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k value/supply water temp question

You can find the exhaustive numbers from commercial food processing. While fairly complex, still <I>nothing</I> like living/working space.

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  • Greennmann
    Greennmann Member Posts: 13
    k value/supply water temp

    is this formula for calculating supply water temp true?
    SWT(*F)=Setpoint(*F)+Heat Required(BTUH/FT2)/K(BTUH/FT2/*F)

    is the k value in this formula when used with a radiant heating system representing everything above the insulation?

    also is the k value is the same as the u value or 1/rvalue?

    is this k value easily calculated like the u value of a wall or should it be supplied by radiant manufactures?

    were can i find radiant k values with different tube spacing, construction method, and floor coverings as variables?

    thanks for you help guys

  • Greennmann
    Greennmann Member Posts: 13


    thanks i'll check them out tomorrow
  • Josh_10
    Josh_10 Member Posts: 787


    You got it my friend! Feel free to email me if you have any more questions. [email protected]

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  • well, that's not a *complete* list of installation options there Josh ;)
  • Josh_10
    Josh_10 Member Posts: 787


  • In very basic terms, the only way to really calculate K-values is by finite element analysis. Which is not easy and is, itself, currently still not perfect.

    Picture this; your average K-value for a floor surface with tubing in the joists. It varies based on the effect of the particular plates you are using (if any) which can vary by thickness and plate size, contact area between pipe and conductive contact members (like plates), tubing spacing, how you model joist temperature rises for different installation methods and how that affects heat transfer through the assembly, etc etc etc....

    Basically the only "exhaustive" numbers on this stuff that I have found out about so far is used by people who program these kinds of design software, and they aren't sharing. And, they all have their own assumptions and biases built in. Though if I was going to trust anyone's assessment of it all, it would be with siegenthaler's Hydronics Design Studio.

    Interesting you bring this up though. I'm thinking of tackling this myself but I'm going to be stuck with rough approximations since I do not have the skills or software (yet) to do true modelling for this stuff...


  • Josh, if you have any way to even attempt this, I would be absolutely psyched to hammer on it with you. But my calculus is a bit rusty these days! I have begged the big boys with cash and promises of NDAs and still they won't share. sniff sniff :(
  • Brad White_9
    Brad White_9 Member Posts: 2,440
    Herr Greenmann,

    This is the part that confuses many. "k" (lower case k) represents "Conductivity", the positive flow of heat.

    It is analogous to the "u" factor but is expressed on a "per inch" basis and of a given material. The lower the better for conservation's sake..

    Thus, an inch of EPS (commonly known to have an R-value of 5.0) has a conductivity or "k" of 0.20

    The "u" factor is analagous to the "k" factor but is used to describe the entire assembly of materials within a wall or ceiling construction.

    "C" (capital C) is similar to the "k" value, also on a per material basis and of a given thickness which may vary.


    Here is another way I describe "R" and "u" values when I was teaching:

    "R" answers the question, "How many square feet of a given material or surface will pass one BTU per Hour per degree F. difference?"

    (Naturally you want this number to be high if you are paying the heating bills.)

    For example, a wall with an R-value of 19 will require 19 square feet of wall to pass one measly little BTU for every degree F difference across that wall.

    Conversely, the "u" factor asks, "How many BTU's will be passed by a square foot of a given surface at a one degree F. differential?"

    (You want this number to be low, bill-payer!)


    Hope this helps.


    Brad



  • Josh_10
    Josh_10 Member Posts: 787


    Brad I think I am going to rename you "The Brains" (as I chuckle to myself) I wish I knew 25% of what is in your cranium!

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  • Brad White_9
    Brad White_9 Member Posts: 2,440
    Oh God No...

    But thanks for the compliment. I do not use a pocket liner nor wear glasses and make obvious and common mistakes routinely. But I will be gracious-

    Thank you!

    Brad
  • Brad White_106
    Brad White_106 Member Posts: 8
    I meant to add as I was running for the 5:55 train...

    That I would wish to someday solder joints like you do, Josh. That is a skill that endures. Folks- check them out.
  • Greennmann
    Greennmann Member Posts: 13


    I saw tab 2 but i needed 7" on center and i wasn't sure if those values would work with wirsbo JT and QT.
  • Greennmann
    Greennmann Member Posts: 13


    I saw tab 2 but i needed 7" on center and i wasn't sure if those values would work with wirsbo JT and QT.
  • Greennmann
    Greennmann Member Posts: 13


    A little clarification are these statements true?

    so what your saying is that it is impossible to calculate the k value for a radiant floor using the r values and other information about the floor unless your like a frecken scientist or genius. both of which i'm not

    the only people that know these values are the people that design the radiant design software and probably the manufactures of radiant heat tubing and products.
  • Brad White_9
    Brad White_9 Member Posts: 2,440
    Are we talking the same \"k\" factor here?

    Just to be clear (and this is for anyone), the "k" factors I am refering to in my post has to do with the transmission of heat through building materials, common ASHRAE stuff. Is there a "k" factor germaine to radiant floors that is not the same thing?

    If someone could set me straight here, that would help all of us. Still learning.

    And Greenmann- Let me know if you need any tutorials; I have some essays from my teaching days (ongoing albeit informally) that I will gladly send to you.

    Thanks!

    Brad


  • It's not as simple as the sum of the K values of the assembly. the "k value" of the whole assembly also has to take into account your heating element position, size, and contact to the assembly or the numbers are just not useful.

    To illustrate, the k value of the floor assembly in a joist application is the same for every inch of floor, so far so good. But if you only have tubing at 16" o.c., the effective k-value of the floor that is 8" from any tube, in terms of how much heat is conducted to the surface of the floor and is thus useful, is much higher than the effective k-value of the floor directly over the pipe.. right?

    Seigenthaler did an article on this in PM magazine in the last year.. maybe this is some misuse of "k-value" as a well defined term, but the concept is sound, I believe.


  • Well, it might be possible to approximate, but it takes some serious knowledge of heat transfer. It's not like adding up the components of a wall assembly to arrive at a total R value though. That calculation assumes the whole wall height is the same temp on both sides. In this case we have a few "points" of heat (the pipes) spreading heat into an emitter. Definitely more complex.
  • Josh_10
    Josh_10 Member Posts: 787


    Whole different ball game with QT man. I don't have any of those statistics. I am pretty sure nobody else in here does either. Do yourself a favor and get a copy of Wirsbo ADS 6.0. It is really good software. Most of us use it. With it you can choose many different floor coverings and choose from several tubing layout methods.

    Cheers

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  • Josh_10
    Josh_10 Member Posts: 787


    Ya, atleast I am Brad. I was joking with Rob about helping me complete my list. I would love to have that info though! I have thought about reverse engineering ADS to find the constants. Maybe I'll get around to it one of these days.

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  • Brad White_9
    Brad White_9 Member Posts: 2,440
    I can see the k value

    being constant, but the Delta-T is variable given the tubing "landscape" issue. Was that the article to which you refer?

    I take my radiant floor temperatures as an average (no matter what you will see some striping at least my IR gun does). Naturally I strive for the lowest water temperature (to minimize that gradient) and within that the tightest tube spacing I can afford. The thinnner the floor depth the tighter the spacing and lower the water temperature essentially.

    Just seeing if our terms and perceptions are aligning.
  • Greennmann
    Greennmann Member Posts: 13


    yeah, please send me them if its not to much trouble

    [email protected]

    Thanks
  • Brad White_9
    Brad White_9 Member Posts: 2,440
    Essays

    are on their way to you. Most all of them, not just heat loss.


  • http://www.pmengineer.com/CDA/Archives/97f2e86f7c298010VgnVCM100000f932a8c0____

    That one. it's a bit different than I remember but he does have an "accurate performance model" in there using what he calls an "A" factor, which I WANT SO BAD.

    I am misusing the term "k-value". If your goal is to find the average surface temperature of a floor for a given set of parameters though, part of that equation must calculate the resistance to transfer FROM THE PIPE to any given part of the floor. This is part of why aluminum plates rock, you get lateral spread at high conductivity to a wider portion of the floor, so your resistance to the floor overall surface is lessened. It makes no difference to the point of floor directly over the pipe, but it makes a big difference to that portion of the floor 2 or 4 inches away from the pipe on either side, right?

    So I don't see this as a question of Delta T, but then, maybe I'm misunderstanding how you approach this?
  • Brad White_9
    Brad White_9 Member Posts: 2,440
    What I see:

    "If your goal is to find the average surface temperature of a floor for a given set of parameters though, part of that equation must calculate the resistance to transfer FROM THE PIPE to any given part of the floor. This is part of why aluminum plates rock, you get lateral spread at high conductivity to a wider portion of the floor, so your resistance to the floor overall surface is lessened. It makes no difference to the point of floor directly over the pipe, but it makes a big difference to that portion of the floor 2 or 4 inches away from the pipe on either side, right?"

    I absolutely agree on the aluminum plate application, well-stated. The part about "from the pipe" is graphically expressed in isotherms, zones of equal temperature. This is where our feet pick up our imperfections if you will.

    The "k" factor is constant and the product of that multiplied by the variable delta-T therefore is always changing. But the constant remains so.

    Thanks for the article! Tubing length. Classic Siggy; My head hurts. In I go....

    Brad


  • well, I think what I would say is that the delta-T differential is *because* of the decreasing conductivity of the assemblage *relative to the position of the pipe*. So I'm seeing that as a flux in k-value for each area of the floor. You can simplify *IF* you can use isotherms to calculate delta-T differentials, but to GET the isotherms, you need the conductivity of the assemblage from the water to each point on the floor; and I think nailing that down is finite element analysis, unless I'm off base there..

    Now if someone has already calculated the isotherms, they have defined that K-value flux already, and you have the end result of it.. hopefully, unbiased. But who knows what assumptions are built into such models.. and output charts!
  • Brad White_9
    Brad White_9 Member Posts: 2,440
    Don't you

    just love dynamic heat transfer equations? :)
This discussion has been closed.