# Boundary Layers Skin - Friction Drag Example Problem

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```					Boundary Layers / Skin - Friction Drag

Example Problem: The skin-friction drag of an airplane wing.

Calculate the skin-friction drag of the Wright Flyer I wings.
Note: The Wright Flyer I is a biplane.

Given:

flight speed:                         V = 50 km /hr
atmospheric conditions:               standard sea level
wing span:                            b = 12.29 m
wing planform area:                   S = 23.69 m2

Solution:
1. We will treat the wing as a flat plate. First, let’s calculate the Re
@ the trailing edge (TE) of the wing. If it is less than Recr = 5x105,
then we may assume a laminar BL throughout the entire wing.

At standard, sea level conditions: ρ ∞ = 1.225 kg / m 3
µ ∞ = 1.785 x10 −5 N . sec/ m 2
T∞ = 15 0 C
50 x10 3          m
V∞ =           = 13.89
3,600           sec
S 23.69
c= =            = 1.93 m
b 12.29
ρ V c 1.225 x13.89 x1.93
Re c = ∞ ∞ =                      = 1.84 x10 6 f Re cr = 5 x10 5
µ∞         1.785 x10 −5
1           1                        N
q∞ = ρ ∞V∞2 = x1.225 x13.89 2 = 118.17 2
2           2                       m

Thus, the BL @ the TE of the wing is turbulent.

2.    Next, we must locate the transition line on the wing. If it turns
out to be very close to the leading edge (LE), we may assume a
turbulent BL throughout the entire wing.

ρ ∞V∞ xcr                   5 x10 5 x1.785 x10 −5
Re cr =             = 5 x10 5 ⇒ xcr =                       = 0.525 m
µ∞                          1.225 x13.89

Well, the transition line is not close to the LE, so we must account for
the fact that part of the BL is laminar and part is turbulent in the
calculation of the skin-friction drag.

3. Calculate the skin-friction drag for the whole wing area assuming a
TBL from LE to TE:

0.074       0.074
Re c p 10 7 ⇒ use Pr andtl ' s eq. C F ,TBL =          =                     = 0.00413
0.2
Re c      (
1.84 x10 6    )
0.2

DSF , A+ B ,TBL = q ∞ SC F ,TBL = 118.17 x 23.69 x 0.00413 = 11.56 N

4. Calculate the turbulent skin-friction drag for area B by calculating
the turbulent skin-friction drag for area A and subtracting it from the
result in step (3) above:

Assuming a TBL over area A:
0.074      0.074
C F ,TBL =           =             = 0.00536
0.2
Re cr    (
5 x10 5 )
0.2

A = bxcr = 12.29 x0.525 = 6.452 m 2

DSF , A,TBL = q ∞ AC F ,TBL = 118.17 x6.452 x0.00536 = 4.09 N

DSF , B = DSF , A+ B ,TBL − DSF , A,TBL = 11.56 − 4.09 = 7.47 N

5. Calculate the laminar skin-friction drag of area A:

1.328     1.328
C F , LBL =          =             = 0.00188
0.5
Re cr   (5 x10 5 )
0.5

DSF , A = q ∞ AC F , LBL = 118.17 x6.452 x0.00188 = 1.43 N

6. The drag for one side of one of the biplane wings is then:

DSF ,oneside = DSF , A + DSF , B = 1.43 + 7.47 = 8.9 N

7. The total drag of the entire wing is:

DSF ,tot = 4 DSF ,oneside = 4 x8.9 ⇒ DSF ,tot = 35.6 N

A different approach, less accurate but much faster, would be to use
the equation for the skin-friction coefficient for combined LBL and
TBL:

0.074 1700         0.074                   1700
CF =          −      =                      −              = 0.00321
0.2
Re c     Re c       (
1.84 x10 6    )0.2
1.84 x10 6

DSF ,oneside = 118.17 x 23.69 x0.00321 = 8.98 N

DSF ,tot = 4 x8.98 = 35.92 N

which is almost identical to the answer in step 7!

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