Convective Heat Transfer Beer Chiller Problem

[killerhertz]

Hello,

I'm an EE with no thermo background, but have done some reading. I need to build an immersion chiller (

http://www.homebrewtalk.com/wiki/index.php/Immersion_Chiller#Immersion_chillers)

to cool my wort for homebrewing. The chiller sits in the boiling wort and the intake connects to a sink. The faster the wort is cooled to ~70F, the more clear comes out, and the less time I have to spend waiting

I plan on doing 5-10 gallon batches, which I've been told requires at least 50' of copper tubing, but I'd like to create a model for this problem and verify it empirically out of curiosity.

Here are some starting input parameters in imperial units. I've created an Excel spreadsheet for my model.

Variable Imperial Units
pipe diameter 0.50 inches
pipe length 900.00 inches
water temp 68.00 °F
water flow 2.00 gal/min
wort temp 220.00 °F
wort specific gravity 1.060 ratio
target temp 70.00 °F
kettle volume 15.00 gals
kettle diameter 19.00 inches
kettle height 15.00 inches

I've read that for laminar flow the Nusselt number should be 3.66. I'm not sure what this should be for my experiment.

int. Nusselt number 3.66 dim-less
water conductivity 0.58 W/(m°C)
water specific heat 4187 J/(kg°C)

Some calculations. I've used // for comments to explain the calculation I used. Note that these calculations use the metric equivalents of the parameters above.

// =int_NuD*inlet_k/pipe_diam
int. heat transfer coefficient 167.1496063 W/(m^2)°C
// =PI()*pipe_diam
pipe perimeter 0.039898227 meters
// =pipe_diam^2*PI()/4
pipe cross-sectional area 0.000126677 m^2
// =wort_sg*1000
wort density 1060 kg/m^3
// =wort_density*kettle_vol
wort mass 60.1880473 kg
wort specific heat 4187 J/(kg°C)

I calculate the energy required to remove:
// =wort_mass*wort_cp/(wort_T-target_T)
7316.342537 J

and apply Newton's Law of Cooling to compute the temperature as a function of cooling time:

Time Init Temp Energy Next Temp
t (s) To (°C) q (W) Q (J) Tn (°C)
0 104.4444444 12939.28219 129392.8219 103.9309958
10 103.9309958 12860.60731 128606.0731 103.4206692
20 103.4206692 12782.4108 127824.108 102.9134455
...
...

// q (W) = kettle_htc*pipe_diam*PI()*pipe_len*(To-inlet_T)
// Q (J) = q*delta_t
// Tn (°C) = To-Q/(wort_mass*wort_cp)

How do these calculations look? My model appears to show that cooling takes much longer than anticipated, so I think something might be wrong. Thanks!

 

[racookpe1978]

Is your laminar flow assumption valid? That might account for the difference in heat transfer and liquid flow.

(Got to admit, the title of the question immediately attracted my attention.)

 

[killerhertz]

That's the thing... I'm not sure I'm getting my information from:

http://en.wikipedia.org/wiki/Nusselt_number

I assumed forced convection in laminar fully developed pipe flow, but this is probably not the case. Should I be using the Dittus-Boelter equation to solve for the Nusselt number?

 

[racookpe1978]

1/2 dia pipe (though you are probably actually using tubing instead) with 2 gpm flow would be turbulent, but that's inside the tubing.

Outside? If I understand the design, the fluid outside will be still, and only convection will be moving the (future) beer.

 

[killerhertz]

That's right. Sorry I assumed that calculation was for the flow of the cooling liquid, not the hot wort.

I stir the wort from time to time, but it mostly moves on it's own from convection. It's quite pretty to watch actually Will the convection induces a perfect laminar flow in this way, assuming no manually stirring?

 

[jmiles]

You can actually calculate the convection coefficient for natural convection, i dont recall the calculation off the top of my head, but taking that and assuming your coils are at least 1-2 diameters apart you could simplify the problem down to a cylinder in a fluid and use the internal and external convection coefficients to build a thermal circuit and get a relatively simple and accurate answer.

To be honest ive enver even considered whether the external flow is laminar or turbulent, but i use equations from Process Heat Transfer by Donald Kern to calculate the external coefficient.

 

[killerhertz]

Unfortunately I don't have any book on heat transfer.

Do you have the names of the equations or could maybe type them up for me?

I'd rather not have to buy a textbook, but do you have any recommendations off of Amazon? Thanks.

 

[vpl]

Here's a link to a free online textbook:

http://web.mit.edu/lienhard/

http://www/ahtt.html

However, especially since this appears to be a home-project versus something you're doing for work, you might want to invest in some books in home brewing.

Patricia Lougheed

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[Jobst]

I do this myself (never simulated it though) and I can tell you that the time to cool the wort is dramatically reduced even with gentle stirring in the boil pot. The natural convection that is set up doesn't do much.

If you've ever used an ice bath to cool the pot you'll know what I mean -- you get a layer of melted ice and hot water around the pot, surrounded by ice water, and stuff all heat transfer (although in this case there are two boundary layers, one inside as well).

I'm upgrading to a counter-current cooler, it promises to be much quicker (and easier to simulate for that matter).

You could do a run, time it without stirring (watch placement of thermometer!) and then use your model to get a U value on the pot side. You could at maybe then do some sensitivity analysis on what degree of stirring would be sufficient to decrease your time to where it is suitable.

Personally, I just drink the beer and forget the calculations

 

[racookpe1978]

Outside (external) pot cooling + stirring the liquid inside is what I was being drawn to as well: something about the numbers sounds funny.

1) initial temp = 220. Above boiling point of (pure) water. Is this what you expect because of the liquid density?

2) kettle vol = 15x3.14xr^2 = 4250 in^3 = 18.40 gals vice 15 gal in the spreadsheet. Is it not filled all the way under normal mixes?

3) pipe length = 900 inches of 1/2 dia tubing.
if the coils were immediately inside the wall radius, coil dia = 19-1 = 18in => length/coil = 18x3.14 = 56.5 in/coil
=> 900 in/56.5 in/coil = 16 coils = about 1/2 in between coils. But then 1/2 of each coil would be against the kettle wall and so wouldn't help cool the mix very well.

Try smaller coil radius = 14 in high of coils at 1/2" per coil = 28 coils. 900 inch/28 coils = 32 in/coil = 10.25 inch dia per coil. This leaves both sides of the coil available to cool the liquid and an inner and outer section of the pot => that both have to be mixed by a spoon or ladle. Cooled liquid can drain out the bottle of the pot because there can room under the lowest coil between the coil and bottom of the kettle.

Alternate: Wrap the coil around the outside of the pot in continuous loop. Insulate the outside of the coils.

Alternate 2. Use small coil inside pot but place the 19" dia kettle in an external pot about 22 in inside dia. Fill outpot with chilled water or brine. This cools the water in the coils, and the kettle at the same time and more doubles your heat transfer area.

 

[killerhertz]

However, especially since this appears to be a home-project versus something you're doing for work, you might want to invest in some books in home brewing.
I have a few home brew texts. I just want to learn something new and apply it to my problem.

I do this myself (never simulated it though) and I can tell you that the time to cool the wort is dramatically reduced even with gentle stirring in the boil pot. The natural convection that is set up doesn't do much.

If you've ever used an ice bath to cool the pot you'll know what I mean -- you get a layer of melted ice and hot water around the pot, surrounded by ice water, and stuff all heat transfer (although in this case there are two boundary layers, one inside as well).

I'm upgrading to a counter-current cooler, it promises to be much quicker (and easier to simulate for that matter).
Yeah I know stirring helps. I normally stir. For the purposes of my calculations, I'll ignore stirring to find the worst case.

I can't use an ice bath because my pot is 15 gal and wouldn't fit in my sink I do plan on building a pre-chiller for this immersion chiller to bring the water down from room temp.

 

[killerhertz]

1) initial temp = 220. Above boiling point of (pure) water. Is this what you expect because of the liquid density?

Yes.

2) kettle vol = 15x3.14xr^2 = 4250 in^3 = 18.40 gals vice 15 gal in the spreadsheet. Is it not filled all the way under normal mixes?

This is definitely an entry error in my calculation. I used 15 gal, which is the total boil volume of the kettle, but this number should really be the gallons of wort = 6 gal for a 5 gal batch typically. Oops!

3) pipe length = 900 inches of 1/2 dia tubing.
if the coils were immediately inside the wall radius, coil dia = 19-1 = 18in => length/coil = 18x3.14 = 56.5 in/coil
=> 900 in/56.5 in/coil = 16 coils = about 1/2 in between coils. But then 1/2 of each coil would be against the kettle wall and so wouldn't help cool the mix very well.

I planned on doing an inner and outer coil. Possibly a 16" outer and 8" inner at 6" high, so even if I do small batches, it's totally submerged. I need to find something that's 8" in diameter to bend the boil around though, which won't be easy...

pi*16"*6"/0.5"+pi*8"*6"/0.5" = ~900"


Oh and thanks for that ebook reference. I'm checking it out now.

 

[Compositepro]

I brew beer. Natural convection will work fine for cooling beer but you must do things correctly to get it to work. A cooling coil in a pot will only cool liquid below the level of the coil. It is very difficult to locate the coil at the top surface of the liquid in the pot. Cooling coils in the bottom of the pot are almost useless.

The simplest approach to cooling wort is to place the hot wort pot into a sink, or other larger pot, full of cold water. Fresh, cool water can be run into the sink and be allowed to overflow. Warm water will overflow over the top. The cooling water level outside the pot must be higher than the wort level in the pot or the wort at the top will cool very slowly, unless stirred. If you want to use a coil (which adds to cost and cleaning headaches) the best approach is to siphon the wort through the coil, which is in a sink of cold water, and into another pot.

 

[racookpe1978]

When I bend multiple pipes and square solid bars for spiral stair rail handrails and for single bends in mild steel for sundials, I've had good results using a large regular schedule pipe for 6, 8, 10, and 14 inch diameters.

Buy 2 to 4 feet of pipe, or salvage some from a constructioin site (with builder's permission!) and weld it firmly to a pad or structure. Be sure the pipe can't move. Clamp the tubing or small steel piece (for me, usually 1/2 x 1/2 bars or 1-1/4 sch 40 pipe) to a fixed point near the vertical pipe then (with a prying level) wrap the tubing around the pipe.

Use a spacer between coils to get the "bootleggers' still" coil you want.

A vertical pipe is easiest to walk around as you apply force for multiple turns: plus, horizonal pipes are harder to use with a long lever arm.