- Aug 31, 2004--- 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. - << Previous post in topic