Heat load calculations Help!

[jbomba]

Ok im having a few dramas calculating airchange load. Product load & defrost load. Here is what i have so far. Im calculating over 24 hours.

The room is loaded with 1500kg of product per day
The storage temp is -35
Temp of product entering the room is 35 degrees
The specific heat of the product above freezing is 3.182kj
The specific heat of the product below freezing is 1.717kj
The freezing temp is is -1.17
The latent heat of fusion is 233.48kj
there are 6 x 20 minute defrosts per day with the heater elements totaling 10Kw
Ineed to allow for 2 complete air changes

here is what i have so far

Airchange Load = IRVxACFxenergyxusage/time

IRV -LxWxH = 15x10x5 = 750m3
Now i have not been given an ACF(or is that 2^ )
Energy - 60%R.H, - 35 degrees storage =0.1045
Usage Factor - im assuming it is 1
1 hour = 3600 seconds x 24 = 86400 seconds

So i have gone :

IRVxACFxEnergyxUsage/time

750m3 x (2?) x 0.1045 x 1/86400= 1.814236111Kw

Now for defrost load

10Kw x 2/24=0.83Kw

Now for the product Load

Q=m x cp x deltaT

m=1500kg
cp=3.182 - Above freezing
1.717 - below freezing

delta T - 35+1.17=36.17 - Delta T above freezing
-35--1.17=33.83 Delta T below freezing

Pull down time 24hours =86400 seconds

so 1500kg x 3.182 x 36.17=172.639/86400=1.998Kw
1500kg x 1.717 x 33.83=87.129/86400=1.008Kw

1.998 + 1.008=3.006Kw

how do i work out the latent heat?

Thanks for your help here guys. I cant be to far off can i?

[jbomba]

anyone?:rolleyes:

[B G Scott]

anyone?:rolleyes:
Google and locate "Coolpack" its a free download and has all the information you need.

[mad fridgie]

1500* 233= 359500Kj/3600=97Kwhrs
97/ 24=4.045Kw. plus all others.
But I would not use the total rating as the rating for your condensing unit.
(Total *24)/18 run hours.
Other issues that need to be considered is this just freezing product or is it also longer term storage.
You need to to size your condensor for a higher evap temp (SST) and/or install CPR.
There are also limits on the product
If it is one big lump then will not freeze in 24hrs, but if very small bit will freeze easy, raising room temp.

[jbomba]

Google and locate "Coolpack" its a free download and has all the information you need.
this is just an exercise im doing at TAFE so i just wanna get my head around the actual calculations. Any help would be greatly appreciated

[jbomba]

can anyone help me nail this one?

[lana]

Hi there,

Latent heat unit is kJ/kg NOT kJ.

So : m x LH /86400 = 1500 x 233.48 /86400 = 4.05 kW

What seems to be the problem?

Cheers

[mad fridgie]

May be your are not understanding Kj/Kg this is energy, there is no time involved, so you have to work out total energy then divide by a time you want this to happen in. But you then have to include defrost and off times, which increases the refrigeration capacity over that of your original calculation