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 990 km/h. Each wing of the airplane can be modeled as a 25-m × 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.The properties of air at –50°C and 1 atm are:cp =0.999 kJ/kg·KPr = 0.7440

Answer 1

h = 110.84 w/m^2.K

Step-by-step explanation:

Given data:

properties of air at - 50 degree celcius

Cp = 0.999 J/kg K

Pr = 0.744

from ideal gas equation we know that

PV = mRT

`(V)/(m) =(RT)/(P)`

`(m)/(V) =(P)/(RT)`

`= (26.5)/(0.287* (-50+ 273))`

= 0.4141 kg/m^3

From modified reynold's analogue

`h = (cf \rho v Cp)/(2 pr^(2/3))`

`h = \frac{0.0016 * 04141 * 990 * (1000)/(3600) * 999}{2* {0.744^(2/3)}}`

h = 110.84 w/m^2.K