Supersonic Thin Airfoil Theory Andrew Ning In class we showed that - PDF document

Supersonic Thin Airfoil Theory Andrew Ning In class we showed that the local pressure coefficient is given by (using small disturbance assumptions): 2 θ C p = (1) � M 2 ∞ − 1 where θ is positive when inclined into the freestream and negative when inclined away from the freestream. I won’t go through that derivation here as it was discussed in class. The part that we glossed over was how to get lift and drag coefficients from that result. As typically done in thin airfoil theory, we will separate an airfoil into a thickness distribution, a camber distribution, and an angle of attack. Specifically, we define the upper and lower surfaces as a superposition of camber and thickness distribution as follows: y u ( x ) = y c ( x ) + y t ( x ) (2) y l ( x ) = y c ( x ) − y t ( x ) (3) Note that y t is not the local thickness τ (it is 1/2 of that). It will be more convenient in the derivation to use this form. Using the formula for the local pressure coefficient: 2 � − α + dy u � C pu = (4) � dx M 2 ∞ − 1 � � 2 α − dy l C pl = (5) � dx M 2 ∞ − 1 The negative sign results from the way that θ is defined. Substituting in the camber and thickness distribu- tions: 2 � − α + dy c dx + dy t � C pu = (6) � dx M 2 ∞ − 1 � � 2 α − dy c dx + dy t C pl = (7) � dx M 2 ∞ − 1 The definition of the (inviscid) normal force coefficient is: � c c n = 1 ( C pl − C pu ) dx (8) c 0 Substituting in the result from above : � c 2 1 (2 α − 2 dy c c n = dx ) dx (9) � M 2 ∞ − 1 c 0 � c � c 2 1 � � dy c = 2 α dx − 2 dx dx (10) � c M 2 ∞ − 1 0 0 2 1 c (2 αc − 2 y c | c = 0 ) (11) � M 2 ∞ − 1 4 α = (12) � M 2 ∞ − 1 1

The definition of the (inviscid) axial force coefficient is: � c c a = 1 � dy u dy l � dx − C pl (13) C pu dx c dx 0 Substituting in: � c � �� � � � dy c � � � � dy c c a = 1 2 − α + dy c dx + dy t dx + dy t 2 α − dy c dx + dy t dx − dy t − dx � � c M 2 dx dx M 2 dx dx ∞ − 1 ∞ − 1 0 (14) Several terms appear in both expressions and so cancel out. Removing those terms leaves us with: � c � � 2 � � 2 2 1 � dy c � dy t − 2 αdy c c a = dx + 2 + 2 (15) dx � c dx dx M 2 ∞ − 1 0 For the first term under the integral, α is a constant and can be taken out. We already saw that � c dy c dx dx = 0 (16) 0 and so that whole term is zero. We are left with: �� dy c � 2 � � 2 4 � dy t c a = + (17) � M 2 ∞ − 1 dx dx where we define � c ζ = 1 ζ ( x ) dx (18) c 0 as a shorthand for convenience. Finally, lift and drag are related to the normal and axial forces as follows: c l = c n cos α − c a sin α (19) c d = c n sin α + c a cos α (20) Using a small angle approximation, consistent with thin airfoil theory, yields: c l ≈ c n − c a α (21) c d ≈ c n α + c a (22) Conventionally, the c a α term is neglected in the lift coefficient because it is of much smaller magnitude than c n ( c a is small and α is small so their product is very small). However, in the drag calculation both c n α and c a are of similar magnitude. c l ≈ c n (23) c d ≈ c n α + c a (24) Thus: 4 α c l = (25) � M 2 ∞ − 1 � � 2 � 2 � 4 � dy c � dy t α 2 + c d = + (26) � dx dx M 2 ∞ − 1 The (inviscid) pitching moment coefficient is: � c c mle = 1 �� dy u dy l � � C pu − C pl x + C pu dx y u − C pl dx y l dx (27) c 2 0 2

Making substitutions: � c �� � � � � dy c � c mle = 1 2 − 2 α + 2 dy c − α + dy c dx + dy t dx + dy t x + ( y c + y t ) − c 2 � M 2 dx dx dx ∞ − 1 0 (28) � � � dy c � � α − dy c dx + dy t dx − dy t ( y c − y t ) dx dx dx Removing terms that cancel: � c 2 2 � c mle = 1 2 � � � �� − 2 αx + 2 dy c − 2 αdy c dx + 2 dy c + 2 dy t − 2 αdy t dx + 4 dy c dy t dx x + y c + y t dx c 2 � dx dx dx dx M 2 ∞ − 1 0 (29) Neglecting all second order terms. � c c mle ≈ 1 2 � � − 2 αx + 2 dy c dx x dx (30) c 2 � M 2 ∞ − 1 0 � c = 1 2 � − 2 αc 2 � 2 dy c 2 + (31) dx xdx c 2 � M 2 ∞ − 1 0 The second term we can integrate by parts: � c c mle = 1 2 � � − αc 2 + 2( xy c ) c 0 − y c dx (32) c 2 � M 2 ∞ − 1 0 The second term is zero, leading to � c 2 � α + 1 � c mle = − y c dx (33) � c 2 M 2 ∞ − 1 0 2 � α + y c � = − (34) � c M 2 ∞ − 1 3