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MAE 150B - 05B - Incompressible Flow over Finite Wings

MAE 150B - 05B - Incompressible Flow over Finite Wings -...

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    MAE 150B - Chap 5 Part B  Incompressible Flow Over Finite Wings
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    Ground effects Ground effects are very complicated phenomena It reduces the downwash velocity magnitude lower α i Lower induced drag
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    Can Spruce Goose (Hughes H-4  Hercules ) fly without ground effects?
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    Summary of Finite Wings: Elliptic Lift Distribution General Lift Distribution ) 43 . 5 ( 2 , AR C C L i D π = ( 29 ) 61 . 5 ( 1 2 , δ π + = AR C C L i D ) 69 . 5 ( / 1 0 0 AR a a a d dC L π α + = = ) 70 . 5 ( ) 1 )( / ( 1 0 0 τ π + + = AR a a a
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    Example 5.1 (pp. 425) Given: AR=8 Taper ratio=0.8 Thin symmetric airfoil section Calculate C L and C D,I when α = 5 o assuming δ = τ Use With a 0 =2 π and δ =0.055 (Fig. 5.70) C L =a α (why?) ) 70 . 5 ( ) 1 )( / ( 1 0 0 τ π + + = AR a a a ( 29 ) 61 . 5 ( 1 2 , δ π + = AR C C L i D
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    Example 5.2 (pp. 426) Given: Rectangular wing AR=6 Induced drag factor δ =0.055 α L=0 = -2 o At α = 3.4 o C D,i =0.01 δ = τ Calculate C D,I for similar wing with AR=10
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    Example 5.2 (pp. 426) From C D,I AR and δ C L (for α = 3.4 o )by using Knowing C L at two angles (3.4 o and -2 o ), we can calculate lift slope a From a we may calculate a 0 a 0 =5.989/rad (note: not 2 π why?) With AR=10 for (AR=10) calculate a With a known for AR=10, CL and C D,I can be computed (note α L=0 = -2 o why?) ( 29 ) 61 . 5 ( 1 2 , δ π + = AR C C L i D ) 70 . 5 ( ) 1 )( / ( 1 0 0 τ π + + = AR a a a
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    Example 5.3 (pp. 427)
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