07-27-2012 09:35 AM
Flow rate converts to 20257 m3/h and temperature 523 C
First you need to "normalize" the exhaust flow rate: 20257*273/(273+523) = 6947 [Normal cubic meters per hour]
Then the g/hph ( let's call it A ) is calculated from :
150 * (Normalized flow rate ) = A * (engine power) * (22.4/30.04) * 1000
A = (150 * 6947 ) / (1818 * (22.4/30.04) * 1000 )
A= 0.76 g/hp-h
07-25-2012 09:10 AM
Is the 18,907#/hour wet or dry? Do you know the complete composition of your exhaust? If you know the complete composition in ppmvd and you know the water content in ppmv wet or in ppmw, take everything to wet conditions. Convert the ppmv wet to mass ratios by multiplying the percent of each constituent by its molecular weight. Total the resultant numbers and divide the mass of formaldehyde by the total to give the weight percent of formaldehyde. Multiply this by the mass flow of the exhaust and divide by the bhp. There is another way working with volumes but this is one of the simplest and more accurate as long as you are sure of the #/hour massflow.
07-10-2012 05:14 PM
I need to convert 150ppmvd of Formaldehyde @ 9.2% exhaust O2 to g/bhp-hr to see if we meet a spec. Anyone have any advise on how to perform this calculation? I did find that the molecular weight of formaldehyde is 30.04 g/mole. Engine in question is a G3516B, 1818bhp, 11,923cfm exhaust at 974F (18,907 lbs/hr). Please help !