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11805Re: Open Flames Burner Power

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  • beerguy84
    Aug 31, 2004
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      --- In new_distillers@yahoogroups.com, BOKAKOB <bokakob@y...> wrote:
      > I would like to ask if someone could post a sequence of
      measurements and calculation formulas for computing power in a boiler
      placed on a burner. I am sure there is a formula to calculate the
      wattage required to heat the given volume of water (say water) from
      one temperature to another with and without losses. I would like
      somehow to calibrate or at least to know the magnitude of my burner.
      For example, fully open it produces 4000 watt and half closed it is
      2300 watts. Are there any mechanical or chemical engineers? Thank you
      in advance, Alex...


      Heating of water or anything for that matter is a function of a
      variable called the specific heat. Values for specific heats can be
      located in any thermodynamics textbook or engineering handbook. The
      science of heat study is often called calorimetry, which is where the
      word calories comes from.

      First some nomenclature

      Kg=Kilograms=1000 grams
      C= Centrigrade or Celsius temperature scale
      M^3= Cubic Meters
      KJ=Kilojoules=1000 J = measurement of energy
      Q=Heat
      K=Kelvin temperature scale
      J=Joules=measurement of energy
      Cp=Specific Heat
      1 Watt=1 J/sec.


      Step one
      The equation to calculate heat is quite simple

      Q=M*Cp*delta T

      The heat, represented by Q, required to increase the temperature of a
      material is a funtion of the mass of the material, for simplicity
      I'll use the metric value of mass, Kg. The density of water at 20
      degees centigrade is 996 Kg/M^3. According to my conversion chart 1
      liter of liquid is approximately 0.001 M^3. , Multiplying the volume
      you have in liters by 0.001 will convert the volume from liters to
      cubic meters (M^3). Multiply this by the density and you will have
      the mass of the volume in question.

      Step two

      Find the specific heat of the material you are dealing with. In this
      case the specific heat of water at 1 atmosphere and 27 degrees
      celsius is 4.179 KJ/Kg*K.

      Step three

      Determine the final temperature and convert this to degrees Kelvin.
      This is done by adding 273.15 to the degree celsius value.

      If you start at room temperature, 20 degrees celcius, and wish to
      boil water, 100 degrees celcius, the difference ( delta T ) in
      temperature is 100-20 or 80 degrees celcius. Add 273.15 to this
      value and you have 353.15 K. Kelvin is the absolute temperature
      scale and it is very rarely referred to as degrees Kelvin, most just
      refer to it as Kelvin.

      Step four

      If you examine the specific heat value and more importantly the units
      associated with it, the next step is self explainatory. Multiply the
      specific heat by the temperature value and the mass value and you
      will end up with a value of heat energy required in KJ. In this case
      the Kg will cancel out and the K will cancel out leaving you with the
      value for energy, KJ.


      A note of caution.............

      Not all engineering textbooks or engineering handbooks are equal, in
      most cases the values for the specific heat may be stated in
      different units. The same is true for the units of density. One of
      the first fundemental principals of engineering is to watch your
      units!

      What remains true are these simple facts:

      Density will always be reported as a mass divided by a volume

      Specific heat will always be reported as a unit of energy divided by
      a mass and a temperature value.

      For us in the states it is even more difficult because the unit of
      mass is not the pound, or Kg, but the slug and that requires another
      iteration to determine the correct answer. When in doubt use the
      metric values, as this will make things easier when you determine the
      wattage.

      Another issue here is heat loss. I have not taken into account heat
      loss during heating and this can be substantial. Heat loss
      calculations are not that easy to calculate and many published
      formulas for heat loss are based on empirically derived test data.
      Each situation will be different. Heat loss is also a specialized
      area of Mechanical Engineering referred to as Heat Transfer and it is
      a upper level engineering course in most curiculums.

      There is also another factor involved here and that is the
      temperature rise of the heating vessel. Copper is an excellent
      conductor of heat and this will have to heat up prior to the water
      heating up. If you use an immersion heater that negates this effect
      somewhat as the heater is in direct contact with the liquid and the
      liquid will heat up first. If your using an open burner then the
      vessel will have to reach temperature first.

      What I have posted here is a very basic outline of entry level
      thermodyamics. You do not need to be a Mechanical Engineer like
      myself to do this stuff, all you need is the basics. Pick up a good
      reference textbook on Thermodynamics or an Engineering handbook and
      have at it. What you want to focus on is the discussion of specific
      heat calculations.
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