ME 438 is a course taught by Dr. Bilal Siddiqui at DHA Suffa University. This set of lectures deals with review of vector calculus, fluid mechanics, circulation, source/sink method, introduction to computational aerodynamics with source panel method and calculation of lift.
9. Review of Fluid Mechanics
Note: We have made no assumption
of inviscidity or compressibility
10. Review of Fluid Mechanics Note: We have assumed inviscid flow
11. Incompressible and Inviscid Flows
• Some very important flows can be solved by neglecting
compressibility (𝑀∞ ≤ 0.3) and viscosity (away from body)
• For incompressibility, the condition (derived from continuity
equation) is
• For inviscid flows (aka potential or irrigational flows), the condition is
12. Laplace Equation for Potential Flows
• The equation 𝛻2 𝜙 = 0 is called the Laplace equation.
• It is one of the most famous and extensively equations in math/physics
• It has well known solutions, therefore it is easier to solve potential
(incompressible inviscid) flows analytically
• Since stream functions and potential functions are cousins, we can show
for 2D flows that
13. Notes on the Laplace Equations
• Any irrotational, incompressible flow has a velocity potential and stream
function (for 2D flow) that both satisfy Laplace’s equation.
• Conversely, any solution of Laplace’s equation represents the velocity
potential or stream function (2D) for an irrotational, incompressible flow.
• Laplace’s equation is a second-order linear partial differential equation.
• The fact that it is linear is important, because the sum of any particular
solutions of a linear differential equation is also a solution of the equation!
• Since irrotational, incompressible flow is governed by Laplace’s equation
and Laplace’s equation is linear, we conclude that a complicated flow
pattern for an irrotational, incompressible flow can be synthesized by
adding together a number of elementary flows that are also irrotational
and incompressible.
14. How to solve the Laplace Equations
• Our strategy is to develop flow solutions for several different
elementary flows, which by themselves may not seem to be practical
flows in real life.
• However, we then proceed to add (i.e., superimpose) these
elementary flows in different ways such that the resulting flow fields
do pertain to practical problems.
• All of these flows have the same governing equation, i.e. 𝛻2 𝜙 = 0
• How, then, do we obtain different flows for the different bodies? The
answer is found in the boundary conditions.
15. Boundary Conditions in Aerodynamics
• There are two boundary conditions in all external flows
• Infinity boundary conditions
Far away from the body (toward infinity), in all directions, the
flow approaches the uniform freestream conditions.
• Wall boundary conditions
It is impossible for the flow to penetrate the body surface.
16. Elementary Potential Flow 1: Uniform Flow
• It is clear that
• Integrating these equations
• In polar coordinates
• The flow is obviously irrotational, therefore circulation Γ = 0
17. Elementary Potential Flow 2: Source/Sink Flow
The source strength is defines as
The potentials, streams and velocities can be
calculated as
Source/Sink flow obeys mass
conservation 𝛻. 𝑉 = 0
everywhere except at origin
since 𝑉𝑟 → ∞. But we accept
it as a ‘singularity’
20. Elementary Potential Flow 3: Doublet
• When source and sink of equal strength are placed at the same point
• 𝜅 is called the doublet strength
• The streamlines are families of circles with diameter
Where C is some constant
22. A bit more on the flow over Cylinder
• We know that pressure coefficient is
• On the surface of the cylinder, we can
show that
• Therefore,
• Cp varies from 1.0 at the stagnation
points to −3.0 at the top/bottom.
23. Elementary Flow 4: Vortex Flow
• All the streamlines are concentric circles about a given point.
• Velocity along any given circular streamline is constant, but vary from
one streamline to another inversely with distance from the center.
• Vortex flow is incompressible 𝛻. 𝑉 = 0 everywhere.
• Vortex flow is irrotational everywhere 𝛻 × 𝑉 = 0 except at origin, but
we except it as a singularity (exception).
• To evaluate the constant, take the circulation around a given circular
streamline of radius r
Therefore, for vortex flow, the
circulation taken about all
streamlines is the same value,
namely, Γ= −2πC.
25. Lifting Flow over Cylinder
• If the cylinder is spinning, it will produce finite (measurable) lift.
• We model this as the superimposition of uniform flow+doublet+vortex
26.
27.
28. Lift due to Spinning of Cylinder
• At the surface of the cylinder (r=R), 𝑉𝑟 = 0 and 𝑉𝜃 = −2𝑉∞ sin 𝜃 −
Γ
2𝜋𝑅
• Since, 𝐶 𝑝 = 1 −
𝑉
𝑉∞
2
= 1 − 4 sin2
𝜃 +
2Γ sin 𝜃
𝜋𝑅𝑉∞
+
Γ
2𝜋𝑅𝑉∞
2
• The drag coefficient is given by
• The lift force can be found by
29. The Kutta-Joukowski Theorem
• Lift per unit span is directly proportional to circulation.
• This is a powerful relation in theoretical aerodynamics called the
Kutta-Joukowski theorem, obtained in ~ 1905.
• Rapidly spinning cylinder can produce a much higher lift than an
airplane wing of same planform area
• However, the drag on the cylinder is also much higher than a well-
designed wing. Hence, no rotating cylinders on aircrafts.
• Although 𝐿′ = 𝜌∞ 𝑉∞Γ was derived for a circular cylinder, it applies in
general to bodies of arbitrary cross section.
30. K-J Theorem and the Generation of Lift
• Let curve A be any curve in the flow enclosing the airfoil.
• If airfoil is producing lift, the velocity field around the airfoil will be such
that the line integral of velocity around A will be finite
• The integral of velocity around any curve not including the body is zero
• Hence, the lift produced by the airfoil is given by 𝑳′ = 𝝆∞ 𝑽∞ 𝚪
• The Kutta-Joukowski theorem states that lift per unit span on a two-
dimensional body is directly proportional to the circulation around the
body.
• Just like we synthesized flow over a spinning cylinder by adding a vortex to
the non-lifting flow, we can synthesize flow over an airfoil by distributing
vortices all over and inside the airfoil.
Note: Lift
produces
circulation, not
the other way
round. Think why?
31. Numerical Source Panel Method: Our first
flavor of CFD
• We added elementary flows in certain ways and discovered that resulting
streamlines turned out to fit certain body shapes.
• But it is not practical to randomly add elementary flows and try to match
the body shape we seek to find the flow around!
• We want to specify the shape of an arbitrary body and solve for the
distribution of singularities which, in combination with a uniform stream,
produce the flow over the given body.
• For the moment, we will concentrate on non-lifting flows….for a reason.
• This technique is called the source-panel method, a standard tool in
aerodynamic industry since the 1960s.
• Numerical solution of potential flows by both source and vortex panel
techniques has revolutionized the analysis of low-speed flows. We will
consider vortex methods later for lifting flows.
32. Recall: Source/Sink Flow
The source strength is defines as
The potentials, streams and velocities can be
calculated as
Now, instead of having a
single source, we want to
have a number of sources
placed side by side along a
contour: “Source Sheet”
33. The Source Sheet
• Define λ = λ(s) to be source strength per unit length
along s. For infinitesimal sources, ds acts like a
regular line source of strength λds and depth l
• Recall that the strength Λ of a single line source
was defined as the volume flow rate per unit depth
in the z direction.
• Typical units for Λ are square meters per second,
but for λ are meters per second.
λ may be negative, so it is
really a combination of
sources and sinks
34. SPM- Derivation
• Consider a point P at distance r from portion ds of the source sheet
• This portion of strength 𝜆𝑑𝑠 produces an infinitesimally potential 𝑑𝜙
𝑑𝜙 =
𝜆𝑑𝑠
2𝜋
ln 𝑟
• Complete velocity potential at point P, induced by the entire source
sheet from a to can be found as 𝜙 = 𝑎
𝑏 𝜆 ln 𝑟
2𝜋
𝑑𝑠
• We can now wrap the entire body with this source sheet and
superimpose uniform flow.
35. SPM-Discretizing the Continuous Equations
• Approximate the source sheet by n straight panels j=1,2,…,n
• Let source strength per panel 𝜆𝑗 be constant.
• The panel strengths 𝜆1, 𝜆2, … , 𝜆𝑗, … , 𝜆 𝑛 are unknown
• We want to iterate till the body surface becomes a streamline of the
flow i.e. 𝑉𝑟 = 0 at the surface.
• ↑THIS is the boundary condition we apply at each control point (i.e.
the middle point of each panel).
• Let us now put this in numbers.
37. The Numerical Recipe (1)
• Since point P is just an arbitrary point, but we are more interested in
what is happening at the surface, let’s move P to the body surface.
• Then the effect of each source panel on a given panel is
• Slope of ith panel is
𝑑𝑦
𝑑𝑥 𝑖
, but its angle with the flow is 𝛽𝑖
• The normal component of free stream flow to each panel is
𝑉𝑖 𝑛∞
= 𝑉∞ cos 𝛽𝑖
38. The Numerical Recipe (2)
• The normal component of velocity induced at (xi , yi ) by the source
panels is
𝑉𝑖 𝑛 𝑠𝑜𝑢𝑟𝑐𝑒𝑠
=
𝜕
𝜕𝑛𝑖
𝜙(𝑥𝑖, 𝑦𝑖) =
1
2𝜋
𝑗=1
𝑛
𝜆𝑗
𝑗
𝜕 ln 𝑟𝑖𝑗
𝜕𝑛𝑖
𝑑𝑠𝑗
• We can show that this can be simply expanded as
𝑉𝑖 𝑛 𝑠𝑜𝑢𝑟𝑐𝑒𝑠
=
𝜆𝑖
2
+
1
2𝜋
𝑗=1
𝑗≠𝑖
𝑛
𝜆𝑗
𝑗
𝜕 ln 𝑟𝑖𝑗
𝜕𝑛𝑖
𝑑𝑠𝑗
39. The Numerical Recipe (3)
• The velocity normal to the ith panel at (xi , yi ) is
𝑉𝑖 𝑛
= 𝑉∞ cos 𝛽𝑖 +
𝜆𝑖
2
+
1
2𝜋
𝑗=1
𝑗≠𝑖
𝑛
𝜆𝑗
𝑗
𝜕 ln 𝑟𝑖𝑗
𝜕𝑛𝑖
𝑑𝑠𝑗
• Therefore, we seek to impose the distribution of sources which gives
us zero normal velocity at each node (i.e. body becomes a streamline)
𝑉∞ cos 𝛽𝑖 +
𝜆𝑖
2
+
1
2𝜋
𝑗=1
𝑗≠𝑖
𝑛
𝜆𝑗
𝑗
𝜕 ln 𝑟𝑖𝑗
𝜕𝑛𝑖
𝑑𝑠𝑗 = 0, ∀𝑖 = 1, … , 𝑛
These are n linear equations with n unknowns (𝝀 𝟏, 𝝀 𝟐, … , 𝝀 𝒏)
40. The Numerical Recipe (3)
• Finally, once we find the source distribution (𝜆𝑖, 𝜆2, … , 𝜆 𝑛), we can find the
actual velocity tangent to the surface
𝑉𝑖 = 𝑉∞ sin 𝛽𝑖 +
1
2𝜋
𝑗=1
𝑗≠𝑖
𝑛
𝜆𝑗
𝑗
𝜕 ln 𝑟𝑖𝑗
𝜕𝑠𝑖
𝑑𝑠𝑗 = 0, ∀𝑖 = 1, … , 𝑛
• We can then find the pressure coefficient at each node
𝐶 𝑝,𝑖 = 1 −
𝑉𝑖
𝑉∞
2
• The pressure forces can then be easily resolved in lift drag and moment!
41. Source Panel Method Applied to a Cylinder
• Let us begin by dividing the
cylinder into 8 panels.
• Coordinates of the ith panel’s
control point are (xi,yi), and the
edge points of the panel are
(Xi,Yi) and (Xi+1,Yi+1).
• Let 𝑉∞ be in the x-direction.
• Let us first evaluate the integrals
𝐼𝑖𝑗 =
𝑗
𝜕 ln 𝑟𝑖𝑗
𝜕𝑛𝑖
𝑑𝑠𝑗
42. SPM applied to the cylinder (2)
• It is easier to
deal with flow
angles with
tangent to the
surface rather
than normal
to it
44. The SPM applied to the cylinder (4)
• To obtain the actual velocity over the cylinder
• The velocity due to upstream flow is 𝑉∞ 𝑠,𝑖
= 𝑉∞ sin 𝛽𝑖
• Velocity induced by sources
• Therefore the total velocity is
• And coefficient of pressure is
45. SPM applied to Cylinder (5)
See Matlab code developed in class