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CV and pressure drop

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Stokehold
Stokehold Member Posts: 43
Hello!

I'm confused again!
If I have a heating manifold rated at a max of 1.3 GPM per loop and a kvs of 1.10, how do I determine my head loss through this manifold based on a flow of .80 GPM per loop or a total of 4 GPM through the floor (5 loops)? The kv value obviously needs to be converted to cv.
If I understand properly from a previous post on mixing valves, the cv value is the max. flow rate to equal 1 PSI drop (2.3 ft. head).
I am just trying to establish total head loss in the zone so I can size a circulator.

Thank you for your expertise!

Comments

  • Gman66
    Gman66 Member Posts: 42
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    Cv is the max flow rate equal to 1 PSI drop.

    Engineering Toolbox website tells me 1.10 Kv = 1.27 Cv (Kv is just the metric version of Cv). My guess is that your "max rating" of manifold is really the Cv. At any rate, with flow of .8 g/min and Cv of 1.27 you get a PSI drop of (.8/1.27)^2 = .4 which is .4 x 2.3 = .91 ft. head.
  • Stokehold
    Stokehold Member Posts: 43
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    Hello Gman66,

    The .8 flow rate is my design per loop. There are a total of (5) loops. Do I figure a loss of .91 ft. per loop for a total loss of 4.55?
    My design flow through the zone is just about 4 GPM.
    I already calculated the head loss on the pex through the floor, I am just trying to establish the loss through the manifold itself. The manufacturer simply states max of 1.32 GPM at 1.10 Kv. Again, I am not sure if this head loss (1.10Kv) is per flow adjuster or the total through put.

    Thanks!
  • Ironman
    Ironman Member Posts: 7,379
    edited March 2017
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    You take the highest head loss of any loop and the total gpm of all loops to determine the circ size. Don't add all the loops head loss together; just use the highest one.

    I would suspect that the Kv rating is per port of the manifold, but again, just use the highest one.

    I don't have the pump curve in front of me, but you're looking at a Grundfos ups15-58 (or equivalent) on low or medium speed. That's assuming your loops don't exceed 300' each of 1/2" pex.
    Bob Boan
    You can choose to do what you want, but you cannot choose the consequences.
  • Alan (California Radiant) Forbes
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    If you have other zones on your system, you might want to get an Alpha pump and let it figure out the pressure drop.
    8.33 lbs./gal. x 60 min./hr. x 20°ΔT = 10,000 BTU's/hour

    Two btu per sq ft for degree difference for a slab
  • Stokehold
    Stokehold Member Posts: 43
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    I have calculated head loss for everything in the system except the manifold itself.
    REHAU, for example, has a very specific chart to figure head loss through their manifolds including the header, isolation valves, circuit balancing valves (wide open), and flow gauges. This does not include the loops themselves.
    For instance, based on my design, using their PROBALANCE five port manifold would give me approx. 3.8 ft. head loss just through the manifold itself.
    My manifold specs only state a maximum of 1.32 GPM at a Kv of 1.10.
    I have to assume since the max. 1.32 GPM is PER circuit, the Kv or Cv is also PER circuit because those specs apply to all of their manifolds, regardless of loops.
    Now, if my flow rate through each circuit is only .8 GPM, I again would assume the Cv/Kv must also be lower.
    I hate to assume anything and please pardon my ignorance.

    Thanks again!

  • EBEBRATT-Ed
    EBEBRATT-Ed Member Posts: 15,607
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    a product with a Cv of 1 = 1 gpm flow with a 1psi pressure drop.

    get a Bell & Gossett "system syzer wheel" or better yet download the electronic version for free. It will do all the calculations for you.
  • Ironman
    Ironman Member Posts: 7,379
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    You don't total the head loss when calculating. You simply use the highest value because that will be what the pump sees. In other words, if I have 4 loops with each having a head loss of 5 ft. oh head, and a 5th loop loop with a head loss of 6 ft., then 6' is my head loss. It's that simple.

    Gpm must be totaled from all loops, but only the highest head loss from any loop is used.
    Bob Boan
    You can choose to do what you want, but you cannot choose the consequences.
  • Gman66
    Gman66 Member Posts: 42
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    Stokehold said:

    I have calculated head loss for everything in the system except the manifold itself.
    REHAU, for example, has a very specific chart to figure head loss through their manifolds including the header, isolation valves, circuit balancing valves (wide open), and flow gauges. This does not include the loops themselves.
    For instance, based on my design, using their PROBALANCE five port manifold would give me approx. 3.8 ft. head loss just through the manifold itself.
    My manifold specs only state a maximum of 1.32 GPM at a Kv of 1.10.
    I have to assume since the max. 1.32 GPM is PER circuit, the Kv or Cv is also PER circuit because those specs apply to all of their manifolds, regardless of loops.
    Now, if my flow rate through each circuit is only .8 GPM, I again would assume the Cv/Kv must also be lower.
    I hate to assume anything and please pardon my ignorance.

    Thanks again!

    I don't think you are reading the Probalance chart correctly. I get a .8 ft head for a total of 4 gallons through 5 outlets. As others have said, the loops off the manifold will be in parallel and their pressure loss is not additive, you just find pressure loss across the longest loop, add the manifold (suspect just under 1 ft for your .8 gallons) and that is what you need. When calculating the pressure drop across the common piping to and from the manifolds you will want to use the aggregate flow (4 gallons in your example).


  • Stokehold
    Stokehold Member Posts: 43
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    I WAS reading the Rehau chart incorrectly! I want to thank everyone who commented here.
    Tinman
  • Alan (California Radiant) Forbes
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    Gman66 is correct. The total head for 5 outlets at 4 gpm is 0.8 feet. See page 6:

    https://www.rehau.com/download/871868/855661-pro-balance-1-in-brass-manifold-product-instructions.pdf

    For pump sizing purposes:
    2.85' - tubing pressure drop
    0.80' - manifold pressure drop
    3.65' - Sub total

    To this, add the pressure drop for the piping to and from the pump to the manifold and the pressure drop through the boiler if you don't have primary-secondary piping.
    8.33 lbs./gal. x 60 min./hr. x 20°ΔT = 10,000 BTU's/hour

    Two btu per sq ft for degree difference for a slab
  • Stokehold
    Stokehold Member Posts: 43
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    That 2.85 ft. loss for tubing has me stumped! Are you referring to the longest pex loop? In my case, that's 275 ft. x 1/2" and would result in approx. 7.5 ft. head loss at 100°.


  • Alan (California Radiant) Forbes
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    I used my RadPad. Set the flow per tube on the left (½" tubing, ½ gpm) and read the pressure drop opposite loop length at the top. I was figuring 300'.
    8.33 lbs./gal. x 60 min./hr. x 20°ΔT = 10,000 BTU's/hour

    Two btu per sq ft for degree difference for a slab
    Ironman
  • Zman
    Zman Member Posts: 7,572
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    Stokehold said:

    That 2.85 ft. loss for tubing has me stumped! Are you referring to the longest pex loop? In my case, that's 275 ft. x 1/2" and would result in approx. 7.5 ft. head loss at 100°.


    You are correct that if you need .8 gpm, your headloss will be right around 7.5'.
    "If you can't explain it simply, you don't understand it well enough"
    Albert Einstein
  • Alan (California Radiant) Forbes
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    My bad. I was reading the pressure drop at .5 gpm.
    8.33 lbs./gal. x 60 min./hr. x 20°ΔT = 10,000 BTU's/hour

    Two btu per sq ft for degree difference for a slab