Drag calculation short write-up. To be tied in with performance…?

Once the configuration, geometry and sizing were set in place, it was necessary to calculate the total drag over the airplane to aid the propulsion team in selecting an engine with an appropriate amount of thrust.

A useful measure of the drag of an airplane is the zero-lift drag coefficient, Cd,0. Accurately estimating the value of Cd,0 is vital in the design of any aircraft. The main component of Cd,0 for most aircraft is the skin friction drag. Appendix A shows the calculation of skin friction drag for the current design, arriving at a value of skin friction drag coefficient of 0.0064.

The other components of Cd,0 include pressure drag, interference drag and drag due to the interactions between the wing and fuselage and the tail and fuselage. Without wind tunnel testing, it is extremely difficult to accurately gauge the effect of these parameters. Thus, reflecting upon the value of skin friction drag obtained and on comparison of similar aircraft the conservative value of Cd,0 = 0.02 has been used throughout the design of this UAV. Output from MATLAB program Tornado has largely agreed with this approximation.

To go in Appendices:

SKIN FRICTION DRAG

Wings

ASSUME: wing can be treated as a flat plate (assumption should hold for a wing of 10% thickness flying at relatively low angles of attack).

Need to determine what portion of the airfoil has a laminar boundary layer, and conversely what portion is turbulent.

ASSUME: transition to turbulence occurs at Re=5x105 for a flat plate. Thus ℜcr=5×105.

Given:

V ∞=150 kmhr

=41.67ms

Mean chord length = 0.72m, wing span = 5m.

At a cruising height of 500ft,

T ∞=287.17K , p∞=99509.54 Pa ,ρ∞=1.2072kg

m3

For air at STP, μ0=1.7894×10−5 , T 0=288.16K

ASSUME: air is an ideal gas.

Using the temperature-viscosity relationship for air as an ideal gas:

μ∞

μ0=(T ∞

T 0)32 ( T 0+110

T ∞+110 )We obtain for air at 500ft:

μ∞=1.7894×10−5( 287.17288.16 )32 ( 398.16387.17 )

μ∞=1.8307×10−5

From definition of Reynolds number:

ℜcr=ρ∞V ∞ xcr

μ∞

=5×105

xcr=ℜcr μ∞

ρ∞ V ∞

=5×105×1.8307×10−5

1.2072×41.67=0.1820m

Therefore from the leading edge to 0.1820m from the leading edge the boundary layer is laminar, and from 0.1820m to the trailing edge at 0.72m from the leading edge the boundary layer is turbulent.

Firstly, assuming that the boundary layer is turbulent over the entire airfoil:

ℜc=ρ∞V ∞ c

μ∞

=1.2072×41.67×0.721.8307×10−5 =1.9784×106

The skin friction drag coefficient for a turbulent boundary layer is given by:

cdsf ,tblc=0.074

ℜc

15

= 0.074

(1.9784×106 )15

=0.004074

Therefore the total drag due to skin friction over an entire turbulent airfoil is given by:

Dsf ,tblc=cdsf , tblc q∞ Sc

Where:

q∞=12

ρ∞V ∞2

Thus:

Dsf ,tblc=0.004074×12

×1.2072×41.672×0.72×52=7.6858N

Secondly, assuming that the boundary layer is turbulent from the leading edge to x = 0.1820m:

ℜx cr=5×105

cdsf ,tblxcr=0.074

ℜxcr

15

= 0.074

(5×105 )15

=0.005363

Dsf ,tblxcr=cdsf ,tblx cr

q∞ Sx cr=0.005363×

12

×1.2072×41.672×0.1820×52

¿2.5575N

Therefore the total drag over the turbulent section is given by:

Dsf ,tbl=Dsf , tblc−D sf ,tblxcr=7.6858−2.5575=5.1283N

Finally, considering the laminar boundary layer section of airfoil:

ℜx cr=5×105

The skin friction drag coefficient for a laminar boundary layer is given by:

cdsf ,lblxcr=1.328

ℜx cr

12

= 1.328

(5×105 )12

=0.001878

Dsf ,lbl=cdsf , lblxcrq∞ S xcr

=0.001878×12

×1.2072×41.672×0.1820×52

¿0.8956N

Therefore the total skin friction drag over the surface of a wing becomes:

Dsf ,surface=Dsf ,lbl+D sf ,tbl=0.8956+5.1283=6.0239N

Over both surfaces of both wings, the total skin friction drag over the wing is given as:

Dsf ,wings=Dsf , surface ×4=6.0239×4=24.0956N

Tail

The skin friction drag over the tail surfaces can be calculated using the same techniques and assumptions as for the wings, with:

Mean chord = 0.5m, tail span = 1.7m.

Firstly, assuming that the boundary layer is turbulent over the entire airfoil:

ℜc=ρ∞V ∞ c

μ∞

=1.2072×41.67×0.51.8307×10−5 =1.3739×106

cdsf ,tblc=0.074

ℜc

15

= 0.074

(1.3739×106 )15

=0.004382

Dsf ,tblc=cdsf , tblc q∞ Sc=0.004382×12

×1.2072×41.672×0.5×1.72

¿1.9519N

Secondly, assuming that the boundary layer is turbulent from the leading edge to x = 0.1820m:

ℜx cr=5×105

cds f , tblxcr=0.074

ℜ xcr

15

= 0.074

(5×105 )15

=0.005363

Dsf ,tblxcr=cdsf ,tblx cr

q∞ Sx cr=0.005363×

12

×1.2072×41.672×0.1820×1.72

¿0.8695N

Therefore the total drag over the turbulent section is given by:

Dsf ,tbl=Dsf , tblc−D sf ,tblx cr=1.9519−0.8695=1.0824N

Finally, considering the laminar boundary layer section of airfoil:

ℜx cr=5×105

cdsf ,lblxcr=1.328

ℜx cr

12

= 1.328

(5×105 )12

=0.001878

Dsf ,lbl=cdsf , lblxcrq∞ S xcr

=0.001878×12

×1.2072×41.672×0.1820×1.72

¿0.3045N

Therefore the total skin friction drag over the surface of a wing becomes:

Dsf ,surface=Dsf ,lbl+D sf ,tbl=0.3045+1.0824=1.3869N

Across the four tail surfaces:

Dsf ,tail=D sf ,surface ×4=1.3869×4=5.5476N

Fuselage

The fuselage is treated basically the same way as wings and tail, assuming that the fuselage is a perfect cylinder, and that the skin friction over the cylinder is the same as the skin friction over the cylinder opened out to a flat plate. This assumption is reasonable as the flow is parallel to the surface at all points, and hence the orientation of the surface in this z direction has little effect. With this simplification:

Fuselage length = 2.6m, fuselage diameter = 0.4m.

Thus, the fuselage span = 2 (0.2) = 1.2566mπ

Firstly, assuming that the boundary layer is turbulent over the entire fuselage:

ℜl=ρ∞V ∞ c

μ∞

=1.2072×41.67×2.61.8307×10−5

=7.1443×106

cdsf ,tbll=0.074

ℜl

15

= 0.074

(7.1443×106 )15

=0.003151

Dsf ,tbll=cdsf , tbll q∞ Sl=0.003151×12

×1.2072×41.672×2.6×1.2566

¿10.7898N

Secondly, assuming that the boundary layer is turbulent from the leading edge to x = 0.1820m:

ℜx cr=5×105

cdsf ,tblxcr=0.074

ℜxcr

15

= 0.074

(5×105 )15

=0.005363

Dsf ,tblxcr=cdsf ,tblx cr

q∞ Sx cr=0.005363×

12

×1.2072×41.672×0.1820×1.2566

¿1.2855N

Therefore the total drag over the turbulent section is given by:

Dsf ,tbl=Dsf , tbll−D sf ,tblxcr=10.7898−1.2855=9.5043N

Finally, considering the laminar boundary layer section of the fuselage:

ℜx cr=5×105

cdsf ,lblxcr=1.328

ℜxcr

12

= 1.328

(5×105 )12

=0.001878

Dsf ,lbl=cdsf , lblxcrq∞ S xcr

=0.001878×12

×1.2072×41.672×0.1820×1.2566

¿0.4502N

Therefore the total skin friction drag over the surface of the fuselage becomes:

Dsf ,surface=Dsf ,lbl+D sf ,tbl=0.4502+9.5043=9.9545N

Since only the outer surface of the fuselage is wetted by the flow, there is no need to multiply this further.

Dsf ,fuselage=Dsf , surface=9.9545N

Whole aircraft

Thus, the skin friction drag over the whole aircraft is given as:

Dsf =Dsf , wings+D sf ,tail+D sf ,fuselage=24.0956+5.5476+9.9545=39.5977N

Corrected Value

Using the skin friction drag correction factor plots from the lecture series, we can calculate a value of whole aircraft skin friction drag corrected for errors incurred when approximating the surfaces as flat plates:

Dsf =Dsf , wings× K w+D sf ,tail × K t+Dsf , fuselage × K f

Where Kw = Kt =1.21 for a 10% thick lifting surface, and Kf = 1.21 also for a body surface with length/diameter ratio of 6.5.

Thus, the corrected whole aircraft skin friction drag value becomes:

Dsf =39.5977×1.21=47.9132N

Dsf =47.9132N

The whole aircraft skin friction drag coefficient can be determined from this value using the whole aircraft wetted surface area as a reference:

Swetted=Swings+Stail+S fuselage=0.72×52

×4+0.5×1.72

×4+2.6×1.2566

Swetted=12.1672m2

cd sf ,wet=

Dsf

q∞ Swetted

= 47.913212

×1.2072×41.672×12.1672=0.0038

Since cd,0 is a parameter based upon wing area, it is this area of reference for which cd,sf will be most useful. Thus,

cd sf=cdsf ,wet

×Swetted

Swings

=0.0038×12.1672

0.72×52

×4=0.0064

cd sf=0.0064

References

Fundamentals of Aerodynamics, Anderson

Loftin, LK, Jr.. "Quest for performance: The evolution of modern aircraft” NASA SP-468

Lecture notes.