Question

Two identical plates $ P $ and $ Q $, radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures $ T_P $ and $ T_Q $, respectively, with $ T_Q<T_P $, as shown in Fig. 1. The radiated power transferred per unit area from $ P $ to $ Q $ is $ W_0 $. Subsequently, two more plates, identical to $ P $ and $ Q $, are introduced between $ P $ and $ Q $, as shown in Fig. 2. Assume that heat transfer takes place only between adjacent plates. If the power transferred per unit area in the direction from $ P $ to $ Q $ (Fig. 2) in the steady state is $ W_S $, then the ratio $ \dfrac{W_0}{W_S} $ is ____. 

Hint:

In multi-body radiation problems, break total energy transfer into segments between adjacent bodies. At thermal steady state, the power flow across each segment must be equal.

Answer 1

Correct Answer: 3

Step 1: Power radiated between two black bodies

For two black bodies at temperatures \( T_P \) and \( T_Q \), the net power radiated per unit area is given by the Stefan-Boltzmann law:

\[ W = \sigma \left( T_P^4 - T_Q^4 \right) \]

So initially (Fig. 1), the power is:

\[ W_0 = \sigma \left( T_P^4 - T_Q^4 \right) \]

Step 2: Insert two intermediate black plates

In Fig. 2, two additional identical black plates are placed between \( P \) and \( Q \). Let the four plates be \( P, A, B, Q \) from left to right.

Assume steady state and let the temperatures of the intermediate plates be \( T_1 \) and \( T_2 \), such that:

\[ T_P>T_1>T_2>T_Q \]

Now, energy is transferred between adjacent pairs only:

• From \( P \) to \( A \): \( W = \sigma (T_P^4 - T_1^4) \)
• From \( A \) to \( B \): \( W = \sigma (T_1^4 - T_2^4) \)
• From \( B \) to \( Q \): \( W = \sigma (T_2^4 - T_Q^4) \)

At steady state, the energy flow rate must be the same through all three segments:

\[ \sigma (T_P^4 - T_1^4) = \sigma (T_1^4 - T_2^4) = \sigma (T_2^4 - T_Q^4) \]

Let this common value be \( W_S \), then:

\[ T_P^4 - T_1^4 = T_1^4 - T_2^4 = T_2^4 - T_Q^4 = \Delta \]

Therefore:

\[ \begin{align} T_P^4 - T_Q^4 &= (T_P^4 - T_1^4) + (T_1^4 - T_2^4) + (T_2^4 - T_Q^4) \\ &= 3\Delta = \frac{3W_S}{\sigma} \Rightarrow W_S = \dfrac{1}{3} \sigma (T_P^4 - T_Q^4) \end{align} \]

Step 3: Compute the ratio

\[ \frac{W_0}{W_S} = \frac{\sigma (T_P^4 - T_Q^4)}{(1/3) \sigma (T_P^4 - T_Q^4)} = \boxed{3} \]