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ByHutton: Fundamentals ofFinite Element Analysis8. Applications in FluidMechanics8.4Text© The McGraw−HillCompanies, 2004The Velocity Potential Function in Two-Dimensional Flow␾ ⫽ Constant␺ ⫽ Constantbcad␾ ⫽ ConstantFigure 8.5 Flow net of lines of constant stream function ␺ andconstant velocity potential ␾.direct analogy with Equations 8.14–8.17, we write␾(x , y) =MN i (x , y)␾i = [N ] {␾}(8.36)i=1N i (x , y)A(e)A(e)[N ] TS (e)NT∂ 2␾∂ 2␾dx dy = 0+∂x2∂ y2∂ 2␾∂ 2␾dx dy = 0+∂x2∂ y2∂␾n x dS −∂x−A(e)i = 1, MA(e)∂ [N ] T ∂ ␾dx dy +∂x ∂x(8.37)(8.38)[N ] TS (e)∂ [N ] T ∂ ␾dx dy = 0∂y ∂y∂␾n y dS∂y(8.39)Utilizing Equation 8.36 in the area integrals of Equation 8.39 and substituting thevelocity components into the boundary integrals, we obtain ∂ [N ] T ∂ [N ]∂ [N ] T ∂ [N ]+dx dy {␾} = − [N ] T (un x + vn y ) dS∂x∂x∂y∂yA(e)S (e)(8.40)ork (e) {␾} = f (e)(8.41)305Hutton: Fundamentals ofFinite Element Analysis3068.

Applications in FluidMechanicsCHAPTER 8Text© The McGraw−HillCompanies, 2004Applications in Fluid MechanicsThe element stiffness matrix is observed to be identical to that of the streamfunction method. The nodal force vector is significantly different, however. Notethat, in the right-hand integral in Equation 8.40, the term in parentheses is thescalar product of the velocity vector and the unit normal to an element boundary.Therefore, the nodal forces are allocations to the nodes of the flow across theelement boundaries. (Recall that we assume unit dimension in the z direction, sothe terms on the right-hand side of Equation 8.40 are volumetric flow rates.) Asusual, on internal element boundaries, the contributions from adjacent elementsare equal and opposite and cancel during the assembly step.

Only elements onglobal boundaries have nonzero nodal force components.EXAMPLE 8.1To illustrate both the stream function and velocity potential methods, we now examinethe case of a cylinder placed transversely to an otherwise uniform stream, as shown inFigure 8.6a. The underlying assumptions are1. Far upstream from the cylinder, the flow field is uniform with u = U = constantand v = 0.2. Dimensions in the z direction are large, so that the flow can be considered twodimensional.3. Far downstream from the cylinder, the flow is again uniform in accordance withassumption 1.yUx(a)b␺ ⫽ Uybc␾⫽0⭸␾⫽ Constant⭸xad␺⫽0e(b)Figure 8.6(a) Circular cylinder in a uniform, ideal flow. (b) Quarter-symmetrymodel of cylinder in a uniform stream.Hutton: Fundamentals ofFinite Element Analysis8.

Applications in FluidMechanics8.4Text© The McGraw−HillCompanies, 2004The Velocity Potential Function in Two-Dimensional Flow■ Velocity PotentialGiven the assumptions and geometry, we need consider only one-fourth of the flow field,as in Figure 8.6b, because of symmetry. The boundary conditions first are stated forthe velocity potential formulation.

Along x = 0 (a-b), we have u = U = constant andv = 0 . So,u(0, y) = U = −v(0, y) = 0 = −∂␾∂x∂␾∂yand the unit (outward) normal vector to this surface is (n x , n y ) = (−1, 0) . Hence,for every element having edges (therefore, nodes) on a-b, the nodal force vector isknown asf(e)=−[N ] (un x + vn y ) dS = UTS (e)[N ] T dSS (e)and the integration path is simply dS = dy between element nodes. Note the change insign, owing to the orientation of the outward normal vector.

Hence, the forces associatedwith flow into the region are positive and the forces associated with outflow are negative.(The sign associated with inflow and outflow forces depend on the choice of signs inEquation 8.33. If, in Equation 8.33, we choose positive signs, the formulation is essentially the same.)The symmetry conditions are such that, on surface (edge) c-d, the y-velocity components are zero and x = x c , so we can writev=−∂␾d␾(x c , y)=−=0∂ydyThis relation can be satisfied if ␾ is independent of the y coordinate or ␾(x c , y) is constant. The first possibility is quite unlikely and requires that we assume the solution form.Hence, the conclusion is that the velocity potential function must take on a constant valueon c-d.

Note, most important, this conclusion does not imply that the x-velocity component is zero.Along b-c, the fluid velocity has only an x component (impenetrability), so we canwrite this boundary condition as∂␾∂␾∂␾=nx +n y = −(un x + vn y ) = 0∂n∂x∂yand since v = 0 and n x = 0 on this edge, we find that all nodal forces are zero along b-c,but the values of the potential function are unknown.The same argument holds for a-e-d. Using the symmetry conditions along this surface, there is no velocity perpendicular to the surface, and we arrive at the same conclusion: element nodes have zero nodal force values but unknown values of the potentialfunction.307Hutton: Fundamentals ofFinite Element Analysis3088.

Applications in FluidMechanicsCHAPTER 8Text© The McGraw−HillCompanies, 2004Applications in Fluid MechanicsIn summary, for the potential function formulation, the boundary conditions are1.2.3.4.Boundary a-b:Boundary b-c:Boundary c-d:Boundary a-e-d:␾ unknown, forces known.␾ unknown, forces = 0 .␾ = constant, forces unknown.␾ unknown, forces = 0 .Now let us consider assembling the global equations. Per the usual assembly procedure, the equations are of the matrix form[K ]{␾} = {F }and the force vector on the right-hand side contains both known and unknown values. Thevector of nodal potential values {␾} is unknown—we have no specified values.

We doknow that, along c-d, the nodal values of the potential function are constant, but we do notknow the value of the constant. However, in light of Equation 8.33, the velocity components are defined in terms of first partial derivatives, so an arbitrary constant in the potential function is of no consequence, as with the stream function formulation. Therefore,we need specify only an arbitrary value of ␾ at nodes on c-d in the model, and the systemof equations becomes solvable.■ Stream Function FormulationDeveloping the finite element model for this particular problem in terms of the streamfunction is a bit simpler than for the velocity potential.

For reasons that become clear whenwe write the boundary conditions, we also need consider only one-quarter of the flow fieldin the stream function approach. The model is also as shown in Figure 8.6b. Along a-e, thesymmetry conditions are such that the y-velocity components are zero. On e-d, the velocity components normal to the cylinder must be zero, as the cylinder is impenetrable.Hence, a-e-d is a streamline and we arbitrarily set ␺ = 0 on that streamline. Clearly, theupper surface b-c is also a streamline and, using previous arguments from the convergentflow example, we have ␺ = U yb along this edge.

(Note that, if we had chosen the value ofthe stream function along a-e-d to be a nonzero value C, the value along b-c would be␺ = U yb + C.) On a-b and c-d, the nodal forces are zero, also per the previous discussion,and the nodal values of the stream function are unknown. Except for the geometricaldifferences, the solution procedure is the same as that for the converging flow.A relatively coarse mesh of four-node quadrilateral elements used for solving thisproblem using the stream function is shown in Figure 8.7a. For computation, the valuesU = 40, distance a-b = yb = 5 , and cylinder radius = 1 are used. The resulting streamlines (lines of constant ␺ ) are shown in Figure 8.7b.

Recalling that the streamlines arelines to which fluid velocity is tangent at all points, the results appear to be correctintuitively. Note that, on the left boundary, the streamlines appear to be very nearly perpendicular to the boundary, as required if the uniform velocity condition on that boundaryis satisfied.For the problem at hand, we have the luxury of comparing the finite element resultswith an “approximately exact” solution, which gives the stream function as␺ =Ux 2 + y2 − R2yx 2 + y240123634(a)4411333946451026289297232374241␺⫽038196305433152 502527352412322212017(b)(a) Coarse, finite element mesh for stream function solution; 40 elements.

(b) Streamlines (␺ = constant) for finite element solutionof Example 8.1.1343494751␺ ⫽ 200TextFigure 8.785348188. Applications in FluidMechanics151614Hutton: Fundamentals ofFinite Element Analysis© The McGraw−HillCompanies, 2004309Hutton: Fundamentals ofFinite Element Analysis3108. Applications in FluidMechanicsCHAPTER 8Text© The McGraw−HillCompanies, 2004Applications in Fluid MechanicsTable 8.1 Selected Nodal Stream Function and Velocity Values for Solution of Example 8.1Node␺ FE␺ ExactVFEVExact1281620212223244546000123.63142.48100.0367.1040.5518.9867.88103.87000122.17137.4099.3764.6739.3618.2865.89100.7475.1841.96338.73540.53344.90347.10951.53557.83668.14241.70642.35980038.440.51042.91445.21549.12155.49965.42540.79941.018This solution is actually for a cylinder in a uniform stream of indefinite extent in both thex and y directions (hence, the use of the oxymoron, approximately exact) but is sufficientfor comparison purposes.

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