Question

1. For the airfoil Cp data shown below; • What is the maximum airspeed in flow just outside the boundary layer if the freestream speed Vo = 120 m/s? • What is the local Mach number at this point if the altitude is 12 km? • What is the approximate value of C, for these conditions? Ans: V = 204.35 m/s, M = 0.693, CL ~ 1.15. -1.00 .. .... -0.50 .... ..... 0 0.1 0.2 0.3 0.4 0.5 0.6 1 0.7 1 0.8 0. 9 x/c 2. Consider a rectangular wing mounted in a low-speed wind tunnel at angle of attack a = 4°. The wing model completely spans the test section and has a NACA 651 - 212 airfoil section (data shown below), chord 0.6m. The airflow freestream pressure, temperature and speed are 100.5kPa, 303K and 42m/s respectively. • Calculate the per-unit-length values of lift, drag and moment about the 1/4-chord location. • What is the corresponding aerodynamic angle of attack aa? • What is the approximate design range of C for this airfoil? Ans: L = 324.0 N/m, D= 3.973 N/m, Mc/4 = -20.17 Nm/m, Caero = 5.3°, 0.1 < < 0.4.

Answer 1

Solution for Question number 1:

i)

Let maximum airspeed = V_max

Op=1-7 or, V = VV1 - Cp

..Vaa

Hence, V is inversely proportional to C_P which means that as V increases, C_P decreases and vice-versa

Hence, for V = maximum, C_P must be minimum

Therefore, V = V_max, C_P = C_P,min

In the C_P data graph, C_P,min = -1.9

Vmax = VV1-CP,min or, mar = 120* V1- (-1.9) = 120 * 12.9 = 204.35

Hence, maximum airspeed = V_max = 204.35 m/s ........................................................... ANSWER (i)

(ii)

At an altitude = 12 km, temperature = T_h = -56.5^o C = 216.5 K

Speed of sound = V_s,h

_{.. Vs.h = VRT..n= V1.4 * 287 * 216.5 = 294.94}

Hence, V_s,h = 294.94 m/s

Mach number = M

M = max _ 204.35 = 0.693 Vs.h 294.94

Hence, Mach number = 0.693 ................................................................ ANSWER (ii)

(iii)

Lift force = F_L

Coefficient of lift = C_L

Surface area = A

or Cermin poveste or, P= Pxo + 0.5pa Vimar Cp, min.

On the other hand, P = F_L/A

Ft = Apex Vibur. On 0r, * = 0.5px Vihar Ct or, P = 0.5px Vihar Cu P or, CL = 0.5pVmax or, Cu= 0.5p Var 0.5pVmar Cp+PX Pa or, CL = Cp +5 0.5pV22 0.0 Pomar PV = nR'T P P or, Pl") = CORT or, p= G) TRT :. Po - 207, P = RT. 1:07 = Cp + TVA or,Cu = Cp + (RT) .5V. ha Poo RT mar or, CL=Cp + 0.5V2 287 * 216.5 or, CL = -1.9+ 0.5 * 204.352 52 ..........[R = 287J/Kg.K or, CL= -1.9 +2.9759 = 1.0759 - 1.08

Hence, C_L = 1.08 ................................................................ ANSWER (iii)