Generator Question

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ELWAY

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Mar 12, 2006
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I have a Montego Bay by KZ that has two 13,500 A/C.? Will an Onan 4.0 be sufficient?? Any help would be appreciated.
 
The simple answer is no, you can't run both at the same time with a 4kW genset. Two 13,500 btu units
would require an 8kW (or more) to run them simultaneously. If you're wired for 30A service, that's your limit, but you could use a load-shedding controller that would alternate running between the two units, and keep the current draw below the 30A main breaker trip level.
 
Karl said:
The simple answer is no, you can't run both at the same time with a 4kW genset.

I suspect you are wrong here Karl,,,, 4kw is (in rough terms) 40 amps, and the full load draw on a Coleman Mach 3 13.5btu unit (the first one I found specs on) is under 14 amps, starting should be around 30 amps, it's likely the 4Kw could run them both, in fact there is a good chance it could kick them both in start mode at the same time (But I'd not guarentee that, starting current on induction motors can be waymore than 2x running current)

The 13.5KBTU in my MH are breakered at 20amps each, that's 2400 watts max, each, and as noted, that's STARTING load, not running load.  The generator I have is a Onan 55oo though 5.5KW 7.2 surge

To be safe, I'd suggest the same (5500) for 2 AC's though as odds are he will have other electronics on line, Television, Microwave, Battery charger (10 amps there, breakered at 20)
 
Karl

I have to agree with John. Our Damon UltraSport DP had 2 A/Cs with a 6.5 KW propane generator. It had a delay to prevent starting both at the same time, but we were able to run both off the genny.
 
Bernie,

The original question was running two 13,500 btu units from a 4 kw genset. At full load, there's no way it will handle both of them. I also mentioned load shedding as an alternative. Certainly it would be unusual to require the full output of two 13.5 units, but rather than having someone sweating bullets based on "normal" a/c needs and not having the power to provide for the extremes would, IMHO, be giving bad information and false hopes. I'd rather be conservate and recommend a larger unit than have someone buy a smaller, less expensive unit and be unhappy with it. With all due respect, gensets are not like cell phones that you replace every two years, so I feel it's better to start out with a little headroom, rather than having to upgrade at a later time and taking a big hit on low or non-existent resale value of the old unit.  :)
 
Karl

Yes, you are right that it is much better to be conservative in your response but, respectfully, I believe your response was much too conservative: "Two 13,500 btu units would require an 8kW (or more) to run them simultaneously."  I still say 6.5kW is more than adequate. What wasn't clear iin the original question was whether the 4kW generator was there already or a considered purchase. If the 4kW genny was there already, the answer might be how to best run the A/Cs using only the 4kW. If one was to be bought, what size should be bought to run both and IMHO, a 6.5kW would be adequate. By the way, my cell phone is 4 years old ;D ;D
 
My two 13,500 btu a/cs run nicely on my 5.5 kw genset, without any load management, so the kw-hour/btu formula that Karl cites is somehow not applicable to this application.    Newer "high efficiency" a/c units don't even require 20A circuits - I can start and run one on 15A shore power, with no more than a slight blip in the voltage when the compressor cycles.

That said, 4 KW is cutting things a bit too fine and if I were specing out a new genset for this rig I would opt for a bit more, at least 5kw.


I found another formula for btus-to-KW that yields an answer more consistent with the observed results:

BTUs x 2.93 ? 10,000 = Kilowatt hours

13,500 x 2 /10,000 = 2.7 KW per a/c unit or 5.4 kw total for two units.

Reference:  http://www.csgnetwork.com/elecnotherformulae.html
 
You folks miss one very important point about air conditioners.

You are not generating heat using a resistive element (except perhaps at night)


YOU ARE PUMPING HEAT

There is something called efficency that has to be factored in,,, An AC unit can easily have an efficency factor far greater than 1 (100%) it can be 2, 3 or even six,

For heating: BTUs x 2.93 ? 10,000 = Kilowatt hours and then you need to add in the fans.

But for cooling it's  BTUs x 2.93 ? (10,000 x [Cooling Effecency]) = Kilowatt hours

And the CE number will be greater than 1 so you use fewer killowatt hours to move 13,500 BTUs than you do to generate them
 
John,

If you can get ANYTHING to operate at greater than 100% efficiency, let me know and I'll finance you to the hilt. You've just come up with a derivation of perpetual motion ;D

And you don't 'move' btu's, a btu is simply a measurement of the heat energy required to heat one pound of water 1 degree F. or cool one pound of water one degree F. This excludes phase transition points where it takes roughly 13 btu's to make the same one degree change.
 
Well,  What I mean by more than 100% is that the amount of heat removed from the RV is greater than the amount of heat that would be generated if you passed the same amount of electricty through a resistor.

In truth, you are right, it's not over 100%, 

Try this... How much enegery would it take to, oh, say, grow a bushel of corn

How much would it take to move that same bushel from the store to your car?

The AC, in cooling mode, does not, as it's job, generate heat (actually it does, but tha't s not what you are paying it electricity to do) it moves it from one spot to another... Very efficently.

GENERATING heat, (A job that, by the way, is always 100% efficent) is another matter... Not moving it, making it. and does, accurately follow the calculations given  I will try to find some web pages between now and next go round

Found it faster than I thought... Here is the relevant quote:

"Energy Efficiency of Room Air Conditioners

A room air conditioner's efficiency is measured by the energy efficiency ratio (EER). The EER is the ratio of the cooling capacity (in British thermal units [Btu] per hour) to the power input (in watts). The higher the EER rating, the more efficient the air conditioner. National appliance standards require room air conditioners built after January 1, 1990, to have an energy efficiency ratio (EER) of 8.0 or greater. "

and here is the URL if you wish to read more:
http://www.eere.energy.gov/consumer/your_home/space_heating_cooling/index.cfm/mytopic=12420 

As you can see, they are talking about room air conditioners here and requireing them to move EIGHT times as much enegery as that formula calculates,,, For every "Btu equivlent" watt they eat, they have to move 8 BTUs of enegery.  Very possible,  because they are not changing it from one form to another, they are moving it from one PLACE to another.  all together different kettle of heat  I'll keep looking
 
John,

I never meant this to get into a contest of any sort, but EER uses and arbritrary 95 deg.F to establish a base number to compare all air conditioners against. In this context, one that has an EER of 10 is more efficient than one that has an EER of 8. The EER was established to give the average consumer some basis of comparison of similar units without having to go thru a lot of meaningless (to them) calculations. Higher the number, less cost for the same amount of cooling.

The AC, in cooling mode, does not, as it's job, generate heat (actually it does, but tha't s not what you are paying it electricity to do)

True, as far as it goes. What you're paying the a/c to do is quite simple - move the heat from one place (inside your coach) to another (outside your coach). To do this, it is necessary to have a suitable medium -  the refrigerant. The main job of the compressor is to put the refrigerant under high pressure which heats it up considerably. It is then fed thru the condensor which cools it and thereby changes it from a gas to a liquid, and then thru the expansion valve where it expands back into a gas that not only decreases its' heat density, but also increases its' ability to absorb heat from the air passing thru the next step in the process, the evaporator. The  evaporator, in turn, absorbs heat from the air inside your coach and passes it on to the refrigerant . This newly heat-charged refrigerant (still in gaseous form) is then passed thru the compressor where it is brought back up to a high pressure, then passed thru the fan-forced condensor where it is cooled back into the liquid state; then the process repeats itself in a continuous, unending loop.

GENERATING heat, (A job that, by the way, is always 100% efficent)

John, How can you say that? Take, for example, the striking of a match. Let's assume that the conversion of the thermochemical reaction into heat IS 100% efficient (it's not; if it was there wouldn't be any ash or other residue). It still takes some energy, (the movement of your hand in the striking process) to generate the friction (heat loss) to initiate the reaction. These are all losses that must be taken into account. Nothing in this world is 100% efficient. We, as humans, may sometimes like to think we are, but the fact is that there is always some loss - even if it's only hysteresis. Now if you want to talk about quantum physics where all commonly accepted physical laws are out the window, we can do that on a personal basis, but I don't think it's relevant to this forum. ;)
 
Absolutely Karl, but the fact is the formula used was "How many BTU's in a KW" and that's good for heaters, which generate heat from electricty.

But it's not valid for an Air Conditioner which simply MOVES that heat from point to point.. AC's move more heat than the same number of watts can generate.  That is my only point.  Thus the formula is flawed and you have to adjust by a number... the exact number will vary depending on many conditions, Temperture, Humidity, how dirty the filters are, how long it's been since serviced, exact line voltage and a number of other factors all affect EER, but the bottom line is it moves more heat than it can generate with the same number of watts.

Air conditioners are more efficent at moving heat than heat pumps,  And one of the web sites I visited was for RV Rooftop Heat pumps.  they claim they can deliver 2.5 times as much heat, per watt, as a straight heat strip.

So, compaired to a resistive power strip they are 250 percent effecent.  This, is a phantom number of course because they do not actually generate all this heat, they simply move it, in this case INTO the rig. 

But the point I'm trying to make is that the formula for heat generation does not apply to heat movement
 
That's the trouble with a slide rule, you have to figure out where to put the decimal points :D
 
karl I have ti disagree, we have a 5500 onan and it runs both of our 13500 AC with no problem
 
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