Bernoulli's equation: 1 p h t p t. near the far from plate the plate. p u

Transcription

1 UNSTEADY FLOW let s re-visit the Kutta condition when the flow is unsteady: p u p l Bernoulli's equation: 2 φ v = () = + p h t p t 2 2 near the far from plate the plate as a statement of the Kutta condition, we require the pressures in the airstreams coming off the upper and lower surfaces at the trailing edge to be equal: p u = p ( φ φ ) ( u u )( u u ) u( u u ) l far from the plate conditions are constant ( φ φ ) ( u u )( u u ) φu uu φl ul φl φu ul u u l u l u l + u + + pu = h= + + pl pu pl = + = + = 0 t 2 t 2 t t 2 t 2 u l l u l + u Δ l + u = = how we can determine φl φu? t at a given instant, the change in φ between positions r and r is given by φ φ = dφ = φidr = vidr r r r r r r referring to path C, shown in blue in the figure above, we find φ φ = vidr =Γ= circulation around the plate u l C ( φ φ ) dγ uu ( u ) u l Δ l + u = = t dt 2 dγ in an unsteady flow 0 so there must be vorticity dt at the trailing edge ( Δu 0) if the pressures are equal

2 before continuoing with the unsteady problem, we re-visit slide 9 of Special topics Flat plate ex soln 5) an interesting observation: for the flow around a single vortex at the one-quarter chord to satisfy the Γ no-penetration condition at the three-quarter chord point = sinα Γ= 4πsinα 4π Γ 1 u y = 2π 2 vortex Γ control point FBD of the plate: x freestream α recall hw4: 1) Consider the two-dimensional flowfield with the velocity components given by y x vx = C and v where is a constant and 2 y = C C r = x + y 2 r r 2 2 2

3 unsteady flow: y xv() 1 xc xc( 3) xc( 2) ( 1) Γ ( 1) Γ ( 2) Γ ( 3) Γ ( 4) Γ ( 5) Γ ( 6) xv( 2) xv( 3) xv( 4) xc( 4) xv( 5) xc( 5) xv( 6) xc( 6) x x x x x x extra vortex to satisfy the Kutta condition Γ 7 Γw 1 x representation of a model of a flat plate with six platelets, each platelet has a concentrated vortex located at xv and a control point x located at xc the spacing of the vortices and control points in each of the panels mimics the entire plate shown on the previous slide. now with an extra vortex at one-quarter chord behind the trailing edge we must take the newly added vortex into account when we impose the the no-penetration condition on the plate. we use the suffix "w" to denote quantities associated with the flowfield behind the plate, that part of the flow called "the wake" ( xw, yw) we denote the coordinates of the new vortex at the trailing edge and the circulation around the new vortex and then add the final cloumn to the influence matrix and the final element of the vector of unknowns: Γw (,7) Ai 1 1 = 2 π xc( i) xw at this point the no-penetration condition becomes (next slide)

4 we must take the newly added vortex into account when we impose the the no-penetration condition on the plate. we use the suffix " w" to denote quantities associated with the flowfield behind the plate, that part of the flow called "the wake" ( xw, yw) we denote the coordinates of the new vortex at the trailing edge and add the final cloumn to the influence matrix: (,7) Ai 1 1 = 2 π xc( i) xw and so at this point the no-penetration condition becomes A( 1,1) A( 1,2 ) A( 1,3) A( 1, 4) A( 1,5) A( 1,6) A( 1,7) Γ(1) sinα A( 2,1) A( 2,2) A( 2,3) A( 2,4) A( 2,5) A( 2,6) A( 2,7) Γ(2) sinα A( 3,1) A( 3, 2) A( 3,3) A( 3, 4) A( 3,5) A( 3,6) A( 3,7) Γ(3) sinα A( 4,1) A( 4,2) A( 4,3) A( 4,4) A( 4,5) A( 4,6) A( 4,7) R sinα A( 5,1) A( 5,2) A( 5,3) A( 5,4) A( 5,5) A( 5,6) A( 5,7) Γ(5) sinα A( 6,1) A( 6,2) A( 6,3) A( 6, 4) A( 6,5) A( 6,6) A( 6,7) Γ (6) sinα??????? Γ ( 7 )? Γ (4) = = where Γ 7 Γw there are six equations and seven unknowns, a hopeless situation because we can arbitrarily assign a value to one of the unknowns and then solve for the other six, getting a different answer for each value we assign to obtain the other equation, we have to consider an initial condition: let's suppose the flow past the plate is suddenly started so that at one instant there is no flow and at the next there is a uniform stream at an angle of attack α (flow impulsively started) C a closed circuit of particles DΓC just before the flow starts, Γ C = 0, after the flow starts, Γ C = 0 and remains = 0 because = 0 Dt according to Helmholtz's temporal conservation of circulation (see slide 17 in "Special topics -- Boundary-layer t heory") it follows from Stokes's theorem that Γ C = Γ j +Γ w= 0 and so for and A 7, j = 1 j = 1,...,7 R 7 = 0 6 j =1

5 the instant following the impulsive start iiii unsteady flow α = 20 degs, 40 panels steady flow Δu is plotted as a function of position on the plate for both the instant following the impulsive start, A, and the steady state, B. B if the velocity remains constant following the implusive start, over time the unsteady flow will approach the steady state? the question is: how does the flow go from A to B? A Or, what happens to the vorticity at the trailing edge? the Kutta condition, as we use it, requires the pressures to be equal at the trailing edge, which leads to DΓ the conclusion that, when viscous effects are ignorable, vorticity moves with the fluid particles = 0 Dt ( 7) Γw( 1) hence, we imagine a small closed loop that encircles the last vortex, Γ in the present example, the one in the wake just just behind the trailing edge the vorticity in this particle will move with the particle and the circulation around the particle will remain constant we "convect" this particle to a new position during the interval Δt, the time step: Δ r = vdt v 0 Δt where v 1 ( 0) = the velocity of the first wake-vortex at time = 0 in its initial position ; we do not change the circulation around this vortex ever: Γ w = a constant during the entire flow 1 Δt t = 0 1

6 now we have a vortex in the wake -- we know the position of this vortex and we know the circulation around it because the wake has changed, the downwash on the wing has changed and will continue to change; so we must add another vortex at the trailing edge to satisfy the Kutta condition and re-impose the no-penetration condition taking both vortices in the wake (the new one at the trailing edge and the one that was there at the previous time step) into account when we calculate the downwash at the control points for the new time step. The first wake vortex is in a known position with a known circulation, and so its contribution becomes part of the right-hand side: () 1 i [ ] [ xc() i xw(1) ] + [ yw(1) ] ()[ ] Γw yw(1) + xc( i) xw(1) j Γw 1 xc( i) xw(1) the velocity at the control point of panel i: v1 () i = v 2 2 = 2 2π 2πr () 1n i A( 1,1) A( 1, 2) A( 1,3) A( 1, 4) A( 1,5) A( 1,6 ) A( 1,7) Γ A( 2,1) A( 2, 2) A( 2,3) A( 2, 4) A( 2,5) A( 2,6) A( 2,7) A( 3,1) A( 3, 2) A( 3,3) A( 3, 4) A( 3,5) A( 3,6 ) A( 3,7) Γ A( 4,1) A( 4, 2) A( 4,3) A( 4, 4) A( 4,5) A( 4,6) A( 4,7) A( 5,1) A( 5, 2) A( 5,3) A( 5, 4) A( 5,5) A( 5,6) A( 5,7) A( 6,1) A( 6, 2) A( 6,3) A( 6, 4) A( 6,5) A( 6,6) A( 6,7) Γ (1) sin α v1n(1) (2) sin α v (2) Γ 1n (3) sin α v1n(3) Γ (4) = R = sin α v1n(4) where Γ( 7 ) Γw(2) Γ(5) sin α v1n(5) Γ (6) sin α v1n(6) 7 Γ w(1) now we can solve the seven equations for the seven unknowns to obtain the solution at the the first time step following the impulsive start

7 y LEΓ x 1 Γ2 Γ3 Γ Γ 4 5 Γ6 Γ TE 7 Γw 1 0,0 1,0 y x 0,0 1,0 Γw Γ x 1 Γ2 Γ3 Γ Γ 4 5 Γ6 Γ7 Γw 2 3 0,0 1,0 y Illustration of the First Three Steps The impulsive start Γ1 Γ2 Γ3 Γ Γ 4 5 Γ6 Γ7 First time step Second time step A new starting vortex Another new starting vortex The starting vortex Γ w1 Γ w 2 Δ r = v 0 Δt 1 1 The starting vortex Δ r1 = v1 Δt Δt Γw 1 Δ r2 = v2 Δt Δt Some numerical examples follow.

8 after n time steps, the solution at n + 1 is obtained from the velocity at the control point of panel i v k () i yw( k) i+ [ xc( i) xw( k) ] [ xc i xw k ] + [ yw k ] Γwk j = π () [ () ] Γwk xci xwk = = v 2 2π r kn () i normal component downwash from the wake n sin α vkn(1) k=1 n sin α vkn(2) A( 1,1) A( 1, 2) A( 1,3) A( 1, 4) A( 1,5) A( 1,6 ) A( 1,7) (1) k=1 Γ ( 2,1) ( 2,2) ( 2,3) ( 2,4) ( 2,5) ( 2,6) ( 2,7) n A A A A A A A Γ (2) sin α vkn(3) A( 3,1) A( 3, 2) A( 3,3) A( 3, 4) A( 3,5) A( 3,6 ) A( 3,7) Γ (3) k=1 n A( 4,1) A( 4,2) A( 4,3) A( 4,4) A( 4,5) A( 4,6) A( 4,7) Γ (4) = R = sin α vkn(4) k=1 A( 5,1) A( 5, 2) A( 5,3) A( 5, 4) A( 5,5) A( 5,6) A( 5,7) Γ(5) n A( 6,1) A( 6, 2) A( 6,3) A( 6, 4) A( 6,5) A( 6,6) A( 6,7) Γ(6) sinα vkn(5) Γ ( 7) k=1 n sin α vkn(6) k=1 n Γwk k=1 where Γ 7 Γ w( n + 1) circulation around the wake minus the circulation around the new vortex at the trailing edge, which is unknown

9 impulsively started motion, α = 20o

10 impulsively started motion, α = 20o

11 vortex control point exact steady-flow solution -i- numerical transient solution more panels give better resolution near the ends at the instant following the impulsive start

12 this series of graphs illustrates how the transient solution approaches the steadystate solution when the velocity remains constant following an impulsive start. the blue line represents the steady-state solution and the red line with the green dots represents the unsteady solution at various times following the impulsive start 100 panels were used in order to capture the rapid changes near the ends

13 Photos from Ideal-Fluid Aerodynamics by K. Karamcheti Flow following an impulsive start Experimental flow visualization.

14 Photos from Ideal-Fluid Aerodynamics by K. Karamcheti Flow following an impulsive start Experimental flow visualization.

15 the figures at the left represent the numerical transient solution: following the impulsive start the freetsream is steady for 250 time steps, then it impulsively stops. vorticity continues to be shed from the trailing edge, but with a different sign (see figure below) note how the regions containing the concentration of vortices migrate downward, each moving in the velocity field associated with the other circulation around the vortex at the trailing edge as a function of timesteps x 100 the plate stopped suddenly at timestep = 250

16 Flow following an impulsive start Comparison with experiment Flow following an impulsive start followed by an impulsive stop Impulsive start followed by an impulsive stop

17 Flow following an impulsive start Comparison with experiment Flow following an impulsive start followed by an impulsive stop Impulsive start followed by an impulsive stop

18 Number of elements on the plate = 20 Length of the plate = 1 Amplitude of the pitching motion = 7 degs. Mean angle of attack = 3 degs Frequency = 0.35 Number of time steps = 200. Vortex-splitting used: 1797 vortices in the wake y oscillating airfoil flow x

19 total velocity field for a plunging airfoil after 450 time steps, disturbance velocity field for a plunging airfoil after 450 time steps,

20 Comparison with Experiment: a pitching airfoil with simple harmonic motion experimental flow M. M. Koochesfahani, AIAA Journal, Sept numerical regions of concentrated vorticity

21 Comparison with Experiment: a pitching airfoil with simple harmonic motion Superimposing the two results: airfoil flow numerical experimental