Electrical load for walk-in freezer

[Eng. AG]

Hi! I'm Eng. AG and I just signed in. I'm a design engineer at a consultancy firm and I'm fairly new. I wanted to ask help from you guys since you are experts in the field of refrigiration. Here is my question:

How can I compute for the electrical load of a walk in freezer?

Here are the facts about the walk in:

Size: 15m (L) x 7.5m (W) x 4m (H)
Stored products: Meat products, fish, poultry etc.
Storage temperature: -18C to -10C
Lights: 10 units of 2 x 40kW lamps

I've computed for the cooling load using Kent's handbook and turned out to be 3.5 TR. My senior said it's ok (then again, he's not an expert in refrigiration so I'm not really sure) and I've assumed that the compressor would have a 60% efficiency therefore 3.5TR=17kW turns into 21kW. Then I assumed that the compressor consumes about 60% of the electrical load, which makes 40% of the total diverted to fans, system pump etc. which turns 21kW into 35kW.

As it turns out, the electrical engineer wants to crucify me for this value. Obviously I did something wrong and I'm not really sure what it is. Please help me.

Best regards,
Eng. AG

[luwajoe]

I'm not believing that your lights are 40kW. I would believe instead 40 watts.

I agree with your electrical engineer that your calculations are high.

[Peter_1]

Most important things we don't know:

product entered....how much and on what temperature and required pull down time?

Outside temperature?

Panel thickness and isolation material?

You first need your cooling load to be correct calculated.
Then you just select a compressor, condenser and evaporators and you take the absorbed power of all those. You don't have to use an efficiency ratio for this because you have the correct figures in the brochures.

Same remark as Lumajoe: 40 kW of lights???:eek:

20 kW refrigeration capacity sounds better if products entered already frozen = +/- 12 kW absorbed and add +/- 5 kW fans = 17 kW total or 50% of your value
This is a rough estimate: (15 x 7 x 4.5 x 50 W/m³)

[smpsmp45]

Typically in a cold store heat load the load due to electrical lights is very small & we by thumb rule say it must be about 2 to 3% of the total load.

The Storage temp must be -18 deg c.

Even you shall need atleast 10 Tube light fixures with 2 x 36 watts tubes for that much area.

[Eng. AG]

Thank you all for your reply. I admit that even in my post I have made an error of stating that the lights are 40 kW when in fact they are 40 W. Thank you for the correction luwajoe.

I've researched into books further and found that scroll compressors have very high efficiency (which makes 60% very unfair to give to a scroll compressor). From what I've read 95% is a safe assumption. I had the instinct of giving 60% efficiency due to past water pump calculation experiences:off topic:.

I did assume that the product was frozen when entering because they (user) dont have a slaughter house. They just really need to feed a lot of people :) that's why they have this large freezer (3 of 'em in fact). I also assumed that the product would be pulled down (-18C) in 24 hours. Regarding the insulaton, I really have no idea what is to be used but I will try to look into that a bit more. Thank you Peter and smpsmp.

I think I went too far in computing with the compressor efficiency though. I should have stopped at the heat load value, yes?