Radiation Exchange Between a Heater and a Conical Shield: 22-Agric-A3 Heat Engineering, May 2018, Q4

Problem 4: Radiation Exchange Between a Heater and a Conical Shield (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. Heater: flat black disk, diameter D_h = 10 cm (r₁ = 0.05 m), temperature T_h = 1200°C = 1473 K, sitting flush with the base of the shield. Shield: black conical frustum (open top), base diameter 10 cm, top-opening diameter D_top = 20 cm (r₂ = 0.10 m), height L = 0.20 m, temperature T_s = 100°C = 373 K.

Find. The net radiant heat transfer rate from the heater to the shield, Q_h→s.

Approach. The heater, the shield's lateral (frustum) surface and the open top together form a 3-surface enclosure, with the open top standing in as a black, non-reflecting fictitious surface facing the surroundings. Because the heater is flat it cannot see itself ($F_{hh}=0$), so its view factor to the shield is $F_{h\to s}=1-F_{h\to top}$, where $F_{h\to top}$ is the coaxial-parallel-disk view factor between the heater and an imaginary flat disk covering the top opening (source's Configuration 3). With both surfaces black, the net exchange follows directly from the two-black-surface relation.

Cross-section (schematic): a straight-up ray from the heater exits through the open top without touching the shield; a shallower ray strikes the sloped shield wall.

View factor from the heater to the (fictitious) top-opening disk. Using the source's Configuration 3 (coaxial parallel disks) with r₁ = 0.05 m, r₂ = 0.10 m, h = 0.20 m: $R_1=r_1/h=0.25$, $R_2=r_2/h=0.5$, $$X = 1+\frac{1+R_2^2}{R_1^2} = 1+\frac{1.25}{0.0625}=21$$ $$F_{h\to top} = \frac12\left[X-\sqrt{X^2-4(R_2/R_1)^2}\right] = \frac12\left[21-\sqrt{441-16}\right] = \boxed{0.1922}$$
View factor from the heater to the shield. The heater is flat, so $F_{hh}=0$ and, by the summation rule over the enclosure (heater, shield, top opening), $$F_{h\to s} = 1 - F_{h\to top} = 1-0.1922 = \boxed{0.8078}$$
Net radiant heat transfer, heater to shield. With both surfaces black, the direct two-surface exchange relation applies without any reflection correction; the heater's area is $A_h=\pi r_1^2 = \pi(0.05)^2 = 7.854\times10^{-3}\ \text{m}^2$, so $$Q_{h\to s} = A_h F_{h\to s}\,\sigma\left(T_h^4-T_s^4\right)$$ $$Q_{h\to s} = (7.854\times10^{-3})(0.8078)(5.67\times10^{-8}) \left(1473^4-373^4\right)$$ $$Q_{h\to s} = \boxed{1{,}690\ \text{W} \approx 1.69\ \text{kW}}$$

Quantity	Result
View factor, heater to top opening, F_h→top	0.1922
View factor, heater to shield, F_h→s	0.8078
Net heat transfer, heater to shield, Q_h→s	1,690 W (1.69 kW)

Concept & Theory

Topic: View factors and net radiant exchange between two black surfaces in an enclosure that includes an opening to the surroundings.

Key relations: Coaxial parallel disks, $F_{1\to2}=\tfrac12\big[X-\sqrt{X^2-4(r_2/r_1)^2}\big]$, $X=1+(1+R_2^2)/R_1^2$, $R_i=r_i/h$; summation rule $\sum_j F_{i\to j}=1$; two-black-surface net exchange $Q_{1\to2}=A_1F_{1\to2}\,\sigma(T_1^4-T_2^4)$.

Why this works: Treating the open top as an imaginary flat black disk lets the summation rule close the enclosure — any radiation from the heater that is not intercepted by the shield must, by conservation of the radiation leaving the heater, exit through that opening, so $F_{h\to s}$ is obtained by subtraction rather than by a direct frustum-surface integral. Because both real surfaces (heater, shield) are black, no reflected radiation needs to be tracked, so the two-surface formula applies to the heater/shield pair even though a third "surface" (the opening) is also present in the enclosure.

Common pitfall: Trying to integrate the view factor to the sloped frustum wall directly (no closed-form chart covers this shape); computing it as 1 minus the opening's view factor sidesteps that entirely. A second pitfall is using the heater's own diameter for $r_2$ in the disk formula instead of the top opening's diameter — the fictitious disk must match the actual opening, not the heater.

Source: J.P. Holman, Heat Transfer, 10th ed., Ch. 8 (radiation shape factors, view-factor algebra); source exam sheet's Configuration 3 (coaxial parallel disks).

Question 4 of 4: Radiation Exchange Between a Heater and a Conical Shield

Problem 4: Radiation Exchange Between a Heater and a Conical Shield (25 points)