can anyone do this?

Question

the source of of emission is the standby diesel set. the generator set is maintained to assure highest efficiency and avoid poor fuel combustion that result to dark emissions. Emission expected from the operation of power generating sets consist of by products from combustion of fossil fuel. these are CO2,Co,NOx,Sox,water vapor, soot particles and traces of unburned hydrocarbons.

pollution loading of generator set
capacity: 550 kw
frequency: 60 Hz
fuel: diesel oil
consumption: 157 liters per hour

assuming that the generator set shall be used for an average of 1 hour per week, its annual fuel consumption(APC) shall be:

AFC = 157 liters/hr * 1 hr/week * 52 weeks/yr
= 8164 liters/yr or 2157 gallon/yr

using table in AP-42 5th edition: item II. internal combustion, sub-item a diesel ; <600 hp or 588.24 kw
the emission factors are:

particulate matters(PM) = 41.49 lbs/1000 gallons of fuel burned
carbon monoxide(CO) = 127.1 lbs/1000 gallons of fuel burned
nitrogen oxide(NO) = 590.2 lbs/1000 gallons of fuel burned
sulfur oxide(SO) = 38.31 lbs/1000 gallons of fuel burned
volatile organic compound(VOC) = 48.18 lbs/1000 gallons of fuel burned

pm loading annually

PM = 2157 gals/yr * ((41.49 lbs)/(1000 gals)) * ((1 metric ton)/(2200 lbs)) = 0.0407 metric ton/yr

CO loading annually

CO = 2157 gals/yr * ((127.1 lbs)/(1000 gals)) * ((1 metric ton)/(2200 lbs)) = 0.125 metric ton/yr

NOx loading annually

NOx = 2157 gals/yr * ((590.2 lbs)/(1000 gals)) * ((1 metric ton)/(2200 lbs)) = 0.579 metric ton/yr

SOx loading annually

SOx = 2157 gals/yr * ((38.31 lbs)/(1000 gals)) * ((1 metric ton)/(2200 lbs)) = 0.037 metric ton/yr

VOC loading annually

VOC = 2157 gals/yr * ((48.18 lbs)/(1000 gals)) * ((1 metric ton)/(2200 lbs)) = 0..040 metric ton/yr


that is my solution and i cant find a way to make its final unit mg/Ncm please help

thanks in advance :) and i wish you all luck who can get the answer :-D

Answer 1

I'm going to do a few conversions, let me know if any of this gets too confusing!


60 Hz frequency converts to 3600 RPM in the engine as 1 rpm is equal to 1/60 Hz

You stated that the capacity of the engine is 550kW, so i'll use that number as it is lower than the 588.24kW you stated later: 

Torque (or "T"), in Nm, is equal to Power ("P") over Angular Velocity aka Radians per Second ("w") in the equation

T = P/w

w can be found using another equation,  w = (2*pi*RPM)/60 , which gives w = approx. 22619.467

Using this, T (in Nm), is equal to about 1458.92 Nm

Now, 1 Nm is equal to 100 Ncm so 1458.92 Nm is 145892 Ncm

Assuming that your initial question is asking you to find the amount in mg of each emission product per torque in Ncm produced over the course of a year, you simply take the values you found earlier, convert from metric tons to mg, and put them over the Ncm value shown above:

So:

PM: 

0.0407 metric tons = 40700000 mg

40700000 / 145892 = 278.973 mg/Ncm

 

CO:
0.125 metric tons = 125000000 mg

125000000 / 145892 = 856.798 mg/Ncm

 

NOx:
0.579 metric tons = 579000000 mg

579000000 / 145892 = 3968.689 mg/Ncm

 

SOx:

0.037 metric tons = 37000000 mg

37000000 / 145892 = 253.612 mg/Ncm

 

VOC:

0.040 metric tons = 40000000 mg

40000000 / 145892 = 274.175 mg/Ncm

 

 

I wasn't quite sure if this was what you were asked to do, but hopefully it helps!