Bernoulli’s Equation

Chapter 6Bernoulli’s equationContents6.16.26.36.46.56.6Bernoulli’s theorem for steady flows . .Bernoulli’s theorem for potential flowsDrag force on a sphere . . . . . . . . . .Separation . . . . . . . . . . . . . . . . . .Unsteady flows . . . . . . . . . . . . . . .Acceleration of a sphere . . . . . . . . .475052545557In section (5.4) we obtained the momentum equation for ideal fluids (i.e. inviscid and withconstant density) in the form 1p u2 u ω kuk g. t2ρSo, since the constant gravity g (g · x), one has u u ω H 0, twhereH(x, t) p 1 kuk2 g · xρ 2(6.1)(6.2)is called the Bernoulli function.16.1Bernoulli’s theorem for steady flowsIn the case of steady flows, i.e. when u/ t 0, taking the scalar product of equation (6.1)with the fluid velocity, u, gives the Bernoulli equation(u · ) H 0,(6.3)since u · (u ω) 0.Hence, for an ideal fluid in steady flow,H(x) p 1 kuk2 g · xρ 2is constant along a streamline.1Not to be mistaken for Bernoulli’s polynomials.47(6.4)

486.1 Bernoulli’s theorem for steady flowsSo, if a streamfunction ψ(x) can be defined, H is a function of ψ (H(x) H(ψ)).Example 6.1 (The Venturi effect.)Consider a flow through a narrow constriction of cross-section area A2 ; upstream and downstream the cross-sectional area is A1 .(a)V1(c)h1(b)S1S3h2A1h3A1S2A2(Three narrow vertical tubes, (a), (b) and (c), are used to measure the pressure at differentpoints.)The fluid velocity is assumedZ inform on cross sections, S. Upstream the fluid velocity is V1 .ρu · n dS constant for any cross-section S, soMass conservation impliesSZρu · n dS S1Zρu · n dS S2Zu · n dS S1 A1 V 1 A2 V 2 V 2 Zu · n dS(ρ constant),S2A1V1 V1A2(since A1 A2 ).Neglecting gravity, we apply Bernoulli’s equation to any streamline, ρp1 1 2 p 2 1 2V22 V12 , V1 V2 p 2 p1 ρ2ρ222 ρV p2 p1 12 A21 A22 p1 .2A2Thus, in the constriction the speed of the flow increases (conservation of mass) and its pressuredecreases (Bernoulli’s equation).This can be measured by the thin tubes where there is fluid but no flow (i.e. fluid in hydrostaticequilibrium). If h1 is the height of fluid in the tube (a) thenp1 p0 ρgh1(p0 patm ).If h2 is the height of fluid in the tube (b) then p1 p0V2p2 p0 1 2 A21 A22 ,ρgρg2gA2 A21 A22 h1 .p2 p0 ρgh2 h2 h2 h1 V122gA22In tube (c), V3 V1 since A3 A1 (mass conservation). So, Bernoulli’s equation gives11p3 ρV32 p1 ρV12 p3 p1 ,22and so h3 h1 . (In practice, h3 will be slightly less than h1 due to viscosity but the effect issmall.)

49Chapter 6 – Bernoulli’s equationExample 6.2 (Flow down a barrel.)How fast does fluid flow out of a barrel?zA(h)gh0a ULet h be the height of fluid level in the barrel above the outlet, which has cross-sectional areaa. If a A(h), then the flow can be treated as approximately steady.dhdh aU (with U 0). So, if a A then U .Mass conservation: AdtdtBernoulli’s theorem: consider a streamline from the surface of the fluid to the outlet,1p ρkuk2 ρgz const.2dh. So,At z 0: p patm and u U ; at z h: p patm and u dt 21dh1patm ρU 2 patm ρ ρgh,22dt 2pdhdh 2gh U 2gh since U . U2 dtdtWe could have guessed this result from conservation of energy)with KE 0 and P E ρgh at z h1 ρU 2 ρgh.1 22and KE ρU and P E 0 at z 02Example 6.3 (Siphon.)A technique for removing fluid from one vessel to another without pouring is to use a siphontube.zBLA0gHC

506.2 Bernoulli’s theorem for potential flowsTo start the siphon we need to fill the tube with fluid, but once it is going, the fluid willcontinue to flow from the upper to the lower container.In order to calculate the flow rate, we can use Bernoulli’s equation along a streamline fromthe surface to the exit of the pipe.At point A: p patm , z 0. We shall assume that the container’s cross-sectional area ismuch larger than that of the pipe. So, UA 0 (from mass conservation; see example 6.2 A dh/dt aU ).At point C: p patm , z H, u Uc U .Bernoulli’s equation:ppatm 1 2patm 1 2 UA U gH U 2gH.ρ2ρ2 0If B is the highest point: (UB UC U from mass conservation)1patm 1 2pB U 2 gL U gH pB patm ρg(L H) patm .ρ2ρ2For pB 0, we need H L 6.2patm105 3 10m.ρg10 10Bernoulli’s theorem for potential flowsIn this section we shall extend Bernoulli’s theorem to the case of irrotational flows.Recall that Euler’s equation can written in the form u u ω H twhereH(x, t) p 1 kuk2 g · x.ρ 2If the fluid flow is irrotational, i.e. if ω u 0, then u ω 0 and u φ; so, theequation above becomes φ H 0, t u φ φsince . t t tThus, for irrotational flows, φ p 1 φ H k φk2 g · x f (t) t tρ 2(6.5)is a function of time, independent of the position, x.If, in addition, the flow is steady,H p 1 k φk2 g · x,ρ 2is constant; H has the same value on all streamlines.(6.6)

51Chapter 6 – Bernoulli’s equationExample 6.4 (Shape of the free surface of a fluid near a rotating rod)We consider a rod of radius a, rotating at constant angular velocity Ω, placed in a fluid.Assuming a potential, axisymmetric and planarzfluid flow, (ur (r), uθ (r)) in cylindrical polar coorΩdinates, we wish to calculate the height of the freesurface of the fluid near to the rod, h(r). We alsoassume that the solid rod is an impenetrable surface on which the fluid does not slip, so that thegh(r)boundary conditions for the velocity field areur 0and uθ aΩatr a.raFrom mass conservation, one has ·u C1 d(rur ) 0 ur (r) ,r drrwhere C is a constant of integration. However, the boundary condition ur C/a 0 at r aimplies that C 0. So, ur 0 and the fluid motion is purely azimuthal.As we assume an irrotational flow, u k1 d(ruθ ) êz 0 uθ (r) ,r drrwhere k is an integration constant to be determined using the second boundary condition. Atr a, uθ k/a aΩ which implies that k a2 Ω. So, the fluid velocity near to the rod isur 0anduθ a2 Ω.rNotice that the velocity potential, function of θ, can be determined usingu φ 1 dφa2 Ω φ(θ) a2 Ωθ.r dθrBy applying Bernoulli’s theorem for steady potential flows to the free surface (which is not astreamline, as streamlines are circles about the rod axis) we obtain,H patmpatm 1 2 uθ (r) gh(r) gh ,ρ2ρ {z} {z}at large rnear rodwhere the constant pressure p patm is the atmospheric pressure and lim h(r) h . (Notice alsor 1/r2that uθ 1/r 0 as r .)Thus, the height of the free surface is1a 4 Ω2h(r) h u2θ (r) h ,2g2gr2h(r)h (6.7)which shows that the free surface dips as 1/r2 nearto the rotating rod.Alternatively, Euler’s equation could be solved directly (i.e. without involving Bernoulli’stheorem) as in § 5.6 with an azimuthal flow, now potential, of the form uθ a2 Ω/r. We

526.3 Drag force on a spherecan then explain the result (6.7) in terms of centripetal acceleration; since the fluid particlesmove in circles, there must be an inwards central force producing the necessary centripetalacceleration (i.e. balancing the centrifugal force). Indeed, from the radial component of themomentum equation, one has ρu2θ p pa 4 Ω2 ρ 3 .r r rrHowever, since the fluid is in vertical hydrostatic equilibrium, the pressure satisfies p ρg p(r, z) patm ρg(z h(r)). zHence, we havedha 4 Ω2a 4 Ω2 p ρg ρ 3 h(r) h , rdrr2gr2as in equation (6.7).6.3Drag force on a sphereWe wish to calculate the pressure force exerted by a steady fluid flow on a solid sphere.nUrraθϕzIn § 4.5.1 we obtained the velocity potential of auniform stream, Uêz , past a stationary sphere ofradius a, a3,φ(r, z) U z 1 2(r2 z 2 )3/2in cylindrical polar coordinates (r, θ, z). In sphericalpolar coordinates, (r, θ, ϕ), this velocity potentialbecomes a3φ(r, θ) U cos θ r 2 .(6.8)2rThe non-zero components of the fluid velocity, u φ, are then a3a31 φ φ U cos θ 1 3 U sin θ 1 3 .and uθ ur rrr θ2r(6.9)Hence, at r a, on the solid sphere’s surface, ur 0 as required by the kinematic boundaryconditions and3uθ (θ) r a U sin θ.2To express the pressure force on the sphere in terms of the fluid velocity, we use Bernoulli’stheorem for steady potential flows, H p/ρ kuk2 /2 constant, ignoring gravity. At r athe fluid pressure, p(θ), therefore satisfiesp(θ) 1 2 uθρ2where p is the pressure as r .r a p 1 2 U ,ρ2

53Chapter 6 – Bernoulli’s equationThus, the pressure distribution on the sphere is 1 292p(θ) p ρU 1 sin θ ,24(6.10)and the total pressure force is the surface integral of p(θ) on the sphere r a,ZZ π Z 2πF p n dS p(θ) êr a2 sin θ dϕdθ,S0(6.11)0where êr sin θ cos ϕ êx sin θ sin ϕ êy cos θ êz .As the flow is axisymmetric, the only non-zero component of the force should be in the axialdirection, z. Indeed,Z 2πZ π2Fx F · êx acos ϕ dϕp(θ) sin2 θ dθ 0,0andFy F · êy a2However, after substituting for p(θ) inZ02πsin ϕ dϕ0Fz F · êz 2πa2ZZπp(θ) sin2 θ dθ 0.0πp(θ) sin θ cos θ dθ,0we find thatFz 2πa2 1p ρU 22 Zπ09sin θ cos θ dθ ρU 28Zπ3 sin θ cos θ dθ 0,0so that the total drag force on the sphere, due to the fluid flow around it, is zero!D’Alembert’s paradox: it can be demonstrated that the drag force on any 3-D solid bodymoving at uniform speed in a potential flow is zero (see, e.g., Paterson, § XI.9, p. 240).This is not true in reality of course, as flows past 3-D solid bodies are not potential.We can see why a potential flow past a sphere gives zero drag by looking at the streamlines.low pUhigh p S1S2 high plow pThe flow is clearly fore-aft symmetric (symmetry about z 0); the front (S1 ) and the back(S2 ) of the sphere are stagnation points at equal pressure, PS1 PS2 p 12 ρU 2 . At theside, ur 0 and u2θ 0, so from Bernoulli’s theorem, the pressure there is lower than atthe stagnation points but it must have the same symmetry as the flow. Notice that, fromBernoulli’s theorem, the pressure does not depend on the direction of the flow, but on itsspeed kuk only.However, the real flow past a sphere is not symmetric and, as a consequence, the fluid exertsa net drag force on the sphere.

546.4 Separation6.4SeparationThe pressure distribution on the surface a solid sphere placed is a uniform stream, 1 292p(θ) p ρU 1 sin θ ,24reaches its minimum, pmin p 5/8 ρU 2 , at θ π/2. So, the pressure gradient in thedirection of the flow, (u· )p, is a positive from θ 0 to θ π/2 and negative beyond.pmin(u· )p 0(u· )p 0Uθpmax(u· )p 0pminpmaxU(u· )p 0An adverse pressure gradient, (u· )p 0 (i.e. pressureincreasing in the direction of the flow along the surface),is “bad news” and causes the flow to separate, leaving aturbulent wake behind the sphere.Very roughly one can estimate the pressure differenceupstream and downstream as 1/2 ρU 2 , so that the dragforce F 1/2 ρU 2 A, where A is the cross-sectionalarea.The ratioFCD 1 2(6.12)2 ρU Ais called drag coefficient and depends, e.g., on the shapeof the body (see Acheson §4.13, p. 150).The way to reduce drag (i.e. resistance) is to reduce separation: Streamlining: separation occurs because of adverse pressure gradients on the surface ofsolid bodies. These can be reduced by using more “streamlined” shapes, that avoid diverging streamlines (e.g., aerodynamic bike helmets (time trial cyclist), ships, aeroplanesand cars). Surface roughness: paradoxically, a rough surface can reduce drag by reducing separation (e.g. dimple pattern of golf balls and shining of cricket ball on one side).