The temperatures of the two outer surfaces of a composite...

The temperatures of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K$ and $2 K$ and thickness $x$ and $4 x$, respectively, are $T_{2}$ and $T_{1}$ $\left(T_{2}>T_{1}\right)$. The rate of heat transfer through the slab in a steady state is $\left[\frac{A\left(T_{2}-T_{1}\right) K}{x}\right] f$ with $f$ equal to (A) 1 (B) $1 / 2$ (C) $2 / 3$ (D) $1 / 3$

Key Concepts

Composite Material Conduction

Composite material conduction deals with calculating the effective thermal conductivity of systems made up of different materials. In such systems, each layer has its own thermal conductivity and thickness, and the overall heat transfer is determined by the sum of the resistances of each individual layer, requiring a careful consideration of series thermal resistances.

Thermal Resistance

Thermal resistance is a concept that quantifies the opposition a material offers to heat flow. In composite systems, individual resistances of layers add in series, making it easier to calculate the overall temperature drop and heat transfer rate through multiple layers, similar to electrical resistances in a series circuit.

Fourier’s Law

Fourier’s law provides the foundation for analyzing heat transfer through materials and states that the heat flux is proportional to the negative of the temperature gradient. This law underpins the calculation of heat transfer rates in both homogeneous and composite media by linking thermal conductivity, temperature difference, and geometry.

Thermal Conduction

Thermal conduction is the process by which heat energy is transmitted through collisions between neighboring atoms or molecules. It is described by Fourier's law, which states that the heat transfer rate is proportional to the temperature gradient and the area through which the heat flows, emphasizing the material's ability to conduct heat.

Transcript

00:01 In the given question, we required to find out the value of f in the equation that has been given to us after comparing the rate of heat transfer using the given scenario to us.

00:12 So there are two slabs.

00:13 The first slab which has thickness x and it has thermal conductivity k.

00:20 The other slab has thickness 4x but it has thermal conductivity as 2k.

00:27 So using this we'll now be calculating what the value of.

00:31 F will be.

00:32 For this we'll be using some basic formulas.

00:34 So we all know that the rate of heat transfer is equal to t minus t0 upon r.

00:45 Here r is a thermal resistivity.

00:48 So now this thermal resistance now to find this we'll be using another relation and that relation is that r, which is a thermal resistance, is equal to l, the length of width of the slab upon k, the thermal conductivity into the area of the slab.

01:07 So using these two formulas, we'll be able to find out and solve the question.

01:12 So, firstly, let us find out are effective of this situation.

01:17 So this is a series combination.

01:20 So for this, in this case, our effective will be r1 plus r2.

01:24 Let's say r1 is the thermal resistance of the first slab and r2 is the thermal resistance of the second slab.

01:32 So r1 is equal to l1 upon k1a1 and r2 is equal to l2 upon k2a2.

01:45 So from this we can say that r1 is equal to x upon ka...

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