(1)ζv2/2g = mi;(2)Q = Ωv.(3)Problem I.—Given the transverse section of stream and discharge, to find the slope. From the dimensions of the section find Ω and m; from (1) find ζ, from (3) find v, and lastly from (2) find i.Problem II.—Given the transverse section and slope, to find the discharge. Find v from (2), then Q from (3).Problem III.—Given the discharge and slope, and either the breadth, depth, or general form of the section of the channel, to determine its remaining dimensions. This must generally be solved by approximations. A breadth or depth or both are chosen, and the discharge calculated. If this is greater than the given discharge, the dimensions are reduced and the discharge recalculated.Fig. 114.Since m lies generally between the limits m = d and m =1⁄2d, where d is the depth of the stream, and since, moreover, the velocity varies as √ (m) so that an error in the value of m leads only to a much less error in the value of the velocity calculated from it, we may proceed thus. Assume a value for m, and calculate v from it. Let v1be this first approximation to v. Then Q/v1is a first approximation to Ω, say Ω1. With this value of Ω design the section of the channel; calculate a second value for m; calculate from it a second value of v, and from that a second value for Ω. Repeat the process till the successive values of m approximately coincide.§ 113.Problem IV. Most Economical Form of Channel for given Side Slopes.—Suppose the channel is to be trapezoidal in section (fig. 114), and that the sides are to have a given slope. Let the longitudinal slope of the stream be given, and also the mean velocity. An infinite number of channels could be found satisfying the foregoing conditions. To render the problem determinate, let it be remembered that, since for a given discharge Ω∞ √χ, other things being the same, the amount of excavation will be least for that channel which has the least wetted perimeter. Let d be the depth and b the bottom width of the channel, and let the sides slope n horizontal to 1 vertical (fig. 114), thenΩ = (b + nd) d;χ = b + 2d √ (n2+ 1).Both Ω and χ are to be minima. Differentiating, and equating to zero.(db/dd + n) d + b + nd = 0,db/dd + 2 √ (n2+ 1) = 0;eliminating db/dd,{n − 2√ (n2+ 1)} d + b + nd = 0;b = 2 {√ (n2+ 1) − n} d.ButΩ / χ = (b + nd) d / {b + 2d √ (n2+ 1)}.Inserting the value of b,m = Ω/χ = {2d √ (n2+ 1) − nd} / {4d √ (n2+ 1) − 2nd} =1⁄2d.That is, with given side slopes, the section is least for a given discharge when the hydraulic mean depth is half the actual depth.A simple construction gives the form of the channel which fulfils this condition, for it can be shown that when m =1⁄2d the sides of the channel are tangential to a semicircle drawn on the water line.SinceΩ / χ =1⁄2d,thereforeΩ =1⁄2χd.(1)Let ABCD be the channel (fig. 115); from E the centre of AD drop perpendiculars EF, EG, EH on the sides.LetAB = CD = a; BC = b; EF = EH = c; and EG = d.Ω = area AEB + BEC + CED,= ac +1⁄2bd.χ = 2a + b.Putting these values in (1),ac +1⁄2bd = (a +1⁄2b) d; and hence c = d.Fig. 115.Fig. 116.That is, EF, EG, EH are all equal, hence a semicircle struck from E with radius equal to the depth of the stream will pass through F and H and be tangential to the sides of the channel.To draw the channel, describe a semicircle on a horizontal line with radius = depth of channel. The bottom will be a horizontal tangent of that semicircle, and the sides tangents drawn at the required side slopes.The above result may be obtained thus (fig. 116):—χ = b + 2d / sin β.(1)Ω = d (b + d cot β);Ω/d = b + d cot β;(2)Ω/d2= b/d + cot β.(3)From (1) and (2),χ = Ω / d − d cot β + 2d / sin β.This will be a minimum fordχ / dd = Ω / d2+ cot β − 2 / sin β = 0,orΩ/d2= 2 cosec. β − cot β.(4)ord = √ {Ω sin β / (2 − cos β)}.From (3) and (4),b/d = 2 (1 − cos β) / sin β = 2 tan1⁄2β.Proportions of Channels of Maximum Discharge for given Area and Side Slopes. Depth of channel = d; Hydraulic mean depth =1⁄2d; Area of section =Ω.Inclinationof Sides toHorizon.Ratio ofSideSlopes.Area ofSection Ω.BottomWidth.Top width =twice lengthof each SideSlope.Semicircle....1.571d202dSemi-hexagon60° 0′3 : 51.732d21.155d2.310dSemi-square90° 0′0 : 12d22d2d75° 58′1 : 41.812d21.562d2.062d63° 26′1 : 21.736d21.236d2.236d53° 8′3 : 41.750d2d2.500d45° 0′1 : 11.828d20.828d2.828d38° 40′11⁄4: 11.952d20.702d3.202d33° 42′11⁄2: 12.106d20.606d3.606d29° 44′13⁄4: 12.282d20.532d4.032d26° 34′2 : 12.472d20.472d4.472d23° 58′21⁄4: 12.674d20.424d4.924d21° 48′21⁄2: 12.885d20.385d5.385d19° 58′23⁄4: 13.104d20.354d5.854d18° 26′3 : 13.325d20.325d6.325dHalf the top width is the length of each side slope. The wettedperimeter is the sum of the top and bottom widths.§ 114.Form of Cross Section of Channel in which the Mean Velocity is Constant with Varying Discharge.—In designing waste channels from canals, and in some other cases, it is desirable that the mean velocity should be restricted within narrow limits with very different volumes of discharge. In channels of trapezoidal form the velocity increases and diminishes with the discharge. Hence when the discharge is large there is danger of erosion, and when it is small of silting or obstruction by weeds. A theoretical form of section for which the mean velocity would be constant can be found, and, although this is not very suitable for practical purposes, it can be more or less approximated to in actual channels.Fig. 117.Let fig. 117 represent the cross section of the channel. From the symmetry of the section, only half the channel need be considered. Let obac be any section suitable for the minimum flow, and let it be required to find the curve beg for the upper part of the channel so that the mean velocity shall be constant. Take o as origin of coordinates, and let de, fg be two levels of the water above ob.Let ob = b/2; de = y, fg = y + dy, od = x, of = x + dx; eg = ds.The condition to be satisfied is thatv = c √ (mi)should be constant, whether the water-level is at ob, de, or fg. Consequentlym = constant = kfor all three sections, and can be found from the section obac. Hence alsoIncrement of section=y dx= k.Increment of perimeterdsy2dx2= k2ds2= k2(dx2+ dy2) and dx = k dy / √ (y2− k2).Integrating,x = k logε{y + √ (y2− k2)} + constant;and, since y = b/2 when x = 0,x = k logε[{y + √ (y2− k2)} / {1⁄2b + √ (1⁄4b2− k2) }].Assuming values for y, the values of x can be found and the curve drawn.The figure has been drawn for a channel the minimum section of which is a half hexagon of 4 ft. depth. Hence k = 2; b = 9.2; the rapid flattening of the side slopes is remarkable.Steady Motion of Water in Open Channels of Varying Cross Section and Slope§ 115. In every stream the discharge of which is constant, or may be regarded as constant for the time considered, the velocity at different places depends on the slope of the bed. Except at certain exceptional points the velocity will be greater as the slope of the bed is greater, and, as the velocity and cross section of the stream vary inversely, the section of the stream will be least where the velocity and slope are greatest. If in a stream of tolerably uniform slope an obstruction such as a weir is built, that will cause an alteration of flow similar to that of an alteration of the slope of the bed for a greater or less distance above the weir, and the originally uniform cross section of the stream will become a varied one. In such cases it is often of much practical importance to determine the longitudinal section of the stream.The cases now considered will be those in which the changes of velocity and cross section are gradual and not abrupt, and in which the only internal work which needs to be taken into account is that due to the friction of the stream bed, as in cases of uniform motion. Further, the motion will be supposed to be steady, the mean velocity at each given cross section remaining constant, though it varies from section to section along the course of the stream.Fig. 118.Let fig. 118 represent a longitudinal section of the stream, A0A1being the water surface, B0B1the stream bed. Let A0B0, A1B1be cross sections normal to the direction of flow. Suppose the mass of water A0B0A1B1comes in a short time θ to C0D0C1D1, and let the work done on the mass be equated to its change of kinetic energy during that period. Let l be the length A0A1of the portion of the stream considered, and z the fall, of surface level in that distance. Let Q be the discharge of the stream per second.Fig. 119.Change of Kinetic Energy.—At the end of the time θ there are as many particles possessing the same velocities in the space C0D0A1B1as at the beginning. The change of kinetic energy is therefore the difference of the kinetic energies of A0B0C0D0and A1B1C1D1.Let fig. 119 represent the cross section A0B0, and let ω be a small element of its area at a point where the velocity is v. Let Ω0be the whole area of the cross section and u0the mean velocity for the whole cross section. From the definition of mean velocity we haveu0= Σ ωv / Ω0.Let v = u0+ w, where w is the difference between the velocity at the small element ω and the mean velocity. For the whole cross section, Σωw = 0.The mass of fluid passing through the element of section ω, in θ seconds, is (G/g) ωvθ, and its kinetic energy is (G/2g) ωv3θ. For the whole section, the kinetic energy of the mass A0B0C0D0passing in θ seconds is(Gθ / 2g) Σωv3= (Gθ/2g) Σω (u03+ 3u02w + 3u02+ w3),= (Gθ / 2g) {u03Ω + Σωw2(3u0+ w)}.The factor 3u0+ w is equal to 2u0+ v, a quantity necessarily positive. Consequently Σωv3> Ω0u03, and consequently the kinetic energy of A0B0C0D0is greater than(Gθ / 2g) Ω0u03or (Gθ) / 2g) Qu02,which would be its value if all the particles passing the section had the same velocity u0. Let the kinetic energy be taken atα (Gθ / 2g) Ω0u03= α (Gθ / 2g) Qu02,where α is a corrective factor, the value of which was estimated by J. B. C. J. Bélanger at 1.1.6Its precise value is not of great importance.In a similar way we should obtain for the kinetic energy of A1B1C1D1the expressionα (Gθ / 2g) Ω1u13= α (Gθ / 2g) Qu12,where Ω1, u1are the section and mean velocity at A1B1, and where a may be taken to have the same value as before without any important error.Hence the change of kinetic energy in the whole mass A0B0A1B1in θ seconds isα (Gθ / 2g) Q (u12− u02).(1)Motive Work of the Weight and Pressures.—Consider a small filament a0a1which comes in θ seconds to c0c1. The work done by gravity during that movement is the same as if the portion a0c0were carried to a1c1. Let dQ θ be the volume of a0c0or a1c1, and y0, y1the depths of a0, a1from the surface of the stream. Then the volumedQ θ or G dQ θ pounds falls through a vertical height z + y1− y0, and the work done by gravity isG dQ θ (z + y1− y0).Putting pafor atmospheric pressure, the whole pressure per unit of area at a0is Gy0+ pa, and that at a1is −(Gy1+ pa). The work of these pressures isG (y0+ pa/G − y1− pa/G) dQ θ = G (y0− y1) dQ θ.Adding this to the work of gravity, the whole work is GzdQθ; or, for the whole cross section,GzQθ.(2)Work expended in Overcoming the Friction of the Stream Bed.—Let A′B′, A″B″ be two cross sections at distances s and s + ds from A0B0. Between these sections the velocity may be treated as uniform, because by hypothesis the changes of velocity from section to section are gradual. Hence, to this short length of stream the equation for uniform motion is applicable. But in that case the work in overcoming the friction of the stream bed between A′B′ and A″B″ isGQθζ (u2/ 2g) (χ / Ω) ds,where u, χ, Ω are the mean velocity, wetted perimeter, and section at A′B′. Hence the whole work lost in friction from A0B0to A1B1will beGQθ∫10ζ (u2/ 2g) (χ / Ω) ds.(3)Equating the work given in (2) and (3) to the change of kinetic energy given in (1),α (GQθ / 2g) (u12− u02) = GQzθ − GQθ∫10ζ (u2/ 2g) (χ / Ω) ds;∴ z = α (u12− u02) / 2g +∫10ζ (u2/ 2g) (χ / Ω) ds.Fig. 120.§ 116.Fundamental Differential Equation of Steady Varied Motion.—Suppose the equation just found to be applied to an indefinitely short length ds of the stream, limited by the end sections ab, a1b1, taken for simplicity normal to the stream bed (fig. 120). For that short length of stream the fall of surface level, or difference of level of a and a1, may be written dz. Also, if we write u for u0, and u + du for u1, the term (u02− u12)/2g becomes udu/g. Hence the equation applicable to an indefinitely short length of the stream isdz = u du/g + (χ/Ω) ζ (u2/2g) ds.(1)From this equation some general conclusions may be arrived at as to the form of the longitudinal section of the stream, but, as the investigation is somewhat complicated, it is convenient to simplify it by restricting the conditions of the problem.Modification of the Formula for the Restricted Case of a Stream flowing in a Prismatic Stream Bed of Constant Slope.—Let i be the constant slope of the bed. Draw ad parallel to the bed, and ac horizontal. Then dz is sensibly equal to a′c. The depths of the stream, h and h + dh, are sensibly equal to ab and a′b′, and therefore dh = a′d. Also cd is the fall of the bed in the distance ds, and is equal to ids. Hencedz = a′c = cd − a′d = i ds − dh.(2)Since the motion is steady—Q = Ωu = constant.Differentiating,Ω du + u dΩ = 0;∴ du = −u dΩ/Ω.Let x be the width of the stream, then dΩ = xdh very nearly. Inserting this value,du = −(ux / Ω) dh.(3)Putting the values of du and dz found in (2) and (3) in equation (1),i ds − dh = −(u2x / gΩ) dh + (χ / Ω) ζ (u2/ 2g) ds.dh/ds = {i − (χ/Ω) ζ (u2/2g)} / {1 − (u2/g) (x/Ω)}.(4)Further Restriction to the Case of a Stream of Rectangular Section and of Indefinite Width.—The equation might be discussed in the form just given, but it becomes a little simpler if restricted in the way just stated. For, if the stream is rectangular, χh = Ω, and if χ is large compared with h, Ω/χ = xh/x = h nearly. Then equation (4) becomesdh/ds = i (1 − ζu2/ 2gih) / (1 − u2/gh).(5)§ 117.General Indications as to the Form of Water Surface furnished by Equation(5).—Let A0A1(fig. 121) be the water surface, B0B1the bed in a longitudinal section of the stream, and ab any section at a distance s from B0, the depth ab being h. Suppose B0B1, B0A0taken as rectangular coordinate axes, then dh/ds is the trigonometric tangent of the angle which the surface of the stream at a makes with the axis B0B1. This tangent dh/ds will be positive, if the stream is increasing in depth in the direction B0B1; negative, if the stream is diminishing in depth from B0towards B1. If dh/ds = 0, the surface of the stream is parallel to the bed, as in cases of uniform motion. But from equation (4)dh/ds = 0, if i − (χ/Ω) ζ (u2/2g) = 0;∴ ζ (u2/2g) = (Ω/χ) i = mi,which is the well-known general equation for uniform motion, based on the same assumptions as the equation for varied steady motion now being considered. The case of uniform motion is therefore a limiting case between two different kinds of varied motion.Fig. 121.Consider the possible changes of value of the fraction(1 − ζu2/ 2gih) / (1 − u2/ gh).As h tends towards the limit 0, and consequently u is large, the numerator tends to the limit −∞. On the other hand if h = ∞, in which case u is small, the numerator becomes equal to 1. For a value H of h given by the equation1 − ζu2/ 2giH = 0,H = ζu2/ 2gi,we fall upon the case of uniform motion. The results just stated may be tabulated thus:—For h = 0, H, > H, ∞,the numerator has the value −∞, 0, > 0, 1.Next consider the denominator. If h becomes very small, in which case u must be very large, the denominator tends to the limit −∞. As h becomes very large and u consequently very small, the denominator tends to the limit 1. For h = u2/g, or u = √ (gh), the denominator becomes zero. Hence, tabulating these results as before:—For h = 0, u2/g, > u2/g, ∞,the denominator becomes −∞, 0, > 0, 1.Fig. 122.§ 118.Case1.—Suppose h > u2/g, and also h > H, or the depth greater than that corresponding to uniform motion. In this case dh/ds is positive, and the stream increases in depth in the direction of flow. In fig. 122 let B0B1be the bed, C0C1a line parallel to the bed and at a height above it equal to H. By hypothesis, the surface A0A1of the stream is above C0C1, and it has just been shown that the depth of the stream increases from B0towards B1. But going up stream h approaches more and more nearly the value H, and therefore dh/ds approaches the limit 0, or the surface of the stream is asymptotic to C0C1. Going down stream h increases and u diminishes, the numerator and denominator of the fraction (1 − ζu2/2gih) / (1 − u2/gh) both tend towards the limit 1, and dh/ds to the limit i. That is, the surface of the stream tends to become asymptotic to a horizontal line D0D1.The form of water surface here discussed is produced when the flow of a stream originally uniform is altered by the construction of a weir. The raising of the water surface above the level C0C1is termed the backwater due to the weir.Fig. 123.Fig. 124.§ 119.Case2.—Suppose h > u2/g, and also h < H. Then dh/ds isnegative, and the stream is diminishing in depth in the direction of flow. In fig. 123 let B0B1be the stream bed as before; C0C1a line drawn parallel to B0B1at a height above it equal to H. By hypothesis the surface A0A1of the stream is below C0C1, and the depth has just been shown to diminish from B0towards B1. Going up stream h approaches the limit H, and dh/ds tends to the limit zero. That is, up stream A0A1is asymptotic to C0C1. Going down stream h diminishes and u increases; the inequality h > u2/g diminishes; the denominator of the fraction (1 − ζu2/2gih) / (1 − u2/gh) tends to the limit zero, and consequently dh/ds tends to ∞. That is, down stream A0A1tends to a direction perpendicular to the bed. Before, however, this limit was reached the assumptions on which the general equation is based would cease to be even approximately true, and the equation would cease to be applicable. The filaments would have a relative motion, which would make the influence of internal friction in the fluid too important to be neglected. A stream surface of this form may be produced if there is an abrupt fall in the bed of the stream (fig. 124).On the Ganges canal, as originally constructed, there were abrupt falls precisely of this kind, and it appears that the lowering of the water surface and increase of velocity which such falls occasion, for a distance of some miles up stream, was not foreseen. The result was that, the velocity above the falls being greater than was intended, the bed was scoured and considerable damage was done to the works. “When the canal was first opened the water was allowed to pass freely over the crests of the overfalls, which were laid on the level of the bed of the earthen channel; erosion of bed and sides for some miles up rapidly followed, and it soon became apparent that means must be adopted for raising the surface of the stream at those points (that is, the crests of the falls). Planks were accordingly fixed in the grooves above the bridge arches, or temporary weirs were formed over which the water was allowed to fall; in some cases the surface of the water was thus raised above its normal height, causing a backwater in the channel above” (Crofton’sReport on the Ganges Canal, p. 14). Fig. 125 represents in an exaggerated form what probably occurred, the diagram being intended to represent some miles’ length of the canal bed above the fall. AA parallel to the canal bed is the level corresponding to uniform motion with the intended velocity of the canal. In consequence of the presence of the ogee fall, however, the water surface would take some such form as BB, corresponding to Case 2 above, and the velocity would be greater than the intended velocity, nearly in the inverse ratio of the actual to the intended depth. By constructing a weir on the crest of the fall, as shown by dotted lines, a new water surface CC corresponding to Case 1 would be produced, and by suitably choosing the height of the weir this might be made to agree approximately with the intended level AA.Fig. 125.§ 120.Case3.—Suppose a stream flowing uniformly with a depth h < u2/g. For a stream in uniform motion ζu2/2g = mi, or if the stream is of indefinitely great width, so that m = H, then ζu2/2g = iH, and H = ζu2/2gi. Consequently the condition stated above involves thatζu2/ 2gi < u2/ g, or that i > ζ/2.If such a stream is interfered with by the construction of a weir which raises its level, so that its depth at the weir becomes h1> u2/g, then for a portion of the stream the depth h will satisfy the conditions h < u2/g and h > H, which are not the same as those assumed in the two previous cases. At some point of the stream above the weir the depth h becomes equal to u2/g, and at that point dh/ds becomes infinite, or the surface of the stream is normal to the bed. It is obvious that at that point the influence of internal friction will be too great to be neglected, and the general equation will cease to represent the true conditions of the motion of the water. It is known that, in cases such as this, there occurs an abrupt rise of the free surface of the stream, or a standing wave is formed, the conditions of motion in which will be examined presently.It appears that the condition necessary to give rise to a standing wave is that i > ζ/2. Now ζ depends for different channels on the roughness of the channel and its hydraulic mean depth. Bazin calculated the values of ζ for channels of different degrees of roughness and different depths given in the following table, and the corresponding minimum values of i for which the exceptional case of the production of a standing wave may occur.Nature of Bed of Stream.Slope belowwhich a StandingWave isimpossible infeet peer foot.Standing Wave Formed.Slope in feetper foot.Least Depthin feet.Very smooth cemented surface0.001470.0020.2620.003.0980.004.065Ashlar or brickwork0.001860.003.3940.004.1970.006.098Rubble masonry0.002350.0041.1810.006.5250.010.262Earth0.002750.0063.4780.0101.5420.015.919Standing Waves§ 121. The formation of a standing wave was first observed by Bidone. Into a small rectangular masonry channel, having a slope of 0.023 ft. per foot, he admitted water till it flowed uniformly with a depth of 0.2 ft. He then placed a plank across the stream which raised the level just above the obstruction to 0.95 ft. He found that the stream above the obstruction was sensibly unaffected up to a point 15 ft. from it. At that point the depth suddenly increased from 0.2 ft. to 0.56 ft. The velocity of the stream in the part unaffected by the obstruction was 5.54 ft. per second. Above the point where the abrupt change of depth occurred u2= 5.542= 30.7, and gh = 32.2 × 0.2 = 6.44; hence u2was > gh. Just below the abrupt change of depth u = 5.54 × 0.2/0.56 = 1.97; u2= 3.88; and gh = 32.2 × 0.56 = 18.03; hence at this point u2< gh. Between these two points, therefore, u2= gh; and the condition for the production of a standing wave occurred.Fig. 126.The change of level at a standing wave may be found thus. Let fig. 126 represent the longitudinal section of a stream and ab, cd cross sections normal to the bed, which for the short distance considered may be assumed horizontal. Suppose the mass of water abcd to come to a′b′c′d′ in a short time t; and let u0, u1be the velocities at ab and cd, Ω0, Ω1the areas of the cross sections. The force causing change of momentum in the mass abcd estimated horizontally is simply the difference of the pressures on ab and cd. Putting h0, h1for the depths of the centres of gravity of ab and cd measured down from the free water surface, the force is G (h0Ω0− h1Ω1) pounds, and the impulse in t seconds is G (h0Ω0− h1Ω1) t second pounds. The horizontal change of momentum is the difference of the momenta of cdc′d′ and aba′b′; that is,(G/g) (Ω1u12− Ω0u02) t.Hence, equating impulse and change of momentum,G (h0Ω0− h1Ω1) t = (G/g) (Ω1u12− Ω0u02) t;∴ h0Ω0− h1Ω1= (Ω1u12− Ω0u02) / g.
(1)
ζv2/2g = mi;
(2)
Q = Ωv.
(3)
Problem I.—Given the transverse section of stream and discharge, to find the slope. From the dimensions of the section find Ω and m; from (1) find ζ, from (3) find v, and lastly from (2) find i.
Problem II.—Given the transverse section and slope, to find the discharge. Find v from (2), then Q from (3).
Problem III.—Given the discharge and slope, and either the breadth, depth, or general form of the section of the channel, to determine its remaining dimensions. This must generally be solved by approximations. A breadth or depth or both are chosen, and the discharge calculated. If this is greater than the given discharge, the dimensions are reduced and the discharge recalculated.
Since m lies generally between the limits m = d and m =1⁄2d, where d is the depth of the stream, and since, moreover, the velocity varies as √ (m) so that an error in the value of m leads only to a much less error in the value of the velocity calculated from it, we may proceed thus. Assume a value for m, and calculate v from it. Let v1be this first approximation to v. Then Q/v1is a first approximation to Ω, say Ω1. With this value of Ω design the section of the channel; calculate a second value for m; calculate from it a second value of v, and from that a second value for Ω. Repeat the process till the successive values of m approximately coincide.
§ 113.Problem IV. Most Economical Form of Channel for given Side Slopes.—Suppose the channel is to be trapezoidal in section (fig. 114), and that the sides are to have a given slope. Let the longitudinal slope of the stream be given, and also the mean velocity. An infinite number of channels could be found satisfying the foregoing conditions. To render the problem determinate, let it be remembered that, since for a given discharge Ω∞ √χ, other things being the same, the amount of excavation will be least for that channel which has the least wetted perimeter. Let d be the depth and b the bottom width of the channel, and let the sides slope n horizontal to 1 vertical (fig. 114), then
Ω = (b + nd) d;
χ = b + 2d √ (n2+ 1).
Both Ω and χ are to be minima. Differentiating, and equating to zero.
(db/dd + n) d + b + nd = 0,db/dd + 2 √ (n2+ 1) = 0;
eliminating db/dd,
{n − 2√ (n2+ 1)} d + b + nd = 0;b = 2 {√ (n2+ 1) − n} d.
But
Ω / χ = (b + nd) d / {b + 2d √ (n2+ 1)}.
Inserting the value of b,
m = Ω/χ = {2d √ (n2+ 1) − nd} / {4d √ (n2+ 1) − 2nd} =1⁄2d.
That is, with given side slopes, the section is least for a given discharge when the hydraulic mean depth is half the actual depth.
A simple construction gives the form of the channel which fulfils this condition, for it can be shown that when m =1⁄2d the sides of the channel are tangential to a semicircle drawn on the water line.
Since
Ω / χ =1⁄2d,
therefore
Ω =1⁄2χd.
(1)
Let ABCD be the channel (fig. 115); from E the centre of AD drop perpendiculars EF, EG, EH on the sides.
Let
AB = CD = a; BC = b; EF = EH = c; and EG = d.
Ω = area AEB + BEC + CED,= ac +1⁄2bd.
χ = 2a + b.
Putting these values in (1),
ac +1⁄2bd = (a +1⁄2b) d; and hence c = d.
That is, EF, EG, EH are all equal, hence a semicircle struck from E with radius equal to the depth of the stream will pass through F and H and be tangential to the sides of the channel.
To draw the channel, describe a semicircle on a horizontal line with radius = depth of channel. The bottom will be a horizontal tangent of that semicircle, and the sides tangents drawn at the required side slopes.
The above result may be obtained thus (fig. 116):—
χ = b + 2d / sin β.
(1)
Ω = d (b + d cot β);
Ω/d = b + d cot β;
(2)
Ω/d2= b/d + cot β.
(3)
From (1) and (2),
χ = Ω / d − d cot β + 2d / sin β.
This will be a minimum for
dχ / dd = Ω / d2+ cot β − 2 / sin β = 0,
or
Ω/d2= 2 cosec. β − cot β.
(4)
or
d = √ {Ω sin β / (2 − cos β)}.
From (3) and (4),
b/d = 2 (1 − cos β) / sin β = 2 tan1⁄2β.
Proportions of Channels of Maximum Discharge for given Area and Side Slopes. Depth of channel = d; Hydraulic mean depth =1⁄2d; Area of section =Ω.
§ 114.Form of Cross Section of Channel in which the Mean Velocity is Constant with Varying Discharge.—In designing waste channels from canals, and in some other cases, it is desirable that the mean velocity should be restricted within narrow limits with very different volumes of discharge. In channels of trapezoidal form the velocity increases and diminishes with the discharge. Hence when the discharge is large there is danger of erosion, and when it is small of silting or obstruction by weeds. A theoretical form of section for which the mean velocity would be constant can be found, and, although this is not very suitable for practical purposes, it can be more or less approximated to in actual channels.
Let fig. 117 represent the cross section of the channel. From the symmetry of the section, only half the channel need be considered. Let obac be any section suitable for the minimum flow, and let it be required to find the curve beg for the upper part of the channel so that the mean velocity shall be constant. Take o as origin of coordinates, and let de, fg be two levels of the water above ob.
Let ob = b/2; de = y, fg = y + dy, od = x, of = x + dx; eg = ds.
The condition to be satisfied is that
v = c √ (mi)
should be constant, whether the water-level is at ob, de, or fg. Consequently
m = constant = k
for all three sections, and can be found from the section obac. Hence also
y2dx2= k2ds2= k2(dx2+ dy2) and dx = k dy / √ (y2− k2).
Integrating,
x = k logε{y + √ (y2− k2)} + constant;
and, since y = b/2 when x = 0,
x = k logε[{y + √ (y2− k2)} / {1⁄2b + √ (1⁄4b2− k2) }].
Assuming values for y, the values of x can be found and the curve drawn.
The figure has been drawn for a channel the minimum section of which is a half hexagon of 4 ft. depth. Hence k = 2; b = 9.2; the rapid flattening of the side slopes is remarkable.
Steady Motion of Water in Open Channels of Varying Cross Section and Slope
§ 115. In every stream the discharge of which is constant, or may be regarded as constant for the time considered, the velocity at different places depends on the slope of the bed. Except at certain exceptional points the velocity will be greater as the slope of the bed is greater, and, as the velocity and cross section of the stream vary inversely, the section of the stream will be least where the velocity and slope are greatest. If in a stream of tolerably uniform slope an obstruction such as a weir is built, that will cause an alteration of flow similar to that of an alteration of the slope of the bed for a greater or less distance above the weir, and the originally uniform cross section of the stream will become a varied one. In such cases it is often of much practical importance to determine the longitudinal section of the stream.
The cases now considered will be those in which the changes of velocity and cross section are gradual and not abrupt, and in which the only internal work which needs to be taken into account is that due to the friction of the stream bed, as in cases of uniform motion. Further, the motion will be supposed to be steady, the mean velocity at each given cross section remaining constant, though it varies from section to section along the course of the stream.
Let fig. 118 represent a longitudinal section of the stream, A0A1being the water surface, B0B1the stream bed. Let A0B0, A1B1be cross sections normal to the direction of flow. Suppose the mass of water A0B0A1B1comes in a short time θ to C0D0C1D1, and let the work done on the mass be equated to its change of kinetic energy during that period. Let l be the length A0A1of the portion of the stream considered, and z the fall, of surface level in that distance. Let Q be the discharge of the stream per second.
Change of Kinetic Energy.—At the end of the time θ there are as many particles possessing the same velocities in the space C0D0A1B1as at the beginning. The change of kinetic energy is therefore the difference of the kinetic energies of A0B0C0D0and A1B1C1D1.
Let fig. 119 represent the cross section A0B0, and let ω be a small element of its area at a point where the velocity is v. Let Ω0be the whole area of the cross section and u0the mean velocity for the whole cross section. From the definition of mean velocity we have
u0= Σ ωv / Ω0.
Let v = u0+ w, where w is the difference between the velocity at the small element ω and the mean velocity. For the whole cross section, Σωw = 0.
The mass of fluid passing through the element of section ω, in θ seconds, is (G/g) ωvθ, and its kinetic energy is (G/2g) ωv3θ. For the whole section, the kinetic energy of the mass A0B0C0D0passing in θ seconds is
(Gθ / 2g) Σωv3= (Gθ/2g) Σω (u03+ 3u02w + 3u02+ w3),= (Gθ / 2g) {u03Ω + Σωw2(3u0+ w)}.
The factor 3u0+ w is equal to 2u0+ v, a quantity necessarily positive. Consequently Σωv3> Ω0u03, and consequently the kinetic energy of A0B0C0D0is greater than
(Gθ / 2g) Ω0u03or (Gθ) / 2g) Qu02,
which would be its value if all the particles passing the section had the same velocity u0. Let the kinetic energy be taken at
α (Gθ / 2g) Ω0u03= α (Gθ / 2g) Qu02,
where α is a corrective factor, the value of which was estimated by J. B. C. J. Bélanger at 1.1.6Its precise value is not of great importance.
In a similar way we should obtain for the kinetic energy of A1B1C1D1the expression
α (Gθ / 2g) Ω1u13= α (Gθ / 2g) Qu12,
where Ω1, u1are the section and mean velocity at A1B1, and where a may be taken to have the same value as before without any important error.
Hence the change of kinetic energy in the whole mass A0B0A1B1in θ seconds is
α (Gθ / 2g) Q (u12− u02).
(1)
Motive Work of the Weight and Pressures.—Consider a small filament a0a1which comes in θ seconds to c0c1. The work done by gravity during that movement is the same as if the portion a0c0were carried to a1c1. Let dQ θ be the volume of a0c0or a1c1, and y0, y1the depths of a0, a1from the surface of the stream. Then the volumedQ θ or G dQ θ pounds falls through a vertical height z + y1− y0, and the work done by gravity is
G dQ θ (z + y1− y0).
Putting pafor atmospheric pressure, the whole pressure per unit of area at a0is Gy0+ pa, and that at a1is −(Gy1+ pa). The work of these pressures is
G (y0+ pa/G − y1− pa/G) dQ θ = G (y0− y1) dQ θ.
Adding this to the work of gravity, the whole work is GzdQθ; or, for the whole cross section,
GzQθ.
(2)
Work expended in Overcoming the Friction of the Stream Bed.—Let A′B′, A″B″ be two cross sections at distances s and s + ds from A0B0. Between these sections the velocity may be treated as uniform, because by hypothesis the changes of velocity from section to section are gradual. Hence, to this short length of stream the equation for uniform motion is applicable. But in that case the work in overcoming the friction of the stream bed between A′B′ and A″B″ is
GQθζ (u2/ 2g) (χ / Ω) ds,
where u, χ, Ω are the mean velocity, wetted perimeter, and section at A′B′. Hence the whole work lost in friction from A0B0to A1B1will be
GQθ∫10ζ (u2/ 2g) (χ / Ω) ds.
(3)
Equating the work given in (2) and (3) to the change of kinetic energy given in (1),
α (GQθ / 2g) (u12− u02) = GQzθ − GQθ∫10ζ (u2/ 2g) (χ / Ω) ds;
∴ z = α (u12− u02) / 2g +∫10ζ (u2/ 2g) (χ / Ω) ds.
§ 116.Fundamental Differential Equation of Steady Varied Motion.—Suppose the equation just found to be applied to an indefinitely short length ds of the stream, limited by the end sections ab, a1b1, taken for simplicity normal to the stream bed (fig. 120). For that short length of stream the fall of surface level, or difference of level of a and a1, may be written dz. Also, if we write u for u0, and u + du for u1, the term (u02− u12)/2g becomes udu/g. Hence the equation applicable to an indefinitely short length of the stream is
dz = u du/g + (χ/Ω) ζ (u2/2g) ds.
(1)
From this equation some general conclusions may be arrived at as to the form of the longitudinal section of the stream, but, as the investigation is somewhat complicated, it is convenient to simplify it by restricting the conditions of the problem.
Modification of the Formula for the Restricted Case of a Stream flowing in a Prismatic Stream Bed of Constant Slope.—Let i be the constant slope of the bed. Draw ad parallel to the bed, and ac horizontal. Then dz is sensibly equal to a′c. The depths of the stream, h and h + dh, are sensibly equal to ab and a′b′, and therefore dh = a′d. Also cd is the fall of the bed in the distance ds, and is equal to ids. Hence
dz = a′c = cd − a′d = i ds − dh.
(2)
Since the motion is steady—
Q = Ωu = constant.
Differentiating,
Ω du + u dΩ = 0;
∴ du = −u dΩ/Ω.
Let x be the width of the stream, then dΩ = xdh very nearly. Inserting this value,
du = −(ux / Ω) dh.
(3)
Putting the values of du and dz found in (2) and (3) in equation (1),
i ds − dh = −(u2x / gΩ) dh + (χ / Ω) ζ (u2/ 2g) ds.
dh/ds = {i − (χ/Ω) ζ (u2/2g)} / {1 − (u2/g) (x/Ω)}.
(4)
Further Restriction to the Case of a Stream of Rectangular Section and of Indefinite Width.—The equation might be discussed in the form just given, but it becomes a little simpler if restricted in the way just stated. For, if the stream is rectangular, χh = Ω, and if χ is large compared with h, Ω/χ = xh/x = h nearly. Then equation (4) becomes
dh/ds = i (1 − ζu2/ 2gih) / (1 − u2/gh).
(5)
§ 117.General Indications as to the Form of Water Surface furnished by Equation(5).—Let A0A1(fig. 121) be the water surface, B0B1the bed in a longitudinal section of the stream, and ab any section at a distance s from B0, the depth ab being h. Suppose B0B1, B0A0taken as rectangular coordinate axes, then dh/ds is the trigonometric tangent of the angle which the surface of the stream at a makes with the axis B0B1. This tangent dh/ds will be positive, if the stream is increasing in depth in the direction B0B1; negative, if the stream is diminishing in depth from B0towards B1. If dh/ds = 0, the surface of the stream is parallel to the bed, as in cases of uniform motion. But from equation (4)
dh/ds = 0, if i − (χ/Ω) ζ (u2/2g) = 0;
∴ ζ (u2/2g) = (Ω/χ) i = mi,
which is the well-known general equation for uniform motion, based on the same assumptions as the equation for varied steady motion now being considered. The case of uniform motion is therefore a limiting case between two different kinds of varied motion.
Consider the possible changes of value of the fraction
(1 − ζu2/ 2gih) / (1 − u2/ gh).
As h tends towards the limit 0, and consequently u is large, the numerator tends to the limit −∞. On the other hand if h = ∞, in which case u is small, the numerator becomes equal to 1. For a value H of h given by the equation
1 − ζu2/ 2giH = 0,H = ζu2/ 2gi,
we fall upon the case of uniform motion. The results just stated may be tabulated thus:—
For h = 0, H, > H, ∞,
the numerator has the value −∞, 0, > 0, 1.
Next consider the denominator. If h becomes very small, in which case u must be very large, the denominator tends to the limit −∞. As h becomes very large and u consequently very small, the denominator tends to the limit 1. For h = u2/g, or u = √ (gh), the denominator becomes zero. Hence, tabulating these results as before:—
For h = 0, u2/g, > u2/g, ∞,
the denominator becomes −∞, 0, > 0, 1.
§ 118.Case1.—Suppose h > u2/g, and also h > H, or the depth greater than that corresponding to uniform motion. In this case dh/ds is positive, and the stream increases in depth in the direction of flow. In fig. 122 let B0B1be the bed, C0C1a line parallel to the bed and at a height above it equal to H. By hypothesis, the surface A0A1of the stream is above C0C1, and it has just been shown that the depth of the stream increases from B0towards B1. But going up stream h approaches more and more nearly the value H, and therefore dh/ds approaches the limit 0, or the surface of the stream is asymptotic to C0C1. Going down stream h increases and u diminishes, the numerator and denominator of the fraction (1 − ζu2/2gih) / (1 − u2/gh) both tend towards the limit 1, and dh/ds to the limit i. That is, the surface of the stream tends to become asymptotic to a horizontal line D0D1.
The form of water surface here discussed is produced when the flow of a stream originally uniform is altered by the construction of a weir. The raising of the water surface above the level C0C1is termed the backwater due to the weir.
§ 119.Case2.—Suppose h > u2/g, and also h < H. Then dh/ds isnegative, and the stream is diminishing in depth in the direction of flow. In fig. 123 let B0B1be the stream bed as before; C0C1a line drawn parallel to B0B1at a height above it equal to H. By hypothesis the surface A0A1of the stream is below C0C1, and the depth has just been shown to diminish from B0towards B1. Going up stream h approaches the limit H, and dh/ds tends to the limit zero. That is, up stream A0A1is asymptotic to C0C1. Going down stream h diminishes and u increases; the inequality h > u2/g diminishes; the denominator of the fraction (1 − ζu2/2gih) / (1 − u2/gh) tends to the limit zero, and consequently dh/ds tends to ∞. That is, down stream A0A1tends to a direction perpendicular to the bed. Before, however, this limit was reached the assumptions on which the general equation is based would cease to be even approximately true, and the equation would cease to be applicable. The filaments would have a relative motion, which would make the influence of internal friction in the fluid too important to be neglected. A stream surface of this form may be produced if there is an abrupt fall in the bed of the stream (fig. 124).
On the Ganges canal, as originally constructed, there were abrupt falls precisely of this kind, and it appears that the lowering of the water surface and increase of velocity which such falls occasion, for a distance of some miles up stream, was not foreseen. The result was that, the velocity above the falls being greater than was intended, the bed was scoured and considerable damage was done to the works. “When the canal was first opened the water was allowed to pass freely over the crests of the overfalls, which were laid on the level of the bed of the earthen channel; erosion of bed and sides for some miles up rapidly followed, and it soon became apparent that means must be adopted for raising the surface of the stream at those points (that is, the crests of the falls). Planks were accordingly fixed in the grooves above the bridge arches, or temporary weirs were formed over which the water was allowed to fall; in some cases the surface of the water was thus raised above its normal height, causing a backwater in the channel above” (Crofton’sReport on the Ganges Canal, p. 14). Fig. 125 represents in an exaggerated form what probably occurred, the diagram being intended to represent some miles’ length of the canal bed above the fall. AA parallel to the canal bed is the level corresponding to uniform motion with the intended velocity of the canal. In consequence of the presence of the ogee fall, however, the water surface would take some such form as BB, corresponding to Case 2 above, and the velocity would be greater than the intended velocity, nearly in the inverse ratio of the actual to the intended depth. By constructing a weir on the crest of the fall, as shown by dotted lines, a new water surface CC corresponding to Case 1 would be produced, and by suitably choosing the height of the weir this might be made to agree approximately with the intended level AA.
§ 120.Case3.—Suppose a stream flowing uniformly with a depth h < u2/g. For a stream in uniform motion ζu2/2g = mi, or if the stream is of indefinitely great width, so that m = H, then ζu2/2g = iH, and H = ζu2/2gi. Consequently the condition stated above involves that
ζu2/ 2gi < u2/ g, or that i > ζ/2.
If such a stream is interfered with by the construction of a weir which raises its level, so that its depth at the weir becomes h1> u2/g, then for a portion of the stream the depth h will satisfy the conditions h < u2/g and h > H, which are not the same as those assumed in the two previous cases. At some point of the stream above the weir the depth h becomes equal to u2/g, and at that point dh/ds becomes infinite, or the surface of the stream is normal to the bed. It is obvious that at that point the influence of internal friction will be too great to be neglected, and the general equation will cease to represent the true conditions of the motion of the water. It is known that, in cases such as this, there occurs an abrupt rise of the free surface of the stream, or a standing wave is formed, the conditions of motion in which will be examined presently.
It appears that the condition necessary to give rise to a standing wave is that i > ζ/2. Now ζ depends for different channels on the roughness of the channel and its hydraulic mean depth. Bazin calculated the values of ζ for channels of different degrees of roughness and different depths given in the following table, and the corresponding minimum values of i for which the exceptional case of the production of a standing wave may occur.
Standing Waves
§ 121. The formation of a standing wave was first observed by Bidone. Into a small rectangular masonry channel, having a slope of 0.023 ft. per foot, he admitted water till it flowed uniformly with a depth of 0.2 ft. He then placed a plank across the stream which raised the level just above the obstruction to 0.95 ft. He found that the stream above the obstruction was sensibly unaffected up to a point 15 ft. from it. At that point the depth suddenly increased from 0.2 ft. to 0.56 ft. The velocity of the stream in the part unaffected by the obstruction was 5.54 ft. per second. Above the point where the abrupt change of depth occurred u2= 5.542= 30.7, and gh = 32.2 × 0.2 = 6.44; hence u2was > gh. Just below the abrupt change of depth u = 5.54 × 0.2/0.56 = 1.97; u2= 3.88; and gh = 32.2 × 0.56 = 18.03; hence at this point u2< gh. Between these two points, therefore, u2= gh; and the condition for the production of a standing wave occurred.
The change of level at a standing wave may be found thus. Let fig. 126 represent the longitudinal section of a stream and ab, cd cross sections normal to the bed, which for the short distance considered may be assumed horizontal. Suppose the mass of water abcd to come to a′b′c′d′ in a short time t; and let u0, u1be the velocities at ab and cd, Ω0, Ω1the areas of the cross sections. The force causing change of momentum in the mass abcd estimated horizontally is simply the difference of the pressures on ab and cd. Putting h0, h1for the depths of the centres of gravity of ab and cd measured down from the free water surface, the force is G (h0Ω0− h1Ω1) pounds, and the impulse in t seconds is G (h0Ω0− h1Ω1) t second pounds. The horizontal change of momentum is the difference of the momenta of cdc′d′ and aba′b′; that is,
(G/g) (Ω1u12− Ω0u02) t.
Hence, equating impulse and change of momentum,
G (h0Ω0− h1Ω1) t = (G/g) (Ω1u12− Ω0u02) t;
∴ h0Ω0− h1Ω1= (Ω1u12− Ω0u02) / g.