Calculating Force Needed to Pull Plunger in Standard Syringe

[Neg. pressure]

I have a standard syringe 1” diameter, with a freely moving plunger requiring uniform force to move the length of the cylinder providing smooth uniform travel.

Cross sectional area = ∏r2 = 3.1415 * 0.5” * 0.5” = 0.785 in2
Negative pressure acting on the piston = 0.758 in2 * 14.7 psi = 11.545 lbs

I start from the plunger fully inserted into the cylinder; therefore, the vacuum chamber is completely empty, except for the nozzle portion. A sealed applied to the nozzle makes it airtight so no air can enter the vacuum chamber.

The force required to pull the plunger out (thus increasing the volume of the vacuum chamber) is initially minimal (@11.545 lbs) but quickly increases as the plunger moves out further and further from point A to B to C and to point D, until the plunger cannot be pull any further, by hand.





Diagram on attachment. Text is the same.








The questions:
1. Why those it get significantly harder to pull the plunger out?
2. How can this be calculated?

 

[russ_watters]

The force required to pull the plunger out (thus increasing the volume of the vacuum chamber) is initially minimal (@11.545 lbs) but quickly increases as the plunger moves out further and further from point A to B to C and to point D, until the plunger cannot be pull any further, by hand.

This isn't correct. The force stays constant at 11.5 lbs.

 

[Naty1]

I don't understand Watters answer...seems like the more volume the lower the pressure. See here: http://en.wikipedia.org/wiki/Universal_gas_law

you can roughly assume pv remains constant...so as the volume doubles so does the negative pressure [vacuum]. As one of the practical matters, the seal on the syringer plunger will leak at some vacuum pressure limiting the vacuum you achieve.

 

[VaqarAhmed]

I don't mean to be Dr. Explainitall but that's called suction. and it's precisely how I get my pepsi out of my cup using a straw.

feel free to pm me for more applications

 

[russ_watters]

I don't understand Watters answer...seems like the more volume the lower the pressure.


See here: http://en.wikipedia.org/wiki/Universal_gas_law

you can roughly assume pv remains constant...so as the volume doubles so does the negative pressure [vacuum]. As one of the practical matters, the seal on the syringer plunger will leak at some vacuum pressure limiting the vacuum you achieve.

The OP's calculation assumes a perfect piston with a starting volume of 0. So the vacuum is a constant -14.7 psig.

 

[bascule]

I have a standard syringe 1” diameter, with a freely moving plunger requiring uniform force to move the length of the cylinder providing smooth uniform travel.

Cross sectional area = ∏r2 = 3.1415 * 0.5” * 0.5” = 0.785 in2
Negative pressure acting on the piston = 0.758 in2 * 14.7 psi = 11.545 lbs

I start from the plunger fully inserted into the cylinder; therefore, the vacuum chamber is completely empty, except for the nozzle portion. A sealed applied to the nozzle makes it airtight so no air can enter the vacuum chamber.

The questions:
1. Why those it get significantly harder to pull the plunger out?
2. How can this be calculated?

You're considering the problem from the incorrect direction. Pressure is always positive. Because pressure is the statistical average of all the fluid particles flying around and bashing against each other or against a surface.

For your problem starting out with a volume of 0 it means that you require 11.545 lbs of force to pull your syringe out because that is the amount of pressure acting on the outside/back edge of the syringe. And this force is the same at each point as the syringe moves further out.

This is slightly counter-intuitive and doesn't exactly match what a person experiences when pulling a real world syringe. The reason for that is because there is always a small amount of air in the syringe to being with. So the first amount of force required is essentially zero because the pressure inside and outside is the same. By the time the volume of air in the syringe has been doubled then the pressure in your syringe is half and the force required to move at that point is 11.545 / 2 lbs.

Does this help?

 

[Neg. pressure]

Dear All,

Thank you for your responses so far. Unfortunately, the answer has not yet surfaced. I hope that the additional information, below, will help us all consolidate of collective brain power and solve the problem.

 Given; ambient pressure is at 14.7 psi (Sea level)
 Given; temperature, ambient, appx. 72° degrees F
 Given; ambient pressure is acting upon all surface areas (Back of the plunger and the syringe walls.)
 Given; ambient pressure cannot change, since our test environment remains constant.


 Pressure inside the syringe becomes 0 psi when 11.545 lbs of force is applied to pull the plunger back.
I have a standard syringe 1” diameter, with a freely moving plunger requiring uniform force to move the length of the cylinder providing smooth uniform travel.

Cross sectional area = ∏r2 = 3.1415 * 0.5” * 0.5” = 0.785 in2
Negative pressure acting on the piston = 0.758 in2 * 14.7 psi = 11.545 lbs (For the sake of this example, that is achieved at point A in the attachment.)

 In order to move the plunger from point A to point B requires a force greater than 11.5 lbs. We can look at it in this way. F1 (pulling) = 11.5 lbs + b.
 In order to move the plunger from point B to point C requires a force greater than 11.5 lbs + b. We can look at it in this way. F2 (pulling) = 11.5 lbs + b + c.
 In order to move the plunger from point C to point D requires a force greater than 11.5 lbs + b + c. We can look at it in this way. F3 (pulling) = 11.5 lbs + b + c + d.

Now going back to the questions:
1. Why does it require more force (11.545 lbs) to move the plunger from point A through point D?
a. Does it have anything to do the ambient pressure acting upon the walls of the syringe? (As in the calculating of the buoyancy of an object in a liquid)
2. How do we calculate for this?

 

[Integral]

What is the syrenge attatched to? Is it open to the amosphere? Is drawing from a closed system?

 

[russ_watters]

If what you are claiming is that this is an actual experimental result, then there is an error in your experimental setup somewhere. If you are just asserting this as a way you think it will work, your understanding of the situation is wrong.

 

[Neg. pressure]

I know this sounds counter intuitive, but those are repeated results done by different individual at different times and locations.

An analogy can be made with the calculating of buoyancy of an object in a liquid. With this buoyancy calculation, you need to include the area of the sides of the object. It does not make sense since the liquid and the object, at that point, have a horizontal interface and the buoyancy is a vertical force.

I am not saying this in a negative manner, as I appreciate your assistance with this problem. Please try it with a syringe and let me understand why this happens. I want to learn…

 

[bascule]

I know this sounds counter intuitive, but those are repeated results done by different individual at different times and locations.

Dear Neg. Pressure,

I'm afraid that when you say that a value of 0 PSI is measured in your experiment then we can see that you are using an inaccurate device that cannot adequately measure the low pressures involved. Low pressure gauges for this experiment would be calibrated in milli or microtorr.

Also in a the real world then the subject of friction would be very important and that would be very dependent on materials and geometry of your seal.

I suggest you study the subject of hydraulic systems and pistons which should help you realize that the length of the cylinder makes no difference to the force on the piston (in a theoretical world)

Regards,

B