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Author Topic: Boiler temp drop math  (Read 3568 times)

heat550

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Boiler temp drop math
« on: February 13, 2016, 02:08:42 AM »

Is there away to figure out how many BTU's are going out of Boiler by using the amount of temp drop in each zone ?
I know Im running close to average 5-10 GPM average on all zones x3 . even at 55 degree temp drop at boiler below 0f  all fan on .
Im not seeing any big difference in wood going throw stove . I did math other way by using radiator out put  and air exchanger out
put to fan size 55 degress temp drop on water was 209,000 BTU's  reason Im asking is Im starting to figure out the problems in my system .
This is the 400dcss .

 I was used to the old 200css just draft if would ever shut off I was amazed below 0f its a big difference with bigger stove
400dcss is a work horse   :thumbup:

Heat550
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heat550

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Re: Boiler temp drop math
« Reply #1 on: February 13, 2016, 02:23:23 AM »

The Formula
Math is the key to understanding how BTUs move through a system. The simple formula for water is System Delivered BTU = 500 x GPM x System Water Temperature Change.

this sound right ? :o

Could a side arm hot water ex changer pull 12,000-15,000 btus ? its only costing me 8 bucks electric a mouth for hot water .

heat550
« Last Edit: February 13, 2016, 02:52:02 AM by heat550 »
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hondaracer2oo4

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Re: Boiler temp drop math
« Reply #2 on: February 13, 2016, 07:25:50 AM »

The problem with that formula is that you really need to know the accurate gpm flowing. I was thinking about this same question about my system. How does this sound? I have monitoring software my system and I can watch the FHA fan kick on and watch the temp before and after it runs. My boiler holds 200 gallons of water. When my system runs for 8 mins my temp drop is between 3.5-4 degrees after it is done running. If we multiply 8.3lb x 200 gallons we get 1660 lbs. if we lose about .5 degrees per min of running we can multiply .5x1660 which equals 830 btus per min. Now this seems low to me since even at its coldest my FHA runs for 30 mins an hour. That would mean that my house only needs 24,900 btus when it's 0 degrees out????
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mlappin

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Re: Boiler temp drop math
« Reply #3 on: February 13, 2016, 11:25:39 AM »

The Formula
Math is the key to understanding how BTUs move through a system. The simple formula for water is System Delivered BTU = 500 x GPM x System Water Temperature Change.

this sound right ? :o

Could a side arm hot water ex changer pull 12,000-15,000 btus ? its only costing me 8 bucks electric a mouth for hot water .

heat550

I redid all my hydronic plumbing, primary/secondary loops, temp gauges after each loop to monitor heat use, etc.

I kept a sidearm as an experiment, normally I see the sidearm pulling 3-5 degrees out of the water after using a lot of domestic hot water, the warmer the water in the water heater is getting the less the sidearm pulls. Can’t tell you my actual flow as it varies as I’m running a delta T pump.
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heat550

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Re: Boiler temp drop math
« Reply #4 on: February 13, 2016, 11:56:35 AM »

I think this is the keey to getting most out of our boilers. I'm figuring it out both ways . I know the cuft of wood burning per hour and that formula is not perfect but its dam close. It's 500 x gpm x temp drop = BTUs  because if load on stove is between 150,000 to 206,000 BTUs at -4f  it burned  23 cuft in 12 hours. The math gets nuts when it comes to boilers. If I get something closer will post it . my math I'm figuring stove is 50% eff. So half woods BTUs are subtracted. But its telling me side arm on water heater and regular hot line bleed off is ,15,000btus that might be closer then I think there's 5 zonevalve in my house so there is a lot of line heat bleed off.

Heat550
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mlappin

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Re: Boiler temp drop math
« Reply #5 on: February 13, 2016, 01:32:09 PM »

Another way, look up the BTU’s of your water heater, figure in the percentage of how well it kept up, then figure out how well your plate or sidearm keeps up, then do the math.
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willieG

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Re: Boiler temp drop math
« Reply #6 on: February 13, 2016, 01:50:06 PM »

I read where you said 8 dollars a month to run your hot water heater
8 dollars a month  and we guess it is 9 cents per KWH means you get about 10 kwh for a buck or 80 per month for 8 bucks. a KWH - 3412 btu x 80 = (rounded) 273,000 btu lets say you are heating water  from 60 degrees to 130 degrees, that's a 70 degree rise and that can be said to be 616 btu per gallon so for the month you could heat about 443 gallons of water. or about 14 gallons per day


but back to the 273,000 btu for 9 cents per kwh for 8 dollars it is said that we can recover about 6,800 btu from a pound of wood (at 100 percent efficient) so lets make a guess you are about 60 percent so you could recoup about 4000 btu per pound so that 273,000 that you paid 8 dollars for will cost you about 68 pounds of wood (more likely 100)
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heat550

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Re: Boiler temp drop math
« Reply #7 on: October 10, 2016, 01:31:30 AM »

Its a old post but great thinking . I can watch it closer because I have electric meter just for my hot water heater :)

Heat550
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