Building a Meat Locker

[johnnyd]

Hey Brody,

Another fan (we all are here) of admirably obsessive projects. Any chance you can snap a picture or two of the fruits of your labor? Thanks to rooftop1000, I remember now how much more engaging Varmint's pig-pickin topic was with Jason & Rachels photography.

Not that you need another chore on the list, though!

Carry on.

[Smithy]

Chris, based on the 10-day forecast, and my basement averaging about 5 degrees cooler, I'd expect my ambient temperature to be 80 F. It will probably hover near this temp for 6 hours. This will be on Friday, and the subsequent days will be 3-5 degrees cooler.

What's the formula?

Been running a bit behind schedule this weekend with a number of other chores, but I have the salvaged lumber and drywall ready to be cut downstairs. Made some adjustments to the dimensions, and am instead going for a 2.5 ft x 2.5 ft x 6 ft space. Just enough space, I hope, to contain the pig carcass and still allow airflow. This should also make it easier to cool. Small will also make it movable to other areas of the basement should I try repurposing it down the line.

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f = A * D / R

where f is going to be the heat flow through the insulation in Btu/hr, A is the surface area of the insulation in square ft, D is the temperature difference in degrees F, and R is the R value in English units (U.S. standard, ft^2 F hr / Btu). To maintain steady-state (i.e. assuming your chamber and everything in it is already at 35 deg F, your "cooling unit" will need to be capable of removing at least this much heat from the chamber. Of course, if your cooling unit is just a tray of ice, it can't remove any heat from the chamber, and then calculation needs to take into account the thermal mass of the ice, and calculate how long it will take to melt. I will think on that for a while... I'm not sure quite how to proceed.

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Actually, the tray of ice will remove heat from the chamber in the sense of absorbing it. As long as the ice is much colder than the basement, you'll get some chilling going on, and with enough ice - if you start out with very cold ice - you should be able to drop the chamber temperature and keep it there. Let's see... figure the 4 sides and the ceiling of the chamber to get the area. Use that in Chris' formula to get the heat flow. The specific heat of ice at these ranges is just under 0.5 btu/lb/degF (ranging from 0.471 at 0*F to 0.504 at 32*F). As that ice changes to water it's going to soak up a great deal of heat - haven't laid my hands on that number yet - and then the liquid water has a specific heat of around 1 btu/lb/deg. It won't want to warm up quickly, but given enough time, of course it will. If I'm approaching this correctly, you can use Chris' formula to figure how much heat needs to be absorbed to get the temperature you want and work that to how many pounds of ice you need to maintain that temperature. This assumes a well-chilled pig to start with.

It seems to me that you'd want a great deal of ice (4 times as much as calculated? 5?) just to save yourself from having to open that chamber and let more heat in. 80*F in the basement? Ye cats, your climate is different than ours!

Hmm. What about putting dry ice in the chamber? Would that affect the curing? You'd for sure have to vent the chamber well (and add a pile of heat) before going in, so you don't asphyxiate. Hmm.

[brody]

Well, unfortunately I got a major setback as I was getting ready to leave work and start building this glorious meat locker. My pig farmer, who I mysteriously hadn't heard from since Thursday...is in Italy for the week.

Thus, meat locker is post-poned until I can source another quailty pig that won't set me back a substantial about of cash. Anyone have recommendations for a pig farmer within 1.5hrs of DC? Definitely still going through with it though, but I won't get to see the fruits of my labors immediately. I will, however, have more time to get this right and consider ways to integrate a dual-use design: meat storage/refrigeration and meat aging. Not at the same time, of course.

It will still get built in the next week or two..and I'll be taking pictures for sure, and hopefully reporting back on it's success (or not).

Chris, based on the 10-day forecast, and my basement averaging about 5 degrees cooler, I'd expect my ambient temperature to be 80 F. It will probably hover near this temp for 6 hours. This will be on Friday, and the subsequent days will be 3-5 degrees cooler.

What's the formula?

Been running a bit behind schedule this weekend with a number of other chores, but I have the salvaged lumber and drywall ready to be cut downstairs. Made some adjustments to the dimensions, and am instead going for a 2.5 ft x 2.5 ft x 6 ft space. Just enough space, I hope, to contain the pig carcass and still allow airflow. This should also make it easier to cool. Small will also make it movable to other areas of the basement should I try repurposing it down the line.

←

f = A * D / R

where f is going to be the heat flow through the insulation in Btu/hr, A is the surface area of the insulation in square ft, D is the temperature difference in degrees F, and R is the R value in English units (U.S. standard, ft^2 F hr / Btu). To maintain steady-state (i.e. assuming your chamber and everything in it is already at 35 deg F, your "cooling unit" will need to be capable of removing at least this much heat from the chamber. Of course, if your cooling unit is just a tray of ice, it can't remove any heat from the chamber, and then calculation needs to take into account the thermal mass of the ice, and calculate how long it will take to melt. I will think on that for a while... I'm not sure quite how to proceed.

←

Actually, the tray of ice will remove heat from the chamber in the sense of absorbing it. As long as the ice is much colder than the basement, you'll get some chilling going on, and with enough ice - if you start out with very cold ice - you should be able to drop the chamber temperature and keep it there. Let's see... figure the 4 sides and the ceiling of the chamber to get the area. Use that in Chris' formula to get the heat flow. The specific heat of ice at these ranges is just under 0.5 btu/lb/degF (ranging from 0.471 at 0*F to 0.504 at 32*F). As that ice changes to water it's going to soak up a great deal of heat - haven't laid my hands on that number yet - and then the liquid water has a specific heat of around 1 btu/lb/deg. It won't want to warm up quickly, but given enough time, of course it will. If I'm approaching this correctly, you can use Chris' formula to figure how much heat needs to be absorbed to get the temperature you want and work that to how many pounds of ice you need to maintain that temperature. This assumes a well-chilled pig to start with.

It seems to me that you'd want a great deal of ice (4 times as much as calculated? 5?) just to save yourself from having to open that chamber and let more heat in. 80*F in the basement? Ye cats, your climate is different than ours!

Hmm. What about putting dry ice in the chamber? Would that affect the curing? You'd for sure have to vent the chamber well (and add a pile of heat) before going in, so you don't asphyxiate. Hmm.

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Chris and Smithy, with the formula what exactly is f again? Is this my BTU/heat loss through the insulation? Thus, to maintain an internal temp of 35 F, when my f value is 65, I would need a cooling solution that removed 65 BTU of heat every hour?

[Chris Hennes]

Chris and Smithy, with the formula what exactly is f again? Is this my BTU/heat loss through the insulation? Thus, to maintain an internal temp of 35 F, when my f value is 65, I would need a cooling solution that removed 65 BTU of  heat every hour?

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Right. For your case I am getting A = 72.5 sq ft and D=45 deg F. R is a variable, I'll choose 10 ft^2 F hr / Btu arbitrarily for now, which would give you 325 BTU/hr is heat coming into the chamber through your insulation. You need to counteract this heat addition by either using that heat to do work (i.e. to melt ice) or you need to remove the heat using a cooling system.

If you are using ice you will need two more equations: one that calculates how much energy it takes to raise the temperature of the ice from whatever your freezer temperature is to 32 deg F, and one that calculates how much energy it takes to convert ice at 32 to water at 32 (melting energy). Theoretically, of course, you could add a third that calculates the energy that it takes to raise the water temp to 35 F, but I'd say that once the ice melts, time to put in a new batch. This calculation will give you the amount of heat that will be added to melt the ice: from this you can calculate how long it will take, based on the BTU/hr calculation, to melt a given mass of ice (by this point we've made quite a few idealizations, but it will give you a rough idea). Et voila! You know how much ice to add and at what frequency to add it to maintain a given temp.

[C. sapidus]

. . . This calculation will give you the amount of heat that will be added to melt the ice: from this you can calculate how long it will take, based on the BTU/hr calculation, to melt a given mass of ice . . .

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Enthalpy of fusion (Wikipedia). Good luck!

[Smithy]

Chris and Smithy, with the formula what exactly is f again? Is this my BTU/heat loss through the insulation? Thus, to maintain an internal temp of 35 F, when my f value is 65, I would need a cooling solution that removed 65 BTU of  heat every hour?

←

Right. For your case I am getting A = 72.5 sq ft and D=45 deg F. R is a variable, I'll choose 10 ft^2 F hr / Btu arbitrarily for now, which would give you 325 BTU/hr is heat coming into the chamber through your insulation. You need to counteract this heat addition by either using that heat to do work (i.e. to melt ice) or you need to remove the heat using a cooling system.

If you are using ice you will need two more equations: one that calculates how much energy it takes to raise the temperature of the ice from whatever your freezer temperature is to 32 deg F, and one that calculates how much energy it takes to convert ice at 32 to water at 32 (melting energy). Theoretically, of course, you could add a third that calculates the energy that it takes to raise the water temp to 35 F, but I'd say that once the ice melts, time to put in a new batch. This calculation will give you the amount of heat that will be added to melt the ice: from this you can calculate how long it will take, based on the BTU/hr calculation, to melt a given mass of ice (by this point we've made quite a few idealizations, but it will give you a rough idea). Et voila! You know how much ice to add and at what frequency to add it to maintain a given temp.

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The energy to raise the ice temperature from, say, 0*F to 32*F is about 16 btu/lb, based on a specific heat of 0.5 btu/lb/Deg. F. (The number changes, as I noted above, but we're using back-of-envelope approximations here.) So if you really need to remove 325 btu/hr you're going to need 20 lbs of ice per hour. Note that this only accounts for the work to warm up the ice. The heat sink is much greater when the ice starts melting. I still haven't laid my hands on that number, but the idea is the same:

lbs of ice needed = f / s

where f is your number from Chris' equation and s is the specific heat of phase change between ice and liquid water, in btu/lb. I'll find that number yet, unless somebody else provides it first.

That's how you figure it. I agree with Chris that, once the ice melts, it's time to change the blocks. That comes to the next point: every time you open the door to change the ice, you're going to let a good deal of heat in and have to start over. You need an air lock. (A chiller is looking better all the time, isn't it? I just caught myself trying to work out a series of tubes through the chamber, circulating very cold water...but then you need a heat exchanger out in the basement someplace, and...suddenly I was trying to build a refrigerator from scratch!)

I went back up to the top of this topic and saw where you're planning on using 2 or 3 layers of fiberglass batting, R value 26 - 39. Arbitrarily choosing 30 as the R value with all that batting but using the same area (A) and temperature differential (D) that Chris used, I come out with f = 72.5 * 45 / 30 = 109 btu/hr to keep the interior temperature at 35F when it's 80F in the basement. That's a bit better than 325, and translates to 7 lb/hr of ice, again assuming that the ice starts at 0F and discounting the extra heat absorption from the ice actually thawing. Clearly, this is where your R value gets critical.

But if you're going to use ice, you really need to be able to change it out without letting the heat back into the chamber.

I hope some engineering prof stumbles over this topic and uses it to inspire her students. Practical applications of engineering principles are so much more fun than abstractions.

I see C. Sapidus posted a link to the missing number while I was writing. Have fun!

[robyn]

Since your whole pig fell through - and the meat locker doesn't look like it will be finished in the near future - I recommend doing pulled pork. There are lots of regional variations - and perhaps you could do it a few ways. No food safety issues - worrying about R values - etc. I make it all the time (almost always start with a Boston butt). It is usually a crowd pleaser. This will undoubtedly sound sexist - but sometimes guys are more into "process" than "result". And it's very hard to go wrong with pulled pork. Robyn

[sparrowgrass]

To find a hog, call the Extension offices in the counties in your area--they will surely know a farmer or a 4-H'er with a pig for sale.

Use this LINK to find the local offices.