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pH Material Balance - Feasible?

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OilBoiler

Chemical
Aug 5, 2003
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Hi everyone,

I'm working on a project to achieve pH control of our storm sewer sump. As you know many streams are the source of waste water, but we have identified the main ones. I'm attempting to demonstrate that if we properly neutralize the largest source of water, then any pH fluctuations of the other two main sources will be dampened.

What I'm doing is the following: I have three sources of water with known volumes and known pHs. I'm doing a component material balance based on the hydrogen ions concentration and the equation looks something like this:

(V1 + [H+]1) + (V2 + [H+]2) + (V3 + [H+]3) = VT + [H+]T

Where stream "T" is the total stream or the sump itself. The unknown would be the hydrogen ion concentration at "T." The pH can be easily found from there with a reverse log.

I did a spreadsheet and changed flows and pHs to predict final pHs and it seems to make sense. I just want to check with you guys and see what you think of this approach, whether it makes sense to you or not.

Again, I appreciate your help, thanks!
 
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Your approach may work for streams where the pH is constant and known in the 2nd two streams. pH, though, varies greatly with relatively small changes in H+ concentration. "Dampen" is a rather mild term for this. In order to change the pH of the water by "1" you need to add 100 times that volume of neutral water. In your scenerio it seems that the dampening wouldn't be noticeable. Your main stream would have to be very much larger than your other two streams in this case. If you were to take this approach in treating only one of the three streams, I would suggest erring on the side of caution, not only nuetralizing the pH of that stream, but also buffering it for local pH changes in the other two streams.

Perhaps it's a typo, but I believe your equation uses a "+" where it should be "*".
Here's how I see it:
V1*[H1] + V2*[H2] + V3*[H3] = Total moles H+
Total [H+]= (Total Moles H+)/(V1+V2+V3)
From this you can predict the final pH.

On another note, if you have sensors in all three streams, you may be able to create a feedforward control loop, where the controller reads the pH from each stream, predicts the final pH, and then adjusts the pH of the main stream to account for the other two streams combined.

ChemE, M.E. EIT
"The only constant in life is change." -Bruce Lee
 
Sorry... misprint. To lower/raise the pH by "1" with neutral water, you need 10 times the volume... not 100... Of course, you probably already caught the mistake.

ChemE, M.E. EIT
"The only constant in life is change." -Bruce Lee
 
aspearin1,

Thanks for the feedback! Yes, I'm sorry, it's a typo. I was trying to retype what I had in Word so I got confused.

Anyway... yes, you're right... The main stream we're trying to control it's about 500 gpm, while the other may be only about 10 gpm at normal conditions or I would say around 30 gpm max under upset conditions.

We already have a feedforward control at the storm sump and it has not worked well at all. Apparently, they have tried to tune it many times before I got here and it hasn't seemed to work. The pH is not constant at all in the other two low-flow streams. It fluctuates quite a bit from 12 to 2 in just a few minutes. In fact, if you look at the trend you could say that it neutralizes itself because of the peaks it has. Anyway... we are considering buffering pH at the other two streams. We are just starting by tackling the main stream, since it's so much larger than the other two.

I appreciate your help, thanks!
 
I don'nt see how the equation can work
if you have three streams
V1=V2=V3
stream 1 pure water
stream 2 strong acid 0.001 N
stream 3 strong nase 0.001 N
pH1=7 pH2=3 pH3=11

[H1]=1.e-7 [H2]=1.e-3 [H3]=1.e-11
(V1*H1+V2*H2+V3*H3)/(V1+V2+V3)=3.33 e-4
yieding pH=3.48 obviously wrong since stream 2 neutralizes stream 3 , PH should be almost 7.

besides, should you have one stream strongly buffered, and the others not, your equation will also fail.
 
I think siretb is entirely right.

When two fully ionized solutions are mixed, and one is of a strong acid with concetration x[sub]A[/sub] with pH=3, and the other is of a strong base with concetration x[sub]B[/sub] with pH=11, the pH of the blend would be estimated from:

x[sub]A[/sub] = [H[sup]+[/sup]] = 10[sup] -pH[/sup], for the acid
x[sub]B[/sub]= [OH[sup]-[/sup]] = 10 [sup]pH-14[/sup], for the base

x[sub]A[/sub]-x[sub]B[/sub] = 10 [sup]-pH[/sup] - 10 [sup]pH-14[/sup]​

In siretb's example x[sub]A[/sub] = x[sub]B[/sub] thus

10[sup] pH-14[/sup] = 10 [sup]-pH[/sup] resulting in pH=7.

When neutralizing weak acids having K[sub]A[/sub] as dissociation constants,

x[sub]B[/sub] - x[sub]A[/sub]/(1+10 [sup]pK[sub]A[/sub]-pH[/sup]) = 10 [sup]pH-14[/sup] - 10 [sup]-pH[/sup]​

When pH>>pK[sub]A[/sub] we approach the previous equality.
When pH=pK[sub]A[/sub] meaning the acid is half dissociated, and x[sub]B[/sub] = 1/2 x[sub]A[/sub], we get pH=7.

Shinskey explains why although feedforward control may be helpful in reacting to flow upsets, it is not sufficient for the required precision needed in neutralizing acidities that can only be approximated with a well-mixed tank provided with a good measuring system on feedback control.

I'd recommend reading the corresponding paragraphs in
Process Control Systems
by F.G. Shinskey (McGraw-Hill) where you'll find a thorough discussion of pH control.

 
I also agree with siretb. You equation does not take into account any buffering materials and the effect that the materials would have on the consumption of neutralizing chemicals.

Your post prompts some additional comments. You are calling this a storm sewer sump. In my experience, storm water is usually non-contaminated storm runoff that has not been in contact with any process. If you indeed are having pH swings in storm water, you have more issues than pH to deal with.
 
Thanks guys for the feedback!

I agree completely and will do more research on it.

bimr, yes, trust me I do know that we have more problems than just pH! That's why it has not been a straight forward approach so far.

Thanks again.
 
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