UY1: Thermal conduction through a compound slab


Often, compound slabs comprising several layers of different materials are used for insulation.

Consider a compound slab comprising two materials of thickness L1, L2 and thermal conductivities k1, k2 respectively. The temperatures of the outer surfaces are T1 and T2, where T2 > T1. Let the temperature at the interface be T. The compound slabs will be at steady state. Hence, the heat flow through the two layers must be the same (otherwise heat will be accumulating somewhere and causing a local rise in temp), we have:

$$\begin{aligned} \left. \frac{dQ}{dt}\right|_{1} &= \left. \frac{dQ}{dt}\right|_{2} \\ \frac{k_{1}A}{L_{1}} (T – T_{1}) &= \frac{k_{2}A}{L_{2}} (T_{2} – T) \end{aligned}$$

 

Radial Heat Flow

Consider a steam pipe of radius a, surrounded by a layer of insulating material of outer radius b, length L and thermal conductivity k. If the temperature of the steam pipe is $T_{2}$ and the air outside the insulating material is $T_{1}$, where $T_{1} < T_{2}$.

We will attempt to find the rate of energy transfer through the insulating material and the temperature at r (a < r < b) when a steady state has been reached.

Firstly, using the equation above for heat conduction through compound slab as a reference and note that the surface of the insulating material has the same temperature as the air. Heat power radiated from the steam pipe is the same in the inner slab and outer slab at steady state.

 

Heat conduction through a cylindrical section in the insulation is given by:

$$\frac{dQ}{dt} = \, – kA \frac{dT}{dr}$$

At steady state, $\frac{dQ}{dt}$ is constant independent of r.

$$\frac{dQ}{dt} = \, – k (2 \pi r L) \frac{dT}{dr}$$

Separating variables and integrating,

$$\int\limits_{a}^{b} \frac{1}{r} \, dr = \int\limits_{T_{2}}^{T_{1}} \frac{- 2 \pi k L}{\frac{dQ}{dt}} \, dT$$

The above integration will yield:

$$\text{ln} \, \frac{b}{a} = \frac{2 \pi k L}{\frac{dQ}{dt}} \left( T_{2} – T_{1} \right)$$

Rearranging the above equation:

$$\frac{dQ}{dt} = \frac{2 \pi k L}{\text{ln} \, \frac{b}{a}} \left( T_{2} – T_{1} \right)$$

 

Side note: This means that the thermal conductance through the insulation is $ \frac{2 \pi k L}{\text{ln} \, \frac{b}{a}}$

 

Since heat flow is constant across r,

$$\frac{2 \pi k L (T_{2} – T_{1} )}{\text{ln} \, \frac{b}{a}} = \frac{2 \pi k L (T_{2} – T)}{\text{ln} \, \frac{r}{a}}$$

Rearranging:

$$T = \frac{ T_{2} \text{ln} \, \frac{b}{r} + T_{1} \text{ln} \, \frac{r}{a} }{ \text{ln} \, \frac{b}{a}}$$

 

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Previous: Build Up Of Temperature Gradient

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