I have a 3000 sf house and the load calculated to 36.5kbtuh heating and 22kbtuh cooling. These numbers are based on R21 walls and R38 roof. If the insulation was then upgraded to R29 walls and R53 roof how much will this affect the load? Reason I ask is we were planning on using a 3 ton heating system but I am now wondering if we could go with a 2 ton system?

That depends on your wall and ceiling areas your window area and window overall U factor is. You should be able to figure this out fairly easilly by converting your insulation R value into U factor to match the windows. R value is the inverse of U factor and vis versa, so to get from R value to U factor, divide it into 1) For example, 1 divided by R21 equals a .048 U factor. Or 1 divided by .19U = R5.3.

U factor = BTU per hour, per sq/ft, of surface area , per degree F difference between your inside and outside temperatures. Total up your surface areas for each U factor. Pick an inside temperature(70F) and an outside temperature and do the math. Your 99% outdoor temps are usually used for the worst case heating and cooling load calculations. Don't forget to include all the wall materials(siding and drywall have r values) for a total overall R value...

OR you can go to the Borst Engineering . com website and use their heat loss calculator(free for personal use). You will need your surface areas and total overall R values, but it will do all the math for you. It also allows you to manipulate air infiltration numbers. Once you get the surface areas loaded in, you can manipulate the R values and infiltration numbers at will to get answers to your questions.

Good Luck.

I assume the "R21 walls" was R21 fiberglass batts in 2x6 studwall construction. How is the R29 wall number achieved?

Was the R38 batts in joists/rafters? How about the R53?

Details matter. R29 done as 5" of closed cell foam has only a small improvement over 2x6/R21, reducing wall losses by only very low double-digit percentage, but R29 as 2" of EPS sheathing on the exterior of a 2x6/R29 wall has a fairly significant effect, cutting the wall losses by about a third.

A 2x6 / R21 wall does NOT have a U-factor of 1/R21= U0.048, and using that number would result in a gross error!!

That would only be correct if it were a continuous homogeneous material that tested a R21. But a typical 16" o.c. studwall has 25% of the face area bridged by R1-R1.2/inch framing (R6.5-ish for 2x6 framing) and is sheathed by ~R0.5 gypsum on one side, and ~0.5 plywood on the other, with a layer of siding, and both interior & exterior air films all contributing to the U-factor calculation. A typical vinyl sided R21 2x6 wall 16" on center with half-inch wallboard and half-inch OSB sheathing comes in at about U0.065, which is substantially lossier than U0.048.

But if you add 2" of Type-II EPS (R8.4) on the exterior to make it R29-ish at center-cavity, the U-factor is about U0.042.

Without knowing what fraction of the total heat load was wall or ceiling losses there's no way to reasonably adjust the whole-house load numbers. In most places most 3 ton air source heat pumps would be oversized for even the 36.5K load (since the tonnage is usually related to the cooling capacity, not heating capacity), and in cold/very-cold locations it might not be big enough, since the heating capacity varies considerably with outdoor temperature. That's also going to be impossible to call without knowing the specific equipment, and the local climate. The output capacity of the equipment at the 99% outside design temperature is critically important, and quite a bit different at -20F than at +20F.

The R21 walls and R38 roof was assuming code minimum (or less) fiberglass batts. The R29 wall is 2x6 rock wool batts (R23) with 1.5" exterior rockwool (R6). The roof is a combination of rockwool batts (cathedral) and blown in fiberglass (vented attic). The heating system is water to water geothermal with a radiant slab main floor and a fan coil for upstairs. I could recalculate it to the nearest BTU but for equipment selection it comes down to the nearest 12,000 BTU. Individual room calcs are another story, but for the whole house load it seems like a lot of work to do a precise calculation only to round it to the nearest +/-12,000.

A slight educational digression… So, how was this 12,000 BTU/hour or “ton” cooling capacity derived?

When ice is at its melting point of 32° F and absorbs heat, the temperature of the ice does not change. Instead, the ice melts to form 32° F liquid water. The amount of heat needed to melt ice into 32° F liquid water is called the latent heat of fusion which has a value of 144 BTUs/pound (or 1201 BTUs/gallon). If you have a pound of ice at 32° F, it takes 144 BTUs of heat to completely melt all of it. If you have one ton of ice, it takes 144 BTUs/pound times 2000 pounds or 288,000 BTUs of heat to completely melt all of it. You could completely melt this one ton of ice in one hour, one day, or one year, depending on how quickly you heat it. At some point, it was decided to use one day for this standard. So, if one ton of ice melts uniformly over one day or 24 hours, it absorbs heat at the rate of 288,000 BTUs divided by 24 hours or 12,000 BTU/hour.

Posted By HP Home on 03 Nov 2015 10:08 PM
The R21 walls and R38 roof was assuming code minimum (or less) fiberglass batts. The R29 wall is 2x6 rock wool batts (R23) with 1.5" exterior rockwool (R6). The roof is a combination of rockwool batts (cathedral) and blown in fiberglass (vented attic). The heating system is water to water geothermal with a radiant slab main floor and a fan coil for upstairs. I could recalculate it to the nearest BTU but for equipment selection it comes down to the nearest 12,000 BTU. Individual room calcs are another story, but for the whole house load it seems like a lot of work to do a precise calculation only to round it to the nearest +/-12,000.

What type of siding, sheathing, interior cladding?

Assuming vinyl, half-inch CDX, half inch gypsum and a 25% framing fraction the U-factor of that "R29" wall is about U0.046 , or a "whole wall R" of about R21.6. That's including the interior & exterior air films.

If the siding is 1x wood ship-lap it's about U0.044 ( R22.5-ish whole-wall.)

That's the range you're looking at.

Calculating the room loads to within a few hundred BTU is important for figuring out the temperature requirements of the water for the radiation to emit that much heat, which affects the capacity of the heat pump at any given ground temperature.  The room with the highest water temp requirement often drives the ultimate size of the equipment.

It's never as simple as a "a 1 ton heat pump will give me 12,000 BTU/hr". If you need 140F water and the local subsoil is 40F, it's going to give you a heluva lot LESS than 12,000 BTU/hr per ton of heat pump.  If the subsoil is 65F and the peak water temp requirements are 85F, it will deliver a lot MORE than 12,000 BTU/hr per ton of heat pump. (Though for a fan coil heating you probably want at least 110F water or higher no matter what to avoid the wind-chill effects, but how many BTUs are needed @ 110F relative to the low temp slab zones still matters.)

So while the heat pumps might be sized in one-ton increments, the system design and individual room loads really do matter, since it affects the system design,  and it may drive you to re-plan how you heat the rooms that aren't on the low temp slab. You may be better off going with low-temp radiator panels for heating those non-slab rooms, even if you still keep the fan-coils for cooling, etc.

Thank you for the schooling. This helped confirm everything I was thinking. The slab will be run off an outdoor reset and I am expecting it to run at a fairly low temp most of the year. I was worried about the fan coil either blowing cold air or killing efficiency by running a higher water temp. So I think the fan coil will be mainly for A/C and the upstairs will get hydronics.

Oh and the wall exterior of the wall has RevealShield over the rockwool, then 1x3 wood furring strips. Open joint cladding over that, some cedar 1x6, corrugated metal, and fiber cement panels. Not sure if that makes much difference, it's probably right there in the same ballpark with vinyl siding. Also some exterior walls are 2x8 thickness and getting R30 batts in the cavities. But either way I need to sharpen my pencil and take another thorough look at the load calc, things have changed significantly since it was first done.

So I think the fan coil will be mainly for A/C and the upstairs will get hydronics.

Unless you put a radiant floor (or other large radiator) upstairs, it will require higher temperatures. But a HP can be set up to produce two temperatures - you don't need to always produce the highest temperature needed.

Radiant floors upstairs would put the tubing under carpet, carpet pad and 3/4 osb. I am thinking about doing radiant wall panels for these rooms. But I need to do some math first before I make a decision.

Radiant walls are probably a good choice - lots of surface area (low temps), easier to add, nothing blocking the heat flow and lots of convective air flow over it. If it's an exterior wall, you could add some rigid foam at the same time. But not so good for AC (most climates).

Dana’s point is well made...you can’t just select a HR heat source simply based on the load calculation estimate. You can only properly select a HR heat source after actually designing the HR system and determining the required supply temp(s) that will be used to satisfy the required design condition.

Another simple example...the estimated floor heat loss based on a 70F heat loss analysis is significantly less than the actual heat loss experienced by a floor that is actually heated and held at a much higher temp.

You might get away with catalog engineering for small residential buildings, but you will fail miserably if you use this approach on larger residences or commercial buildings. There are no shortage of stories about HR systems that are grossly inefficient, short cycle, or simply just don't keep the building warm in colder weather.