Fluid Phsics 8.292J/12.0J Problem Set 4 Solutions 1. Consider the problem of a two-dimensional (infinitel long) airplane wing traeling in the negatie x direction at a speed c through an Euler fluid. In the frame of reference of the airplane, the stead flow around the wind looks like The wing has width L and the flow oer its upper surface can be characterized b a speed u t, while under the lower surface it has a speed u b. Since the upper surface is more cured than the lower surface, u t > u b. The flow has a uniform densit ρ and ou ma neglect grait in this problem. a.) Derie an expression for the lift on the wing, per unit length in the direction. (Hint: Consider the pressure acting on the lower and upper surfaces of the wing.) Solution: The force per unit length in the direction is just the pressure multiplied b L. Thus the net upward force is just From Bernoulli's equation for an Euler fluid, F = ( p p ) L b 1 2 1 2 p + ρu = p0 + ρc, 2 2 where p 0 is the ambient fluid pressure. Combining the two expression aboe gies 1 2 2 F = ρl( ut ub ). 2 t b.) In the limit that the flow speeds u t and u b are not er different from c, show that the lift per unit length is proportional to the circulation around the wing, with circulation defined, as usual, C V dl, where l is a unit ector along the wing surface.
Solution: From the answer to a, we can write 1 1 F = ρl u u = 2 2 L u + u u u = ρluc, 2 2 ( t b ) ρ ( t b)( t b) where u is the aerage elocit, which, if u t and u b are not er different from c, is itself approximatel equal to c. c.) The oncoming flow is irrotational. What can ou deduce about the lift of an airplane wing moing through an Euler fluid? Solution: B Stokes theorem, the circulation is the integral of the orticit oer the area enclosed b the cure. If the oncoming flow is irrotational, it follows that it has no orticit and, in an iniscid fluid, incompressible fluid, there is no wa for it to acquire an. We hae apparentl proen that airplanes and birds cannot fl: No circulation, no lift. This is known as D'Alembert's Paradox, named after Jean Le Rond d'alembert (1717-178), who performed a series of experiments to measure the drag on a sphere in a flowing fluid, and on the basis of the potential flow analsis he expected that the force would approach zero as the iscosit of the fluid approached zero. His experiments disproed this idea. The trick here is to realize that real fluids alwas contain some iscosit, which, no matter how small, causes the fluid in immediate contact with a rigid surface (such as an airplane wing) to come to rest with respect to that surface. Since the fluid far from the surface is moing, this implies the existence of orticit in the interening laer. As the iscosit anishes, so too does the thickness of this boundar laer, and the orticit within the boundar laer becomes er large, so that the product of the orticit and the boundar laer depth remains finite. Thus real wings do generate orticit and circulation. 2. Much of the circulation of the ocean is drien b the frictional stress exerted on its surface b the wind. The essential dnamics can be understood using the orticit equation for an incompressible fluid. Including the effect of a turbulent momentum flux, τ, in the ertical direction, this equation is dω dt τ = ( ω ) V +, (1) z where ω V is the orticit. In the ocean, the orticit is strongl dominated b its ertical component, and the ertical component of (1) can be written, to a er good approximation, as dωz w ω ˆ τ z + k, dt (2) where w is the ertical component of elocit and ˆk is a unit ector in the ertical direction.
If we approximate the flow of the ocean as stead, then and using this, (2) becomes dωz dt V ω, () z ω w ˆ z ω τ V z + k (4) Now we are interested in flow relatie to the rotating earth. From the point of iew of an obserer in an inertial reference frame, the ertical component of the orticit can be written ( ) ω = kˆ V + 2Ωsin( θ ), (5) z where V r is the earth-relatie elocit (i.e. the elocit we are interested in), Ω is the angular elocit of the earth, and θ is latitude. The second term in (5) is just twice the projection of the earth's angular elocit ector onto the local ertical plane, and oer most of the ocean, it is substantiall larger in magnitude than the first term on the right of (5). r Next we substitute (5) into (4) and linearize the result about a state of rest ( V r =0), with the result that 2Ωcos( θ ) w 2 sin( ) kˆ τ Ω θ +, a (6) where a is the radius of the earth, and is the south-to-north earth-relatie elocit component. Finall, for simplicit, we consider a range of latitude that is small enough that we can neglect the ariabilit of the coefficients of (6) with latitude. Using the definitions β 2Ωcos( θ0), a f 2Ω sin( θ ), 0 where θ 0 is some mean latitude, we can further approximate (6) as w f kˆ τ β +. (7) Equation (7) sas that sources of orticit owing to stretching and to wind stress at the surface must be balanced b north-south flow, bringing in water with higher or lower
rotation rates coming from higher or lower latitude. (Water at rest at the equator has no ertical component of rotation, You are going to make some deductions using (7) and the incompressible mass conseration equation. Consider an idealized rectangular ocean basin of dimensions Lx L and depth H, phrased in Cartesian coordinates x, and z. The basin is assumed to be in the Northern Hemisphere, so that both β and f are positie. Take the southern boundar of the basin to lie along = 0, the northern boundar to lie along = L, the western boundar to lie along x = 0, and the eastern boundar to lie along x = Lx. a.) Assuming that the ertical elocit, w, anishes at both the top and bottom of the ocean, and that the wind stress also anishes at the bottom (but not the top!), integrate (7) oer the whole depth of the ocean to find an expression for the depth-aeraged south-to-north elocit,. Solution: The first term on the right of (7) integrates to zero, leaing kˆ τ s =, (8) β H where τ s is the surface stress. b.) Integrate the mass continuit equation for an incompressible fluid through the depth of the ocean to find a differential relation between and the depth-aeraged west-toeast elocit component, u. Solution: The mass continuit equation for an incompressible fluid is just Integrating this with depth gies u w + + = 0. u =, (9) where u is the depth-aeraged west-to-east elocit component. c.) The wind flow oer the North Atlantic ocean is dominated b an anticclone (sometimes called the Bermuda High or the Azores High) which is associated with clockwise flow. Thus the curl of the wind stress on the ocean is almost eerwhere negatie, reaching a maximum amplitude in middle latitudes. For our simple ocean model, we are going to idealize this wind stress as
kˆ ( τs ) = ηsin π, L (10) where τ s is the wind stress at the ocean surface and η is the amplitude of the wind stress curl. Using the result of (a) aboe, find an expression for. Now using the result of (b) aboe, find an expression for u. Note that the boundar condition on u is that it must anish at both the eastern and western boundaries. You will not be able to satisf both of these conditions. So find two solutions: one that satisfies u =0 on the eastern boundar, and one that satisfies u =0 on the western boundar. Solution: Using (10) in (8) gies η sin = π, β H L (11) while using this in (9) gies u ηπ = cos π x β HL L. (12) Integrating this, subject to u = 0 on the western boundar gies u ηπ cos = π x, β HL L (1) while if u = 0 on the eastern boundar, we get ηπ cos u = π ( x Lx ). β HL L (14) d.) The reason that both boundar conditions on u cannot be satisfied is that there exists at either the western or the eastern boundar a thin boundar laer in which all of the "return flow" is concentrated and in which some of the approximations we used break down. The most important approximation is the neglect of the friction of the ocean flowing along the side boundaries. One effect of this friction is to change the orticit of the fluid. Note that the wind stress in our problem proides a negatie definite basinaeraged orticit tendenc. In the stead state, this will hae to be balanced b a source. Considering qualitatiel how the side boundar friction might affect orticit, at which boundar (east or west) do ou think the return flow is concentrated? Explain our reasoning, and sketch what the flow might look like (including the boundar laer).
Solution: If we add a boundar laer on the eastern boundar to the flow gien b (1), the result looks something like This solution cannot work, because friction along the eastern boundar would impart een more clockwise rotation to the fluid, in addition to that imparted b the clockwise wind in the interior of the ocean. Instead, the correct solution is (14), with a western boundar laer: This wa, the western boundar imparts counterclockwise rotation to the fluid, balancing that imparted b the wind. This is a simple model of the Gulf Stream. Extra Credit: The effect of side boundar friction can be modeled b adding a term υ to the right-hand side of (2), where υ is a coefficient which ou ma assume is constant. This term is negligible in the interior, where the solution we obtained before is alid. But it dominates the right side of (2) in the thin boundar laer. B obtaining a solution for the thin boundar laer and matching this to the interior solution at the outer edge of the
boundar laer, find a solution that satisfies both boundar conditions on u and also satisfies 0 = on the continental side of the boundar laer. Solution: When the iscous stress term gien aboe is added to the right-hand side of (2), the deriation leading to (7) instead gies w ˆ τ β f + k + ν. (15) In the thin western boundar current, the last term aboe dominates oer the wind stress, so that, when aeraged oer depth, (15) becomes, approximatel, β b b ν (16) If we define a iscous length scale Lν ( ) 1 ν, β then the general solution of (16) is x x L 2L x x ν b = Ae + e Bcos + Csin, 2Lν 2Lν (17) where A, B and C are integration constants. These are determined b appling the following boundar conditions. First, at the western boundar (x=o) itself, the elocit of a iscous fluid must anish. Second, at the outer edge of the boundar laer, x L ν, the elocit must match η the interior elocit, sin = π β H L. Finall, the flow integrated oer the with of the boundar laer must be equal in magnitude an opposite in sign to the elocit integrated oer the interior of the ocean basin, so that there is no net northward or southward flow of mass: ( ) Lν Lx Lν η dx sin. 0 b = π β H L (18) Appling these three conditions ields the integration constants. The solution looks qualitatiel like the second figure aboe.