Download presentation

Presentation is loading. Please wait.

Published byAbigail Patterson Modified over 3 years ago

1
Chapter 3: Steady uniform flow in open channels

2
Learning outcomes By the end of this lesson, students should be able to: – Understand the concepts and equations used in open channel flow – Determine the velocity and discharge using Chezys & Mannings equation – Able to solve problems related to optimum cross section in both conduits and open channel 2 UiTMSarawak/ FCE/ BCBidaun/ ECW301

3
Introduction Comparison between full flow in closed conduit and flow in open channel 3 Full flow in closed conduitOpen channel flow No free surface and pressure in the pipe is not constant Existence of free water surface through out the length of flow in the channel. Pressure at the free surface remains constant, with value equal to atmospheric pressure. Flow cross sectional area remains constant and it is equal to the cross sectional area of the conduit (pipe). Flow cross sectional area may change throughout the length depending on the depth of flow. UiTMSarawak/ FCE/ BCBidaun/ ECW301

4
4 Flow classification based on fluid particles motion Turbulent flow Laminar Flow UiTMSarawak/ FCE/ BCBidaun/ ECW301

5
Flow classification Turbulent flow: – Characterized by the random and irregular movement of fluid particles. – Movement of fluid particles in turbulent flow is accompanied by small fluctuations in pressure. – Flows in open channel are mainly turbulent. – E.g. Hydraulic jump from spillway, Flow in fast flowing river 5 UiTMSarawak/ FCE/ BCBidaun/ ECW301

6
Flow classification Laminar flow: – Flow characterized by orderly movement of fluid particles in well defined paths. – Tends to move in layers. – May be found close to the boundaries of open channel. 6 UiTMSarawak/ FCE/ BCBidaun/ ECW301

7
For flow in pipes UiTMSarawak/ FCE/ BCBidaun/ ECW301 7 Turbulent Re > 2000 Laminar Re < 2000

8
Flow in open channel UiTMSarawak/ FCE/ BCBidaun/ ECW301 8 Turbulent Re > 500 Laminar Re < 500

9
9 Flow classification wrt space Uniform flow Flow parameters (velocity & pressure) remain constant wrt space at any point in the flow. E.g. Flow with a fixed discharge in a channel of constant geometrical shape and slope Non-uniform flow Flow parameters (velocity &pressure) change wrt time & space. E.g. Flow of fixed discharge through a channel which changes either in geometrical shape or slope of the channel UiTMSarawak/ FCE/ BCBidaun/ ECW301

10
10 Flow classification wrt time Steady flow Flow parameters remain constant over a specified time interval Unsteady flow Flow parameters vary over time UiTMSarawak/ FCE/ BCBidaun/ ECW301

11
Steady uniform flow Flow parameters do not change wrt space (position) or time. Velocity and cross-sectional area of the stream of fluid are the same at each cross-section. E.g. flow of liquid through a pipe of uniform bore running completely full at constant velocity. UiTMSarawak/ FCE/ BCBidaun/ ECW301 11

12
Steady non-uniform flow Flow parameters change with respect to space but remain constant with time. Velocity and cross-sectional area of the stream may vary from cross-section to cross-section, but, for each cross-section, they will not vary with time. E.g. flow of a liquid at a constant rate through a tapering pipe running completely full. UiTMSarawak/ FCE/ BCBidaun/ ECW301 12

13
Unsteady uniform flow Flow parameters remain constant wrt space but change with time. At a given instant of time the velocity at every point is the same, but this velocity will change with time. E.g. accelerating flow of a liquid through a pipe of uniform bore running full, such as would occur when a pump is started. 13 UiTMSarawak/ FCE/ BCBidaun/ ECW301

14
Unsteady non-uniform flow Flow parameters change wrt to both time & space. The cross-sectional area and velocity vary from point to point and also change with time. E.g. a wave travelling along a channel. UiTMSarawak/ FCE/ BCBidaun/ ECW301 14

15
Flow classification Normal depth – depth of flow under steady uniform condition. Steady uniform condition – long channels with constant cross-sectional area & constant channel slope. – Constant Q – Constant terminal v – Therefore depth of flow is constant (y n or D n ) UiTMSarawak/ FCE/ BCBidaun/ ECW301 15

16
Total energy line for flow in open channel 16 UiTMSarawak/ FCE/ BCBidaun/ ECW301

17
From the figure : – Water depth is constant slope of total energy line = slope of channel When velocity and depth of flow in an open channel change, then non-uniform flow will occur. 17 UiTMSarawak/ FCE/ BCBidaun/ ECW301

18
Flow classification Non uniform flow in open channels can be divided into two types: – Gradually varied flow, GVF Where changes in velocity and depth of flow take place over a long distance of the channel – Rapidly varied flow, RVF Where changes in velocity and depth of flow occur over short distance in the channel UiTMSarawak/ FCE/ BCBidaun/ ECW301 18

19
Non-uniform flow UiTMSarawak/ FCE/ BCBidaun/ ECW301 19

20
UiTMSarawak/ FCE/ BCBidaun/ ECW301 20

21
UiTMSarawak/ FCE/ BCBidaun/ ECW301 21

22
UiTMSarawak/ FCE/ BCBidaun/ ECW Analysis of flow in open channel Continuity equation Momentum equation Energy equation

23
Total energy line for flow in open channel 23 UiTMSarawak/ FCE/ BCBidaun/ ECW301

24
Continuity equation: For rectangular channel: Express as flow per unit width, q: 24 UiTMSarawak/ FCE/ BCBidaun/ ECW301

25
Momentum equation Produced by the difference in hydrostatic forces at section 1 and 2: Resultant force, 25 UiTMSarawak/ FCE/ BCBidaun/ ECW301

26
Energy equation But hydrostatic pressure at a depth x below free surface, Therefore, Energy equation rewritten as, UiTMSarawak/ FCE/ BCBidaun/ ECW301 26

27
Energy equation For steady uniform flow, Therefore the head loss is, And energy equation reduces to, Known as specific energy, E, (total energy per unit weight measured above bed level), 27 UiTMSarawak/ FCE/ BCBidaun/ ECW301

28
Geometrical properties of open channels Geometrical properties of open channels: – Flow cross-sectional area, A – Wetted perimeter, P – Hydraulic mean depth, m 28 UiTMSarawak/ FCE/ BCBidaun/ ECW301

29
A – covers the area where fluid takes place. P – total length of sides of the channel cross- section which is in contact with the flow. 29 D B UiTMSarawak/ FCE/ BCBidaun/ ECW301

30
Example 3.1 Determine the hydraulic mean depth, m, for the trapezoidal channel shown below. 30 UiTMSarawak/ FCE/ BCBidaun/ ECW301

31
31 Roughness coefficient Chezy, CManning, n

32
Chezys coefficient, C From chapter 1, Rearranging to fit for open channel, the velocity: Where Chezy roughness coefficient, For open channel i can be taken as equal to the gradient of the channel bed slope s. Therefore, 32 UiTMSarawak/ FCE/ BCBidaun/ ECW301

33
Mannings n Introduced roughness coefficient n of the channel boundaries. UiTMSarawak/ FCE/ BCBidaun/ ECW301 33

34
Example 3.2 Calculate the flow rate, Q in the channel shown in Figure 3.5, if the roughness coefficient n = and the slope of the channel is 1: UiTMSarawak/ FCE/ BCBidaun/ ECW301

35
Example 3.3 Determine the flow velocity, v and the flow rate for the flow in open channel shown in the figure. The channel has a Mannings roughness n = and a bed slope of 1: UiTMSarawak/ FCE/ BCBidaun/ ECW301 B 2.75 m 90 0

36
Example 3.4 (Douglas, 2006) An open channel has a cross section in the form of trapezium with the bottom width B of 4 m and side slopes of 1 vertical to 11/2 horizontal. Assuming that the roughness coefficient n is 0.025, the bed slope is 1/1800 and the depth of the water is 1.2 m, find the volume rate of flow Q using a. Chezy formula (C=38.6) b. Manning formula 36 UiTMSarawak/ FCE/ BCBidaun/ ECW301

37
Example 3.5 (Munson, 2010) Water flows along the drainage canal having the properties shown in figure. The bottom slope s o = Estimate the flow rate when the depth is 0.42 m. UiTMSarawak/ FCE/ BCBidaun/ ECW301 37

38
Example 3.6 (Bansal, 2003) Find the discharge of water through the channel shown in figure. Take the value of Chezys constant = 60 and slope of the bed as 1 in C AD B E 2.7 m 1.2 m 1.5 m UiTMSarawak/ FCE/ BCBidaun/ ECW301

39
Example 3.8 (Bansal, 2003) Find the diameter of a circular sewer pipe which is laid at a slope of 1 in 8000 and carries a discharge of 800 L/s when flowing half full. Take the value of Mannings n = d D UiTMSarawak/ FCE/ BCBidaun/ ECW301

40
Optimum cross-sections for open channels Optimum cross section – producing Q max for a given area, bed slope and surface roughness, which would be that with P min and A min therefore tend to be the cheapest. Q max : A min, P min 40 UiTMSarawak/ FCE/ BCBidaun/ ECW301

41
Example 3.9 Given that the flow in the channel shown in figure is a maximum, determine the dimensions of the channel. 41 UiTMSarawak/ FCE/ BCBidaun/ ECW301

42
Optimum depth for non-full flow in closed conduits Partially full in pipes can be treated same as flow in an open channel due to presence of a free water surface. 42 UiTMSarawak/ FCE/ BCBidaun/ ECW301

43
Optimum depth for non-full flow in closed conduits Flow cross sectional area, Wetted perimeter, 43 UiTMSarawak/ FCE/ BCBidaun/ ECW301

44
Optimum depth for non-full flow in closed conduits Under optimum condition, v max, Substituting & simplifying, UiTMSarawak/ FCE/ BCBidaun/ ECW301 44

45
Optimum depth for non-full flow in closed conduits Hence depth, D, at v max, Using Chezy equation, UiTMSarawak/ FCE/ BCBidaun/ ECW301 45

46
Optimum depth for non-full flow in closed conduits Q max occurs when (A 3 /P) is maximum, Substituting & simplifying, Therefore depth at Q max, UiTMSarawak/ FCE/ BCBidaun/ ECW301 46

47
Review of past semesters questions

48
OCT 2010 Analysis of flow in open channels is based on equations established in the study of fluid mechanics. State the equations. UiTMSarawak/ FCE/ BCBidaun/ ECW301 48

49
OCT 2010 Determine the discharge in the channel (n = 0.013) as shown in Figure Q3(b). The channel has side slopes of 2 : 3 (vertical to horizontal) and a slope of 1 : Determine also the discharge if the depth increases by 0.1 m by using Manning's equation. UiTMSarawak/ FCE/ BCBidaun/ ECW301 49

50
OCT 2010 State the differences between : i) Steady and unsteady flow ii) Uniform and non-uniform flow UiTMSarawak/ FCE/ BCBidaun/ ECW301 50

51
OCT 2010 Figure Q4(b) shows the channel. Prove that for a channel, the optimum cross-section occurs when the width is 4 times its depth (B = 4D). (Hint: A = 4/3BD and P = B + 4D) UiTMSarawak/ FCE/ BCBidaun/ ECW301 51

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google