 5.1.1
Continuity Equation
 Inflow = Outflow


Area; A
1
V
2 and A
2 and Velocity; V
Area x Velocity (A . V) = Discharge, Q
1 and
 ie. A
1
V
1
= A
2
V
2
= Q

 Energy is Capacity to do work
 Work done = Force x Distance moved
 Forms of Energy
 Kinetic Energy - velocity


Pressure Energy - pressure
Potential Energy - Height or elevation

 Energy possessed by Moving objects.
 Solid Mechanics : KE = 1/2 m V 2
 But Mass = W/g where W is the weight
 In hydraulics: KE = 1/2 . W/g . V 2 =
W V 2 /2g
 KE per unit weight =
 ( W V 2 /2g) / W = V 2 /2g







Fluid flow under pressure has ability to do work and so possesses energy by virtue of its pressure.
Pressure force = P. a where P is pressure.
w = specific volume
If W is the weight of water flowing, then Volume =
W/w
Distance moved by flow = W/w.a
(Recall
Volume/area = distance
Work done = Force x distance = P.a x W/wa =
WP/w
Pressure Energy per unit weight = P/w

 Energy Related to Position
 Wt. of Fluid W at a height Z
 Then Potential Energy = W Z
 Potential Energy per unit wt. = Z

 Total Energy available is the sum of the three:
 E = P/w + V 2 /2g + Z : The
Bernoulli Equation.

 Total Energy of Each Particle of a Body of
Fluid is the Same Provided That No Energy
Enters or Leaves the System at Any Point.
 Division of Available Energy Between
Pressure, Kinetic and Position May Change but Total Energy Remains Constant.


Bernoulli Equation Is Generally Used to
Determine Pressures and Velocities at
Different Positions in a System.
Z
1
+ V
1
2 /2g + P
1
/w = Z
2
+ V
2
2 /2g + P
2
/w


 5.2.1
Definitions
 a) Open Channel: Duct through which
Liquid Flows with a Free Surface River,
Canal

 b) Steady and Non- Steady Flow: In
Steady Flows, all the characteristics of flow are constant with time.
In unsteady flows, there are variations with time.

Uniform and Non-Uniform Flow
 In Uniform Flow, All Characteristics of Flow
Are Same Along the Whole Length of Flow.
 Ie. Velocity, V
1
= A
2
= V
2
; Flow Areas, A
1
 In Uniform Channel Flow, Water Surface is
Parallel to Channel Bed.
In Non-uniform
Flow, Characteristics of Flow Vary along the
Whole Length.

 d) Normal Flow: Occurs when the
Total Energy line is parallel to the bed of the Channel.
 f)Uniform Steady Flow: All characteristics of flow remain constant and do not vary with time.


 a) Wetted Perimeter, P : The Length of contact between Liquid and sides and base of
Channel
P = b + 2 d ; d = normal depth
Area, A d
Wetted Perimeter b b)Hydraulic Mean Depth or Hydraulic Radius (R): If cross sectional area is A, then R = A/P, e.g. for rectangular channel, A = b d, P = b + 2 d

Empirical Flow Equations for
Estimating Normal Flow Velocities
 a) Chezy Formula (1775): Can be derived from basic principles. It states that: ;

V

C R S
 Where: V is velocity; R is hydraulic radius and S is slope of the channel.
C is Chezy coefficient and is a function of hydraulic radius and channel roughness.

 Empirical Formula based on analysis of various discharge data.
The formula is the most widely used.

V

1
 n
R
2 / 3
S
1 / 2
 'n' is called the Manning's Roughness
Coefficient found in textbooks. It is a function of vegetation growth, channel irregularities, obstructions and shape and size of channel.

 For a given Q, there are many channel shapes.
There is the need to find the best proportions of B and D which will make discharge a maximum for a given area, A.


Using Chezy's formula:
Flow rate, Q = A C
V

R S
C
=
R
A C
S
A
S
P
.....( 1 )
 For a rectangular Channel: P = b +2d
 A = b d and therefore: b = A/d i.e. P = A/d + 2 d

 For a given Area, A, Q will be maximum when P is minimum (from equation 1)
 Differentiate P with respect to d
 dp/dd = - A/d 2 + 2




For minimum P i.e. P min
, - A/d 2 + 2 = 0
A = 2 d 2 , d
 A
2
Since A = b d ie. b d = 2 d 2 ie. b = 2 d i.e. for maximum discharge, b = 2 d OR d
 A
2

Zd
Zd d
1
Z b
Area of cross section(A) = b d + Z d 2
Width , b = A/d - Z d ...........................(1)
Perimeter = b + 2 d ( 1 + Z 2 ) 1/2
From (1), Perimeter = A/d - Z d + 2 d(1 + Z 2 ) 1/2

For maximum flow, P has to be a minimum i.e dp/dd = - A/d 2 - Z + 2 (1 + Z 2 ) 1/2
For P min
, - A/d 2 - Z + 2 (1 + Z 2 ) 1/2 = 0
A/d 2 = 2 (1 + Z 2 ) - Z
A = 2 d 2 ( 1 + Z 2 ) 1/2 - Z d 2
But Area = b d + Z d 2 ie. bd + Z d 2 = 2 d 2 (1 + Z 2 ) - Z d 2
For maximum discharge, b = 2 d (1 + Z 2 ) 1/2 - 2 Z d or: d


Z
A
)

Z
Try: Show that for the best hydraulic section: b
 d
2
1
( cos



 5.3.1
Definition: By non-uniform flow, we mean that the velocity varies at each section of the channel.
 Velocities at Sections 1 to 4 vary (
Next Slide)
 Non-uniform flow can be caused by i)Differences in depth of channel and

 ii) Differences in width of channel.
iii) Differences in the nature of bed

 iv) Differences in slope of channel and v) Obstruction in the direction of flow.

1
2
3 4

 In the non-uniform flow, the Energy Line is not parallel to the bed of the channel.
 The study of non-uniform flow is primarily concerned with the analysis of
Surface profiles and Energy Gradients.

 For the Energy Line, total head is equal to the depth above datum plus energy due to velocity plus the depth of the channel.
Pressure energy is not included because we are working with atmospheric pressure.
 ie. H = Z + d + V 2 / 2 g






When we neglect Z, the energy obtained is called specific energy (Es).
Specific energy (Es) = d + V 2 /2g
Non-uniform flow analysis usually involves the energy measured from the bed only, the bed forming the datum, and this is called specific energy.
In Uniform flow, the specific energy is constant and the energy grade line is parallel to the bed.
In non-uniform flow, although the energy grade line always slopes downwards in the direction of flow, the specific energy may increase or decrease according to the particular channel's flow conditions.

Variation of Specific Energy( Es) with depth (d) Contd.


Es = d + V 2 /2g, since q = v d ie.
v = q/d, Es = d + q 2 /2 g d 2
 For a given q, we can plot the variation of Es
(specific energy) with flow depth, and use the graph to solve the cubic equation above. For a given value of Es, there are two values of d, indicating two different flow regimes.
 Flow at A is slow and deep (sub-critical) while flow at B is fast and shallow (supercritical)

Depth, d
Increasing
Flow per unit
Width, q
A
C.
B
Minimum Es
Specific Energy, Es

 Critical Depth we observe from the graph is the depth at which the hydraulic specific energy possessed by a given quantity of flowing water is minimum.
 CRITICAL FLOW occurs at CRITICAL
DEPTH and CRITICAL VELOCITY.
 At Critical point C, the value of Es is minimum for a given flow rate q.

b) Some Properties of Critical Flow
Es (specific energy) = d + q 2 /2gd 2 ..............(1)
At critical flow, E has a minimum value obtain minimum value by differentiation: dEs/dd = 1 - q 2 /gd 3 = 0 ie. q 2 /g d d c c
3 = 1 , d = d c
- critical depth.
= q 2 /g (q 2 = g d 3 ) d c
 3 q
2 g
This means that critical depth, d is a function of flow per unit width, q only.
From above: q 2 = g d c
3 but q = V c and V c
2 ie. d c d c ie. V
/2g = 1/2 d c
= V c
2 /g c
2 d c
2 = g d c
3
- kinetic energy and V c
2 = g d c
This means that when the value of velocity head is double the depth of flow, the depth is critical.
The specific energy equation (1), now becomes:
E = d c
+ 1/2 d c
= 3/2 d c d c
= 2/3 E c ie. d c
= q 2 /g = V c
2 /g = 2/3 E c

Some Properties of Critical Flow
Contd.
 It is also interesting to see how discharge q varies with depth, d for a given amount of specific energy, E d
Max Discharge dc q

Es = d + q 2 /2g d 2 ie. q 2 = 2g d 2 (Es - d) = 2g d 2 Es - 2g d 3
For maximum q, dq/dd = 0 ie. 2 dq/dd = 4 Es g d - 6 g d 2 dq/dd = (4 Es g d - 6 g d 2 )/2 = 0 ie. 2 Es g d - 3 g d 2 = 0 ie. d c
= 2 Es/3
This means that maximum flow occurs when d = 2/3 Es. This equation is similar to the one above for critical flow. Therefore, if the specific energy available is Ec, then maximum flow occurs at
2/3 Ec ie. at the critical depth.
Summarising: When discharge is constant, critical depth is the depth at minimum specific energy and when the specific energy is constant, critical depth is the depth at maximum discharge .

Sub-Critical and Super-Critical
Flows
 At the increase of slope of a bed of flow, the level of flow drops and velocity of flow increases.
 Where a condition exists such that the depth of flow is below critical depth, the flow is referred to as super critical.
 Super-critical velocity refers to the velocity above critical velocity.


Similarly, sub-critical velocity refers to velocity below critical velocity.
These flow regimes can be represented by the two limbs of the depth-specific energy curve.

Sub-Critical and Super-Critical
Flows Contd.
d
Sub-critical
Critical
Super-critical
Es

 This is a Dimensionless Ratio Characterizing
Open Channel Flow.
 Froude Number, F = V/ gd
 = Stream velocity/wave velocity
 When F = 1, Flow is critical (d = d c and V = Vc)
 F < 1, Flow is sub-critical (d > d c and V < Vc)
 F > 1, Flow is super-critical(d< d c and V > Vc)

 If a super-critical flow suddenly changes to a sub-critical flow, a hydraulic jump is said to have occurred.
 The change from super-critical to sub-critical flow may occur as a result of an obstruction placed in the passage of the flow or the slope of the bed provided is not adequate to maintain super-critical flow eg. water falling from a spillway.

d d sub dc
Water Falling From a Spillway

 As energy is lost in the hydraulic jump, we cannot use the Bernoulli equation to analyse.
 Momentum equation is suited to this case - no mention of energy. The aim is to find expression for d
2

d
1
V
1
V
2 d
2

It can be shown that: d
2
  d d
2
1

4
1
2

2 d V
1 1
2 g
Also: the loss of energy (m) during a hydraulic jump can be derived as:

E

( d
2
 d
1
)
3
4 d d
1 2
The depths are in m and:
Power loss (kW) = 9.81 x Flow rate , m 3 /s x Energy loss (m)

 A hydraulic jump is use
 ful when we require:
 i) Dissipation of energy e.g. at the foot of a spillway
 ii) When mixing of fluids is required e.g. in chemical and processing plants.
 iii) Reduction of velocity e.g. at the base of a dam where large velocities will result in scouring.
 It is, however, undesirable and should not be allowed to occur where energy dissipation and turbulence are intolerable.