CHAPTER FIVE PRINCIPLES OF OPEN CHANNEL FLOW

5.1 FUNDAMENTAL EQUATIONS OF FLOW



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

5.1.2 Energy Equation

 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

Kinetic Energy (KE):

 Energy possessed by Moving objects.

 Solid Mechanics : KE = 1/2 m V 2  But Mass = weight W/g where W is the  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

Pressure Energy

      Fluid flow under pressure has ability to do work and so possesses energy by virtue of its pressure.

Pressure force = P. a specific volume where P is pressure.

w = If W is the weight of water flowing, then Volume = W/w Distance moved by Volume/area = distance flow = W/w.a

(Recall Work done = Force x distance = P.a x W/wa = WP/w Pressure Energy per unit weight = P/w

Potential Energy

 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 Bernoulli Equation.

/2g + Z : The

Bernoulli Theorem

 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 UNIFORM FLOW IN OPEN CHANNELS



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.

Flow, Characteristics of Flow Vary along the Whole Length.

In Non-uniform

More Open Channel Terms



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.

Parameters of Open Channels

 

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):

cross sectional area is A, then R = A/P, If 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.

coefficient and is a function of hydraulic radius and channel roughness.

C is Chezy

Manning Formula (1889)

 Empirical Formula based on analysis of various discharge data.

most widely used.

The formula is the 

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.

Best Hydraulic Section or Economic Channel Section

 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

Best Hydraulic Section Contd.

 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 A = 2 d 2 ,

d



A

2 Since A = b d ie. b d = 2 d 2 2 + 2 = 0 ie. b = 2 d i.e. for maximum discharge, b = 2 d

OR

d



A

2

For a Trapezoidal Section

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 A/d 2 - Z + 2 (1 + Z = 2 (1 + Z 2 ) 2 - Z ) 1/2 = 0 A = 2 d 2 ( 1 + Z 2 But Area = b d + Z d 2 ) 1/2 - Z d ie. bd + Z d 2 2 = 2 d 2 For maximum discharge, b = 2 d (1 + Z 2 ) 1/2 (1 + Z 2 - 2 Z d ) - Z d 2 or:

d

 

Z A

) 

Z

Try: Show that for the best hydraulic section:

b



d

2 1 ( cos  

NON-UNIFORM FLOW IN OPEN CHANNELS

 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) by  Non-uniform flow can be caused 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.

Variations of Flow Velocities

1 2 3 4

Non-uniform Flow In Open Channels

Contd.

 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.

Energy Line Analysis

Energy Line Analysis Contd

 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

Specific Energy, Es

     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)

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 (super critical)

Variation of Spcific Energy, Es with depth, d

Depth, d Increasing Flow per unit Width, q A C.

B Minimum Es Specific Energy, Es

Critical Depth

 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 DEPTH and CRITICAL VELOCITY.

CRITICAL  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 differentiation: dEs/dd = 1 - q 2 ie. q d c 2 /g d c 3 = q 2 = 1 /g /gd (q 3 = 0 2 , d = d c = g d 3 ) obtain minimum value by - critical depth.

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 but q = V c and V c 2 ie. d c 2 = g d d c c 3 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 discharge interesting to see given amount of specific energy, E how q varies with depth, d for a d Max Discharge dc q

Variation of Es with d Concluded

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

Froude Number

 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)

Hydraulic Jump

 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.

Diagram of Hydraulic Jump

d d sub dc Water Falling From a Spillway

Depth and Energy Loss of Hydraulic Jump

 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

Depth and Energy Loss of Hydraulic Jump Contd.

d 1 V 1 V 2 d 2

Depth and Energy Loss of Hydraulic Jump Concluded

It can be shown that:

d

2  

d

1 2 

d

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)

Hydraulic Jump Concluded

 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.