Question 63: Thermodynamics & KTG

Question

Two slabs are of the thicknesses $$d_1$$ and $$d_2$$. Their thermal conductivities are $$K_1$$ and $$K_2$$, respectively. They are in series. The free ends of the combination of these two slabs are kept at temperatures $$	heta_1$$ and $$	heta_2$$. Assume $$	heta_1 > 	heta_2$$. The temperature $$	heta$$ of their common junction is

Accepted Answer

For first slab,
Heat current, $$H_1=\frac{K_1\left(	heta_1-	hetaight) A}{d_1}$$
For second slab,
Heat current, $$H_2=\frac{K_2\left(	heta-	heta_2ight) A}{d_2}$$
As slabs are in series, $$H_1=H_2$$
$$\begin{aligned}
	herefore \quad & \frac{K_1\left(	heta_1-	hetaight) A}{d_1}=\frac{K_2\left(	heta-	heta_2ight) A}{d_2} \
\Rightarrow \quad & 	heta=\frac{K_1 	heta_1 d_2+K_2 	heta_2 d_1}{K_2 d_1+K_1 d_2}
\end{aligned}$$

Answer

$$\frac{K_1 	heta_1+K_2 	heta_2}{	heta_1+	heta_2}$$

Answer

$$\frac{K_1 	heta_1 d_1+K_2 	heta_2 d_2}{K_1 d_2+K_2 d_1}$$

Answer

$$\frac{K_1 	heta_1 d_2+K_2 	heta_2 d_1}{K_1 d_2+K_2 d_1}$$

Answer

$$\frac{K_1 	heta_1+K_2 	heta_2}{K_1+K_2}$$

Thermodynamics & KTG

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