--- In firstname.lastname@example.org
, 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
C= Centrigrade or Celsius temperature scale
M^3= Cubic Meters
KJ=Kilojoules=1000 J = measurement of energy
K=Kelvin temperature scale
J=Joules=measurement of energy
1 Watt=1 J/sec.
The equation to calculate heat is quite simple
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.
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.
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.
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
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
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