X-Message-Number: 2005 Date: Mon, 22 Mar 93 00:41:24 CST From: Brian Wowk <> Subject: CRYONICS Reply to Ettinger II Since my last posting I have been increasingly bothered by the non-linear temperature gradient issue raised by Robert Ettinger. It seems to me that this issue is really too complex to just wave away by saying I'm going to use room-temp K values throughout the foam. To get a better handle on this problem, I wrote down the differential equation governing heat flow into a spherical cold room with thick walls. The equation is J*(R1/r)^2 = K * sqrt(T/T2) * dT/dr (1) where R1 is the inner radius of the room, R2 is the outer radius, T1 is the temperature inside, and T2 is the temperature outside, and J is the heat flow per unit area of the inside wall. R2 and T1 do not appear explicitly, but they must be included in the boundary conditions that determine the final solution to the equation. The term that J is multiplied by is a geometrical factor that accounts for the reduction in heat flow per unit area as you move away from the interior of the sphere. K is the thermal conductivity at temperature T2 (room temperature), and the factor beside it accounts for the variation of K with the square root of absolute temperature. As you can see, this equation is complete, and accounts for all contributing factors to the non-linear gradient that were raised by Professor Ettinger. It turns out that this equation can be solved by straight- foward integration, yielding the solution: J=2/3*K/sqrt(T2)*(T2^1.5 - T1^1.5)/(R2-R1) * R2/R1 (2) Interested parties can verify that in the limit of small differences between R2 and R1 (thin insulation), and small differences between T2 and T1 (constant K value), this formula reduces to the simple J=K*(T2-T1)/(R2-R1) (3) we expect. If we assume that our rectangular cold room is approximately equivalent to a spherical cold room with inside radius given by the average of the length, width, and height, then we can proceed to use equation (2) to solve for the heat flow. Replace line 10 in my BASIC program with 10 DEF FNB= IK*47.6*A/T*( ((L+W+H)/6 +T)/((L+W+H)/6) ) and you will be all set. Well, it turns out that everything Bob wrote was right, including his last sentence. By not considering the variable temperature gradient, I had underestimated the heat flow (and LN2 boiloff) by about 25%. The new formula in the computer program now predicts an optimum foam thickness of 1.3 meters. The new 1 meter and 1.3 meter comparisons are shown below. INSULATION THICKNESS: 1.00 INSULATION CAPITAL COST: 34000.00 INSULATION AMORTIZATION: 3400.00 LN2 LITERS PER DAY: 160.70 ANNUAL LN2 COST: 17597.19 ANNUAL FLOOR SPACE CHARGE: 2450.00 ANNUAL TOTAL OPERATING COST: 23447.19 INSULATION THICKNESS: 1.30 INSULATION CAPITAL COST: 49691.20 INSULATION AMORTIZATION: 4969.12 LN2 LITERS PER DAY: 135.33 ANNUAL LN2 COST: 14818.69 ANNUAL FLOOR SPACE CHARGE: 2888.00 ANNUAL TOTAL OPERATING COST: 22675.81 As you can see, there is still only a small difference between 1 meter and 1.3 meters of insulation, so I still suggest that 1 meter be used. By the way, the program vindicates my earliest analytic formulas by predicting that that 2 meters is optimum for a "large" room. The problem is the room we are considering (5m x 5m x 3m) is not large compared to the insulation thickness, so this more sophisticated treatment must be used. A way to nail down these figures even more solidly would be to write a finite element heat flow simulation for the exact room geometry in question. However someone would have to really twist my arm to do that one. --- Brian Wowk Rate This Message: http://www.cryonet.org/cgi-bin/rate.cgi?msg=2005