BTU Output of A Pellet Stove

Fire_Man_2014

Hello!
This problem has been presented to me by my co-op company and I am seeking verification or guidance in the right direction.
We are a green energy company seeking to be Net Zero Energy by Summer of 2011. In order to do this, I need to find out (within some relative error) how much power the two pellet stoves in our building are actually producing. My attempt at this is below using the specific heat equation (which I'm not even sure if that is the correct way to go at this).
Note: I know that some heat is produced through thermal radiation as well, but this thread is ONLY focusing on the heat being pushed out of the ventilation by the blower.

Given/measurements:
Rated Air flow Rate: 135 (CFM) = 3.823(m^3/min)
Ti (initial)= Ambient Room Temp= 62 (F) = 16.66(C)
Tf (final) = Vent Air Temp.= 200(F) =93.33(C)
DeltaT=(Tf-Ti)=76.67 Kelvin

Energy=c*m*(DeltaT)
where..
c=specific heat of air= 1.005 kJ/(kg*K)
m=mass of air=Density*Rate*Time=1.29(kg/m^3)*135(CFM) (only using rate here, time can be taken into account later)
DeltaT= Tf-Ti
Converting cubic feet/min to cubic meters/min = 3.823(m^3/min)
I’m using the Rated CFM here because I have yet to measure the actual CFM. Eventually I will.
so..
1.005(kJ/(kg*K))*1.29(kg/m^3)*3.823(m^3/min)*(Tf-Ti)
Everything is held constant with the assumptions made above except the Temperatures.
Assuming that the Tf=200 F = 93.33 C
and Ti=62 F= 16.66 C
DeltaT=76.67 K (change in degrees kelvin= change in degrees Celsius)


1.005(kJ/(kg*K))*1.29(kg/m^3)*3.823(m^3/min)*(76.67K)=380(kJ/min)
Conversion factor for kJ >>>> BTU
1kJ=.947817 BTU
380 kJ/min= 21,600 BTU/hr
Conversion for kW
1 kW= 3412.14 BTU/hr
21,600 BTU/hr=6.33 kW

So, this is my attempt, any input on this is welcomed and appreciated.