Therminator, Ice Water, Whirlpool... question

beerthirty said:

It does. Just make sure you keep the crap out of it.

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How do most do this? I already use SS mesh pads.

I bought 30x30 mesh screen from mcmaster and made a hop stopping cylinder, no pellet hops at all get through. I just used stainless lock wire to sew it all up.

I pump 10 gallons of wort through my therminator at full speed with just tap water and it gets to pitch temps.

Some more pics here

https://www.homebrewtalk.com/f85/8-22-09-brew-day-pumpkin-ale-133537/

I cannot use tap water, this must be self contianed.

I used to use the scrubbies. Now I bag the hops to keep most of them from the diptube and use a hop stopper modeled after Bobby_M's. https://www.homebrewtalk.com/f51/diy-hop-filter-video-102697/
I tried the hopstopper by itself with 2.5 oz pellet and it really slowed down flow and left some wort in the kettle. Since combining the two, I have had great success. The 5G paint strainer bag allows plenty of flow through the bag, the hopstopper does a very good job of removing any hops that get out of the bag and keeps a large amount of break material out while still maintaining good flow. there is very little wort left in the kettle at the end.

Edit; I forgot that you use electric. Forget the bags. Instead build a ss bag 6-8" in diameter that extend out the top of the kettle. It wont interfere with the boil and you will still get good utilization.

The Pol said:

Well, here is the "idea"...

Utilize (2) march pumps
(1) Therminator
(1) 10 gallons of ICE water

During the boil I clean my MLT.
Refill MLT with water and (3) gallons of ICE

After boil is complete operate ONE pump to recirc boiling wort through THERMINATOR back to the kettle to help sanitize therminator.

To chill, run both pumps at full flow. One to recirc ice water from MLT to the chiller and back to the MLT. The other pump will recirc the wort through the Therminator to the kettle through a QD fitting on the side of the kettle that is attached to a length of LocLine on the INside of the BK.

The PID thermocouple in the BK would monitor temp.

Would this work? I am looking to spend some money is all, is this an effective way to spend money?

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Not sure if I am reading this correctly Pol, but I have been doing exactly this for about 10 years. About 5 of that with the therminator. Guess you have never read anything about Brutus Ten! LOL

Just kidding my friend...

In my case, and I think in any case, ice alone will not knock the boil down from boiling temp to pitching temp unless you use a monstrous amount if ice. I have never ran any numbers but I can say that I still have to use hose water to get it down to around a hundred or so just so my 4-5 (7 lb) bags of ice will take it the rest of the way down to pitch temp. You want to see some ice melt fast, just recirculate boiling wort through it!

Hmmm.... interesting Lonnie. So you need to use 4-5 seven pound bags of ice (i.e., 28-35 pounds) to bring 10g from 100 to pitching? That is way more than I would have guessed. Weird.

I have no problem with the duda chillers, I was just saying to besure you are getting the surface area you are looking for. Just becasue it has x-plates doesn't mean it is awesome, you need to know how big the plates are. You obviously know that, I was just making sure

I still dont get it...

If it will cool 10 gallons from boiling to 68F with 58F water in 5 minutes...

I am cooling 5 gallons from boiling to 68F with what starts as 32F with 20 pounds of ice augmenting the cooling. You are telling me that in 5 minutes I will have NO ice and my water will exceed 58F?

I wish there were a way to calcualte this accurately before I spend $400 to do it

Well, what will a 5 gallon volume (25 pounds) of ice do when dropped into 5 gallons of boiling wort (water)?

I can at least test that.

The biggest problem with recirculating the exhaust coolant back into the reservoir is that it takes a bit of time for the ice to cool it. You'll start out by pumping 32F water into the chiller, but within 2 minutes, you'll be pumping something like 70F as the ice tries to cool the incoming HOT water. The ice will melt pretty quickly.

What I'd recommend doing is filling the MLT with tap water first and run that out through the chiller and into buckets. Then chuck the ice in and add another batch of tap water. The water that comes out of the plate chiller in the first few minutes will be like 180F.

The idea is to remove most of the heat out of the system immediately. The closed loop thing screws that up.

Yes, a 3 gallon block of ice has a lot more cooling potential than 3 gallons of 33F water because it takes a lot of energy to freeze the water from 33.

The Pol said:

I still dont get it...

If it will cool 10 gallons from boiling to 68F with 58F water in 5 minutes...

I am cooling 5 gallons from boiling to 68F with what starts as 32F with 20 pounds of ice augmenting the cooling. You are telling me that in 5 minutes I will have NO ice and my water will exceed 58F?

I wish there were a way to calcualte this accurately before I spend $400 to do it

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There is.

Come on- there has to be some Mech Eng. out there bored ina cubicle. Help a brewer out!!!

I would run it for you Pol but I am slammed right now. Maybe later this week I could spare some time and figure it out.

EDIT!!

I just looked it up, I am talking about adding 33 pounds of ice (solid block) to the 5 gallons of water. That is what I am proposing to use for my chilling meduim.

I must say, for $400, you could spend an extra $50 and install a hose bib in your garage, then you could use that and drain into your garage drain.

I still think the contained system is sexy. That was my initial plan to until I was shot down by naysayers... perhaps my pipedream lives???

The physics problem would be stated "what would the resulting temp be if I combined 5 gallons of water at 212F with 5 gallons of water at 60F (guess) and 33 pounds of ice?". Of course, sugar water has a different density so it would skew the answer a bit but I would think you'd be close enough.

Of course, if you do this during the winter when you potentially have free access to ice (leaving a few buckets outside), it doesn't matter how many times you have to add ice.

I'm setting my system up to do a variant on this theme. However, I WILL be using the hose, even in the winter. Just keep the hose empty and run as needed for brewday then store in the garage, empty.

March pump will recirc wort through Schirron plate chiller using hose water. This will give me a nice whirlpool and keep the little break and trub in the middle. I use a large grain bag and small muslin bags (no more than one ounce per bag) for hops. Keeps the bulk of the debris out chiller and fermenter. Once the wort drops to ~120F, I'll kick in the 20lbs of ice and ~2G of water pumped with the undergravel fish tank pump.

I've found I can get 10.75G of wort to around 70F within 30min with constant swirling with my long handle spoon. I don't like standing and stirring my wort for 30min! So, I'm going to try the two chiller system and whirlpool the wort.

Frankly, the advantage of plates/CFC external exchangers is the ability to just push the wort into your fermenter in one quick pass so I don't really see the advantage of a recirculate. You might say it's to chill the whole wort down quickly, but with icewater as the coolant, you can run the March full bore and drain 5 gallons within 5 minutes.

Pol, you might want to check this build out before dropping all that cash.


Scroll down to post #5.

https://www.homebrewtalk.com/f51/shell-tube-heat-exchanger-122655/

Inexpensive, efficient, no need to worry about clogging it with hops.

Bobby_M said:

Frankly, the advantage of plates/CFC external exchangers is the ability to just push the wort into your fermenter in one quick pass so I don't really see the advantage of a recirculate. You might say it's to chill the whole wort down quickly, but with icewater as the coolant, you can run the March full bore and drain 5 gallons within 5 minutes.

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Personally this is exactly why I do it. Not withstanding, my trub and cold break is left in the kettle and not in my fermenter.

In Texas, nothing comes out of the hose unless it's about 85 degrees!

Boerderij_Kabouter said:

I must say, for $400, you could spend an extra $50 and install a hose bib in your garage, then you could use that and drain into your garage drain.

I still think the contained system is sexy. That was my initial plan to until I was shot down by naysayers... perhaps my pipedream lives???

Click to expand...

No drain in garage...

Prefer it to be self contained

Boerderij_Kabouter said:

Hmmm.... interesting Lonnie. So you need to use 4-5 seven pound bags of ice (i.e., 28-35 pounds) to bring 10g from 100 to pitching? That is way more than I would have guessed. Weird.

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That's about what I use, too, and I first use tap water to chill the wort before it exchanges heat with the ice water.

The Pol said:

I am cooling 5 gallons from boiling to 68F with what starts as 32F with 20 pounds of ice augmenting the cooling. You are telling me that in 5 minutes I will have NO ice and my water will exceed 58F?

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Yes, the HEX exhaust will melt that ice fast. Once it's gone, the entire cooling tank will heat up. I don't know how much "extra" cooling capacity the ice holds, but remember, it won't be 20 lbs after it chills 5 gals down to 32º - maybe only 1/3 or 1/2 of that.

The Pol said:

No drain in garage...

Prefer it to be self contained

Click to expand...

For the longest time, I ran my exhaust hose right into my apartment dryer drain in the wall. I got my cold water from my washer supply with a "y" hose fitting...

Here's a similar "problem" I found and how it's worked out:

5. The temperature in the freezer compartment of a typical household refrigerator is about
– 15oC. Thus, an ice cube taken from the freezer must first warm up to 0oC before it will begin to melt. The specific heat of ice is 0.51 cal/g-oC and its latent heat of fusion is 80 cal/g.

Imagine that a 10 g cube of ice at –15oC is taken from the freezer and added to 100 g of water in a Styrofoam cup. The temperature of the 100 g of water is initially 22oC. What is the final temperature of the water after all the ice has melted? Show all your work.

a. Begin by determining the amount of energy required to heat the ice cube from –15oC to 0oC. Assume that none of the ice melts.
The heat to warm the ice to 0C is:
Heat = 10 grams X 15 C X 0.51 cal/gC = 76.5 cal

b. Next, find the amount of energy required to melt the ice cube. Show your work.
10 grams X 80 cal/g= 800 cal

c. The sum of your answers to parts a and b represents the total amount of heat required to warm the ice cube up to 0oC and melt it. Since the ice cube was added to a cup of water, this energy will come from the water, causing its temperature to decrease. Calculate the temperature of the cup of water after the ice cube has melted.
876.5 cal = 100 grams of water X T X 1 cal/gC
T= 8.8C
The water started at 22C, so the final temperature would be 22C-8.8C= 13.2C

d. Your answer to part c is not the final temperature of the mixture, because after the ice cube has melted it is still at 0oC. Thus the cup now contains 10 g of water at 0oC and 100 g of water at the temperature found in part c. These two samples of water will exchange heat until they reach an equilibrium condition. Set up a table like the one we used last quarter to work out the final temperature. To make the calculations easier, determine the amount of heat necessary to lower the temperature of the 100g of water by 0.2oC. Then determine the temperature of the 10 g of cold water when it receives this heat. Continue until you have determined the equilibrium temperature.\


For your example it wouldn't be 10 grams of ice, it would be 15,000 grams. Your wort would be about 19,000 grams at 100c and your coolant water is 19,000 grams at (tap temp). Since the wort and water are both liquid, you might just say that your "liquid" is 38,000 grams at (the average between your 100c wort and tap temp water).

Well, I can get this far.

18,820g (5 gal volume) of ice at -7C(20F)
It will take 1,572,787 cal. to heat this ice to 0C AND melt it.(leaving it at 0C still)

The cooling water(4 gal at 65F) + wort(5.5 gal at 208F) will equal 35,789g (9.5 gal) at an average temp of 66C(151F)

EDIT... okay, we are getting there....

1,572,787 cal / 35,789g = 44C

The wort+cooling water resultant temp was 66C, so now we can subtract 44C which places us at 22C!

22C is the resulting temp (72F) Now per the equation above, this is assuming we HEATED the ice to 0C and then melted it. This then leaves the resultant water from the melted cube at 0C as we did not compute the calories to HEAT it above 0C. We only calculated the cal. needed to HEAT it to 0C and MELT it.

Per this equation, this would work. By the time the wort was at 72F, we would have JUST melted the 43lb ice block in the MLT. The resultant water (from the melted ice) would still be 0C.

Since the melted cube would actually be in 4 gallons of what began as 18C water we can go further.

4 x 18C = 72
5 x 0C = 0

72/9 = 8C

The resultant temp. in the MLT after the ice has melted, and the wort is at 72F (22C) should be 8C or 46F

Anyone care to check my math?

Pol. I have to go back to work but this document can help you I think.

http://www.zamorascience-enhs.com/files/1.4_Temperature_in_Thermal_Systems0001.pdf

Q=mc(T2-T1)

1572787 cal x (1 Joule / .2390 cal) = 6580699 J

6580699J = 35789g x 4.1813J/(g·K) x [ 66 - T1 ]

44 = 66 - T1

T1 = 22C = 72F

That is assuming 100% efficiency, but I think this idea has merit.

Boerderij_Kabouter said:

Q=mc(T2-T1)

1572787 cal x (1 Joule / .2390 cal) = 6580699 J

6580699J = 35789g x 4.1813J/(g·K) x [ 66 - T1 ]

44 = 66 - T1

T1 = 22C = 72F

That is assuming 100% efficiency, but I think this idea has merit.

Click to expand...

Right 72F, but that is what it takes to MELT the ice... it is still at 0C per what BobbyM posted, meaning that my MLT water is still in the 40s, providing plenty of cooling power still, right?

Bobby_M said:

Here's a similar "problem" I found and how it's worked out:

5. The temperature in the freezer compartment of a typical household refrigerator is about
– 15oC. Thus, an ice cube taken from the freezer must first warm up to 0oC before it will begin to melt. The specific heat of ice is 0.51 cal/g-oC and its latent heat of fusion is 80 cal/g.

Imagine that a 10 g cube of ice at –15oC is taken from the freezer and added to 100 g of water in a Styrofoam cup. The temperature of the 100 g of water is initially 22oC. What is the final temperature of the water after all the ice has melted? Show all your work.

a. Begin by determining the amount of energy required to heat the ice cube from –15oC to 0oC. Assume that none of the ice melts.
The heat to warm the ice to 0C is:
Heat = 10 grams X 15 C X 0.51 cal/gC = 76.5 cal

b. Next, find the amount of energy required to melt the ice cube. Show your work.
10 grams X 80 cal/g= 800 cal

c. The sum of your answers to parts a and b represents the total amount of heat required to warm the ice cube up to 0oC and melt it. Since the ice cube was added to a cup of water, this energy will come from the water, causing its temperature to decrease. Calculate the temperature of the cup of water after the ice cube has melted.
876.5 cal = 100 grams of water X T X 1 cal/gC
T= 8.8C
The water started at 22C, so the final temperature would be 22C-8.8C= 13.2C

d. Your answer to part c is not the final temperature of the mixture, because after the ice cube has melted it is still at 0oC. Thus the cup now contains 10 g of water at 0oC and 100 g of water at the temperature found in part c. These two samples of water will exchange heat until they reach an equilibrium condition. Set up a table like the one we used last quarter to work out the final temperature. To make the calculations easier, determine the amount of heat necessary to lower the temperature of the 100g of water by 0.2oC. Then determine the temperature of the 10 g of cold water when it receives this heat. Continue until you have determined the equilibrium temperature.\


For your example it wouldn't be 10 grams of ice, it would be 15,000 grams. Your wort would be about 19,000 grams at 100c and your coolant water is 19,000 grams at (tap temp). Since the wort and water are both liquid, you might just say that your "liquid" is 38,000 grams at (the average between your 100c wort and tap temp water).

Click to expand...

According to this, my wort would be at 72F but I would still have plenty of cooling power in the MLT yet. These equations take into consideration heating the ice to 32F and then melting it, still at 32F. This would then translate into much more cooling power being available due to the differential between the MLT cooling water and the 72F wort.

Bobby_M said:

Here's a similar "problem" I found and how it's worked out:

5. The temperature in the freezer compartment of a typical household refrigerator is about
– 15oC. Thus, an ice cube taken from the freezer must first warm up to 0oC before it will begin to melt. The specific heat of ice is 0.51 cal/g-oC and its latent heat of fusion is 80 cal/g.

Imagine that a 10 g cube of ice at –15oC is taken from the freezer and added to 100 g of water in a Styrofoam cup. The temperature of the 100 g of water is initially 22oC. What is the final temperature of the water after all the ice has melted? Show all your work.

a. Begin by determining the amount of energy required to heat the ice cube from –15oC to 0oC. Assume that none of the ice melts.
The heat to warm the ice to 0C is:
Heat = 10 grams X 15 C X 0.51 cal/gC = 76.5 cal

b. Next, find the amount of energy required to melt the ice cube. Show your work.
10 grams X 80 cal/g= 800 cal

c. The sum of your answers to parts a and b represents the total amount of heat required to warm the ice cube up to 0oC and melt it. Since the ice cube was added to a cup of water, this energy will come from the water, causing its temperature to decrease. Calculate the temperature of the cup of water after the ice cube has melted.
876.5 cal = 100 grams of water X T X 1 cal/gC
T= 8.8C
The water started at 22C, so the final temperature would be 22C-8.8C= 13.2C

d. Your answer to part c is not the final temperature of the mixture, because after the ice cube has melted it is still at 0oC. Thus the cup now contains 10 g of water at 0oC and 100 g of water at the temperature found in part c. These two samples of water will exchange heat until they reach an equilibrium condition. Set up a table like the one we used last quarter to work out the final temperature. To make the calculations easier, determine the amount of heat necessary to lower the temperature of the 100g of water by 0.2oC. Then determine the temperature of the 10 g of cold water when it receives this heat. Continue until you have determined the equilibrium temperature.\


For your example it wouldn't be 10 grams of ice, it would be 15,000 grams. Your wort would be about 19,000 grams at 100c and your coolant water is 19,000 grams at (tap temp). Since the wort and water are both liquid, you might just say that your "liquid" is 38,000 grams at (the average between your 100c wort and tap temp water).

Click to expand...

This is why ICE is much more powerful than 32F water (ice water)

So basically, you need twice as much ice to chill your wort to avoid using tap water from a hose... and all of this from one of the forum's big proponent of no-chill brewing?

If you consider the wort at 22C after melting the ice...

Then if you took that 5 gallons of water from the melted ice that is at 0C and found the resultant temp of that AND the wort...

The max cooling power in this scenario would be:

5.5 x 22c = 121
5 x 0c = 0

121/10.5 = 11.5c or 53F if the temps were allowed to equalize.

Theoretically, per the scenario that BobbyM posted above, you could use this scenario:

4 gallons of water at 65F, 43 pounds of ice (5 gal) at 20F and it would have the potential to bring boiling wort to 53F when the temps of the cooling medium and the wort equalized.

Lil' Sparky said:

So basically, you need twice as much ice to chill your wort to avoid using tap water from a hose... and all of this from one of the forum's big proponent of no-chill brewing?

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Twice as much ice... as what?

Lil' Sparky said:

So basically, you need twice as much ice to chill your wort to avoid using tap water from a hose... and all of this from one of the forum's big proponent of no-chill brewing?

Click to expand...

Well in the winter it is much more environmentally friendly to use ice than to run much more tap water through your system.

I am still trying to figure out... twice as much ice as what?

I am trying to prove that I can build a self contained rig, without dragging frozen hoses in the middle of winter. Also, this water isnt wasted. This 9 gallons can be recirculated directly from my MLT through my system for cleaning, then transferred to my BK for the same purpose after it has been used to cool the wort.

People say it cant be done... so, well I thought it would be neat to actually see if they knew what they were talking about. Thanks to BobbyM for that post with some equations to use, that was invaluable.

Let me run the #s for 10 gallons... my guess is that it simply will not work at that volume, there is too much thermal mass. I will get that posted in a bit.

Lil' Sparky said:

So basically, you need twice as much ice to chill your wort to avoid using tap water from a hose... and all of this from one of the forum's big proponent of no-chill brewing?

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Sparks, I may have missed something in all of those calculations. But, at 7-10 GPM, a hose uses at least 100 gallons, more likely double.

arturo7 said:

Sparks, I may have missed something in all of those calculations. But, at 7-10 GPM, a hose uses at least 100 gallons, more likely double.

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I am calculating this using 5 gallons of water (frozen) and 4 gallons of water at 65F (tap).

Total of 9 gallons, which would then be used for cleaning.

Pol, I was referring to the amount of water needed from a tap using no recirculation. Guess I didn't make that very clear.

For 10 gallons, it doesnt work nearly as well.

1,572,787 calories to melt the ice to a 0C liquid.

The cooling water and wort equalize as:

54,590g at 78C

1,572,787 / 54,590 = 29C

79C - 29C = 49C(120F)

At this time the MLT cooling water would still be 8C (46F)

arturo7 said:

Pol, I was referring to the amount of water needed from a tap using no recirculation. Guess I didn't make that very clear.

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I know, I am agreeing with you, I am using 9 gallons in this scenario... which is very little water. The only reason it works is

#1 it is cooling 5.5 gallons, not 10.5
#2 it is using ICE as about 60% of the medium, which requires a lot more energy to heat it to 0C and then to actually melt it, even while keeping it at 0C.

I am about to do an experiemnt that will prove these equations to be correct. It will be verified by tomorrow PM. As long as the practical application works tomorrow, this will be built and I will have a self contained wort cooling apparatus.