SO2 Scrubber

[aschaefer023]

I would just like to know what equations I would need to answer a question like this. I am sure once I have the equation the math will be very simple. Any help or advice would be greatly appreciated.

 

"When using a recirculating packed column scrubber that is 100% efficient how much sodium hydroxide at 20% would be used per hour if the flow rate of 300,000 ppm SO2 was 15 cfm? How many hours could we run the scrubber until the caustic had neutralized all of the SO2?"

 

I assume 100% efficient means the outlet gas stream has a SO2 concentration of 0% but I could be wrong. Can anyone give me any help setting up the equations for these two problems?

[John Cuthber]

Well, let's start with the good news. You are right about what they mean by 100%.

 

The bad news is that there's not enough information to answer the question.

But you can make a start on it (if there's more information in the question, you can probably answer it)

How much SO2 reacts with 1 mole of NaOH?

How SO2 much is there in 15cf at 30,000 ppm?

and so on...

[aschaefer023]

I have seen different equations for SO2 reacting with NaOH but this site:

 

http://www.monroeenvironmental.com/air-and-gas-cleaning-systems/packed-bed-fume-scrubbers/packed-bed-scrubbers-applications-and-engineering

 

suggests one mole of NaOH reacts with one mole of SO2 to form NaHSO3 and Na2SO3 (in the presence of excess NaOH)

 

I know the constant of 379 SCF per lbmol of gas. So 15cf X 1lbmol/379cf = 0.0396 lbmol gas total (or 17.952 mol)

 

the molar mass of air is 28.97 g/mol and the molar mass of SO2 is 64.07 g/mol so in 1 million grams of gas we have

 

(300,000g SO2 X 1 mol SO2/64.07g) + (700,000 g air X 1 mol air/28.97 g) = 28,845.31 mols (please dont comment on my sig figs I'm just quickly crunching numbers)

 

so 1,000,000 grams / 28,845.31 mols = 34.67 g/mol for the inlet gas

 

34.67 g/mol X 17.952 mols/15cf = 622.35 grams gas per 15 cf

 

622.35 g gas X 0.3 g SO2/g gas = 186.71 g SO2 per 15 cf

 

186.71 g SO2 X 1 mol/64.07 g SO2 = 2.91 mol SO2 per 15 cf of inlet gas

 

if we need one mol NaOH per one mol SO2 (please correct me if you believe this is wrong) then we also need 2.91 mol NaOH per 15 cf of inlet gas to react

 

according to chembuddy.com a solution of 20% NaOH has a density of 1.2219 g/mL

 

so in 1 mL we have 1.2219 g solution X 0.2 g NaOH/g solution = 0.2444 g NaOH per mL

 

molar mass of NaOH is 39.997 g/mol

 

need 2.91 mol NaOH

 

2.91 mol NaOH X 39.997 g/mol X 1mL/0.2444 g NaOH = 476.27 mL of 20% NaOH solution for every 15 cf

 

converting back to american units and changing so the answer is in hours and not minutes:

 

476.27 mL/minute X 60 minutes/hr X 1 gallon/3785.41mL = 7.55 gallons of 20% NaOH solution per hour

 

Does this seem right?

 

For the second part of the question I guess you are saying there isn't enough information because we do not know the initial volume of the inlet gas to be neutralized. But I could calculate how much time per cf of initial gas like:

 

15 ft³/min X 60 min/hr = 900 ft³/hr or that it will be 0.0011 hours per cf of the initial 300,000 ppm SO2

 

please comment back to let me know what you think of my calculations. Am I going about this correctly? Or am I way off?

 

Thanks

[John Cuthber]

You seem to have the right approach. I haven't time at the moment to check the details.