 2
57
Manning-Strickler Coefficient for Natural Streams: Wenatchee River at Plain, Washington
Photo 169: Downstream from above
section 1, Wenatchee River at Plain,
Washington
•Station name
Photo 173: Downstream from section
2, Wenatchee River at Plain,
Washington
Wenatchee River at Plain, Washington
•Station number
12-4570
•Gage location
Lat 47°45'50'', long 120°39'30'', in lot 8, sec.
12, T. 26 N., R. 17 E., on left bank at Plain, 0.25
mile downstream from Beaver Creek, 7.5 miles
downstream from Nason Creek, and 12 miles
north of Leavensworth. Section 1 is about
1,360 ft upstream from gage.
•Drainage area
591 sq mi.
•Date of flood
May 29, 1948
•Gage height
12.43 ft at gage: 16.50 ft at section 1
•Peak discharge
22,700 cfs
A
(ft2)
B
(ft)
h
(ft)
Rh
(ft)
U
(ft/s)
Lenght (ft)
between
sections
Fall (ft)
between
sections
1
2,480
224
11.1
10.86
9.15
.......
.....
•Computed roughness
coefficient
Manning n = 0.037
2
2,470
228
10.8
10.58
9.19
311
0.75
•Description of channel
3
2,440
237
10.3
10.05
9.30
325
.75
Bed is boulders; d50 = 162 mm, d85 = 320 mm.
Bank are lined with trees and bushes.
Sect.
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
58
Grain Size Distribution in Natural Channel Bed
Measurement of grain size distribution of sediments by sieving:
Weight
Percent
Cumulative
Weight %Retained
0.01
0.6
0.02
100
0
0.063
0.5
0.02
99.98
0.02
0.125
0.6
0.02
99.96
0.04
0.25
1.2
0.04
99.94
0.06
0.354
1.8
0.06
99.9
0.1
0.5
4.7
0.16
99.84
0.16
0.707
17.5
0.59
99.68
0.32
1
172.2
5.81
99.09
0.91
1.41
2570.1
86.74
93.28
6.72
2
152.7
5.15
6.54
93.46
2.83
41.2
1.39
1.39
98.61
4
0
0
0
100
2963.1
100
Weight of Size Fraction (g)
Sieve Shaker AS 400
control by Retsch
Weight of Size
Fraction (g)
Cumulative Weight %Retained
http://www.retsch.com/279.0.html?&L=0
Cumulative
Weight %Passed
Grain Size
(mm)
d (mm)
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
59
Manning Coefficients for Channel with Mobile Bed Made of Granular Material
Manning-Strickler Coefficient for Mobile Bed Made of Granular Material
n
d 50
1/ 6
where
21 . 1
n
d 90
d 50 Diameter of the holes of a sieve which would pass 50% of a sediment sample taken
from the bed of the channel
1/ 6
where
26 . 0
d 90 Diameter of the holes of a sieve which would pass 90% of a sediment sample taken
from the bed of the channel
Other expressions that are used in practice:
According to “River Mechanics” by Pierre Julien:
n
d 50
1/ 6
n
16 . 1
n  0 . 062 d 50
Lecture 03.
d 75
1/ 6
n
21 . 7
1/ 6
n  0 . 046 d 75
d 90
1/ 6
26 . 3
1/ 6
n  0 . 038 d 90
1/ 6
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
60
Dependence of Manning Coefficient on Relative Depth!
C

g
8
f

Rh
n
1/ 6

g
U
gR h S o
At this point, we are only discussing the “grain roughness”. In natural
channels, the friction is also caused by bed forms. This will be
discussed in detail later. The figure compares the Manning-Strickler
and logarithmic law relationships with measured data as a function of
the relative depth (flow depth divided by the characteristic height of
the sediment grains, such as ds = d50).
h
ds
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels

h
d 50
61
Composite Roughness
Consider an open channel cross section with different roughness types.
One way to handle the situation of composite
roughness is to divide the flow area into parts, each
of which is influenced by a single type of roughness.
A1
A3
A2
Each part will have its own flow area (A1, A2, … AN),
its own perimeter (P1, P2, … PN), thus its own
hydraulic radius (Rh1, Ah2, … AhN), and its own
roughness (n1, n2, … nN). We will assume that the
velocity through each individual flow area is the
same and is equal to the average value through the
cross section (U1 = U2 = … = UN = U = Q/A).
P3
P1
P2
N
Note that the individual flow areas sum up to give the total flow area:
A1  A 2  ....  A N 
A
i
 A
1
Assuming that the Manning-Strickler equation is valid for each individual flow area, we can write:
1  A1 
U 


n1  P1 
2/3
Sf
1/ 2
1  A2 



n 2  P2 
2/3
Sf
1/ 2
1  AN 
 .... 


n N  PN 
After various simplifying assumptions,
Einstein and Horton has suggested to use:
Lecture 03.
n eq
2/3
Sf
1/ 2
1  A

 
n eq  P 
 N
3/2 
  Pi n i 

 1
P




2/3
Sf
1/ 2
2/3
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
62
Composite Roughness
References for Einstein Horton equation are:
A1
A3
A2
P3
P1
Horton R. E., “Separate roughness coefficients for channel bottom and
sides.” Engineering News-Record, vol iii, No. 22 (30 November
1933), pp 652-653
Einstein.. H. A.. “Der hvdraulische oder Profil-Radius.”
Schweizerische Bauzeitung ,“vol 103~ No. 8 (24 February 1934), pp
89-91.
P2
In fact, in addition to Einstein-Horton equation, there are other expressions proposed*:
N
n eq 
Los Angeles District equation
Colebatch equation
n eq
n
i
U. S . Army, Office, Chief of Engineers, Hydraulic Design of
Flood Control Channels. EM 1110-2-1601 (unpublished
Engineer Manual draft )
Ai
1
A
 N 3/2
  n i Ai
 1
A







2/3
Colebatch, G. T., “Model tests on Liawenee Canal roughness
coefficients.” Transactions of the Institution, Journal of the
Institution of Engineers, vol 13, No. 2, Australia (February
1941), pp 27-32.
In general, Einstein-Horton equation gives a more conservative estimation. It is preferred for design purposes.
(*) Hydraulic design criteria, SHEETS 610-1 to 610-7, TRAPEZOIDAL CHANNELS (http://chl.erdc.usace.army.mil/Media/2/8/3/600.pdf)
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
63
Concept of Conveyance
Let us study the concept of conveyance.
Consider a prismatic open channel with a fixed bed. The uniform
flow discharge can be calculated using Manning-Strickler Equation:
Q  AU 
A
n
Rh
2/3
So
1/ 2
The parameters characterizing the cross section of the channel are:
A  f (h )
R h  f (h )
n
Flow area
Hydraulic radius
Manning coefficient
Together they form a term that can be interpreted as
a measure of the ability of the cross section to
convey flow; it is therefore called conveyance.
Conveyance is a function of depth:
The Manning-Strickler equation , therefore reduces to:
Lecture 03.
Q  AU  K ( h ) S o
K (h) 
A
n
Rh
2/3
1/ 2
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
64
Best Hydraulic Section (Most Efficient Cross-Section)
What is best hydraulic section ?
A section that gives the largest flow area for the smallest perimeter !
For a rectangular channel
A  Bz
P 
P  B  2z
A
 2z
z
Let us keep A as constant. P is then only a function of z. Let us vary z to minimize the perimeter
dP
dz

A
z
2
A
20
z
2
Bz
2
z
2
B
2
 z
2
To obtain best rectangular hydraulic section the depth must equal half of the width.
r
r
h
B
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
65
Compound Channel
Method : Solve for each flow region separately and add the discharges.
Am
A1
P1
So
n1
nm
Pm
Channel slope is the same for main channel and the flood plain
R h1 
A1
R hm 
Am
P1
Pm
V1 
Vm 
1
n1
1
nm
R h1
2/3
R hm
So
2/3
1/ 2
So
1/ 2
Q 1  A1 V1 
Q m  Am V m 
A1
n1
Am
nm
R h1
2/3
R hm
So
2/3
1/ 2
So
1/ 2
Q T  Q m  Q1
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
66
Example: Computation of Uniform Flow
The triangular drainage ditch shown in the figure has a side slope of m = 2:
a) Find the critical depth, hc, for a discharge of Q = 0.35 m3/s and the
corresponding minimum specific energy.
b) Calculate the discharge if the flow depth is h = 0.6m. The channel has a
Manning coefficient of n = 0.025m-1/3/s and a bed slope of So = 0.001.
Froude number for a
triangular channel is
given by:
V
Fr 
gD h
A
gD h
A  mh
For a triangular channel, we have:
Q
2

 mh
g

2 2
2
h

m h
2
When flow is
2
Fr
critical, we have Fr
= 1:
Q

5
h
2
2Q
5
2
Dh 
1

Q

 A gD
h

2
2

Q
 
1
2

A
gD
h

Q
g
2
 A Dh
2
h
2
2
hc 
2
gm
5
2  0 . 35
2
9 . 81  2
2
0 . 35
2
 0 . 362 m
Corresponding minimum specific energy is:
H s min  h c 
Vc
2
2g
 hc 
Q
2
 hc 
2
Ac 2 g
Q
2
mh  2 g
2 2
 hc 
c
Q
2
2
4
 0 . 362 
m hc 2 g
2  0 . 362  2  9 . 81
2
4
Q 
Uniform flow discharge is calculated using Manning-Strickler equation:
h  0 .6 m
Q 
A
n
Rh
Lecture 03.
A  mh  2  0 . 6  0 . 72 m
2
2/3
S
1/ 2

0 . 72
0 . 268
2
2/3
0 . 001
1/ 2
2
Rh 
mh
2 1 m

2
n
2  0 .6
2 1 2
A
Rh
 0 . 453 m
2/3
S
1/ 2
 0 . 268 m
2
 0 . 379 m / s
3
0 . 025
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
67
Equation of Energy for Open Channel Flow
Solved Problem 14.4
A trapezoidal channel having a bottom width of b = 2.0 m and side slopes of m = 1.5 carries a uniform flow with a
depth of h = 0. 557m. The channel has a bed slope of S = 0.005 and the coefficient of Manning is n = 0.030 m -1/3s.
a) What is the discharge of the uniform flow?
b) What is the regime of flow?
For a trapezoidal channel
the geometric relationships
can be calculated as
follows:
B  b  2 mh  2  2  1 . 5  0 . 557  3 . 671 m
P  b  2h
2
 2  2  0 . 557
1  1 .5
2
 4 . 008 m
A   b  mh  h   2  1 . 5  0 . 557  0 . 557  1 . 579 m
R
The uniform flow discharge can
be calculated using ManningStrickler equation
1 m
h
2
 A / P  1 . 579 / 4 . 008  0 . 394 m
Q 
A
n
Rh
2/3
S
1/ 2

1 . 579
( 0 . 394 )
2/3
( 0 . 005 )
1/ 2
 2 . 001 m / s
3
0 . 030
To determine the regime of the flow we need calculate Froude number
Fr 
V
g Dh
Hydraulic depth for a trapezoidal channel:
Q
Fr 
A
Lecture 03.
g Dh
2 .0

1 . 579
9 . 81  0 . 43
 0 . 617
Q

A
g Dh
D h  A / B  1 . 579 / 3 . 671  0 . 43 m
Fr  0 . 617  1
The uniform flow is subcritical
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
68
Example: Computation of Critical Slope
A rectangular laboratory channel has a width of B = 2.0m. The Manning coefficient of for this channel
is n = 0.020 m-1/3/s. What should be the bed slope to achieve a critical uniform flow in this channel for
a discharge of Q = 3.0m3/s ?
Hint: critical uniform flow is achieved when uniform flow depth for a given discharge is equal to the
critical depth for that same discharge.
Fr 
Froude number for a rectangular channel is given by:
V
gh
Bh
Q
2
2
2
B h gh

Q
2
2
B gh
3
1
h
Q
3
2
hc 
2
B g
Uniform flow discharge is given by the Manning-Strickler equation.
 Qn  Bh   2 / 3 
From which we obtain an equation
S  



for the slope S,
Bh
B

2
h




If the uniform flow is going to be
critical, its depth should be hc =
0.612m, thus we get.
Lecture 03.
A
Q
Fr 
When flow is critical Froude number is equal to 1:
Q

gh
Bh

3
Q

gh
Bh
gh
1
gh
Q
3
2
2
B g
Q 
A
n
2
Q

3
2
2  9 . 81
2
Rh
2/3
2
S
1/ 2
 0 . 612 m
Bh  Bh 



n  B  2h 
2/3
S
1/ 2
2
 Qn  Bh   2 / 3 
 3  0 . 020  2  0 . 612   2 / 3 
S  




  
  0 . 00874
Bh
B

2
h
2

0
.
612
2

2

0
.
612








Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
69
ANGLE OF REPOSE
A pile of sediment at resting at the angle of repose jr represents
a threshold condition; any slight disturbance causes a failure.
Here the pile of sediment is under water. Consider the indicated
grain. The net downslope gravitational force acting on the grain
(gravitational force – buoyancy force) is
3
Fc
3
4
D
D
  s  g   sin j r   g   sin j r 
3
3
 2 
 2 
4
F gt
3
D
 Rg   sin j r
3
 2 
4
,
The net normal force is
F gn
s
R 

j
r
r
3
3
D
Fc   c  Rg   cos j r
3
 2 
Force balance requires that
Fgt
D
  Rg   cos j r
3
 2 
4
The net Coulomb resistive force to motion is
4
1
Fgn
F gt  Fc  0
or thus:
tan j r   c
which is how c is measured (note that it is
dimensionless).
For natural sediments, jr ~ 30 ~ 40 and
c ~ 0.58 ~ 0.84.
1D Sediment Transport Morphodynamics with applications to Rivers and Turbidity Currents, Gary Parker, http://cee.uiuc.edu/people/parkerg/morphodynamics_e-book.htm
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
70
Hjulström curve
Hjulstrom has provided a chart for the initiation of motion and sedimentation as a function of average velocity. This
brings in the notion of critical velocity, or erosion velocity, Ucr=UE.
This chart shows that the velocity for eroding the bed is greater than the velocity for sedimentation, i.e. Ucr=UE > UD.
This indicates that, once the particle is eroded it may stay in suspension even at lower velocities.
Henning Filip Hjulström (October 6, 1902–March 26, 1982) was a Swedish geographer. Hjulström was professor of geography at
Uppsala University from 1944, and in 1949, when the subject of geography was split, he became professor of Physical Geography.
Lecture 03.
Engr 691-73 Introduction to Free-Surface Hydraulics in Open Channels
71
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