By Vedat S. Arpacı
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Extra info for Conduction Heat Transfer
Example text
2-4 The electrical power P,,,,;, being identical to the electrical internal energy generated in the volume AVi, must be included in the internal energy flow across the boundaries of the control volume. If the friction is negligible, the mass flow power Piureduces to N Piu= - lim At-0 N C (pi AWilAt) = -Ci= PiViWi, 1 where A'Ui is the volume and v i the specific volume of the mass Ami, and pi is the pressure on the boundaries of A'Ui. Similarly, the rate of heat q received by the system may be conveniently expressed in terms of q, received by the control volume.
Hence in terms of isotropic continua, for example, introducing Eq. (2-75) into Eq. (2-69), we have So far we have been able to derive expressions, namely, the equations of conduction given by Eqs. (2-88), (2-91), (2-93), and (2-94), which satisfy a partial differential equation in terms of the unlinown temperature, rather than an algebraic equation. Since the solution of a differential equation involves a number of integration constants, the completion of the formulation requires that we state an equal number of appropriate conditions in space and time to determine these constants.
I n solids the pressure dependency, being very small, is always neglected. For narrow temperature intervals the temperature dependency may also be negligible. Otherwise a linear relation is assumed in the form where p is small and negative for most solids. To illustrate the numerical values, the thermal conductivities of some gases, liquids, and solids are given in Fig. 2-12* as functions of the temperature. The experimental methods for determining the thermal conductivity of continua are many and varied.
