Open Channel Flow
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Open Channel Flow
 Liquid (water) flow with a ____
________
free surface
(interface between water and air)
 relevant for
 natural channels: rivers, streams
 engineered channels: canals, sewer
lines or culverts (partially full), storm drains
 of interest to hydraulic engineers
 location of free surface
 velocity distribution
 discharge - stage (______)
depth relationships
 optimal channel design
Topics in Open Channel Flow
 Uniform Flow
normal depth
 Discharge-Depth relationships
 Channel transitions
 Control structures (sluice gates, weirs…)
 Rapid changes in bottom elevation or cross section
 Critical, Subcritical and Supercritical Flow
 Hydraulic Jump
 Gradually Varied Flow
 Classification of flows
 Surface profiles
Classification of Flows
 Steady and Unsteady
(Temporal)
 Steady: velocity at a given point does not change with
time
 Uniform, Gradually Varied, and Rapidly Varied (Spatial)
 Uniform: velocity at a given time does not change
within a given length of a channel
 Gradually varied: gradual changes in velocity with
distance
 Laminar and Turbulent
 Laminar: flow appears to be as a movement of thin
layers on top of each other
 Turbulent: packets of liquid move in irregular paths
Momentum and Energy
Equations
 Conservation of Energy
 “losses” due to conversion of turbulence to heat
 useful when energy losses are known or small
 ____________
Contractions
 Must account for losses if applied over long distances
We need an equation for losses
 _______________________________________________
 Conservation of Momentum
 “losses” due to shear at the boundaries
 useful when energy losses are unknown
Expansion
 ____________
Open Channel Flow:
Discharge/Depth Relationship
 Given a long channel of
constant slope and cross
section find the relationship
A
between discharge and depth
P
 Assume
 Steady Uniform Flow - ___
_____________
no acceleration
 prismatic channel (no change in _________
geometry with distance)
 Use Energy, Momentum, Empirical or Dimensional
Analysis?
hl d
 What controls depth given a discharge?
0  
4l
 Why doesn’t the flow accelerate? Force balance
Steady-Uniform Flow: Force
Balance
V2
oP D x
Shear force =________
2g
P
Wetted perimeter = __
Energy grade line
Hydraulic grade line
b
Dx sin
Gravitational force = A
________
Dx
c
ADx sin    o PDx  0
o  
A
sin 
P
A
= Rh
a
d
Shear force
Hydraulic radius
W cos  

W
P
S 
sin 
cos 
Turbulence
Relationship between shear and velocity? ___________
t
o
= g Rh S
W sin 
 sin 
Open Conduits:
Dimensional Analysis
Geometric parameters
Hydraulic radius (Rh)
___________________
Channel length (l)
___________________
Roughness (e)
___________________
Rh 
A
P
Write the functional relationship
æl e
ö
C p = f ç , , Re, Fr , M, W÷
èRh Rh
ø
No!
Does Fr affect shear? _________
Fr =
V
yg
Pressure Coefficient for Open
Channel Flow?
Cp 
 2Dp
V 2
Ch 
l
2 ghl
V2
Pressure Coefficient
(Energy Loss Coefficient)
Head loss coefficient
 Dp  hl
hl = S f l
Friction slope
CS f =
2 gS f l
V2
Friction slope coefficient
Slope of EGL
Dimensional Analysis
CS f
æl e
ö
= f ç , , Re÷
èRh Rh
ø
CS f
ö
l æe
=
f ç , Re÷
Rh èRh
ø
CS f =
2 gS f l
V2
Head loss  length of channel
Rh
æe
ö
Rh
C
=l
CS f
= f ç , Re÷ = l (like f in Darcy-Weisbach)
Sf
l
l
èRh
ø
2
LV
hl  f
D 2g
2 gS f l Rh
2g
l V2
2
gS
R
f
h
V
=
S f Rh
Sf =
=l
V
=
2
l
Rh 2 g
V
l
l
Chezy Equation (1768)
Introduced by the French engineer Antoine
Chezy in 1768 while designing a canal for
the water-supply system of Paris
V = C Rh S f
compare
2g
V=
S f Rh
l
where C = Chezy coefficient
0.0054 > l > 0.00087
m
m
60
< C < 150
0.022 > f > 0.0035
s
s
where 60 is for rough and 150 is for smooth
also a function of R (like f in Darcy-Weisbach)
For a pipe
d  4 Rh
Darcy-Weisbach Equation (1840)
f = Darcy-Weisbach friction factor
l V2
hl = f
d 2g
l V2
Sfl = f
4 Rh 2 g
l V2
hl = f
4 Rh 2 g
V2
S f Rh = f
8g
 d 2 



A  4  d
Rh  

P
d
4
V=
8g
S f Rh
f
 e
1
2.5 
 2 log 

 Similar to Colebrook
f
Re f 
 12 Rh
1
For rock-bedded streams
f
2
where d84 = rock size larger than 84% of the

 Rh  
1.2  2.03log   
rocks in a random sample
 d84  

Manning Equation (1891)
 Most popular in U.S. for open channels
1 2/3 1/2
(MKS units!)
V  R h So
1/3
n
Dimensions of n? T /L
V 
1.49
Is n only a function of roughness? NO!
1/2
R 2/3
S
h
o
n
Q  VA
Q 
1
n
ARh2 / 3 S o1 / 2
(English system)
Bottom slope
very sensitive to n
Values of Manning n
Lined Canals
Cement plaster
Untreated gu nite
Wood , planed
Wood , u nplaned
Concrete, trow led
Concrete, w ood forms, u nfinished
Ru bble in cement
Asphalt, smooth
Asphalt, rou gh
N atural Channels
Gravel bed s, straight
Gravel bed s plu s large bou ld ers
Earth, straight, w ith some grass
Earth, w ind ing, no vegetation
Earth , w ind ing w ith vegetation
n  0.031d 1 / 6 d in ft
n  0.038d 1 / 6 d in m
n
0.011
0.016
0.012
0.013
0.012
0.015
0.020
0.013
0.016
n = f(surface
roughness,
channel
irregularity,
stage...)
0.025
0.040
0.026
0.030
0.050
d = median size of bed material
Trapezoidal Channel
Q 
1
ARh2 / 3 S o1 / 2
n
Derive P = f(y) and A = f(y) for a
trapezoidal channel
How would you obtain y = f(Q)?
A  yb  y 2 z
2 1/ 2
2
ù +b
P = 2é
y
+
yz
(
)
ë
û
2 1/ 2
P = 2y é
ë1 + z ù
û +b
Use Solver!
1
y
z
b
Flow in Round Conduits
= (r sin q )(r cosq )
r  y

  arccos
 r 
radians
A  r 2   sin  cos 
r
T  2r sin 
P  2r
Maximum discharge
0.938d
when y = ______

A
y
T
Velocity Distribution
v y  V 
k » 0.4
1

y

gdS0 1  ln 
d

For channels wider than 10d
Von Kármán constant
depth [m]
V = average velocity
d = channel depth
At what elevation does the
velocity equal the average
velocity?
y
- 1 = ln
d
1
y= d
e
0.8d
2
0.368d
1.5
0.4d
0.2d
1
0.5
0
0
1
2
3
v(y) [m/s]
4
V
5
Open Channel Flow: Energy
Relations
velocity head
1
V12
hL = S f Dx
2g
2
V22
2g
energy
______
grade line
hydraulic
_______
grade line
y1
y2
S o Dx
Dx
Bottom slope (So) not necessarily equal to EGL slope (Sf)
Energy Relationships
p1
V12 p2
V22
+ z1 + a 1
=
+ z2 + a 2
+ hL
g
2g g
2g
Pipe flow
z - measured from
horizontal datum
From diagram on previous slide...
y1 + So Dx +
2
1
2
2
V
V
= y2 +
+ S f Dx
2g
2g
Turbulent flow (  1)
y - depth of flow
Energy Equation for Open Channel Flow
V12
V22
y1 +
+ So Dx = y2 +
+ S f Dx
2g
2g
Specific Energy
The sum of the depth of flow and the
velocity head is the specific energy:
E  y 
V2
2g
E1  So Dx  E2  Sf Dx
+ pressure
y - _______
potential energy
V2
- _______
kinetic energy
2g
If channel bottom is horizontal and no head loss
E1  E2
For a change in bottom elevation
E1 - Dy = E2
y
Specific Energy
In a channel with constant discharge, Q
Q  A1V1  A2V2
V2
Q2
E  y 
E  y 
where A=f(y)
2
2g
2gA
Consider rectangular channel (A = By) and Q = qB
q is the discharge per unit width of channel
q2
E  y 
y
2gy 2
A
3 roots (one is negative)
B
2
How many possible depths given a specific energy? _____
Specific Energy: Sluice Gate
10
9
y1 8
7
6
5
4
3
2
1
y2
0
sluice gate
E  y 
q2
y
2gy 2
1
q = 5.5 m2/s
EGL y2 = 0.45 m
V2 = 12.2 m/s
E2 = 8 m
vena contracta
2
0
1
2
3
4
5
6
7
8
9 10
E1  E2
Given downstream depth and discharge, find upstream depth.
alternate depths (same specific energy)
y1 and y2 are ___________
E
Why not use momentum conservation to find y1?
Specific Energy: Raise the Sluice
Gate
sluice gate
4
3
EGL
y
y1
2
2
1
y2 1
E1  E2
0
0
1
2
E
3
4
E  y 
q2
2gy 2
as sluice gate is raised y1 approaches y2 and E is minimized:
Maximum discharge for given energy.
Step Up with Subcritical Flow
Short, smooth step with rise Dy in channel
Given upstream depth and discharge find y2
4
3
y
4
3
y
Energy conserved
2
2
1
1
0
0
1
Dy
2
E
0
0
1
2
E
3
3
4
E1 = E2 + Dy
4
Is alternate depth possible? __________________________
NO! Calculate depth along step.
Max Step Up
Short, smooth step with maximum rise Dy in channel
What happens if the step is
y1 increases
increased further?___________
4
3
y
4
3
2
y
1
2
0
1
0
0
0
1
2
E
3
4
1
Dy
2
E
3
4
E1 = E2 + Dy
Step Up with Supercritical flow
Short, smooth step with rise Dy in channel
Given upstream depth and discharge find y2
4
3
y
4
y
3
2
2
1
1
0
0
1
Dy
2
E
0
0
1
2
E
3
3
4
E1 = E2 + Dy
4
What happened to the water depth?______________________________
Increased! Expansion! Energy Loss
4
y
3
yc
Critical Flow
2
1
0
0
1
2
3
E
Arbitrary cross-section
Find critical depth, yc
dE
 0
dy
Q2
E  y 
A=f(y)
2gA2
Q 2 dA
dE
= 1=0
3
dy
gA dy
1 
Q 2Tc
QT
gAc3
gA3
2
 Fr
T
y
dy
dA
A
P
dA = Tdy
T=surface width
More general definition of Fr
2
V 2T
gA
 Fr
2
A
=D
T
Hydraulic Depth
4
Critical Flow:
Rectangular channel
1 
Q 2Tc
T
T  Tc
3
c
gA
Q  qT
1
q 2T 3
3
c
gy T
q
yc  
g

Ac
q2

3
gyc3
1/ 3




Only for rectangular channels!
gyc3
Given the depth we can find the flow!
2
q 
Ac  ycT
yc
Critical Flow Relationships:
Rectangular Channels
q
yc  
g

2
Vc
 Vc2 yc2 

yc3  
 g 


1/ 3




 1
inertial force
gravity force
Froude number
yc g
yc 
Vc2
yc
g
2
E  y 
V2
2g

Vc2
q  Vc yc
because
Kinetic energy
Potential energy
velocity head = 0.5 (depth)
2g
E  yc 
yc
2
yc 
2
3
E
Critical Depth
Minimum energy for a given q
dE
dy
0
Occurs when
=___
Vc2 yc
=
2g 2
When kinetic = potential! ________
Fr=1
4
Fr>1 = ______critical
Super
Fr<1 = ______critical
Sub
3
y
q
T
Vc
=
=
Q
Fr =
3
3
gA
yc g
gyc
2
1
0
0
1
2
E
3
4
4
y
3
Critical Flow
2
1
0
0
1
2
E
 Characteristics
 Unstable surface
 Series of standing waves
dE
 0
dy
Difficult to measure depth
 Occurrence
 Broad crested weir (and other weirs)
 Channel Controls (rapid changes in cross-section)
 Over falls
 Changes in channel slope from mild to steep
 Used for flow measurements
 ___________________________________________
Unique relationship between depth and discharge
3
4
Broad-Crested Weir
q
yc  
g

2
q 
gy
3
c
yc 
yc
1/ 3




H
P
Q = b gyc3
2
E
Hard to measure yc
3
2
Qb g 
3
yc
Broad-crested
weir
3/ 2
E 3/ 2
2 
Q  Cd b g  H 
3 
E measured from top of weir
3/ 2
Cd corrects for using H rather
than E.
E
Broad-crested Weir: Example
Calculate the flow and the depth upstream.
The channel is 3 m wide. Is H approximately
equal to E?
H
0.5
E
yc m
yc=0.3
Broad-crested
weir
How do you find flow?____________________
Critical flow relation
Energy equation
How do you find H?______________________
Solution
Could a hydraulic jump be laminar?
Hydraulic Jump
Used for energy dissipation
Occurs when flow transitions from
supercritical to subcritical
base of spillway
Steep slope to mild slope
We would like to know depth of water
downstream from jump as well as the
location of the jump
Which equation, Energy or Momentum?
Hydraulic Jump
M1  M 2  W  Fp1  Fp2  Fss Conservation of Momentum
hL
EGL
M 1x  M 2 x  Fp  Fp
1x
2x
M 1x   V12 A1
y2
y1
M 2 x  V A2
2
2
L
 QV1  QV2  p1 A1  p2 A2

Q
2
A1

Q
2
A2

gy1 A1
2

gy2 A2
2
r gy
p=
2
V 
Q
A
Hydraulic Jump:
Conjugate Depths
For a rectangular channel make the following substitutions
A  By
Fr1 =
V1
gy1
Much algebra
Q  By1V1
Froude number
y2 
y1
 1 
1  8Fr12
2
y2 - 1 + 1 + 8Fr12
=
y1
2
valid for slopes < 0.02

Hydraulic Jump:
Energy Loss and Length
Energy
Loss
E  y 
q2
E1  E2  hL
algebra
hL 
 y2  y1 3
4 y1 y2
2gy 2
significant energy loss (to turbulence) in jump
Length
of jump
No general theoretical solution
Experiments show
L  6 y2 for 4.5 < Fr1 < 13
Specific Momentum
5
gy1 A1 Q 2 gy2 A2 Q 2



2
A1
2
A2
y1 A1 Q 2
y2 A2 Q 2



2
A1 g
2
A2 g
y
2
1
When is M minimum?
dM
q
 y 2
dy
y g
4
3
y12 q 2
y22 q 2



2 y1 g 2 y2 g
2
1
0
2
q  3
y    Critical depth!
 g 
2
E
1:1 slope
M
DE
3
4
5
E or M
6
7
Hydraulic Jump Location
Suppose a sluice gate is located in a long
channel with a mild slope. Where will the
hydraulic jump be located?
Outline your solution scheme
reservoir
2m
10 cm
Sluice gate
Gradually Varied Flow:
Find Change in Depth wrt x
V12
V22
y1 
 So Dx  y2 
 S f Dx
2g
2g
Energy equation for nonuniform, steady flow
 V22 V12 
So dx   y2  y1   

  S f dx
 2g 2g 
dy  y2  y1
Shrink control volume
T
V 
dy  d 
  S f dx  So dx
 2g 
dy d  V 2 
dx
dx

Sf
 So


dy dy  2 g 
dy
dy
dy
2
y
A
P
T
Gradually Varied Flow:
Derivative of KE wrt Depth
dy
y
dA
A
P
  2Q 2  dA
  Q 2T 
d  V 2 
d  Q 2 

   Fr 2

 
 
 2 gA3  dy
 gA3 
dy  2 g 
dy  2 gA2 




Change in KE
2

dA  Tdy
dy d  V 
dx
dx
Change in PE

Sf
 So
dy dy  2 g 
dy
dy
1  Fr  S f
2
dy
dx

dx
dy
 So
So  S f
1  Fr 2
dx
dy
We are holding Q constant!
Is V ┴ A?
Does V=Q/A?_______________
The water surface slope is a function of:
bottom slope, friction slope, Froude number
Gradually Varied Flow:
Governing equation
dy
dx

So  S f
1  Fr 2
Governing equation for
gradually varied flow
 Gives change of water depth with distance along
channel
 Note
So and Sf are positive when sloping down in
direction of flow
y is measured from channel bottom
constant
dy/dx =0 means water depth is _______
yn is when So = S f
Surface Profiles
 Mild slope (yn>yc)
 in a long channel subcritical flow will occur
 Steep slope (yn<yc)
 in a long channel supercritical flow will occur
 Critical slope (yn=yc)
 in a long channel unstable flow will occur
 Horizontal slope (So=0)
 yn undefined
 Adverse slope (So<0)
 yn undefined
Note: These slopes are f(Q)!
Surface Profiles
Normal depth
Obstruction
Steep slope (S2)
Sluice gate
Steep slope
dx

So  S f
1  Fr
2
S0 - Sf
1 - Fr2
dy/dx
4
+
-
+
+
+
-
yn
yc
3
y
dy
Hydraulic Jump
2
1
-
-
+
0
0
1
2
E
3
4
More Surface Profiles
dy
S0 - Sf
1 - Fr2
dy/dx
4
+
+
2 +
-
-
dx
1  Fr 2
3
y
1 +

So  S f
2
yc
yn
1
3 -
-
+
0
0
1
2
E
3
4
Direct Step Method
y1 
V12
 So Dx  y2 
2g
V22
energy equation
2g
y1  y2 
V12

2g
Dx 
V22
2g
S f  So
rectangular channel
V1 
 S f Dx
q
y1
V2 
solve for Dx
prismatic channel
q
y2
V2 
Q
A2
V1 
Q
A1
Direct Step Method
Friction Slope
Manning
Darcy-Weisbach
n 2V 2
S f = 4/3
Rh
V2
Sf = f
8 gRh
n 2V 2
Sf =
2.22 Rh4 / 3
SI units
English units
Direct Step
 Limitation: channel must be prismatic
_________
(channel geometry is independent of x so that
velocity is a function of depth only and not a
function of x)
 Method
 identify type of profile (determines whether Dy is +
or -)
 choose Dy and thus yi+1
 calculate hydraulic radius and velocity at yi and yi+1
 calculate friction slope given yi and yi+1
 calculate average friction slope
 calculate Dx
Direct Step Method
=y*b+y^2*z
y1  y2 
=2*y*(1+z^2)^0.5 +b
=A/P
V12
2g
Dx 
=Q/A
=(n*V)^2/Rh^(4/3)
=y+(V^2)/(2*g)

V22
2g
S f  So
=(G16-G15)/((F15+F16)/2-So)
A
B
C
D
E
F
G
H
I
y
A
P
Rh
V
Sf
E
Dx
x
0.900 1.799 4.223
0.426 0.139
0.00004 0.901
0
0.870 1.687 4.089
0.412 0.148
0.00005 0.871
0.498 0.5
J
K
L
M
T
Fr
bottom surface
3.799
0.065 0.000
0.900
3.679
0.070 0.030
0.900
Standard Step
 Given a depth at one location, determine the depth at a
second given location
 Step size (Dx) must be small enough so that changes in
water depth aren’t very large. Otherwise estimates of the
friction slope and the velocity head are inaccurate
 Can solve in upstream or downstream direction
 Usually solved upstream for subcritical
 Usually solved downstream for supercritical
 Find a depth that satisfies the energy equation
y1 
V12
2g
 So Dx  y2 
V22
2g
 S f Dx
What curves are available?
Steep Slope
1.4
1.2
S1
0.8
S3
0.6
bottom
surface
yc
yn
0.4
0.2
0.0
20
15
10
5
distance upstream (m)
Is there a curve between yc and yn that increases in
depth in the downstream direction? ______
NO!
0
elevation (m)
1.0
Mild Slope
If the slope is mild, the depth is less than the
critical depth, and a hydraulic jump occurs,
what happens next?
0.6
bottom
surface
yc
yn
When dy/dx is large then
V isn’t normal to cs
0.4
0.3
0.2
0.1
Hydraulic jump! Check
conjugate depths
0.0
40
35
30
25
20
15
distance upstream (m)
10
5
0
elevation (m)
0.5
Rapidly varied flow!
Water Surface Profiles:
Putting It All Together
reservoir
Sluice gate
2m
10 cm
1 km downstream from gate there is a broad crested
weir with P = 1 m. Draw the water surface profile.
Wave Celerity
Vw
y
V
V+V
y+y
unsteady flow
M1 + M 2 = W + Fp1 + Fp2 + Fss
Per unit width
1
1
2
Fp1 = r gy 2
Fp2 = - r g ( y + d y )
2
2
1
2
2
ù
Fp1 + Fp2 = r g é
y
y
+
d
y
(
)
ë
û
2
y
V-Vw V+V-Vw
y+y
steady flow
Fp1
Fp2
V-Vw V+V-Vw
Wave Celerity:
Momentum Conservation
M 1    V  Vw  y
2
M 2 = r (V + dV - Vw )(V - Vw ) y
Per unit width
M1 + M 2 = r y (V - Vw )[(V + dV - Vw ) - (V - Vw )]
1
2
2
é
F
+
F
=
r
g
y
y
+
d
y
(
) ùû
M1 + M 2 = r y (V - Vw )dV
p1
p2
ë
2
Now equate pressure and momentum
1
2
2
2
ù
r gé
y
y
2
y
d
y
d
y
= r y (V - Vw )dV
ë
û
2
 gy  V  Vw V
y
V-Vw V+V-Vw
steady flow
y+y
Wave Celerity
yV  Vw    y  y V  V  Vw 
Mass conservation
yV  yVw  yV  yV  yV  yV  yVw  yVw
V  V  Vw 
y
y
 gy  V  Vw V
gy  V  Vw 
2
Momentum
y
V-Vw V+V-Vw
y
steady flow
y
gy  V  Vw 
2
y+y
c  V  Vw
c 
V
gy
yg
 Fr 
V
c
Wave Propagation
 Supercritical flow
 c<V
 waves only propagate downstream
 water doesn’t “know” what is happening downstream
upstream control
 _________
 Critical flow
 c=V
 Subcritical flow
 c>V
 waves propagate both upstream and downstream
Discharge Measurements
Sharp-Crested Weir
2
Q  Cd b 2 g H 3/ 2
3
V-Notch Weir
Q
Broad-Crested Weir
Sluice Gate
Explain the exponents of H!
8
 
Cd 2 g tan   H 5/ 2
15
2
2 
Q  Cd b g  H 
3 
Q  Cd byg 2 gy1
V  2 gH
3/ 2
Summary (1)
All the complications of pipe flow plus
additional parameter... _________________
free surface location
Various descriptions of energy loss
Chezy, Manning, Darcy-Weisbach
4
Importance of Froude Number
3
y
Fr>1 decrease in E gives increase in y
Fr<1 decrease in E gives decrease in y
Fr=1 standing waves (also min E given Q)
2
1
0
0
1
2
E
3
4
Summary (2)
Methods of calculating location of free
surface (Gradually varying)
Direct step (prismatic channel)
Standard step (iterative)
dy
So  S f

Differential equation
Rapidly varying
Hydraulic jump
dx
1  Fr 2
Broad-crested Weir: Solution
q 
E
gyc3
q  (9.8m / s )0.3m 
2
3
0.5
q  0.5144m 2 / s
Q  qL  1.54m3 / s
2
yc  E
3
q2
E1 @y1
2
2 gE1
yc m
yc=0.3
Broad-crested
weir
E2 
3
yc  0.45m
2
E1  E2  P  0.95m
q2
E1  y1 
2gy12
y1  0.935
H1  y1  0.5m  0.435
Summary/Overview
Energy losses
Dimensional Analysis
Empirical
8g
V=
S f Rh
f
1 2/3 1/2
V  R h So
n
Energy Equation
V12
V22
y1 +
+ So Dx = y2 +
+ S f Dx
2g
2g
V2
q2
Q2
= y+
E  y 
2 = y+
2
2
gy
2
gA
2g
Specific Energy
Two depths with same energy!
y
How do we know which depth4
is the right one?
3
Is the path to the new depth
2
possible?
1
0
0
1
2
E
3
4
What next?
Water surface profiles
Rapidly varied flow
A way to move from supercritical to subcritical flow
(Hydraulic Jump)
Gradually varied flow equations
Surface profiles
Direct step
Standard step
Hydraulic Jump!
Open Channel Reflections
 Why isn’t Froude number important for describing
the relationship between channel slope, discharge,
and depth for uniform flow?
 Under what conditions are the energy and
hydraulic grade lines parallel in open channel
flow?
 Give two examples of how the specific energy
could increase in the direction of flow.