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Mixing valves

I am trying to figure out the math for using a mixing valve on a water heater to increase the tank temp to 150 degrees and keep 120 degrees out to fixtures to see how many GPM of hot water are actually being used for every gallon of tempered water so we can see exactly how much volume is increase. Any takers?


  • Mark Eatherton
    Mark Eatherton Member Posts: 5,853
    Do the math...

    One gallon raised from 40 to 120 degrees F requires 666 btu's/

    An 80 gallon storage tank full of 150 degree F water represents 73,300 btu's.

    Assuming you can only draw 80% of the tanks stored volume, then 73,300 btu's times .8 = 58,640 available STORAGE btu's. Add to this the capacity of the boiler/heat exchanger assembly and you have your hourly capacity, but the storage tank being held at 150 will change 80 gallons into 88 gallons of 120 degree F water. (58,640 divided by 666 = 88 gallons)

    If you are using a tank with internal baffles that will enhance the tanks output prior to mix and dilution, then your output will also change. 73,300 divided by 666 = 110 gallons.

    The net throughput of the boiler can then be divided by 666 to see what it's contribution would be.

    You can adjust the math to your particular situation as it pertains to starting temperature, storage tank temperature, throughput etc.



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  • WaterHeaterGuy
    WaterHeaterGuy Member Posts: 80
    Missing a value

    Need to get an idea of the cold water temperature feeding your mixing valve.

    Try this little calculator I made up for figuring mixing percentages of static flows:

  • EricAune
    EricAune Member Posts: 432
    Any other format for link?

    I don't think that format is supported here.  All I get is some HTML, no formula.

    There was an error rendering this rich post.

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