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Boiler firing rate

kcopp
kcopp Member Posts: 3,827
Looking to replace a Peerless EC-05 steam boiler.
I measured out the radiation as best I could (some really different CI convectors) and got a EDR number....it seemed low. I want to see what they were using for a firing rate.... boiler has a .50 gph 70W nozzle. Seemed way undersized for the boiler. I asked about the pressure firing rate and it was 1200 psi. Apparently this was an experimental burner. Anyone have an idea what the firing rate on a 1200psi half gallon/ hr is?
They are going to convert to gas so am I'm looking to eliminate the burner.

Comments

  • HVACNUT
    HVACNUT Member Posts: 4,209
    Pump pressure is not 1200 psi. 120 psi maybe. 
  • STEVEusaPA
    STEVEusaPA Member Posts: 5,322
    Depending on the burner, that boiler should be fired on or over 1.5 gph @ 140 psi.
    There's almost no reason to underfire a properly sized steam boiler.

    But if you have a burner that fires at 1200 psi, I'd like to see it, and I'll come get it off of you.

    Are they converting the boiler to gas, or replacing the boiler with gas?
    steve
  • kcopp
    kcopp Member Posts: 3,827
    1200 psi... not a typo.
    It was an experimental burner a few years back. Preheat.
    Happy to get pix.
    Im sure he would be willing to sell it.
    He actually invested money to get this onto the open market but it went bust.
    The boiler is shot, back wall toast.
    They are going to nat gas. Want a Nat gas unit now.
    Big place.
  • HVACNUT
    HVACNUT Member Posts: 4,209
    edited September 21
    I don't know how to do the math, but...
    .50 nozzle @ 200 PSI is .71 GPH for a 42% increase. 
    .50 nozzle @ 300 PSI is .87 GPH for a 74% increase. 
    Is it safe to say that a .50 nozzle @ 1200 PSI is a tad over fired for the boiler?
    I would like to see this burner. Maybe it needs a really long chamber. 
  • Steamhead
    Steamhead Member Posts: 14,857
    Is that a "Burner Booster"?
    All Steamed Up, Inc.
    Steam, Vapor & Hot-Water Heating Specialists
    Oil & Gas Burner Service
    Consulting
  • kcopp
    kcopp Member Posts: 3,827
    I believe so. It was a heck of a set up. Burned super clean he said.
  • Steamhead
    Steamhead Member Posts: 14,857
    I went to see this when it was first advertised. The principle appeared sound, and probably would have gained traction 30 or so years ago. But at that time, gas was a lot less expensive and still is.
    All Steamed Up, Inc.
    Steam, Vapor & Hot-Water Heating Specialists
    Oil & Gas Burner Service
    Consulting
    kcopp
  • EBEBRATT-Ed
    EBEBRATT-Ed Member Posts: 10,355
    here's the formula:

    F1= nozzle flow rate at (100psi)

    F2=nozzle flow rate at (1200psi)

    P1=100 psi nozzle pressure

    P2= 1200 psi new nozzle pressure


    F2=F1 x (P2/P1) .5 (exponent)


    No math wiz here I use a calculator or count on my fingers LOL

    I'll bet a smat guy like @Jamie Hall could whip this out


  • mattmia2
    mattmia2 Member Posts: 3,991
    edited September 22
    I have 2 engineering degrees and i'm not sure where .5(exponent) is supposed to go

    Is it:

    F2=F1 * ((P2/P1)^.5)

    note that X^.5=squareroot(X)
  • EBEBRATT-Ed
    EBEBRATT-Ed Member Posts: 10,355
    @mattmia2
    .5
    Yes I think so Exponent applies to the P2
    P1

    They had it written like above with brackets around the P1 & P2

    I guess some things you can't do on a keyboard

  • STEVEusaPA
    STEVEusaPA Member Posts: 5,322
    mattmia2 said:

    I have 2 engineering degrees and i'm not sure where .5(exponent) is supposed to go

    Is it:

    F2=F1 * ((P2/P1)^.5)

    note that X^.5=squareroot(X)

    Did you pay off both of those degrees or is there some kind of refund available...
    Of course I'm teasing you...easy...easy everyone...lol
    steve
  • EBEBRATT-Ed
    EBEBRATT-Ed Member Posts: 10,355
    I found what I was looking for on a Beckett nozzle chart.

    Divide new oil pressure (1200) new pressure) by (100 std nozzle pressure=12

    Find the square root of 12 (you can google it)=3.46410162

    Multiply the square root you found (3.46410162) x nozzle rating @100psi in this case .50gph

    3.46410162 x .50 (nozzle size) + a firing rate of 1.73205081 or 1.73 gph


    The one I always have stuck in my head for some reason is a 12 gph nozzle @300psi will give you about 20gph

    Answer: from Beckett Chart:20.79gph

    Answer from calculation: 300/100=3, square root of 3 is 1.74something 1.74 x 12=20.88
  • EdTheHeaterMan
    EdTheHeaterMan Member Posts: 2,397
    here is the correct way to write the equation

    Edward Young
    Retired HVAC Contractor from So. Jersey Shore.
    Cleaned & services first oil heating system at age 16
    mattmia2
  • mattmia2
    mattmia2 Member Posts: 3,991
    so it is this:
    F2=F1 * ((P2/P1)^.5)
    which is this:
    F2=F1 * sqrt(P2/P1)
  • STEVEusaPA
    STEVEusaPA Member Posts: 5,322
    According to the little app I made, I agree with @EBEBRATT-Ed. However when I looked up that formula that @EdTheHeaterMan posted, one of the notes was that it's not as accurate at pressures over 300 psi.

    steve
  • kcopp
    kcopp Member Posts: 3,827
    Wow! Thanks!
  • EBEBRATT-Ed
    EBEBRATT-Ed Member Posts: 10,355
    I also tried the formula that @EdTheHeaterMan posted but I don't digest exponents very well
    kcopp
  • STEVEusaPA
    STEVEusaPA Member Posts: 5,322
    edited October 21
    Parentheses, exponents, multiplication & division, addition & subtraction.
    P1=100
    P2=1200
    P2/P1=12
    12^.5= 3.46410161514 or the square root of 12
    F2=F1 * 3.46410161514
    F2= .5 * 3.46410161514
    F2= 1.73205080757
    steve
  • STEVEusaPA
    STEVEusaPA Member Posts: 5,322
    mattmia2 said:

    so it is this:
    F2=F1 * ((P2/P1)^.5)
    which is this:
    F2=F1 * sqrt(P2/P1)

    You don’t need the extra parentheses.

    F2=F1 * (P2/P1)^.5
    Either way ya need a calculator…

    steve
  • mattmia2
    mattmia2 Member Posts: 3,991
    don't need them but it make it more clear when you aren't typesetting it the way you would write it
  • EBEBRATT-Ed
    EBEBRATT-Ed Member Posts: 10,355
    I like the Beckett formula, much simpler for me at least