Finding particulate matter emissions

[chemman218]

Homework Statement

A power plant burns oil that is 4% ash and 3% Sulfur. At 50% excess air, what particulate (mg/m^3) emissions would you expect?

Homework Equations

PM in grams = 2.8*.04=.112 grams of PM per 1000 kg of oil
Start with 1000 g of oil
oil = 4% ash, 3%sulfur, and 93% carbon
oxygen is 20% of air
PM will be found in units of mg/m^3
PV = nrt
where p = 1 atm, r= .08206 l*atm/mole*k, t= 298K

The Attempt at a Solution

I started with the simple combustion equation provided by the professor:

C + O2 ---> CO2

The mole ratio of each part is 1.
I then found the amount in grams or carbon. Since we started with 1000g of oil, we would have 930g of Carbon.

930g C x (1 mole carbon/12g of carbon)x(1 moleO2/1mole C) = 77.5 moles of carbon.

Since the mole ratio was one to one. Then there would be 77.5 moles of oxygen as well.

77.5 moles of carbon x (1 mole of O2/ 1 mole of C) = 77.5 moles of O2. Then found the amount of O2 burned intially.

77.5 moles of O2 X (32g of O2/1 mole of O2) = 2480 g of O2. Now to find the moles of air used.

77.5 moles of O2 = .20 x (moles of air)
x = 387.5 moles of air

Now that the moles have been found, you can find the volume of air using
Pv=nrt
v=nrt/p
v= ((387.5 moles of air)*(.08206 L*atm/mol*K)*(273K))/ 1atm
v= 8680.92 L

air was in 50% excess in the reaction so there needs to be a conversion.

8680.92 L x ( 1.5)= 13021.4 L of air
volume needs to be in m^3

13021.4 L x (1 m^3/1000 L) = 13.021 m^3

The amount of PM is equal to: 2.8 x .04 = .112 grams of PM
This needs to be in mg

.112 grams of PM x (1000 mg/ 1 gram) = 112 grams of PM

Final solution of PM: 112 grams of PM / 13.021 m^3 = 8.60 mg/m^3 of PM

***** I am not sure if i correctly interpreted the excess air part of the formula. Is my method correct for finding moles of O2 to then find moles of air and using the ideal gas law? Thanks.

 

[chemman218]

I actually went back and calculated the grams of O2 and used the molar mass of air which was:

77.5 moles of Carbon x (1 mole of O2 / 1 mole of C) x ( 32 grams of 02 / 1 mole of 02) = 2480 g of O2

2480g of O2 = .20 x ( grams of Air)
grams of air = 12400

12400 grams of air x (1 mole of air / 28.967 g of Air ) = 428.073 moles of air

Pv=nrt
v=((428.073 moles of air)*(.08206 L*atm/mol*K)*(273K))/ 1atm
v=9589.85 L of air

air is in excess of 50%...so 9589.85 X 1.5 = 14384.5 L
converting 14384.5 L to m^3 is 14.3845 m^3

Final solution of PM: 112 grams of PM / 14.385 m^3 = 7.79 mg/m^3 of PM

**** I am not sure which route is correct in solving the problem****

 

[Astronuc]

Well, excess air should an amount of air beyond that which is required for stoichiometric combustion.

Since this is oil rather than coal, any thoughts about the amount of hydrogen in the oil? Combustion of oil (some heavy alkane?) would produce CO₂ and H₂O.

See this example - http://www.ohio.edu/mechanical/thermo/Applied/Chapt.7_11/Chapter11.html

Are there similar examples in one's text?

 

[chemman218]

Our professor just told us to use the simple combustion formula... C+ O2 ---> CO2
He gave us some hints saying that we needed to solve for the mass of carbon and oxygen initally burned in order to find the amount of air.
He just told us to use the fact that oxygen makes up 20% of air.
The breakdown of the oil was: 4% ash, 3% sulfur, and 93% carbon.

That link was helpful, but the only example in the link I could find was similar was 11.1, but still did not really help me.

 

[chemman218]

There are no other problems like this one in the text.

 

[niazkilam]

"PM in grams = 2.8*.04=.112 grams of PM per 1000 kg of oil"
where did 2.8 come from?

 

[SteamKing]

"PM in grams = 2.8*.04=.112 grams of PM per 1000 kg of oil"
where did 2.8 come from?

chemman218 has dropped out of sight from PF, unfortunately. You may not receive a reply from him.

 

[Astronuc]

The breakdown of the oil was: 4% ash, 3% sulfur, and 93% carbon.

"PM in grams = 2.8*.04=.112 grams of PM per 1000 kg of oil"
where did 2.8 come from?

One would have to know if the 0.04 = 4% by mass of the PM in the oil, or some other basis, e.g., by volume.

If 0.04 is dimensionless, then the units of 2.8 would be the same as the units of 0.112, but it's not clear what the 0.04 represents. I would think 4 kg ash/100 kg would be quite high. The result, 0.112 gm PM in 1000 kg of oil seems to represent a very low ash content. So, something seems amiss. I wonder if the results is gms of PM per kg (1000 g) of oil.