Wednesday, 11 May 2016

Thermal Resistance Circuits

Thermal Resistance Circuits

There is an electrical analogy with conduction heat transfer that can be exploited in problem solving. The analog of $ \dot{Q}$ is current, and the analog of the temperature difference, $ T_1 - T_2$ , is voltage difference. From this perspective the slab is a pure resistance to heat transfer and we can define
$\displaystyle \dot{Q} = \frac{T_1 - T_2}{R}$

where $ R = L/kA$ , the thermal resistance. The thermal resistance $ R$ increases as $ L$ increases, as $ A$ decreases, and as $ k$ decreases.
Figure : Heat transfer across a composite slab (series thermal resistance)
Image fig8CompositeSlabSeriesResistance_web
The concept of a thermal resistance circuit allows ready analysis of problems such as a composite slab (composite planar heat transfer surface). In the composite slab shown in Figure 16.6, the heat flux is constant with $ x$ . The resistances are in series and sum to $ R = R_1 + R_2$ . If $ T_L$ is the temperature at the left, and $ T_R$ is the temperature at the right, the heat transfer rate is given by
$\displaystyle \dot{Q} = \frac{T_L -T_R}{R} = \frac{T_L-T_R}{R_1+R_2}.$(

Figure : Heat transfer for a wall with dissimilar materials (parallel thermal resistance)
Image fig8BoltInWallParallelResistance_web
Another example is a wall with a dissimilar material such as a bolt in an insulating layer. In this case, the heat transfer resistances are in parallel. Figure 16.7 shows the physical configuration, the heat transfer paths and the thermal resistance circuit.
For this situation, the total heat flow $ \dot{Q}$ is made up of the heat flow in the two parallel paths, $ \dot{Q} = \dot{Q}_1 + \dot{Q}_2$ , with the total resistance given by
$\displaystyle \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}.$(16..23)

More complex configurations can also be examined; for example, a brick wall with insulation on both sides (Figure 16.8).
Figure : Heat transfer through an insulated wall
Image fig8BrickWallSeriesResistance_web
The overall thermal resistance is given by
$\displaystyle R = R_1 + R_2 + R_3 = \frac{L_1}{k_1A_1} + \frac{L_2}{k_2A_2}+\frac{L_3}{k_3A_3}.$(16..24)

Some representative values for the brick and insulation thermal conductivity are:
$\displaystyle k_\textrm{brick}$$\displaystyle = k_2$$\displaystyle = 0.7\textrm{ W/m-K}$   
$\displaystyle k_\textrm{insulation}$$\displaystyle = k_1 = k_3$$\displaystyle = 0.07\textrm{ W/m-K}$   

Using these values, and noting that $ A_1 = A_2 = A_3 = A$ , we obtain
$\displaystyle AR_1 = AR_3 = \frac{L_1}{k_1} = \frac{0.03\textrm{ m}}{0.07\textrm{ W/m-K}} = 0.42\textrm{ m\textsuperscript{2} K/W}$
$\displaystyle AR_2 = \frac{L_2}{k_2} = \frac{0.1\textrm{ m}}{0.7\textrm{ W/m-K}} = 0.14\textrm{ m\textsuperscript{2} K/W}.$
This is a series circuit so
$\displaystyle \dot{q} = \frac{\dot{Q}}{A} =\textrm{ constant throughout }= \fra... ...{0.98\textrm{ m\textsuperscript{2}K/W}} = 142\textrm{ W/m\textsuperscript{2}}. $
Figure : Temperature distribution through an insulated wall
Image fig8BrickWallTemperatureDistribution_web
The temperature is continuous in the wall and the intermediate temperatures can be found from applying the resistance equation across each slab, since $ \dot{Q}$ is constant across the slab. For example, to find $ T_2$ :
$\displaystyle \dot{q} =\frac{T_1 - T_2}{R_1A} = 142\textrm{ W/m\textsuperscript{2}}.$
This yields $ T_1 - T_2 = 60\textrm{ K}$ or $ T_2 = 90^\circ\textrm{C}$ .
The same procedure gives $ T_3 = 70^\circ\textrm{C}$ . As sketched in Figure , the larger drop is across the insulating layer even though the brick layer is much thicker.
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