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Ductless Heat Pump - Which one?

I am purchasing a ductless mini-split heat pump. I have had several estimates and in all cases the companies recommend a 12,000 or 15,000 BTU unit. Apparently the area I want to heat/cool falls just in the middle for their calculations. I have a 1,500 sq. ft. bungalow with a large open concept living room, dining room, kitchen and front entry which is where the head unit will be. Off the living room are 3 bedrooms (no hallways) 2 on one side and one on the other. I'm told that the 12,000 BTU unit is more than capable of heating the open area, which is my main concern, but if I want to also provide at least moderate heating to the bedrooms I would be best to go with the 15,000 BTU unit. There is less than $200 in cost between the two units. I have decided to go with a specific brand and here are the specs for both models.

The 12,000 BTU unit has a SEER of 25, an EER of 13.80 and an HSPF (Region 5) of 10.435.

The 15,000 BTU unit has a SEER of 21.5, an EER of 12.5 and an HSPF of 9.565. Both are Energy Star rated.

I know higher numbers are better, but how much better? I also know that bigger is not necessarily better but does it make sense given the information I have provided?

Thanks in advance!

Asked by Robb Francis
Posted Dec 6, 2012 9:15 AM ET
Edited Dec 6, 2012 10:47 AM ET


6 Answers

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The question of whether the unit will heat the bedrooms has more to do with the layout of the building, the tightness of the thermal envelope, the number of windows in each bedroom, and whether or not the bedroom doors will be open during the day time than it does with the capacity of the minisplit.

Most heat loss calculations overestimate the design heat load. It is always possible (although unlikely) that the person who did your heat loss calculation is one of the rare individuals who actually did an accurate calculation. It is far more likely that the 12,000 Btuh unit will serve your needs.

Answered by Martin Holladay
Posted Dec 6, 2012 9:42 AM ET
Edited Dec 6, 2012 9:44 AM ET.


Thanks Martin.

All of the heat loss calculations were done at my kitchen table in less than 5 minutes. Only one of the three asked about window sizes, house orientation, r-values for wall and ceiling insulation etc., so I doubt if they were very accurate.

The house is not R-2000 but it's a new house by a reputable builder with higher quality windows and I'm hoping the thermal envelope is fairly tight.

Answered by Robb Francis
Posted Dec 6, 2012 12:38 PM ET


It's almost certain that you'd be better off with the 12000BTU/hr unit. Almost all napkin-math heat load calcs overshoot reality by quite a bit, and the output of a mini-split is typically higher than the nameplate spec until you get down to single-digit outdoor temps.

While there is SOME boost in operating efficiency when the unit is oversized somewhat for the heat load, if you asseme the 5-minute heat load calc overshoots by at least 25% (usually does- it's often more than 50%), upsizing by very much from there would likely lower the average operating efficiency, as well as deliver lower comfort from the higher air volume rates. HSPF numbers aren't the whole story on heating efficiency with a ductless, since the part-load efficiency is QUITE high as long as it's in a modulating range, but falls off at the lightest loads when it's cycling on & off regularly.

What's your 99% design condition temp? If you're not sure, guesstimate based on nearby locations:


Then what manufacturer & model #s were you looking at?

If you know your wall & ceiling construction and the U-factors of the windows and the outside design temp you can hit pretty close to reality with an I=B=R spreadsheet heat load model and a WAG about the infiltration rate. If that comes up sufficiently under the lowest "pro" heat load you might even back off to a 3/4 ton mini-split rather than the 1-ton.

Answered by Dana Dorsett
Posted Dec 6, 2012 1:13 PM ET


Thanks Dana,

I've got to admit that you lost me with the second half of the last paragraph - sorry.

The two units I am looking at are the Fujitsu 12RLS2 and 15RLS2 the heating BTU/h ratings according to their specs are 16,000 and 18,000 respectively. They both have variable speed compressors. They also make an 9RLS2 with a BTU/h rating of 12,000.

Based on the link you provided the "heating 99% dry bulb" is -7. The "cooling 1% dry bulb" is 83.
The u-factor of the windows is .3. Walls are r-20 and ceiling is r-50 and the house has a heated basement.

Answered by Robb Francis
Posted Dec 6, 2012 3:40 PM ET


At -7F these units will be putting out less than the nominal heating ratings. The derating with temperatures will be pretty much the same within the model series, but the output of the 12RLS at -7F is about 13,000BTU/hr. See the measured output and trend line on figure 3 page 16 (.pdf pagination) of this document:


The RLS15 would be putting out something like (13K x 18K/16K=) 14,600BTU/hr @ -7F.

For a crude I=B=R spreadsheet heat calc, assuming it's OK if the house is at 68F at dawn on the coldest days of the year, the delta-T between indoors & outdoors at the 99% condition is 75F. If you multiply the square footage of the difference surfaces by it's U-factor and by it's delta-T the result is BTU/hr. eg:

Assuming you have 100 square feet of U0.30 window, the heat loss out the windows at a 75F delta-T is:

100 x 0.30 x 75= 2250 BTU/hr.

For 2x6 construction with a 25% framing fraction (pretty typical 16" o.c. framing, not some newer lower-bridging methods) with R20 cavity insulation, wood siding & gypsum interior the heat loss per square foot of the wall area per degree delta is about U 0.076. Calculate your wall area (less windows & doors) and run the same arithmetic:

(wall area) x 0.076 x 75F = (heat loss from walls)

For an R50 trussed roof with that is full depth out to above the top plates of the studwalls (a "energy heel" trusses) with the chords buried in the insulation the U-factor of the ceiling is about 0.022. Same thing:

(ceiling area ) x 0.022 x 75F= (ceiling losses.)

You'll have to do the same thing for the doors- if it's a solid 2" wood door rather than an insulated the U-factor will be about 0.5. If it's a panelized wood door bump that to U0.75.

Losses through the floor aren't zero, but are close to zero if the basement is sealed and insulated even if it's not heated. But since it's being heated, call it zero.

Infiltration/ventilation losses aren't zero either, but if it's a pretty-tight house that maintains an interior relative humidity of >30% all winter without active humidification call it zero. If you want to take a WAG at the natural ventilation rates, use a specific heat of 0.018 BTU/ cubic foot per degree F. eg:

If you think you're leaking a full 50 cfm (about a small bath-fan's worth), that's 3000 cubic feet per hour, so at a delta-T of 75F you'd get:

3000 x 0.018 x 75F= 4050 BTU/hr.

Real leakage rates are all over the place, but 50cfm might be on the high side for what's perceived to be a tight single story house without active ventilation (unless the wind is really blowing.) But you can see how air leakage can be a substantial fraction of the total, and there's no such thing as a house that's "too tight".

Build a room by room spreadsheet and get heat loads for each room, and add it all up. Use a fudge-factor of say, 15% to account for the stuff we've called zero for lack of better info, and for any sins of compression/omission on the insulation package, and that will be pretty close to your real heat load at -7F. The real heat load may be slightly more, but it won't be 2x more, or even 1.5x more (unless it's a lot draftier than you imply.) But it won't be a lot less either, unless you have three Tivos running 24/365 (or some other substantial continuous plug load.)

You could also subtract out 250BTU/hr for every sleeping human, and another 200BTU/hr per 'merican sized refrigerator if you want to fine tune it.

If your goal is to keep the main space at 70F and let the adjacent rooms run a bit lower in temp it's reasonable to use a lower room temp and the correspondingly lower delta-T for coming up with a heat load. If those rooms have a relatively high heat load at design condition they could easily run 10F cooler than the main room at design condition with the doors closed, but would likely stay within 10F of the main zone temp with the doors open, and probably within 5F. But it really depends on the layout the door area, and the actual heat load per degree. (A 5F difference can induce significant convection currents between the rooms, but a 10Fdelta moves a lot more air.)

Since U0.3 windows lose about 4x the amount of heat than the adjacent walls, don't be surprised if the window losses dominate the heat load. This can be improved significantly with low-E exterior storm windows or tight fitting sealed interior shades. One aspect of the exercise of running a quick & dirty I=B=R type heat loss spreadsheet is that it can point to where the better bang/buck for reducing the loads might be. You'll probably find that taking the attic up to R100 won't reduce the total load very much, but adding low-E storms (which would take the total U-factor for a U0.30 window down to the U0.22-U0.24 range) could do quite a bit.

Answered by Dana Dorsett
Posted Dec 6, 2012 5:48 PM ET


Thanks Dana (and Martin). I'll sharpen my pencil and do some figuring. You have given me lots of homework!

Answered by Robb Francis
Posted Dec 7, 2012 10:07 AM ET

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