Name: Dan
Status: other
Age: 50s
Location: N/A
Country: N/A
Date: 2000-2001
Question:
1) How do I calculate the amount of energy achieved by
burning of Nagtural and LPG Gas.
Assuming I have a HOt air output in SCFM (1500) and the temperature
(1200F). How do I calculate the energy (in 1 hour)
Replies:
This question can be answered at several levels of sophistication, taking
into account heat transfer, the second law... I chose the simplest approach.
There is so much air that needs to be heated from 25C. to ~650 C. (1500 F.)
every hour.
This demands a certain amount of heat Q = C*(delta T) where C is the heat
capacity expressed in Kj/deg K*Kgm.
There is a fuel natural gas or LPG whatever. The fuel supplies a certain
amount of heat per mass, expressed as Kj/Kgm.
The heat demand expressed in Kj is then equated to the equivalent mass of
fuel (in Kgm).
The problem is simple, but it requires a number of conversion factors. I
hope I got them strung together correctly.
We need certain data from the literature:
The fuel energy of methane, ethane, and propane are all about the same
when expressed as heat per unit mass. This is 0.0301*10^3 Kj / Kgm
fuel.
The amount of air to be heated per hour is SCFM * 60 min = 9*10^4
SCF/hr. There are
2.832*10^4 liters/ft^3. So the amount of air is SL/hr = 255*10^4 SL/hr.
There are 22.4 SL[air] / mol[air] @ S.T.P. and the molecular weight of
air can be approximated by the mol weighted average of that of .
Assuming a weight percentage of N2 and O2 = 79 /21. The mol weight of
air is 28.84 gm/ mol. These can be combined to give the air flow in Kgm/hr,
which is: 32.812 Kg / Hr.
The heat capacity of air can also be approximated by the mol fraction
weighted average of N2 and O2. This turns out to be: N2 [29.124
j/K*mol] and O2 [54.317 j/K*mol]. This is:
1.1747 Kj/C.*Kg The temperature difference expressed in Celsius is
(650 - 25)= 625 C.
So the total heat demand is: Q = air flow(Kg/Hr)*heat capacity
(Kj/KgC.)* temp difference =
625 C. This equals: 32.812 (Kg/Hr) * 1.1747(Kj/Kg*C.) * 625 (C.) =
24.1*10^3 (Kj / Hr).
Equating the heat demand per Hr and the energy supplied by the fuel
(Kj/Kgm) you get:
NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.
