Heat Transfer Solved Problems

Heat Transfer Solved Problems-12
The thermodynamic and transport properties ρ, C, λ and η of most materials are in general dependent on temperature and (particularly in the case of ρ) also on pressure.It is often possible to neglect this dependency, with considerable simplification to the computational procedure.Thus, a full solution of the energy equation and perhaps also the equations of motion is required.

The thermodynamic and transport properties ρ, C, λ and η of most materials are in general dependent on temperature and (particularly in the case of ρ) also on pressure.It is often possible to neglect this dependency, with considerable simplification to the computational procedure.Thus, a full solution of the energy equation and perhaps also the equations of motion is required.

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The usual objective in any heat transfer calculation is the determination of the rate of heat transfer to or from some surface or object.

In conduction problems, this requires finding the temperature gradient in the material at its surface.

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.

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).In this entry, emphasis will be given to methods for conduction and convection problems, with only a brief mention of radiation.Attention will be limited to incompressible fluids except when buoyancy is important, in which case the Boussinesq approximation will be made.Calculate the time required for the temperature to drop to 150°C when h = 25 W/m2K and density p = 7800 kg/m3. Numerical heat transfer is a broad term denoting the procedures for the solution, on a computer, of a set of algebraic equations that approximate the differential (and, occasionally, integral) equations describing conduction, convection and/or radiation heat transfer.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.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.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.

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