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herefishy
25-02-2002, 07:02 PM
When calculating new load requirements for a space to be maintained @ 28degF, which currently is being amply maintained at 35degF, a factory engineer and I were comparing notes and scenario-building in regard to usage of entries.

I had stated that I had derived a refrigeration load of 150,000 btuh. In one scenario, the factory came up with 200,000btuh in a case regarding entry scenarios (time doors are open). During this exchange, I stated that the space is currently being maintained at 35degF with approx. 95,000btuh refrigerating capacity in service.

At that point, the (factory) said, "WEll, if your maintaing 35degF with 95,000 btuh capacity, you are certainly in the ballpark at 150,000btuh for 28degF. Furthermore, with that information (he) felt very comfortable with our conclusion resting at the 150K mark.

Do you think that you can base loads upon existing operating conditions and extrapulate with some accuracy, the load at another condition, or requirement as illustrated above?

herefishy
25-02-2002, 07:39 PM
Performing the load calc from scratch you still recommend. However you are suggesting that several scenarios be constructed, and then graph them and check the curve?

Could you merely plot an air enthalpy diagram (perhaps) and come to some result?

herefishy
25-02-2002, 08:25 PM
Okay, Now your heading the way that I am interested. In regard to freezing, the room temp is to be 28degF, but the product is beef, which we do not intend to freeze (what is freeze point, 26degF, or less), so "degF below freeze point", would not apply (all load sensible). But I presume, as you seem to have hit on, that the air enthalpy would change, regardless.


(90-25)/(90-35)*90K = 106,363

Let's try it -

(50-28)/(50-35)*95,000 = 139,270

50 = infiltration air @ doorway (adjacent refrigerated rooms). Actually, the rooms adjacent to the space are kept at less that 45degF. The space is about 35' x 45' x 25' high ceiling.

herefishy
25-02-2002, 08:28 PM
product load = approx. 30,000 pounds beef daily. My customer assures me that it comes in at 32deg. I called it 40degF. You know, the customer is usually very optimistic about his product conditions!

herefishy
25-02-2002, 08:43 PM
LOL!!!.....

Usually, I type, post, and then edit fourteen times!!!:p

I'm going to try out your little formula there, and I'll get back with you so we can see how it works out.

I like it.:D

Andy
25-02-2002, 09:19 PM
Hi, Folks sorry for butting in. I fired up an program called Calcrite, this gave me the following.

Wall load 4" poly 95F ambient = 39766 Btu's

Air/infiltration load 93F ambient = 19862 Btu's

Product load @.77 Btus/Lb/F 50F to 28F = 25410 Btu's

Lights/fans/2 fork truck 4 people 12Hrs = 44573 Btu's

Add 10%

Total Load 142572 Btus for a 16 Hrs run time .
Regards. Andy.:)

Andy
25-02-2002, 11:22 PM
Hi, I used the load program, load programs only estimate laods when you input data. A true room laoding is only arrived at by experience. If the customer has a similar room under similar conditions adding to the room, or in this case dropping the temperature a rule of thumb is the way to go. When calculating postmorten carcase chills for the meat trade I may use a load program to model values and different conditions, but I usually check my load by saying 1.3Kw for every carcase. A 200 carcase chill will give me a loading of 260Kw to be pulled down every hour for 22 Hrs. I then size the coolers for a 6C td, this gives me a nice capacity and a large cooler surface area to play with minimising carcase weight loss.
Regards. Andy.

herefishy
26-02-2002, 01:15 AM
Okay, fellers...

I had to reduplicate my figures from the initial evaluation that I made 3 years ago for this customer with his 95,000btuh application @35degF. this is what it WAS:

old = 35degF room new=28degF room
all figures @ 18 hr runtime (if you want to know why, ask)

Wall heat gain (old = 20,880 btuh) (new 24,669 btuh)
sensible infiltration (old = 2,789 btuh) (new = 2,511 btuh)
Latent infiltration (old = 1,331 btuh) (new = 1,697 btuh)
Product Load (old = 5,833 btuh) (new = 47,060 btuh)*
*more entering volume included
misc load (old = 47,531 btuh) (new = 54,840 btuh)

total old load = 78,364 btuh
total new load = 130,777 btuh

Okay, this is basic, I've had to reconstruct it, and everytime I re-entered the data into my "program", I always had something not exactly right.

Oh?, you ask how my latent infiltration can be so low? One entire wall (40,000 sq. ft.) adjoins a 0 to -10degF Freezer.

Sensible infiltration seems way too low to you? All entries are into 40degF processing and Dock areas (plus one entry into freezer).

Fridgetechs proposed formula -



Load A = [Amb-RoomTemp-new] / [Amb-Roomtemp-old]*(Xmissionload-old+infiltrationload-old]

load A (28/35)*72,531 = 58,024 ??? maybe it should be...
(35/28)*72,531 = 90,663 :D that sounds better to me.



Load B = [product arrival temp-room temp-new]/[ProductArrivalTemp-RoomTemp-old]*Product Load-old

Load B 12/5*5,833 = 14,000 [3x for increase in product entering] = 42,000 btuh

Load A + Load B = 90,663 + 42,000 = 132,663 btuh

Fridgetech!!!!..... you're TOO COOL FOR SCHOOL!!!!!:cool:

I know that I jumbled a few variables like what's up with the damn 3x increase in product load... etc. But let's consider this for a moment...

herefishy
26-02-2002, 02:47 AM
Originally posted by Andy
I usually check my load by saying 1.3Kw for every carcase

Just for clarity, the meat entering the plant is semi-processed. NO CARCASSES, and the meat entering the room is packaged. In my program, I kept running into the Latent load of (freezing) beef (freeze point @29degF), and we aren't freezing the meat. So, I plugged "meat", instead of "Beef, Fresh", in order to eliminate the latent load of freezing (default @ 27degF Freeze point), in order to derive my calculated load.

I like that you (Andy) pointed out the T.D. in regard to the product weight loss thingy, however My design criteria is 10degF T.D. to maintain approx 90% R.H. (standard miscellaneous applic.).

herefishy
26-02-2002, 03:50 PM
Originally posted by Fridgetech
[36-28]/[36-35] = 8 = 800% = true increase in product load.

[50-28]/[50-35] = 22/15 = 1.46 = 146% = unrealistic estimate of increased load.

You see what I'm saying, it's the opposite to the conventional thinking with load calcs.


Oh, I see.....

Well, I did say I was specing for 40degF entering product, but I lowered that to 35degF in the above formula. I didn't realize that I was fattening it, that wasn't my intent

The more capacity I spec., the more conservative I am (in saving my butt). The cheaper that I spec., the more liberal I am (in losing my butt).

Again, I found it necessary to adjust certain variables to stay within reason. I like to "fatten" the load to add extra cushion for safety and the reality that I cannot control, however I must balance that by rationalizing the expense to the customer if I spec 200,000 btuh systems for the 150,000 btuh applications.

On face value, I can understand the concern regarding the results. The criteria for the (old) load was different. What changed? The United States Department of Agriculture now has an office inside this plant. New Food regulations (and a specific law, thereof) has more stringent requirements which I understand to be more stringently enforced.... so I decreased the entering temperature for the (new) load.

I do not think that I polluted the formula however, because I adjusted the product load accordingly.

How about this:

(35-28)/(40-35) = 1.40 ?????

Andy
28-02-2002, 10:41 PM
Hi, Marc the 1.3 Kw thing is all encompasing 200 carcases would be 260 Kw cooler duty at a -6c evaporation. Here is a load calc done the long way, which can be then checked by the above. This is for a chill I priced today, one of a complete project which has a total cooling load in excess of 1.5Mw

Wall Load
24m*6m*6m(height)
5kw

Infiltration Load
25.2kw

Product load
150 Cattle Carcases @410kg each 61500kg
135.2kw

Lights/fans/personell
2.7kw

Total including 10% safety factor
185kw

Check by rule of thumb 150 Cattle*1.1kw to 1.3kw=165 to 195kw

The above is for a -1c room 22Hr pull-down and a product brought from 37c to 4c deep ham temperature.

Regards. Andy.:)

Andy
04-03-2002, 10:21 PM
Hi, Marc average carcase dead weight is 370-410Kg live weight is 620-680Kg. The dead weight is a product of the live weight aprox 60% kill out. Lambs are generally taken as 20Kg dead 40Kg live weight.
Regards. Andy.