Which quantity signifies the ratio of temperature gradient at the surface to a reference temperature gradient?
It is given by h l/k.
The determination of value of Nusselt number or the convective film coefficient forms a basis for the computation of heat transfer by convection. Towards that end, following approaches have been suggested
(i) Nondimensional analysis and experimental correlations
(ii) Hydrodynamic concept of velocity boundary layer
(iii) Reynolds similarity between the mechanism of fluid friction in the boundary layer and the transfer of heat by convection
It should be dimensional analysis and experimental correlations.
Nusselt number is given by
The length parameter l specifies the geometry of solid body.
The temperature profile at a particular location in a thermal boundary layer is prescribed by n expression of the form
t (y) = A B y + C y ^{2}
Where, A, B and Care constants. What is the value of heat transfer coefficient?
h = – k/ (t _{s} – t_{ infinity}) [d t/d y]_{ y = 0} and (d t/d y) _{y = 0 }= – B.
The temperature profile at a particular location on a surface is prescribed by the identity
(t _{s} – t) / (t _{s} – t_{ infinity}) = sin (π y/0.015)
If thermal conductivity of air is stated to be 0.03 W/m K, determine the value of convective heat transfer coefficient
h = – k/ (t _{s} – t_{ infinity}) [d t/d y]_{ y = 0}. Therefore h = – k/ (t _{s} – t_{ infinity}) [π (t _{s} – t_{infinity})/0.015].
Air at 20 degree Celsius flows over a flat plate maintained at 75 degree Celsius. Measurements shows that temperature at a distance of 0.5 mm from the surface of plate is 50 degree Celsius. Presuming thermal conductivity of air is 0.0266 W/m K, estimate the value of local heat transfer coefficient
h = – k/ (t _{s} – t_{ infinity}) [d t/d y]_{ y = 0} and d t/d y = – 50 * 10 ^{3} degree Celsius/m.
Air at 20 degree Celsius flows over a flat surface maintained at 80 degree Celsius. Estimate the value of local heat transfer coefficient if the local heat flow at a point was measured as 1250 W/m^{2}. Take thermal conductivity of air as 0.028 W/m K
Q = h A (t _{s} – t_{ infinity}).
Consider the above problem, calculate the temperature gradient at the surface
(d t/d y) _{y = 0} = – h (t _{s} – t_{ infinity})/k
At the interface of solid body, heat flows by conduction and is given by
Q = – k A (d t/d y)_{ y = 0}.
For a given value of Nusselt number, the convective surface coefficient h is directly proportional to
It is directly proportional to k and inversely proportional to significant length.
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