ALP vacuum notes

Today Damian and I got to play around with the vacuum. We were brainstorming how to increase the amount of air flow through the inhaler using the in-vitro lung and I must admit I was quite stumped when we were discussing it. Never having worked with vacuum before, I lacked the basic knowledge of how to think about it. But, thanks to the work of Danielle and her post about flow through an orifice, I think I understand what is going. Also thanks go to Damian for speaking his mind and indicating that what we were wanting to do may not be necessary. Good job! Below are my notes on what Danielle and Damian taught me.

Danielle wrote in her post that the volumetric flow rate can be written as
Q_v=\sqrt{\frac{2\Delta P}{\rho}}\times\left(\frac{A_2}{\sqrt{1-\left(\frac{A_2}{A_1}\right)^2}}\right),
where Q_v is the volumetric flow rate, \Delta P is the change in pressure across an orifice, \rho is the density of the air, and A_i are the areas of the pipe around the orifice. I think it is important to understand where this equation is coming from because it says a lot about how fluid flows through something. It of course hinges completely on Bernoulli’s equation.

P = \frac{1}{2}\rho v^2,

where again P is pressure, \rho is the density of air and, v is velocity. Excuse the horrible ascii art but below is a picture of our setup.

			Tube
		=========================	       ________
Vacuum	       |			|Inlet _______/        |
	    ___|			|_____/	               |
       [--] ___		      A2	 _____	       Inhaler |
	       |			|A1   \_______	       |
	       |			|	      \________|
		=========================

The inlet is what is connected to the inhaler on the right side of the tube and the left most side of the tube is where we pull the vacuum. Now, we need the air flow through the inhaler to be at least 90L/min. But, the inlet serves as an obstruction, a resistance if you will. I will get back to the idea of resistance in a moment. As we know, Bernoulli’s equation will tell us what the fluid flow through the inlet will be if we know the following: pressure inside the tube, the pressure outside the tube, and the velocity of air in the tube.

Now, if we look at Bernoulli’s equation in the tube (2) and it in the inlet (1) we obtain two sets of equations.

P_2 = \frac{1}{2}\rho v_2^{\ 2}
P_1 = \frac{1}{2}\rho v_1^{\ 2}

which can be written as

P_2 - \frac{1}{2}\rho v_2^{\ 2} = 0
P_1 - \frac{1}{2}\rho v_1^{\ 2} = 0

and thus setting them equal to each other gives

P_2 - \frac{1}{2}\rho v_2^{\ 2} = P_1 - \frac{1}{2}\rho v_1^{\ 2}
P_2 - P_1 = \frac{1}{2}\rho v_2^{\ 2} - \frac{1}{2}\rho v_1^{\ 2}.

We can equate these together because of conservation of mass which just means that the air that flows through the inlet isn’t going to disappear in the tube because well, it can’t. Now, if we replace the velocities with a volumetric velocity, i.e. Q_V = v\cdot A then we have that

P_2 - P_1 = \frac{1}{2}\rho\left(\frac{Q_V^{\ 2}}{A_2^{\ 2}} - \frac{Q_V^{\ 2}}{A_1^{\ 2}}\right),

and then

\Delta P = \frac{1}{2}\rho Q_v^{\ 2}\left(\frac{1}{A_2^{\ 2}} - \frac{1}{A_1^{\ 2}}\right)

giving

\Delta P = \frac{1}{2}\rho Q_v^{\ 2}\frac{1}{A_2^{\ 2}}\left(1 - \frac{A_2^{\ 2}}{A_1^{\ 2}}\right).

This is readily solved for Q_V

Q_V = \sqrt{\frac{2\Delta P}{\rho}}\times\left(\frac{A_2}{\sqrt{1-\left(\frac{A_2}{A_1}\right)^2}}\right).

Going through this hoopla teaches us a couple things. One important thing is that the volumetric flow rate—the quantity we need to increase—depends on the pressure difference between the tube and the orifice in a nonlinear fashion. It also tells us that the only way to increase the flow rate is to either increase the pressure differential or, make A_1 = A_2. Making the cross-sectional areas the same is impossible due to our valve constraints however, we can make them as close as possible. Danielle, can you do some calculations on comparing  the difference between varying the pressure differential the areas and see which coefficient gives us the greatest boost in volumetric flow?

Inspecting the above equation also taught me an analog to vacuum and current. If I have a wire and I put current through it, the electrons will happily move along the wire with no problem. But, if I add a resistor to the wire, the rate at which the electrons move down the wire is changed. The same thing happens in our vacuum tube when air flow meets a change in diameter. The fluid flow hits a resistance and thus slows the bulk fluid motion down if the diameter is decreased.

This is good. And yes, you were right Damian. This of course means that you (Damian) are going to need to figure out a way to pull a better vacuum on the tube.

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  1. #1 by Damian on November 6, 2012 - 12:15 am

    I don’t understand why volumetric flow rate approaches infinity as A1 approaches the size of A2. What are the implications of this?

    • #2 by andymaloney on November 6, 2012 - 8:36 am

      In that situation you have a straight pipe with no orifice and wouldn’t need to consider this at all. The last equation I have up in the post isn’t the entire story as can be gleaned from Danielle’s references in her post. It’s merely a means to indicate how we increase the volumetric flow rate. Either decrease the difference between the orifice diameters or increase the pressure differential. The above description is simplistic and I don’t think it should be applied for all situation, e.g. the one you just described.

  2. #3 by daniellestolley on November 6, 2012 - 12:55 am

    I will have to look at my equations again, but when I went into lab this morning I was getting a reading of about 65 L/min with about 600- 700 mmHg vac which I used as the delta P.. I was having trouble getting my units to work out.. I will revisit it and attempt to get it to work.

    I believe I got it to work once because I got a flow of around 60 L/min using .5 inches as the inlet and .25 inches diameter for the orifice ( flowing from the inhaler to the tube) – I decided to treat the system as the inhaler side and the side between the tube and the shut off valve, since this is where the air will flow through first ( since we will have the same pressure here as in the tube since it is connected)

    However I went to calculate again and I could not get it to work, I seemed to keep getting different answers, having a bit of trouble with equivalence in my units but I am sure i will be able to make it work.

    From what I can see the best way to get this to work is to either increase the pressure, or decrease the change in orifice area. The shut off valve ( I am not sure this is the correct term, I am sure you know what i am talking about though) measures at 3/8 i believe which is fairly close to the 1/2 diameter of the tubing. If we could get it to this area from the shutoff valve to the tube, it should help quite a bit.

    I am assuming that since the Pressure change is inside the square root and the Area value is outside the square root that we will get greater change if we change the cross sectional areas than if we change the pressure, though we may actually have to do both to get what we need. It will be easier to vary the pressure during experiments so I suggest we get the least resistance through the inlet as possible so that it will make our lives easier when it comes to testing.

    In summary we need to
    1) find the maximum vacuum that we can pull on the tube
    2) try to get the inlet running from the shutoff valve to the tube as close to 3/8 to 1/2 inch diameter as possible

    That should give us the best possible chance to get 95 L/min

    This is not taking into account the flow meter which we have, which probably reduces the flow rate by being there, but for simplicity’s sake I will just ignore its existence in my calculations.

    Also I will look up that vacuum you told me to find and see if I can figure out the power source. I am 85% sure that the black wire is ground, this is usually the case ( though it is quite odd that the other wire is blue) I will check as soon as I can find the correct one.

    — I will repost this in my blog as well since I have yet to do that —

    • #4 by andymaloney on November 6, 2012 - 8:46 am

      Eeek. The black wire is not always ground! When dealing with AC, the color codes are: black=live, white=neutral, and green=earth.

      We are all in agreement that we need to increase the pressure differential and/or decrease the inlet/tube diameter differences. Great! Let’s do it.

  3. #5 by Tyler Fields on November 6, 2012 - 11:56 pm

    Have you guys ever considered reversing the setup and, instead of creating a vacuum, using something like an air compressor to push air through the inhaler? You might have so much pressure to work with that small diameters wouldn’t be an issue. It would be silent too.

    • #6 by andymaloney on November 7, 2012 - 7:41 am

      This is true and I’m actually using the pushing of air instead of pulling in a different experiment. Having a pressurized tube may be easier to fit into the NGI setup as well. I’ll have to think about this.

      • #7 by Tyler Fields on November 7, 2012 - 9:07 am

        NGI?

      • #8 by andymaloney on November 7, 2012 - 9:15 am

        NGI = Next Generation Impactor.

        http://www.appliedphysicsusa.com/ngi_moudi.html

      • #9 by hughsmyth on November 7, 2012 - 9:44 am

        It would be nicer if we could simulate the inspiration breathing event by pulling a vacuum. When we move to the next phase of testing the inhalers the pressure/pushing air method may interfere with our data collection in subtle ways. Good idea though and we may have to head that direction.

  4. #10 by Tyler Fields on November 7, 2012 - 11:38 pm

    Another unsolicited thought: you could convert the PVC pipe into a giant syringe and create a sustained vacuum by withdrawing a plunger from one end, like filling a hypodermic needle. A steady rate could be maintained by using a heavy weight and gravity to withdraw the plunger.

    • #11 by Tyler Fields on November 7, 2012 - 11:47 pm

      Now that I think about it you probably wouldn’t be able to get the flow rate high enough unless you used a heavy spring or some elastic or something, and even then you’d have to figure out how to “cock” it, which is probably more trouble than it’s worth.

  1. ALP 11 « Damian Jimenez

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