CONDUCTION, CONVECTION & RADIATION - con'd THE ELECTRICITY THROUGH A WIRE ANALOGY We can draw parallels between these systems in the following approximate way.
Heat Electricity Fluid Driving agent temperature voltage pressure D difference difference difference Resisting agent thermal electrical pipe/fluid R impedance resistance resistance Flow rate power (Watts) current (amps) flow (litres/sec) F (energy/sec) (charge/sec) (fluid vol/sec) If we consider this flow to be constrained by say a solid cylinder, a wire, or a pipe respectively, then generally we can say that : R=C1 x A/l Where A is cross sectional area, l is length and C1 is a constant of proportionality and that D = C2 x FxR Where C2 is another constant. The obvious example of this is Ohm's Law for electricity which states that V=IR Click here to return Let us say that we need to achieve a detector temperature of less than 100K and that we only have enough room to have a cooling radiator of 0.5 m2. The detector dissipates 1W and we estimate that the parasitic heat load from the warm surround will be about 9W. If we opt for a single radiator (0.5 m2) to radiate the total power of 10W we can apply Stefan's law P=AT4 to show that , for a perfectly radiating surface, the temperature we can achieve is 136K. This does not meet the requirement. If we have a 2-stage radiator where the 1st stage intercepts and radiates away the parasitic 9W and the 2nd stage radiates away the dissipative load of the detector, then we use Stefan's law twice for each radiator. If we split the area 50:50 then we can show that T1 = 159K and T2 = 92K thus achieving our objective. If we split the area 10:90 then we achieve something apparently better i.e. T1 = 237K and T2 = 79K Clearly this is not the whole story because we have ignored the heat transfer between the shrouds which do the intercepting and which are coupled to the radiators. The magnitude of the transfer will be greater in the second case because there is a temperature difference of 158K driving it compared to that of 67K in the first case. So there is a limit to which we can divide up the area. It can be shown that there is an optimum solution and this confirms the common sense approach that if we have an n-stage radiator the dedicated areas should progressively increase from the outer radiator to the inner. Click here to return |
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