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Q.10
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 ______.

Figures:

Fig. 1:
$P \overset{W_{0}}{\to} Q$
Temperatures: $T_{p}$ (left), $T_{q}$ (right).
Fig. 2:
$P \overset{W_{s}}{\to} Plate 1 \overset{W_{s}}{\to} Plate 2 \overset{W_{s}}{\to} Q$
Temperatures: $T_{p}$ (left), $T_{1}$ (middle 1), $T_{2}$ (middle 2), $T_{q}$ (right).

Given:

All plates are perfect black bodies.
Heat transfer occurs only between adjacent plates.
Steady-state conditions apply in Fig. 2.

Answer: 3

Solution:

1. Key Concepts

Stefan-Boltzmann Law: The power radiated per unit area by a black body is $W = σ T^{4}$ , where $σ$ is the Stefan-Boltzmann constant.
Net Power Transfer: Between two plates at temperatures $T_{1}$ and $T_{2}$ ( $T_{1} > T_{2}$ ), the net power transferred per unit area is:
$W = σ (T_{1}^{4} - T_{2}^{4}) .$

2. Calculate $W_{0}$ (Fig. 1)

For the two-plate system (Fig. 1), the power transferred from $P$ to $Q$ is:

$W_{0} = σ (T_{p}^{4} - T_{q}^{4}) .$

3. Calculate $W_{s}$ (Fig. 2)

In the four-plate system (Fig. 2), let the steady-state temperatures of the two intermediate plates be $T_{1}$ and $T_{2}$ . Since heat transfer occurs only between adjacent plates, the power transferred per unit area between each pair must be equal in steady state:

$W_{s} = σ (T_{p}^{4} - T_{1}^{4}) = σ (T_{1}^{4} - T_{2}^{4}) = σ (T_{2}^{4} - T_{q}^{4}) .$

This implies:

$T_{p}^{4} - T_{1}^{4} = T_{1}^{4} - T_{2}^{4} = T_{2}^{4} - T_{q}^{4} .$

Let $T_{p}^{4} - T_{1}^{4} = Δ$ . Then:

$T_{1}^{4} = T_{p}^{4} - Δ, T_{2}^{4} = T_{1}^{4} - Δ = T_{p}^{4} - 2 Δ, T_{q}^{4} = T_{2}^{4} - Δ = T_{p}^{4} - 3 Δ .$

Thus:

$Δ = \frac{T_{p}^{4} - T_{q}^{4}}{3} .$

The power transferred per unit area $W_{s}$ is:

$W_{s} = σ Δ = σ (\frac{T_{p}^{4} - T_{q}^{4}}{3}) = \frac{W_{0}}{3} .$

4. Compute the Ratio $\frac{W_{0}}{W_{s}}$

From the above results:

$\frac{W_{0}}{W_{s}} = \frac{σ (T_{p}^{4} - T_{q}^{4})}{\frac{σ (T_{p}^{4} - T_{q}^{4})}{3}} = 3.$

5. Final Answer

The ratio $\frac{W_{0}}{W_{s}}$ is:

\boxed{3}

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Solution:

1. Key Concepts

2. Calculate $W_{0}$ (Fig. 1)

3. Calculate $W_{s}$ (Fig. 2)

4. Compute the Ratio $\frac{W_{0}}{W_{s}}$

5. Final Answer

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Solution:

1. Key Concepts

2. Calculate W0W0​ (Fig. 1)

3. Calculate WsWs​ (Fig. 2)

4. Compute the Ratio W0WsWs​W0​​

5. Final Answer

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2. Calculate $W_{0}$ (Fig. 1)

3. Calculate $W_{s}$ (Fig. 2)

4. Compute the Ratio $\frac{W_{0}}{W_{s}}$