Heat gain associated with refrigerators

[MechScotty]

I am doing my first restaurant and have a question about the amount of heat gain to figure for the coolers and refrigerators.

At first I thought it might be tied to the watts consumed but then the cooler absorbs heat from the surrounding area. Then I thought it might be the wattage divided by the EER of the system. If the EER is 5, the BTU's added by the cooler can be calculated by the wattage used divided by the EER. Any tips ?

What is the typical EER of coolers in commercial kitchens ?
Scott

 

[cdxx139]

We covered this in
thread403-298715

knowledge is power

 

[MechScotty]

Thank you but let's back up. I am not a mechanical engineer, rather an architectural engineer (not a choice when signing up)doing an energy audit for a friends restaurant.

I haven't access to ASHRAE stuff. It would seem to be a simple calculation since I've measured the wattage of the units over a period of an hour. Pardon the use of the EER, COP is the appropriate term.

Is there a simple relationship between the energy consumed, the COP and the heat gain ?

 

[DRWeig]

MechScotty,

The watts you measured = cooling load. Think of it this way, the power came into the refrigerator via some wires. How does it leave the refrigerator? As heat.

There's no bonus because the insides are cool, and no bonus for COP (or EER). Watts in = heat out. Just multiply the hour's watt-hours you measured by 3.413 and you'll have BTUs.

Note: your measurements may not be typical. Refrigerator rarely runs when it stays shut with cold food in it. However, when opened and warm stuff placed inside, it'll run continuously for a while. Internal refrigerator load has an effect on input power.

Good on ya,

Goober Dave

 

[MechScotty]

Thanks Dave,
So all the carnot magic happens inside the box. That makes sense. I measured during the heat of the day and full business activity (doors open a lot).
Scott

 

[DRWeig]

Sounds good MechScotty, you had it loaded pretty well.

If you do find the ASHRAE handbook, it'll give you a bit of a break over what you probably measured (that is, a lower heat load on average). They have done lots of study on diversity in kitchen equipment.

Sorry I can't quote you the ASHRAE answer, I'm nowhere near the darned thing this week.

Good luck,

Goober Dave

 

[MintJulep]

Are the condensers in the conditioned space?

 

[DRWeig]

Ouch!

Good question that I neglected, Mint.

Goober Dave

 

[Waramanga]

Well if the condensers are outside then the heat 'gain' has a -ive in front of it!!

 

[KiwiMace]

Some scratchy thermo going on here, but Mint has saved the thread.

A normal refrigerator has the condenser integrated with the unit meaning that the heat load to the space is the sum of the power used and the heat removed from the inside. All this heat is dumped out the condenser into the space.

What makes this a reasonably pointless exercise is the usage factor - what you really want is reasonable data for how often the compressor is actually running in real life.

ASHRAE lists very little for refrigerators, and is not specific if this is a remote condenser, but I would assume not.

Chap 18, Table 5A (2009 Fund.) has a 1407W (Nameplate) reach-in fridge emitting 352W, making a 25% usage factor (based on nameplate).

 

[DRWeig]

Kiwi,

You're right in a much more thorough way -- as usual. To simplify, I've always just used motor power times a usage factor (such as your 25% figure), since in steady-state (food all cool, door not opening a lot), the heat rejected by the condenser in excess of the motor energy is (approximately) equal to the heat gained by the interior of the box from the room. It's when you load the box with warm food that the condenser heat becomes a significant room load for a while.

I still back up to the basic (ignoring all transients): The electric watt-hours expended by the machine all end up in the room eventually. The small bit of cooled space (and food) is a one-time difference.

If I'm missing something significant, please lead me toward the light! I'm a perpetual student, after all. Plus, since I learned all my thermo after college (I studied electrical engineering), someone may have led me astray...

Good on ya,

Goober Dave

 

[KiwiMace]

In steady state, the heat comes from the room and is discharged back to the room as you say. My high food replacement rate scenario is, on reflection, a poor assumption too. Most likely to be somewhere in the middle.

It's too hard to guess, you need field data - most people are unlikely to want to determine this themselves. I would take 25% of nameplate and run with it. The fridge is such a minor part of a kitchen load that I doubt any reasonable +/- error or padding would make a difference to the overall load and margin.

 

[DRWeig]

Thanks Kiwi, and I'll echo your comment about refrigerators being a small portion of the cooling load.

Goober Dave