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How would you go about solving the following problem. I believe Bernoulli's equation needs to be employed, but I'm not sure how.

Find the magnitude of the force that needs to be applied to a piston of a 20ml syringe with 1cm diameter tube to drain it in 20 seconds through a 40mm length needle of 0.2mm inner diameter. The fluid inside the syringe is water.

Force = ?

Volume of syringe = 20 ml = 0.00002 m^3

Syringe diameter = 0.01 m

Length of needle = 0.04 m

Needle diameter = 0.0002 m

Time to drain syringe = 20 s

Fluid density of water at 20 degrees celcius = 998.21 kg/m^3

Dynamic viscosity of water at 20 degrees celcius = 0.001002 Pa.s

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You can get a minimum bound from energy balance alone. This is as if the fluid has no viscosity, so the force you have to apply over the distance is only due to the kinetic energy required to expell the fluid.

The diameter of the tube is 1 cm, so the area is .785 cm². That means the plunger travel distance is 25.5 cm = 0.255 m.

The fluid is squeezed down to 200 µm diameter, which is a crossectional area of 31.42x10-9 m². The volume of fluid is 20 ml = 20x10-6

(20x10-6 m³)/(31.42x10-9 m²) = 637 m

That is how far the 200 µm stream has to travel in 20 seconds, for a speed of 31.8 m/s. 20 ml of water has a mass of 20 g, or 0.020 kg. The total kinetic energy that is therefore imparted on the fluid is

½(0.020 kg)(31.8 m/s)² = 10.1 J

Now we can solve for the force required over the plunger travel distance to impart this energy:

(10.1 J)/(0.255 m) = 39.8 N = 8.95 pounds

That's actually a lot more than I expected before working it out. It would be interesting to see how much higher the force is when viscosity of the fluid is taken into account. It might be possible that kinetic energy is actually the dominant effect for something with relatively low viscocity like water. Obviously the force would go way up for something thick and gloppy, probably to the point where a typical syringe couldn't handle the pressure to acheive the expulsion time of 20 seconds.

Hmm, that's a interesting point. Let's see what the pressure is. The area of 0.785 cm² is 0.123 in²

(8.95 pounds)/(0.123 in²) = 73 PSI

Which is the pressure inside the syringe required to expell the fluid just due to the kinetic energy requirement alone.

Added

There is yet another effect at work that makes the minimum required force higher, still without invoking viscosity. The speed won't be the same for every part of the flow thru the narrow tube of the needle. The flow will be laminar, so the outer edges will be slower with the highest speed in the center. The average still needs to be what was calculated above, but the power will be higher because it scales with the square of the speed.

The difference is the same as the ratio between the RMS flow rate and the average flow rate. For example, for a linear profile from edge to center, RMS is 22.5% higher than the average. Of course that's a rather unreasonable profile, but it illustrates the concept. I picked the shape of a half-sine as a close-enough profile. This means the flow speed is 0 at the edges and smootly peaks in the middle. Perhaps someone more familiar with fluid dynamics can tell us what the real profile is, but I expect this will come close enough for the purpose of increases the energy requirement due to the spread of flow speeds.

I was too lazy to do the 2D integrals, so I had the computer do the integrals numerically for me. The RMS of the sine peak profile is 17.9% higher than the average. That means the 10.1 J calculated before needs to be increased by this amount. That comes out to:

Force = 46.9 N = 10.5 pounds

Pressure = 86 PSI

As before, this is without the additional force needed to overcome viscosity of the liquid. The only properties of the liquid this relies on are its density, and that flow thru a 200 µm diameter pipe will be laminar.

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  • $\begingroup$ Had the same idea until I realized that you already have answered it this way. $\endgroup$ – MrYouMath Apr 28 '17 at 9:15
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You can use Bernoulli, for example:

$$\dfrac{P_1}{\gamma} + \dfrac{V_1^2}{2g} + z_1 = \dfrac{P_2}{\gamma} + \dfrac{v_2^2}{2g} + z_2 + h_f$$

$P_1$ = pressure after the plunge

$P_2$ = pressure at the exit of the needle (atmospheric)...

$z_1 = z_2$ ...for a horizontal setup

$\gamma = $ density times gravitational acceleration $=pg$

$v_1 =$ the speed of the plunger is pretty much negligible when compared to the speed of the fluid expelled through the needle

$h_f =$ all the frictions due to diameter reduction etc. For this case, I will consider $h_f =0$ (ideal case)

Hence if we solve for $P_1$:

$$\dfrac{P_1}{pg} = \dfrac{P_2}{pg} + \dfrac{1}{2}\dfrac{V_2^2}{g}$$

then

$$(P_1-P_2) = \delta P = \dfrac{1}{2}pV_2^2$$

Hence, the minimum force will be calculated as follows:

$$\text{Force} = \text{Area of plunger} \cdot \dfrac{1}{2}pV_2^2$$

For the calculation of the velocity $v_2$:

$v_2 =$ (volume of liquid in the syringe)/(time to empty the syringe)/(Area of the needle inner diameter)

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  • $\begingroup$ Welcome on the site! Note, we support Latex formatting. Enter $P_1-P_2$ and you will get $P_1-P_2$. $\endgroup$ – peterh May 31 '19 at 18:01
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You could try to analyze this problem with doing very complicated fluid dynamics analytically or numerically. The problem is nonstationary and the convective terms do not vanish, hence very difficult to treat analytically.

The inviscid approximation by Olin Lathrop seems do be a good model for this problem.

An alternative way and in my opinion the most reliable way would be just to use a simple experiment. Fix the syringe vertically so that the needle shows downward. Then use small weights as force and measure the time it takes to drain the syringe. Then vary the weight until you achieve the drain time of $ 20 \text{ s}$. If you don't quite get to this time, because you don't have this specific weight, you can interpolate your measurements (for example with: Excel, R, MATLAB) to estimate the appropriate weight.

You might need to add an additional plate on the pushing side so that you can place your weights. If it is not too heavy then you don't have to take this also into account. The piston weight should also be negligible.

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  • $\begingroup$ -1. Even though the convective terms don't identically vanish, it's often possible to argue that they are negligible compared to a dominant viscous term. $\endgroup$ – Max Apr 28 '17 at 11:32
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    $\begingroup$ Are you just using downvotes to take revenge? I gave a clear explanation why I downvoted your answer. Downvoting is there to prevent false/incomplete answers, if you are too childish to cope with constructive criticism then you should not be on this site. If you downvoted this answer then you should at least give an explanation why you downvoted. $\endgroup$ – MrYouMath Apr 28 '17 at 22:15

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