Tips on getting the most from your wort chiller. - Page 3 - Home Brew Forums

 Home Brew Forums > Tips on getting the most from your wort chiller.

08-18-2012, 01:33 PM   #21
Pilgarlic

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The science has long ago been settled. It supports the intuitive, that a faster flow rate removes heat faster by creating a greater differential. Details (with all appropriate formulas, if you prefer) in July/August issue of BYO Magazine, Science column by Chris Bible.

08-18-2012, 02:55 PM   #22
macleod319
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Before I waste my time then, is there any way to show those of us w/out a hard copy? Thanks - Jeremey
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08-18-2012, 03:03 PM   #23
passedpawn
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To repeat all the sane people here...
If you are recirculating back to the kettle:
• Maximize the cooling water flow. Period.
• Maximize the hot wort flow
If you are trying to do a single pass chill, where the wort comes out at pitching temp:
• Maximize the cooling water flow. Period.
• Choke the hot wort flow until you achieve the output temp you want.
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08-18-2012, 03:08 PM   #24
passedpawn
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Quote:
 Originally Posted by macleod319 Q = mc(T2-T1) Q=overall heat transfer in BTU/time m= the mass flow rate of your cooling medium in vol/time c= the specific heat transfer capability of your chilling surface (i.e. the copper surface of the wort chiller in this case) T2= wort temp (higher temp substance) T1= water temp (lower temp substance) The only way to increase the heat transfer capability of your cooler (without structural modification) is to either: INCREASE the mass flow rate of your cooling medium (not decrease) by turning the water flow up or INCREASE the difference in the two temperatures, by using a pre-chiller or some other way of getting colder water inside the chiller.
Perfect.
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08-18-2012, 10:08 PM   #25
JeffersonJ

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Quote:
 Originally Posted by macleod319 *Crazy scientific discussion ahead* Alright, a little science from a power plant mechanic. Heat transfer in a condenser/cooler/wort chiller/etc. is boiled down to the basic following equation: Q = mc(T2-T1) Q=overall heat transfer in BTU/time m= the mass flow rate of your cooling medium in vol/time c= the specific heat transfer capability of your chilling surface (i.e. the copper surface of the wort chiller in this case) T2= wort temp (higher temp substance) T1= water temp (lower temp substance) The only way to increase the heat transfer capability of your cooler (without structural modification) is to either: INCREASE the mass flow rate of your cooling medium (not decrease) by turning the water flow up or INCREASE the difference in the two temperatures, by using a pre-chiller or some other way of getting colder water inside the chiller. This can be proven with further explanation if required. Anyone that disputes this fact is temping the laws of thermodynamics, and I would like to hear an arguement against. Discuss.
You're absolutely right that increasing the flow rate will increase heat transfer. However, your use of that particular equation is absolutely wrong. This equation can be used to measure the heat transfer of the water only.

The proper use of the equation's variable is this:

Q = heat transfer rate
m = mass flow rate of water
c = specific heat capacity of the water
T2 = the leaving temperature of the water
T1 = the entering temperature of the water

You can calculate your heat transfer rate if you know all of your variables. In an immersion chiller scenario, increasing mass flow rate will tend to decrease T2. Decreasing mass flow rate will tend to increase T2. However, knowing that still doesn't tell you what conditions will maximize your heat transfer rate. The best way to determine your ideal conditions is by experiment.

Obviously, your calculated heat transfer rate of the water (Q from above) will also be the amount of heat removed from the wort by the water. Again though, the only way to properly calculate the conditions at which the heat transfer rate will be highest is by experiment (or by modeling with FEA/CFD software).

The being said, I intrinsically know that your heat transfer will be at its maximum when the flow rate is at its maximum. For a moment ignoring any knowledge of the water in the chiller, the way in which you'd calculate heat transfer from the chiller to the wort is very complicated (and nearly impossible by hand). But, I can tell you that the heat transfer rate will largely be proportional to two things:

• The surface area of the chiller
• The temperature difference between the chiller and the wort

Since the temperature of the chiller is a gradient throughout the length of the chiller, any method of calculation involves calculus (and a lot of assumptions and Reynolds numbers and Prandtl numbers and blah blah blah). But think of it this way: you want a maximum surface area of the chiller to be as cool as possible. A lower flow rate will cause the latter half of the chiller to become warmer, thereby decreasing the heat transfer from the chiller to the wort. A higher flow rate will keep a larger portion of the chiller at a lower temperature, thereby increasing heat transfer from the chiller to the wort.

Also, efficiency in this context is very wishy-washy term. If your goal of "efficiency" is to decrease water consumption, then there will be an ideal flow rate that balances heat transfer and flow rate. On the other hand, if your goal of "efficiency" is to decrease cooling time, then you want the flow rate to be the maximum possible.

(Edited for clarity)

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08-18-2012, 11:11 PM   #26
JeffersonJ

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Also, sorry if that came across as condescending in any way.

When I was in grad school, I would often substitute for a professor and teach a Heat Transfer course for mechanical engineering students (it was my specialty, hence my passion for it). I can tell you that the equation above (there's a name for it but it escapes me) is very often misapplied even by 3rd-and-4th year engineering students. It's extremely useful but only applies to very specific applications.

Since we're so used to it, the concept of an immersion chiller in hot wort seems really simple. But trying to characterize with equations is extremely complicated. First, everything is changing...

• The temperature of the water increases as it goes through the length of the chiller
• The temperature of the wort changes by height due to stratification
• Of course the temperature of both the wort and water change over time as you remove heat

On top of that, we're assuming that the liquid is stagnant when in reality most of either stir or full-on whirlpool! On top of that, the brewpot is transferring heat to/from the ambient air. And so on and so on.

08-19-2012, 12:06 AM   #27
amandabab
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if high flow always cools better why do we need flow restrictors in the thermostat housing on our circle track car and drag boat? wide open flow they over heat, 50% flow restriction and they don't. the water absorbing heat isn't a myth

08-19-2012, 12:34 AM   #28
JeffersonJ

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Quote:
 Originally Posted by amandabab if high flow always cools better why do we need flow restrictors in the thermostat housing on our circle track car and drag boat? wide open flow they over heat, 50% flow restriction and they don't. the water absorbing heat isn't a myth
Because the system and end goal is different in those cases (and more complicated). You've essentially got two different heat exchangers - your radiator and your engine. The engine is actively generating heat that must be removed and the cylinder walls need to maintain a certain temperature. The radiator in your car uses air to transfer heat from the recirculating water. The water has to get down to a certain temperature in order to effectively cool the engine. If you run it through your radiator too quickly you'll end up with a leaving water temperature that is too high to do that. There are two big differences here: You're recirculating the cooling water in your car and your engine is actively generating heat.

A radiator needs a certain lag time of the water to transfer enough heat. It's much more comparable to a plate or counterflow chiller than an immersion chiller.

Beyond all that, the main issue with overheating is not the heat transfer rate but rather the fact that the recirculating water is reaching its boiling point!

08-19-2012, 12:48 AM   #29
amandabab
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Quote:
 Originally Posted by JeffersonJ Because the system and end goal is different in those cases (and more complicated). You've essentially got two different heat exchangers - your radiator and your engine. The engine is actively generating heat that must be removed and the cylinder walls need to maintain a certain temperature. The radiator in your car uses air to reduce the transfer heat from the recirculating water. The water has to get down to a certain temperature in order to effectively cool the engine. If you run it through your radiator too quickly you'll end up with a leaving water temperature that is too high to do that. There are two big differences here: You're recirculating the cooling water in your car and your engine is actively generating heat. A radiator needs a certain lag time of the water to transfer enough heat. It's much more comparable to a plate chiller than an immersion chiller. Beyond all that, the main issue with overheating is not the heat transfer rate but rather the fact that the recirculating water is reaching its boiling point!
I'll give you the radiator in the circle track car. the boat is 100% draw and dump.

08-19-2012, 12:56 AM   #30
JeffersonJ

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Quote:
 Originally Posted by amandabab I'll give you the radiator in the circle track car. the boat is 100% draw and dump.
I'm not entirely sure how most boat engines reject heat. Do they actually run seawater (or freshwater) through/around the engine? Or is there a recirculating loop and then seawater cools that loop with a heat exchanger?

If I had to guess I would say it's the latter, but I could be wrong. In that case it's essentially the same situation as a car radiator.