Lower surface area would have to be insulated against the hot funnel gasses, and I'll show you how:
We will treat each pair of funnels independently from the other pair.
The shape with the lowest perimeter for a given area is the circle - and you have two of them in each pair. Each one of yours right now has an outside diameter of 10 feet, with an 8 foot gap between them. This now gives us a rectangular box of 10 feet by 28 feet. Storing something between the funnels is a bit out of the question - the space is small (eight feet by ten feet), hard to use, and even if fully insulated will be rather hot If we assume say 6 inches of insulation around the inside each funnel, we have a functional exhaust area of about 63.6 square feet per funnel, or 127 square feet for both combined.
Now, let us evaluate an elliptical trunked funnel. If we assume that we sill need the same amount of exhaust air flow, we need 127.2 square feet of exhaust. The formula for the area of a circle is πr², and the area for an ellipse is παβ where α and β are one half the major and minor axises of the ellipse. If we hold α to r (we're not going to make the total exhaust duct any wider) we find that to make get the same area for the single ellipse and the two circles, we need to make β equal 2r. This gives us an outside size of 10 feet by 20 feet - or 8 feet shorter than the original design.
Now, let us look at the perimeter of the shapes - the circumference of a circle is 2πr - because these are back of the envelope level calcs, we can assume that r is half way between the inside of the funnel and the outside of the funnel - or about 4.75 feet. This gives us a length of 29.85 feet around each funnel. We have two funnels so we need to double this for the purposes of comparison with the ellipse - or a total of 59.7 feet around both funnels.
The formula for the circumference of an ellipse is very complicated and calculus dependent and here, I'll use the approximation of 2π√((½)(α²+β²)). Now remember, we held α to the earlier r. β is no longer held to 2r because we added 0.25 feet to the original r to get the new r' for the circumference calculation (to account for the depth of the insulation). It follows thus that the new β equals 2r'+.025, or in this case 9.75 feet. Plugging these values into the given equation we find that the circumference of the ellipse is about 48.2 feet, or a difference of 11.5 feet from the two circular funnel set up. now, if we assume a deck height of 9 feet (from your drawing) and 5 decks to insulate plus another 15 feet in the funnels above the superstructure, we discover that we've saved 690 square feet of insulation.
That doesn't sound like a lot but if you consider that the density of asbestos is 15lbm per cubic foot, and we've got oh, 250+ cubic feet of insulation, we've saved over three quarters of a ton per funnel pair (or over a ton and a half total). We've also improved deck layout and flow by decreasing the deck footprint of the funnels from 260 to 160 square feet - a major improvement.
_________________ 𝐌𝐀𝐓𝐇𝐍𝐄𝐓- 𝑻𝒐 𝑪𝒐𝒈𝒊𝒕𝒂𝒕𝒆 𝒂𝒏𝒅 𝒕𝒐 𝑺𝒐𝒍𝒗𝒆
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