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Answer:
h = 375 KW/m^2K
Explanation:
Given:
Thermo-couple distances: L_1 = 10 mm , L_2 = 20 mm
steel thermal conductivity k = 15 W / mK
Thermo-couple temperature measurements: T_1 = 50 C , T_2 = 40 C
Air Temp T_∞ = 100 C
Assuming there are no other energy sources, energy balance equation is:
E_in = E_out
q"_cond = q"_conv
Since, its a case 1-D steady state conduction, the total heat transfer rate can be found from Fourier's Law for surfaces 1 and 2
q"_cond = k * (T_1 - T_2) / (L_2 - L_1) = 15 * (50 - 40) / (0.02 - 0.01)
=15KW/m^2
Assuming SS is solid, temperature at the surface exposed to air will be 60 C since its gradient is linear in the case of conduction, and there are two temperatures given in the problem. Convection coefficient can be found from Newton's Law of cooling:
q"_conv = h * ( T_∞ - T_s ) ----> h = q"_conv / ( T_∞ - T_s )
h = 15000 W / (100 - 60 ) C = 375 KW/m^2K
Answer:
(b)False
Explanation:
Given:
Prandtl number(Pr) =1000.
We know that 
Where
is the molecular diffusivity of momentum
is the molecular diffusivity of heat.
Prandtl number(Pr) can also be defined as

Where
is the hydrodynamic boundary layer thickness and
is the thermal boundary layer thickness.
So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
In given question Pr>1 so hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
So hydrodynamic layer will be thicker than the thermal boundary layer.
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