Product Load

[scottmax]

When working out the product load for a freezer is the following method correct-

(SA x TD) + L + (SB X TD) x PW

PD X 3.6

Where-
SA = sensible heat above freezing
TD = Temperature above/below freezing
L= LAtente heat
SB= sensible below freezing
PW= product weight
PD = pull down period of product in hours

By my recall the top line give me the load in kj/kg and by dividing it by the time give me an answer in watts.

Please any information is greatly appreciated.

[scottmax]

When working out the product load for a freezer is the following method correct-

(SA x TD) + L + (SB X TD) x PW

PD X 3.6

Where-
SA = sensible heat above freezing
TD = Temperature above/below freezing
L= LAtente heat
SB= sensible below freezing
PW= product weight
PD = pull down period of product in hours

By my recall the top line give me the load in kj/kg and by dividing it by the time give me an answer in watts.

Please any information is greatly appreciated.

[QUOTE=scottmax;187373]When working out the product load for a freezer is the following method correct-

(SA x TD) + L + (SB X TD) x PW

PD X 3.6

[Magoo]

Change the 3.6 to 3600,

[lana]

Hi there,

There is a problem with your equation :

Product weight must be "Product weight loaded per day" and therefore; you must replace 3.6 with 86400 which is " 60x60x24

cheers

[scottmax]

Change the 3.6 to 3600,

What if my weight measurement is in kg's. Don't I have to divide both the top and bottom by 1000?

[scottmax]

I am working out a load for a spiral freezer so I only need 1 hour for the time.

[Aik]

(SA x TD) + L + (SB X TD) x PW

PD X 3.6

Where-
SA = sensible heat above freezing
TD = Temperature above/below freezing
L= LAtente heat
SB= sensible below freezing
PW= product weight
PD = pull down period of product in hours

By my recall the top line give me the load in kj/kg and by dividing it by the time give me an answer in watts.

Please any information is greatly appreciated.

There is a mistake, (SA x TD) + L also must multiplied by PW i.e. formula is to be:
((SA x TD) + L + (SB X TD))x PW

PD X 3.6

What if my weight measurement is in kg's. Don't I have to divide both the top and bottom by 1000?

If Latente heat is in kJ/kg, sensible heat above/below in kJ/(kg*Cdeg), then result will be in W.

[Magoo]

Hi scottmax.
Specific heat values are expressed as Kj/Kg/ 'K, using the 3.6 factor gives answer in watts, and 3600 gives answer in Kw.
For a spiral the pull down factor has to be expressed as hours or fraction of an hour if retention time in spiral is in minutes.
Don't forget the base load with spiral, that is fans, drives and insulation. Also depending on production runs the evaporator size for snow/ice building is important as well. Add a safety factor on to refrig capacity as well. Retention time, air direction and velocity, belt coverage, sealed product or open product, thickness, cooked or raw, wet or dry. All things to be taken into account.
Example : 750kg/ hour of fish product, retention 35 minutes, exit temp -18'C, fan Kw 30, refrig capacity 135Kw @ -40 evaporating +35 condensing.

hope this helps. Magoo

[sendhilkmar]

If it is a repeated failure then it might be related to system processing/ charging. Slow vapor charging results in scroll failure after 3 to 4 months. Set the LP not less than 25 Psig. Flush out the system with N2 and charge 70% of the refrigerant in the vacuum then charge the balance in suction side while running.

[sendhilkmar]

Please igore my previous reply.

[ref717]

To clarify more, the correct equation for Product Load calculation should be :

[(SA x TDa) + L + (SB X TDb)] xPW x SLF
PDt X 3600

where : SA & SB are Specific Heats, units are KJ
Kg.deg.K

: TDa = entering product temp. - product
freezing temp. (deg.K)
: TDb = product freezing temp. - final
product temp. (deg.K)
: PW = Prod. weight (Kg.)
: SLF = Safety load factor(1.2),depends on
your design considerations.
: PDt = Pull down (hr.)

After inputing all above parameters, your Product Load shall be in the Kilowatts unit.

In Spiral Freezers heat load calculations, you have to consider many factors as what Magoo have said.

goodluck..

[scottmax]

Thanks for the info.
Would it be fair to say with the experience on this board that a core temp of around -3 C is achievable if I use my calculation to -10C exit temp?
I am cooling small meat pies (its an Australian thing) that weigh around 220grams in total. There are going in nude directly on the belt. Entering @50C. I have worked out a Product Load of 142kW. We are cooling 5106p/hr
The client will be happy with a core of -3 we will get the rest over the next 24 hours in the main freezer.

[Magoo]

Hi Scottmax,
the core temp will drop after a short period after leaving spiral. The temp equibulates, ie evens out. Pies are a strange thing to freeze due to air gap under the lid of pie.
You can reduce refrig load if you have space for an ambient cooler using fresh air to drop temp lower than +50'C after leaving the oven.

[scottmax]

We have made allowance for ambient cooling but in summer up here 35 C is not uncommon. They are only able to stay in the ambient cooler for 35min max before they are shipped out and run through the spiral. They enter the ambient cooler @ approx 85 C and we are guestimating they will loose around 35K in this time. The ambient cooler will be force fed air via a mechanical means. This serves as a double purpose to feed oxygen to the oven also.

[shafaqat ali]

Kool sharing

[scorpx]

Hi all
firstly it is really good topic.

I try to use it but I don't understand some things.
Pls correct my mistakes

I need 750 kg Lamb freezer in 8 hour. To find How many kw I need for freezing ?

SA = I don't know
TDa = 40 - (-18) = 58 C
L = I don't know
SB = I don't know
TDb = -18C
PW = 750
SLF = 1.2

Thanks for help

[scorpx]

Hi again I found all but can't find K ?
What is L ?

[ref717]

K is only a temp. unit meaning deg. Kelvin.
L is the specific heat of the product at freezing temp. or more commonly known as Latent Heat of the product. L varies with product type and can be found on ref. design tables and charts.