Somewhat off topic....

[tgmb]

A dilution question. (That indirectly relates to pool turnover math, I suppose..http://ppoa.org/?page_id=463)

Given: A standard I.V. tube (4mm internal diameter, with drip chamber, 60" length) containing 20ml of water, to the midline of the drip chamber. (Volume A.) Tubing oriented vertically. Infuse a second, 20ml volume of water (Volume B), allowing drainage from the bottom of the tube. Clamp tube upon completion of infusion, such that the 20ml volume contained in the tube is unchanged.

Question: What percentage of volume A remains in the tube? And if any remains, what amount would volume B have to be to completely eliminate volume A from the tubing? (A colleague and I are having a debate...)

Thanks for your help...

Tom

 

[JasonLion]

The answer depends greatly on all kinds of details that I don't know for I.V. drip chambers and the two solutions (A & B) which you have not described. Do A and B have the same temperature/density, to what extent is the drip chamber setup to promote mixing, and on and on. The answer is fairly easy to calculate if you assume through instantaneous mixing. That seems like an unlikely assumption, but often it is at least somewhat close to reality.

Only one part of your question appears to have a clear answer. In nearly any situation, to completely eliminate volume A requires an nearly infinite amount of replacement water.

 

[tgmb]

Assume identical temp and density, with the concern being how much of volume A is replaced on the first pass...

 

[chem geek]

Continuous dilution would be much worse than what actually occurs since in practice the water mostly flows from top to bottom in the tube and mixes some from diffusion (dependent on time, temperature, viscosity, size of particles) and maybe some from turbulence/adhesion at the surface of the tube. [EDIT] This assumes no turbulence from the drip itself -- that, of course, creates turbulence and some mixing. [END-EDIT] Continuous dilution assumes that even the tiniest amount of the new liquid immediately dilutes into the full volume of the tube. So the infinitesimal amount of B added to a tube of A would have as a result:

(1-f)*A + f*B

where "f" is an infinitesimal amount. This repeats with a new amount of B mixing with the above result:

(1-f)*[(1-f)*A + f*B] + f*B = (1-f)^2 * A + f*B*[(1-f)+1]

repeating again and you'll see the pattern:

(1-f)^n * A + f*B*[(1-f)^n + (1-f)^(n-1) + ... + (1-f) + 1]

If you choose f and n such that f*n=1 then that says the added volume equals the volume in the tube. Take the limit of f-->0 and n-->infinity and you get (1/e) * A + (1-1/e) * B = 0.36788 * A + 0.63212 * B

This is just the standard dilution formula where a single water replacement dilutes to e^-1 or down to 37%, two water volumes dilutes down to e^-2 or 14%, three down to 5%, four down to 2% and five down to 0.7%.

 

[tgmb]

So, in summary, it would take 100 ML to dilute down to .7%? In a straight vertical run of small bore tubing, that seems somewhat counterintuitive. After all, wouldn't the vast majority of the existing column of water evacuate on first pass, simply through gravity?

 

[chem geek]

That's why the assumption of continuous dilution isn't necessarily a good one. It would be a worst-case assumption assuming a very slow drip rate where the diffusion of mixing was faster than the flow rate. Clearly, if you imagined a stirrer in the drip tube then the assumption of continuous dilution in that tube would not be so far-fetched, but in a typical IV drip there probably isn't completely continuous dilution.

On the other hand, the drop of liquid does tend to mix the contents more than a slow flow without any dripping. If the amount of liquid in the tube is rather small compared to the size of the drop and the distance it falls, then the continuous dilution model would not be so bad, at least for the drip tube. In other words, in the thin tube carrying liquid, your "push" or gravity model is closer to reality where there is a lot of dilution with a single volume, but in a "drip" chamber, the turbulence from the drop would have the result be closer to continuous dilution.

As Jason said, you really didn't give enough info to be able to know how to calculate this properly. I think one of the easiest ways to see which model is most valid is to do a dye test. You can then see how much in volume it takes for dye to start coming through the IV and to see how diluted it is and how slowly or quickly it grows in saturation to get close to the level of the feed color. Then switch back to clear water and see how long that takes. For more precise measurements, you can use a colorimeter to measure the dye saturation or you can use a chemical you can otherwise measure.

4 mm diameter, 60" (1524 mm) length, would be pi*(4/2)^2 * 1524 = 19151 mm^3 = 19.151 ml. You said 20 ml to the midline of the drip chamber, but I suspect that the drip chamber has more than 1 ml, doesn't it? Isn't it more like 5 ml or 10 ml? Anyway, it does seem that in your situation most of the liquid is in the line and not in the drip chamber so there shouldn't be a lot of mixing and therefore the dilution would be more complete after just one volume. Of course, if the chemicals were attracted to the material of the line itself, then there can be more remaining.

 

[tgmb]

Yes, the drip chamber is about 5 ml, so that's where all your initial mixing would occur. We do a wide open flush, about 10 seconds for 20 MLS to run through.