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 $\frac{W_0}{W_S}$ is ___.

Answer 1

3

Answer 2

In Fig. 1, 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 $T_Q < T_P$. The radiated power transferred per unit area from P to Q is $W_0$.

Since the plates are perfect black bodies, their emissivity is $e=1$. The power radiated per unit area by a black body at temperature $T$ is $\sigma T^4$, where $\sigma$ is the Stefan-Boltzmann constant.

The net power transferred per unit area from P to Q is the difference between the power radiated by P towards Q and the power radiated by Q towards P. Assuming they are parallel plates with large area, the net power transferred per unit area from P to Q is given by: $W_0 = \sigma T_P^4 - \sigma T_Q^4$

In Fig. 2, two more plates, identical to P and Q, are introduced between P and Q. Let's call these intermediate plates R and S, placed between P and Q in the order P, R, S, Q. All plates are perfect black bodies. Let the temperatures of the intermediate plates in the steady state be $T_R$ and $T_S$.

In the steady state, the net power transferred per unit area between any two adjacent plates is the same, say $W_S$. Heat transfer takes place only between adjacent plates.

The net power transferred per unit area from P to R is $W_S = \sigma T_P^4 - \sigma T_R^4$. The net power transferred per unit area from R to S is $W_S = \sigma T_R^4 - \sigma T_S^4$. The net power transferred per unit area from S to Q is $W_S = \sigma T_S^4 - \sigma T_Q^4$.

From these equations, we have: $\sigma T_P^4 - \sigma T_R^4 = W_S$ $\sigma T_R^4 - \sigma T_S^4 = W_S$ $\sigma T_S^4 - \sigma T_Q^4 = W_S$

Adding these three equations, we get: $(\sigma T_P^4 - \sigma T_R^4) + (\sigma T_R^4 - \sigma T_S^4) + (\sigma T_S^4 - \sigma T_Q^4) = W_S + W_S + W_S$ $\sigma T_P^4 - \sigma T_Q^4 = 3W_S$

We know that $W_0 = \sigma T_P^4 - \sigma T_Q^4$. So, $W_0 = 3W_S$.

The ratio $\frac{W_0}{W_S} = \frac{3W_S}{W_S} = 3$.