Questions regarding air vents and heatloss calcs

Apprentice_3

The company I work for just finished a big heating installation. Got me thinking about a couple of things,

1- Automatic air vents on hot-water systems. How many is enough? i know a basic rule of thumb is to locate a vent at each high point in the system, but what about on a particularly long main? Should there be one every certain number of feet? Does pipe size change anything? Maybe there is a formula someone can direct me to.

2- Heat Loss calculations. None of the guys I work with seem to know anything about it, maybe someone here can help. I researched the basic formula (heatloss = AxUxT), but I don't know what to do with it. I'm not sure wether the product of that equation gives me the heatloss of BTU/Hr for a given room, or what. And for that matter, is the heatloss of the house just the heat loss of all the rooms, added together? To select the proper-sized boiler and heaters, do I just take the heatloss of the building and find a boiler that matches or exceeds the number?

Thanks!

Brad White

Vents and BTU's

Hi Addam.

The vents are to me the easier of the two. High points should do it. If you have very large piping there is plenty of room for that to scoot along. If small diameter piping, velocity during filling or purge or even proper operation will move the air along to a point where it can be vented. If any small bubbles get missed and trapped somewhere (such as in a large fitting), it is not enough to worry about. It will get out eventually.

Heat loss: Yes, your A x U x Delta-T covers part of it, what is known as transmission. Heat transfered through solid materials or different materials in series (such as a wall, roof, glass, doors or floors over unheated spaces.

Your areas really only need to cover those surfaces keeping you from cooler places (the great outdoors, a cooler unheated basement, unused rooms closed off, for example.)

The other part is infiltration. It can be a big one, often 1/3 on insulated construction and oddly 1/4 on not so well insulated construction because it can be fairly constant despite insulation and without judicious sealing.

For infiltration, I would start at 1.0 to 1.2 air changes per hour (ACH) for an older house that has been tightened. Modern SIP construction could be 0.30 or less. An older Victorian without any insulation could be 2.0 ACH.

The other way is the crack method which considers the length of operable sash moving joints or door perimeter and assigning a CFM value to each foot of crack. Typically I would assume 0.50 CFM per foot of window and 1.0 per foot of door as conservative numbers.

Absent a blower-door test, it is still a guess.

Each method will yield a CFM value of airflow which is multiplied by a constant (I use 1.08) and the delta-T to get BTU's per hour.

ADD both transmission and infiltration numbers together and you have a net heat loss figure for the space in question. Some designers add 10% safety factor to account for insulation gaps, things you cannot see.

The total heat loss per room is divided by, for example, the baseboard radiation rating you are using. A room with an 8,000 BTUH heat loss and radiation at 550 BTUH per LF would require 14.5 call it 15 LF of element as one example.

MUCH more to it than that, but it gives you an idea.

EDIT: I have attached an essay I wrote when teaching at the Boston Architectural Center (now Boston Architectural College) which you are free to use. It was written for architecture students so is directed at a wider audience.

Brad

ALH_4

Resources

One automatic air vent can handle a typical residential system, unless there is a special circumstance that requires a manual vent at a high spot during filling. I am reluctant to leave auto air vents in areas where water damage could occur if the vent fails.

You are miles ahead of your coworkers in realizing that heat loss calculations are important. A good place to start is the Air Conditioning Contractors of America's website. Manual J is the standard method for calculating residential heat loss. It is worth investing in a copy and studying it. Your employer absolutely should own a copy for reference if heating work is part of the business.

Heat loss calculations are generally done with computer software. With this software, it is much easier to run a detailed heat loss calculation for every job without a lot of effort. Most heat loss software is based on Manual J calculations. There is a "Free Heat Loss Calcs" link at the top of this page. Download the Slant Fin software and give it a try. John Siegenthaler also has a heat loss program available on his website. The combo of his book and Hydronics Design Studio is probably the best investment you could make. Ideally, your employer would make this small investment to improve his business. Even if your employer does not see the value, it's worth it to you professionally to invest your own $$ in John S's book and download the free Slant Fin software as a starter.

Apprentice_3

Thanks for the thorough responses guys! I think I've got a better understanding of it now, and I'll definetly be looking into all the resources mentioned.
Reading what Brad wrote brings up another question though ---> how does boiler size (based on the BTU rating stamped on the boiler's rating plate) relate to the heat loss of a house. After finding the heatloss of the house or building, do you simply select a boiler whose rating matches/is slightly above that number, or is there more to it?
Thanks again!

Brad White

From Heat Loss to Boiler Size

Decent question- entirely so.

The short answer is, you are correct, the smallest boiler which will meet the heating load of the entire house. The more accurately you calculate that number, the better we all are.

The "more to it" part:

Infiltration: That component will really occur 100% at one time or another for any one room at some point. But the house as a whole? It will be about 1/2 to 2/3 of the total room-by-room infiltration. Air can only leak in or out on two sides of a house. When air leaks into a room, naturally it imposes a loss on that room. But when air leaks out of a room on the opposite side of a house or on an upper floor, that air is in effect, leaking into the room in a preheated condition. The "loss" was paid for by the previous room into which the air first leaked. Does this make sense?

My point here is that, once you calculate a heat loss, back out the infiltration totals of the rooms. Take the entire house "transmission losses" and add back in about 67% of the infiltration number. I think that more accurately reflects real heat loss. Minor point but something to consider.

Here is an example: Say you have a house with a group of rooms, the sum of which total 65,000 BTUH. Of that number, 45,000 is transmission loss, conductance through surfaces of all types. The remaining 20,000 is infiltration but the sum of the totals of all rooms. Instead of taking the 65,000 BTUH to size the boiler, take the 45,000 fully but add in about 2/3 of the infiltration number (0.67x20,000)=13,400 for a total calculated heat loss of 58,400. Not a huge difference but it saved that much over-sizing.

Now, when picking a boiler, you will see different ratings. Say a boiler has 75 MBH input of gas. The net AFUE might be, say, 83% and the net AGA output would be 62,250. There may also be an I=B=R rating of say 54,130 BTUH. This number is artificially de-rated by 15% for piping and pickup losses. Personally, I use the net larger number. The house will not be warmed from nothing, it will have some basis from which to start. If you DID take the I=B=R number as gospel you would wind up taking the next larger boiler and a loss of efficiency.

Glenn Sossin_2

Brad

I tried to download your file but it said it was damaged. Could you repost and/or email to me. I am very interested in reading it.
Thanks
Glenn E Sossin

Tried to download again and it worked.

Apprentice_3

use the larger number

All clear now. It's not as complicated as I thought it would be, but there are a lot of factors to consider.

Thanks for clearing it all up!

mtfallsmikey

Awesome!

Nice work on the paper Brad!..I had no idea that the conductivity of metal stud walls was as high as it is. In most commercial interior construction, my 2 buildings included, do not use the fly-by method.(costs too much!), but as pressure builds to construct to LEED, I hope to see this method used more often. Now the dumb question: what is the formula to convert U factor to BTU?

Brad White

Whatsamatta- U

Hi Mike-

Not a dumb question at all.

The U factor is the inverse of the R-Value. (Gosh, Brad, that really helps a lot- get to the freakin' point!)

To get BTU's you take the number of SF of a given construction, multiply it by the U-factor then multiply that by the Delta-T.

For example, one SF of 2x4 insulated wall at a U-factor of 0.075 with a 70-degree difference across it would come out to 5.25 BTU's per hour at design.


Now, back to R-values and U-factors.

R-value answers the question, "how many SF of a surface does it take to transmit one BTU per hour with one degree F. temperature difference across it?"

An R-19 wall takes 19 square feet. The more wall for that one BTU per degree, the better.

The U, being the inverse of this, answers the question, "How many BTU's per hour will one SF of a surface transmit with one degree F. temperature difference across it?"

Naturally, you want that number to be low....