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

Answer:

h = 110.84 w/m^2.K

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

$\frac{V}{m} =\frac{RT}{P}$

$\frac{m}{V} =\frac{P}{RT}$

                 $= \frac{26.5}{0.287\times (-50+ 273)}$

                  = 0.4141 kg/m^3

From modified reynold's analogue

$h = \frac{cf \rho v Cp}{2 pr^{2/3}}$

$h = \frac{0.0016 \times 04141 \times 990 \times \frac{1000}{3600} \times 999}{2\times {0.744^{2/3}}}$

h = 110.84 w/m^2.K