Fundamentals of Aerodynamics Reviewer Part 3

VISCOUS EFFECTS, THE BOUNDARY LAYER AND FLOE SEPARATION

Skin Friction – Air resistance and it is the tangential component of force on the surface of a body due to the friction between the two particles.

Stream line and Turbulent flow – A stream line flow may be defined as a smooth non turbulent flow. A turbulent flow is defined as a flow characterized by turbulence, that is, a flow in which the velocity varies erratically in both magnitude and direction with time.

Laminar flow – The word laminar is derived from the latin word “lamina” meaning a thin plate of metal or some other material. Laminar flows employs, the concept that air is flowing in thin sheets or layers close to the surface of a wing with no disturbance between the layers of air.

Boundary Layer – A boundary layer is that layer of air adjacent to the airfoil surface. The cause of the boundary layer is the friction between the surface of the wing and the air.

Laminar Boundary Layer – Is the laminar boundary layers the flow is steady and smooth. As a result, the layer is very thin and so the form drag is very small. Also, the velocity gradient at the walls through large enough to give significant viscous stress is yet only moderate, so that the skin friction, though not negligible, is also very small.

𝛿=5.2 𝑥

√𝑅𝑁 𝑋

h𝑊 𝑒𝑟𝑒 :𝛿=𝐵𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑙𝑎𝑦𝑒𝑟 h𝑡 𝑖𝑐𝑘𝑛𝑒𝑠𝑠𝑥=𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛𝑝𝑜𝑖𝑛𝑡𝑅𝑁 𝑋

=𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛 𝑅𝑎𝑦𝑛𝑜𝑙𝑑𝑠𝑁𝑢𝑚𝑏𝑒𝑟

The rubbing of the boundary layer on the flat plate gives rise to friction forces of: friction drag. The skin friction drag coefficient for one side of a plate in laminar flow is given by:

𝐶 𝑓=𝐷 𝑓

12𝜌𝑉 2𝑆

=1.328√𝑅𝑁

h𝑊 𝑒𝑟𝑒 :𝑆=𝑎𝑟𝑒𝑎𝑜𝑓 𝑜𝑛𝑒𝑠𝑖𝑑𝑒𝑜𝑓 𝑎𝑝𝑙𝑎𝑡𝑒

𝑅𝑁=𝑅𝑎𝑦𝑛𝑜𝑙𝑑𝑠𝑛𝑢𝑚𝑏𝑒𝑟 𝑏𝑎𝑠𝑒𝑑𝑜𝑛 h𝑡 𝑒𝑡𝑜𝑡𝑎𝑙𝑝𝑙𝑎𝑡𝑒 h𝑙𝑒𝑛𝑔𝑡

Turbulent Boundary Layer – In a turbulent boundary layer, the flow is unsteady and not smooth, but eddying.

When the flow is transitioned to turbulent flow, the boundary layer thickness will be increased. In fact, this phenomenon is often used to determine the location of the transition region. The boundary layer thickness can be determined by:

𝛿=0.37 𝑥

(𝑅𝑁 𝑋 )15

The skin friction drag coefficient for a flat plate can be calculated with formula:

𝐶 𝑓=𝐷 𝑓

12𝜌𝑉 2𝑆

= 0.455

(log 10𝑅𝑁 )2.58

h𝑊 𝑒𝑟𝑒 :𝜌=𝑎𝑖𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑛

𝑠𝑙𝑢𝑔𝑓𝑡 3

𝑜𝑟𝑘𝑔𝑚3

Critical Raynolds Number – Experimentally value for which when the values of R.N. less than critical, the flow is smooth or laminar; for values greater than the critical R.N., the flow is turbulent.

Transition takes place on a flat plate at point x determined by:

(𝑅𝑁𝑋 )𝑐𝑟𝑖𝑡=( 𝜌𝑉𝑥µ )𝑐𝑟𝑖𝑡

=3.5𝑥105𝑡𝑜106

𝑅𝑁 𝑋=𝜌𝑉𝐿µ

=𝜌𝑉𝐶µ

h𝑊 𝑒𝑟𝑒 :𝜌= 𝑓𝑟𝑒𝑒𝑠𝑡𝑟𝑒𝑎𝑚𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑛

𝑠𝑙𝑢𝑔𝑓𝑡3


𝑉= 𝑓𝑟𝑒𝑒𝑠𝑡𝑟𝑒𝑎𝑚𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛𝑓𝑡𝑠𝑜𝑟

𝑚𝑠

𝐿=𝐶= h𝐶 𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 h𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 h𝑡 𝑒𝑏𝑜𝑑𝑦𝑖𝑛 h𝑡 𝑒 𝑓𝑙𝑜𝑤𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛𝑖𝑠 𝑒𝑞𝑢𝑎𝑙𝑡𝑜 h𝑡 𝑒h𝑐 𝑜𝑟𝑑 h𝑙𝑒𝑛𝑔𝑡 𝑓𝑜𝑟 𝑎𝑛𝑎𝑖𝑟𝑓𝑜𝑖𝑙 𝑖𝑛 𝑓𝑡 𝑜𝑟 𝑚 .

µ=𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑜𝑓 𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑖𝑛𝑠𝑙𝑢𝑔𝑓𝑡 .𝑠𝑒𝑐

𝑜𝑟𝑘𝑔

𝑚 .𝑠𝑒𝑐

For air µ increases with temperature and can be calculated by the following approximate formula for the standard atmosphere.

µ=2.329 𝑥10− 8𝑇

32

𝑇+216,𝑠𝑙𝑢𝑔𝑓𝑡 .𝑠𝑒𝑐

,𝑇 𝑖𝑛 ᵒ𝑅

Or

µ=1.458 𝑥10− 6𝑇

32

𝑇 +110.4,

𝑘𝑔𝑚 .𝑠𝑒𝑐

,𝑇 𝑖𝑛𝐾

Example # 1: Two plates, one having 6 ft span and 3 ft chord, the other having 9ft span and 6ft chord are placed in different airstream. The free stream velocity for the smaller plate is 100ft per sec. It is found that the total skin friction drag for the two plate is the same. Find the airspeed for the larger plate. Assume the laminar flow at standard sea level conditions.

𝐺𝑖𝑣𝑒𝑛 :𝐴𝑠𝑠𝑢𝑚𝑒𝑙𝑎𝑚𝑖𝑛𝑎𝑟 𝑓𝑙𝑜𝑤𝑎𝑡 𝑆𝑆𝐿𝐶

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :

𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑎𝑙𝑎𝑟𝑔𝑒𝑝𝑙𝑎𝑡𝑒

Small plate

Large plate

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :𝐷 𝑓 𝑠=𝐷 𝑓 𝑙

𝐶 𝑓=𝐷 𝑓

12ρ𝑉 2𝑆

𝐷 𝑓 =𝐶 𝑓12ρ𝑉 2𝑆

𝐵𝑢𝑡𝐶 𝑓 𝑓𝑜𝑟 𝑙𝑎𝑚𝑖𝑛𝑎𝑟 𝑓𝑙𝑜𝑤𝑠

𝐶 𝑓=1.328

√𝑅𝑁

1.328

√𝑅𝑁𝑠

𝑉 𝑠2𝑏𝑠𝐶𝑠=

1.328

√𝑅𝑁𝑙

𝑉 𝑙2𝑏𝑙𝐶𝑙

𝐵𝑢𝑡 𝑅𝑁

𝑅𝑁 𝑋=𝜌𝑉𝐶µ

𝑉 𝑠2𝑏𝑠𝐶𝑠

√ ρ𝑉 𝑠𝐶 𝑠

µ

=𝑉 𝑙

2𝑏𝑙𝐶𝑙

√ ρ𝑉 𝑙𝐶𝑙

µ

𝑉 𝑠

32𝑏𝑠𝐶𝑠

12=𝑉 𝑙

32 𝑏𝑙𝐶𝑙

12

𝑉 𝑙

32=𝑉 𝑠

32 𝑏𝑠𝑏𝑙 (

𝐶𝑠

𝐶𝑙)12

𝑉 𝑙=𝑉 𝑠(𝑏𝑠

𝑏𝑙)23 (𝐶𝑠

𝐶𝑙)13

𝑉 𝑙=(100 𝑓𝑡𝑠 )( 6 𝑖𝑛9 𝑖𝑛)32 ( 3 𝑖𝑛6 𝑖𝑛)

13

𝑉 𝑙=60.57𝑓𝑡𝑠

Example # 2: An airplane is flying at a density altitude of 15,000ft. At an ambient temperature a -39ᵒF. If the wing chord is 6ft and the equivalent airspeed is 200knots. What is the overall Raynolds number of the wing?

𝐺𝑖𝑣𝑒𝑛 :h𝑑=15,000 𝑓𝑡 ( 𝑚

3.28 𝑓𝑡 )=4,573.17𝑚𝑇=−39˚ 𝐹 +460=421 ˚𝑅 ( 288𝐾519 ˚𝑅 )=233.62𝐾𝐶=6 𝑓𝑡( 𝑚

3.28 𝑓𝑡 )=1.83𝑚𝑉 𝑒=200

𝑛 .𝑚𝑖𝑙𝑒𝑠h𝑜𝑢𝑟 ( 6,080 𝑓𝑡𝑛 .𝑚𝑖𝑙𝑒𝑠 )( 𝑚

3.28 𝑓𝑡 )( h𝑜𝑢𝑟3,600𝑠𝑒𝑐 )=102.99𝑚𝑠

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝑅𝑁

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :

𝑅𝑁=𝜌𝑉𝐶µ

ρ=ρ𝑂 (1+𝑎h𝑑𝑇𝑂)4.26

ρ=(1.225 𝑘𝑔𝑚3 )[1+ (−0.00651 𝐾𝑚 ) (4,573.17𝑚 )

288𝐾 ]4.26

σ= ρρ𝑜

=0.770

𝑘𝑔𝑚3

1.225𝑘𝑔𝑚3

σ=0.629

𝑉=𝑉 𝑒

√σ=102.99

𝑚𝑠

√0.629

𝑉=129.86𝑚𝑠

𝐹𝑜𝑟 µ

µ=1.458 𝑥10− 6𝑇

32

𝑇 +110.4,


,𝑇 𝑖𝑛𝐾

µ=1.458 𝑥10− 6 (233.62𝐾 )

32

233.63𝐾 +110.4,


,𝑇 𝑖𝑛𝐾

µ=1.5133𝑥10−5𝑘𝑔𝑚 .𝑠

𝑅𝑁=ρ𝑉𝐶µ

𝑅𝑁=(0.770 𝑘𝑔𝑚3 )(129.86𝑚𝑠 ) (1.83𝑚)

1.5133 𝑥10− 5 𝑘𝑔𝑚 .𝑠

𝑅𝑁=12,091,834.14

WIND TUNNEL

A device for testing aircraft and its force components in a controlled airstream under laboratory conditions.

TYPES OF LOW-SPEED WIND TUNNELS

OPEN-CIRCUIT WIND TUNNELS

CLOSED CIRCUIT TUNNEL

Advantages of Closed circuit with comparison with Open Circuit Tunnels

1.) Power Requirement for a given speed is lower.

2.) Particulate matter can be contained within the circuit.

3.) Noise is significantly lower.

4.) Laboratory air movement (air vents, doors, windows, etc.) does not affect wind tunnel flow.

5.) Air entering the test section is free of laboratory dust.

6.) Fan blades are not as vulnerable to damage from model failure.

Disadvantages of Closed circuit with Open Circuit Tunnels

1.) Cost is generally three times greater for a given test section size.

2.) Air supply is recycled which can be prohibitive when working with combustion engines.

3.) Footprint is much larger and requires more overall space.

4.) Increasing air temperature can become an issue during prolonged use.

FORCES AND MOTION OF AIRPLANE UNDER TESTING

1.) Lift

2.) Drag

3.) Side Force

4.) Pitching Moment

5.) Yawing Movement

6.) Rolling Movement

Importance of

Used in comparison on flow pattern of on theorem bodies which are geometrically similar but not in dimension.

Flow pattern similarity at a particular point

1.) Magnitude of velocity at constant proportion.

2.) Direction of flow is the same.

3.) Both bodies must be oriented or positioned in similar altitudes.

4.) Both bodies must be positioned at the same angle of attack.

Example # 1: Find the Raynold Number for an airplane wing 4ft chord, moving at 130mph through standard atmosphere.

𝐺𝑖𝑣𝑒𝑛 :𝐶=4 𝑓𝑡𝑉=130 h𝑚𝑝 ( 2215 )=190.67 𝑓𝑡𝑠𝑆𝑆𝐿𝐶




𝑅𝑁=(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡 3 )(190.67 𝑓𝑡𝑠 ) (4 𝑓𝑡 )

0.00000037372𝑠𝑙𝑢𝑔𝑓𝑡 . 𝑠

𝑅𝑁=4,850,932.14

Example # 2: Find the Raynold Number for an airplane wing with 3ft 6inches chord moving at 180mph through standard air.

𝐺𝑖𝑣𝑒𝑛 :𝐶=3 𝑓𝑡 𝑎𝑛𝑑6 𝑖𝑛=3.5 𝑓𝑡𝑉=180 h𝑚𝑝 ( 2215 )=264 𝑓𝑡𝑠𝑆𝑆𝐿𝐶


𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :𝑅𝑁=

𝜌𝑉𝐶µ

𝑅𝑁=(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡 3 )(264 𝑓𝑡𝑠 ) (3.5 𝑓𝑡 )

0.00000037372𝑠𝑙𝑢𝑔𝑓𝑡 .𝑠

𝑅𝑁=5,876,988.12

Example # 3: Find the Raynold Number for an airplane wing, 4ft chord, moving at 150mph. Air is +40°C, barometer,21in Hg.

𝐺𝑖𝑣𝑒𝑛 :𝐶=4 𝑓𝑡 ( 𝑚

3.28 𝑓𝑡 )=1.22𝑚𝑉=150 h𝑚𝑝 ( 2215 )=220 𝑓𝑡𝑠 ( 𝑚

3.28 𝑓𝑡 )=67.07𝑚𝑠𝑇=40 °𝐶+273=313𝐾𝑃=21𝑖𝑛 .𝐻𝑔( 101325𝑃𝑎29.92 𝑖𝑛 .𝐻𝑔 )=71,117.14 𝑁𝑚2




𝐹𝑜𝑟 ρ𝑃=ρ𝑅𝑇ρ=

𝑃𝑅𝑇

ρ=71,117.14

𝑁𝑚2

[287.08 (𝑁 ) (𝑚 )𝑘𝑔 .𝐾 ] (313𝐾 )

ρ=0.791𝑘𝑔𝑚3

𝐹𝑜𝑟 𝜇

𝜇=1.458𝑥 10− 6𝑇

32

𝑇 +110.4

𝜇=1.458𝑥 10− 6 (313𝐾 )

32

313+110.4

𝜇=0.0000191𝑘𝑔𝑚 .𝑠


𝑅𝑁=(0.791 𝑘𝑔𝑚3 )(67.07𝑚𝑠 ) (1.22𝑚 )

0.0000191𝑘𝑔𝑚 . 𝑠

𝑅𝑁=3,388,685.41

Example # 4: Find the velocity at which test should be run in a wind tunnel on a model wing of 4in chord in order that the Raynold Number shall be the same as for a wing with 4 ft chord at 100mph. Air under standard conditions in both cases.

𝐺𝑖𝑣𝑒𝑛 :𝐶1=4 𝑖𝑛( 𝑓𝑡

12𝑖𝑛 )=0.33 𝑓𝑡𝐶2=4 𝑓𝑡𝑅𝑁 1

=𝑅𝑁 2

𝑉 2=100 h𝑚𝑝 ( 2215 )=146.67 𝑓𝑡𝑠𝑆𝑆𝐿𝐶𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝑉 1

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :𝑅𝑁 1

=𝑅𝑁 2

𝑅𝑁 2=

(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡3 )(146.67 𝑓𝑡𝑠 ) (4 𝑓𝑡 )

3.7372𝑥10−7 𝑠𝑙𝑢𝑔𝑓𝑡 .𝑠

𝑅𝑁 2=3,731,505.833

𝑅𝑁 1=ρ𝑉 1𝐶1

𝜇

𝑉 1=𝑅𝑁1

𝜇

ρ𝐶1

𝑉 1=(3,731,505.833 )(3.7372𝑥10−7 𝑠𝑙𝑢𝑔𝑓𝑡 .𝑠 )

(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡3 ) (0.33 𝑓𝑡 )

𝑉 1=1,777.82𝑓𝑡𝑠 ( 1522 )=1,212.15 h𝑚𝑝

Example # 5: In a variable density wind tunnel, what pressure should test be run on a model with a 3in. chord, air velocity being 60mph in order that the Raynold Number shall be the same as for the full size wing of 4ft chord, moving at 100mph through the air? Air temperatures are the same in each case.

𝐺𝑖𝑣𝑒𝑛 :𝐶1=3 𝑖𝑛( 𝑓𝑡

12 𝑖𝑛)=0.25 𝑓𝑡𝑉 1=60 h𝑚𝑝 ( 2215 )=88 𝑓𝑡𝑠𝐶2=4 𝑓𝑡𝑉 2=100 h𝑚𝑝 ( 2215 )=146.67 𝑓𝑡𝑠𝑇 1=𝑇 2

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝑃𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :𝑅𝑁 1

=𝑅𝑁 2


𝜇

ρ=𝑅𝑁2

𝜇

𝑉 1𝐶1

ρ=(3,731,505.833 )(0.00000037372 𝑠𝑙𝑢𝑔𝑓𝑡 . 𝑠 )

(88 𝑓𝑡𝑠 )(0.25 𝑓𝑡 )

𝑅𝑁 2=

(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡3 )(146.67 𝑓𝑡𝑠 ) (4 𝑓𝑡 )

0.00000037372𝑠𝑙𝑢𝑔𝑓𝑡 . 𝑠

𝑅𝑁 2=3,731,505.833


𝜇

ρ=0.06339𝑠𝑙𝑢𝑔𝑓𝑡3

𝑃= ρ𝑔𝑅𝑇

𝑃=(0.06339 𝑠𝑙𝑢𝑔𝑓𝑡3 )(32,174 𝑓𝑡𝑠2 )(53.342 𝑓𝑡 . 𝑙𝑏𝑙𝑏°𝑅 ) (519° 𝑅 )

𝑃=56,462.81𝑙𝑏𝑓𝑡2 ( 1𝑎𝑡𝑚

2116.8𝑙𝑏𝑓𝑡 2 )=26.67 𝑎𝑡𝑚

Variable Density Wind Tunnel

A wind tunnel in which the air density can be increased by means of compressed air.

Flat Plates

Flat plates perpendicular to the airstream

𝐹=1.28𝜌2𝐴𝑉 2

h𝑊 𝑒𝑟𝑒 :𝐹=𝐹𝑜𝑟𝑐𝑒𝑜𝑛 𝑓𝑙𝑎𝑡 𝑝𝑙𝑎𝑡𝑒𝑛𝑜𝑟𝑚𝑎𝑙𝑡𝑜𝑎𝑖𝑟𝑠𝑡𝑟𝑒𝑎𝑚𝑖𝑛𝑙𝑏𝑜𝑟 𝑁

𝜌=𝑎𝑖𝑟𝑠𝑡𝑟𝑒𝑎𝑚𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑛𝑠𝑙𝑢𝑔𝑓𝑡 3


𝐴=𝑐𝑟𝑜𝑠𝑠𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙𝑎𝑟𝑒𝑎𝑖𝑛 𝑓𝑡 2𝑜𝑟 𝑚2

𝑉= 𝑓𝑟𝑒𝑒𝑠𝑡𝑟𝑒𝑎𝑚𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑓𝑝𝑠𝑜𝑟𝑚𝑝𝑠

Example# 1: What is the total force of a 45mph wind on a hangar door 40ft by 25ft?

𝐺𝑖𝑣𝑒𝑛 :𝑉=45 h𝑚𝑝 ( 2215 )=66 𝑓𝑡𝑠𝐴=40 𝑓𝑡 𝑥25 𝑓𝑡=1,000 𝑓𝑡2

𝑆𝑆𝐿𝐶𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝐹𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :

𝐹=1.28ρ20𝐴𝑉 2

𝐹=1.28 ( 0.002377 𝑠𝑙𝑢𝑔𝑓𝑡 32 ) (1,000 𝑓𝑡2 ) (66 𝑓𝑡𝑠 )2

𝐹=6,626.7 𝑙𝑏

Example # 2: What is the force against the side of the building 70ft long and 40ft high in a 90mph wind?

𝐺𝑖𝑣𝑒𝑛 :𝑉=90 h𝑚𝑝 ( 2215 )=132 𝑓𝑡𝑠𝐴=70 𝑓𝑡 𝑥 40 𝑓𝑡=2,800 𝑓𝑡2

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝐹𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :

𝐹=1.28ρ20𝐴𝑉 2

𝐹=1.28 ( 0.002377 𝑠𝑙𝑢𝑔𝑓𝑡 32 ) (2,800 𝑓𝑡2 ) (132 𝑓𝑡𝑠 )2

𝐹=74,219 𝑙𝑏

Example # 3: What force is required to push a flat plate, 3ft by 2ft at a speed of 35 fps in a direction perpendicular to its surface?

𝐺𝑖𝑣𝑒𝑛 :𝑉=35

𝑓𝑡𝑠

𝐴=3 𝑓𝑡 𝑥2 𝑓𝑡=6 𝑓𝑡 2

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝐹𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :

𝐹=1.28ρ20𝐴𝑉 2

𝐹=1.28 ( 0.002377 𝑠𝑙𝑢𝑔𝑓𝑡 32 ) (6 𝑓𝑡 2) (35 𝑓𝑡𝑠 )2

𝐹=11.18 𝑙𝑏

Example # 4: An auto windshield is 40in wide by 15in high and is vertical. What is the force against the windshield at 60mph?

𝐺𝑖𝑣𝑒𝑛 :𝑉=60 h𝑚𝑝 ( 2215 )=88 𝑓𝑡𝑠𝐴=40 𝑖𝑛 𝑥15 𝑖𝑛=600 𝑖𝑛2( 𝑓𝑡 2

144 𝑖𝑛2 )=4.17 𝑓𝑡 2𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝐹𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :

𝐹=1.28ρ20𝐴𝑉 2

𝐹=1.28 ( 0.002377 𝑠𝑙𝑢𝑔𝑓𝑡 32 ) (4.17 𝑓𝑡2 )(88 𝑓𝑡𝑠 )2

𝐹=49.12 𝑙𝑏

Curved Deflecting Surfaces

𝐹𝑉=𝜌 𝐴𝑉2sin 𝜃

𝐹𝐻=𝜌 𝐴𝑉 2 (1− cos𝜃 )The Resultant of these two components is

𝐹=√𝐹𝐻2+𝐹𝑉

2

𝐹=𝜌 𝐴𝑉 2√2 (1− cos𝜃 )

h𝑊 𝑒𝑟𝑒 :𝐹𝐻=h𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒𝐹 𝑖𝑛 𝑙𝑏𝑜𝑟 𝑁

𝐹𝑉=𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒𝐹 𝑖𝑛 𝑙𝑏𝑜𝑟 𝑁

𝐹=𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒𝑖𝑛𝑙𝑏𝑜𝑟 𝑁𝑉= 𝑓𝑟𝑒𝑒𝑠𝑡𝑟𝑒𝑎𝑚𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦𝜃=𝑎𝑛𝑔𝑙𝑒𝑜𝑓 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑔 .

𝐴=𝑐𝑟𝑜𝑠𝑠−𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙𝑎𝑟𝑒𝑎𝑜𝑓 𝑎𝑖𝑟𝑠𝑡𝑟𝑒𝑎𝑚𝑖𝑛 𝑓𝑡2𝑜𝑟𝑚2

Example # 1: A stream of air 50 ft wide and 10 ft high is moving horizontally at a speed of 60mph. What is the magnitude of these force required to deflect it movement 4downward without loss in speed?

𝐺𝑖𝑣𝑒𝑛 :𝐴=50 𝑓𝑡 𝑥10 𝑓𝑡=500 𝑓𝑡2

𝑉=60 h𝑚𝑝 ( 2215 )=88 𝑓𝑡𝑠θ=4 °𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝐹𝑜𝑟𝑐𝑒𝑎𝑛𝑑 h𝑡 𝑒𝑖𝑟𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :

𝐹=𝜌 𝐴𝑉 2√2 (1− cos𝜃 )

𝐹=(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡3 ) (500 𝑓𝑡2 )(88 𝑓𝑡𝑠 )2

√2 (1− cos4 ° )

𝐹=642.41 𝑙𝑏

𝐹𝑜𝑟 h𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 :𝐹𝐻=𝜌 𝐴𝑉 2 (1− cos𝜃 )

𝐹𝐻=(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡3 )(500 𝑓𝑡 2 )(88 𝑓𝑡𝑠 )2

(1− cos4 ° )

𝐹𝐻=22.42 𝑙𝑏

𝐹𝑜𝑟 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 :

𝐹𝑉=𝜌 𝐴𝑉2sin 𝜃

𝐹𝑉=(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡 3 ) (500 𝑓𝑡2 ) (88 𝑓𝑡𝑠 )2

sin 4 °

𝐹𝑉=642.02𝑙𝑏

Example # 2: A stream of air 72 sq ft in cross section is moving horizontally at a speed of 100mph. What force is required to deflect it downward 10 without loss in speed?

𝐺𝑖𝑣𝑒𝑛 :𝐴=72 𝑓𝑡2

𝑉=100 h𝑚𝑝 ( 2215 )=146.67 𝑓𝑡𝑠θ=10 °𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝐹𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :𝐹=𝜌 𝐴𝑉 2√2 (1− cos𝜃 )

𝐹=(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡3 ) (72 𝑓𝑡 2 )(146.67 𝑓𝑡𝑠 )2

√2 (1−cos10 ° )

𝐹=641.76 𝑙𝑏

Example # 3: A stream of air 60ft wide and 8ft high is moving horizontally at a speed of 75mph. What force is required to deflect it downward 8deg?

𝐺𝑖𝑣𝑒𝑛 :𝐴=60 𝑓𝑡 𝑥 8 𝑓𝑡=480 𝑓𝑡2

𝑉=75 h𝑚𝑝 ( 2215 )=110 𝑓𝑡𝑠θ=8 °

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝐹𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :𝐹=𝜌 𝐴𝑉 2√2 (1− cos𝜃 )

𝐹=(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡3 ) (480 𝑓𝑡 2 )(110 𝑓𝑡𝑠 )2

√2 (1−cos 8 ° )

𝐹=1,926.06 𝑙𝑏

Example # 4: A stream of air 100sq.ft. in cross section is moving horizontally at a speed of 150mph. It strikes tangentially against the interior wall of semi-circular cylinder so that it is deflected through 180°. What is total force against the cylinder?

𝐺𝑖𝑣𝑒𝑛 :𝐴=100 𝑓𝑡 2

𝑉=150 h𝑚𝑝 ( 2215 )=220 𝑓𝑡𝑠θ=180 °𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝐹𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :𝐹=𝜌 𝐴𝑉 2√2 (1− cos𝜃 )

𝐹=(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡3 ) (100 𝑓𝑡2 )(220 𝑓𝑡𝑠 )2

√2 (1−cos 180 ° )

𝐹=23,009.36 𝑙𝑏

AIRFOIL THEORY

Airfoil-is a streamline body which when set at a suitable angle of attack produces more lift than drag. Any surface such as an airplane wing, aileron, elevator.

DEFINITION OF AIRFOIL GEOMETRY

Mean Camber Line – is the line joining the midpoints between the upper and lower surfaces of an airfoil and measured normal to the mean camber line.

Chord Line – Is the line joining the end points of the mean camber line.

Thickness – Is the height of profile measured normal to the chordline.

Thickness Ratio – Is the maximum thickness to the chord ratio,.

Camber – Is the maximum distance of the mean camber line from the chordline.

Leading Edge Radius – Is the radius of a circle tangent to the upper and lower surfaces, with its center located on a tangent to the mean camber line drawn through the leading edge of this line.

DEFINITION OF SECTION FORCES AND MOMENT

FACTORS AFFECTING THE AERODYNAMIC FORCES

1.) Velocity of air, V

2.) Air density, ρ

3.) Characteristic area or size, S

4.) Coefficient of dynamic viscosity, 𝜇5.) Speed of sound,(compressibility effect),

FORMULAS

Lift Force

𝑙=𝐶𝑙12𝜌𝑉 2𝑐

Drag Force

𝑑=𝐶𝑑12𝜌𝑉 2𝑐

Pitching Moment

𝑚=𝐶𝑚12𝜌𝑉 2𝑐 .𝑐

Where:

𝑙=𝑙𝑖𝑓𝑡 𝑓𝑜𝑟𝑐𝑒 𝑖𝑛𝑁𝑚

𝑑=𝑑𝑟𝑎𝑔 𝑓𝑜𝑟𝑐𝑒𝑖𝑛𝑁𝑚

𝑚= h𝑝𝑖𝑡𝑐 𝑖𝑛𝑔𝑚𝑜𝑚𝑒𝑛𝑡 𝑖𝑛𝑁 .𝑚𝑚

𝑐= h𝑐 𝑜𝑟𝑑 h𝑙𝑒𝑛𝑔𝑡

Important Airfoil Characteristics

The following relationship are of fundamental importance to airplane design and airplane analysis.

𝐿𝑖𝑓𝑡 𝐶𝑢𝑟𝑣𝑒 :𝐶𝑙𝑣𝑒𝑟𝑠𝑢𝑠𝛼

The linear portion of the lift curve can be represented mathematically by the equation.

𝐶𝑙=𝑎 (𝛼−𝛼𝑜 )𝑜𝑟𝐶𝑙=𝐶𝑙𝛼 (𝛼−𝛼𝑜 )

Where “a” or is the lift curve slope and the angle of attack for zero lift. The theoretical value of “a” is 2π per radian.

𝐷𝑟𝑎𝑔𝑃𝑜𝑙𝑎𝑟 :𝐶𝑙𝑣𝑒𝑟𝑠𝑢𝑠𝐶𝑑

h𝑃𝑖𝑡𝑐 𝑖𝑛𝑔𝑚𝑜𝑚𝑒𝑛𝑡𝐶𝑢𝑟𝑣𝑒 :𝐶𝑙 𝑣𝑒𝑟𝑠𝑢𝑠𝐶𝑚

AIRFOIL PRESSURE DISTRIBUTION

The pressure distribution is normally expressed in terms of the pressure coefficient,

𝐶𝑝=𝑃−𝑃∞

𝑞=

𝑃−𝑃∞

12𝜌∞𝑉 ∞

2𝑆

At low speeds, according to the incompressible Bernoulli Equation,

𝐶𝑝=1−( 𝑉𝑉 ∞)2

Critical Pressure Coefficient

𝐶𝑝𝑟=2

𝛾𝑀∞2 {[ 2+ (𝛾−1 )𝑀∞

2

𝛾+1 ]𝛾𝛾−1−1}

Critical Pressure

Is the local pressure at the point in the air flow where M=1.0 and the velocity is critical.

𝑝𝑐𝑟=𝑃∞[ 2+(𝛾−1 )𝑀∞2

𝛾+1 ]𝛾

𝛾− 1

Critical Velocity

𝑉 𝑐𝑟=𝑉𝑎∞ [ (𝛾−1 )𝑀∞2+2

𝛾+1 ]12

Example # 1: An Airplane is flying at 480mph at an altitude of 30,000ft.What is the critical Velocity?

𝐺𝑖𝑣𝑒𝑛 :𝑉=480 h𝑚𝑝 ( 2215 )=704 𝑓𝑡𝑠𝑎=30,000 𝑓𝑡

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝑉 𝑐𝑟


𝑉 𝑐𝑟=𝑉𝑎∞ [ (𝛾−1 )𝑀∞2+2

𝛾+1 ]12

𝐹𝑜𝑟 𝑉 𝑎∞

𝑉 𝑎∞=√ γ 𝑃ρ

𝑃=𝑃𝑜(1+ h𝑎𝑇𝑜

)5.26

𝑃=(2116.8 𝑙𝑏𝑓𝑡 2 )[1+ (30,000 𝑓𝑡 )(−0.003566 °𝑅𝑓𝑡 )519 °𝑅 ]

5.26

𝑃=628.60𝑙𝑏𝑓𝑡 2

𝐹𝑜𝑟 ρ

ρ=ρ𝑜(1+ h𝑎𝑇𝑜

)4.26

ρ=(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡3 ) [1+ (30,000 𝑓𝑡 )(−0.003566 ° 𝑅𝑓𝑡 )519°𝑅 ]

4.26

ρ=0.000889𝑠𝑙𝑢𝑔𝑓𝑡3

𝐹𝑜𝑟 𝑉 𝑎∞

𝑉 𝑎∞=√ γ 𝑃ρ

𝑉 𝑎∞=√ (1.4 )(628.60 𝑙𝑏𝑓𝑡 2 )

0.000889𝑠𝑙𝑢𝑔𝑓𝑡3

𝑉 𝑎∞=994.95

𝑓𝑡𝑠

𝐹𝑜𝑟 𝑀𝑀=

𝑉𝑉 𝑎∞

𝑀=704

𝑓𝑡𝑠

994.95𝑓𝑡𝑠

𝑀=0.708

( 𝑙𝑏𝑓𝑡2 )( 𝑓𝑡 3

𝑠𝑙𝑢𝑔 )( 𝑠𝑙𝑢𝑔. 𝑓𝑡𝑠2 )( 𝑓𝑡𝑠𝑙𝑢𝑔 )

𝑉 𝑐𝑟=𝑉𝑎∞ [ (𝛾−1 )𝑀∞2+2

𝛾+1 ]12

𝑉 𝑐𝑟=(994.95 𝑓𝑡𝑠 )[ (1.4−1 ) (0.708 )2+21.4+1 ]

12

𝑉 𝑐𝑟=952.70𝑓𝑡𝑠

Example # 2: An Airplane is flying at 500knots in air at -50°F. What is the critical Velocity?

𝐺𝑖𝑣𝑒𝑛 :

𝑉=500𝑁 .𝑚=844𝑓𝑡𝑠

𝑇=−50 °𝐹 +460=410 ° 𝑅

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝑉 𝑐𝑟


𝑉 𝑐𝑟=𝑉𝑎∞ [ (𝛾−1 )𝑀∞2+2

𝛾+1 ]12

𝑉 𝑐𝑟=969.35𝑓𝑡𝑠

Example # 3: What is the critical value of pressure and pressure coefficient for an airplane flying at 500knots in air at 25°F.

𝐺𝑖𝑣𝑒𝑛 :𝑉=500𝐾𝑛𝑜𝑡𝑠=844.44

𝑓𝑡𝑠

𝑇=25 ° 𝐹 +460=485 °𝑅

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝐶𝑝𝑟


𝑉 𝑎∞=49.02√ 485° 𝑅

𝑉 𝑎∞=1,079.55

𝑓𝑡𝑠

𝐶𝑝𝑟=2

𝛾𝑀∞2 {[ 2+ (𝛾−1 )𝑀∞

2

𝛾+1 ]𝛾𝛾−1−1}

𝑀=844.44

𝑓𝑡𝑠

1,079.55𝑓𝑡𝑠

𝑀=0.782

𝐶𝑝𝑟=2

𝛾𝑀∞2 {[ 2+ (𝛾−1 )𝑀∞

2

𝛾+1 ]𝛾𝛾−1−1}

𝐶𝑝𝑟=2

(1.4 ) (0.782 )2 {[ 2+(1.4−1 ) (0.782 )2

1.4+1 ]1.41.4 −1

−1}𝐶𝑝𝑟=−0.488

Example # 4: An airfoil has a lift curve slope of 6.3 per radian and angle of attack zero lift of -2°. At what angle of attack will the airfoil developed a lift of 140 lb/ft at 100mph under standard sea level condition. Assume c=8ft.

𝐺𝑖𝑣𝑒𝑛 :𝑎= 6.3

𝑟𝑎𝑑 ( π180 )= 0.11𝑑𝑒𝑔𝛼𝑜=−2°𝑙=140

𝑙𝑏𝑓𝑡

𝑉=100 h𝑚𝑝 ( 2215 )=146.67 𝑓𝑡𝑠𝑐=8 𝑓𝑡

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :𝛼𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :𝐶𝑙=𝑎 (𝛼−𝛼𝑜 )

𝛼−𝛼𝑜=𝐶𝑙

𝑎

𝛼=𝐶𝑙

𝑎+𝛼𝑜

𝐵𝑢𝑡𝑙=𝐶𝑙

12ρ𝑉 2𝑐

𝐶𝑙=2 𝑙ρ𝑉 2𝑐

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 :𝛼=

2 𝑙𝑎 ρ𝑉 2𝑐

+𝛼𝑂

𝛼=2(140 𝑙𝑏𝑓𝑡 )

( 0.11𝑑𝑒𝑔 )(0.002377 𝑠𝑙𝑢𝑔𝑓𝑡3 )(146.67 𝑓𝑡𝑠 )2

(8 𝑓𝑡 )+(−2𝑑𝑒𝑔 )

𝛼=4.22𝑑𝑒𝑔

NACA AIRFOILS DESIGNATION

4 –DIGIT AIRFOILS: Example NACA 4412

4=Camber 0.04c

4=Position of the camber at 0.4c from L.E.

12=Maximum thickness 0.12c

5-DIGIT AIRFOILS: Example NACA 23012

2=camber 0.02c=design lift coefficient is 0.15 times the first digit for this series

30=position of camber at =0.15c

12=Maximum thickness 0.12c


6-SERIES AIRFOILS: Example NACA 653-421

6=series designation

5=min. pressure at 0.5c

3=The drag coefficient is near its minimum value over a range of lift coefficient of 0.3 above and below the design lift coefficient.

4=design lift coefficient 0.4

21=max. thickness 0.21c


7-SERIES AIRFOILS: Example NACA 747A315

7=series designation

4=favourable pressure gradient on the upper surface from L.E. to 0.7c at the design lift coefficient.

7=favourable pressure gradient on the lower surface from L.E. to 0.7c at the design lift coefficient.

A=a serial letter to distinguish different section having the same numerical designation but different mean line or thickness distribution.

3=design lift coefficient 0.3

15=max thickness 0.15c

Example # 1: NACA 4412, c=100cm, find the camber, position of camber and max. thickness.

𝐺𝑖𝑣𝑒𝑛 :𝑁𝐴𝐶𝐴 4412

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :

(𝑎 )𝑐𝑎𝑚𝑏𝑒𝑟(𝑏)𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑜𝑓 𝑐𝑎𝑚𝑏𝑒𝑟(𝑐 )𝑚𝑎𝑥 . h𝑡 𝑖𝑐𝑘𝑛𝑒𝑠𝑠

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :(𝑎 )𝑐𝑎𝑚𝑏𝑒𝑟=0.04 𝑐𝑐𝑎𝑚𝑏𝑒𝑟=0.04 (100𝑐𝑚 )𝑐𝑎𝑚𝑏𝑒𝑟=4𝑐𝑚

(𝑏)𝑝𝑜𝑠 .𝑜𝑓 𝑐𝑎𝑚𝑏𝑒𝑟=0.4𝑐𝑝𝑜𝑠 .𝑜𝑓 𝑐𝑎𝑚𝑏𝑒𝑟=0.4 (100𝑐𝑚 )

𝑝𝑜𝑠 .𝑜𝑓 𝑐𝑎𝑚𝑏𝑒𝑟=40 𝑐𝑚

(𝑐 )max h𝑡 𝑖𝑐𝑘𝑛𝑒𝑠𝑠=0.12𝑐max h𝑡 𝑖𝑐𝑘𝑛𝑒𝑠𝑠=0.12 (100𝑐𝑚 )

max h𝑡 𝑖𝑐𝑘𝑛𝑒𝑠𝑠=12𝑐𝑚

Example # 2: NACA 23015,c=48 in., find the camber, position of camber and max. thickness.

𝐺𝑖𝑣𝑒𝑛 :𝑁𝐴𝐶𝐴23015𝑐=48 𝑖𝑛𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 :(𝑎 )𝑐𝑎𝑚𝑏𝑒𝑟(𝑏)𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑜𝑓 𝑐𝑎𝑚𝑏𝑒𝑟(𝑐 )𝑚𝑎𝑥 . h𝑡 𝑖𝑐𝑘𝑛𝑒𝑠𝑠

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :(𝑎 )𝑐𝑎𝑚𝑏𝑒𝑟=0.02𝑐𝑐𝑎𝑚𝑏𝑒𝑟=0.02 (48 𝑖𝑛)

𝑐𝑎𝑚𝑏𝑒𝑟=0.96 𝑖𝑛

(𝑏)𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑜𝑓 𝑐𝑎𝑚𝑏𝑒𝑟=0.302

𝑐=0.15𝑐

𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑜𝑓 𝑐𝑎𝑚𝑏𝑒𝑟=0.15 (48 𝑖𝑛)

𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑜𝑓 𝑐𝑎𝑚𝑏𝑒𝑟=7.2 𝑖𝑛

(𝑐 )max h𝑡 𝑖𝑐𝑘𝑛𝑒𝑠𝑠=0.15𝑐max h𝑡 𝑖𝑐𝑘𝑛𝑒𝑠𝑠=0.15 (48 𝑖𝑛 )

max h𝑡 𝑖𝑐𝑘𝑛𝑒𝑠𝑠=7.2 𝑖𝑛