By A. J. Ede
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Extra info for An Introduction to Heat Transfer Principles and Calculations
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.