Cryogenic Dewar with a Radiation Shield: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2017, Q7

Question 7: Cryogenic Dewar with a Radiation Shield (equal value)

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Surface	Value
Inner shell (liquid O₂)	$r_1=255$ mm, $T_1=90$ K, $A_1=0.817$ m²
Radiation shield (floating)	$r_s=280$ mm, $A_s=0.985$ m²
Outer shell	$r_2=305$ mm, $T_2=45$ °C $=318.15$ K, $A_2=1.169$ m²
Surfaces	All $\varepsilon=0.3$; vacuum (radiation only)

Given. The three-shell geometry, temperatures and emissivity above. Find. the rate of heat gain by the liquid oxygen with the radiation shield in place.

Figure 7 — Three concentric spheres. The floating shield at $T_s$ splits the radiation path into two series gaps; at steady state the same heat crosses both, which fixes $T_s$ and the leak.

Approach. In the vacuum only radiation transfers heat; with a floating shield the inner gap (inner shell ↔ shield) and the outer gap (shield ↔ outer shell) carry the same heat in series, so equate them to find the shield temperature, then evaluate the leak.

Concentric-sphere radiation resistance. For inner surface $i$ and outer surface $o$, $\;\dot Q=\dfrac{\sigma A_i(T_o^4-T_i^4)}{\dfrac1{\varepsilon_i}+\dfrac{A_i}{A_o}\!\left(\dfrac1{\varepsilon_o}-1\right)}$. Denominators: inner gap $=3.333+\dfrac{r_1^2}{r_s^2}(2.333)=5.269$; outer gap $=3.333+\dfrac{r_s^2}{r_2^2}(2.333)=5.300$.
Series (steady) condition. Writing $\dot Q=c_A(T_s^4-T_1^4)=c_B(T_2^4-T_s^4)$ with $c_A=\dfrac{\sigma A_1}{5.269}=8.793\times10^{-9}$ and $c_B=\dfrac{\sigma A_s}{5.300}=1.054\times10^{-8}$, solve for the shield temperature: $$T_s^4=\frac{c_BT_2^4+c_AT_1^4}{c_A+c_B}\ \Rightarrow\ T_s\approx273.7\ \text{K}$$
Heat gained by the liquid oxygen (with shield). $\dot Q=c_B(T_2^4-T_s^4)=1.054\times10^{-8}(318.15^4-273.7^4)$. $\dot Q_{shield}\approx48.8$ W
Comparison with no shield. Removing the shield, $\dot Q_{no}=\dfrac{\sigma A_1(T_2^4-T_1^4)}{3.333+\dfrac{r_1^2}{r_2^2}(2.333)}\approx95.0$ W, so the single shield cuts the leak by about half. Shield reduces the heat gain from ≈ 95 W to ≈ 49 W — a ≈ 49 % reduction (the shielded leak is ≈ 51 % of the unshielded one)

Quantity	Result
Shield equilibrium temperature	≈ 273.7 K
Heat gain with shield	≈ 48.8 W
Heat gain without shield	≈ 95.0 W
Reduction	≈ 49 %

Concept & Theory

Topic. Radiation exchange between concentric spheres in vacuum, with a floating radiation shield.

Key relations. $\dot Q=\dfrac{\sigma A_1(T_1^4-T_2^4)}{\frac1{\varepsilon_1}+\frac{A_1}{A_2}(\frac1{\varepsilon_2}-1)}$ for concentric spheres; a floating shield gives two such resistances in series with a common (unknown) $T_s$ set by $\dot Q_{in}=\dot Q_{out}$.

Why this works. With the space evacuated, conduction and convection vanish and only surface-to-surface radiation remains. The shield reaches a steady temperature at which it re-radiates inward exactly what it absorbs from outside; inserting it adds a large surface resistance in the path, and for equal emissivities a single shield roughly halves the leak. The exact split is weighted by the sphere areas, so the shield sits nearer the cold side in temperature-to-the-fourth space.

Common pitfall. Using temperatures in °C instead of kelvin in the $T^4$ terms; mixing up which area ($A_1$ vs $A_s$ vs $A_2$) belongs in each gap's resistance; and assuming the shield temperature is the arithmetic mean of 90 K and 318 K (it is not — the fourth-power weighting and area ratios shift it).

Source. Çengel & Ghajar, Heat and Mass Transfer §13 (radiation heat transfer — the radiation-shield analysis and concentric enclosures).

Question 7 of 8: Cryogenic Dewar with a Radiation Shield

Question 7: Cryogenic Dewar with a Radiation Shield (equal value)