How do you calculate the work reqd for freezing meat from normal temperature? What would be the capacity of a freezer to freeze 1500 kg. meat?
3
-
1You should post this on physics.stackexchange.com, listing all the necessary details (starting and end temperature, thermal capacity and the mass of your meat, freezer specs etc.).– RykerMar 9, 2015 at 7:18
-
2Oi, that's not an easy one to answer. If you wanted to make a precise calculation on this you have need to specify the size of the different cuts, since smaller cuts would increase the total surface of the meat and thus facilitate the transfer of heat. I think you need to supply more data before you can get a truly helpful asnwer tot his one.– Richard ten BrinkMar 9, 2015 at 7:19
-
Richard is correct. Beside the heat transfer part, the cut shape will determine the packing density and so the final volume needed. If this is a theoretical physics problem, it will be very hard to calculate. If you are running a business and so don't need an exact figure, but a ballpark one with some puffer, it is answerable here.– rumtscho ♦Mar 9, 2015 at 9:38
Add a comment
|
2 Answers
For water it's 1 Kg = 1 litre. Meat is around 75% water. So for most foods you can roughly say 1 Kg = 1 litre. Give or take some, plus packaging materials and voids in packing
So 1500 Kg meat is going to be up to 2000 to 3000 litres!
Freezers compressors are rated by their ability to cool quickly only at much smaller capacity than the total capacity of the freezer. e.g. A typical 350 l freezer would only be able to cool 30 ish Kg of fridge cold meat to frozen in a day, many freezers are not even this good!
Racks and shelves help with uniform cooling, but shifting 1500 Kg of meat from 4°C to -20°C is going to take some serious work
Roughly; shifting 1 Kg of water -20C uses about 1.4 KW of heat removal. So 1500 Kg is 2100 KW. A freezer is 2.5 to 3.5 efficient due to heat pump, so 2100 KW / 3 = 700 KW of electricity. Your typical domestic freezer is rated at 1 KW, so it's going to take 29 days to freeze!
1 calorie is the energy needs to heat or cool 1 ccm of water 1 degree C in one second. Let's simplify this question by just saying meat is water...no it's not correct, but I don't know how to handle meat as far as how much meat can a calorie warm or cool by 1 degree, and I think it's close enough to at least get a practical even if not mathematically accurate.
1500 kg of meat is a LOT of meat. that would ordinarily be kept in a walk-in.
assuming 1 cc of meat = 1 ml = 1 gm (see my meat and water discussion above), 1 kg of meat will take 1000 calories to cool by 1 degree. 1500 kg would require 1,500,000 calories to cool by 1 degree.
How many degrees of cooling do you need to freeze your meat? Shall we assume the meat is at 4 degrees celcius (40 Deg F) for chilled meat or 15 degrees celcius (60 degrees F) for room temp meat?
Lets start with chilled refrig temp meat.
Freeazing = 0 Deg, but to freeze something we need it to be below 0, usually something like -2 or -3, but we'll stick with 0 since that's what the "meat" needs to get to.
40 degrees X 1,500,000 calories = 60,000,000 Calories...that's right sixty million calories.
1 calorie = approximately, but not accurately 4 watt seconds (I am rounding for ease since this is an order of magnitude discussion really than an exact math).
4 x 60,000,000 = 240,000,000 watt seconds = 4,000,000 watt minutes = 66666 watt hours = 66.6 KW/hrs
This problem is also why meat is usually kept frozen or chilled until needed because freezing that much solid meat takes quite a bit of time and energy.
