Re: calculat hot air unit to heat 4000 cfm of air to 100 degrees cent.

Question:
sir,please let me know how to calculate kilo watt rating of hot air unit to 
heat 400cfm of air from ambient temp. to 100 degrees centigrade.

Response,

Prasad,

Thank you for your question. I have to make a couple of assumptions in 
answering your question.  This is not unusual since people learning science 
are also still learning that even the maddest of scientists, because they 
are scientists, need to be careful about the way we communicate 
information.

The first assumption is that the term "cfm" means cubic feet per minute.  
This is my guess because I am not familiar with it.  The second assumption 
is that "ambient" temperature is 20 degrees Celsius.  The word ambient has 
no defined scientific meaning and can vary so we need to use a known number 
in the calculation.  

We will remind ourselves of some definitions along the way.  A good web 
site for definitions is: 
http://physics.nist.gov.cuu/Units/html 

You want to know how many kilowatts are needed to heat 400 cubic feet of air 
per minute from  20 Celcius to 100 Celcius.

Our first definition is "Heat Capacity" which is the amount of energy 
needed to raise the temperature of an amount of a substance.  Using grams 
and degrees Celsius, we define a calorie as "the amount of heat needed to 
raise a gram of water by a degree Celsius.  Other substances need different 
amounts of heat to raise their temperature.  The more heat they need the 
more heat energy they store when they are hot, that is the more capacity 
for heat they have.

Sometimes the term "specific heat" is used and this is more or less the 
same thing except that it relates the heat needed to raise a substance's 
temperature to the amount needed for water.  In the case of nitrogen the 
specific heat is approximately 0.25 ( at constant pressure ) so nitrogen 
needs 0.25 times the amount of heat that water needs to raise its 
temperature by one degree, that is 0.25 calories.

We now need to find out how much a cubic foot of air is in terms we can use 
to do the arithmetic.  I am going to make a few approximations to make this 
answer shorter and simpler.  First, I'm going to ignore the difference 
between oxygen and nitrogen and assume air behaves like nitrogen.  We will 
come back to this assumption.

Next we can work out how much a cubic foot of air weighs.  Here we must use 
the fact that a mole of gas occupies 22.4 litres at zero degrees Clesius 
and one atmosphere pressure.  A mole of nitrogen weighs 28 grams.  A foot 
is about 30 centimetres so a cubic foot is 30 x 30 x 30 cubic centimetres 
or 27 litres.  This means that the weight of a cubic foot of nitrogen is 28 
x 27 / 22.4 grams or 33.75 grams.  This is at zero Celsius not the 20 
degrees we have taken as our ambient temperature.  The higher temperature 
means that the volume of a mole of gas is higher so the weight of a cubic 
foot is lower.  Charles' Law allows us to estimate the volume increase due 
to the extra 20 degrees as 22.4 litres x 293 degrees / 273 degrees where 
the degrees are now in Kelvin so zero Celcius becomes 273 and 20 becomes 
293.  The difference is only about 7% but this adjustment means that our 
cubic foot weighs 31.44 grams.

To raise 400 cubic feet from 20 to 100 Celsius therefore takes
400 cubic feet x 31.44 grams x 0.25 calories per gram x 80 degrees which 
works out to be 251570 calories. To convert this to kilowatts we need first 
to divide by 4.2 to convert it to joules and this works out at 59898 
joules.  A watt is a unit of power or the rate of expenditure of energy ( 
which as Mr Joule demonstrated by stirring up liquids in his father's 
brewery in England, is equivalent to heat ).  We define a watt as a joule 
per second.  How many joules per second do we need ?  Every minute we need 
59898 joules which is very close to 1000 joules per second or watts.  How 
convenient !  This is one kilowatt.

To refine this calculation, we need to take into account that a fifth of 
the air is oxygen with a weight of 32 grams per mole and a specific heat of 
0.219.  I'll leave you to make the adjustments.  They will spoil the nice 
round number we have reached.

The web-site below also deals with this subject
antoine.fsu


 Best wishes