Question 173: Thermodynamics & KTG

Question

Rods $x$ and $y$ of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points $A$ and $F$ are maintained at $100^{\circ} \mathrm{C}$ and $40^{\circ} \mathrm{C}$ respectively. Given the thermal conductivity of $\operatorname{rod} x$ is three times of that of $\operatorname{rod} y$, the temperature at junction points $B$ and $E$ are (close to):

Accepted Answer

Using law of heat conduction the rate of heat flow $\left(H=\frac{d Q}{d t}ight)$ through a material is proportional to the crosssectional area (A) and the temperature difference ( $\Delta \mathrm{T}$ ), and inversely proportional to the length (L) :
$$ H=\frac{k A \Delta T}{L} $$
Where k is the thermal conductivity of the material.
We can rearrange this equation to look like Ohm's Law for electricity $\left(\mathrm{I}=\frac{\Delta \mathrm{V}}{\mathrm{R}}ight)$ :
$$ H=\frac{\Delta T}{L / k A} $$
Here, the term $L / k A$ acts as the thermal resistance $\left(R_{t h}ight)$ of the rod. Heat current $(H)$ is analogous to electrical current (I), and temperature difference ( $\Delta \mathrm{T}$ ) is analogous to potential difference ( $\Delta \mathrm{V}$ ).
$$ \mathrm{R}_{\mathrm{th}}=\frac{\mathrm{L}}{\mathrm{kA}} $$
It is given that rods $x$ and $y$ have equal dimensions, meaning they have the same length $(L)$ and same crosssectional area (A).
We are also given that the thermal conductivity of $\operatorname{rod} \mathrm{x}\left(\mathrm{k}_{\mathrm{x}}ight)$ is three times that of $\operatorname{rod} \mathrm{y}\left(\mathrm{k}_{\mathrm{y}}ight)$.
Let $\mathrm{k}_{\mathrm{y}}=\mathrm{k}$, then $\mathrm{k}_{\mathrm{x}}=3 \mathrm{k}$.
So, the thermal resistances of rod x and y will be,
$$ \mathrm{R}_{\mathrm{y}}=\frac{\mathrm{L}}{\mathrm{k}_{\mathrm{y}} \mathrm{~A}}=\frac{\mathrm{L}}{\mathrm{kA}}=\mathrm{R} $$
$$\Rightarrow $$ $$ \mathrm{R}_{\mathrm{x}}=\frac{\mathrm{L}}{\mathrm{k}_{\mathrm{x}} \mathrm{~A}}=\frac{\mathrm{L}}{3 \mathrm{kA}}=\frac{1}{3}\left(\frac{\mathrm{~L}}{\mathrm{kA}}ight)=\frac{\mathrm{R}}{3} $$
We can now treat the given structure as an electrical circuit with specific resistances.
The heat current splits into two parallel paths BC and BD at B .
$$ \mathrm{R}_{\mathrm{BCE}}=\mathrm{R}_{\mathrm{BC}}+\mathrm{R}_{\mathrm{CE}}=\mathrm{R}+\mathrm{R}=2 \mathrm{R} $$
$$\Rightarrow $$ $$ \mathrm{R}_{\mathrm{BDE}}=\mathrm{R}_{\mathrm{BD}}+\mathrm{R}_{\mathrm{DE}}=\frac{\mathrm{R}}{3}+\frac{\mathrm{R}}{3}=\frac{2 \mathrm{R}}{3} $$
For parallel resistors $R_1$ and $R_2$, the equivalent resistance is $\frac{R_1 R_2}{R_1+R_2}$ :
$$ \mathrm{R}_{\mathrm{BE}}=\frac{(2 \mathrm{R}) 	imes\left(\frac{2 \mathrm{R}}{3}ight)}{(2 \mathrm{R})+\left(\frac{2 \mathrm{R}}{3}ight)} $$
$$\Rightarrow $$ $$ R_{B E}=\frac{4 R^2 / 3}{\frac{6 R+2 R}{3}}=\frac{4 R^2}{8 R}=\frac{R}{2} $$
From $A$ to $F$ with three equivalent resistors: $R_{A B}, R_{B E}$, and $R_{E F}$.
The total equivalent resistance ( $R_{	ext {total }}$ ) is :
$$ \mathrm{R}_{	ext {total }}=\mathrm{R}_{\mathrm{AB}}+\mathrm{R}_{\mathrm{BE}}+\mathrm{R}_{\mathrm{EF}} $$
$$\Rightarrow $$ $\mathrm{R}_{	ext {total }}=\frac{\mathrm{R}}{3}+\frac{\mathrm{R}}{2}+\mathrm{R}$
$$\Rightarrow $$ $$ \mathrm{R}_{	ext {total }}=\frac{11 \mathrm{R}}{6} $$
We know the overall temperature difference between A and F :
$$ \Delta \mathrm{T}_{	ext {total }}=\mathrm{T}_{\mathrm{A}}-\mathrm{T}_{\mathrm{F}}=100-40=60^{\circ} \mathrm{C} $$
The total steady-state heat current (H) flowing through the entire system is :
$$ H=\frac{\Delta T_{	ext {total }}}{R_{	ext {total }}}=\frac{60}{11 R / 6}=\frac{360}{11 R} $$
Consider the segment from A to B :
$$ H=\frac{T_A-T_B}{R_{A B}} $$
$\begin{aligned} \Rightarrow \frac{360}{11 \mathrm{R}} & =\frac{100-\mathrm{T}_{\mathrm{B}}}{\mathrm{R} / 3} \ \Rightarrow \frac{360}{11 \mathrm{R}} & =\frac{3\left(100-\mathrm{T}_{\mathrm{B}}ight)}{\mathrm{R}} \ \Rightarrow \frac{120}{11} & =100-\mathrm{T}_{\mathrm{B}}\end{aligned}$
$$\Rightarrow $$ $\mathrm{T}_{\mathrm{B}}=100-\frac{120}{11}=\frac{1100-120}{11}=\frac{980}{11}$
$$\Rightarrow $$ $$ \mathrm{T}_{\mathrm{B}} \approx 89.09^{\circ} \mathrm{C} $$
Consider the segment from E to F :
$$ \mathrm{H}=\frac{\mathrm{T}_{\mathrm{E}}-\mathrm{T}_{\mathrm{F}}}{\mathrm{R}_{\mathrm{EF}}} $$
$\begin{aligned} & \Rightarrow \frac{360}{11 R}=\frac{T_E-40}{R} \ & \Rightarrow \frac{360}{11}=T_E-40\end{aligned}$
$$\Rightarrow $$ $\mathrm{T}_{\mathrm{E}}=40+\frac{360}{11}=\frac{440+360}{11}=\frac{800}{11}$
$$\Rightarrow $$ $$ \mathrm{T}_{\mathrm{E}} \approx 72.72^{\circ} \mathrm{C} $$
Therefore, the temperatures at junction points B and E are close to $89^{\circ} \mathrm{C}$ and $73^{\circ} \mathrm{C}$ respectively.
Hence, the correct option is (C).

Answer

$60^{\circ} \mathrm{C}$ and $45^{\circ} \mathrm{C}$ respectively

Answer

$80^{\circ} \mathrm{C}$ and $70^{\circ} \mathrm{C}$ respectively

Answer

$89^{\circ} \mathrm{C}$ and $73^{\circ} \mathrm{C}$ respectively

Answer

$80^{\circ} \mathrm{C}$ and $60^{\circ} \mathrm{C}$ respectively

Thermodynamics & KTG

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