Question

Consider an airplane cruising at an altitude of 10 km where standard atmospheric conditions are െ50°c and 26.5 kpa at a speed of 800 km/h. each wing of the airplane can be modeled as a 25 m x 3 m flat plate, and the friction coefficient of the wings is 0.0016. using the momentum-heat transfer analogy, determine the heat transfer coefficient for the wings at cruising conditions.

Answer 1

h = 89.6 W/m^2 K

Step-by-step explanation:

Given data:

altitude is 10 km

speed of airplane is 800 km/h

coeeficent of frcition is 0.0016

From standard table

for - 50 degree C and 1 atm pressure

Cp= 999 J/kg K

P_r = 0.744 for reynold's analogy

From Gas law we have

`\rho = (P)/(RT) = (26.5)/(.287 * (-50 + 273))`

`\rho = .414 kg/m^3`

from Modified Reynolds analogy we have

`(Cf)/(2) = (h)/(\rho v Cp) P_r^(2/3)`

solving for h so we have

`h = (Cf)/(2) (\rho Cp v)/(P_r^(2/3))`

`h = (0.0016)/(2) (0.414 * 9999 * 800 * (10^3)/(3600))/(.744^(2/3))`

h = 89.6 W/m^2 K