How much heat for a room?

rich67

I want to install new hot water baseboard in my living room after reinsulating the walls and around the windows. Can you tell me how to calculate the heat loss for the room. I read that baseboard puts out approx. 600 BTU/foot, but I can't find anywhere how to figure the heat loss. House is a ranch and the living room is 24 x12. Thanks.

kcopp

need more info.

window size, doors, outside walls, heated below, insulation up and down and walls....?

Brad White

Short Way/Long Way

Normally, I would direct you to Slant-Fin which has a free-ware heat loss calculation software offering. For one room or heck, a lot if you have time, here is a basic primer on calculating heat loss.



There are two basic components to heat loss, transmission and infiltration. Transmission is the conductance through solid surfaces and materials from warm to cold, from inside to outside. Infiltration is the exchange of air in and through the space, from outside to inside and out again. Basically air leakage.



Transmission- It all gets down to this:  A x u x Delta-T.



Area times the u-factor times the indoor-outdoor temperature difference.



Each surface, be it wall, glass, door, roof, floor, has a certain rate at which it transfers heat. This is called the "u-factor", which expresses how many Btu's per hour the surface will pass, per square foot, per degree F.  The lower the u factor the better.



The u-factor is the inverse of the "R-value" which is touted by a large pink panther and people wearing orange aprons. The R-value answers the question, "how many SF of this material at thickness will pass one BTU per hour, per degree F. difference?". So an R-19 surface takes 19 SF to pass that one BTU, while an R-13 surface takes 13 SF. You can see, a higher R is better.



The area is just that, the net surface in square feet. The Delta-T is the design indoor temperature minus the design outdoor temperature.



Some typical u-factors, adjusted for framing and the like:



2x4 insulated wall with R-13 insulation: u=0.07 to 0.09

(Variations I can explain later if you like.)



2x6 insulated wall with R-19 insulation: u= 0.05 to 0.06



Single-pane glass: 1.04 to 1.13



Double-pane glass: 0.65 to 0.70 (includes single pane with storm window).



Low-E glass: Often given as 0.25 to 0.33, this is "center of glass". Factor in the framing, edge conductance and other variables, the overall area might average 0.38 to 0.45, still better than regular glass.



Roofs and ceilings vary widely. R values can be R19 to R75. Batts do not perform as well as open-blown cellulose, so hard to compare. A good default might be to take a u-factor of 0.05, about as good as most 2x6 walls, assuming you have any insulation.



Let's assume I live in the Boston area. (Hey, I do!). The outdoor design is 6 degrees in my area outside the city. I keep the indoors at design at 68 degrees, so my delta-T is 62 degrees.



So let's say I have a room with one exposed wall, ten feet high (include floor to floor dimensions, not floor to ceiling!). The wall is 20 feet long or 200 gross SF and is 2x4 construction with a u-factor of 0.08.  (All of the other walls are bounded by heated interior space, so we do not count them. We will also assume that there is heated space above to keep this simple.)

In this wall is a window, 60 inches high and 36 inches wide, 15 SF, single pane with a storm window. That leaves 185 net SF of wall.



Wall Loss: 185 SF x 0.08 x 62F = 918 BTUs per Hour (BTUH)

Glass Loss: 15 SF x 0.68 x 62F = 632 BTUH

Total Wall Transmission =          1,550 BTUH



Infiltration- This can be a huge wild-card. A new tight house might leak 0.25 air changes per hour (ACH) or less, sometimes much less. An old Victorian with chimneys, poor dampers and no insulation, might run 1.5 to 2.0 ACH.



Let's say for discussion that this room with one exposed wall, might have 0.75 ACH.



The room is 20x10 with a 9 foot ceiling, so 1800 cubic feet. Every hour, 0.75 or 3/4ths of that, 22.5 call it 23 cubic feet per minute, enters as cold air and leaves as warm air.



Another simplified formula: CFM x 1.085 x Delta-T. CFM is cubic feet per minute, based on the air change rate. The 1.085 is a constant which takes into account the specific heat of air and mass flow rate in pounds, over an hours time. The delta-T is 62 degrees, same as for transmission.



So, 23 cfm x 1.085 x 62F = 1,547 BTUH (greater than your transmission and fully half of your total heat loss!)



Add your transmission total and infiltration, (1,550 + 1,547) = 3,097 BTUH.

You may want to add a 5% or 10% safety factor to this, most designers and engineers do, to cover unknown variables or a total lack of self-confidence



Call the design heat loss 3,400 BTUH.  So to heat such a room, with fin tube emitting 600 BTUH per foot, I would need at least  5'-8" of element, call it six feet.



OK, here is where it gets tricky and interesting:

You have in our example, 20 feet of wall. You could and probably should cover that wall with radiation to spread it out, but you also want to put the heat below the windows primarily to counter down-drafts.



But ask yourself, does the rest of the house have a similar "Radiation to Heat Loss" ratio?



Say your other rooms are not insulated and have radiation sufficient to heat the spaces with 180 degree supply water. You intend to insulate them eventually, just not now. But say an equivalent room to the one you just insulated has 12 feet of radiation. 



Your new insulated room needs the 6 feet we established. That will work with 180F water. But if you gave that new insulated room 12 feet, you might heat it with, oh, 140 degree water.



When you insulate the rest of your house and find you can heat the existing rooms with original radiation, with 140F water, your just-insulated room will not have enough radiation to meet the heat loss with that cooler water temperature. You will either forget about dropping your water temperature and run 180F water for that one room, or suffer a cold room while the rest of the house is comfortable. I am suggesting that you give your insulated room more radiation to make the eventual ratios similar in all rooms.



Do give that some thought!

Hollis

Great post

Brad,..but on the edge of my understanding. So is E glass about as well as as an extra pane? At least comparing a single to a double?  Do you think that the radiant bounchback is as effective as the conduction benefit of two panes?



Re U factors,..seems to me to turn it around to an R factor is  easier for non engineer to understand but anyway  from graphs I've seen that and of r factor of say about 8 gives a pretty big bang and what? maybe about 80% of what an R factor of 30 would give? (just a guess as its been a while since I looked at this)



But I remember esp looking at big diminishing returns of adding r 60 to my western Ma building compared to the r-40 that was there,..such a small apparent increase  of heat savings.

As far as infiltration,..Ummm I thought that MOST houses really lost about 2  air exchange every hour and EPA thought that one ACH,every two hours  is good ventilation .Again I might well be wrong about this,..its been a while.

Lastly have you ever used or seen any real independent test on products NANSULATE or  HY-TECH,..coatings (both seem very different modes of insulation of each other ) but they make some pretty big claims and claim to have their "(OWN independent tests) ,..I guess that would be an oxymoron.

I have seen some Infrared pics that "look" impressive but not sure they can't be orchestrated to give  false results.

 

Brad White

Low E Glass, etc.

Hi Hollis- thanks.

In my opinion and experience, Low-E glass is essentially a radiant barrier as you note, a bounce-back effect that, when seen under an IR camera, shows up as warmer temperatures.



(All glass does this and you can see your thermal reflection even in single-pane glass when seen through the viewfinder.)



But when tested by a contact thermistor, the actual surface temperature is just a degree or two higher than standard dual pane glass. (The low-E itself in question is dual pane too.)  I mention this specifically because using low-E glass in a swimming pool application does not buy you much margin against condensation compared to standard dual pane glass. The benefit is in the Mean Radiant Temperature area, with the body feeling less chill facing low-E glass than standard glazing systems. It shows up as IR energy instantaneously, but indirectly over time this warms the objects back in the room or rather, slows the re-emission to the outside.



The laws of diminishing returns are at work with insulation thickness.  What I seek in insulation is integrity over R value. By this I mean, I would rather have R10 in a wall that is verifiable, vs. R20 with 10% gaps or air voids. From then-on, thickness works this way:



Take a 600 SF attic with no insulation, just wood lath plaster ceiling, joists, air films. The R value might be 1.75 total including air films. The u-factor is about 0.57 and the heat loss at say 65F difference,  22,230 BTUH.



Add a one-inch "flash coat" of urethane foam as an air-sealing layer if nothing else.  At R-6 value added (total R at about 7.5), and the heat loss drops to 5,200 BTUH, about 24% of what it was.



Add 8 inches of cellulose at R3.5 per inch, enough to cover the joists, to a total R value of about 36, and the heat loss drops to 1,085 BTUH, less than 5% of what it was. The 8 inches of cellulose without the flash coat of foam would get you down to 1,300 BTUH, so in essence, that flash coat of R6 nets barely 215 BTUH difference compared to cellulose alone.



Double the cellulose to 16 inches, total R of about 63.5 including the foam layer, and the heat loss drops to 615 BTUH. And on it goes. But the key point I would make is, the flash coat and if the cellulose top layer has a water mist spray to form a crust, the air sealing properties are outstanding and the integrity of the insulation pretty much to specifications in the lab.



The "paint" approaches you mentioned, the coatings? Have never seen them pan out by independent tests. They may in fact be high radiant reflectors/emitters, which when using IR thermography, "cheat" the measurements much as the reflective Low-E effects, but cannot yield the true insulative values they cite. I am open to persuasion, but so far, nothing I have seen about them seems to live up to the talk.