Question 3 of 4: Radiation Shielding of a Liquid-Nitrogen Line
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 04-Agric-A3 Heat Engineering, National Exams
May 2013 — a three-hour open-book examination; any non-communicating
calculator is permitted. The cover page states that four (4) questions constitute a
complete exam paper and that only the first four as they appear in the answer book are
marked, that each question is of equal value, and that all questions require calculation.
All four printed problems are worked here.
Reference texts. Y.A. Çengel & A.J. Ghajar, Heat and Mass
Transfer: Fundamentals and Applications, 5th ed. (conduction, flat-plate solar
collectors); J.P. Holman, Heat Transfer, 10th ed. (natural-convection
correlations and boundary-layer thickness); R.F. Barron, Cryogenic Heat Transfer,
2nd ed. (radiation shields for cryogenic lines).
Problem 3: Radiation Shielding of a Liquid-Nitrogen Line (25 points)
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 Tc.
Check: the shield tube's emissivity is not given in the source. It is
assumed equal to the line's, εs = 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 $T_1^4=80^4=4.096\times10^{7}\ \text{K}^4$,
$$\frac{Q}{\ell} = 0.2\times(5.67\times10^{-8})\times(1.995\times10^{-2})\times
\left(2.798\times10^{9}-4.096\times10^{7}\right)$$
$$\frac{Q}{\ell} = \boxed{0.624\ \text{W/m}}$$
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 (the shield's own
surface-to-surface terms combine because 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 very 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.
