Open channel hydraulics for engineers. Chapter 2 uniform flow
lượt xem 16
download
Open channel hydraulics for engineers. Chapter 2 uniform flow
The chapter on uniform flow in open channels is basic knowledge required for all hydraulics students. In this chapter, we shall assume the flow to be uniform, unless specified otherwise. This chapter guides students how to determine the rate of discharge, the depth of flow, and the velocity. The slope of the bed and the crosssectional area remain constant over the given length of the channel under the uniformflow conditions. The same holds for the computation of the most economical cross section when designing the channel. The concept of permissible velocity against erosion and sedimentation is introduced.......
Bình luận(0) Đăng nhập để gửi bình luận!
Nội dung Text: Open channel hydraulics for engineers. Chapter 2 uniform flow
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  Chapter UNIFORM FLOW 2.1. Introduction 2.2. Basic equations in uniform openchannel flow 2.3. Most economical crosssection 2.4. Channel with compound crosssection 2.5. Permissible velocity against erosion and sedimentation Summary The chapter on uniform flow in open channels is basic knowledge required for all hydraulics students. In this chapter, we shall assume the flow to be uniform, unless specified otherwise. This chapter guides students how to determine the rate of discharge, the depth of flow, and the velocity. The slope of the bed and the crosssectional area remain constant over the given length of the channel under the uniformflow conditions. The same holds for the computation of the most economical cross section when designing the channel. The concept of permissible velocity against erosion and sedimentation is introduced. Key words Uniform flow; most economical crosssection; discharge; velocity; erosion; sedimentation 2.1. INTRODUCTION 2.1.1. Definition Uniform flow relates to a flow condition over a certain length or reach of a stream and can occur only during steady flow conditions. Uniform flow may be also defined as the flow occurring in a channel in which equilibrium has been reached between gravitational force and shear force. Many irrigation and drainage canals and other artificial channels are designed to carry water at uniform depth and cross section all along their lengths. Natural channels as rivers and creeks are seldom of uniform shape. The design discharge is set by considerations of acceptable risk and frequency analysis, whereas the channel slope and the crosssectional shape are determined by topography, and soil and land conditions. Uniform equilibrium openchannel flows are characterized by a constant depth and a constant mean flow velocity: h V 0 and 0 (21) s s where s is the coordinate in the flow direction, h the flow depth and V the flow velocity. Uniform equilibrium openchannel flows are commonly called “uniform flows” or “normal flows”. Note: The velocity distribution in fullydeveloped turbulent open channel flows is given approximately by Prandtl’s power law (Fig. 2.1): y N 1 V (22) Vmax h where the exponent 1/N varies from ¼ down to ½ depending on the boundary friction and the crosssection shape. The most commonlyused power law formulae are the onesixth  Chapter 2: UNIFORM FLOW 25
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  power (1/6) and the oneseventh power (1/7) formulas. It should be noted that the velocity in openchannel flow is assumed constant over the entire crosssection. Vmax V velocity h distribution v y Fig. 2.1. Velocity distribution profile in turbulent flow Such flow conditions are represented schematically in Fig. 2.2. Considering Bernoulli’s theorem of the conservation of energy, between crosssections 1 and 2, leads to the expression: V2 V2 E1 z1 1 1 1 E 2 h L z 2 2 2 2 h L p p (23) 2g 2g where 1 and 2 are the Corrioliscoefficients corresponding to the velocities V1 and V2, respectively. They are also called the kineticenergy correction coefficients. is equal to or larger than 1 but rarely exceeds 1.1. (Li and Hager, 1991). For a uniform velocity distribution, = 1. The slope of the energy gradient line S is equal to the bed slope i of the channel, or: S= i L h (24) L energygradient line V12 hL 2g V 22 S 2g h1 1 p hydraulicgradient line E1 g h2 p2 E2 g i z1 L z2 Datum Fig. 2.2. Energy and hydraulic gradient in uniformflow channel If the flow is uniform, the cross sections at points 1 and 2 must be constant. Consequently, the velocity and the depth of flow must also remain constant, or: V1 = V2 and h1 = h2 (25) The flow resistance in an open channel is more difficult to quantify. The importance of the resistance coefficient goes beyond its use in channel design for uniform flow.  Chapter 2: UNIFORM FLOW 26
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  2.1.2. Momentum analysis Consider a control volume of length L in uniform flow, as shown in Fig. 2.3. L FP1 A Wsin h FP2 = FP1 oPL P W = gAL Fig. 2.3. Force balance in uniform flow By definition, the hydrostatic forces, Fp1 and Fp2, are equal and opposite. In addition, the mean velocity is invariant in the flow direction, so that the change in momentum flux is zero. Thus, the momentum equation reduces to a balance between the gravity force component in the flow direction and the resisting shear force: A L sin = o P L (26) in which = g = specific weight of the fluid, A = crosssectional area of flow, o = mean boundary shear stress, and P = wetted perimeter of the boundary on which the shear stress acts. If Eq. (26) is divided by PL, the hydraulic radius R = A/P appears as an intrinsic variable. Physically, Eq. (26) represents the ratio of flow volume to boundary surface area, or shear stress to unit weight, in the flow direction. Eq. (26) can be written as: o = R sin RS (27) if we replace sin with S = tan for small values of . Furthermore, if we solve Eq. (27) for the bed slope, which equals the slope of the energy grade line, hL/L, and express the shear stress in terms of the friction factor f for uniform pipe flow according Darcy Weisbach: o f (28) V 2 8 we have the DarcyWeisbach equation (for uniform pipe flow): h f o f ..V 2 f V2 i=S= . (29) L R 8 R 4R 2g from which it is evident that the appropriate length scale, when applied to openchannel flow, is 4R. It seems reasonable to use 4R as the length scale in the Reynoldsnumber and the relative roughness as well. Before applying uniform flow formulas to the design of open channels, the background of Chezy’s as well as Manning’s formulas for steady, uniform in open channels are presented in the next section.  Chapter 2: UNIFORM FLOW 27
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  2.2. Basic equations in uniform openchannel flow 2.2.1. Chezy’s formula Consider an open channel of uniform crosssection and bed slope as shown in Fig. 2.4: L Q VA Q i Fig.2.4. Sloping bed of a channel Let L = length of the channel; A = crosssectional area of flow; V = velocity of water; P = wetted perimeter of the crosssection; f = friction coefficient according to DarcyWeisbach; and i = uniform slope of the bed. It has been experimentally found, that the total frictional resistance along the length L of the channel, follows the law: Frictional resistance = contact area (velocity)2 f 8 = P.L Vn f (210) 8 The exponent n has been experimentally found to be nearly equal to 2. But for all practical purposes, its value is taken to be 2. Therefore, P.L V2 f Frictional resistance = (211) 8 Since the water moves over a distance V in 1 second, therefore, the work done in overcoming the friction reads as: Frictional resistance distance V in 1 second = PLV2V = PL V3 (212) f f 8 8 The weight of the water, W, in the channel over a length of L is: W = .A.L (213) This water “falls” vertically down over a distance V.i in 1 second, so Loss of potential energy = Weight of water Height = .A.L.V.i (214) We know that work done in overcoming friction = Loss of potential energy P.L V3 = .A.L.V.i f i.e. (215) 8  Chapter 2: UNIFORM FLOW 28
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  .A.i V2 f . .P =8 8 A .f or V = i (216) P 8 is known as Chezy’s coefficeint and R = 8g A where C = as hydraulic radius. .f f P The discharge of flow then is Q = A V = AC Ri (217) Note: Unlike the DarcyWeisbach coefficient f , which is dimensionless, the Chezy coefficient C has the dimension, [L1/2T1], as mentioned in Chapter 1. Chezy’s coefficient C depends on the mean velocity V, the hydraulic radius R, the kinematic viscosity and the relative roughness. There is experimental evidence that the value of the resistance coefficient does vary with the shape of the channel and therefore with R and possibly also with the bed slope i, which for uniform flow will be equal to the slope of the energyhead line io, yielding a relationship for the velocity of the form: V = K. Rx.ioy (218) where K, x and y are constants. Example 2.1: A rectangular channel is 4 m deep and 6 m wide. Find the discharge through the channel, when it runs full. Take the slope of the bed as 1:1000 and Chezy’s coefficient as 50 m1/2s1. Solution: Given: Depth h = 4 m, Width b = 6 m, h Bed slope i = 1/1000 = 0.001, Chezy’s coefficient C = 50 m1/2s1 Q = ? (m3/s) b Area of the rectangular channel: A=hb = 24 m2 Perimeter of the rectangular channel: P = b + 2h = 14 m A Hydraulic radius of the flow: R = = 1.71 m P Discharge through the channel: Q = AC Ri = 49.62 m3s1 Ans. Example 2.2 : Water is flowing at the rate of 8.5 m3s1 in an earthen trapezoidal channel with a bed width 9 m, a water depth 1.2 m and side slope 2:1. Calculate the bed slope, if the value of C in Chezy’s formula be 49.5 m1/2s1. Solution: Given: Discharge Q = 8,5 m3/s, Bed width b = 9 m, B Depth h = 1.2 m, Side slope m = 2, Chezy’s coefficient C = 49.5 m1/2s1, h 2 Bed slope = ? h/2 1 b h Surface width of the trapezoidal channel B = b + 2( ) = 10.2 m 2  Chapter 2: UNIFORM FLOW 29
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  Area of the trapezoidal channel: A = b B h = 11.52 m2 2 P = b 2 h 2 h 2 2 Wetted perimeter: = 11.68 m A Hydraulic radius: R = = 0.986 m P Now using the relation: Q = AC Ri Q2 1 i = = Ans. R(AC)2 4440 2.2.2. Manning’s formula Manning, after carrying out a series of experiments, deduced the following relation for the value of C in Chezy’s formula: = R 6 1 1 C (219) n where n is the Manning constant in metric units, n = [m1/3s]. n is expressing the channel’s relative roughness properties and values are given in Table 2.1 Now we see that the velocity: = C Ri = R 6 Ri = R 6 R 2 i 2 1 1 1 1 1 1 V n n = R 3 i 2 1 2 1 V (220) n Now, the discharge is: Q = AV = A R 3 i 2 1 2 1 (221) n Table 2.1: Values of Manning coefficient n [m1/3s] Wetted perimeter n Wetted perimeter n A. Natural channel D. Artificially lined channel Clean and straight 0.030 Glass 0.010 Sluggish with deep pools 0.040 Brass 0.011 Major rivers 0.035 Steel, smooth 0.012 B. Flood plain Steel, painted 0.014 Pasture, farmland 0.035 Steel, riveted 0.015 Light brush 0.050 Cast iron 0.013 Heavy brush 0.075 Concrete, finished 0.012 Trees 0.150 Concrete, unfinished 0.014 C. Excavated earth channels Planned wood 0.012 Clean 0.022 Clay tile 0.014 Gravelly 0.025 Brickwork 0.015 Weedy 0.030 Asphalt 0.016 Stony, cobbles 0.035 Corrugated metal 0.022 Rubble masonry 0.025 For a more detailed description, we can take the value of Manning’s n from the Table at the end of this chapter.  Chapter 2: UNIFORM FLOW 30
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  Example 2.3: An earthen trapezoidal channel with a 3 m wide base and side slopes 1:1 carries water with a depth of 1 m. The bed slope is 1/1600. Estimate the discharge. Take the value of n in Manning’s formula as 0.04 m1/3s. Solution: B Given: Base width b = 3 m, Side slope = 1:1, h Water depth h = 1 m, 1 Bed slope 1/1600, 1 Manning’s coefficient n = 0.04 m1/3s h b Discharge Q = ? (m3/s) Surface width of the trapezoidal channel B = b + 2h = 5 m Area of the trapezoidal channel: A= b B = 4 m2 h 2 Wetted perimeter: P = b 2 h2 h2 = 5.828 m A Hydraulic radius: R = = 0.686 m P Q = A R 3 i 2 1 2 1 Now using the relation: = 1.94 m3/s Ans. n Example 2.4 : Water at the rate of 0.1 m3/s flows through a vitrified sewer with a diameter of 1 m with the sewer pipe half full. Find the slope of the water surface, if Manning’s n is 0.013 m1/3s. Solution: Given: Discharge Q = 0.1 m3/s, Diameter of pipe D = 1 m, Manning’s n = 0.013 m1/3s Sewer slope i = ? 1 D2 = 0.393 m 2 Area of the flow: A= D 2 4 D Wetted perimeter: P = = 1.57 m 2 A Hydraulic radius: R= = 0.25 m P Using Manning’s formula: Q = A R 3 i 2 1 2 1 n Qn 2 Water surface slope: 1 i = 2 = Ans. A R 3 1430  Chapter 2: UNIFORM FLOW 31
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  2.2.3. Discussion of factors affecting f and n The dependence of f on the relative roughness in open channel flow is not as well known as in pipe flow, because it is difficult to assign an equivalent sandgrain roughness to the large values of the absolute roughness height typically found in open channels. The dependence of flow resistance on the crosssectional shape occurs as a result of changes of both the channel hydraulic radius, R, and the crosssectional distribution of velocity and shear. There is no substitute for experience in the selection of Manning’s n for natural channels. Table 2.2. (at the end of this chapter) from Ven Te Chow (1959) gives an idea of the variability to be expected in Manning‘s n. 2.3. MOST ECONOMICAL CROSSSECTION 2.3.1. Concept A typical uniform flow problem in the design of an artificial canal is the economical proportioning of the crosssection. A canal, having a given Manning coefficient n and a slope i, is to carry a certain discharge Q, and the designer’s aim is to minimize the crosssectional area. Clearly, if A is to be a minimum, the velocity V is to be a maximum. The Chezy and Manning formulas indicate, therefore, that the hydraulic radius R = A/P must be a maximum. It can be shown that the problem is equivalent to that of minimizing P for a given constant value of A. This concept has a practical application in estimating the cost for a canal excavation and /or lining. From economic considerations of minimizing the flow crosssectional area for a given design discharge, a theoretically optimum crosssection will be introduced. 2.3.2. Conditions for maximum discharge The conditions for maximum discharge for the following crosssections will be dealt with: (a). Rectangular crosssection, and (b). Trapezoidal crosssection. (a). Channel with rectangular crosssection Consider a channel of rectangular crosssection as shown in Fig. 2.5. h Let b = breadth of the channel, and b h = depth of the channel. A=b h b= A Area of flow: (220) Fig.2.5. A rectangular channel h Q=AV = AC Ri AC A Discharge: i (222) P Keeping A, C and i constant in the above equation, the discharge will be maximum when A/P is maximum or the wetted perimeter P is minimum. Or in other words, when: 0 dP (223) dh P = b + 2h = 2h A We know that (224) h  Chapter 2: UNIFORM FLOW 32
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  Differentiating the above equation with respect to h and equating to zero yields: 2 20 dP A (225) dh h A = 2h = b h 2 or b = 2h i.e. the breadth is two times the depth (226) In this case, the hydraulic radius is: 2h h 2h 2 h R A bh (227) P b 2h 2h 2h 4h 2 Hence, we can say that for the maximum discharge or the maximum velocity, these two conditions (i.e. b = 2h and R = h/2) should be used for solving the problem of optimizing channels of rectangular crosssection. 1.00 0.95 Q Q max 0.90 h 0.85 b A = bh = constant b 0 h 1 2 3 4 5 Q b Fig. 2.6. Experimental relationship between and Qmax h As can be seen in Fig. 2.6, the maximum represented by this optimal configuration is a b rather weak one. For example, for aspect ratios, , between 1 and 4, the flow rate is within h 96% of the maximum flow rate obtained with the same area and by b/h = 2. Example 2.5.: Find the most economical crosssection of a rectangular channel to carry 0.3 m3/s of water, when the bed slope is 1/1000. Assume Chezy’s C = 60 m1/3s1. Solution: Given: Discharge Q = 0.3 m3/s, Bed slope i = 1/1000, Chezy coefficient C = 60 m1/3s1 Breadth of channel b = ? (m) and depth of the channel h = ? (m) We know that for the most economical rectangular section: b = 2h Area: A = b h = 2h h = 2h2 and hydraulic radius: R = h/2 = 0.5 h Using the relation: Q = AC Ri  Chapter 2: UNIFORM FLOW 33
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  and squaring both sides yields: 0.09 m5 7.2h 5 or h 5 0.0125 h 5 0.0125 m Depth of the channel: h = 0.416 m Ans. And breadth: b = 2d = 0.832 m Ans. (a). Channel with trapezoidal crosssection Consider a channel of trapezoidal crosssection ABCD as shown in Fig. 2.7. B = b + 2nh D C 1 h n A b B nh Fig.2.7. A trapezoidal channel Let b = breadth of the channel at the bottom, h = depth of the channel, and 1 = side slope (i.e. 1 vertical to n horizontal) n Area of flow: A = h(b + nh) = b + nh b = A A or  nh (228) h h Q = A V AC Ri AC A Discharge: i (229) P Keeping A, C and i constant in the above equation, the discharge will be maximum, when A/P is maximum or the wetted perimeter P is minimum. Or in other words: 0 dP dh We know that: P = b 2 n 2 h 2 h 2 b 2h n 2 1 (230) Substituting the value of b from equation (228) yields: nh 2h n 2 1 A P= (231) h Differentiating the above equation with respect to d and equating to zero results into: 2 n 2 n2 1 0 dP A (232) dh h n 2 n2 1 A h2  Chapter 2: UNIFORM FLOW 34
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  h(b nh) 2 n 2 n2 1 [A = h(b + nh)] h b nh h n 2 n2 1 h h b 2nh or h n2 1 (233) 2 We see that b + 2nh = B is the top width of the channel and h n 2 1 is the length of the sloping side, i.e. the length of the sloping side is equal to half the top width In this case, the hydraulic radius: h(b nh) h(b nh) h(b nh) h A R= (234) P b 2h n 1 b (b 2nh) 2(b nh) 2 2 Hence, we can say that for the maximum discharge or the maximum velocity, these two b 2nh h n 2 1 and/or R = ) should be used for solving the problems h conditions (i.e. 2 2 in the case of channels of trapezoidal crosssection. Example 2.6: A canal of trapezoidal crosssection has to be excavated through hard clay at the least cost. Determine the dimensions of the channel for a discharge equal to 14 m3/s, a side slope for hard clay n = 1:1, a bed slope 1:2500 and Manning’s n = 0.02 m1/3s. Solution: Given: Discharge Q = 14 m3/s; Bed slope i = 1/2500; h 1 Side slope n = 1:1 Manning’s n = 0.02 m1/3s h b 1 Breath b = ? (m) Depth h = ? (m) We know that for the least cost: half of the top width = length of sloping side b 2nh h n 1 2 2 with n =1 b 2h 2h 2 2.83h b = 0.83 h Area of flow: A = h(b + nh) = 1.83 h2 Using Manning’s formula: Q = A R 3 i 2 1 2 1 n h 3 12.142 m8/3 8 yields: h 2.55 m Ans. And b = 0.83 h = 2.12 m Ans.  Chapter 2: UNIFORM FLOW 35
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  Note: The semicircular section (the semicircle having its center in the surface) is the best hydraulic section. The best hydraulic crosssection for other shapes can be drawn as presented in Fig. 2.8. D R R R D R 2 D D 2 2 2 60 90 circular channel rectangular channel trapezoidal triangular channel Fig. 2.8. Crosssections of maximum flow rate: i.e. “optimum design” channel Students should try to proof the conditions for circular and triangular channels for the best A Q.n = constant hydraulic crosssection based on the below relationship: 5 Q 1 2 1 1 A 1 3 P i 2 V .R 3 .i 2 . .i 2 3 (225) A n n P 2 3 1 2 Table 2.2.: The “best” design summary for several channel crosssections Crosssection Optimum Optimum Optimum Optimum width B crosssection A wetted perimeter P hydraulic radius R Semicircle D D2 D D 8 2 4 Rectangular D D2 2D D 2 4 Trapezoidal 2D 3 2 3D D D 3 4 4 Triangular 2D D2 2D D 2 4 2.3.3. Problems of uniformflow computation The computation of uniform flow may be performed by the use of two equations: the continuity equation and a uniformflow formula. When the Manning formula is used as the uniformflow formula, the computation will involve the following six variables: (1) the normal discharge Q (4) the mean velocity of flow V (2) the normal depth h (5) the coefficient of roughness n (3) the channel slope i (6) the geometric elements that depend on the shape of the channel section, such as A,R, etc, …  Chapter 2: UNIFORM FLOW 36
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  When any four of the above six variables are given, the remaining two unknowns can be determined by the two equations. The following represents some types of problems of uniform flow computation: A. to compute the normal discharge: In practical applications, this computation is required for the determination of the capacity of a given channel or for the construction of a sysnthetic rating curve of the channel. B. to determine the velocity of the flow: This computation has many applications. For example, it is often required for the study of scouring and silting effects in a given channel. C. to compute the normal depth: This computation is required for the determination of the stage of flow in a given channel. D. to determine the channel roughness: This computation is used to ascertain the roughness coefficient in a given channel; the coefficient thus determined may be used in other similar channels. E. to compute the channel slope: This computation is required for adjusting the slope of a given channel. F. to determine the dimensions of the channel section: This computation is required mainly for design purposes. Table 2.3. lists the known and unknown variables involved in each of the six types of problems mentioned above. Table 2.3: Problems of uniformflow computation Type of Discharge Velocity Depth Roughness Slope Geometric problem Q V d n i elements A ?  B  ? C  ? D  ? E  ? F  ? Notes: The known variables are indicated by the check mark () and the unknown required in the problems by the question mark (?). The unknown variable(s) that can be determined from the known variables is(are) indicated by a dash (). Table 2.3. does not include all types of problems. By varying combinations of various known and unknown variables, more types of problems can be formed. In design problems, the use of the best hydraulic section and of empirical rules is generally introduced and thus new types of problems are created.  Chapter 2: UNIFORM FLOW 37
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  2. 4. CHANNEL WITH COMPOUND CROSSSECTION A compound channel consists of a main channel, which carries the base flow (frequently running off up to bankfull conditions), and a floodplain on one or both sides that carries overbank flow during the time of flooding as sketched in Fig. 2.9. The compound crosssection of a channel may be composed of several distinct subsections with each subsection different in roughness form the others. canal bank high water level dyke low water level (1): A1, P1, n1 (2): A2, P2, n2 (3): A3, P3, n3 Fig.2.9. Overbank flow in a compound channel The roughness of the side channels will be different (generally rougher) from that of the main channel and the method of analysis is to consider the total discharge to be the sum of the component discharges computed by the Manning equation. The mean velocity for the whole channel section is equal to the total discharge divided by the total water area. The classical method of computation of discharge, as presented by Chow in 1959, consisted in subdividing the composite crosssection into subareas with vertical interfaces in which the shear stresses are neglected. The discharge for each subarea is calculated by assuming a common friction slope i for the whole channel. Thus in the channel, as shown in Fig. 2.9, assuming that the bed slope is the same for the three subareas, it holds: Q Qi i i i 2 m m AR 3 (226) i 1 i 1 ni The division of the channel by these artificial vertical boundaries assumes implicity that the shear stress on these interfaces is relatively small with respect to the shear stress acting on the wetted perimeter of the channel. Note: It has been found by more recent experimentation that this hypothesis is incorrect and that it leads to a considerable overestimation of the discharge in the compound channel. Example 2.7: Water flows along a drainage canal having the properties shown in the figure below. If the bottom slope i = 1/500=0.002, estimate the discharge. 3m 2m 3m 0.6 m n1 = 0.020 (1) (2) n3 = 0.030 (3) n2 = 0.015 0.8 m [ni] = m1/3s  Chapter 2: UNIFORM FLOW 38
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  Solution: We divide the crosssection into three subsections as is indicated in the figure and write the discharge as Q = Q1 + Q2 + Q3, where for each section, it holds: Qi =AiVi = A i 1 2 3 12 Ri i ni The appropriate values of Ai, Pi, Ri and ni are listed in the table below: i Ai (m2) Pi (m) Ri (m) ni (m1/3s) (1) 1.8 3.6 0.500 0.020 (2)* 2.8 3.6 0.778 0.015 (3) 1.8 3.6 0.500 0.020 Note that the imaginary portions of the wetted perimeter between the sections (denoted by the dashed lines in the figure) are not implemented in Pi. That is, for section (2): A2 = 2 (0.8 + 0.6) m2 = 2.8 m2 P2 = {2 + 2(0.8)} m = 3.6 m R2 2 A 2.8 So that m = 0.778 m P2 3.6 Thus the total discharge is: Q = Q1 + Q2 + Q3 = 1.8 0.500 3 2.8 0.778 3 1.8 0.500 2 2 2 Q = 0.002 3 2 m3s1 1 0.020 0.015 0.030 Q = 11.275 m3/s Ans. If the entire channel crosssection were considered as one flow area, then : A = A1 + A2 + A3 = 6.4 m2 P = P1 + P2 + P3 = 10.8 m A Then R= = 0.593 m P The total discharge can be written as A R 3i 2 1 1 2 1 Q= n eff n where neff is the effective value of n for the whole compound channel. With Q = 11.275 m3/s, as determined above, the value of neff is found to be: A R 3 i 2 2 1 n eff = 0.01681 m1/3s Q As expected, the effective roughness (Manning’s n) is between the minimum (n2 = 0.015 m1/3s) and maximum (n3 = 0.030 m1/3s) values for the individual subsections.  Chapter 2: UNIFORM FLOW 39
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  2.5. PERMISSIBLE VELOCITY AGAINST EROSION AND SEDIMENTATION The excavation and lining cost of open channels or conduits varies with their size. With respect to waterresourcessystem economics, erosion of and sedimentation in channels are problems in hydraulic engineering. Erosion and sedimentation must be predicted because they can change the bed slope, the channel width and therefore the flow conditions. So, if the available slope permits, the cost of the initial construction may be reduced by using the highest velocity. However, if the velocity is becoming too high, the channel may be damaged or destroyed by erosion. This must be avoided by limiting the velocities according to the boundary materials. For clear water in hardsurfaced water conductors, the limiting velocity is beyond practical requirements. Velocities above 10 m/s for clear water in concrete channels have been found to do no harm. If the water carries abrasive material, damage may occur at lower velocities. No definite relation has been established between the nature of abrasive materials, the material of channel bank and bed, and a permissible velocity. In unlined earthen channels, the limiting velocity involves many factors. Generally, a fine soil is eroded more easily than a coarse one, but the effect of the grain size may be obscured by the presence or absence of a cementing or binding material. The tendency to erode is reduced by seasoning. Groundwater conditions can exert an important influence. Seepage out of the channel, particularly if the water is turbid, tends to toughen the banks, whereas infiltration reduces the resistance to erosion. Erosion can be reduced or avoided by designing for low velocities. If the water carries an appreciable amount of silt in suspension, too low a velocity will cause the canal to fill up until its capacity is impaired. It is necessary to choose a velocity that will keep the silt in motion but that will not erode the bank or bottom of the canal. The margin of permissible velocities depends on the amount and nature of the silt in the water, the nature of the bank material, the size and shape of the canal, and many other factors. The silt content of most turbid water varies with the season, as does also the demand for water and the resultant velocity of the flow. The determination of nonscouring, nonsilting velocities for earthen canals has attracted the attention of many investigators over a long period of time, and a considerable mass of data and formulas have been accumulated. However, for preliminary purposes, and for design in many cases, use may be made of the approximate values purposed by Fortie and Scobey, in 1926, as shown in Table 2.4. Where the silt is important, it is better to make the slope a little too steep rather than a little too flat. A gradient that proves to be too steep can be controlled by checks. In hardsurfaced channels, silting is easily controlled if fall for scouring velocity is available.  Chapter 2: UNIFORM FLOW 40
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  Table 2.4: Permissible canal velocities (Fortier and Scobey, 1926) Velocity, m/s, after aging, of canal carrying: Clear Water Water transporting water, no transporting noncolloidal silts, Original material excavated for canal detritus colloidal silts sands, gravels, or rock fragments (1) (2) (3) (4) Fine sand (noncolloidal) 0.46 0.76 0.46 Sandy loam (noncolloidal) 0.53 0.76 0.61 Sandy loam (noncolloidal) 0.61 0.91 0.61 Alluvial silts when noncolloidal 0.61 1.07 0.61 Ordinary firm loam 0.76 1.07 0.69 Volcanic ash 0.76 1.07 0.61 Fine gravel 0.76 1.52 1.14 Stiff clay (very colloidal) 1.14 1.52 0.91 Graded, loam to cobbles, when noncolloidal 1.14 1.52 1.52 Alluvial silts when colloidal 1.14 1.52 0.91 Graded, loam to cobbles, when colloidal 1.22 1.68 1.52 Coarse gravel (noncolloidal) 1.22 1.83 1.98 Cobbles and shingles 1.52 1.68 1.98 Shales and hardpans 1.83 1.98 1.52  Chapter 2: UNIFORM FLOW 41
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  Table 2.5: Value of Manning’s Roughness Coefficient n [m1/3s] (Ven Te Chow, 1973) Type of channel and description Minimum Normal Maximum A. Closed Conduits Flowing Party Full A.1. Metal a. Brass, smooth 0.009 0.010 0.013 b. Steel 1. Lock bar and welded 0.010 0.012 0.014 2. Riveted and spiral 0.013 0.016 0.017 c. Cast iron 1. Coated 0.010 0.013 0.014 2. Uncoated 0.011 0.014 0.016 d. Wrought iron 1. Black 0.012 0.014 0.015 2. Galvanized 0.013 0.016 0.017 e. Corrugated metal 1. Subdrain 0.017 0.019 0.021 2. Storm drain 0.021 0.024 0.030 A.2. Nonmetal a. Lucite 0.008 0.009 0.010 b. Glass 0.009 0.010 0.013 c. Cement 1. Neat surface 0.010 0.011 0.013 2. Mortar 0.011 0.013 0.015 d. Concrete 1. Culvert, straight and free of debris 0.010 0.011 0.013 2. Culvert with bends, connections, and some 0.011 0.013 0.014 debris 3. Finished 0.011 0.012 0.014 4. Sewer with manholes, inlet, etc., straight 0.013 0.015 0.017 5. Unfinished, steel form 0.012 0.013 0.014 6. Unfinished, smooth wood form 0.012 0.014 0.016 7. Unfinished, rough wood form 0.015 0.017 0.020 e. Wood 1. Stave 0.010 0.012 0.014 2. Laminated, treated 0.015 0.017 0.020 f. Clay 1. Common drainage tile 0.011 0.013 0.017 2. Vitrified sewer 0.011 0.014 0.017 3. Vitrified sewer with manholes, inlet, etc. 0.013 0.015 0.017 4. Vitrified subdrain with open joint 0.014 0.016 0.018 g. Brickwork 1. Glazed 0.011 0.013 0.015 2. Lined with cement mortar 0.012 0.015 0.017 h. Sanitary sewers coated with sewage slimes, with 0.012 0.013 0.016 bends and connections i. Paved invert, sewer, smooth bottom 0.016 0.019 0.020 j. Rubble masonry, cemented 0.018 0.025 0.030  Chapter 2: UNIFORM FLOW 42
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  Type of channel and description Minimum Normal Maximum B. Lined or builtup channels B.1. Metal a. Smooth steel surface 1. Unpainted 0.011 0.012 0.014 2. Painted 0.012 0.013 0.017 b. Corrugated 0.021 0.025 0.030 B.2. Nonmetal a. Cement 1. Neat surface 0.010 0.011 0.013 2. Mortar 0.011 0.013 0.015 b. Wood 3. Planed, untreated 0.010 0.012 0.014 4. Planed, creosoted 0.011 0.012 0.015 5. Unplanted 0.011 0.013 0.015 6. Plank with battens 0.012 0.015 0.018 7. Lined with roofing paper 0.010 0.014 0.017 c. Concrete 1. Trowel finish 0.011 0.013 0.015 2. Float finish 0.013 0.015 0.016 3. Finished, with gravel on bottom 0.015 0.017 0.020 4. Unfinished 0.014 0.017 0.020 5. Gunite, good section 0.016 0.019 0.023 6. Gunite, navy section 0.018 0.022 0.025 7. On good excavated rock 0.017 0.020 8. On irregular excavated rock 0.022 0.027 d. Concrete bottom float finished with sides of 1. Dressed stone in mortar 0.015 0.017 0.020 2. Random stone in mortar 0.017 0.020 0.024 3. Cement rubble masonry, plastered 0.016 0.020 0.024 4. Cement rubble masonry 0.020 0.025 0.030 5. Dry rubble or riprap 0.020 0.030 0.035 e. Gravel bottom with sides of 1. Formed concrete 0.017 0.020 0.025 2. Random stone in mortar 0.020 0.023 0.026 3. Dry rubble or riprap 0.023 0.033 0.036 f. Brick 1. Glazed 0.011 0.013 0.015 2. In cement mortar 0.012 0.015 0.018 g. Masonry 1. Cemented rubble 0.017 0.025 0.030 2. Dry rubble 0.023 0.032 0.035 h. Dressed ashlar 0.013 0.015 0.017 i. Asphalt 1. Smooth 0.013 0.013 2. Rough 0.016 0.016 j. Vegetal lining 0.030 …. 0.500  Chapter 2: UNIFORM FLOW 43
 OPEN CHANNEL HYDRAULICS FOR ENGINEERS  Type of channel and description Minimum Normal Maximum C. Excavated or dredged a. Earth, straight and uniform 1. Clean, recently completed 0.016 0.018 0.020 2. Clean, after weathering 0.018 0.022 0.025 3. Gravel, uniform section, clean 0.022 0.025 0.030 4. With short grass, few weeds 0.022 0.027 0.033 b. Earth, winding and sluggish 1. No vegetation 0.023 0.025 0.030 2. Grass, some weeds 0.025 0.030 0.033 3. Dense weeds or aquatic plants in deep channels 0.030 0.035 0.040 4. Earth bottom and rubble sides 0.028 0.030 0.035 5. Stony bottom and weedy banks 0.025 0.035 0.040 6. Cobble bottom and clean sides 0.030 0.040 0.050 c. Draglineexcavated or dredged 1. No vegetation 0.025 0.028 0.033 2. Light brush on banks 0.035 0.050 0.060 d. Rock cuts 1. Smooth and uniform 0.025 0.035 0.040 2. Jagged and irregular 0.035 0.040 0.050 e. Channels not maintained, weeds and brush uncut 1. Dense weeds, high as flow depth 0.050 0.080 0.120 2. Clean bottom, brush on sides 0.040 0.050 0.080 3. Same, highest stage of flow 0.045 0.070 0.110 4. Dense brush, high stage 0.080 0.100 0.140 D. Natural streams D.1. Minor stream (top width at flood stage < 100 ft) a. Streams on plain 1. Clean, straight, full stage, no rifts or deep pools 0.025 0.030 0.033 2. Same as above, but no more stones and weeds 0.030 0.035 0.040 3. Clean, winding, some pools and shoals 0.033 0.040 0.045 4. Same as above, but some weeds and stones 0.035 0.045 0.050 5. Same as above, lower stages, more ineffective 0.040 0.048 0.055 slopes and sections 6. Same as 4, but more stones 0.045 0.050 0.060 7. Sluggish reaches, weedy, deep pools 0.050 0.070 0.080 8. Very weedy reaches, deep pools, or floodways 0.075 0.100 0.150 with heavy stand of timber and underbrush b. Mountain stream, no vegetation in channel, banks usually steep, trees and brush along banks submerged at high stages 1. Bottom: gravels, cobbles, and few boulders 0.030 0.040 0.050 2. Bottom: cobbles with large boulders 0.040 0.050 0.070  Chapter 2: UNIFORM FLOW 44
CÓ THỂ BẠN MUỐN DOWNLOAD

Open channel hydraulics for engineers. Chapter 3 hydraulics jump
0 p  125  27

Open channel hydraulics for engineers. Chapter 1 introduction
0 p  103  20

Nonlinear Optics  Chapter 2
65 p  76  20

Ebook Physics for Scientists and Engineers
1307 p  41  20

Open channel hydraulíc for engineers
0 p  168  15

Geostatistics with Applications In Earth Sciences
220 p  47  13

Open channel hydraulics for engineers. Chapter 4 non uniform flow
0 p  89  13

Open channel hydraulics for engineers. Chapter 5 spillways
0 p  94  12

Open channel hydraulics for engineers. Chapter 7 unsteady flow
0 p  112  12

Open channel hydraulics for engineers. Chapter 6 transitions and energy dissipators
0 p  117  10

Nonlinear optics, Third edition
595 p  59  10

PROBLEMS IN LASER PHYSICS
323 p  43  8

Magnetic Material Engineering  Chapter 6: Applications in Medical and Biology
33 p  20  4

Electromagnetic Waves and Antennas combined  Chapter 2
23 p  25  4

Ebook Schaums theory and problems of engineering thermodynamics for engineers
386 p  16  4

CHAPTER 2: COULOMB'S LAW AND ELECTRIC FIELD INTENSITY
26 p  41  3

Altruistically Inclined?
391 p  13  2