Is it possible to convert tonnage to KW?
Say I’ve got a rooftop unit that is listed as a 55 ton unit. The only other number I have is that in the winter the unit draws an average of 12KW per hour to heat with natural gas. The 12 KW is mostly just the main blower and some controls.
Is it possible to estimate the KW output of that unit in the cooling mode?
Thanks.
Multiply tons by 3.517 to get kilowatts.
That seems about what I’ve found on the internet. Thanks for confirming!
Another question, is this based on the unit running at 100% 24/7?
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Why not E=mc ?
Sure, if you convert your AC directly into pure energy.
This is the cooling ‘ton’, which is not a measure of weight or mass.
Wikipedia on ‘ton’: a confusing mess, to be sure:
Highlighting mine.
No. In fact, I would expect that the KW load would be substantially higher when cooling----whether winter or summer.
A unit that large often has an “economizer”; a device that measures the outdoor temperature in the winter and opens a damper to provide “free” air conditioning----like when the temperature outside is 35 F and the building needs cooling. When in the economizer mode, the KW output may be close to the load when heating. (you’re still using the main blower—the largest source of electricity) In some cases the KW will be higher if the unit uses a “power exhaust”; when you go into economizer the building will be pressurized and so the building needs to be “relieved” or the doors will be blown open. Some units use a “Barometric relief” and others use an exhaust fan to force air out of the building to keep the building neutral—and that fan uses more energy.
Now, if I understand you correctly, this unit uses 12 KW to heat using natural gas. If this is so, the ovwewhelming amount of that energy will used by the main blower—the circuits powering the gas train/valve use very little current.
But you will need that same blower to cool----the bulk of that 12 KW. On top of that, if this units uses mechanical cooling (compressors) to cool I would expect that you not only can’t eliminate the 12 KW, it will be higher due to the compressor load.
IOW, if this is packaged gas powered RTU, it will use more electricity in the cooling mode that in the heating mode.
3.5 KW/ton? That seems high.
It’s been a while since I’ve needed to know such things, but I thought that about 1.6 KW/ton was a good estimate to use for air cooled equipment (like an RTU). Of course, everything depends on the operating conditions, but one could expect about 1.1 or 1.2 KW/ton for the compressors, and the balance for the condensor fans and supply fan (the 12 KW fan load is reasonable for a 55 ton unit).
No, it’s basically a units conversion dealing with cooling capacity. Now if you’re dealing with the electricity required to make that ton of A/C, then you can have less electricity per ton (such as in a heat pump). I think the latter is what you’re thinking of.
Yeah, because the OP talked about the 12 KW draw of the supply fan. But I see why I was confused. Thanks.
Um, so maybe the OP could clarify what he wants to know? We can probably give him an answer.
Are you interested in electrical demand of the equipment, or the actual heat removed (or added) to the conditioned space?
Electrical demand of the unit. For addressing the cost to run the thing in the summer.
raindog I appreciate your input and I was hoping you would see this thread!
[everyone else has been helpful too, thanks!!]
However, I think you were preaching to the choir, so to speak, in your reply. I realized that the unit will draw much more in cooling mode than heat mode in KW.
I’ve had a recording meter on the unit for a few weeks now to get an average KW reading for heating. What I wanted to know is;* can we estimate the cost to cool only knowing the tonnage and the KW of the main blower fan?*
Lets say that the unit (at 55 tons) could run at 100% 24/7.
If we convert directly to KW then we should see something like [55 x 12,000btu = 660,000btu] [660,000 x .293 = 193,380 watts] [or 193.38kw]
Since (i don’t believe) these units are designed to run at 100% for long, more like 80% (again an assumption) and then even at 80% they cycle on and off (let’s assume 50% on and 50% off, correct me if I’m wrong, maybe it’s more like 30% on and 70% off in a properly designed facility?) we can take 80% of 193kw which is 154kw and then half of that to get 77kw. With no losses!
Even at 77kw, this is more than 6 times the draw of the heating mode, seems high. Is it possible that maybe there is a simple multiplier that we can use? 6 seems kind of high, I would have guessed maybe more like 2.5 or something along those lines.
Thanks again!
For that, use 1.6 KW per ton as a starting point. I’ve used that in initial designs as a starting point for the electrical engineer to size the transformers required, and I’ve never ended up with an undersized transformer. Thus for a 55 ton unit you are looking at about 88KW.
As you note, these things do not usually run 100% of the time, and we have no way of knowing if the actual heat load of the building is 55 tons or not (from the information given). In fact, the 55 tons is only the nominal rating of the unit, the actual operating conditions (such as outside air conditions, operating setpoints, humidity control, etc.) all have an effect on the actual capacity. One way of determining the actual electrical demand would be to know the the heat load of the building. Another way to do it is to look at the historical electrical demand for a given period of time. Or, if your calculation is not too critical, use 75-80% of the 88 KW, or about 66-70KW.
I wouldn’t just use the electrical demand of the supply fan as a starting point because the fan demand is dependent on the static pressure the fan operates at, and this is a function of the duct system and not the heat load of the building.
Any estmate is going to be a guesstament.
What type of building? What are your outside air temps and how constant.
You can look at the name plate on the compressors, it should give you full load amps. If your building is running 24/7 so you will have the 12KW for the fan. Then you know your building so guess at the average load and run times. Do not forget the condencer fan.
If you need this information now you can only guess. If you can collect the information as you go then install a Nemon Demon to record the energy used.