Radiation Heat Leak into a Liquid-Oxygen Dewar: 23-Chem-A3 Heat and Mass Transfer, December 2018, Q3

Question 3: Radiation Heat Leak into a Liquid-Oxygen Dewar (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Inner sphere $D_1=0.30$ m at the LOX boiling point $T_1=-183$ °C $=90.15$ K; outer sphere $D_2=0.50$ m at $T_2=40$ °C $=313.15$ K; evacuated gap (radiation only); both aluminium, $\varepsilon=0.3$ in part (a) and $0.05$ in part (b).

Quantity	Value
Inner sphere $D_1$ (LOX)	0.30 m at −183 °C (90.15 K)
Outer sphere $D_2$	0.50 m at 40 °C (313.15 K)
Emissivity (a) / (b)	0.3 / 0.05 (both walls)
Gap	evacuated → radiation only

Find. (a) the radiative heat leak; (b) the reduction when both walls are polished to $\varepsilon=0.05$.

Figure 3 — Concentric-sphere (dewar) geometry. Heat radiates across the evacuated gap from the warm 40 °C outer wall to the cold liquid-oxygen container; the area ratio $A_1/A_2=(D_1/D_2)^2=0.36$ weights the outer-surface resistance.

Approach. Apply the two-surface enclosure result for concentric spheres, whose radiation network reduces to a single expression with an area-ratio term, then evaluate it for the two emissivities and compare.

Concentric-sphere exchange formula. For a small grey sphere inside a large grey sphere,$$Q=\frac{\sigma A_1\,(T_2^{4}-T_1^{4})}{\dfrac{1}{\varepsilon_1}+\dfrac{A_1}{A_2}\left(\dfrac{1}{\varepsilon_2}-1\right)},\qquad A_1=\pi D_1^{2}=0.2827\ \text{m}^2,\quad \frac{A_1}{A_2}=\left(\frac{0.30}{0.50}\right)^2=0.36.$$The temperature term is $T_2^4-T_1^4=313.15^4-90.15^4=9.55\times10^{9}\ \text{K}^4$ — the warm wall dominates, so the exact LOX temperature barely matters.
Part (a): both walls $\varepsilon=0.3$. The denominator is $1/0.3+0.36(1/0.3-1)=3.333+0.840=4.173$, so$$Q_a=\frac{(5.67\times10^{-8})(0.2827)(9.55\times10^{9})}{4.173}=\boxed{36.7\ \text{W}}.$$
Part (b): both walls polished, $\varepsilon=0.05$. The denominator becomes $1/0.05+0.36(1/0.05-1)=20+6.84=26.84$, so$$Q_b=\frac{(5.67\times10^{-8})(0.2827)(9.55\times10^{9})}{26.84}=\boxed{5.70\ \text{W}}.$$
Reduction.$$\frac{Q_a-Q_b}{Q_a}=\frac{36.7-5.70}{36.7}=\boxed{84.5\%}.$$Reading "container walls" as both aluminium surfaces polished; if only the inner container were polished the reduction would be 80.0%. The huge gain comes because each surface contributes a $1/\varepsilon$ resistance, so dropping $\varepsilon$ from 0.3 to 0.05 multiplies both resistances roughly six-fold.

Quantity	Result
(a) Radiative heat leak, $\varepsilon=0.3$	36.7 W
(b) Heat leak, polished $\varepsilon=0.05$	5.70 W
(b) Reduction in heat flow	84.5%

Concept & Theory

Topic. Radiation exchange in a two-surface concentric-sphere enclosure (cryogenic dewar).

Key relations. Grey-body network $Q=\sigma A_1(T_2^4-T_1^4)/[1/\varepsilon_1+(A_1/A_2)(1/\varepsilon_2-1)]$; Stefan–Boltzmann $\sigma=5.67\times10^{-8}\ \text{W/m}^2\text{K}^4$.

Why this works. With the gap evacuated, conduction and convection vanish and only radiation crosses. The enclosure has three series resistances — two surface resistances $(1-\varepsilon)/(\varepsilon A)$ and one space resistance $1/(A_1F_{12})$ with $F_{12}=1$. Combining them gives the compact form above; low emissivity attacks both surface resistances, which is why polishing is so effective.

Common pitfall. Using $Q=\sigma A_1\varepsilon(T^4)$ (single-surface) and ignoring the area-ratio term, or forgetting to convert °C to K before taking the fourth power.

Source. Incropera & DeWitt §13.3 (radiation exchange between diffuse-grey surfaces), concentric spheres/cylinders.

Question 3 of 6: Radiation Heat Leak into a Liquid-Oxygen Dewar

Part A — Heat Transfer

Question 3: Radiation Heat Leak into a Liquid-Oxygen Dewar (25 points)