What is the formulae to calculate the refrigeration KW to freez food 1kg, +35Deg C to -18Deg c in 3.5 Hours?

The question you are asking is moderately complex. I am in the U.S., and am familiar with non-SI units.

First, in freezing your product, you must determine its freezing temperature, which will be somewhat less than 0 dC, owing to non-water components of the product. This temperature might be about -2 dC (depending on the particulars). You will be cooling the product in three stages: sensible cooling from 35 dC to the freezing point, latent (change-of-state) freezing through this transition, and finally sensible-heat subcooling to -18 dC. These values you will need to look up in a table or other data base. Add up the total joules/kg or whatever other units; multiply by the kg/hour to be processed; this will give you the total heat removal required per hour.

Then there will be the design of the refrigeration system. You are aiming for -18 dC final temperature of the product; thus the evaporation temperature of your refrigeration system must be less than that. As a practical matter, this means about -26 dC. Then you need to determine the condensing temperature of the refrigeration system. If air-cooled, this will be about 15 dC higher than the design dry-bulb temperature of the location of installation; if evaporatively-cooled, it will be about 10 dC higher than the design wet-bulb temperature; if water-cooled, it will be about 7 dC higher than the water temperature available.

Then you must examine the compressor manufacturer's literature to find a compressor (or combination of compressors) that equal or exceed this capacity. This literature will tell you the kW required. To this you need to add similar information from the condenser manufacturer (for additional pumps and/or fans). And you must also select some air units (evaporators) of suitable capacity, and add the kW of their fans. Only then can you determine the total kW required for the process.

That's rather abstract, I must admit, but that's what refrigeration is all about. (And there are some other details I have omitted, such as air infiltration and wall insulation leakages into the freezing chamber.)

Dear Tornado

Thanks for the detailed answer. One thing I left to mention. Actually the chicken is pre chilled down to +4 Deg C prior to Freezing. So As I understand from your advice for freezing I must do a calorie calculation from +4 to 0, then Latent heat, then 0 to-18 Deg C. Then make a calculation for an hour with the total weight to be frozen (in an Hour)! Kindly comment.

Please could you give me the formulas to calculate these three steps? No problem if the units are US.

Highly appreciating the support

Hari

Thanks again Tornado

It is very clear now. I will get into detailed study now following your suggestion. Great answer.

Hari

Poultry meat is preserved both by chilling and freezing. Former requires lower capital investment and later offers quality and flexibility of operation. For freezing use of air-blast tunnels operating at air temperatures of -29 to -40 C and air velocities of 2.5 m/s or more is recommended. For load calculation you have to calculate sensible cooling and latent cooling. Sensible cooling is calculated by mCDT. M is mass in kg. C is SP heat and DT is the temperature difference. Latent heat is by mL. L is the latent heat of fusion. You can refer any refrigeration book these simple formula are given.

Informative answers by Tornado and mrswamy, since I've done comparative calculation I'll post it anyway. If you consider large part of most foods is essentially water and since Specific heat (C) of water is higher than most if not all food stuff this an assumption on safe size, ie answer will be higher than what may be needed in reality and will give margin for error. Specific heat of water C(H2Oliq) ≈ 4.19 kJ/kg/K, Latent heat of fusion (water changing from liquid at 0 deg to ice at 0 deg) = 334 kJ/kg, Cice = 2.05 kJ/kg/K, Energy to be removed is given by Q=m*Cliq*DT+m*Lf+m*Cice*T [J] Or calculated on a per kg basis Q/m = Cliq*DT+Lf+Cice*DT [J/kg] Thus 4.19x35+334+2.05x18 = 146.65+334+36.9 = 517 kJ/kg Interesting to note that cooling ice below 0 deg requires comparatively little energy to be removed To remove said amount of heat assuming (wrongly) that there is no additional heat ingress, you want to get rid of this heat in 3.5 hours which is 12600 s. Required heat removal rate will be approximately 517/12600 = 41.03 J/s/kg or W/kg. Good luck. WR

thanks to White Rock

I will revise again according to this now. Meanwhile I am trying to get more practical figures from ASHRAE too.

Regards

Hari

Starting from the simple, 1 calorie is defined as the energy required to heat 1 g of water by 1 degree Celsius. Food is not all water, but the other components have a lesser heat capacity, so designing the freezer for a 1000 g water volume is erring on the side of slight overcapacity. The energy required is 1000 x 53 calories or 61.6 watt-hours. Over 3.5 hours the power required is 17.6 watts. Not a lot, in other words, even allowing for an inefficiency in the freezing process. Peter Harris

Thanks a lot Peter Harris

I am opt to go for such a good margin for safety

Regards

Hari

Please refer previous comments. You can not simply multiply the temp difference with mass.During the phase change from liquid to solid, energy will have to be removed without any change in temperature which is known as latent heat of freezing. Again you may refer the previous threads- the specific heat is different in liquid and solid state of water. Therefore need to be calculated separately and then added to get the result.

Toatally agreed and understood. The above suggestion already inclusive of the latent heat as I observe. Correct me if I am wrong

Thanks for responding and correcting

Hari

This has become an excellent discussion, and everyone has added good input. Your own acknowledgements will have been gratefully received by all contributors. By asking good questions (as you have done), getting good answers, and digesting them well, you should progress nicely in your work. If you have further questions, please feel free also to email me at tornado[at]kpunet.net. (It is a bad idea to give email addresses in forums like this, so I have expressed this in an artificial format. Replace [at] with @ to get the correct address.) Good luck!

Another thing I will mention (following "mrswamy") is that it is a good idea to chill your product to about 4 dC before sending it to the freezer. For best quality, you want the product to transit the freezing part of the process as rapidly as possible. This prevents the formation of large ice crystals as the product freezes, thereby preserving its texture. If possible, arrange the freezer so that air can flow around each individual piece (in other words, avoid putting the product in pans or boxes). This is termed "IQF," meaning "individually quick-frozen."

Thanks Dear Tornado

Ours will be mass production. Two box freezers (May be Eurotech, Jackston or some other supplier) each with a capacity of 4000Kg/Hr trhough put. Each shelf will be ten cartons (uncovered. 1 X 10 bagged chicken/box).

So individual will be difficult in such case I believe. Also the handling before and after will be easy in this case, till market sales.

If there are other solutions, we are well interested.

Chilling from +38 Deg C down to +4 Deg C is already by a Air spray Chill tunnel about 90 Minutes process, before entering the Freezer.

If there are more thoughts please add ( All those CR4 friends and Gurus)

Regards Hari