Manufacturer Published Electrical Data

[fiberstress]

I'm still relatively new to the field, but for a while now I've wondered how manufacturer electrical data seems to defy the law of conservation of energy for A/C units.

For example, a 20 ton split system we just put in has a published MCA of approx 60 amps. That equates to about 50kva, which in no way can do the 20 tons the unit is rated. It'd be lucky to hit 15 tons. They're all like this; heat pumps, split units, Trane, Carrier, whatever.

What secret haven't I learned yet?

 

[MintJulep]

http://en.wikipedia.org/wiki/Coefficient_of_performance

 

[fiberstress]

Whoops! Forgot about the energy received from the cold space...

A better one-word or one-link response would have been 1st law of thermo, but I got it, thanks.

 

[fiberstress]

As a follow up, in my example you could say that the work input is about 75% of the tonnage of the unit. So the heat rejected in the condenser coil should be about 1.75x the tonnage of the unit. Can anyone add to my confidence by saying they've actually measured this?

 

[DRWeig]

fiberstress,

I think you missed a bit. In your tonnage range, you should expect about 1.2 kW input for each ton output. That's an EER of 9.5 or a COP of 2.8. I'm working with the code minimums here, so yours might be less than 1.2 kW/ton.

In your 20-ton case, you'd have 24 kW (81,912 btu/h) input and 240,000 btu/h cooling. So input power is only about 34 percent of cooling capacity, not 75 percentl.

Your condenser has to reject the 240,000 btu/h plus the compressor heat (heat of compression plus inefficiency due to friction, etc...) that goes into the refrigerant stream. Ambient air takes care of the motor's waste heat, so about 7 to 8 percent of the 1.2 kW is lost in that manner. The rest of the work goes into the refrigerant and must be rejected by the condenser along with the heat gained in the space. In the 2.8 COP case, you'll have 1.2 kW x 0.92-ish or 1.1 kW to reject along with the space heat contribution. That gives 1.1 kW x 0.92 x 20T x 3413 btu/kW / 12000 btu/T or 5.76T. Your condenser has to reject 25.76T in this case. That's only 1.29 times the cooling capacity.

Large, efficient centrifugal refrigeration compressors get down into the 0.6 kW/T range, so a 1000T unit may need a cooling tower that can reject 1000T from the space plus 0.6 kW x 1000T x 3413 btu/kW / 12000 btu/T or 170.6T. (These machines have the motor in the refrigerant stream, so all heat goes there). End result is a cooling tower of 1170.6T or 1.17 times the cooling capacity.

I hope I haven't goofed that up. Wait for a few critics to reply before accepting my answer. I'm working on a Saturday with a body full of Red Bull and my mind is sometimes not right in these circumstances.





Best to you,

Goober Dave

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[EnergyProfessional]

Not sure what the question is. A device with 20 tons cooling is rated at 60A (50KVA) and all manufacturers have similar numbers? Maybe they are onto something.

If you do the math and have 1.2 kVA/ton you get 24 kVA. Now you wonder why you need 50 kVA per manufacturer?
Well, the manufacturer can rate the device whatever way they want. They also include safety, startup current (when the compressor starts it draws more than just 24 kVA like your car needs 150 hp to accelerate, but when you reach the speed it only needs 30 hp).

The electrical rating doesn't have much to do with actual consumption.