

08192010, 11:02 PM

#21

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Quote:
Originally Posted by prosper
the simple answer is that it's all about surface area. More = faster cooling, provided you're able to flow enough coolant water over it to absorb the heat.
There are other factors, but mainly it comes down to surface area and coolant capacity.

I disagree that it is "all about surface area" but acknowledge that surface area is a factor. It's just that the price of copper and ability to work with larger copper tubing limits just what the average person can do to increase surface area. But the average person can work to ensure the greatest temp difference between the wort and chiller water. A Google search for heat transfer rates shows numerous formulas based on the particular application, and a common exponent in these formulas is the temperature differential. The general experience of homebrewers is that the wort temp drop from boil down to 100 or so occurs relatively quickly as compared to the drop from 100 down to 70. This is simply because temp differential between 212 degree wort and 70 degree tapwater is greater than that between 100 degree wort and 70 degree tapwater.
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08202010, 12:36 AM

#22

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Wow, I just had a flashback to my days as a chemical engineering student. I now know why I got into sales and marketing instead fo staying on the technical side of the chemical business.
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08202010, 12:56 AM

#23

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I didn't read all the way through, so if this is redundant, sorry.
Heat transfer is greater when the temp difference is high. So, to make a simple point, 212º water in a pot, with a 70º coil immersed, regardless of length, there is a big difference in temp. As the cooler water travels through the coils, it heats up, while the surrounding water cools. So, as the water temp drops, and the coil heats up, heat transfer is slowed.
Which brings me to my preferred method of cooling: a Counter Flow Chiller (CFC). I've been getting temps going into the fermenter right at ground water temp with my CFC. My well water is right around 60º right now. I have to throttle back my coolant flow just to keep temps at pitching temp. 60º water entering at the out flow of the wort, traveling the 25' of hose, and out the other end, does wonders.
YMMV, but I'd suggest building a CFC. Bobby_M has a good one on here, search for it.
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08202010, 01:59 AM

#24

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I can totally appreciate the idea of a CFC to maintain the highest temp differential throughout the flow process. I just figured a dual coil immersion chiller was easier for me and still brought 5 gallons down in under 14 minutes.
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08202010, 02:01 AM

#25

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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.
On the above graph, the x axis is distance along the chiller from the entry  10 units could be, for example 50 feet. The yaxis 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 xaxis 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*(1exp(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*(1exp(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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Last edited by DeafSmith; 08202010 at 04:13 AM.



06242011, 03:49 PM

#26

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That's a pretty sweet explanation, Deaf.
I have to admit it's not intuitive, as i would guess that the 2 parallel chillers would be faster. The "intuitiveness" of this may relate to the likely faster flow you'd get in reality, as you point out.
I'll have to think about your equations a bit more (and your assumption of exponential decline), but at first glance this seems reasonable in the theoretical (though perhaps not the practical) sense.
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06252011, 03:55 AM

#27

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From the practical side I just went from 20 to 40 feet of coil and on a 6 gallon boil I can go from 212 to 130F in under 3 minutes and to 70F is 7=8 minutes total on the output temp is still cooler than I would like but I don't want to add any more weight or cost. Input temp is between 38 and 40 F because I have a small IC in an ice bath.
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06252011, 03:03 PM

#28

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You can approach this by modeling the chiller as a pair of channels in contact with one another. A length, dx, of the contact area is assumed to have an area per unit length of A so that the area in contact is A*dx, and a thermal conductivity G so that the heat flow between the two channels in time dt is
dq = (Tw(x)  Tc(x))A*G*dx*dt where Tw(x) is the temperature of the wort and Tc(x) is the temperature of the coolant as a function of distance, x, along the chiller. If the flow rate of coolant is Fc then in time dt an amount of coolant Fc*dt with mass pc*Fc*dt will flow past x where pc is the density and be subject to temperature rise dTx = dq/(pc*Fc*Cc*dt) = (Tw(x)  Tc(x))A*G*dx*dt/(p*Fc*Cw*dt) where Cc is the specific heat of the coolant. Thus dTc/dx = (Tw(x)  Tc(x))A*G/(pc*Fc*Cc) . Similar reasoning gives another equation for the rate of temperature loss in the wort as you move along the chiller
dTw/dx = (Tw(x)  Tc(x))A*G*/(pw*Fw*Cw)
Subtracting these two gives the equation
dTw/dx  dTc/dx = [(Tw(x)  Tc(x))]*(A*G)*[1/(pw*Fw*Cw)  1/(pc*Fc*Cc)]
Thus d(Tw(x)  Tx(x))/dx = a*[Tw(x)  Tc(x)]
and it's clear that the solution will be exponential in form i.e. the temperature difference along the chiller is an exponentially decreasing function of x (assuming that the origin is where the beer enters and the coolant comes out). I'll leave the rest to the curious as it gets into some rather tiresome algebra.
This should be enough to illustrate that:
1. Enhanced cooling depends on the length of the chiller but the returns are exponentially diminishing with length.
2. The faster the coolant flow the more efficient (with 100% efficiency being defined as the wort leaving the chiller at the same temperature at which the coolant enters) the chiller. More broadly, the larger flow rate of coolant thermal mass, the more efficient the cooler.
3. The slower the wort flow rate, the more efficient the cooling will be. Again, the broader statement is the less the thermal mass passing through the wort channel per unit time, the better the efficiency. Thus a wort of high specific gravity will not be cooled as effectively as one of lower specific gravity.
4. The thermal conductivity (A*G) between the channels is the major driver in performance. Increasing G (as by putting fins on the wort tube in a coaxial design) always works. Increasing A may not be so effective because, where the flows are laminar, the wort in the center of the channel is insulated from the walls where the heat exchange takes place. One way around this is to be sure that the flows are rapid enough that they are turbulent. Fins not only help heat transfer but they can break up laminar flow and benefit in that way too.
It's probably clear, given particular length, A, and G that the faster you run coolant, the colder the wort will be but that you will consume more coolant. It's also probably clear that the slower you run the wort the cooler it will be but that as the process will take longer more coolant will be required. One can trade time, efficiency and coolant consumption but to do so requires thorough knowledge of the chiller. One can make measurements on a chiller to determine its A and G (or at least A*G) and thus model its cooling capabilities but one must be careful in doing this as the effective G changes as the flow transitions from laminar to turbulent.
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06252011, 11:12 PM

#29

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Your post is for a counterflow chiller, right ajdelange?
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06252011, 11:52 PM

#30

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Without the benefit of all the math and formulas I add my observation. I use 50' of ½" copper tube Immersion chiller. I had the opportunity to brew with a guy that uses a 25' IC. This was a situation where we had to share a hose for chilling. The hose had a "Y" so that we could share a single hose. The other guy began chilling about 15 minutes before I finished my boil. When I started chilling we were using the exact same ground water. We had to finish at the same time because other people needed to use the hose to start their chill. Even though I started 15 minutes later I was chilled to a lower temperature at the end.
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