Question 1: Thermodynamics & KTG

Question

If two metallic plates of equal thickness and thermal conductivities $K_1$ and $K_2$ are put together face to face and a common plate is constructed, then the equivalent thermal conductivity of this plate will be

Accepted Answer

Let the temperature of interface be $T$.
Let the temperature of two sides be $T_1$ and $T_2$, then
$\frac{K A\left(T_1-T_2ight) t}{2 l}=\frac{K_1 A\left(T_1-Tight) t}{I}=\frac{K_2 A\left(T-T_2ight) t}{I}$
($\because$ The rate of each flow in each plate is same)
$\begin{array}{rlrl}
	ext { Here } K_1 T_1-K_1 T & =K_2 T-K_2 T_2 \
K_1 T_1+K_2 T_2 & =T\left(K_1+K_2ight) \
	herefore \quad & T =\frac{K_1 T_1+K_2 T_2}{K_1+K_2} \quad 	ext{..... (i)}
\end{array}$
$\begin{aligned}
	ext { Further, } \frac{K A\left(T_1-T_2ight) t}{2} & =K_1 A\left(T_1-Tight) t \
K & =2 K_1\left(\frac{T_1-T}{T_1-T_2}ight) \quad 	ext{..... (ii)}
\end{aligned}$
Substituting the value of $T$ from Eq. (i) in Eq. (ii), we get
$K=\frac{2 K_1}{T_1-T_2}\left[T_1-\left(\frac{K_1 T_1+K_2 T_2}{K_1+K_2}ight)ight]$
Solving it, we get
$K=\frac{2 K_1 K_2}{K_1+K_2}$

Answer

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

Answer

$\frac{2 K_1 K_2}{K_1+K_2}$

Answer

$\frac{\left(K_1^2+K_2^2\right)^{3 / 2}}{K_1 K_2}$

Answer

$\frac{\left(K_1^2+K_2^2\right)^{3 / 2}}{2 K_1 K_2}$

Thermodynamics & KTG

If two metallic plates of equal thickness and thermal conductivities K_1 and K_2 are put together face to face and a common plate is constructed, then the equivalent thermal conductivity of this plate will be

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