CHAPTER 11 SUBSONIC COMPRESSIBLE FLOW OVER AIRFOILS LINEAR THEORY 11. 1 Introduction This chapter mainly deal with the properties of two-dimensional airfoils at mach number above 0.3 but below 1, where the compressibility must be considered
CHAPTER 11 SUBSONIC COMPRESSIBLE FLOW OVER AIRFOILS: LINEAR THEORY 11.1 Introduction This chapter mainly deal with the properties of two- dimensional airfoils at Mach number above 0.3 but below 1, where the compressibility must be considered
city potential equation PrandtI-Glauet Compressibilty correction Linearized velocity potential equation Improved compressibilty Correction Critical mach number The area rule for transonic flow Figure 11.1 Supercritical air foils Road Map for Chap. 11
Velocity potential equation Linearized velocity potential equation Prandtl-Glauet Compressibilty correction Improved compressibilty Correction Critical Mach number The area rule for transonic flow Supercritical airfoils Figure 11.1 Road Map for Chap.11
1.2 The Velocity Potential Equation For two-dimensional, steady, irrotational, isentropic flow,a velocity potential =o(x, y) can be defined such that V=VO The introduction of velocity potential can greatly simplify the governing equations, we can derive the velocity potential equation from continuity, momentum, energy equations
11.2 The Velocity Potential Equation For two-dimensional , steady, irrotational , isentropic flow, a velocity potential can be defined such that : = (x, y) V = The introduction of velocity potential can greatly simplify the governing equations, we can derive the velocity potential equation from continuity, momentum, energy equations:
The continuity equation for steady two-dimensional flow 无法显示该图片 O(m)+/m)=0 ax 十+1—+ 0 or Ox Substituting u== v= p Into it. we get 0、0 p 0 Or or
The continuity equation for steady,two-dimensional flow is : 0 ( ) ( ) = + y v x u = 0 + + + y v y v x u x u or Substituting into it, we get y v x u = = , ( ) 0 2 2 2 2 = + + + x y x x y y
To eliminate p from above equation, we consider the momentum equation 中p=-pd pldr=p u+y 2 2 02,02 Since the flow we are considering is isentropic, so dp dp
To eliminate from above equation, we consider the momentum equation : dp = −VdV ( ) 2 2 2 2 2 dp = − VdV = − dV = − d u + v + = − 2 2 ( ) ( ) 2 x y dp d Since the flow we are considering is isentropic, so 2 a p d dp s = =
02,0 2a Ox SO a ox Pa(0+ 2a ayl a P( 83,2、oOp,Op=0 ax Ox Ox ay ay
+ = − 2 2 2 ( ) ( ) 2 x y d a d + = − 2 2 2 ( ) ( ) x 2a x x y so + = − 2 2 2 ( ) ( ) y 2a y x y ( ) 0 2 2 2 2 = + + + x y x x y y
We get the velocity potential equation 102|0 2 X 200、O =0 a ox In this equation, the speed of sound is also the function ofφ y
( )( ) 0 2 ( ) 1 ( ) 1 1 1 2 2 2 2 2 2 2 2 2 2 = − + − − a x y x y a x x a y y We get the velocity potential equation: In this equation , the speed of sound is also the function of : + − = − 2 2 2 0 2 ( ) ( ) 2 1 x y a a (11.12)
For subsonic flow, Eq. 11 12 is an elliptic partial differential equation. For supersonic flow, Eq 11 12 is a hyperbolic partial differential equation. For transonic flow, Eq 11 12 is mixed type equation
For subsonic flow, Eq. 11.12 is an elliptic partial differential equation. For supersonic flow, Eq.11.12 is a hyperbolic partial differential equation. For transonic flow, Eq.11.12 is mixed type equation
Eq. 11 12 represents a combination of continuity, momentum, energy equations. In principle, it can be solved to obtain for the flow field around any two-dimensional flow The infinite boundary condition is ao=vo ax 0p=0 The wall boundary condition is 0p=0 an
Eq. 11.12 represents a combination of continuity, momentum, energy equations. In principle, it can be solved to obtain for the flow field around any two-dimensional flow. The infinite boundary condition is = = V x u = 0 = y v The wall boundary condition is = 0 n
Once is known, all the other value flow variables are directl obtained as follows Calculate u and v: l ao p and d x 2 Calculate a 0p2, x 3. Calculate M: M √u2+y 4. Calculate Tp,p: T=10(+M) B=P0(1+2 M2) 2 0(1+ M2)
Once is known, all the other value flow variables are directly obtained as follows: 1. Calculate u and v: and x u = y v = 2.Calculate a: + − = − 2 2 2 0 ( ) ( ) 2 1 x y a a 3.Calculate M: a u v a V M 2 2 + = = 4. Calculate T,p, : 1 1 2 0 2 1 0 2 1 0 ) 2 1 (1 ) 2 1 (1 ) 2 1 (1 − − − − − − = + − = + − = + M p p M T T M