Therminator, Ice Water, Whirlpool... question

[Lonnie Mac]

No drain in garage...

Prefer it to be self contained

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...

 

[Bobby_M]

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).

 

[The Pol]

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?

 

[Sawdustguy]

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

 

[Boerderij_Kabouter]

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.

 

[The Pol]

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.

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?

 

[The Pol]

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).

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.

 

[The Pol]

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).

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

 

[Lil' Sparky]

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?

 

[The Pol]

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.

 

[The Pol]

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?

Twice as much ice... as what?

 

[conpewter]

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?

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

 

[The Pol]

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.

 

[The Pol]

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.

 

[arturo7]

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?

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.

 

[The Pol]

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.

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.

 

[arturo7]

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

 

[The Pol]

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)

 

[The Pol]

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

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.

 

[The Pol]

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.

 

[arturo7]

Pol, not to change the subject (much), but if your washer and dryer is in your garage it is pretty easy to install a sink with a drain. I'm not advocating the use of a CFC, just an FYI.

 

[Lil' Sparky]

Twice as much ice... as what?

... as what I would use to chill 5 gals, though like I said I use a water hose to do part of the chilling (like most everyone else).

I get what you're trying to do. For me, I couldn't bring myself to use 70+ lbs of ice to chill a 10 gal batch. I wonder if that much ice would even fit in my sanke HLT.

More than anything, I thought this idea was interesting coming from you with some many pro-no-chill posts lately.

I've got no problem with brewers trying new things. I actually applaud you and a bunch of others on here for the way you pioneer new stuff.

 

[The Pol]

Pol, not to change the subject (much), but if your washer and dryer is in your garage it is pretty easy to install a sink with a drain. I'm not advocating the use of a CFC, just an FYI.

Our washer and dryer are in a dedicated laundry room... I really need to make it self contined if I want to chill. Now, I dont HAVE to, but it is my desire.

 

[The Pol]

... as what I would use to chill 5 gals, though like I said I use a water hose to do part of the chilling (like most everyone else).

I get what you're trying to do. For me, I couldn't bring myself to use 70+ lbs of ice to chill a 10 gal batch. I wonder if that much ice would even fit in my sanke HLT.

More than anything, I thought this idea was interesting coming from you with some many pro-no-chill posts lately.

I've got no problem with brewers trying new things. I actually applaud you and a bunch of others on here for the way you pioneer new stuff.

Well, I am not doing 10 gal. batches, and it is easy for me to make a 43lb cube of ice essentially free.

I have done plenty to further the use and exploration of no chill. It works, and well, but one reason I do it is to avoid copious amounts of water and keep my rig self contained. It appears that I can do the same with this method here.

No chill works, but I have never thought it is the ONLY way or BEST way... just equally as good. I want to give myself the option to chill while keeping my rig completely self contained. I am trying to find a way to spend some brewing dollars and this seemed like a neat idea. A plate chiller and a pump and it is ready to go... I have the extra outlets on my control panel to accomodate it currently.

This fulfills many of the benefits (to me) of no chill, while still chilling.

 

[Lil' Sparky]

So what's the plan for free 43 lb ice cubes? Listeners want to know.

 

[lamarguy]

So what's the plan for free 43 lb ice cubes? Listeners want to know.

I believe he plans to freeze a couple of water buckets outside in the winter. Obviously, that wouldn't work in TX.

 

[The Pol]

Making a 43lb block in a 5 gallon bucket. No need for winter, I have a freezer.

 

[arturo7]

Probably don't need to say this but just in case...

Be sure to break up that block before you use it. Heat transfer is facilitated by surface area. A 5 pound block of ice doesn't cool nearly as well as 5 pounds of cubes.

 

[Lonnie Mac]

Probably don't need to say this but just in case...

Be sure to break up that block before you use it. Heat transfer is facilitated by surface area. A 5 pound block of ice doesn't cool nearly as well as 5 pounds of cubes.

Yes, it is all about surface area... A block may not do so well... Yet cubes melt VERY fast... This is the quandary my friend... All the math in the world wont figure that out...

 

[The Pol]

This doesnt seem like a quandry, it seems simple to me....

LARGE ice block will melt slowly, increasing chilling time.

ICE CUBES will melt faster, allowing for a faster heat exchange.

BOTH have the same cooling potential, as they will BOTH require the same 1,500,000 calories to melt them. The cubes will simply absorb those calories faster than the block.

BOTH will have the same cooling POWER as they would be identical amounts of ice, so identical numbers of calories per g to melt it. The difference it would seem would be in the RATE of the heat exchange due to melting, and thus the rate of wort cooling.

I would compare this to using a set volume of water (cooling medium) and using two different plate chillers. One with 2' of surface area, the other with 8' of surface area. By recirculating this set volume of water, you will have the SAME final temperature... the difference will be in the rate of cooling since you are using the same cooling medium, you are only changing the surface area, and thusly the rate of cooling, not cooling potential.

I need to get my hands on cubes, about 40 pounds.