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7/26/2019 Aerodynamics Review Homework http://slidepdf.com/reader/full/aerodynamics-review-homework 1/8 AE 6015 Homework 1 Name: Edwin Goh GTID: 902931454 1. NACA 2412 - Inviscid, Incompressible Analysis a-i.) What is the lift curve slope for the NACA 2412 airfoil? Between -13° and 17°, the lift curve slope of the NACA 2412 airfoil has been empirically found to be 5.56 rad -1 . a-ii.) What influence does the Reynolds number have on the lift-curve slope?  According to Abbott and Von Doenhoff, variations of Reynolds number between 3 and 9 million and variation of camber up to 0.04c appear to have no systematic effect on the lift curve slope.  Instead, it was found that variations in Reynolds number affect the maximum achievable lift coefficient, which, according to Anderson, is a difficult viscous flow problem.  This indicates that, up until the point of flow separation at high angles of attack, an airfoil s lift-curve slope is independent of the relative strengths of the viscous and inertial forces in the flow. Consequently, we can see infer the independence of the generation of lift under attached flow on viscous effects, thus providing a valid situation for one of the

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7/26/2019 Aerodynamics Review Homework

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AE 6015 Homework 1

Name: Edwin Goh

GTID: 902931454

1. NACA 2412 - Inviscid, Incompressible Analysis

a-i.) What is the lift curve slope for the NACA 2412 airfoil?

Between -13° and 17°, the lift curve slope of the NACA 2412 airfoil has been empirically found to be

5.56 rad-1. 

a-ii.) What influence does the Reynolds number have on the lift-curve slope?

 According to Abbott and Von Doenhoff, “variations of Reynolds number between 3 and 9 million andvariation of camber up to 0.04c appear to have no systematic effect on the lift curve slope. ” Instead, it

was found that variations in Reynolds number affect the maximum achievable lift coefficient, which,

according to Anderson, “is a difficult viscous flow problem.” This indicates that, up until the point of flow

separation at high angles of attack, an airfoil ’s lift-curve slope is independent of the relative strengths of

the viscous and inertial forces in the flow. Consequently, we can see infer the independence of the

generation of lift under attached flow on viscous effects, thus providing a valid situation for one of the

7/26/2019 Aerodynamics Review Homework

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underlying assumptions of thin-airfoil theory --- inviscid flow.

a-iii.) What is the angle of zero-lift?

From the figure above, the angle of zero-lift, α L=0  = -1.93 °. Note that this is a non-zero number because

the NACA 2412 airfoil is indeed a cambered airfoil as opposed to the NACA 0012, the zero-lift angle ofattack of which would be 0°.

b-i.) What is the moment slope at the quarter-chord point?

From the figure above, it can be seen that the moment slope has a value of 0.0037 rad-1.

b-ii.) Is this the aerodynamic center? Why or why not?For this airfoil, the quarter-chord point can be considered the aerodynamic center because the change

of moment coefficient with respect to the section angle of attack is on the order of 0.0037, which essen-

tially means that the c m,c /4 is constant with angle of attack.

c-i.) What is the angle of attack at which minimum drag is achieved?

abc

c-ii.) Is this true for all Reynolds numbers? What influence does Reynolds

number have on the drag?

abc

2. NACA 2412 Transonic Analysis

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a.) Reproduce experimental data for the NACA 2412 airfoil under

incompressible conditions.

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b.) Plot [c L vs  M∞], [c D vs  M∞], [c m,c /4 vs  M∞], and [c L,α  vs  M∞] for the NACA

2412 airfoil.

c.) Compute the  Mcr and  Mdd from the drag plots, giving a short explanation of

how you chose these values.

 Mcr

The M cr  was found to be:

 MDD

One common definition of the M DD is: the Mach number at whichdCd 

dM∞ first exceeds 0.1. Using the

c d vs.α  curve, M DD was found to be equal to

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d.) Using Prandtl-Glauert, Karman-Tsien, and  Laitone compressibility corrections,

compute the  Mcr using equations and the incompressible drag values from

Abbott and von Doenhoff. Compare with values obtained in part (c.).

i.) Prandtl-Glauert

In[1]:=   c p0 = -0.8239;  (*   Obtained from XFOIL *)c p[ M ∞] =   c p0

1 - M ∞2

;  (*   For any given FREESTEAM Mach number,

this will give the LOCAL Cp based on P-G rule *)

c p,critical[ M ∞] =   2

γ M ∞2

1 + 0.5 γ- 1 M ∞2

1 + 0.5 γ- 1

γγ-1

- 1   /. γ  7 / 5;

(*   For any FREESTREAM Mach number,

this gives us the LOCAL c p   for a point where the LOCAL Mach number

is 1. Note that this point is not necessarily on the airfoil. *) McrPG2412 =  M ∞ /. Solve[c p[ M ∞]  c p,critical[ M ∞], M ∞, Reals][[2]] //

Quiet (*Solving the two equations for M ∞,

 we find the FREESTREAM Mach number at the intersection point where

the C p   obtained from P-G rule is equal to the  (local)   critical C p   *)Out[1]=   0.638208

ii.) Karman-Tsien

In[2]:=   c p[ M ∞] =   c p0

1 - M ∞2 +   M ∞2

1+   1- M ∞2 c p0

2

;  (*   For any given FREESTEAM Mach number,

this will give the LOCAL Cp based on K -T rule *)

c p,critical[ M ∞] =   2

γ M ∞21 + 0.5 γ- 1 M ∞2

1 + 0.5 γ- 1

γγ-1

- 1   /. γ  7 / 5;

(*  For any FREESTREAM Mach number,

this gives us the LOCAL c p   for a point where the LOCAL Mach number is 1. *) McrKT2412 =  M ∞ /. Solve[c p[ M ∞]  c p,critical[ M ∞], M ∞, Reals][[2]] //

Quiet (*Solving the two equations for M ∞,

 we find the FREESTREAM Mach number at the intersection point where

the C p   obtained from K -T rule is equal to the  (local)   critical C p   *)Out[2]=   0.618459

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iii.) Laitone

In[3]:=   c p[ M ∞] =   c p0

1 - M ∞2 1 + M ∞2 1 + γ-1

2 M ∞2 c p0

2

/. γ  7 / 5;

(*   For any given FREESTEAM Mach number,

this will give the LOCAL Cp based on P-G rule *)

c p,critical[ M ∞] =   2

γ M ∞21 + 0.5 γ- 1 M ∞2

1 + 0.5 γ- 1

γγ-1

- 1   /. γ  7 / 5;

(*   For any FREESTREAM Mach number,

this gives us the LOCAL c p   for a point where the LOCAL Mach number is 1. *) McrLT2412 =  M ∞ /. Solve[c p[ M ∞]  c p,critical[ M ∞], M ∞, Reals][[2]] //

Quiet (*Solving the two equations for M ∞,

 we find the FREESTREAM Mach number at the intersection point where theC p   obtained from Laiton's rule is equal to the (local)   critical C p   *)

Out[3]=   0.608414

3. NASA SC(2)-0412 Versus NACA 2412

a.) Plot the coordinates of the NACA 2412 and the NASA SC(2)0412 airfoil

together on one plot, and compare the two geometries.

Similarities: 

- Both are cambered airfoils.

- Both leading edges have approximately the same curvature.

Differences: 

- The NASA SC(2)-0412 airfoil has a flatter upper surface.  delays the drag-divergence Mach

number by delaying the shocks associated with locally supersonic flow.

- The NASA SC(2)-0412 airfoil has a cusp near its trailing edge. This cusp was added to compensate

for the lost in lift caused by the flatter upper surface.

- The NACA 2412 has a larger thickness than the NASA SC(2)-0412.

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b.) Plot [c L vs  M∞], [c D vs  M∞], [c m,c /4 vs  M∞], and [c L,α  vs  M∞] for both the NASA

SC(2)-0412 and NACA 2412 airfoils.

c.) Compute the  Mcr and  Mdd from the drag plots for the supercritical airfoil,

explain your procedure, and compare these values with that obtained for the

NACA 2412.

blah blah blah

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4. NACA 2412 - Fully Turbulent Flow

a.) Plot [c L vs  M∞], [c D vs  M∞], [c m,c /4 vs  M∞], and [c L,α  vs  M∞] for both the

turbulent and laminar cases.

abc

b.) Compute the  Mcr and  Mdd from the drag plots, giving a short explanation of

how you chose these values.

It was found from the C D vs M ∞ graph that, for the current definition of drag-divergence Mach number,

M DD = 0.74

c.) Plot the c p distribution for M = 0.1, 0.4,  Mcr and  MDD.

i.) Compute and show Cpcrit on the plots

ii.) List the location and strength of any shocks, separated flows, and the location

of the stagnation pressure for each (shock/separated flow).

d.) Compute and plot the drag polars for this airfoil with and without transition.

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