### Reshaping heat transfer

The last step in production from my home bioreactor is to heat the mixture to a specified internal temperature, about 95-98C. The first few times I tried, I followed the instructions exactly and it worked great.

Then I decided that the final shape of the product when following the instructions was not as convenient to me as I would like. Using a different shaped container was in order. However, this different container was not only a different shape, it was not conducive to pre-heating and had nowhere near the capacity for storing heat and holding its temperature as the original, inconveniently-shaped container did.

I tested it out anyway, and while it did produce a final product of about the shape I wanted, the characteristics of the material in the centre had changed. It was usable, but not as easily so. I eventually figured out that it hadn't heated up right through as it should have. I tried heating it for a slightly longer period, but that didn't seem to make much of a difference.

Trial and error will take far too long. Clearly, this calls for some math.

So the basic heat transfer equation is:

$Q = {kA(T_1 - T_2)t\over d}$

Where d is the distance the heat has to travel, A is the surface area the heat is in contact with, the two Ts are the temperatures inside and outside, t is the time, Q is the heat that travels, and k is the heat transfer coefficient, which is not only unknown but quite likely changes as the product heats up and sets in its final shape. Well, since $$T_2$$ is the centre of the product, it will change with time as well, from room temperature to its final temperature.

However, to simplify the math, I will deem both k and $$T_2$$ to be constant. That should get my desired heating time close enough for my purposes; this isn't an ISO9001 process, so it doesn't have to be exactly the same every time.

So, the temperature $$T_1$$, the hot side, is 245C, while I will declare $$T_2$$, the cool side (or the centre) to be 98C, or roughly the final temperature I want.

I'm also going to ignore the heat transfer of the container. The inconveniently shaped container has thick heavy walls and takes a long time to heat up, however it is pre-heated to full temperature before the product is placed inside. The more convenient container has thin walls and heats up fairly quickly, but starts at room temperature with the product already inside it. The thin-walled container will slightly slow the heat transfer, but given how many assumptions and approximations I'm making, I'm going to ignore that for now. I'm also not going to worry about k, that unknown (non-?)constant, because I'm comparing two different conditions with the same constant and I can cancel it out.

Then, because I'm making so many approximations, I'm going to be rounding these numbers off quite a lot. There's no sense in reporting the 8 digits the calculator gives me when I have, at best, 2 significant figures due to all my approximations.

I can easily get A and d, as they're characteristic of the shapes. For the inconvenient shape, A is the area of the circular container, 25.4cm in diameter, or 507cm2, while d is about 2.5cm. The heating time is 30min. As the inconvenient shape is a wide, low circle, I am ignoring the heat transfer in from the sides, a relatively small area that is quite far from the centre of the product, as well as only considering the heat transfer in from one of the two large circular sides because I am assuming it's symmetrical.

${Q\over k} = {507 \times (245 - 98) \times 30 \over 2.5} = 912,000$

The more convenient shape is a rectangular container. I can no longer ignore heat transfer in from the sides, because two of the sides are roughly the same distance from the centre and the same size as the top and bottom are. I can ignore the two ends, however, as they are, as before, relatively smaller and farther away from the centre. The cross-section of this shape is roughly a square, about 7.5cm on a side, and the container is about 22.5cm long. Since the above calculation used only one side of the product, or half the actual surface area (heat was entering from all directions) I will here again use half the actual surface area, or two of the four sides, for an area of about 350cm2.

Then, using the heat required to get the temperature up, I solve for t:

$t = {{Q\over k} d \over A ( T_1 - T_2 )} = {912,000 \times 5 \over 350 \times (245 - 98)} = 87$

Okay, the math says I need almost triple the heating time to reach the same temperature. Now to test this.

I actually chickened out on the first trial and only doubled the heating time; the product had only reached 95C at the centre, and the properties of the centre were still not what they should have been.

On the second trial I trusted the math, and heated the product for 85 minutes.

It worked. 98C, and the properties of the centre were about what they should have been. To my surprise, the edges, exposed to 245C for about triple the normal time, were not notably damaged by the heat.