Question

Consider an aircraft traveling at high speed. At a point on its wing, the local shear stress is 312 N/m2, and the local conductive heat transfer to the wing is 450 kW/m2. Calculate the air velocity and temperature gradients normal to the surface assuming that the surface has a temperature of 330 K.

Answer 1

The air velocity is 1442.3m/s

The temperature gradient is 0.00311K/m

Step-by-step explanation:

Rate of heat transfer = local conductive heat transfer × area

Rate of heat transfer = force × distance/time (distance/time = velocity)

Therefore, rate of heat transfer = force × velocity

Force (F) × velocity (v) = local conductive heat transfer coefficient (k) × Area (A)

F/A × v = k

Shear stress = F/A = 312N/m^2

k = 450kW/m^2 = 450×1000W/m^2 = 450000W/m^2

312 × v = 450000

v = 450000/312 = 1442.3m/s

Air velocity (v) = 1442.3m/s

Temperature gradient = Temperature (T)/distance (s)

From equations of motion

v^2 = u^2 + 2gs

u = 0m/s, v = 1442.3m/s, g = 9.8m/s^2

1442.3^2 = 2×9.8×s

s = 2080229.29/19.6 = 106134.15m

Temperature gradient = 330K/106134.15m = 0.00311K/m

Answer 2

The solution is shown in the images attached with the answer.

Step-by-step explanation: