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Recall that the steady-state heat equation takes the form:
This equation dictates the shape of the temperature profile over the material for a given temperature difference. This differential equation will always produce a linear (or, if dT/dx = 0, a homogenous) relationship between temperature, T and distance away from the heat source, x in a plane wall. Therefore, if one can determine the temperature at two values of x (i.e. have 2 sets of boundary conditions), then one can fully determine the temperature profile across the wall if the wall is made of one material.
If the plane wall is made of multiple materials, as in a composite wall, then the temperature profile across the wall is likely to consist of many sections with different linear temperature gradients, as can be seen below:
These temperature gradients depend upon the each of the materials' thermal conductivity, k. To understand the temperature profile across a composite wall one needs knowledge of all the boundary conditions. For example, in the composite wall above, knowing Ts1, T2, T3 and Ts4 and using the steady-state conduction equation gives a full understanding of the wall's temperature profile.
But what if the question in an exam isn't that simple? What if you were expected to design a system which resists heat flow, rather than to just provide a simple conduction steady-state analysis for a number of given boundary conditions? In these circumstances, it is useful to represent the heat flux-temperature system as an electrical system, where thermal resistance is represented by electrical resistance, heat rate is represented by current and temperature difference is represented by potential difference.
Before attempting to tackle a thermal resistance analysis, it is important to define heat flow, heat flux and the difference between the two terms.
The crucial difference between the two is that heat rate refers to actual energetic transfers across a material, whereas heat flux refers to a heat per unit area. The gradient of heat rate uses the materials' thermal conductivity and the geometry of the system (cross sectional area, A and length, L) to define the section's conductivity = kA/L. The reciprocal of this is equal to the section's thermal resistance, such that L/kA = R. Using this principle, we can define a heat-flow system as a network of 'resistors' to heat flow for a given temperature difference across the system. Whereas V = IR in an electrical system, Tdifference = qR in a thermal system.
For example, the composite wall above can be represented as 3 resistors in series, such that:
In this system, we could play with the system parameters of the materials of conductivity kA, kB and kC and their section's geometries to manipulate the relationship between temperature difference and heat flow.
As an example, lets study the effect of having multiple layers of clothes on during a standard English spring day. Assume Tatmos = 5 degrees, Tskin = 35 degrees, Rtshirt = 0.1 K/W, Rjumper = 0.3 K/W, Rcoat = 1 K/W.
This can be computed by considering the t-shirt, jumper and coat as resistors which impede a heat flow, caused by a 30 degrees temperature difference between the skin and the environment.With just a t-shirt on and using T/R = q, a person will suffer a heat loss of 300W. When the person starts introducing layers, the person experiences dramatically less heat loss due to the effect of having these thermal 'resistors' in series. With a jumper and a t-shirt on, the person experiences a heat loss of 75W and a person with a tshirt, jumper and coat on only experiences a 21W heat loss!
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