By A. J. Ede

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11) must be used. For the particular case of coaxial cylinders it is exact. The heat flux from the pipe towards the casing is thus 5-67xlQ- 8 (473 4 -T 4 ) ~~ (1/0-95) + (0·25/0·50)[(1/0·20) - 1 ] Φ = 1-86 x Κ Γ 8 ( 4 7 3 4 - Γ 4 ) W/m 2 . 25(4734-r4)W. Solving for Tand substituting in either of the formulae gives the heat loss from the pipe as 4=0« 11 kW. 27 RADIATION How would these calculations be affected if one part of the room were at a different temperature from the rest? If the temperature of the room could not be regarded as uniform the simple formulae would no longer apply.

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. The internal convective coefficient is treated in the same way.

In turn, each tem­ perature is adjusted or "relaxed" towards the equilibrium value; this will affect the values of neighbouring temperatures, and these are adjusted in accordance with another simple rule, so that a new temperature distribution is built up which is a little nearer to the true distribution. The process is repeated as often as necessary until a true steady-state distribution has been determined. The heat flow can then readily be calculated. For further information see Refs. 9, 12, 43.

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