The velocity of air at the conditions is approximately as v

The velocity of air at the conditions is approximately as v = 165 times 10^4 m^2/s, and the thermal conductivity is approximate and heat transfer in the boundary layer? What is the air velocity (m/s) required for case 2 in order to get the similarity in momentum end heat transfer in the boundary layer? What is the heat transfer rate (W/m) to the blade surface per unit depth (into the page) in case 2?

Solution

First obtain the heat transfer from surface-air by

q= h( Tinf- Ts) ,

h= q/(deltaT)

Also Nu= Nusselt number = hL/K

for the first foil: Internal heat flux in the foil: h( 1150- 800) =1000 w/m^2 (given)

h= 1000/350 = 2.857

SImilarity of flow implies Re is same:

V₁L₁/nu = V₂L₂/nu

Since viscosity is unchanged, the velocity must vary st V₁/V₂ = L₂/L₁

Hence V₂ = 40 m/s

Since the thermal diffusivity of air is unchnaged Prandtl number Cp mu/k = const.

for similarity use Nusselt number hL/K

h1L1= h2L2

h2 = h1 L1/L2 = 1/2 h1, where h1 = 1000/(1150-800) from the first case = 1000/350

h2 = 1,428

heat flux = h2(DT) = 1.428* 450 =642.9 w/m^\\

or per unit length 642.9 W/m