Two identical plates and , radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures and , respectively, with , as shown in Fig. 1. The radiated power transferred per unit area from to is . Subsequently, [2025]

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 ______. [2025]

Smart Achievers · Accepted Answer

(3)

$Initially the radiated power transferred per unit area from P to Q$

$Finally in steady state$

$T_{Q} < T_{P}$

$Q$

$Adding we get$

$T_{P}^{4} - T_{1}^{4} = T_{2}^{4} - T_{Q}^{4} \Rightarrow T_{1}^{4} + T_{2}^{4} = T_{P}^{4} + T_{Q}^{4}$

$or, T_{1}^{4} - T_{2}^{4} = T_{P}^{4} - T_{1}^{4}$

$Adding: T_{1}^{4} = \frac{2 T_{P}^{4} + T_{Q}^{4}}{3}$

$So, W_{S} = σ (T_{P}^{4} - T_{1}^{4})$

$= σ [T_{P}^{4} - (\frac{2 T_{P}^{4} + T_{Q}^{4}}{3})] = σ (\frac{T_{P}^{4} - T_{Q}^{4}}{3})$

$∴$ $\frac{W_{0}}{W_{S}} = \frac{σ (T_{P}^{4} - T_{Q}^{4})}{σ (\frac{T_{P}^{4} - T_{Q}^{4}}{3})} = 3$

Solution

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