Energy Calculation ```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: 800 Kgm fuel / hr V. Calder Click here to return to the Physics Archives

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.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012