# Heat Transfer Solved Problems

It could be said that an exact analytical solution will be obtained for an approximate problem.The solution will, to some extent, be in error, and it will not normally be possible to estimate the magnitude of this error without recourse to external information such as an experimental result. To do this, the continuous solution region is, in most methods, replaced by a net or grid of lines and elements.

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The solution variables—temperature, velocity, etc.—are not obtained at all of the infinite number of points in the solution region, but only at the finite number of nodes of the grid, or at points within the finite number of elements.

The differential equations are replaced by set of linear (or, rarely, nonlinear) algebraic equations, which must and can be solved on a computer.

The heat transfer by conduction in solids can only take place when there is a variation of temperature, in both space and time.

Let us consider a small volume of a solid It should be noted that Fourier law can always be used to compute the rate of heat transfer by conduction from the knowledge of temperature distribution even for unsteady condition and with internal heat generation. Due to roughness, 40 percent of the area is in direct contact and the gap (0.0002 m) is filled with air (k = 0.032 W/m K).

Radiation is somewhat different, involving surfaces separated (in general) by a fluid which may or may not participate in the radiation.

If it is transparent, and if the temperatures of the surfaces are known, the radiation and convection phenomena are uncoupled and can be solved separately.The difference in temperature between the two outside surfaces of the plate is 200°C Estimate (i) the heat flow rate, (ii) the contact resistance, and (iii) the drop in temperature at the interface.4. Inside diameter 100 mm, outside diameter 120 mm (k 50 W/m K) IS Insulated with a40mm thick hightemperature Insulation(k = 0.09 W/m K) and another Insulation 60 mm thick (k = 0.07 W/m K). The heattransfer coefficient for the inside and outside surfaces are 550 and 15 W/m5.Steel balls 10 mm in diameter (k = 48 W/m K), (C = 600 J/kg K) are cooled in air at temperature 35°C from an initial temperature of 750°C.These equations represent a system of five equations in three dimensions, or four equations in two dimensions.For conduction in solids, DT/Dt in (3) is replaced by ∂T/∂t, and (1) and (2) are not relevant.For a Boussinesq fluid, it is assumed that ρ = f(T).The usual assumption is a linear relationship: where β is the volumetric expansion coefficient, and the subscript denotes a reference state.There is also an error in this approach: for example, if derivatives are replaced by finite differences, only an approximate value for the derivative will be obtained, in principle, if the problem is well-posed and if the solution method is well-designed, this error will approach zero as the grid is made increasingly fine.In practice, a fine but not infinitesimal grid must be used.In convection problems, the temperature gradient in a fluid flowing over a surface is needed to find the heat flux at that surface.In both cases, the determination of the complete temperature distribution in the region of interest is needed as a first step, and in convection one must also find the velocity distribution.

## Comments Heat Transfer Solved Problems

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