Choosing between Mitsu and Fujitsu mini split, need advice
I’m debating between two contractors. One sells mitsubishi mini splits and the other sells fujitsu
I’m heating a finished room over my garage which is about 750 sqft and has awful(!) insulation in the roof. The Fujitsu guy said that an 18K unit should suffice but I’ve plugged numbers into a sizing calculator and I’m pretty sure that I need a 24K unit (~26K BTU heating required per the online calculator I used)
The mitsubishi dealer is actually the cheaper of the two, however they do not seem to have any 24K BTU single head units that are actually rated for temperature ranges as low as -15F, and I live in Massachusetts so it gets pretty cold here. He quoted me on a MUZ-GL24NA-U1 unit. There is, however, an 18K BTU unit MSZ-FH18NA2 (or -NAH2) that is rated for the lower temperature range.
There is currently a gas imitation wood-burning oven in the room which I have used for heat up to now, so I would have supplemental heat as well if needed.
For a >$1200 savings would choosing the Mitsubishi GL24NA-U1 unit over the Fujitsu AOU24RLXFWH / ASU24RLF be the wiser decision regardless of the difference in efficiency? Or for even possibly greater savings, going with the MSZ-FH18NA2 and potentially relying on some supplemental heating?
Thanks!
Mark
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Replies
Even with awful insulation in the roof it's unlikely that you'll need even a 1.5 tonner, let alone a 2 ton. Run a more realistic Manual-J or I=B=R type load calculation, not some silly sizing tool off the web.
There are no locations in MA with 99% outside design temperatures anywhere near as cold as -15F (or -13F, the coldest temp at which Mitsubishi specified an output.) Most locations have 99th percentile temperature bins north of 0F. Even though it gets colder than that, it doesn't much matter, especially since you have the backup gas-burner for the Polar Vortex extremes. If I had to guess, I suspect the FH12NA, FH15NA, 12RLS3H or 15RLS3H are a better fit for the real loads at your actual winter temperatures than the FH18NA or any 2-tonner. (Seriously- an FH18 would cover more than half the 99% heat load of my 2400' 2x4 framed house with lousy insulation in the cathedralized ceilngs in Worcester.)
Got a ZIP code? (for weather data and design temperature estimation)
How many windows, of what size & type?
Wall construction & insulation is...?
Dimensions of the room are...?
Thanks Dana. That confirms what a friend of mine told me as well. 01776 is my zip (Sudbury). Probably about the same as Worcester.
The construction is post and beam with wooden walls on two sides and drywall on one side. Aesthetically it looks very nice, but unfortunately it also lends itself to many air-leaks and drafts.
The dimensions are roughly 40x20. Walls and ceiling both appear to only have some thin polyiso insulation, hard to tell exactly how thick it is since I can only barely see it through some places where there are knots in the wood.
There are two normal sized windows on one wall, the other two walls that are exposed to the outside have 3 small windows on each side. They all are double-paned.
Find out how thick the foam is, even if you have to drill a hole with hole saw (for ease of repair) to get an accurate measurement.
Dana, assuming it extends to the roof-line, it appears to be 4" thick. Could be 2" if I'm unlucky.
Even 2" of foil faced continuous polyiso on the interior of a 2x4 studwall with empty stud bays would come close to meeting current code-minimum for the walls on a U-factor basis. If it's 4" it beats code minimums. (Your description was ".. only have some thin polyiso insulation...", but 4" or even 2" isn't thin at all!)
Run an I=B=R type load calculation using U0.07 for walls/ceiing with 2" of polyiso, U0.035 for wall/ceiling assemblies with 4" continuous polyiso.
For clear-glass wood or vinyl double panes assume U0.50. If low-E, assume U0.35. Measure the width & height of the whole window units, not just the glass, but don't include the casings.
For the area of exposed beams that extend from the interior to the exterior, measure the thickness and use a U-factor = 1/(thickness x 1.2)
For how to run these numbers, see:
https://www.greenbuildingadvisor.com/blogs/dept/musings/how-perform-heat-loss-calculation-part-1
https://www.greenbuildingadvisor.com/blogs/dept/musings/how-perform-heat-loss-calculation-part-2
For Sudbury, use +6F for the 99% outside design temperature (same as Framingham's.)
Run a sub-total heat load, assuming ZERO for air infiltration & ventilation- we can work those issues separately.
For illustration purposes, assuming two clear glass double pane windows at 12 square feet,an six that are 5 square feet each you have 54 square feet of U0.5 windows. Assuming a code minimum 68F indoor design temperature and +6F outside design temp that's a difference of 62F, and window losses of
U0.50 x 54 sq.ft.' x 62F= 1674 BTU/hr of window losses.
Assuming 2" foam on the walls and 10' tall walls, with 120' of perimeter you have 1200 gross square feet, less 54' of window area leaves 1146' of U0.06 wall. So the wall losses will be:
U0.06 x 1146 sq. ft. x 62F= 4263 BTU/hr of wall losses.
Assuming you have some slope to the ceilings you're probably looking at about 1200' of ceiling. Assuming 4" of foam that would be
U0.035 x 1200 sq.ft. x 62F= 2604 BTU/hr of ceiling losses.
Assuming the floor joists have at least some insulation the floor losses to the cold but warmer than outdoors garage will be at most 2000 BTU/hr.
Add it up and you're at 10,571 BTU/hr before infiltration & ventilation is added on, and before plug loads & warm body heat is subtracted off. The real heat load might be as high as 1.5x that if it's pretty air-leaky, but even that's within the +6F heating capacity of a 15RLS3H or FH15NA, and even the 1.5 tonner would be a bit overkill.
http://www.fujitsugeneral.com/us/resources/pdf/support/downloads/submittal-sheets/15RLS3H.pdf
http://meus1.mylinkdrive.com/files/MSZ-FH15NA_MUZ-FH15NA_Submittal.pdf
You can probably do just fin with a 1 tonner, but the 1.25 tonners may or may not be overkill. Measure it all up and let us know where the raw I=B=R numbers are prior to infiltration & plug load adjustments.
I get very different numbers using IBR and your calcuation. I think the air flow (ACH) is what's really creating for different numbers:
For I=B=R:
Room is roughly a giant triangle on top of 3 foot walls.
Width =~ 24
Length =~ 30
Height (at center) ~ 12ft
So, volume should be half of the volume of the prism/triangle plus the 3' x 24' x 30' remainder;
volume = .5 x (9'h x 24'w ) x 30'l + (3'h x 24'w x 30'l) = 5400 ft^3
Rooms with windows or exterior doors on 3 sides (leakier) from the page you linked gives ACH = 2 and I = 0.036
5400 x 62 (deltaT, from you) x 2 x 0.036 = 24,105
This seems way higher than your other numbers seem to add up to.
I have about 62 sqft of window space (including a cupola at the top of the roof with 4 windows that I put plastic tape over to reduce heat loss)
The roof should be 900 sqft ( using a^2+b^2=c^2 where a=9' and b=12' , c is the length of the roof from peak down to where it meets the 3' knee wall, 15')
540 sqft of wall
900*0.035*62 = 1953 ceiling loss
0.06*540*62 = 2008.8 wall loss
.5*62x62 = 1,922 window loss
2000 sqft floor loss
I suggested to ignore the ventilation for reason. A leakage rate of 2ACH/natural is probably on the order 4-10x reality, given the layers of extremely air-retardent rigid polyiso. If it leaks anywhere NEAR that much it's fixable for cheap- cheaper than the increased cost of a bigger mini-split and 5 years of extra power use. The I=B=R air leakage methodology is just plain terrible, which is why I recommend just ignoring it for now.
So what you have is about 2000 BTU/hr each for walls, roof, floor, and window for about 8000 BTU/hr of conducted losses, plus whatever you're going to use for air infiltration.
An open window or door on a calm not-so-windy day is worth 100-200 cfm, or 6000-12000 cubic feet per hour. Is your leakage REALLY as big as an open window? (Back, in the 1940 when I=B=R was invented, in a plank sheathed house with rattly single-panes it might have been that high, or even higher.) Even taking the high side of 200cfm, 12000 cubic feet per hour
12000cubic feet per hour x 62F x 0.018 BTU per degree per cubic foot = 13,392 BTU/hr. That's probably higher than reality now, and much higher than reality after one round of air sealing. Add that to the 8K of conducted load and your'e at 21,392 BTU/hr. The 1.5 ton Mitsubishi FH18NA will deliver that at +5F.
A the (probably more realistic) 100 cfm number, it's only 6696 BTU/hr of infiltration heat loss, for a total of 14,696 BTU/hr, which is only slightly more than the 13,600 BTU/hr that a 1-ton FH12NA will deliver.
And you haven't begun to subtract warm bodies and plug loads. If you have one conscious live adult human with a pulse sitting, not standing that's about 250 BTU/hr of heat. A room that size probably also has at least a couple hundred watts of fluorescent lights that would be in use at 7AM on the coldest morning of year, for another ~700 BTU/hr, call it 1000 BTU/hr of bodies & plug loads whenever the room is occupied. That would make the 1-ton nearly exactly sized for the load at +6F, but it wouldn't be insane to go for the 1.25 ton FH15NA instead, which can deliver 18,000 BTU/hr.
Given that you have a wood burner to back it up during the Polar Vortex temperature extremes, either the 1-ton or 1.25 ton would do just fine and the 1.5 tonner would be edging into the overkill zone.
The cupelo acts a bit like a solar chimney and drives air infiltration. Take the time to really air seal those windows, since they are the most important air leaks in the building from a stack-effect infiltration drive point of view. If the wood burner vents through the roof it too could also be a real contributor, but air sealing around flues safely has more details to contend with.