By A. J. Ede

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Supposing that it has been estimated that the external convective heat transfer coefficient is h! [H/L 2 T] so that the heat transfer between the external face and the surrounding air is q=h'A(9h — 9')9 where 0' is the surface temperature and 9h the ambient air temperature. This still involves the unknown surface temperature; but for determining the overall heat transfer the difficulty can be avoided, as for the multiple slab, by taking the reciprocal of the heat transfer coefficient and regarding it as a resist ance, to be added to the other conductive resistances.

STEADY-STATE HEAT TRANSFER THE PLANE SLAB It is assumed to be large enough for edge effects to be neglected, so that the flow of heat is along parallel straight lines normal to the surface. 6) where A is the area, d the thickness, and 6h and 0C the temperatures of the two faces. An important variant arises when the slab consists of layers of two or more different substances. The heat transfer cannot be calculated directly, because the temperatures of the interfaces be tween the different substances are not known.

7). In all methods of this type, in which a continuous temperature distribution is divided into discontinuous "lumps", the larger the number of sections the greater the accuracy and the greater the amount of work involved. The number shown is a reasonable compromise. Since α=0·01 m (this is, of course, a different a) At=pea2/2k = 7900 x 480 x 0-01 2 /2x 45 = 4-2 s. The initial temperature distribution is represented by A-A. The points O correspond to the temperature of the surrounding oil, namely 20°C, and are at a distance kjh=45/6000 =0-0075 m from the surface in the diagram.