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DeafSmith 08-20-2010 03:01 AM

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Here's what a little theoretical investigation revealed - it's all about flow volume through the chiller (gallons per minute) - at least under the assumptions I made, which, of course are a gross simplification of reality.

The heat flow into the chiller is a quantity which declines as we proceed along the chiller from entry to exit. I'm pretty sure this is an exponential decrease with length (I'm pretty sure I'm right because using the exponential equations to figure the average heat loss yields the same results as using the well known "log mean temperature difference" equation (LMTD). So the equation for heat loss per unit length of chiller is of the form: h=h0*exp(-ax) where x is the distance from the chiller input to the point where the heat loss is calculated. h0 is the heat loss per unit length at the chiller entry; h is proportional to the temp. difference between wort and chiller water at the point in the chiller where h is calculated.

Attachment 16943

On the above graph, the x axis is distance along the chiller from the entry - 10 units could be, for example 50 feet. The y-axis is the heat loss per unit length (arbitrary units) at each point along the chiller and is directly proportional to the temperature difference at that point.

The black curve extending to 10 on the x-axis represents a chiller 10 units long with a given flow (some number of gallons per minute) through it. The total heat flow into the chiller is the area under the curve. Curve equation is 100*exp(-x/5), so the area under it is the integral from 0 to 10 of that function which is (-100/.2)*exp(-x/5) evaluated from 0 to 10 which is 500*(1-exp(-10/5)) = 432.332 arbitrary units.

If the chiller tube is split into two parallel pieces of half the length, then the cross sectional area of both tubes together is double that of the single tube. If the tubes are filled with water in all cases, and if the total water flow into the chiller is the same with the parallel tubes as for the single long tube, then the flow rate through the parallel tubes must be only half as fast as the rate through the single tube, causing the water in each tube to heat up more rapidly than in the faster flowing single tube; therefore the temperature difference and heat flow drops off more rapidly with the shorter, slower flowing tubes. So we can model the heat transfer for one of the short tubes by h=h0*exp(-2ax) (the magenta curve) and the sum of both short tubes is 2*ho*exp(-2ax), or inserting arbitrary units = 200*exp(-x/2.5) (the red curve). The total heat being transferred at any given time is given by the area under the red curve from 0 to 5. This area is THE SAME as the area under the black curve from 0 to 10 ! Mathematically, the curve equation is 200*exp(-x/2.5), so the area under it is the integral from 0 to 5 of that function which is (-200/.4)*exp(-x/2.5) evaluated from 0 to 5 which is 500*(1-exp(-10/5)) = 432.332 arbitrary units.

In reality, you can probably get more total flow with two parallel tubes than with a single long tube because the resistance to flow will be less. I have made the assumption that the tubes are always completely filled, and of course I have no way to predict the details of the fluid flow (laminar, turbulent) and various other complicating factors. It's still interesting, nonetheless that the heat flow into the chiller under these assumptions seems to just depend on the total rate of chiller water flow. Even though the parallel tubes start out by absorbing a lot more heat, the water in them heats up very quickly because the water flow through them is slow, thus lowering the temperature difference quickly.

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