Help about thermodinamics!

[nike123]

Guys (and girls, if any), I need to calculate how much heat is extracted from, say, pork carcass which is cooled from 10°C to -24°C in, say, 24 hrs.
I know specific heat of 0,62 Btu/lbs above freezing and 0,31 Btu/lbs below freezing and freezing point of -2,22°C.

What is mathematical expression for that calculation.

Sorry, I am little rattle brained at thermodynamics.
I searching net last hour, but not finding what I need.

[US Iceman]

Sensible cooling load above freezing = M X cp X ΔT (cooling above freeze point)

Latent cooling load at freezing = M X Δh

Sensible cooling load below freezing = M X cp X ΔT (cooling below freeze point)

M = Mass of product being cooled/frozen

cp = Specific heat of product being cooled

ΔT (above freeze point) = initial product temperature - product temperature at freezing point

ΔT (below freeze point) = product temperature at freeze point - desired final product temperature below freezing point

Happy calculations...

[nike123]

Thanks Us Iceman. I have book on thermodynamics of 1000 pages and this way is much easier!

[US Iceman]

You had a book!

Well, then it that case forget what I wrote.

Really, I'm glad to help.

ps. was this homework?

[nike123]

You had a book!

Well, then it that case forget what I wrote.

Really, I'm glad to help.

ps. was this homework?

No, it is not.
I have some discussion on local newsgroup and I needed to calculate that for discussion purpose.

[US Iceman]

Make sure you charge them for the engineering time....

[Peter_1]

Guys (and girls, if any), I need to calculate how much heat is extracted from, say, pork carcass which is cooled from 10°C to -24°C in, say, 24 hrs.
I know specific heat of 0,62 Btu/lbs above freezing and 0,31 Btu/lbs below freezing and freezing point of -2,22°C.

What is mathematical expression for that calculation.

Sorry, I am little rattle brained at thermodynamics.
I searching net last hour, but not finding what I need.

Rough calculation in SI units
10 K x weight in kg x 1 kcal x 1.16 (kcal ->W)
Weight x 80 x 1.16
24 K x weight x 0.8 x 1.16

Result in Watt if you want to cool this it in 1 hour.

These are figures for water, therefore this will give you a very good estimate of the total heat load of the product.

So 1000 kg pork meat
(10 x 1000 x 1 x 1.16) + (1000 x 80) + (1000 x 24 *0.8 x 1.16) = +/- 114 kW, so in a night of 8 hours you will need 14 kW to cool this down.

I use for a rough estimate multiplying this figure x 1.25 to have an idea of the total cooling I need to install.
So for the 1 ton @8 hours , I should install something in the range of +/- 18 kW@-10°C to -15°C .
I will recalculate this accurately but I'm having already a very good idea what to expect

[nike123]

Rough calculation in SI units
10 K x weight in kg x 1 kcal x 1.16 (kcal ->W)
Weight x 80 x 1.16
24 K x weight x 0.8 x 1.16

Result in Watt if you want to cool this it in 1 hour.

These are figures for water, therefore this will give you a very good estimate of the total heat load of the product.

So 1000 kg pork meat
(10 x 1000 x 1 x 1.16) + (1000 x 80) + (1000 x 24 *0.8 x 1.16) = +/- 114 kW, so in a night of 8 hours you will need 14 kW to cool this down.

I use for a rough estimate multiplying this figure x 1.25 to have an idea of the total cooling I need to install.
So for the 1 ton @8 hours , I should install something in the range of +/- 18 kW@-10°C to -15°C .
I will recalculate this accurately but I'm having already a very good idea what to expect

Thank you Peter_1!
I was calculating that in somehow smaller quantities.
Discussion, in what I need that, was about domestic freezer and how much of influence to his annual power consumption have loading of fresh goods to it.

How did you got this 80 number?

Weight x 80 x 1.16

Also, when I convert Btu/lib*°F to kcal/kg*°C
i got same numbers 0,62 and 0,32.

1 Btu (th)/pound/°F = 1 kilocalorie (th)/kilogram/°C

Here are conversions page.

Using Keeprite Calc-Rite it returns 2617 W for product load for pull down time of 24 hours or 62805 w for 1 hour.
I still looking to find enthalpy's for pork carcasses (moisture 49.8%).

[Peter_1]

If you do the calculation once with Coolpack, you will see that my estimate is pretty correct.
For 80, I multiply with 100 and subtract 20% of this 0 number.
80 comes from the latent heat of fusion of water:http://en.wikipedia.org/wiki/Enthalpy_of_fusion see applications

[icecube51]

yo guys, be aware for meat rot. when you freeze to fast and to deep at once it become frozen on the outside but still soft on the inside.you get dry meat and making a isolation around the inner meat and it starts to get bad. each product has his delay time for freezing.

Ice