Radiation Shielding of a Liquid-Nitrogen Line: 22-Agric-A3 Heat Engineering, December 2014, Q3

Problem 3: Radiation Shielding of a Liquid-Nitrogen Line (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.

Quantity	Value
LN₂ line saturation temperature, T₁	80 K
Line outside diameter, D₁	6.35 mm
Line emissivity, ε₁	0.2
Shield tube diameter, D₂	12.7 mm
Vacuum chamber wall temperature, T_c	230 K
Chamber area vs. line	large (chamber acts as an infinite black surroundings)

Find. The radiant heat gain per unit length of line, (a) bare, and (b) with the concentric shield tube fitted.

Approach. With the chamber wall area far larger than the line, radiation exchange with it reduces to the small-object-in-a-large-enclosure result. Adding a floating concentric shield inserts two more grey-surface resistances (shield inner face to the line, shield outer face to the chamber) in series with the line's own surface resistance; combine them in a per-unit-length radiation network and solve for Q between the fixed temperatures T₁ and T_c.

Check: the shield tube's emissivity is not given in the source. It is assumed equal to the line's, ε_s = 0.2, since both are stated to be stainless steel tubing with no finish specified otherwise — the standard assumption for this class of problem.

Cross-section (schematic, not to scale): the LN₂ line radiates to the floating shield tube, which in turn radiates to the far, large-area vacuum chamber wall.

Bare-line heat gain (no shield). A small grey cylinder of area $A_1=\pi D_1$ (per unit length) inside a much larger enclosure exchanges radiation as $$\frac{Q}{\ell} = \varepsilon_1\sigma A_1\left(T_c^4-T_1^4\right)$$ With $A_1 = \pi(6.35\times10^{-3}) = 1.995\times10^{-2}\ \text{m}^2/\text{m}$, $T_c^4=230^4=2.798\times10^{9}\ \text{K}^4$ and $T_1^4=80^4=4.096\times10^{7}\ \text{K}^4$, $$\frac{Q}{\ell} = 0.2\times(5.67\times10^{-8})\times(1.995\times10^{-2})\times\left(2.798\times10^{9}-4.096\times10^{7}\right) = \boxed{0.624\ \text{W/m}}$$
Radiation network with the shield fitted. Per unit length, with $A_2=\pi D_2 = 3.990\times10^{-2}\ \text{m}^2/\text{m}$, the resistance path line → shield inner face → shield outer face → chamber collapses (both shield faces share the same emissivity $\varepsilon_s$) to $$R_{tot} = \frac{1}{\varepsilon_1A_1} + \frac{2/\varepsilon_s - 1}{A_2}$$ $$R_{tot} = \frac{1}{0.2(1.995\times10^{-2})} + \frac{2/0.2-1}{3.990\times10^{-2}} = 250.7 + 225.6 = 476.2\ \text{m}^{-1}$$
Revised heat gain. The same temperature difference now drives flow through the larger resistance, $$\frac{Q}{\ell} = \frac{\sigma\left(T_c^4-T_1^4\right)}{R_{tot}} = \frac{5.67\times10^{-8}\times 2.757\times10^{9}}{476.2} = \boxed{0.328\ \text{W/m}}$$ The single shield nearly halves the radiant heat gain (0.328/0.624 = 0.53), even though it has the same emissivity as the line itself — the reduction comes from inserting an extra pair of surface resistances and a larger radiating area at the shield, not from a lower emissivity.

Quantity	Result
Heat gain per unit length, bare line	0.624 W/m
Heat gain per unit length, with shield	0.328 W/m
Reduction	≈47%

Concept & Theory

Topic: Radiation shielding of a cryogenic line using a floating concentric cylindrical shield.

Key relations: Small object in a large enclosure, $Q=\varepsilon_1\sigma A_1(T_c^4-T_1^4)$; with one floating shield of area $A_2$ and emissivity $\varepsilon_s$ on both faces, $Q=\sigma(T_c^4-T_1^4)\big/\big[\tfrac{1}{\varepsilon_1A_1}+\tfrac{2/\varepsilon_s-1}{A_2}\big]$.

Why this works: Each grey surface contributes a resistance $(1-\varepsilon)/(\varepsilon A)$ in the radiation network, and a floating shield adds its own inner-face and outer-face resistances in series between the line and the chamber because the shield reaches its own equilibrium temperature (heat in from the line equals heat out to the chamber). Since the shield's two space-resistance and two surface-resistance terms combine algebraically to $(2/\varepsilon_s-1)/A_2$, the shield need not have low emissivity to help — the larger diameter also lowers each individual resistance term, but the dominant effect is simply adding resistance terms in series.

Common pitfall: Using the two-parallel-plate shield formula $1/(1/\varepsilon_1+1/\varepsilon_2-1)$ here — that reduction factor assumes equal areas and is NOT valid for concentric cylinders of different diameter; the area ratio $A_1/A_2$ must appear explicitly, as it does in the resistance-network form used above.

Source: R.F. Barron, Cryogenic Heat Transfer, Ch. 7 (radiation heat transfer, cylindrical radiation shields); Çengel & Ghajar, Heat and Mass Transfer, Ch. 13 (radiation shields, network method).

Question 3 of 4: Radiation Shielding of a Liquid-Nitrogen Line

Problem 3: Radiation Shielding of a Liquid-Nitrogen Line (25 points)