Quick Visit to Bernoulli Land

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1 Although we have een the Bernoull equaton and een t derved before, th next note how t dervaton for an uncopreble & nvcd flow. The dervaton follow that of Kuethe &Chow ot cloely (I lke t better than Anderon). tart fro nvcd, ncopreble oentu equaton u + u u = p t ρ There a vector calculu dentty: u u = u u ( u) ω, vortcty u + u + p = u ω t ρ Fro here, we can ake the fnal re-arrangeent: p+ u = u u t ρ ρ ω ρ Two coon applcaton:. teady rrotatonal flow u = ω = t teady Irrotatonal p+ ρ u = p + ρ u = cont. for entre flow Kuethe and Chow, 5 th Ed. ec

2 . teady but rotatonal flow u = ω t Rotatonal teady p+ ρ u = ρu ω Th a vector equaton. If we dot product th nto the treawe drecton: u treawe drecton u p+ ρ u = ρ ( u ω) =, u ( u ω) d p+ ρ u = d p + ρ u = cont. along trealne Vortex Panel Method tep#: Replace arfol urface wth panel Orgnal arfol panel (+ node) tep #: Dtrbute ngularte on each panel wth unknown trength In our cae we wll ue vortce dtrbuted uch that ther trength vare lnearly fro node to node: Recall a pont vortex at the orgn : φ Γ Γ y π π x = θ = tan Kuethe and Chow, 5 th Ed. ec

3 A pont vortex at xˆ, yˆ : y yˆ Γ φ = tan π x xˆ Next, conder an arbtrary panel: Vortex of trength (x,y ) ( )d ( expanded ) vew d Mdpont wll be (x,y ) + (x +,y + ) At any, we wll place a vortex wth trength ( ) ( ˆ ) d y y dφ ( x, y) = tan π x xˆ where xˆ x + ( x+ x) ( + ) yˆ y + y y d: Thu, the potental at any ( x, y) due to the entre panel : φ ( x y) ( ) ˆ y y, tan d π x xˆ We wll aue lnear varyng on each panel: ( ) = + ( + ) 6. 3

4 Wth th type of panel, we have + unknown =,, 3...,, +, o we need + equaton. tep#3: Enforce Flow Tangency at Panel Mdpont The next tep to enforce oe approxaton of the boundary condton at the arfol urface. To do th, we wll enforce flow tangency at the dpont of each panel. Panel ethod lngo: control pont a locaton where u n = o enforced. To do th, we need to fnd the potental and the velocty at each control pont. The potental ha the followng for: freetrea ndvdual panel φ = potental + # panel potental uppoe freetrea ha angle α : ( ) y yˆ φ( x, y) = V ( xcoα + ynα) tan d = π x xˆ φ n The requred boundary condton ( ) o, let carry th out a lttle further: x, y = for all = ( ) ˆ y y φ ( xy, ) = V ( coα + nα ) n tan d n = π n x xˆ x, y = coponent of freetrea noral to urface of panel And recall ( ) = + ( + ). We can re-wrte thee ntegral n a copact notaton: ( ) ˆ y y tan π n ˆ x x x, d = C + C n n + n y x, y C = Influence of panel due to node on control pont of panel noral velocty due to panel at control pont of panel.e. C = noral velocty fro panel due to node on control pont of panel. n 6. 4

5 C n = Influence of panel due to node + on control pont at panel Total noral velocty at control pont of panel due to panel = Cn + Cn + o, let look at the control pont noral velocty o, for panel, flow tangency look lke: = ( ) ( α α ) + Cn + Cn = V co + n n, for all = V n We can wrte th a a et of equaton for + unknown. Queton: What can we do for one ore equaton? tep#4: Apply Kutta condton We need to relate Kutta condton to the unknown vortex trength. To do th, conder a porton of a vortex panel. d 6. 5

6 Put a contour about dfferental eleent d U V dn d V U d Γ= u d Recall: d = [ Vdn Ud Vdn + Ud] = ( V V ) dn ( U U ) d Now let dn & d : dn O, d = U U d = U U, or = U U top botto ( ), n general o, nce the Kutta condton requre Utop = Ubotto at TE: te.. =, Kutta condton For the vortex panel ethod, th ean: + = + tep#5: et-up yte of Equaton & olve 6. 6

7 un = I I I9 I I un = = V I3 I3 I33 3. I4.. I 5. =. I6. I 7 un = I 8 8 V Kutta 9 Where I = total nfluence of node at control pont For exaple: I 37 = C n 37 + Cn36 I n n n8 V n C n C n Node Control pont The proble thu reduce to Ax = I =V n, whch we olve to fnd the vector of! b, or, ung our notaton 6. 7

8 tep #6: Pot-proceng The fnal tep to pot-proce the reult to fnd the preure and the lft actng on the arfol. L = ρv Γ= ρv d arfol o, for our ethod, th reduce to: L = ρv ( + ) + = L Cl = = + ρvc = V V c + Vortex Panel Method uary In practce, the vortex panel ethod ued for arfol flow a lttle dfferent than the trategy ued n the wndy cty proble. Here a uary: tep #: Replace arfol urface wth panel Note: the tralng edge double-nubered + pont, panel. tep #: Dtrbute vortex ngularte wth lnear trength varable on each panel ( ) ( ) = + ( + ) () (+) d d

9 Th ean we have + unknown:,, 3,..., + tep #3: Enforce flow tangency at panel dpont u n = at dpont of every panel equaton tep#4: Apply Kutta condton Kutta condton becoe: te.. = + + = + equaton & + unknown 6. 9