Problem 2 : Tandem Airfoil ProblemGiven: A tandem airfoil (drawn below), in which the angle of attack is 5 degrees. Each airfoil is

modeled by 1 panel.

Find: Using the 6-step procedure decribed in class...

(1) Draw the system and place the vortices and control points

(2) Perform step 3 of the procedure

(3) Perform step 4 of the procedure

(4) Perform step 5 of the procedure

(5a) Calculate the lift per unit span for the first airfoil

(5b) Calculate the lift per unit span for the second airfoil

(5c) Calculate the lift per unit span for the system

(5d) Calculate the sum of a + b, and compare with c

Assumptions: Steady, Incompressible, Negligable Body Effects, and Irrotational

Small angle assumption for calculating distantaces between points (ie: the distance between γ1 and

Cp2 is (10/4)*c

Solution:

----------------------------------------------------------------------- Part 1 -------------------------------------------------------------------------

See solution at

http://www.phaux.org/School_Documents/Principles_Of_Aerodynamics/Official_Solutions/

----------------------------------------------------------------------- Part 2 -------------------------------------------------------------------------

First, consider the Cp1

Vcp1n

γ1−

2 π⋅c

2⋅

γ2

2 π⋅6

4c⋅

⋅

+ Vinf sin α( )⋅+=

In the same fashion, we'll express the normal velocity at Cp2

Vcp2n

γ2−

2 π⋅c

2

⋅

γ1

2 π⋅10

4c⋅

⋅

− Vinf sin α( )⋅+=

----------------------------------------------------------------------- Part 3 -------------------------------------------------------------------------

Setting both expressions to zero, combine both equations to solve simultaneously.

0γ1−

2 π⋅c

2⋅

γ2

2 π⋅6

4c⋅

⋅

+ Vinf sin α( )⋅+=

Solve for γ1 in the first equation yields..

γ1

γ2

3π Vinf⋅ c⋅ sin α( )⋅+

=

Substituting this into the velocity equation at Cp2 (as well as setting that equation to zero) yields..

0γ2−

2 π⋅c

2

⋅

γ2

3π Vinf⋅ c⋅ sin α( )⋅+

2 π⋅10

4c⋅

⋅

− Vinf sin α( )⋅+=

Solving for γ2 using the MathCAD solve function..

γ2

3 π⋅ Vinf⋅ c⋅ sin α( )⋅

4=

γ1

3 π⋅ Vinf⋅ c⋅ sin α( )⋅

4

3π Vinf⋅ c⋅ sin α( )⋅+

=

----------------------------------------------------------------------- Part 4 -------------------------------------------------------------------------

Per step 5, we will define the circulation as the sum of the γ terms.

Γ γ1 γ2+=

3 π⋅ Vinf⋅ c⋅ sin α( )⋅

4

3π Vinf⋅ c⋅ sin α( )⋅+

3 π⋅ Vinf⋅ c⋅ sin α( )⋅

4+=

Doing a little bit of cleanup, step 5 comples with the circulation for the system, which can be expressed as..

3 π⋅ Vinf⋅ c⋅ sin α( )⋅

4

3π Vinf⋅ c⋅ sin α( )⋅+

3 π⋅ Vinf⋅ c⋅ sin α( )⋅

4+ simplify 2 π⋅ Vinf⋅ c⋅ sin α( )⋅→

----------------------------------------------------------------------- Part 5a -------------------------------------------------------------------------

Now, to study the lift per unit span for the first airfoil, we'll backtrack to the normal component at Cp1.

Vcp1n

γ1−

2 π⋅c

2⋅

γ2

2 π⋅6

4c⋅

⋅

+ Vinf sin α( )⋅+=

Since we're neglecting the effects of the second air foil, we'll neglect the second term. At the same time, we'll set the

equation to zero and solve for γ1.

0γ1−

2 π⋅c

2⋅

Vinf sin α( )⋅+= solve γ1, π Vinf⋅ c⋅ sin α( )⋅→

The circulation is equal to the sum of the γ, which in this case is the only γ we have...

Γa π Vinf⋅ c⋅ sin α( )⋅=

The lift per unit span can be found using equation 3.140.

Lp ρinf Vinf⋅ Γ⋅=

For the first foil, the result is..

Lpa ρinf V.inf⋅ π Vinf⋅ c⋅ sin α( )⋅( )⋅= π Vinf2

⋅ c⋅ ρinf⋅ sin α( )⋅=

----------------------------------------------------------------------- Part 5b -------------------------------------------------------------------------

Since the second foil is geometrically the same as the first foil, as well as exhibit full flow similarity, the lift per unit span

for the second foil will be equal to that of the first foil.

Lpb ρinf V.inf⋅ π Vinf⋅ c⋅ sin α( )⋅( )⋅= π Vinf2

⋅ c⋅ ρinf⋅ sin α( )⋅=

----------------------------------------------------------------------- Part 5c -------------------------------------------------------------------------

For the entire system, we'll use the result we gathered earlier when we performed steps 3, 4, and 5 for the whole

system...

Recall we found the circulation to be..

Γsys 2 π⋅ Vinf⋅ c⋅ sin α( )⋅=

The lift per unit span, therefore, is..

Lpsys ρinf Vinf⋅ 2 π⋅ Vinf⋅ c⋅ sin α( )⋅( )⋅= simplify Lpsys 2 π⋅ Vinf2

⋅ c⋅ ρinf⋅ sin α( )⋅=→

----------------------------------------------------------------------- Part 5d -------------------------------------------------------------------------

If we add the lift per unit span for the first foil, as well as the lift per unit span for the second foil, we find it does indeed

equal the lift per unit span for the entire system.

π Vinf2

⋅ c⋅ ρinf⋅ sin α( )⋅ π Vinf2

⋅ c⋅ ρinf⋅ sin α( )⋅+ simplify 2 π⋅ Vinf2

⋅ c⋅ ρinf⋅ sin α( )⋅→

Lpa Lpb+ Lpsys=

----------------------------------------------------------------------- Part 6a -------------------------------------------------------------------------

The coefficient of lift can be expressed as..

clΓ

1

2ρinf Vinf

2⋅ c⋅

=

Revisiting part 5a, recall the circulation was..

Γa π Vinf⋅ c⋅ sin α( )⋅=

Plugging the circulation into the lift coefficient equation, as well as simplify a little, the result is..

cla

π Vinf⋅ c⋅ sin α( )⋅

1

2ρinf Vinf

2⋅ c⋅

= simplify cla2 π⋅ sin α( )⋅

Vinf ρinf⋅=→

----------------------------------------------------------------------- Part 6b -------------------------------------------------------------------------

Using the same technique in part 6a, as well as remembering the circulation for foil A and B independantly is the

same, the lift coefficient for the second foil is again equal to that of the first foil.

clb

π Vinf⋅ c⋅ sin α( )⋅

1

2ρinf Vinf

2c⋅⋅

= simplify clb2 π⋅ sin α( )⋅

Vinf ρinf⋅=→

----------------------------------------------------------------------- Part 6c -------------------------------------------------------------------------

Now we'll find the lift coefficient for the entire system. Again we'll use our previous results. The circulation for the

system was determined in part 4.

Γsys 2 π⋅ Vinf⋅ c⋅ sin α( )⋅=

Entering this circulation into the coefficient of lift equation, the result is..

clsys

2π Vinf⋅ c⋅ sin α( )⋅

1

2ρinf Vinf

2⋅ 2c( )⋅

= simplify clsys2 π⋅ sin α( )⋅

Vinf ρinf⋅=→

----------------------------------------------------------------------- Part 6d -------------------------------------------------------------------------

Again, we'll add parts a with b and compare to c.

2 π⋅ sin α( )⋅

Vinf Vinf⋅

2 π⋅ sin α( )⋅

Vinf ρinf⋅+ 2

2 π⋅ sin α( )⋅

Vinf ρinf⋅

⋅=

So (assuming no errors were made.. which is entirely possible), the lift coefficients do not exhibit superposition ( a + b

does not equal c)

References:

Anderson, John D. Jr. Fundamentals of Aerodynamics. 4th ed. New York: McGraw-Hill, 2007.