Flow in Open Channel Flow Conditions

Transcription

1 Civil Engineering Hydraulics Flow The graduate with a Science degree asks, "Why does it work?" The graduate with an Engineering degree asks, "How does it work?" The graduate with an Accounting degree asks, "How much will it cost?" The graduate with a Liberal Arts degree asks, "Do you want fries with that?" Flow in Flow Conditions 2 While we used the Froude Number to calculate the critical flow conditions, the Reynolds number is still used to characterize the flow as laminar, transitional, or turbulent 1

2 Flow in Flow Conditions 3 The calculation of the Reynolds Number for flow is a bit different than for closed conduits. Flow in Flow Conditions For our purposes, we will assume that transition occurs at a Reynolds number of Most common open-channel flows are turbulent. 4 2

3 If we consider flow in an open channel Some of the notation in this section may be different from that we have used earlier. This material is taken from a different text. I will try and point out where the differences are. 5 If we consider flow in an open channel In this case, the z is taken to the bottom of the channel from some reference datum and y is the depth of flow from the bottom of the channel. 6 3

4 7 The bottom of the channel is sloped at some angle α in the direction of flow 8 It is very common to have the slope of the channel expressed as a fall of so many feet per 100 or 1000 feet 4

5 9 If we choose two points in the flow (1) and (2) we can write a continuity expression at the two points. 10 At point 1, we have a velocity v1 and a depth of flow (relative to the channel bottom) of y1 5

6 So we can write the specific energy as v12 y There is also a potential energy term that is the elevation of the channel bottom above the datum Adding this we have v12 y1 + + z1 12 6

7 We can follow the same steps at point 2 to determine the energy there v12 y1 + + z1 v 22 y2 + + z2 13 The energy line in the drawing represents the sum of the specific and potential energy with respect to the datum. v12 y1 + + z1 v 22 y2 + + z2 14 7

8 If we consider friction losses between the two points Remember hl is negative v12 y1 + + z1 + hl v 22 y2 + + z2 15 For continuity, we can equate the energy values at the two points. v12 y1 + + z1 + hl = v 22 y2 + + z2 16 8

9 z2 can be expressed in terms of z1, the length of the channel between points 1 and 2, and the slope of the channel. v12 v2 + z1 + hl = y z2 z2 = z1 L sin α y Substituting for z2 in the energy equation v12 v2 + z1 + hl = y z1 L sin α z2 = z1 L sin α y

10 Manipulating the expression we then have v12 v 22 y1 + + hl = y 2 + L sin α 19 If we constrain α to very small values, we can use the approximation v12 v 22 y1 + + hl = y 2 + L sin α sin α tan α 20 10

11 So we can move the L sin α to the other side of the expression and make the substitution of tan for sin v12 v 22 y1 + + hl + L tan α = y The slope, and therefore the tangent of α, is given as S0 so the expression becomes v12 v 22 y1 + + hl + S0L = y

12 If the rate of head loss due to friction is constant, we can define a slope for the energy line, Sf v12 v 22 y1 + + hl + S0L = y Assuming a constant rate of head loss with distance the slope of the energy line would be hl L v12 v 22 y1 + + Sf L + S0L = y 2 + Sf = 24 12

13 Now if we consider the flow to be uniform hl L v12 v 22 y1 + + Sf L + S0L = y 2 + Sf = 25 Uniform flow exists if the depth is uniform (i.e., unchanging) at every section of the channel. Uniform flow occurs when volume flow rate does not change during the interval of interest. hl L v12 v 22 y1 + + Sf L + S0L = y 2 + Sf = 26 13

14 Forces in 27 If we consider the forces acting on an element of a uniform flow. We will have diagram similar to what is shown below. Forces in The element goes from the free surface to the contact with the channel It has a cross-sectional area in the zy plane of A 28 14

15 Forces in There are forces developed by the pressure acting on the upstream and downstream faces of the element. The partial differential represents the change in pressure in the downstream direction. 29 Forces in There is no shear force acting at the top of the element as it is only in contact with a free surface (air) The shear force on the bottom is from the contact with the channel bottom and sides

16 Forces in 31 The final forces on the diagram represent the weight on the element and how it is broken into components along the z and x axis. Forces in Since uniform flow is considered as a steady condition, there is no acceleration to the element. Also there is no change in the pressure in the x direction

17 Forces in Therefore 33 p =0 x Fx = 0 Forces in So () τ w dac + ρ gadx sin θ = 0 Ac = Area of contact with the channel 34 17

18 Forces in The area of contact is the wetted perimeter times the distance along the channel. () τ w dac + ρ gadx sin θ = 0 Ac = Px () τ w Pdxc + ρ gadx sin θ = 0 35 Forces in Isolating the shear τw = τw = () ρ gadx sin θ Pdx ρ gasin θ () P ( ) PA = ρ g sin (θ ) R τ w = ρ g sin θ τw 36 h 18

19 Forces in 37 The shear can be expressed in terms of the velocity of the flow, the density, and the friction at the contact surface () τ w = ρ g sin θ Rh τw = f ρv f ρv 2 = ρ g sin θ Rh 4 2 () Forces in Isolating the velocity term f ρv 2 = ρ g sin θ Rh ρ g sin θ Rh v2 = fρ () () v= 38 () 8g sin θ Rh f 19

20 Forces in The θ in this expression is the same as the α used earlier. So the sin(θ) is equal to the channel slope. 39 f ρv 2 = ρ g sin θ Rh ρ g sin θ Rh v2 = fρ () () v= () 8g sin θ Rh f Forces in In pipe flow, the f is dependent only on the material of the pipe. Open channels are more complex. 40 f ρv 2 = ρ g sin θ Rh ρ g sin θ Rh v2 = fρ () () v= () 8g sin θ Rh f 20

21 Forces in The f can depend of the variable geometry of the channel, the roughness of the surface, and changes in alignment of the channel. () 8g sin θ Rh v= 41 f Forces in To reflect this, a different term was developed for the roughness It is known as the Manning roughness coefficient and is related to f by 8g 1 61 = Rh f n 42 21

22 Returning to the channel with uniform flow both the depth of flow and the velocity are the same at points 1 and 2 v12 v 22 y1 + + Sf L + S0L = y 2 + y1 = y 2 v1 = v 2 43 In this case, the right and left hand sides of the expression we developed reduces to Sf L + S0L = 0 Sf L = S0L 44 22

23 Normal Flow The slope of the head loss due to friction is exactly equal to the slope of the channel Remember this is for normal/uniform flow Sf L + S0L = 0 Sf L = S0L Sf = S0 45 Normal Flow We calculated the head loss over a length of conveyance in the work we did on pipes That hasn t changed L v2 hl = f Dh 46 23

24 Normal Flow Dh is the hydraulic diameter and was defined as 4 times the wetted area divided by the wetted perimeter S0L = Sf L S0 = Sf L v2 hl = f Dh 47 Normal Flow Dh is the same as 4 times the hydraulic radius, Rh Sf = SO L v2 L v2 hl = f =f Dh 4Rh 48 24

25 Normal Flow Sf was the ratio of hl to L, so we have Sf = SO L v2 L v2 hl = f =f Dh 4Rh L v2 f hl 4Rh = L L 49 Normal Flow So the slope can be equated to Sf = SO L v2 f hl 4Rh SO = = L L 50 25

26 Normal Flow Combining the two expression we derived S0 = f v2 8gRh 8g 1 61 = Rh f n 8g 1 31 = 2 Rh f n S0 8gRh 2 v = f S0 8gRh v = R 8g n 2 h v 2 = S0 v= R n2 h R S n h 0 Normal Flow The final expression is known as the Manning Equation v = Rh S0 n 52 26

27 Normal Flow R is given in meters, and v is in meters/sec To achieve dimensional consistency, the units of 1/n are m1/3/s You text gives units of n as L1/6 but the units of 1/n above are consistent 2 3 h v= R S n Normal Flow The Manning equation in USCS units is v= Rh S0 n 54 27

28 Normal Flow The units of R are feet and v are in feet/ second The units of 1.486/n are ft1/3/s v= Rh S0 n 55 Normal Flow Values for n are found in tables such as Table 7.2 in your text v= Rh S0 n v = Rh S0 n 56 28

29 13 54 A trapezoidal channel with a bottom width of 5 m, free surface width of 10 m, and flow depth of 2.2 m discharges water at a rate of 120 m3/s. If the surfaces of the channel are lined with asphalt (n=0.016), determine the elevation drop of the channel per km. We know the geometry of the channel, the flow depth, and the rate of flow along with the Manning coefficient for the wetted surfaces. We can start by writing the Manning equation and see is we can isolate the slope, S A trapezoidal channel with a bottom width of 5 m, free surface width of 10 m, and flow depth of 2.2 m discharges water at a rate of 120 m3/s. If the surfaces of the channel are lined with asphalt (n=0.016), determine the elevation drop of the channel per km. Since the units are in SI, we will use the SI version of the Manning equation v = Rh S0 n 58 29

30 13 54 A trapezoidal channel with a bottom width of 5 m, free surface width of 10 m, and flow depth of 2.2 m discharges water at a rate of 120 m3/s. If the surfaces of the channel are lined with asphalt (n=0.016), determine the elevation drop of the channel per km. Isolating S0 we have. 2 nv = S 0 2 R 3 h A trapezoidal channel with a bottom width of 5 m, free surface width of 10 m, and flow depth of 2.2 m discharges water at a rate of 120 m3/s. If the surfaces of the channel are lined with asphalt (n=0.016), determine the elevation drop of the channel per km. We can calculate the velocity from the flow rate if we first calculate the cross sectional area. 2 widthtop := 10m widthbottom := 5m yn := 2.2m 3 Q := 120 Area := 60 m n := sec widthtop + widthbottom 2 yn 2 Area = 16.5 m Q n A 2 = S0 Rh3 30

31 13 54 A trapezoidal channel with a bottom width of 5 m, free surface width of 10 m, and flow depth of 2.2 m discharges water at a rate of 120 m3/s. If the surfaces of the channel are lined with asphalt (n=0.016), determine the elevation drop of the channel per km. Since we know the flow area, all we need is the wetted perimeter to calculate the hydraulic radius. 2 Assume that the channel slopes are the same on both sides of the channel. 2 Lengthside := 61 widthtop widthbottom + yn2 2 Lengthside = 3.33 m Q n A 2 = S0 Rh A trapezoidal channel with a bottom width of 5 m, free surface width of 10 m, and flow depth of 2.2 m discharges water at a rate of 120 m3/s. If the surfaces of the channel are lined with asphalt (n=0.016), determine the elevation drop of the channel per km. 2 WP := 2 Lengthside + widthbottom 62 WP = m Q n A 2 = S0 Rh3 31

32 13 54 A trapezoidal channel with a bottom width of 5 m, free surface width of 10 m, and flow depth of 2.2 m discharges water at a rate of 120 m3/s. If the surfaces of the channel are lined with asphalt (n=0.016), determine the elevation drop of the channel per km. 2 Q n Area S0 := 2 1 Area 3 m3 WP sec S0 = Q n A 2 = S0 Rh A trapezoidal channel with a bottom width of 5 m, free surface width of 10 m, and flow depth of 2.2 m discharges water at a rate of 120 m3/s. If the surfaces of the channel are lined with asphalt (n=0.016), determine the elevation drop of the channel per km. This is the elevation drop per meter of length, the question asked for the drop per kilometer. 2 Q n A 2 = S0 Rh

33 Homework 23-1 Cooling water used in condensers of power plants is sometimes conveyed to cooling ponds before being returned to the body of water from which it came. One such pond is fed by a channel that is rectangular in cross section, 3 m wide, and formed by dredging. The channel is well maintained, uniform, and weathered. If the water depth is 2 m and the channel slope is 0.09, calculate the flow rate in the channel. 65 Homework 23-2 A trapezoidal channel carries water at a flow of 200 ft3/s. Its width is 8 ft, and m = 1.0. The channel is made of riveted steel. l l 66 Calculate the slope necessary to maintain a uniform depth of 3 ft. Calculate the critical depth of the flow. 33

34 Homework Water is flowing in a 6 m diameter culvert at a depth of 4.7 m.the culvert is uniform, made of concrete, and laid on a slope of 1. Determine the volume flow rate through the culvert. HINT(The expressions from Wednesday s class might be useful here). 34