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CHAPTER FIVE CHAPTER FIVE PRINCIPLES OF OPEN CHANNEL FLOW

CHAPTER FIVE

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CHAPTER FIVE. PRINCIPLES OF OPEN CHANNEL FLOW. 5.1 FUNDAMENTAL EQUATIONS OF FLOW. 5.1.1 Continuity Equation Inflow = Outflow Area; A 1 and A 2 and Velocity; V 1 and V 2 Area x Velocity (A . V) = Discharge, Q ie. A 1 V 1 = A 2 V 2 = Q. - PowerPoint PPT Presentation

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Page 1: CHAPTER  FIVE

CHAPTER FIVECHAPTER FIVE

PRINCIPLES OF OPEN CHANNEL FLOW

Page 2: CHAPTER  FIVE

5.1 FUNDAMENTAL 5.1 FUNDAMENTAL EQUATIONS OF FLOWEQUATIONS OF FLOW

5.1.1 Continuity Equation Inflow = Outflow Area; A1 and A2 and Velocity; V1 and

V2

Area x Velocity (A . V) = Discharge, Q ie. A1 V1 = A2 V2 = Q

Page 3: CHAPTER  FIVE

5.1.2 Energy Equation5.1.2 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

Page 4: CHAPTER  FIVE

Kinetic Energy (KE):Kinetic Energy (KE):

Energy possessed by Moving objects.Solid Mechanics : KE = 1/2 m V2

But Mass = W/g where W is the weight

In hydraulics: KE = 1/2 . W/g . V2 = W V2 /2g

KE per unit weight = ( W V2 /2g) / W = V2 /2g

Page 5: CHAPTER  FIVE

Pressure EnergyPressure Energy

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

Page 6: CHAPTER  FIVE

Potential EnergyPotential 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 + V2 /2g + Z : The

Bernoulli Equation.

Page 7: CHAPTER  FIVE

Bernoulli TheoremBernoulli 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.

Z1 + V12/2g + P1 /w = Z2 + V2

2 /2g + P2 /w

Page 8: CHAPTER  FIVE

5.2 UNIFORM FLOW IN 5.2 UNIFORM FLOW IN OPEN CHANNELSOPEN 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.

Page 9: CHAPTER  FIVE

Uniform and Non-Uniform FlowUniform and Non-Uniform Flow

In Uniform Flow, All Characteristics of Flow Are Same Along the Whole Length of Flow.

Ie. Velocity, V1 = V2 ; Flow Areas, A1 = A2

In Uniform Channel Flow, Water Surface is Parallel to Channel Bed. In Non-uniform Flow, Characteristics of Flow Vary along the Whole Length.

Page 10: CHAPTER  FIVE

More Open Channel TermsMore 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.

Page 11: CHAPTER  FIVE

Parameters of Open ChannelsParameters 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

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

Area, A

Wetted Perimeter

d

b

Page 12: CHAPTER  FIVE

Empirical Flow Equations for Empirical Flow Equations for Estimating Normal Flow VelocitiesEstimating Normal Flow Velocities

a) Chezy Formula (1775): Can be derived from basic principles. It states that: ;

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.

SRCV

Page 13: CHAPTER  FIVE

Manning Formula (1889)Manning Formula (1889)

Empirical Formula based on analysis of various discharge data. The formula is the most widely used.

  '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.

2/13/21SR

nV

Page 14: CHAPTER  FIVE

Best Hydraulic Section or Best Hydraulic Section or Economic Channel SectionEconomic 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 =

For a rectangular Channel: P = b +2d A = b d and therefore: b = A/di.e. P = A/d + 2 d

SRCV SRC )1.....(S

P

ACA

Page 15: CHAPTER  FIVE

Best Hydraulic Section Contd.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/d2 + 2 For minimum P i.e. Pmin , - A/d2 + 2 = 0

A = 2 d2 , Since A = b d ie. b d = 2 d2 ie. b = 2 d i.e. for maximum discharge, b = 2 d OR

2Ad

2Ad

Page 16: CHAPTER  FIVE

For a Trapezoidal SectionFor a Trapezoidal Section

Zd Zd

d

b

1

Z

Area of cross section(A) = b d + Z d2

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 + Z2 )1/2

Page 17: CHAPTER  FIVE

For maximum flow, P has to be a minimumi.e dp/dd = - A/d2 - Z + 2 (1 + Z2 )1/2

For Pmin, - A/d2 - Z + 2 (1 + Z2)1/2 = 0

A/d2 = 2 (1 + Z2 ) - Z A = 2 d2 ( 1 + Z2 )1/2 - Z d2 But Area = b d + Z d2 ie. bd + Z d2 = 2 d2 (1 + Z2 ) - Z d2

For maximum discharge, b = 2 d (1 + Z2)1/2 - 2 Z d or:  

Try: Show that for the best hydraulic section: 

dA

Z Z

2 1 2 1 2( ) /

b

d 2

1(cos

tan )

Page 18: CHAPTER  FIVE

          NON-UNIFORM FLOW IN NON-UNIFORM FLOW IN OPEN CHANNELSOPEN 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)

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.

Page 19: CHAPTER  FIVE

Variations of Flow VelocitiesVariations of Flow Velocities

1 2 3 4

Page 20: CHAPTER  FIVE

Non-uniform Flow In Open Non-uniform Flow In Open ChannelsChannels Contd. 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.

Page 21: CHAPTER  FIVE

Energy Line AnalysisEnergy Line Analysis

Page 22: CHAPTER  FIVE

Energy Line Analysis ContdEnergy 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

Page 23: CHAPTER  FIVE

Specific Energy, EsSpecific Energy, Es

When we neglect Z, the energy obtained is called specific energy (Es).

Specific energy (Es) = d + V2 /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.

Page 24: CHAPTER  FIVE

Variation of Specific Variation of Specific Energy( Es) with depth (d)Energy( Es) with depth (d)

Page 25: CHAPTER  FIVE

Variation of Specific Energy( Es) Variation of Specific Energy( Es) with depth (d)with depth (d) Contd.Contd.

Es = d + V 2 /2g, since q = v d ie. v = q/d, Es = d + q 2 /2 g d2

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)

Page 26: CHAPTER  FIVE

Variation of Spcific Energy, Es Variation of Spcific Energy, Es with depth, dwith depth, d

A

B

Depth, d

Specific Energy, Es

IncreasingFlow per unitWidth, q

C.

Minimum Es

Page 27: CHAPTER  FIVE

Critical DepthCritical 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 CRITICAL DEPTH and CRITICAL VELOCITY.

At Critical point C, the value of Es is minimum for a given flow rate q.

Page 28: CHAPTER  FIVE

b) Some Properties of Critical FlowEs (specific energy) = d + q 2 /2gd 2 ..............(1)

At critical flow, E has a minimum value - obtain minimum value by differentiation:

dEs/dd = 1 - q2 /gd 3 = 0ie. q 2 /g dc

3 = 1 , d = dc - critical depth.

dc = q 2 /g (q 2 = g d3 )

 

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 = Vc dc ie. Vc2 dc

2 = g dc3 and V c

2 = g dc

and Vc2 /2g = 1/2 dc - kinetic energy

ie. dc = Vc2 /g

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 = dc + 1/2 dc = 3/2 dc dc = 2/3 Ec

  ie. dc = q2 /g = Vc

2 /g = 2/3 Ec

  

dq

gc 32

Page 29: CHAPTER  FIVE

Some Properties of Critical Flow Some Properties of Critical Flow Contd.Contd.

It is also interesting to see how discharge q varies with depth, d for a given amount of specific energy, E

dc

Max Discharge

d

q

Page 30: CHAPTER  FIVE

Variation of Es with d Variation of Es with d ConcludedConcluded

Es = d + q 2 /2g d2 ie. q2 = 2g d 2 (Es - d) = 2g d2 Es - 2g d3

For maximum q, dq/dd = 0ie. 2 dq/dd = 4 Es g d - 6 g d2

dq/dd = (4 Es g d - 6 g d2)/2 = 0ie. 2 Es g d - 3 g d 2 = 0 ie. dc = 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.

Page 31: CHAPTER  FIVE

            Sub-Critical and Super-Critical Sub-Critical and Super-Critical FlowsFlows

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.

Page 32: CHAPTER  FIVE

Sub-Critical and Super-Critical Sub-Critical and Super-Critical Flows Contd.Flows Contd.

Super-critical

Sub-critical

d

Es

Critical

Page 33: CHAPTER  FIVE

Froude NumberFroude 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 = dc and V = Vc)

F < 1, Flow is sub-critical (d > dc and V < Vc)

F > 1, Flow is super-critical(d< dc and V > Vc)

Page 34: CHAPTER  FIVE

Hydraulic JumpHydraulic 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.

Page 35: CHAPTER  FIVE

Diagram of Hydraulic JumpDiagram of Hydraulic Jump

dc

dsub

d

Water Falling From a Spillway

Page 36: CHAPTER  FIVE

Depth and Energy Loss of Depth and Energy Loss of Hydraulic JumpHydraulic 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 d2

Page 37: CHAPTER  FIVE

Depth and Energy Loss of Depth and Energy Loss of Hydraulic Jump Contd.Hydraulic Jump Contd.

d1

d2

V1

V2

Page 38: CHAPTER  FIVE

Depth and Energy Loss of Depth and Energy Loss of Hydraulic Jump ConcludedHydraulic Jump Concluded

It can be shown that: 

Also: the loss of energy (m) during a hydraulic jump can be derived as: 

The depths are in m and:

Power loss (kW) = 9.81 x Flow rate , m3/s x Energy loss (m)

 

 

dd d d V

g21 1

21 1

2

2 4

2

Ed d

d d

( )2 13

1 24

Page 39: CHAPTER  FIVE

Hydraulic Jump ConcludedHydraulic 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.