Question 4 of 4: Radiation Exchange Between Two Black Disks
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 04-Agric-A3 Heat Engineering, National Exams
December 2017 — 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. (specific heat, calorimetry, conduction);
J.P. Holman, Heat Transfer, 10th ed. (radiation shape factors, radiation networks with
reradiating surfaces); R.F. Barron, Cryogenic Heat Transfer, 2nd ed. (radiation shields
for cryogenic lines); F.P. Incropera & D.P. DeWitt, Fundamentals of Heat and Mass
Transfer, 7th ed. (coaxial-disk view-factor relation).
Problem 4: Radiation Exchange Between Two Black Disks (25 points)
Find. The radiant heat flow Q between the disks (1) with no other
surfaces present, and (2) with the disks connected by an adiabatic (reradiating) cylindrical
side wall.
Approach. Case 1 is a direct two-black-surface exchange, $Q=A_1F_{12}
\sigma(T_1^4-T_2^4)$, with $F_{12}$ the coaxial-parallel-disk view factor (the source's
Figure 1 chart, D/L = 0.5). Case 2 turns the pair into a three-surface enclosure (disk 1,
disk 2, and the now-present cylindrical wall, which is adiabatic/reradiating); because a flat
disk cannot see itself, $F_{1R}=1-F_{12}$, and with both disks black the radiation network
reduces to a direct path $A_1F_{12}$ in parallel with a two-resistor path through the
reradiating wall.
Case 1: the two black disks exchange radiation directly with no
enclosing wall (any radiation missing the far disk is lost). Case 2: an adiabatic
cylindrical wall now closes the enclosure and reradiates energy back and forth between the
disks, raising the effective coupling well above the direct view factor.
Coaxial-disk view factor. With equal radii $r=1$ ft and separation
$L=4$ ft, $R=r/L=0.25$. Using the analytic coaxial-parallel-disk relation (equivalent to
reading D/L = 0.5 off the source's Figure 1, curve 1 "Disks"),
$$S = 1+\frac{1+R^2}{R^2} = 1+\frac{1.0625}{0.0625}=18,\qquad
F_{12}=\tfrac12\Big[S-\sqrt{S^2-4}\Big]=\boxed{0.0557}$$
Case 1 — no other surfaces. Each disk's area is
$A_1=\pi r^2=3.1416\ \text{ft}^2$; with both disks black, radiation leaving disk 1 that
misses disk 2 is simply lost (no wall to return it), so
$$Q_{12}=A_1F_{12}\,\sigma\left(T_1^4-T_2^4\right)
=3.1416(0.0557)(0.1714\times10^{-8})\left(2000^4-1000^4\right)$$
$$Q_{12}=\boxed{4{,}500\ \text{Btu/hr}}$$
Case 2 — adiabatic reradiating side wall. A flat disk cannot see
itself, so $F_{1R}=1-F_{12}=0.9443=F_{2R}$ by symmetry ($A_1=A_2$). With both disks black
(zero surface resistance) the radiation network between disk 1 and disk 2 is the direct path
$A_1F_{12}$ in parallel with the two-resistor path through the floating reradiating wall,
$$\frac{1}{R_{via}}=\left[\frac{1}{A_1F_{1R}}+\frac{1}{A_2F_{2R}}\right]^{-1}
=\frac{A_1F_{1R}}{2}=1.4834\ \text{ft}^2$$
$$Q_{12}=\sigma\left(T_1^4-T_2^4\right)\Big[A_1F_{12}+\tfrac{1}{R_{via}}\Big]
=(0.1714\times10^{-8})(1.5\times10^{13})\big[0.1751+1.4834\big]$$
$$Q_{12}=\boxed{42{,}600\ \text{Btu/hr}}$$
Closing the enclosure with the reradiating wall increases the exchange by almost a factor of
ten: nearly all the radiation that used to escape past the far disk now reflects/reradiates
off the adiabatic wall until it eventually reaches the other disk.
