Transcript AE 301 Aerodynamics I

Compressible Potential Equation
•
Although we have seemed to have strayed – remember
that we are still solving a PDE. In particular:
2
2
2






2
   2  2  2 0
x
y
z
•
This potential equation was derived under the
assumption of incompressible, irrotational flow.
•
Let’s know revisit the original equations and consider
how this relation would be differ with compressibility.
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We will also see how to apply the potential flow concept
to high speed (transonic and supersonic) flows.
•
Technically, flow with shock waves is rotational, but only
slightly if the shocks are weak.
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Compressible Potential Equation [2]
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Our starting point is the Euler conservation equations
we derived at the start of the semester.
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If we assume steady flow, mass and momentum
conservation can be written as:
  V  0
  V V  p  0
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In incompressible flow, the first equation was enough to
derive our potential flow relation.
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This time, we will concentrate our attention primarily on
the second equation, momentum conservation.
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We will also need the energy equation, but let’s
consider that a little later.
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Compressible Potential Equation [3]
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We can “combine” mass and momentum conservation
through chain rule expansion of momentum:
V   V   V  V  p  0
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The first term, by continuity, vanishes. The remaining
terms, expanded in 2-D become:
u
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u
u
p
 v

x
y
x
v
v
p
 v

x
y
y
If the potential function exists, then:
V   
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u
 ˆ  ˆ  ˆ
i
j
k
x
y
y
Note, we have switched to using capital phi for our
symbol – the reason will be apparent later.
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Compressible Potential Equation [4]
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The definition of the potential function gives:
u
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
x
•

y
With these, momentum conservation becomes:
  2
  2
p





x x 2
y xy
x
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v
  2
  2
p




x xy
y y 2
y
Since these equations are getting long, let’s introduce a
common notational simplification – indicate the partial
differentiation through subscripts:
p
  x  xx   y  xy   
x
  x  xy   y  yy   
p
y
Next, since we are compressible, we know we should
introduce some thermodynamics. Let’s do that with the
definition of the speed of sound.
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Compressible Potential Equation [5]
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The speed of sound is:
a2 
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p

If we allow ourselves to be a little loose with our
calculus, this can be written as:
dp
d  2
a
Or, since density and pressure are functions of location:
 1 p



 
  y  xy 
x a x
a
 1 p

 2
  2  x  xy   y  yy 
y a y
a
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2
2
x
xx
To finish up, return and consider the continuity again.
AE 401 Advanced Aerodynamics
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Compressible Potential Equation [6]
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If we expand the continuity equation:
  V    V  V  
 u v 


      u
v
0
x
y
 x y 
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Introducing the potential function gives:
 x   y 
 xx   yy 

0
 x  y
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Finally, inserting our previous results from the
momentum/speed of sound equation:
y
x
 xx   yy  2  x  xx   y  xy   2  x  xy   y  yy   0
a
a
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203
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Compressible Potential Equation [7]
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Or, after combining like 2nd order differentials:
  2y 
2 x  y
  2x 
1  2  xx  1  2  yy 
 xy  0
2


a 
a 
a


•
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It is fairly easy to see that as the Mach number goes to
zero, this equation reduced to the Laplace equation:
2
 2x  y u 2  v 2
2



M
a2 a2
a2
However, this new equation, unfortunately, is a highly
non-linear PDE with no easy solutions.
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This equation is also not yet complete – the speed of
sound terms are not constants but vary with speed.
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Thus we need the energy equation.
AE 401 Advanced Aerodynamics
204
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Compressible Potential Equation [8]
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Rather than use the differential form of the energy
equation, lets use the algebraic equation from Aero 2:
V2
c pT 
 c pT0
2
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Remember those cp’s are specific heats, NOT pressure
coefficients!
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This can be using our equations for cp and the a:
RT V 2 RT0


 1 2  1
a02
a2 V 2


 1 2  1
a a 
2
2
0
AE 401 Advanced Aerodynamics
 1
2
V a 
2
2
0
 1
2

205
2
x
  2y

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Compressible Potential Equation [9]
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To solve these equations, we could use a numerical or
Computational Fluid Dynamics (CFD) approach:
– Express the 2nd derivatives terms using some numerical method
like Finite Differences, Finite Volumes, or Finite Elements.
– Asssume values of velocity (1st derivatives) and a, solve for .
– Update the velocities and a and start iterating.
– (Easier said then done, believe me!)
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Or, we can look at approximate solutions obtained by
“linearizing” the equations.
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This process involves expanding the terms we have
using a perturbation potential, and then throwing away
terms of lower magnitude.
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That is where we are going next.
AE 401 Advanced Aerodynamics
206
7/18/2015