To determine the component acceleration of a particle, suppose F to denote any function of x, y, z, t, and investigate the time rate of F for a moving particle; denoting the change by DF/dt,DF= lt·F(x + uδt, y + vδt, z + wδt, t + δt) − F(x, y, z, t)dtδt=dF+ udF+ vdF+ wdF;dtdxdydz(1)and D/dt is called particle differentiation, because it follows the rate of change of a particle as it leaves the point x, y, z; butdF/dt, dF/dx, dF/dy, dF/dz(2)represent the rate of change of F at the time t, at the point, x, y, z, fixed in space.The components of acceleration of a particle of fluid are consequentlyDu=du+ udu+ vdu+ wdu,dtdtdxdydz(3)Dv=dv+ udv+ vdv+ wdv,dtdtdxdydz(4)Dw=dw+ udw+ vdw+ wdw,dtdtdxdydz(5)leading to the equations of motion above.If F (x, y, z, t) = 0 represents the equation of a surface containing always the same particles of fluid,DF= 0, ordF+ udF+ vdF+ wdF= 0,dtdtdxdydz(6)which is called the differential equation of thebounding surface. A bounding surface is such that there is no flow of fluid across it, as expressed by equation (6). The surface always contains the same fluid inside it, and condition (6) is satisfied over the complete surface, as well as any part of it.But turbulence in the motion will vitiate the principle that a bounding surface will always consist of the same fluid particles, as we see on the surface of turbulent water.24. To integrate the equations of motion, suppose the impressed force is due to a potential V, such that the force in any direction is the rate of diminution of V, or its downward gradient; and thenX = −dV/dx, Y = −dV/dy, Z = −dV/dz;(1)and puttingdw−dv= 2ξ,du−dw= 2η,dv−du= 2ζ,dydzdzdxdxdy(2)dξ+dη+dζ= 0,dxdydz(3)the equations of motion may be writtendu− 2vζ + 2wη +dH= 0,dtdx(4)dv− 2wξ + 2uζ +dH= 0,dtdy(5)dw− 2uη + 2wξ +dH= 0,dtdz(6)whereH = ∫ dp/ρ + V + ½q2,(7)q2= u2+ v2+ w2,(8)and the three terms in H may be called the pressure head, potential head, and head of velocity, when the gravitation unit is employed and ½q2is replaced by ½q2/g.Eliminating H between (5) and (6)Dξ− ξdu− ηdw− ζdv+ ξ(du+dv+dw)= 0,dtdxdxdxdxdydz(9)and combining this with the equation of continuity1Dρ+du+dv+dw= 0,ρdtdxdydz(10)we haveD(ξ)−ξdu−ηdv−ζdw= 0,dtρρdxρdxρdx(11)with two similar equations.Puttingω2= ξ2+ η2+ ζ2,(12)avortex lineis defined to be such that the tangent is in the direction of ω, the resultant of ξ, η, ζ, called the components of molecular rotation. A small sphere of the fluid, if frozen suddenly, would retain this angular velocity.If ω vanishes throughout the fluid at any instant, equation (11) shows that it will always be zero, and the fluid motion is then called irrotational; and a function φ exists, called thevelocity function, such thatu dx + v dy + w dz = −dφ,(13)and then the velocity in any direction is the space-decrease or downward gradient of φ.25. But in the most general case it is possible to have three functions φ, ψ, m of x, y, z, such thatu dx + v dy + w dz = −dφ − m dψ,(1)as A. Clebsch has shown, from purely analytical considerations (Crelle, lvi.); and thenξ = ½d(ψ, m), η = ½d(ψ, m), ζ = ½d(ψ, m),d(y, z)d(z, x)d(x, y)(2)andξdψ+ ηdψ+ ζdψ= 0, ξdm+ ηdm+ ζdm= 0,dxdydzdxdydz(3)so that, at any instant, the surfaces over which ψ and m are constant intersect in the vortex lines.PuttingH −dφ− mdψ= K,dtdt(4)the equations of motion (4), (5), (6) § 24 can be writtendK− 2uζ + 2wη −d(ψ,m)= 0, ..., ...;dxd(x,t)(5)and thereforeξdK+ ηdK+ ζdK= 0.dxdydz(6)Equation (5) becomes, by a rearrangement,dK−dψ(dm+ udm+ vdm+ wdm)dxdxdtdxdydz+dm(dψ+ udψ+ vdψ+ wdψ)= 0, ..., ...,dxdtdxdydz(7)dK−dψDm+dmDψ= 0, ..., ...,dxdxdtdxdt(8)and as we prove subsequently (§ 37) that the vortex lines are composed of the same fluid particles throughout the motion, the surface m and ψ satisfies the condition of (6) § 23; so that K is uniform throughout the fluid at any instant, and changes with the time only, and so may be replaced by F(t).26. When the motion issteady, that is, when the velocity at any point of space does not change with the time,dK− 2vζ + 2wη = 0, ..., ...dx(1)ξdK+ ηdK+ ζdK= 0, udK+ vdK+ wdK= 0,dxdydzdxdydz(2)andK = ∫ dp/ρ + V + ½q2= H(3)is constant along a vortex line, and astream line, the path of a fluid particle, so that the fluid is traversed by a series of H surfaces, each covered by a network of stream lines and vortex lines; and if the motion is irrotational H is a constant throughout the fluid.Taking the axis of x for an instant in the normal through a point on the surface H = constant, this makes u = 0, ξ = 0; and in steady motion the equations reduce todH/dν = 2vζ − 2wη = 2qω sin θ,(4)where θ is the angle between the stream line and vortex line; and this holds for their projection on any plane to which dν is drawn perpendicular.In plane motion (4) reduces todH= 2qζ = q(dQ+q),dνdvr(5)if r denotes the radius of curvature of the stream line, so that1dp+dV=dH−d ½q2=q2,ρdνdνdνdνr(6)the normal acceleration.The osculating plane of a stream line in steady motion contains the resultant acceleration, the direction ratios of which areudu+ vdu+ wdu=d ½q2− 2vζ + 2wη =d ½q2−dH, ...,dxdydzdxdxdx(7)and when q is stationary, the acceleration is normal to the surface H = constant, and the stream line is a geodesic.Calling the sum of the pressure and potential head the statical head, surfaces of constant statical and dynamical head intersect in lines on H, and the three surfaces touch where the velocity is stationary.Equation (3) is called Bernoulli’s equation, and may be interpreted as the balance-sheet of the energy which enters and leaves a given tube of flow.If homogeneous liquid is drawn off from a vessel so large that the motion at the free surface at a distance may be neglected, then Bernoulli’s equation may be writtenH = p/ρ + z + q2/2g = P/ρ + h,(8)where P denotes the atmospheric pressure and h the height of the free surface, a fundamental equation in hydraulics; a return has been made here to the gravitation unit of hydrostatics, and Oz is taken vertically upward.In particular, for a jet issuing into the atmosphere, where p = P,q2/2g = h − z,(9)or the velocity of the jet is due to the head k − z of the still free surface above the orifice; this is Torricelli’s theorem (1643), the foundation of the science of hydrodynamics.27.Uniplanar Motion.—In the uniplanar motion of a homogeneous liquid the equation of continuity reduces todu+dv= 0,dxdy(1)so that we can putu = −dψ/dy, v = dψ/dx,(2)where ψ is a function of x, y, called the stream- or current-function; interpreted physically, ψ − ψ0, the difference of the value of ψ at a fixed point A and a variable point P is the flow, in ft.3/second, across any curved line AP from A to P, this being the same for all lines in accordance with the continuity.Thus if dψ is the increase of ψ due to a displacement from P to P′, and k is the component of velocity normal to PP′, the flow across PP′ is dψ = k·PP′; and taking PP′ parallel to Ox, dψ = v dx; and similarly dψ= −u dy with PP′ parallel to Oy; and generally dψ/ds is the velocity across ds, in a direction turned through a right angle forward, against the clock.In the equations of uniplanar motion2ζ =dv−du=d2ψ+d2ψ= −∇2ψ, suppose,dxdydx2dy2(3)so that in steady motiondH+ ∇2ψdψ= 0,dH+ ∇2ψdψ= 0,dH+ ∇2ψ = 0,dxdxdydydψ(4)and ∇2ψ must be a function of ψ.If the motion ia irrotational,u = −dφ= −dψ, v = −dφ=dψ,dxdydydx(5)so that ψ and φ are conjugate functions of x and y,φ + ψi = ƒ(x + yi), ∇2ψ = 0, ∇2φ = 0;(6)or puttingφ + ψi = w, x + yi = z, w = ƒ(z).The curves φ = constant and ψ = constant form an orthogonal system; and the interchange of φ and ψ will give a new state of uniplanar motion, in which the velocity at every point is turned through a right angle without alteration of magnitude.For instance, in a uniplanar flow, radially inward towards O, the flow across any circle of radius r being the same and denoted by 2πm, the velocity must be m/r, andφ = m log r, ψ = mθ, φ + ψi = m log reiθ, w = m log z.(7)Interchanging these valuesψ = m log r, φ = mθ, ψ + φi = m log reiθ(8)gives a state of vortex motion, circulating round Oz, called a straight or columnar vortex.A single vortex will remain at rest, and cause a velocity at any point inversely as the distance from the axis and perpendicular to its direction; analogous to the magnetic field of a straight electric current.If other vortices are present, any one may be supposed to move with the velocity due to the others, the resultant stream-function beingψ = Σm log r = log Πrm;(9)the path of a vortex is obtained by equating the value of ψ at the vortex to a constant, omitting the rmof the vortex itself.When the liquid is bounded by a cylindrical surface, the motion of a vortex inside may be determined as due to a series of vortex-images, so arranged as to make the flow zero across the boundary.For a plane boundary the image is the optical reflection of the vortex. For example, a pair of equal opposite vortices, moving on a line parallel to a plane boundary, will have a corresponding pair of images, forming a rectangle of vortices, and the path of a vortex will be the Cotes’ spiralr sin 2θ = 2a, or x−2+ y−2= a−2;(10)this is therefore the path of a single vortex in a right-angled corner; and generally, if the angle of the corner is π/n, the path is the Cotes’ spiralr sin nθ = na.(11)A single vortex in a circular cylinder of radius a at a distance c from the centre will move with the velocity due to an equal opposite image at a distance a2/c, and so describe a circle with velocitymc/(a2− c2) in the periodic time 2π (a2− c2)/m.(12)Conjugate functions can be employed also for the motion of liquid in a thin sheet between two concentric spherical surfaces; the components of velocity along the meridian and parallel in colatitude θ and longitude λ can be writtendφ=1dψ,1dψ= −dψ,dθsin θdλsin θdλdθ(13)and thenφ + ψi = F (tan ½θ·eλi).(14)28.Uniplanar Motion of a Liquid due to the Passage of a Cylinder through it.—A stream-function ψ must be determined to satisfy the conditions∇2ψ = 0, throughout the liquid;(1)ψ = constant, over any fixed boundary;(2)dψ/ds = normal velocity reversed over a solid boundary,(3)so that, if the solid is moving with velocity U in the direction Ox, dψ/ds = −U dy/ds, or ψ + Uy = constant over the moving cylinder; and ψ + Uy = ψ′ is the stream function of the relative motion of the liquid past the cylinder, and similarly ψ − Vx for the component velocity V along Oy; and generallyψ′ = ψ + Uy − Vx(4)is the relative stream-function, constant over a solid boundary moving with components U and V of velocity.If the liquid is stirred up by the rotation R of a cylindrical body,dψ/ds = normal velocity reversed= −Rxdx− Rydy,dsds(5)ψ + ½R (x2+ y2) = ψ′,(6)a constant over the boundary; and ψ′ is the current-function of the relative motion past the cylinder, but nowV2ψ′ + 2R = 0,(7)throughout the liquid.Inside an equilateral triangle, for instance, of height h,ψ′ = −2Rαβγ/h,(8)where α, β, γ are the perpendiculars on the sides of the triangle.In the general case ψ′ = ψ + Uy − Vx + ½R (x2+ y2) is the relative stream function for velocity components, U, V, R.29.Example 1.—Liquid motion past a circular cylinder.Consider the motion given byω = U (z + a2/z),(1)so thatψ = U(r +a2)cos θ = U(1 +a2)x,rr2(2)φ = U(r +a2)sin θ = U(1 +a2)y.rr2(2)Then ψ = 0 over the cylinder r = a, which may be considered a fixed post; and a stream line past it along which ψ = Uc, a constant, is the curve(r −a2)sin θ = c, (x2+ y2) (y − c) − a2y = 0r(3)a cubic curve (C3).Over a concentric cylinder, external or internal, of radius r = b,ψ′ = ψ + U1y = [ U(1 −a2)+ U1] y,b2(4)and ψ′ is zero ifU1/U = (a2− b2)/b2;(5)so that the cylinder may swim for an instant in the liquid without distortion, with this velocity U1, and ω in (1) will give the liquid motion in the interspace between the fixed cylinder r = a and the concentric cylinder r = b, moving with velocity U1.When b = 0, U1= ∞; and when b = ∞, U1= −U, so that at infinity the liquid is streaming in the direction xO with velocity U.If the liquid is reduced to rest at infinity by the superposition of an opposite stream given by ω = −Uz, we are left withω = Ua2/z,(6)φ = U (a2/r) cos θ = Ua2x/(x2+ y2),(7)ψ = −U (a2/r) sin θ = −Ua2y/(x2+ y2),(8)giving the motion due to the passage of the Cylinder r = a with velocity U through the origin O in the direction Ox.If the direction of motion makes an angle θ′ with Ox,tan θ′ =dφ/dφ=2xy= tan 2θ, θ = ½θ′,dydxx2− y2(9)and the velocity is Ua2/r2.Along the path of a particle, defined by the C3of (3),sin2½θ′ =y2=y (y − c),x2+ y2a2(10)½ sin θ′dθ′=2y − cdy,dsa2ds(11)on the radius of curvature is ¼a2/(y − ½c), which shows that the curve is an Elastica or Lintearia. (J. C. Maxwell,Collected Works, ii. 208.)If φ1denotes the velocity function of the liquid filling the cylinder r = b, and moving bodily with it with velocity U1,φ1= −U1x,(12)and over the separating surface r = bφ= −U(1 +a2)=a2+ b2,φ1U1b2a2− b2(13)and this, by § 36, is also the ratio of the kinetic energy in the annular interspace between the two cylinders to the kinetic energy of the liquid moving bodily inside r = b.Consequently the inertia to overcome in moving the cylinder r = b, solid or liquid, is its own inertia, increased by the inertia of liquid (a2+ b2)/(a2~ b2) times the volume of the cylinder r = b; this total inertia is called the effective inertia of the cylinder r = b, at the instant the two cylinders are concentric.With liquid of density ρ, this gives rise to a kinetic reaction to acceleration dU/dt, given byπρb2a2+ b2dU=a2+ b2M′dU,a2− b2dta2− b2dt(14)if M′ denotes the mass of liquid displaced by unit length of the cylinder r = b. In particular, when a = ∞, the extra inertia is M′.When the cylinder r = a is moved with velocity U and r = b with velocity U1along Ox,φ = Ua2(b2+ r)cos θ − U1b2(r +a2)cos θ,b2− a2rb2− a2r(15)ψ = −Ua2(b2− r)sin θ − U1b2(r −a2)sin θ,b2− a2rb2− a2r(16)and similarly, with velocity components V and V1along Oyφ = Va2(b2+ r)cos θ − V1b2(r +a2)cos θ,b2− a2rb2− a2r(17)ψ = Va2(b2− r)sin θ + V1b2(r −a2)sin θ,b2− a2rb2− a2r(18)and then for the resultant motionw = (U2+ V2)a2z+a2b2U + Vib2− a2U + Vib2− a2z−(U12+ V12)b2z−a2b2U1+ V1i.b2− a2U1+ V1ib2− a2z(19)The resultant impulse of the liquid on the cylinder is given by the component, over r = a (§ 36),X = ∫ ρφ cos θ·a dθ = πρa2(Ub2+ a2− U12b2);b2− a2b2− a2(20)and over r = bX1= ∫ ρφ cos θ·b dθ = πρb2(U2a2− U1b2+ a2),b2− a2b2− a2(21)and the difference X − X1is the component momentum of the liquid in the interspace; with similar expressions for Y and Y1.Then, if the outside cylinder is free to moveX1= 0,V1=2a2, X = πρa2Ub2− a2.Ub2+ a2b2+ a2(22)But if the outside cylinder is moved with velocity U1, and the inside cylinder is solid or filled with liquid of density σ,X = −πρa2U,U1=2ρb2,Uρ (b2+ a2) + σ (b2− a2)U − U1=(ρ − σ) (b2− a2),U1ρ (b2+ a2) + σ (b2− a2)(23)and the inside cylinder starts forward or backward with respect to the outside cylinder, according as ρ > or < σ.30. The expression for ω in (1) § 29 may be increased by the addition of the termim log z = −mθ + im log r,(1)representing vortex motion circulating round the annulus of liquid.Considered by itself, with the cylinders held fixed, the vortex sets up a circumferential velocity m/r on a radius r, so that the angular momentum of a circular filament of annular cross section dA is ρm dA, and of the whole vortex is ρmπ (b2− a2).Any circular filament can be started from rest by the application of a circumferential impulse πρm dr at each end of a diameter; so that a mechanism attached to the cylinders, which can set up a uniform distributed impulse πρm across the two parts of a diameter in the liquid, will generate the vortex motion, and react on the cylinder with an impulse couple −ρmπa2and ρmπb2, having resultant ρmπ (b2− a2), and this couple is infinite when b = ∞, as the angular momentum of the vortex is infinite. Round the cylinder r = a held fixed in the U current the liquid streams past with velocityq′ = 2U sin θ + m/a;(2)and the loss of head due to this increase of velocity from U to q′ isq′2− U2=(2U sin θ + m/a)2− U2,2g2g(3)so that cavitation will take place, unless the head at a great distance exceeds this loss.The resultant hydrostatic thrust across any diametral plane of the cylinder will be modified, but the only term in the loss of head which exerts a resultant thrust on the whole cylinder is 2mU sin θ/ga, and its thrust is 2πρmU absolute units in the direction Cy, to be counteracted by a support at the centre C; the liquid is streaming past r = a with velocity U reversed, and the cylinder is surrounded by a vortex. Similarly, the streaming velocity V reversed will give rise to a thrust 2πρmV in the direction xC.Now if the cylinder is released, and the components U and V are reversed so as to become the velocity of the cylinder with respect to space filled with liquid, and at rest at infinity, the cylinder will experience components of force per unit length(i.) − 2πρmV, 2πρmU, due to the vortex motion;(ii.) − πρa2dU/dt, − πρa2dV/dt, due to the kinetic reaction of the liquid;(iii.) 0, −π(σ − ρ) a2g, due to gravity,taking Oy vertically upward, and denoting the density of the cylinder by σ; so that the equations of motion areπρa2dU= − πρa2dU− 2πρmV,dtdt(4)πρa2dV= − πρa2dV+ 2πρmV − π (σ − ρ) a2g,dtdt(5)or, putting m = a2ω, so that the vortex velocity is due to an angular velocity ω at a radius a,(σ + ρ) dU/dt + 2ρωV = 0,(6)(σ + ρ) dV/dt − 2ρωU + (σ-ρ) g = 0.(7)Thus with g = 0, the cylinder will describe a circle with angular velocity 2ρω/(σ + ρ), so that the radius is (σ + ρ) v/2ρω, if the velocity is v. With σ = 0, the angular velocity of the cylinder is 2ω; in this way the velocity may be calculated of the propagation of ripples and waves on the surface of a vertical whirlpool in a sink.Restoring σ will make the path of the cylinder a trochoid; and so the swerve can be explained of the ball in tennis, cricket, baseball, or golf.Another explanation may be given of the sidelong force, arising from the velocity of liquid past a cylinder, which is encircled by a vortex. Taking two planes x = ± b, and considering the increase of momentum in the liquid between them, due to the entry and exit of liquid momentum, the increase across dy in the direction Oy, due to elements at P and P′ at opposite ends of the diameter PP′, isρ dy (U − Ua2r−2cos 2θ + mr−1sin θ) (Ua2r−2sin 2θ + mr−1cos θ)+ ρ dy ( −U + Ua2r−2cos 2θ + mr−1sin θ) (Ua2r−2sin 2θ − mr−1cos θ)= 2ρdymUr−1(cos θ − a2r−2cos 3θ),(8)and with y = b tan θ, r = b sec θ, this is2ρmU dθ (1 − a2b−2cos 3θ cos θ),(9)and integrating between the limits θ = ±½π, the resultant, as before, is 2πρmU.31.Example 2.—Confocal Elliptic Cylinders.—Employ the elliptic coordinates η, ξ, and ζ = η + ξi, such thatz = c ch ζ, x = c ch η cos ξ, y = c sh η sin ζ;(1)then the curves for which η and ξ are constant are confocal ellipses and hyperbolas, andJ =d(x, y)= c2(ch2η − cos2ξ)d(η, ξ)= (1/2)c2(ch 2η − cos 2ξ) = r1r2= OD2,(2)if OD is the semi-diameter conjugate to OP, and r1, r2the focal distances,r1, r2= c (ch η ± cos ξ);(3)r2= x2+ y2= c2(ch2η − sin2ξ)= ½c2(ch 2η + cos 2ξ).(4)Consider the streaming motion given byw = m ch (ζ − γ), γ = α + βi,(5)φ = m ch (η − α) cos (ξ − β), ψ = m sh (η − α) sin (ξ − β).(6)Then ψ = 0 over the ellipse η = α, and the hyperbola ξ = β, so that these may be taken as fixed boundaries; and ψ is a constant on a C4.Over any ellipse η, moving with components U and V of velocity,ψ′ = ψ + Uy − Vx = [ m sh (η − α) cos β + Uc sh η ] sin ξ- [ m sh (η − α) sin β + Vc ch η ] cos ξ;(7)so that ψ′ = 0, ifU = −msh (η − α)cos β, V = −msh (η − α)sin β,csh ηcch η(8)having a resultant in the direction PO, where P is the intersection of an ellipse η with the hyperbola β; and with this velocity the ellipse η can be swimming in the liquid, without distortion for an instant.At infinityU = −me−acos β = −mcos β,ca − bV = −me−asin β = −msin β,ca − b(9)a and b denoting the semi-axes of the ellipse α; so that the liquid is streaming at infinity with velocity Q = m/(a + b) in the direction of the asymptote of the hyperbola β.An ellipse interior to η = α will move in a direction opposite to the exterior current; and when η = 0, U = ∞, but V = (m/c) sh α sin β.Negative values of η must be interpreted by a streaming motion on a parallel plane at a level slightly different, as on a double Riemann sheet, the stream passing from one sheet to the other across a cut SS′ joining the foci S, S′. A diagram has been drawn by Col. R. L. Hippisley.The components of the liquid velocity q, in the direction of the normal of the ellipse η and hyperbola ξ, are−mJ−1sh (η − α) cos (ξ − β), mJ−1ch (η − α) sin (ξ − β).(10)The velocity q is zero in a corner where the hyperbola β cuts the ellipse α; and round the ellipse α the velocity q reaches a maximum when the tangent has turned through a right angle, and thenq = Qea√(ch 2α − cos 2β);sh 2α(11)and the condition can be inferred when cavitation begins.With β = 0, the stream is parallel to x0, andφ = m ch (η − α) cos ξ= −Uc ch (η − α) sh η cos ξ/sh (η − α)(12)over the cylinder η, and as in (12) § 29,φ1= −Ux = −Uc ch η cos ξ,(13)for liquid filling the cylinder; andφ=th η,φ1th (η − α)(14)over the surface of η; so that parallel to Ox, the effective inertia of the cylinder η, displacing M′ liquid, is increased by M′th η/th(η- α), reducing when α = ∞ to M′ th η = M′ (b/a).Similarly, parallel to Oy, the increase of effective inertia is M′/th η th (η − α), reducing to M′/th η = M′ (a/b), when α = ∞, and the liquid extends to infinity.32. Next consider the motion given byφ = m ch 2 (η − α) sin 2ξ, ψ = −m sh 2 (η − α) cos 2ξ;(1)in which ψ = 0 over the ellipse α, andψ′ = ψ + ½R (x2+ y2) = [ −m sh 2 (η − α) + ¼Rc2] cos 2ξ + ¼Rc2ch 2η,(2)which is constant over the ellipse η if¼ Rc2= m sh 2 (η − α);(3)so that this ellipse can be rotating with this angular velocity R for an instant without distortion, the ellipse α being fixed.For the liquid filling the interior of a rotating elliptic cylinder of cross sectionx2/a2+ y2/b2= 1,(4)ψ1′ = m1(x2/a2+ y2/b2)(5)with∇2ψ1′ = −2R = −2m1(1/a2+ 1/b2),ψ1= m1(x2/a2+ y2/b2) − ½R (x2+ y2)= −½R (x2− y2) (a2− b2) / (a2+ b2),(6)φ1= Rxy (a2− b2) / (a2+ b2),w1= φ1+ ψ1i = −½iR (x + yi)2(a2− b2) / (a2+ b2).The velocity of a liquid particle is thus (a2− b2)/(a2+ b2) of what it would be if the liquid was frozen and rotating bodily with the ellipse; and so the effective angular inertia of the liquid is (a2− b2)2/(a2+ b2)2of the solid; and the effective radius of gyration, solid and liquid, is given byk2= ¼(a2+ b2), and ¼(a2− b2)2/ (a2+ b2).(7)For the liquid in the interspace between α and η,φ=m ch 2 (η − α) sin 2ξφ1¼ Rc2sh 2η sin 2ξ (a2− b2) / (a2+ b2)= 1/th 2 (η − α) th 2η;(8)and the effective k2of the liquid is reduced to¼ c2/th 2 (η − α) sh 2η,(9)which becomes ¼ c2/sh 2η =1⁄8(a2− b2)/ab, when α = ∞, and the liquid surrounds the ellipse η to infinity.An angular velocity R, which gives components −Ry, Rx of velocity to a body, can be resolved into two shearing velocities, −R parallel to Ox, and R parallel to Oy; and then ψ is resolved into ψ1+ ψ2, such that ψ1+ ½Rx2and ψ2+ ½Ry2is constant over the boundary.Inside a cylinderφ1+ ψ1i = −½ iR (x + yi)2a2/ (a2+ b2),(10)φ2+ ψ2i = ½ iR (x + yi)2b2/ (a2+ b2),(11)and for the interspace, the ellipse α being fixed, and α1revolving with angular velocity Rφ1+ ψ1i = −1⁄8iRc2sh 2 (η − α + ξi) (ch 2α + 1) / sh 2 (α1− α),(12)φ2+ ψ2i =1⁄8iRc2sh 2 (η − α + ξi) (ch 2α − 1) / sh 2 (α1− α),(13)satisfying the condition that ψ1and ψ2are zero over η = α, and over η = α1ψ1+ ½ Rx2=1⁄8Rc2(ch 2α1+ 1),(14)ψ2+ ½ Ry2=1⁄8Rc2(ch 2α1− 1),(15)constant values.In a similar way the more general state of motion may be analysed, given byw = m ch 2 (ζ − γ), γ = α + βi,(16)as giving a homogeneous strain velocity to the confocal system; to which may be added a circulation, represented by an additional term mζ in w.Similarly, withx + yi = c√[ sin (ξ + ηi) ](17)the functionψ = Qc sh ½ (η − α) sin ½ (ξ − β)(18)will give motion streaming past the fixed cylinder η = α, and dividing along ξ = β; and thenx2− y2= c2sin ξ ch η, 2xy = c2cos ξ sh η.(19)In particular, with sh α = 1, the cross-section of η = α isx4+ 6x2y2+ y4= 2c4, or x4+ y4= c4(20)when the axes are turned through 45°.33.Example 3.—Analysing in this way the rotation of a rectangle filled with liquid into the two components of shear, the stream function ψ1is to be made to satisfy the conditions(i.) ∇2ψ1= 0,(ii.) ψ1+ ½Rx2= ½Ra2, or ψ1= 0 when x = ±a,(iii.) ψ1+ ½Rx2= ½Ra2, ψ1= ½R (a2− x2), when y = ± b.Expanded in a Fourier series,a2− x2=32a2Σcos (2n + 1) ½ πx/a,π3(2n + 1)3(1)so thatψ1= R16a2Σcos (2n + 1) ½πx/a · ch (2n + 1) ½πy/a),π3(2n + 1)3· ch (2n + 1) ½πb/aw1= φ1+ ψ1i = iR16a2Σcos (2n + 1) ½πz/a,π3(2n + 1)3ch (2n + 1) ½πb/a(2)an elliptic-function Fourier series; with a similar expression for ψ2with x and y, a and b interchanged; and thence ψ = ψ1+ ψ2.Example 4.—Parabolic cylinder, axial advance, and liquid streaming past.The polar equation of the cross-section beingr1/2cos ½θ = a1/2, or r + x = 2a,(3)the conditions are satisfied byψ′ = Ur sin θ − 2Ua1/2r1/2sin ½θ = 2Ur1/2sin ½θ (r1/2cos ½θ − a1/2),(4)ψ = 2Ua1/2r1/2sin ½θ = −U √ [ 2a (r − x) ],(5)w = −2Ua1/2z1/2,(6)and the resistance of the liquid is 2πρaV2/2g.A relative stream line, along which ψ′ = Uc, is the quartic curvey − c = √ [ 2a (r − x) ], x =(4a2y2− (y − c)4, r =4a2y2+ (y − c)4,4a (y − c)24a (y − c)2(7)and in the absolute space curve given by ψ,dy= −(y − c)2, x =2ac− 2a log (y − c).dx2ayy − c(8)34.Motion symmetrical about an Axis.—When the motion of a liquid is the same for any plane passing through Ox, and lies in the plane, a function ψ can be found analogous to that employed in plane motion, such that the flux across the surface generated by the revolution of any curve AP from A to P is the same, and represented by 2π (ψ − ψ0); and, as before, if dψ is the increase in ψ due to a displacement of P to P′, then k the component of velocity normal to the surface swept out by PP′ is such that 2πdψ = 2πyk·PP′; and taking PP′ parallel to Oy and Ox,u = −dψ/ydy, v = dψ/ydx,(1)and ψ is called after the inventor, “Stokes’s stream or current function,” as it is constant along a stream line (Trans. Camb. Phil. Soc., 1842; “Stokes’s Current Function,” R. A. Sampson,Phil. Trans., 1892); and dψ/yds is the component velocity across ds in a direction turned through a right angle forward.In this symmetrical motionξ = 0, η = 0, 2ζ =d(1dψ)+d(1dψ)dxydxdyydy=1(d2ψ+d2ψ−1dψ)= −1∇2ψ,ydx2dy2ydyy(2)suppose; and in steady motion,dH+1dψ∇2ψ = 0,dH+1dψ∇2ψ = 0,dxy2dxdyy2dy(3)so that2ζ/y = −y−2∇2ψ = dH/dψ(4)is a function of ψ, say ƒ′(ψ), and constant along a stream line;dH/dv = 2qζ, H − ƒ(ψ) = constant,(5)throughout the liquid.When the motion is irrotational,ζ = 0, u = −dφ= −1dψ, v = −dφ=1dψ,dxydydyydx(6)∇2ψ = 0, ord2ψ+d2ψ−1dψ= 0.dx2dy2ydy(7)Changing to polar coordinates, x = r cos θ, y = r sin θ, the equation (2) becomes, with cos θ = μ,r2d2ψ+ (1 − μ2)d2ψ= 2 ζr3sin θ,dr2dμ(8)of which a solution, when ζ = 0, isψ =(Arn+1+B)(1 − μ2)dPn=(Arn − 1+B)y2dPn,rndμrn+2dμ(9)φ = { (n + 1) Arn− nBr−n−1} Pn,(10)where Pndenotes the zonal harmonic of the nth order; also, in the exceptional case ofψ = A0cos θ, φ = A0/r;ψ = B0r, φ = −B0log tan ½θ = −½B0sh−1x/y.(11)Thus cos θ is the Stokes’ function of a point source at O, and PA − PB of a line source AB.The stream function ψ of the liquid motion set up by the passage of a solid of revolution, moving with axial velocity U, is such that1dψ= −Udy, ψ + ½Uy2= constant,ydsds(12)over the surface of the solid; and ψ must be replaced by ψ′ = ψ + ½Uy2in the general equations of steady motion above to obtain the steady relative motion of the liquid past the solid.For instance, with n = 1 in equation (9), the relative stream function is obtained for a sphere of radius a, by making itψ′ = ψ + ½Uy2= ½U (r2− a3/r) sin2θ, ψ = −½Ua3sin2θ/r;(13)and thenφ′ = Ux (1 + ½a3/r2), φ = ½Ua3cos θ/r2,(14)−dφ= Ua3cos θ, −dφ= ½Ua3sin θ,drr3r dθr3(15)so that, if the direction of motion makes an angle ψ with Ox,tan (ψ − θ) = ½ tan θ, tan ψ = 3 tan θ/(2 − tan2θ),(16)Along the path of a liquid particle ψ′ is constant, and putting it equal to ½Uc2,(r2− a3/r) sin2θ = c2, sin2θ = c2r / (r3− a3),(17)the polar equation; ory2= c2r3/ (r3− a3), r3= a3y2/ (y2− c2),(18)a curve of the 10th degree (C10).In the absolute path in spacecos ψ = (2 − 3 sin2θ) / √ (4 − sin2θ), and sin3θ = (y3− c2y) / a3,(19)which leads to no simple relation.The velocity past the surface of the sphere is1dψ′= ½U(2r +a3)sin2θ=3⁄2U sin θ, when r = a;r sin θdrr2r sin θ(20)so that the loss of head is(9⁄4sin2θ − 1) U2/2g, having a maximum5⁄4U2/2g,(21)which must be less than the head at infinite distance to avoid cavitation at the surface of the sphere.With n = 2, a state of motion is given byψ = −½ Uy2a4μ/r4, ψ′ = ½ Uy2(1 − a4μ/r4),(22)φ′ = Ux + φ, φ = −1⁄3U (a4/ r3) P2, P2=3⁄2μ2− ½,(23)representing a stream past the surface r4= a4μ.
To determine the component acceleration of a particle, suppose F to denote any function of x, y, z, t, and investigate the time rate of F for a moving particle; denoting the change by DF/dt,
(1)
and D/dt is called particle differentiation, because it follows the rate of change of a particle as it leaves the point x, y, z; but
dF/dt, dF/dx, dF/dy, dF/dz
(2)
represent the rate of change of F at the time t, at the point, x, y, z, fixed in space.
The components of acceleration of a particle of fluid are consequently
(3)
(4)
(5)
leading to the equations of motion above.
If F (x, y, z, t) = 0 represents the equation of a surface containing always the same particles of fluid,
(6)
which is called the differential equation of thebounding surface. A bounding surface is such that there is no flow of fluid across it, as expressed by equation (6). The surface always contains the same fluid inside it, and condition (6) is satisfied over the complete surface, as well as any part of it.
But turbulence in the motion will vitiate the principle that a bounding surface will always consist of the same fluid particles, as we see on the surface of turbulent water.
24. To integrate the equations of motion, suppose the impressed force is due to a potential V, such that the force in any direction is the rate of diminution of V, or its downward gradient; and then
X = −dV/dx, Y = −dV/dy, Z = −dV/dz;
(1)
and putting
(2)
(3)
the equations of motion may be written
(4)
(5)
(6)
where
H = ∫ dp/ρ + V + ½q2,
(7)
q2= u2+ v2+ w2,
(8)
and the three terms in H may be called the pressure head, potential head, and head of velocity, when the gravitation unit is employed and ½q2is replaced by ½q2/g.
Eliminating H between (5) and (6)
(9)
and combining this with the equation of continuity
(10)
we have
(11)
with two similar equations.
Putting
ω2= ξ2+ η2+ ζ2,
(12)
avortex lineis defined to be such that the tangent is in the direction of ω, the resultant of ξ, η, ζ, called the components of molecular rotation. A small sphere of the fluid, if frozen suddenly, would retain this angular velocity.
If ω vanishes throughout the fluid at any instant, equation (11) shows that it will always be zero, and the fluid motion is then called irrotational; and a function φ exists, called thevelocity function, such that
u dx + v dy + w dz = −dφ,
(13)
and then the velocity in any direction is the space-decrease or downward gradient of φ.
25. But in the most general case it is possible to have three functions φ, ψ, m of x, y, z, such that
u dx + v dy + w dz = −dφ − m dψ,
(1)
as A. Clebsch has shown, from purely analytical considerations (Crelle, lvi.); and then
(2)
and
(3)
so that, at any instant, the surfaces over which ψ and m are constant intersect in the vortex lines.
Putting
(4)
the equations of motion (4), (5), (6) § 24 can be written
(5)
and therefore
(6)
Equation (5) becomes, by a rearrangement,
(7)
(8)
and as we prove subsequently (§ 37) that the vortex lines are composed of the same fluid particles throughout the motion, the surface m and ψ satisfies the condition of (6) § 23; so that K is uniform throughout the fluid at any instant, and changes with the time only, and so may be replaced by F(t).
26. When the motion issteady, that is, when the velocity at any point of space does not change with the time,
(1)
(2)
and
K = ∫ dp/ρ + V + ½q2= H
(3)
is constant along a vortex line, and astream line, the path of a fluid particle, so that the fluid is traversed by a series of H surfaces, each covered by a network of stream lines and vortex lines; and if the motion is irrotational H is a constant throughout the fluid.
Taking the axis of x for an instant in the normal through a point on the surface H = constant, this makes u = 0, ξ = 0; and in steady motion the equations reduce to
dH/dν = 2vζ − 2wη = 2qω sin θ,
(4)
where θ is the angle between the stream line and vortex line; and this holds for their projection on any plane to which dν is drawn perpendicular.
In plane motion (4) reduces to
(5)
if r denotes the radius of curvature of the stream line, so that
(6)
the normal acceleration.
The osculating plane of a stream line in steady motion contains the resultant acceleration, the direction ratios of which are
(7)
and when q is stationary, the acceleration is normal to the surface H = constant, and the stream line is a geodesic.
Calling the sum of the pressure and potential head the statical head, surfaces of constant statical and dynamical head intersect in lines on H, and the three surfaces touch where the velocity is stationary.
Equation (3) is called Bernoulli’s equation, and may be interpreted as the balance-sheet of the energy which enters and leaves a given tube of flow.
If homogeneous liquid is drawn off from a vessel so large that the motion at the free surface at a distance may be neglected, then Bernoulli’s equation may be written
H = p/ρ + z + q2/2g = P/ρ + h,
(8)
where P denotes the atmospheric pressure and h the height of the free surface, a fundamental equation in hydraulics; a return has been made here to the gravitation unit of hydrostatics, and Oz is taken vertically upward.
In particular, for a jet issuing into the atmosphere, where p = P,
q2/2g = h − z,
(9)
or the velocity of the jet is due to the head k − z of the still free surface above the orifice; this is Torricelli’s theorem (1643), the foundation of the science of hydrodynamics.
27.Uniplanar Motion.—In the uniplanar motion of a homogeneous liquid the equation of continuity reduces to
(1)
so that we can put
u = −dψ/dy, v = dψ/dx,
(2)
where ψ is a function of x, y, called the stream- or current-function; interpreted physically, ψ − ψ0, the difference of the value of ψ at a fixed point A and a variable point P is the flow, in ft.3/second, across any curved line AP from A to P, this being the same for all lines in accordance with the continuity.
Thus if dψ is the increase of ψ due to a displacement from P to P′, and k is the component of velocity normal to PP′, the flow across PP′ is dψ = k·PP′; and taking PP′ parallel to Ox, dψ = v dx; and similarly dψ= −u dy with PP′ parallel to Oy; and generally dψ/ds is the velocity across ds, in a direction turned through a right angle forward, against the clock.
In the equations of uniplanar motion
(3)
so that in steady motion
(4)
and ∇2ψ must be a function of ψ.
If the motion ia irrotational,
(5)
so that ψ and φ are conjugate functions of x and y,
φ + ψi = ƒ(x + yi), ∇2ψ = 0, ∇2φ = 0;
(6)
or putting
φ + ψi = w, x + yi = z, w = ƒ(z).
The curves φ = constant and ψ = constant form an orthogonal system; and the interchange of φ and ψ will give a new state of uniplanar motion, in which the velocity at every point is turned through a right angle without alteration of magnitude.
For instance, in a uniplanar flow, radially inward towards O, the flow across any circle of radius r being the same and denoted by 2πm, the velocity must be m/r, and
φ = m log r, ψ = mθ, φ + ψi = m log reiθ, w = m log z.
(7)
Interchanging these values
ψ = m log r, φ = mθ, ψ + φi = m log reiθ
(8)
gives a state of vortex motion, circulating round Oz, called a straight or columnar vortex.
A single vortex will remain at rest, and cause a velocity at any point inversely as the distance from the axis and perpendicular to its direction; analogous to the magnetic field of a straight electric current.
If other vortices are present, any one may be supposed to move with the velocity due to the others, the resultant stream-function being
ψ = Σm log r = log Πrm;
(9)
the path of a vortex is obtained by equating the value of ψ at the vortex to a constant, omitting the rmof the vortex itself.
When the liquid is bounded by a cylindrical surface, the motion of a vortex inside may be determined as due to a series of vortex-images, so arranged as to make the flow zero across the boundary.
For a plane boundary the image is the optical reflection of the vortex. For example, a pair of equal opposite vortices, moving on a line parallel to a plane boundary, will have a corresponding pair of images, forming a rectangle of vortices, and the path of a vortex will be the Cotes’ spiral
r sin 2θ = 2a, or x−2+ y−2= a−2;
(10)
this is therefore the path of a single vortex in a right-angled corner; and generally, if the angle of the corner is π/n, the path is the Cotes’ spiral
r sin nθ = na.
(11)
A single vortex in a circular cylinder of radius a at a distance c from the centre will move with the velocity due to an equal opposite image at a distance a2/c, and so describe a circle with velocity
mc/(a2− c2) in the periodic time 2π (a2− c2)/m.
(12)
Conjugate functions can be employed also for the motion of liquid in a thin sheet between two concentric spherical surfaces; the components of velocity along the meridian and parallel in colatitude θ and longitude λ can be written
(13)
and then
φ + ψi = F (tan ½θ·eλi).
(14)
28.Uniplanar Motion of a Liquid due to the Passage of a Cylinder through it.—A stream-function ψ must be determined to satisfy the conditions
∇2ψ = 0, throughout the liquid;
(1)
ψ = constant, over any fixed boundary;
(2)
dψ/ds = normal velocity reversed over a solid boundary,
(3)
so that, if the solid is moving with velocity U in the direction Ox, dψ/ds = −U dy/ds, or ψ + Uy = constant over the moving cylinder; and ψ + Uy = ψ′ is the stream function of the relative motion of the liquid past the cylinder, and similarly ψ − Vx for the component velocity V along Oy; and generally
ψ′ = ψ + Uy − Vx
(4)
is the relative stream-function, constant over a solid boundary moving with components U and V of velocity.
If the liquid is stirred up by the rotation R of a cylindrical body,
dψ/ds = normal velocity reversed
(5)
ψ + ½R (x2+ y2) = ψ′,
(6)
a constant over the boundary; and ψ′ is the current-function of the relative motion past the cylinder, but now
V2ψ′ + 2R = 0,
(7)
throughout the liquid.
Inside an equilateral triangle, for instance, of height h,
ψ′ = −2Rαβγ/h,
(8)
where α, β, γ are the perpendiculars on the sides of the triangle.
In the general case ψ′ = ψ + Uy − Vx + ½R (x2+ y2) is the relative stream function for velocity components, U, V, R.
29.Example 1.—Liquid motion past a circular cylinder.
Consider the motion given by
ω = U (z + a2/z),
(1)
so that
(2)
(2)
Then ψ = 0 over the cylinder r = a, which may be considered a fixed post; and a stream line past it along which ψ = Uc, a constant, is the curve
(3)
a cubic curve (C3).
Over a concentric cylinder, external or internal, of radius r = b,
(4)
and ψ′ is zero if
U1/U = (a2− b2)/b2;
(5)
so that the cylinder may swim for an instant in the liquid without distortion, with this velocity U1, and ω in (1) will give the liquid motion in the interspace between the fixed cylinder r = a and the concentric cylinder r = b, moving with velocity U1.
When b = 0, U1= ∞; and when b = ∞, U1= −U, so that at infinity the liquid is streaming in the direction xO with velocity U.
If the liquid is reduced to rest at infinity by the superposition of an opposite stream given by ω = −Uz, we are left with
ω = Ua2/z,
(6)
φ = U (a2/r) cos θ = Ua2x/(x2+ y2),
(7)
ψ = −U (a2/r) sin θ = −Ua2y/(x2+ y2),
(8)
giving the motion due to the passage of the Cylinder r = a with velocity U through the origin O in the direction Ox.
If the direction of motion makes an angle θ′ with Ox,
(9)
and the velocity is Ua2/r2.
Along the path of a particle, defined by the C3of (3),
(10)
(11)
on the radius of curvature is ¼a2/(y − ½c), which shows that the curve is an Elastica or Lintearia. (J. C. Maxwell,Collected Works, ii. 208.)
If φ1denotes the velocity function of the liquid filling the cylinder r = b, and moving bodily with it with velocity U1,
φ1= −U1x,
(12)
and over the separating surface r = b
(13)
and this, by § 36, is also the ratio of the kinetic energy in the annular interspace between the two cylinders to the kinetic energy of the liquid moving bodily inside r = b.
Consequently the inertia to overcome in moving the cylinder r = b, solid or liquid, is its own inertia, increased by the inertia of liquid (a2+ b2)/(a2~ b2) times the volume of the cylinder r = b; this total inertia is called the effective inertia of the cylinder r = b, at the instant the two cylinders are concentric.
With liquid of density ρ, this gives rise to a kinetic reaction to acceleration dU/dt, given by
(14)
if M′ denotes the mass of liquid displaced by unit length of the cylinder r = b. In particular, when a = ∞, the extra inertia is M′.
When the cylinder r = a is moved with velocity U and r = b with velocity U1along Ox,
(15)
(16)
and similarly, with velocity components V and V1along Oy
(17)
(18)
and then for the resultant motion
(19)
The resultant impulse of the liquid on the cylinder is given by the component, over r = a (§ 36),
(20)
and over r = b
(21)
and the difference X − X1is the component momentum of the liquid in the interspace; with similar expressions for Y and Y1.
Then, if the outside cylinder is free to move
(22)
But if the outside cylinder is moved with velocity U1, and the inside cylinder is solid or filled with liquid of density σ,
(23)
and the inside cylinder starts forward or backward with respect to the outside cylinder, according as ρ > or < σ.
30. The expression for ω in (1) § 29 may be increased by the addition of the term
im log z = −mθ + im log r,
(1)
representing vortex motion circulating round the annulus of liquid.
Considered by itself, with the cylinders held fixed, the vortex sets up a circumferential velocity m/r on a radius r, so that the angular momentum of a circular filament of annular cross section dA is ρm dA, and of the whole vortex is ρmπ (b2− a2).
Any circular filament can be started from rest by the application of a circumferential impulse πρm dr at each end of a diameter; so that a mechanism attached to the cylinders, which can set up a uniform distributed impulse πρm across the two parts of a diameter in the liquid, will generate the vortex motion, and react on the cylinder with an impulse couple −ρmπa2and ρmπb2, having resultant ρmπ (b2− a2), and this couple is infinite when b = ∞, as the angular momentum of the vortex is infinite. Round the cylinder r = a held fixed in the U current the liquid streams past with velocity
q′ = 2U sin θ + m/a;
(2)
and the loss of head due to this increase of velocity from U to q′ is
(3)
so that cavitation will take place, unless the head at a great distance exceeds this loss.
The resultant hydrostatic thrust across any diametral plane of the cylinder will be modified, but the only term in the loss of head which exerts a resultant thrust on the whole cylinder is 2mU sin θ/ga, and its thrust is 2πρmU absolute units in the direction Cy, to be counteracted by a support at the centre C; the liquid is streaming past r = a with velocity U reversed, and the cylinder is surrounded by a vortex. Similarly, the streaming velocity V reversed will give rise to a thrust 2πρmV in the direction xC.
Now if the cylinder is released, and the components U and V are reversed so as to become the velocity of the cylinder with respect to space filled with liquid, and at rest at infinity, the cylinder will experience components of force per unit length
(i.) − 2πρmV, 2πρmU, due to the vortex motion;
(ii.) − πρa2dU/dt, − πρa2dV/dt, due to the kinetic reaction of the liquid;
(iii.) 0, −π(σ − ρ) a2g, due to gravity,
taking Oy vertically upward, and denoting the density of the cylinder by σ; so that the equations of motion are
(4)
(5)
or, putting m = a2ω, so that the vortex velocity is due to an angular velocity ω at a radius a,
(σ + ρ) dU/dt + 2ρωV = 0,
(6)
(σ + ρ) dV/dt − 2ρωU + (σ-ρ) g = 0.
(7)
Thus with g = 0, the cylinder will describe a circle with angular velocity 2ρω/(σ + ρ), so that the radius is (σ + ρ) v/2ρω, if the velocity is v. With σ = 0, the angular velocity of the cylinder is 2ω; in this way the velocity may be calculated of the propagation of ripples and waves on the surface of a vertical whirlpool in a sink.
Restoring σ will make the path of the cylinder a trochoid; and so the swerve can be explained of the ball in tennis, cricket, baseball, or golf.
Another explanation may be given of the sidelong force, arising from the velocity of liquid past a cylinder, which is encircled by a vortex. Taking two planes x = ± b, and considering the increase of momentum in the liquid between them, due to the entry and exit of liquid momentum, the increase across dy in the direction Oy, due to elements at P and P′ at opposite ends of the diameter PP′, is
ρ dy (U − Ua2r−2cos 2θ + mr−1sin θ) (Ua2r−2sin 2θ + mr−1cos θ)+ ρ dy ( −U + Ua2r−2cos 2θ + mr−1sin θ) (Ua2r−2sin 2θ − mr−1cos θ)= 2ρdymUr−1(cos θ − a2r−2cos 3θ),
ρ dy (U − Ua2r−2cos 2θ + mr−1sin θ) (Ua2r−2sin 2θ + mr−1cos θ)
+ ρ dy ( −U + Ua2r−2cos 2θ + mr−1sin θ) (Ua2r−2sin 2θ − mr−1cos θ)
= 2ρdymUr−1(cos θ − a2r−2cos 3θ),
(8)
and with y = b tan θ, r = b sec θ, this is
2ρmU dθ (1 − a2b−2cos 3θ cos θ),
(9)
and integrating between the limits θ = ±½π, the resultant, as before, is 2πρmU.
31.Example 2.—Confocal Elliptic Cylinders.—Employ the elliptic coordinates η, ξ, and ζ = η + ξi, such that
z = c ch ζ, x = c ch η cos ξ, y = c sh η sin ζ;
(1)
then the curves for which η and ξ are constant are confocal ellipses and hyperbolas, and
= (1/2)c2(ch 2η − cos 2ξ) = r1r2= OD2,
(2)
if OD is the semi-diameter conjugate to OP, and r1, r2the focal distances,
r1, r2= c (ch η ± cos ξ);
(3)
r2= x2+ y2= c2(ch2η − sin2ξ)
= ½c2(ch 2η + cos 2ξ).
(4)
Consider the streaming motion given by
w = m ch (ζ − γ), γ = α + βi,
(5)
φ = m ch (η − α) cos (ξ − β), ψ = m sh (η − α) sin (ξ − β).
(6)
Then ψ = 0 over the ellipse η = α, and the hyperbola ξ = β, so that these may be taken as fixed boundaries; and ψ is a constant on a C4.
Over any ellipse η, moving with components U and V of velocity,
ψ′ = ψ + Uy − Vx = [ m sh (η − α) cos β + Uc sh η ] sin ξ- [ m sh (η − α) sin β + Vc ch η ] cos ξ;
(7)
so that ψ′ = 0, if
(8)
having a resultant in the direction PO, where P is the intersection of an ellipse η with the hyperbola β; and with this velocity the ellipse η can be swimming in the liquid, without distortion for an instant.
At infinity
(9)
a and b denoting the semi-axes of the ellipse α; so that the liquid is streaming at infinity with velocity Q = m/(a + b) in the direction of the asymptote of the hyperbola β.
An ellipse interior to η = α will move in a direction opposite to the exterior current; and when η = 0, U = ∞, but V = (m/c) sh α sin β.
Negative values of η must be interpreted by a streaming motion on a parallel plane at a level slightly different, as on a double Riemann sheet, the stream passing from one sheet to the other across a cut SS′ joining the foci S, S′. A diagram has been drawn by Col. R. L. Hippisley.
The components of the liquid velocity q, in the direction of the normal of the ellipse η and hyperbola ξ, are
−mJ−1sh (η − α) cos (ξ − β), mJ−1ch (η − α) sin (ξ − β).
(10)
The velocity q is zero in a corner where the hyperbola β cuts the ellipse α; and round the ellipse α the velocity q reaches a maximum when the tangent has turned through a right angle, and then
(11)
and the condition can be inferred when cavitation begins.
With β = 0, the stream is parallel to x0, and
φ = m ch (η − α) cos ξ= −Uc ch (η − α) sh η cos ξ/sh (η − α)
(12)
over the cylinder η, and as in (12) § 29,
φ1= −Ux = −Uc ch η cos ξ,
(13)
for liquid filling the cylinder; and
(14)
over the surface of η; so that parallel to Ox, the effective inertia of the cylinder η, displacing M′ liquid, is increased by M′th η/th(η- α), reducing when α = ∞ to M′ th η = M′ (b/a).
Similarly, parallel to Oy, the increase of effective inertia is M′/th η th (η − α), reducing to M′/th η = M′ (a/b), when α = ∞, and the liquid extends to infinity.
32. Next consider the motion given by
φ = m ch 2 (η − α) sin 2ξ, ψ = −m sh 2 (η − α) cos 2ξ;
(1)
in which ψ = 0 over the ellipse α, and
ψ′ = ψ + ½R (x2+ y2) = [ −m sh 2 (η − α) + ¼Rc2] cos 2ξ + ¼Rc2ch 2η,
(2)
which is constant over the ellipse η if
¼ Rc2= m sh 2 (η − α);
(3)
so that this ellipse can be rotating with this angular velocity R for an instant without distortion, the ellipse α being fixed.
For the liquid filling the interior of a rotating elliptic cylinder of cross section
x2/a2+ y2/b2= 1,
(4)
ψ1′ = m1(x2/a2+ y2/b2)
(5)
with
∇2ψ1′ = −2R = −2m1(1/a2+ 1/b2),
ψ1= m1(x2/a2+ y2/b2) − ½R (x2+ y2)= −½R (x2− y2) (a2− b2) / (a2+ b2),
(6)
φ1= Rxy (a2− b2) / (a2+ b2),
w1= φ1+ ψ1i = −½iR (x + yi)2(a2− b2) / (a2+ b2).
The velocity of a liquid particle is thus (a2− b2)/(a2+ b2) of what it would be if the liquid was frozen and rotating bodily with the ellipse; and so the effective angular inertia of the liquid is (a2− b2)2/(a2+ b2)2of the solid; and the effective radius of gyration, solid and liquid, is given by
k2= ¼(a2+ b2), and ¼(a2− b2)2/ (a2+ b2).
(7)
For the liquid in the interspace between α and η,
= 1/th 2 (η − α) th 2η;
(8)
and the effective k2of the liquid is reduced to
¼ c2/th 2 (η − α) sh 2η,
(9)
which becomes ¼ c2/sh 2η =1⁄8(a2− b2)/ab, when α = ∞, and the liquid surrounds the ellipse η to infinity.
An angular velocity R, which gives components −Ry, Rx of velocity to a body, can be resolved into two shearing velocities, −R parallel to Ox, and R parallel to Oy; and then ψ is resolved into ψ1+ ψ2, such that ψ1+ ½Rx2and ψ2+ ½Ry2is constant over the boundary.
Inside a cylinder
φ1+ ψ1i = −½ iR (x + yi)2a2/ (a2+ b2),
(10)
φ2+ ψ2i = ½ iR (x + yi)2b2/ (a2+ b2),
(11)
and for the interspace, the ellipse α being fixed, and α1revolving with angular velocity R
φ1+ ψ1i = −1⁄8iRc2sh 2 (η − α + ξi) (ch 2α + 1) / sh 2 (α1− α),
(12)
φ2+ ψ2i =1⁄8iRc2sh 2 (η − α + ξi) (ch 2α − 1) / sh 2 (α1− α),
(13)
satisfying the condition that ψ1and ψ2are zero over η = α, and over η = α1
ψ1+ ½ Rx2=1⁄8Rc2(ch 2α1+ 1),
(14)
ψ2+ ½ Ry2=1⁄8Rc2(ch 2α1− 1),
(15)
constant values.
In a similar way the more general state of motion may be analysed, given by
w = m ch 2 (ζ − γ), γ = α + βi,
(16)
as giving a homogeneous strain velocity to the confocal system; to which may be added a circulation, represented by an additional term mζ in w.
Similarly, with
x + yi = c√[ sin (ξ + ηi) ]
(17)
the function
ψ = Qc sh ½ (η − α) sin ½ (ξ − β)
(18)
will give motion streaming past the fixed cylinder η = α, and dividing along ξ = β; and then
x2− y2= c2sin ξ ch η, 2xy = c2cos ξ sh η.
(19)
In particular, with sh α = 1, the cross-section of η = α is
x4+ 6x2y2+ y4= 2c4, or x4+ y4= c4
(20)
when the axes are turned through 45°.
33.Example 3.—Analysing in this way the rotation of a rectangle filled with liquid into the two components of shear, the stream function ψ1is to be made to satisfy the conditions
(i.) ∇2ψ1= 0,(ii.) ψ1+ ½Rx2= ½Ra2, or ψ1= 0 when x = ±a,(iii.) ψ1+ ½Rx2= ½Ra2, ψ1= ½R (a2− x2), when y = ± b.
(i.) ∇2ψ1= 0,
(ii.) ψ1+ ½Rx2= ½Ra2, or ψ1= 0 when x = ±a,
(iii.) ψ1+ ½Rx2= ½Ra2, ψ1= ½R (a2− x2), when y = ± b.
Expanded in a Fourier series,
(1)
so that
(2)
an elliptic-function Fourier series; with a similar expression for ψ2with x and y, a and b interchanged; and thence ψ = ψ1+ ψ2.
Example 4.—Parabolic cylinder, axial advance, and liquid streaming past.
The polar equation of the cross-section being
r1/2cos ½θ = a1/2, or r + x = 2a,
(3)
the conditions are satisfied by
ψ′ = Ur sin θ − 2Ua1/2r1/2sin ½θ = 2Ur1/2sin ½θ (r1/2cos ½θ − a1/2),
(4)
ψ = 2Ua1/2r1/2sin ½θ = −U √ [ 2a (r − x) ],
(5)
w = −2Ua1/2z1/2,
(6)
and the resistance of the liquid is 2πρaV2/2g.
A relative stream line, along which ψ′ = Uc, is the quartic curve
(7)
and in the absolute space curve given by ψ,
(8)
34.Motion symmetrical about an Axis.—When the motion of a liquid is the same for any plane passing through Ox, and lies in the plane, a function ψ can be found analogous to that employed in plane motion, such that the flux across the surface generated by the revolution of any curve AP from A to P is the same, and represented by 2π (ψ − ψ0); and, as before, if dψ is the increase in ψ due to a displacement of P to P′, then k the component of velocity normal to the surface swept out by PP′ is such that 2πdψ = 2πyk·PP′; and taking PP′ parallel to Oy and Ox,
u = −dψ/ydy, v = dψ/ydx,
(1)
and ψ is called after the inventor, “Stokes’s stream or current function,” as it is constant along a stream line (Trans. Camb. Phil. Soc., 1842; “Stokes’s Current Function,” R. A. Sampson,Phil. Trans., 1892); and dψ/yds is the component velocity across ds in a direction turned through a right angle forward.
In this symmetrical motion
(2)
suppose; and in steady motion,
(3)
so that
2ζ/y = −y−2∇2ψ = dH/dψ
(4)
is a function of ψ, say ƒ′(ψ), and constant along a stream line;
dH/dv = 2qζ, H − ƒ(ψ) = constant,
(5)
throughout the liquid.
When the motion is irrotational,
(6)
(7)
Changing to polar coordinates, x = r cos θ, y = r sin θ, the equation (2) becomes, with cos θ = μ,
(8)
of which a solution, when ζ = 0, is
(9)
φ = { (n + 1) Arn− nBr−n−1} Pn,
(10)
where Pndenotes the zonal harmonic of the nth order; also, in the exceptional case of
ψ = A0cos θ, φ = A0/r;ψ = B0r, φ = −B0log tan ½θ = −½B0sh−1x/y.
ψ = A0cos θ, φ = A0/r;
ψ = B0r, φ = −B0log tan ½θ = −½B0sh−1x/y.
(11)
Thus cos θ is the Stokes’ function of a point source at O, and PA − PB of a line source AB.
The stream function ψ of the liquid motion set up by the passage of a solid of revolution, moving with axial velocity U, is such that
(12)
over the surface of the solid; and ψ must be replaced by ψ′ = ψ + ½Uy2in the general equations of steady motion above to obtain the steady relative motion of the liquid past the solid.
For instance, with n = 1 in equation (9), the relative stream function is obtained for a sphere of radius a, by making it
ψ′ = ψ + ½Uy2= ½U (r2− a3/r) sin2θ, ψ = −½Ua3sin2θ/r;
(13)
and then
φ′ = Ux (1 + ½a3/r2), φ = ½Ua3cos θ/r2,
(14)
(15)
so that, if the direction of motion makes an angle ψ with Ox,
tan (ψ − θ) = ½ tan θ, tan ψ = 3 tan θ/(2 − tan2θ),
(16)
Along the path of a liquid particle ψ′ is constant, and putting it equal to ½Uc2,
(r2− a3/r) sin2θ = c2, sin2θ = c2r / (r3− a3),
(17)
the polar equation; or
y2= c2r3/ (r3− a3), r3= a3y2/ (y2− c2),
(18)
a curve of the 10th degree (C10).
In the absolute path in space
cos ψ = (2 − 3 sin2θ) / √ (4 − sin2θ), and sin3θ = (y3− c2y) / a3,
(19)
which leads to no simple relation.
The velocity past the surface of the sphere is
(20)
so that the loss of head is
(9⁄4sin2θ − 1) U2/2g, having a maximum5⁄4U2/2g,
(21)
which must be less than the head at infinite distance to avoid cavitation at the surface of the sphere.
With n = 2, a state of motion is given by
ψ = −½ Uy2a4μ/r4, ψ′ = ½ Uy2(1 − a4μ/r4),
(22)
φ′ = Ux + φ, φ = −1⁄3U (a4/ r3) P2, P2=3⁄2μ2− ½,
(23)
representing a stream past the surface r4= a4μ.
35. A circular vortex, such as a smoke ring, will set up motion symmetrical about an axis, and provide an illustration; a half vortex ring can be generated in water by drawing a semicircular blade a short distance forward, the tip of a spoon for instance. The vortex advances with a certain velocity; and if an equal circular vortex is generated coaxially with the first, the mutual influence can be observed. The first vortex dilates and moves slower, while the second contracts and shoots through the first; after which the motion is reversed periodically, as if in a game of leap-frog. Projected perpendicularly against a plane boundary, the motion is determined by an equal opposite vortex ring, the optical image; the vortex ring spreads out and moves more slowly as it approaches the wall; at the same time the molecular rotation, inversely as the cross-section of the vortex, is seen to increase. The analytical treatment of such vortex rings is the same as for the electro-magnetic effect of a current circulating in each ring.