tHERMAL CYU 12

Solution 12: Temperature of a Double-Thickness Plate

Let $I$ be the solar intensity, $\alpha$ be the heat loss coefficient to the air, and $d$ be the thickness of the plate. The surroundings are at $T_0$. We analyze the steady state where Heat Absorbed = Total Heat Lost.

1. Conservation of Energy (Sum of Temperatures):
The total heat lost to the air is the sum of losses from the front face ($T_1$) and the back face ($T_2$). $$ I = \alpha(T_1 – T_0) + \alpha(T_2 – T_0) $$ For the double thickness plate ($2d$) with new temperatures $T_1’$ and $T_2’$, the same balance applies: $$ I = \alpha(T_1′ – T_0) + \alpha(T_2′ – T_0) $$ Equating these yields: $$ (T_1 – T_0) + (T_2 – T_0) = (T_1′ – T_0) + (T_2′ – T_0) $$ $$ T_1 + T_2 = T_1′ + T_2′ \quad \dots(1) $$

2. Conduction Constraint (Difference of Temperatures):
Heat conducted through the plate equals heat lost from the back face. For thickness $d$: $$ \frac{K}{d}(T_1 – T_2) = \alpha(T_2 – T_0) $$ For thickness $2d$: $$ \frac{K}{2d}(T_1′ – T_2′) = \alpha(T_2′ – T_0) $$ Dividing the second equation by the first: $$ \frac{1}{2} \left( \frac{T_1′ – T_2′}{T_1 – T_2} \right) = \frac{T_2′ – T_0}{T_2 – T_0} $$ $$ T_1′ – T_2′ = 2 \left( \frac{T_1 – T_2}{T_2 – T_0} \right) (T_2′ – T_0) \quad \dots(2) $$

3. Solving for $T_2’$:
From eq (1), substitute $T_1′ = (T_1 + T_2) – T_2’$ into eq (2): $$ (T_1 + T_2 – T_2′) – T_2′ = 2 \left( \frac{T_1 – T_2}{T_2 – T_0} \right) (T_2′ – T_0) $$ $$ T_1 + T_2 – 2T_2′ = 2 \left( \frac{T_1 – T_2}{T_2 – T_0} \right) (T_2′ – T_0) $$ Let $C = \frac{T_1 – T_2}{T_2 – T_0}$. $$ T_1 + T_2 + 2C T_0 = T_2′(2 + 2C) $$ $$ T_2′ = \frac{T_1 + T_2 + 2C T_0}{2(1+C)} $$ Substitute $C$ back. The denominator $2(1+C) = 2 \left( \frac{T_2 – T_0 + T_1 – T_2}{T_2 – T_0} \right) = \frac{2(T_1 – T_0)}{T_2 – T_0}$. The numerator simplifies to $\frac{T_1 T_2 + T_0(T_1 – 3T_2) + T_2^2}{T_2 – T_0}$. Result for $T_2’$: $$ T_2′ = \frac{T_1 T_2 + T_0(T_1 – 3T_2) + T_2^2}{2(T_1 – T_0)} $$