The normal pressures on the surface of the mass (excluding the ends A, B) are at each point normal to the direction of motion, and do no work. Hence the only external forces to be reckoned are gravity and the pressures on the ends of the stream.The work of gravity when AB falls to A′B′ is the same as that of transferring AA′ to BB′; that is, GQt (z − z1). The work of the pressures on the ends, reckoning that at B negative, because it is opposite to the direction of motion, is (pω × vt) − (p1ω1× v1t) = Qt(p − p1). The change of kinetic energy in the time t is the difference of the kinetic energy originally possessed by AA′ and that finally acquired by BB′, for in the intermediate part A′B there is no change of kinetic energy, in consequence of the steadiness of the motion. But the mass of AA′ and BB′ is GQt/g, and the change of kinetic energy is therefore (GQt/g) (v12/2 − v2/2). Equating this to the work done on the mass AB,GQt (z − z1) + Qt (p − p1) = (GQt/g) (v12/2 − v2/2).Dividing by GQt and rearranging the terms,v2/2g + p/G + z = v12/2g + p1/G + z1;(1)or, as A and B are any two points,v2/2g + p/G + z = constant = H.(2)Now v2/2g is the head due to the velocity v, p/G is the head equivalent to the pressure, and z is the elevation above the datum (see § 16). Hence the terms on the left are the total head due to velocity, pressure, and elevation at a given cross section of the filament, z is easily seen to be the work in foot-pounds which would be done by 1 ℔ of fluid falling to the datum line, and similarly p/G and v2/2g are the quantities of work which would be done by 1 ℔ of fluid due to the pressure p and velocity v. The expression on the left of the equation is, therefore, the total energy of the stream at the section considered, per ℔ of fluid, estimated with reference to thedatum line XX. Hence we see that in stream line motion, under the restrictions named above, the total energy per ℔ of fluid is uniformly distributed along the stream line. If the free surface of the fluid OO is taken as the datum, and −h, −h1are the depths of A and B measured down from the free surface, the equation takes the formv2/2g + p/G − h = v12/2g + p1/G − h1;(3)or generallyv2/2g + p/G − h = constant.(3a)Fig. 26.§ 30.Second Form of the Theorem of Bernoulli.—Suppose at the two sections A, B (fig. 26) of an elementary stream small vertical pipes are introduced, which may be termed pressure columns (§ 8), having their lower ends accurately parallel to the direction of flow. In such tubes the water will rise to heights corresponding to the pressures at A and B. Hence b = p/G, and b′ = p1/G. Consequently the tops of the pressure columns A′ and B′ will be at total heights b + c = p/G + z and b′ + c′ = p1/G + z1above the datum line XX. The difference of level of the pressure column tops, or the fall of free surface level between A and B, is thereforeξ = (p − p1) / G + (z − z1);and this by equation (1), § 29 is (v12− v2)/2g. That is, the fall of free, surface level between two sections is equal to the difference of the heights due to the velocities at the sections. The line A′B′ is sometimes called the line of hydraulic gradient, though this term is also used in cases where friction needs to be taken into account. It is the line the height of which above datum is the sum of the elevation and pressure head at that point, and it falls below a horizontal line A″B″ drawn at H ft. above XX by the quantities a = v2/2g and a′ = v12/2g, when friction is absent.§ 31.Illustrations of the Theorem of Bernoulli.In a lecture to the mechanical section of the British Association in 1875, W. Froude gave some experimental illustrations of the principle of Bernoulli. He remarked that it was a common but erroneous impression that a fluid exercises in a contracting pipe A (fig. 27) an excess of pressure against the entire converging surface which it meets, and that, conversely, as it enters an enlargement B, a relief of pressure is experienced by the entire diverging surface of the pipe. Further it is commonly assumed that when passing through a contraction C, there is in the narrow neck an excess of pressure due to the squeezing together of the liquid at that point. These impressions are in no respect correct; the pressure is smaller as the section of the pipe is smaller and conversely.Fig. 27.Fig. 28 shows a pipe so formed that a contraction is followed by an enlargement, and fig. 29 one in which an enlargement is followed by a contraction. The vertical pressure columns show the decrease of pressure at the contraction and increase of pressure at the enlargement. The line abc in both figures shows the variation of free surface level, supposing the pipe frictionless. In actual pipes, however, work is expended in friction against the pipe; the total head diminishes in proceeding along the pipe, and the free surface level is a line such as ab1c1, falling below abc.Froude further pointed out that, if a pipe contracts and enlarges again to the same size, the resultant pressure on the converging part exactly balances the resultant pressure on the diverging part so that there is no tendency to move the pipe bodily when water flows through it. Thus the conical part AB (fig. 30) presents the same projected surface as HI, and the pressures parallel to the axis of the pipe, normal to these projected surfaces, balance each other. Similarly the pressures on BC, CD balance those on GH, EG. In the same way, in any combination of enlargements and contractions, a balance of pressures, due to the flow of liquid parallel to the axis of the pipe, will be found, provided the sectional area and direction of the ends are the same.Fig. 28.Fig. 29.The following experiment is interesting. Two cisterns provided with converging pipes were placed so that the jet from one was exactly opposite the entrance to the other. The cisterns being filled very nearly to the same level, the jet from the left-hand cistern A entered the right-hand cistern B (fig. 31), shooting across the free space between them without any waste, except that due to indirectness of aim and want of exact correspondence in the form of the orifices. In the actual experiment there was 18 in. of head in the right and 201⁄2in. of head in the left-hand cistern, so that about 21⁄2in. were wasted in friction. It will be seen that in the open space between the orifices there was no pressure, except the atmospheric pressure acting uniformly throughout the system.Fig. 30.Fig. 31.§ 32.Venturi Meter.—An ingenious application of the variation of pressure and velocity in a converging and diverging pipe has been made by Clemens Herschel in the construction of what he terms a Venturi Meter for measuring the flow in water mains. Suppose that, as in fig. 32, a contraction is made in a water main, the change of section being gradual to avoid the production of eddies. The ratio ρof the cross sections at A and B, that is at inlet and throat, is in actual meters 5 to 1 to 20 to 1, and is very carefully determined by the maker of the meter. Then, if v and u are the velocities at A and B, u = ρv. Let pressure pipes be introduced at A, B and C, and let H1, H, H2be the pressure heads at those points. Since the velocity at B is greater than at A the pressure will be less. Neglecting frictionH1+ v2/2g = H + u2/2g,H1− H = (u2− v2) / 2g = (ρ2− 1) v22g.Let h = H1− H be termed the Venturi head, thenu = √ { ρ2.2gh / (ρ2− 1) },from which the velocity through the throat and the discharge of the main can be calculated if the areas at A and B are known and h observed. Thus if the diameters at A and B are 4 and 12 in., the areas are 12.57 and 113.1 sq. in., and ρ = 9,u = √ 81/80 √ (2gh) = 1.007 √ (2gh).If the observed Venturi head is 12 ft.,u = 28 ft. per sec.,and the discharge of the main is28 × 12.57 = 351 cub. ft. per sec.Fig. 32.Fig. 33.Hence by a simple observation of pressure difference, the flow in the main at any moment can be determined. Notice that the pressure height at C will be the same as at A except for a small loss hfdue to friction and eddying between A and B. To get the pressure at the throat very exactly Herschel surrounds it by an annular passage communicating with the throat by several small holes, sometimes formed in vulcanite to prevent corrosion. Though constructed to prevent eddying as much as possible there is some eddy loss. The main effect of this is to cause a loss of head between A and C which may vary from a fraction of a foot to perhaps 5 ft. at the highest velocities at which a meter can be used. The eddying also affects a little the Venturi head h. Consequently an experimental coefficient must be determined for each meter by tank measurement. The range of this coefficient is, however, surprisingly small. If to allow for friction, u = k √ {ρ2/(ρ2− 1)} √(2gh), then Herschel found values of k from 0.97 to 1.0 for throat velocities varying from 8 to 28 ft. per sec. The meter is extremely convenient. At Staines reservoirs there are two meters of this type on mains 94 in. in diameter. Herschel contrived a recording arrangement which records the variation of flow from hour to hour and also the total flow in any given time. In Great Britain the meter is constructed by G. Kent, who has made improvements in the recording arrangement.In the Deacon Waste Water Meter (fig. 33) a different principle is used. A disk D, partly counter-balanced by a weight, is suspended in the water flowing through the main in a conical chamber. The unbalanced weight of the disk is supported by the impact of the water. If the discharge of the main increases the disk rises, but as it rises its position in the chamber is such that in consequence of the larger area the velocity is less. It finds, therefore, a new position of equilibrium. A pencil P records on a drum moved by clockwork the position of the disk, and from this the variation of flow is inferred.§ 33.Pressure, Velocity and Energy in Different Stream Lines.—The equation of Bernoulli gives the variation of pressure and velocity from point to point along a stream line, and shows that the total energy of the flow across any two sections is the same. Two other directions may be defined, one normal to the stream line and in the plane containing its radius of curvature at any point, the other normal to the stream line and the radius of curvature. For the problems most practically useful it will be sufficient to consider the stream lines as parallel to a vertical or horizontal plane. If the motion is in a vertical plane, the action of gravity must be taken into the reckoning; if the motion is in a horizontal plane, the terms expressing variation of elevation of the filament will disappear.3Fig. 34.Let AB, CD (fig. 34) be two consecutive stream lines, at present assumed to be in a vertical plane, and PQ a normal to these lines making an angle φ with the vertical. Let P, Q be two particles moving along these lines at a distance PQ = ds, and let z be the height of Q above the horizontal plane with reference to which the energy is measured, v its velocity, and p its pressure. Then, if H is the total energy at Q per unit of weight of fluid,H = z + p/G + v2/2g.Differentiating, we getdH = dz + dp/G + v dv/g,(1)for the increment of energy between Q and P. Butdz = PQ cos φ = ds cos φ;∴ dH = dp/G + v dv/g + ds cos φ,(1a)where the last term disappears if the motion is in a horizontal plane.Now imagine a small cylinder of section ω described round PQ as an axis. This will be in equilibrium under the action of its centrifugal force, its weight and the pressure on its ends. But its volume is ωds and its weight Gω ds. Hence, taking the components of the forces parallel to PQ—ω dp = Gv2ω ds/gρ − Gω cos φ ds,where ρ is the radius of curvature of the stream line at Q. Consequently, introducing these values in (1),dH = v2ds/gρ + v dv/g = (v/g) (v/ρ + dv/ds) ds.(2)Currents§ 34.Rectilinear Current.—Suppose the motion is in parallel straight stream lines (fig. 35) in a vertical plane. Then ρ is infinite, and from eq. (2), § 33,dH = v dv/g.Comparing this with (1) we see thatdz + dp/G = 0;∴ z + p/G = constant;(3)Fig. 35.or the pressure varies hydrostatically as in a fluid at rest. For two stream lines in a horizontal plane, z is constant, and therefore p is constant.Radiating Current.—Suppose water flowing radially between horizontal parallel planes, at a distance apart = δ. Conceive two cylindrical sections of the current at radii r1and r2, where the velocities are v1and v2, and the pressures p1and p2. Since the flow across each cylindrical section of the current is the same,Q = 2πr1δv1= 2πr2δv2r1v1= r2v2r1/r2= v2/v1.(4)The velocity would be infinite at radius 0, if the current could be conceived to extend to the axis. Now, if the motion is steady,H = p1/G + v12/2g = p2/G + v22/2g;= p2/G + r12+ v12/ r222g;(p2− p1) / G = v12(1 − r12/r22) / 2g;(5)p2/G = H − r12v12/ r222g.(6)Hence the pressure increases from the interior outwards, in a way indicated by the pressure columns in fig. 36, the curve through the free surfaces of the pressure columns being, in a radial section, the quasi-hyperbola of the form xy2= c3. This curve is asymptotic to a horizontal line, H ft. above the line from which the pressures are measured, and to the axis of the current.Fig. 36.Free Circular Vortex.—A free circular vortex is a revolving mass of water, in which the stream lines are concentric circles, and in which the total head for each stream line is the same. Hence, if by any slow radial motion portions of the water strayed from one stream line to another, they would take freely the velocities proper to their new positions under the action of the existing fluid pressures only.For such a current, the motion being horizontal, we have for all the circular elementary streamsH = p/G + v2/2g = constant;∴ dH = dp/G + v dv/g = 0.(7)Consider two stream lines at radii r and r + dr (fig. 36). Then in (2), § 33, ρ = r and ds = dr,v2dr/gr + v dv/g = 0,dv/v = −dr/r,v ∞ 1/r,(8)precisely as in a radiating current; and hence the distribution of pressure is the same, and formulae 5 and 6 are applicable to this case.Free Spiral Vortex.—As in a radiating and circular current the equations of motion are the same, they will also apply to a vortex in which the motion is compounded of these motions in any proportions, provided the radial component of the motion varies inversely as the radius as in a radial current, and the tangential component varies inversely as the radius as in a free vortex. Then the whole velocity at any point will be inversely proportional to the radius of the point, and the fluid will describe stream lines having a constant inclination to the radius drawn to the axis of the current. That is, the stream lines will be logarithmic spirals. When water is delivered from the circumference of a centrifugal pump or turbine into a chamber, it forms a free vortex of this kind. The water flows spirally outwards, its velocity diminishing and its pressure increasing according to the law stated above, and the head along each spiral stream line is constant.§ 35.Forced Vortex.—If the law of motion in a rotating current is different from that in a free vortex, some force must be applied to cause the variation of velocity. The simplest case is that of a rotating current in which all the particles have equal angular velocity, as for instance when they are driven round by radiating paddles revolving uniformly. Then in equation (2), § 33, considering two circular stream lines of radii r and r + dr (fig. 37), we have ρ = r, ds = dr. If the angular velocity is α, then v = αr and dv = αdr. HencedH = α2r dr/g + α2r dr/g = 2α2r dr/g.Comparing this with (1), § 33, and putting dz = 0, because the motion is horizontal,dp/G + α2r dr/g = 2α2r dr/g,dp/G = α2r dr/g,p/G = α2/2g + constant.(9)Let p1, r1, v1be the pressure, radius and velocity of one cylindrical section, p2, r2, v2those of another; thenp1/G − α2r12/ 2g = p2/G − α2r22/2g;(p2− p1) / G = α2(r22− r12) / 2g = (v22− v12) / 2g.(10)That is, the pressure increases from within outwards in a curve which in radial sections is a parabola, and surfaces of equal pressure are paraboloids of revolution (fig. 37).Fig. 37.Dissipation of Head in Shock§ 36.Relation of Pressure and Velocity in a Stream in Steady Motion when the Changes of Section of the Stream are Abrupt.—When a stream changes section abruptly, rotating eddies are formed which dissipate energy. The energy absorbed in producing rotation is at once abstracted from that effective in causing the flow, and sooner or later it is wasted by frictional resistances due to the rapid relative motion of the eddying parts of the fluid. In such cases the work thus expended internally in the fluid is too important to be neglected, and the energy thus lost is commonly termed energy lost in shock. Suppose fig. 38 to represent a stream having such an abrupt change of section. Let AB, CD be normal sections at points where ordinary stream line motion has not been disturbed and where it has been re-established. Let ω, p, v be the area of section, pressure and velocity at AB, and ω1, p1, v1corresponding quantities at CD. Then if no work were expended internally, and assuming the stream horizontal, we should havep/G + v2/2g = p1/G + v12/2g.(1)But if work is expended in producing irregular eddying motion, the head at the section CD will be diminished.Suppose the mass ABCD comes in a short time t to A′B′C′D′. The resultant force parallel to the axis of the stream ispω + p0(ω1− ω) − p1ω1,where p0is put for the unknown pressure on the annular space between AB and EF. The impulse of that force is{ pω + p0(ω1− ω) − p1ω1} t.Fig. 38.The horizontal change of momentum in the same time is the difference of the momenta of CDC′D′ and ABA′B′, because the amount of momentum between A′B′ and CD remains unchanged if the motion is steady. The volume of ABA′B′ or CDC′D′, being the inflow and outflow in the time t, is Qt = ωvt = ω1v1t, and the momentum of these masses is (G/g) Qvt and (G/g) Qv1t. The change of momentum is therefore (G/g) Qt (v1− v). Equating this to the impulse,{ pω + p0(ω1− ω) − p1ω1} t = (G/g) Qt (v1− v).Assume that p0= p, the pressure at AB extending unchanged through the portions of fluid in contact with AE, BF which lie out of the path of the stream. Then (since Q = ω1v1)(p − p1) = (G/g) v1(v1− v);p/G − p1/G = v1(v1− v) / g;(2)p/G + v2/2g = p1/G + v12/2g + (v − v1)2/ 2g.(3)This differs from the expression (1), § 29, obtained for cases where no sensible internal work is done, by the last term on the right. That is, (v − v1)2/ 2g has to be added to the total head at CD, which is p1/G + v12/2g, to make it equal to the total head at AB, or (v − v1)2/ 2g is the head lost in shock at the abrupt change of section. But (v − v1) is the relative velocity of the two parts of the stream. Hence, when an abrupt change of section occurs, the head due to the relative velocity is lost in shock, or (v − v1)2/2g foot-pounds of energy is wasted for each pound of fluid. Experiment verifies this result, so that the assumption that p0= p appears to be admissible.If there is no shock,p1/G = p/G + (v2− v12) / 2g.If there is shock,p1/G = p/G − v1(v1− v) / g.Hence the pressure head at CD in the second case is less than in the former by the quantity (v − v1)2/ 2g, or, putting ω1v1= ωv, by the quantity(v2/2g) (1 − ω/ω1)2.(4)V. THEORY OF THE DISCHARGE FROM ORIFICES AND MOUTHPIECESFig. 39.§ 37.Minimum Coefficient of Contraction. Re-entrant Mouthpiece of Borda.—In one special case the coefficient of contraction can be determined theoretically, and, as it is the case where the convergence of the streams approaching the orifice takes place through the greatest possible angle, the coefficient thus determined is the minimum coefficient.Let fig. 39 represent a vessel with vertical sides, OO being the free water surface, at which the pressure is pa. Suppose the liquid issues by a horizontal mouthpiece, which is re-entrant and of the greatest length which permits the jet to spring clear from the inner end of the orifice, without adhering to its sides. With such an orifice the velocity near the points CD is negligible, and the pressure at those points may be taken equal to the hydrostatic pressure due to the depth from the free surface. Let Ω be the area of the mouthpiece AB, ω that of the contracted jet aa Suppose that in a short time t, the mass OOaa comes to the position O′O′ a′a′; the impulse of the horizontal external forces acting on the mass during that time is equal to the horizontal change of momentum.The pressure on the side OC of the mass will be balanced by the pressure on the opposite side OE, and so for all other portions of the vertical surfaces of the mass, excepting the portion EF opposite the mouthpiece and the surface AaaB of the jet. On EF the pressure is simply the hydrostatic pressure due to the depth, that is, (pa+ Gh). On the surface and section AaaB of the jet, the horizontal resultant of the pressure is equal to the atmospheric pressure paacting on the vertical projection AB of the jet; that is, the resultant pressure is −paΩ. Hence the resultant horizontal force for the whole mass OOaa is (pa+ Gh) Ω − paΩ = GhΩ. Its impulse in the time t is GhΩt. Since the motion is steady there is no change of momentum between O′O′ and aa. The change of horizontal momentum is, therefore, the difference of the horizontal momentum lost in the space OOO′O′ and gained in the space aaa′a′. In the former space there is no horizontal momentum.The volume of the space aaa′a′ is ωvt; the mass of liquid in that space is (G/g)ωvt; its momentum is (G/g)ωv2t. Equating impulse to momentum gained,GhΩt = (G/g) ωv2t;∴ ω/Ω = gh/v2Butv2= 2gh, and ω/Ω = cc;∴ ω/Ω =1⁄2= cc;a result confirmed by experiment with mouthpieces of this kind. A similar theoretical investigation is not possible for orifices in plane surfaces, because the velocity along the sides of the vessel in the neighbourhood of the orifice is not so small that it can be neglected. The resultant horizontal pressure is therefore greater than GhΩ, and the contraction is less. The experimental values of the coefficient of discharge for a re-entrant mouthpiece are 0.5149 (Borda), 0.5547 (Bidone), 0.5324 (Weisbach), values which differ little from the theoretical value, 0.5, given above.Fig. 40.Fig. 41.§ 38.Velocity of Filaments issuing in a Jet.—A jet is composed of fluid filaments or elementary streams, which start into motion at some point in the interior of the vessel from which the fluid is discharged, and gradually acquire the velocity of the jet. Let Mm, fig. 40 be such a filament, the point M being taken where the velocity is insensibly small, and m at the most contracted section of the jet, where the filaments have become parallel and exercise uniform mutual pressure. Take the free surface AB for datum line, and let p1, v1, h1, be the pressure, velocity and depth below datum at M; p, v, h, the corresponding quantities at m. Then § 29, eq. (3a),v12/2g + p1/G − h1= v2/2g + p/G − h(1)But at M, since the velocity is insensible, the pressure is the hydrostatic pressure due to the depth; that is v1= 0, p1= pa+ Gh1. At m, p = pa, the atmospheric pressure round the jet. Hence, inserting these values,0 + pa/G + h1− h1= v2/2g + pa/ G − h;v2/2g = h;(2)orv = √ (2gh) = 8.025V √ h.(2a)That is, neglecting the viscosity of the fluid, the velocity of filaments at the contracted section of the jet is simply the velocity due to the difference of level of the free surface in the reservoir and the orifice. If the orifice is small in dimensions compared with h, the filaments will all have nearly the same velocity, and if h is measured to the centre of the orifice, the equation above gives the mean velocity of the jet.Case of a Submerged Orifice.—Let the orifice discharge below the level of the tail water. Then using the notation shown in fig. 41, we have at M, v1= 0, p1= Gh; + paat m, p = Gh3+ pa. Inserting these values in (3), § 29,0 + h1+ pa/G − h1= v2/2g + h3− h22 + pa/G;v2/2g = h2− h3= h,(3)where h is the difference of level of the head and tail water, and may be termed theeffective headproducing flow.Fig. 42.Case where the Pressures are different on the Free Surface and at the Orifice.—Let the fluid flow from a vessel in which the pressure is p0into a vessel in which the pressure is p, fig. 42. The pressure p0will produce the same effect as a layer of fluid of thickness p0/G added to the head water; and the pressure p, will produce the same effect as a layer of thickness p/G added to the tail water. Hence the effective difference of level, or effective head producing flow, will beh = h0+ p0/G − p/G;and the velocity of discharge will bev = √ [ 2g { h0+ (p0− p) / G } ].(4)We may express this result by saying that differences of pressure at the free surface and at the orifice are to be reckoned as part of the effective head.Hence in all cases thus far treated the velocity of the jet is the velocity due to the effective head, and the discharge, allowing for contraction of the jet, isQ = cωv = cω √ (2gh),(5)where ω is the area of the orifice, cω the area of the contracted section of the jet, and h the effective head measured to the centre of the orifice. If h and ω are taken in feet, Q is in cubic feet per second.It is obvious, however, that this formula assumes that all the filaments have sensibly the same velocity. That will be true for horizontal orifices, and very approximately true in other cases, if the dimensions of the orifice are not large compared with the head h. In large orifices in say a vertical surface, the value of h is different for different filaments, and then the velocity of different filaments is not sensibly the same.Simple Orifices—Head ConstantFig. 43.§ 39.Large Rectangular Jets from Orifices in Vertical Plane Surfaces.—Let an orifice in a vertical plane surface be so formed that it produces a jet having a rectangular contracted section with vertical and horizontal sides. Let b (fig. 43) be the breadth of the jet, h1and h2the depths below the free surface of its upper and lower surfaces. Consider a lamina of the jet between the depths h and h + dh. Its normal section is bdh, and the velocity of discharge √2gh. The discharge per second in this lamina is therefore b√2ghdh, and that of the whole jet is thereforeQ =∫h2h1b √ (2gh) dh=2⁄3b √2g{ h23/2− h13/2},(6)where the first factor on the right is a coefficient depending on the form of the orifice.Now an orifice producing a rectangular jet must itself be very approximately rectangular. Let B be the breadth, H1, H2, the depths to the upper and lower edges of the orifice. Putb (h23/2− h13/2) / B (H23/2− H13/2) = c.(7)Then the discharge, in terms of the dimensions of the orifice, instead of those of the jet, isQ =2⁄3cB √2g(H23/2− H13/2),(8)the formula commonly given for the discharge of rectangular orifices. The coefficient c is not, however, simply the coefficient of contraction, the value of which isb (h2− h1) / B (H2− H1),and not that given in (7). It cannot be assumed, therefore, that c in equation (8) is constant, and in fact it is found to vary for different values of B/H2and B/H1, and must be ascertained experimentally.Relation between the Expressions(5)and(8).—For a rectangular orifice the area of the orifice is ω = B(H2− H1), and the depth measured to its centre is1⁄2(H2+ H1). Putting these values in (5),Q1= cB (H2− H1) √ {g (H2+ H1) }.From (8) the discharge isQ2=2⁄3cB √2g(H23/2− H13/2).Hence, for the same value of c in the two cases,Q2/Q1=2⁄3(H23/2− H13/2) / [ (H2− H1) √ { (H2+ H1)/2} ].Let H1/H2= σ, thenQ2/Q1= 0.9427 (1 − σ3/2) / {1 − σ √ (1 + σ) }.
The normal pressures on the surface of the mass (excluding the ends A, B) are at each point normal to the direction of motion, and do no work. Hence the only external forces to be reckoned are gravity and the pressures on the ends of the stream.
The work of gravity when AB falls to A′B′ is the same as that of transferring AA′ to BB′; that is, GQt (z − z1). The work of the pressures on the ends, reckoning that at B negative, because it is opposite to the direction of motion, is (pω × vt) − (p1ω1× v1t) = Qt(p − p1). The change of kinetic energy in the time t is the difference of the kinetic energy originally possessed by AA′ and that finally acquired by BB′, for in the intermediate part A′B there is no change of kinetic energy, in consequence of the steadiness of the motion. But the mass of AA′ and BB′ is GQt/g, and the change of kinetic energy is therefore (GQt/g) (v12/2 − v2/2). Equating this to the work done on the mass AB,
GQt (z − z1) + Qt (p − p1) = (GQt/g) (v12/2 − v2/2).
Dividing by GQt and rearranging the terms,
v2/2g + p/G + z = v12/2g + p1/G + z1;
(1)
or, as A and B are any two points,
v2/2g + p/G + z = constant = H.
(2)
Now v2/2g is the head due to the velocity v, p/G is the head equivalent to the pressure, and z is the elevation above the datum (see § 16). Hence the terms on the left are the total head due to velocity, pressure, and elevation at a given cross section of the filament, z is easily seen to be the work in foot-pounds which would be done by 1 ℔ of fluid falling to the datum line, and similarly p/G and v2/2g are the quantities of work which would be done by 1 ℔ of fluid due to the pressure p and velocity v. The expression on the left of the equation is, therefore, the total energy of the stream at the section considered, per ℔ of fluid, estimated with reference to thedatum line XX. Hence we see that in stream line motion, under the restrictions named above, the total energy per ℔ of fluid is uniformly distributed along the stream line. If the free surface of the fluid OO is taken as the datum, and −h, −h1are the depths of A and B measured down from the free surface, the equation takes the form
v2/2g + p/G − h = v12/2g + p1/G − h1;
(3)
or generally
v2/2g + p/G − h = constant.
(3a)
§ 30.Second Form of the Theorem of Bernoulli.—Suppose at the two sections A, B (fig. 26) of an elementary stream small vertical pipes are introduced, which may be termed pressure columns (§ 8), having their lower ends accurately parallel to the direction of flow. In such tubes the water will rise to heights corresponding to the pressures at A and B. Hence b = p/G, and b′ = p1/G. Consequently the tops of the pressure columns A′ and B′ will be at total heights b + c = p/G + z and b′ + c′ = p1/G + z1above the datum line XX. The difference of level of the pressure column tops, or the fall of free surface level between A and B, is therefore
ξ = (p − p1) / G + (z − z1);
and this by equation (1), § 29 is (v12− v2)/2g. That is, the fall of free, surface level between two sections is equal to the difference of the heights due to the velocities at the sections. The line A′B′ is sometimes called the line of hydraulic gradient, though this term is also used in cases where friction needs to be taken into account. It is the line the height of which above datum is the sum of the elevation and pressure head at that point, and it falls below a horizontal line A″B″ drawn at H ft. above XX by the quantities a = v2/2g and a′ = v12/2g, when friction is absent.
§ 31.Illustrations of the Theorem of Bernoulli.In a lecture to the mechanical section of the British Association in 1875, W. Froude gave some experimental illustrations of the principle of Bernoulli. He remarked that it was a common but erroneous impression that a fluid exercises in a contracting pipe A (fig. 27) an excess of pressure against the entire converging surface which it meets, and that, conversely, as it enters an enlargement B, a relief of pressure is experienced by the entire diverging surface of the pipe. Further it is commonly assumed that when passing through a contraction C, there is in the narrow neck an excess of pressure due to the squeezing together of the liquid at that point. These impressions are in no respect correct; the pressure is smaller as the section of the pipe is smaller and conversely.
Fig. 28 shows a pipe so formed that a contraction is followed by an enlargement, and fig. 29 one in which an enlargement is followed by a contraction. The vertical pressure columns show the decrease of pressure at the contraction and increase of pressure at the enlargement. The line abc in both figures shows the variation of free surface level, supposing the pipe frictionless. In actual pipes, however, work is expended in friction against the pipe; the total head diminishes in proceeding along the pipe, and the free surface level is a line such as ab1c1, falling below abc.
Froude further pointed out that, if a pipe contracts and enlarges again to the same size, the resultant pressure on the converging part exactly balances the resultant pressure on the diverging part so that there is no tendency to move the pipe bodily when water flows through it. Thus the conical part AB (fig. 30) presents the same projected surface as HI, and the pressures parallel to the axis of the pipe, normal to these projected surfaces, balance each other. Similarly the pressures on BC, CD balance those on GH, EG. In the same way, in any combination of enlargements and contractions, a balance of pressures, due to the flow of liquid parallel to the axis of the pipe, will be found, provided the sectional area and direction of the ends are the same.
The following experiment is interesting. Two cisterns provided with converging pipes were placed so that the jet from one was exactly opposite the entrance to the other. The cisterns being filled very nearly to the same level, the jet from the left-hand cistern A entered the right-hand cistern B (fig. 31), shooting across the free space between them without any waste, except that due to indirectness of aim and want of exact correspondence in the form of the orifices. In the actual experiment there was 18 in. of head in the right and 201⁄2in. of head in the left-hand cistern, so that about 21⁄2in. were wasted in friction. It will be seen that in the open space between the orifices there was no pressure, except the atmospheric pressure acting uniformly throughout the system.
§ 32.Venturi Meter.—An ingenious application of the variation of pressure and velocity in a converging and diverging pipe has been made by Clemens Herschel in the construction of what he terms a Venturi Meter for measuring the flow in water mains. Suppose that, as in fig. 32, a contraction is made in a water main, the change of section being gradual to avoid the production of eddies. The ratio ρof the cross sections at A and B, that is at inlet and throat, is in actual meters 5 to 1 to 20 to 1, and is very carefully determined by the maker of the meter. Then, if v and u are the velocities at A and B, u = ρv. Let pressure pipes be introduced at A, B and C, and let H1, H, H2be the pressure heads at those points. Since the velocity at B is greater than at A the pressure will be less. Neglecting friction
H1+ v2/2g = H + u2/2g,
H1− H = (u2− v2) / 2g = (ρ2− 1) v22g.
Let h = H1− H be termed the Venturi head, then
u = √ { ρ2.2gh / (ρ2− 1) },
from which the velocity through the throat and the discharge of the main can be calculated if the areas at A and B are known and h observed. Thus if the diameters at A and B are 4 and 12 in., the areas are 12.57 and 113.1 sq. in., and ρ = 9,
u = √ 81/80 √ (2gh) = 1.007 √ (2gh).
If the observed Venturi head is 12 ft.,
u = 28 ft. per sec.,
and the discharge of the main is
28 × 12.57 = 351 cub. ft. per sec.
Hence by a simple observation of pressure difference, the flow in the main at any moment can be determined. Notice that the pressure height at C will be the same as at A except for a small loss hfdue to friction and eddying between A and B. To get the pressure at the throat very exactly Herschel surrounds it by an annular passage communicating with the throat by several small holes, sometimes formed in vulcanite to prevent corrosion. Though constructed to prevent eddying as much as possible there is some eddy loss. The main effect of this is to cause a loss of head between A and C which may vary from a fraction of a foot to perhaps 5 ft. at the highest velocities at which a meter can be used. The eddying also affects a little the Venturi head h. Consequently an experimental coefficient must be determined for each meter by tank measurement. The range of this coefficient is, however, surprisingly small. If to allow for friction, u = k √ {ρ2/(ρ2− 1)} √(2gh), then Herschel found values of k from 0.97 to 1.0 for throat velocities varying from 8 to 28 ft. per sec. The meter is extremely convenient. At Staines reservoirs there are two meters of this type on mains 94 in. in diameter. Herschel contrived a recording arrangement which records the variation of flow from hour to hour and also the total flow in any given time. In Great Britain the meter is constructed by G. Kent, who has made improvements in the recording arrangement.
In the Deacon Waste Water Meter (fig. 33) a different principle is used. A disk D, partly counter-balanced by a weight, is suspended in the water flowing through the main in a conical chamber. The unbalanced weight of the disk is supported by the impact of the water. If the discharge of the main increases the disk rises, but as it rises its position in the chamber is such that in consequence of the larger area the velocity is less. It finds, therefore, a new position of equilibrium. A pencil P records on a drum moved by clockwork the position of the disk, and from this the variation of flow is inferred.
§ 33.Pressure, Velocity and Energy in Different Stream Lines.—The equation of Bernoulli gives the variation of pressure and velocity from point to point along a stream line, and shows that the total energy of the flow across any two sections is the same. Two other directions may be defined, one normal to the stream line and in the plane containing its radius of curvature at any point, the other normal to the stream line and the radius of curvature. For the problems most practically useful it will be sufficient to consider the stream lines as parallel to a vertical or horizontal plane. If the motion is in a vertical plane, the action of gravity must be taken into the reckoning; if the motion is in a horizontal plane, the terms expressing variation of elevation of the filament will disappear.3
Let AB, CD (fig. 34) be two consecutive stream lines, at present assumed to be in a vertical plane, and PQ a normal to these lines making an angle φ with the vertical. Let P, Q be two particles moving along these lines at a distance PQ = ds, and let z be the height of Q above the horizontal plane with reference to which the energy is measured, v its velocity, and p its pressure. Then, if H is the total energy at Q per unit of weight of fluid,
H = z + p/G + v2/2g.
Differentiating, we get
dH = dz + dp/G + v dv/g,
(1)
for the increment of energy between Q and P. But
dz = PQ cos φ = ds cos φ;
∴ dH = dp/G + v dv/g + ds cos φ,
(1a)
where the last term disappears if the motion is in a horizontal plane.
Now imagine a small cylinder of section ω described round PQ as an axis. This will be in equilibrium under the action of its centrifugal force, its weight and the pressure on its ends. But its volume is ωds and its weight Gω ds. Hence, taking the components of the forces parallel to PQ—
ω dp = Gv2ω ds/gρ − Gω cos φ ds,
where ρ is the radius of curvature of the stream line at Q. Consequently, introducing these values in (1),
dH = v2ds/gρ + v dv/g = (v/g) (v/ρ + dv/ds) ds.
(2)
Currents
§ 34.Rectilinear Current.—Suppose the motion is in parallel straight stream lines (fig. 35) in a vertical plane. Then ρ is infinite, and from eq. (2), § 33,
dH = v dv/g.
Comparing this with (1) we see that
dz + dp/G = 0;
∴ z + p/G = constant;
(3)
or the pressure varies hydrostatically as in a fluid at rest. For two stream lines in a horizontal plane, z is constant, and therefore p is constant.
Radiating Current.—Suppose water flowing radially between horizontal parallel planes, at a distance apart = δ. Conceive two cylindrical sections of the current at radii r1and r2, where the velocities are v1and v2, and the pressures p1and p2. Since the flow across each cylindrical section of the current is the same,
Q = 2πr1δv1= 2πr2δv2
r1v1= r2v2
r1/r2= v2/v1.
(4)
The velocity would be infinite at radius 0, if the current could be conceived to extend to the axis. Now, if the motion is steady,
H = p1/G + v12/2g = p2/G + v22/2g;
= p2/G + r12+ v12/ r222g;
(p2− p1) / G = v12(1 − r12/r22) / 2g;
(5)
p2/G = H − r12v12/ r222g.
(6)
Hence the pressure increases from the interior outwards, in a way indicated by the pressure columns in fig. 36, the curve through the free surfaces of the pressure columns being, in a radial section, the quasi-hyperbola of the form xy2= c3. This curve is asymptotic to a horizontal line, H ft. above the line from which the pressures are measured, and to the axis of the current.
Free Circular Vortex.—A free circular vortex is a revolving mass of water, in which the stream lines are concentric circles, and in which the total head for each stream line is the same. Hence, if by any slow radial motion portions of the water strayed from one stream line to another, they would take freely the velocities proper to their new positions under the action of the existing fluid pressures only.
For such a current, the motion being horizontal, we have for all the circular elementary streams
H = p/G + v2/2g = constant;
∴ dH = dp/G + v dv/g = 0.
(7)
Consider two stream lines at radii r and r + dr (fig. 36). Then in (2), § 33, ρ = r and ds = dr,
v2dr/gr + v dv/g = 0,
dv/v = −dr/r,
v ∞ 1/r,
(8)
precisely as in a radiating current; and hence the distribution of pressure is the same, and formulae 5 and 6 are applicable to this case.
Free Spiral Vortex.—As in a radiating and circular current the equations of motion are the same, they will also apply to a vortex in which the motion is compounded of these motions in any proportions, provided the radial component of the motion varies inversely as the radius as in a radial current, and the tangential component varies inversely as the radius as in a free vortex. Then the whole velocity at any point will be inversely proportional to the radius of the point, and the fluid will describe stream lines having a constant inclination to the radius drawn to the axis of the current. That is, the stream lines will be logarithmic spirals. When water is delivered from the circumference of a centrifugal pump or turbine into a chamber, it forms a free vortex of this kind. The water flows spirally outwards, its velocity diminishing and its pressure increasing according to the law stated above, and the head along each spiral stream line is constant.
§ 35.Forced Vortex.—If the law of motion in a rotating current is different from that in a free vortex, some force must be applied to cause the variation of velocity. The simplest case is that of a rotating current in which all the particles have equal angular velocity, as for instance when they are driven round by radiating paddles revolving uniformly. Then in equation (2), § 33, considering two circular stream lines of radii r and r + dr (fig. 37), we have ρ = r, ds = dr. If the angular velocity is α, then v = αr and dv = αdr. Hence
dH = α2r dr/g + α2r dr/g = 2α2r dr/g.
Comparing this with (1), § 33, and putting dz = 0, because the motion is horizontal,
dp/G + α2r dr/g = 2α2r dr/g,
dp/G = α2r dr/g,
p/G = α2/2g + constant.
(9)
Let p1, r1, v1be the pressure, radius and velocity of one cylindrical section, p2, r2, v2those of another; then
p1/G − α2r12/ 2g = p2/G − α2r22/2g;
(p2− p1) / G = α2(r22− r12) / 2g = (v22− v12) / 2g.
(10)
That is, the pressure increases from within outwards in a curve which in radial sections is a parabola, and surfaces of equal pressure are paraboloids of revolution (fig. 37).
Dissipation of Head in Shock
§ 36.Relation of Pressure and Velocity in a Stream in Steady Motion when the Changes of Section of the Stream are Abrupt.—When a stream changes section abruptly, rotating eddies are formed which dissipate energy. The energy absorbed in producing rotation is at once abstracted from that effective in causing the flow, and sooner or later it is wasted by frictional resistances due to the rapid relative motion of the eddying parts of the fluid. In such cases the work thus expended internally in the fluid is too important to be neglected, and the energy thus lost is commonly termed energy lost in shock. Suppose fig. 38 to represent a stream having such an abrupt change of section. Let AB, CD be normal sections at points where ordinary stream line motion has not been disturbed and where it has been re-established. Let ω, p, v be the area of section, pressure and velocity at AB, and ω1, p1, v1corresponding quantities at CD. Then if no work were expended internally, and assuming the stream horizontal, we should have
p/G + v2/2g = p1/G + v12/2g.
(1)
But if work is expended in producing irregular eddying motion, the head at the section CD will be diminished.
Suppose the mass ABCD comes in a short time t to A′B′C′D′. The resultant force parallel to the axis of the stream is
pω + p0(ω1− ω) − p1ω1,
where p0is put for the unknown pressure on the annular space between AB and EF. The impulse of that force is
{ pω + p0(ω1− ω) − p1ω1} t.
The horizontal change of momentum in the same time is the difference of the momenta of CDC′D′ and ABA′B′, because the amount of momentum between A′B′ and CD remains unchanged if the motion is steady. The volume of ABA′B′ or CDC′D′, being the inflow and outflow in the time t, is Qt = ωvt = ω1v1t, and the momentum of these masses is (G/g) Qvt and (G/g) Qv1t. The change of momentum is therefore (G/g) Qt (v1− v). Equating this to the impulse,
{ pω + p0(ω1− ω) − p1ω1} t = (G/g) Qt (v1− v).
Assume that p0= p, the pressure at AB extending unchanged through the portions of fluid in contact with AE, BF which lie out of the path of the stream. Then (since Q = ω1v1)
(p − p1) = (G/g) v1(v1− v);
p/G − p1/G = v1(v1− v) / g;
(2)
p/G + v2/2g = p1/G + v12/2g + (v − v1)2/ 2g.
(3)
This differs from the expression (1), § 29, obtained for cases where no sensible internal work is done, by the last term on the right. That is, (v − v1)2/ 2g has to be added to the total head at CD, which is p1/G + v12/2g, to make it equal to the total head at AB, or (v − v1)2/ 2g is the head lost in shock at the abrupt change of section. But (v − v1) is the relative velocity of the two parts of the stream. Hence, when an abrupt change of section occurs, the head due to the relative velocity is lost in shock, or (v − v1)2/2g foot-pounds of energy is wasted for each pound of fluid. Experiment verifies this result, so that the assumption that p0= p appears to be admissible.
If there is no shock,
p1/G = p/G + (v2− v12) / 2g.
If there is shock,
p1/G = p/G − v1(v1− v) / g.
Hence the pressure head at CD in the second case is less than in the former by the quantity (v − v1)2/ 2g, or, putting ω1v1= ωv, by the quantity
(v2/2g) (1 − ω/ω1)2.
(4)
V. THEORY OF THE DISCHARGE FROM ORIFICES AND MOUTHPIECES
§ 37.Minimum Coefficient of Contraction. Re-entrant Mouthpiece of Borda.—In one special case the coefficient of contraction can be determined theoretically, and, as it is the case where the convergence of the streams approaching the orifice takes place through the greatest possible angle, the coefficient thus determined is the minimum coefficient.
Let fig. 39 represent a vessel with vertical sides, OO being the free water surface, at which the pressure is pa. Suppose the liquid issues by a horizontal mouthpiece, which is re-entrant and of the greatest length which permits the jet to spring clear from the inner end of the orifice, without adhering to its sides. With such an orifice the velocity near the points CD is negligible, and the pressure at those points may be taken equal to the hydrostatic pressure due to the depth from the free surface. Let Ω be the area of the mouthpiece AB, ω that of the contracted jet aa Suppose that in a short time t, the mass OOaa comes to the position O′O′ a′a′; the impulse of the horizontal external forces acting on the mass during that time is equal to the horizontal change of momentum.
The pressure on the side OC of the mass will be balanced by the pressure on the opposite side OE, and so for all other portions of the vertical surfaces of the mass, excepting the portion EF opposite the mouthpiece and the surface AaaB of the jet. On EF the pressure is simply the hydrostatic pressure due to the depth, that is, (pa+ Gh). On the surface and section AaaB of the jet, the horizontal resultant of the pressure is equal to the atmospheric pressure paacting on the vertical projection AB of the jet; that is, the resultant pressure is −paΩ. Hence the resultant horizontal force for the whole mass OOaa is (pa+ Gh) Ω − paΩ = GhΩ. Its impulse in the time t is GhΩt. Since the motion is steady there is no change of momentum between O′O′ and aa. The change of horizontal momentum is, therefore, the difference of the horizontal momentum lost in the space OOO′O′ and gained in the space aaa′a′. In the former space there is no horizontal momentum.
The volume of the space aaa′a′ is ωvt; the mass of liquid in that space is (G/g)ωvt; its momentum is (G/g)ωv2t. Equating impulse to momentum gained,
GhΩt = (G/g) ωv2t;
∴ ω/Ω = gh/v2
But
v2= 2gh, and ω/Ω = cc;
∴ ω/Ω =1⁄2= cc;
a result confirmed by experiment with mouthpieces of this kind. A similar theoretical investigation is not possible for orifices in plane surfaces, because the velocity along the sides of the vessel in the neighbourhood of the orifice is not so small that it can be neglected. The resultant horizontal pressure is therefore greater than GhΩ, and the contraction is less. The experimental values of the coefficient of discharge for a re-entrant mouthpiece are 0.5149 (Borda), 0.5547 (Bidone), 0.5324 (Weisbach), values which differ little from the theoretical value, 0.5, given above.
§ 38.Velocity of Filaments issuing in a Jet.—A jet is composed of fluid filaments or elementary streams, which start into motion at some point in the interior of the vessel from which the fluid is discharged, and gradually acquire the velocity of the jet. Let Mm, fig. 40 be such a filament, the point M being taken where the velocity is insensibly small, and m at the most contracted section of the jet, where the filaments have become parallel and exercise uniform mutual pressure. Take the free surface AB for datum line, and let p1, v1, h1, be the pressure, velocity and depth below datum at M; p, v, h, the corresponding quantities at m. Then § 29, eq. (3a),
v12/2g + p1/G − h1= v2/2g + p/G − h
(1)
But at M, since the velocity is insensible, the pressure is the hydrostatic pressure due to the depth; that is v1= 0, p1= pa+ Gh1. At m, p = pa, the atmospheric pressure round the jet. Hence, inserting these values,
0 + pa/G + h1− h1= v2/2g + pa/ G − h;
v2/2g = h;
(2)
or
v = √ (2gh) = 8.025V √ h.
(2a)
That is, neglecting the viscosity of the fluid, the velocity of filaments at the contracted section of the jet is simply the velocity due to the difference of level of the free surface in the reservoir and the orifice. If the orifice is small in dimensions compared with h, the filaments will all have nearly the same velocity, and if h is measured to the centre of the orifice, the equation above gives the mean velocity of the jet.
Case of a Submerged Orifice.—Let the orifice discharge below the level of the tail water. Then using the notation shown in fig. 41, we have at M, v1= 0, p1= Gh; + paat m, p = Gh3+ pa. Inserting these values in (3), § 29,
0 + h1+ pa/G − h1= v2/2g + h3− h22 + pa/G;
v2/2g = h2− h3= h,
(3)
where h is the difference of level of the head and tail water, and may be termed theeffective headproducing flow.
Case where the Pressures are different on the Free Surface and at the Orifice.—Let the fluid flow from a vessel in which the pressure is p0into a vessel in which the pressure is p, fig. 42. The pressure p0will produce the same effect as a layer of fluid of thickness p0/G added to the head water; and the pressure p, will produce the same effect as a layer of thickness p/G added to the tail water. Hence the effective difference of level, or effective head producing flow, will be
h = h0+ p0/G − p/G;
and the velocity of discharge will be
v = √ [ 2g { h0+ (p0− p) / G } ].
(4)
We may express this result by saying that differences of pressure at the free surface and at the orifice are to be reckoned as part of the effective head.
Hence in all cases thus far treated the velocity of the jet is the velocity due to the effective head, and the discharge, allowing for contraction of the jet, is
Q = cωv = cω √ (2gh),
(5)
where ω is the area of the orifice, cω the area of the contracted section of the jet, and h the effective head measured to the centre of the orifice. If h and ω are taken in feet, Q is in cubic feet per second.
It is obvious, however, that this formula assumes that all the filaments have sensibly the same velocity. That will be true for horizontal orifices, and very approximately true in other cases, if the dimensions of the orifice are not large compared with the head h. In large orifices in say a vertical surface, the value of h is different for different filaments, and then the velocity of different filaments is not sensibly the same.
Simple Orifices—Head Constant
§ 39.Large Rectangular Jets from Orifices in Vertical Plane Surfaces.—Let an orifice in a vertical plane surface be so formed that it produces a jet having a rectangular contracted section with vertical and horizontal sides. Let b (fig. 43) be the breadth of the jet, h1and h2the depths below the free surface of its upper and lower surfaces. Consider a lamina of the jet between the depths h and h + dh. Its normal section is bdh, and the velocity of discharge √2gh. The discharge per second in this lamina is therefore b√2ghdh, and that of the whole jet is therefore
Q =∫h2h1b √ (2gh) dh
=2⁄3b √2g{ h23/2− h13/2},
(6)
where the first factor on the right is a coefficient depending on the form of the orifice.
Now an orifice producing a rectangular jet must itself be very approximately rectangular. Let B be the breadth, H1, H2, the depths to the upper and lower edges of the orifice. Put
b (h23/2− h13/2) / B (H23/2− H13/2) = c.
(7)
Then the discharge, in terms of the dimensions of the orifice, instead of those of the jet, is
Q =2⁄3cB √2g(H23/2− H13/2),
(8)
the formula commonly given for the discharge of rectangular orifices. The coefficient c is not, however, simply the coefficient of contraction, the value of which is
b (h2− h1) / B (H2− H1),
and not that given in (7). It cannot be assumed, therefore, that c in equation (8) is constant, and in fact it is found to vary for different values of B/H2and B/H1, and must be ascertained experimentally.
Relation between the Expressions(5)and(8).—For a rectangular orifice the area of the orifice is ω = B(H2− H1), and the depth measured to its centre is1⁄2(H2+ H1). Putting these values in (5),
Q1= cB (H2− H1) √ {g (H2+ H1) }.
From (8) the discharge is
Q2=2⁄3cB √2g(H23/2− H13/2).
Hence, for the same value of c in the two cases,
Q2/Q1=2⁄3(H23/2− H13/2) / [ (H2− H1) √ { (H2+ H1)/2} ].
Let H1/H2= σ, then
Q2/Q1= 0.9427 (1 − σ3/2) / {1 − σ √ (1 + σ) }.