# Calculating time to eject liquid from cylinder with nozzle

## Main Question or Discussion Point

Say I have a device that looks like a large syringe, filled with some liquid (water). On the "plunger" side I have a flat plate say, that holds the water in the cylinder. On the "nozzle" side, the water is hold in by some cover. I eject the water by striking the flat plate with a weight of mass "m" moving at a velocity "v" (which knocks off the cover). Say the plate is of diameter "D" and the nozzle is of diameter "d". Is it possible to calculate the time it takes to eject all the water from the cylinder. Does it depend on the viscosity?

Roger

Related General Engineering News on Phys.org
FredGarvin
Sure. It all depends on how in depth you want to get. As a first, very rough pass, you can simply use continuity to calculate the volumetric flow rate.

If the plate is moving with $$V$$ velocity (and you can make the assumption that the fluid is moving with the same velocity) and the large area has an area of $$a$$ then the volumetric flow $$Q$$ in that section is
$$Q=V*a$$

Through continuity that is how much should be flowing out of the nozzle section. To calculate the time, divide the volume of the syringe by the volumetric flow rate.

Again, this is a very rough approximation. You can get pretty complicated if you want.

How does the mass of the weight, and the fact that it impacts the "syringe" affect the calculation?

AlephZero
Homework Helper
Assuming the syringe is rigid and rigidly fixed to something, and the fluid is incompressible, when the mass hits the plate it decelerates rapidly. The force which decelerates it will produce a high pressure in the fluid. That will knock off the cap and force some fluid through the nozzle.

However this seems a very inefficient way to try to empty the syringe (assuming it's horizontal and gravity is irrelevant). The reason is that to keep the flow going, the flow through the nozzle must be faster than the flow down the syringe (the area is smaller, so the velocity must be higher to move the same volume of fluid per second). So you need to accelerate the fluid through the nozzle. To do that you need to maintain some pressure inside the syringe for the whole time it is emptying.

I expect what would actually happen is that as soon as a small amount of liquid flows out, the pressure in the syringe will drop to zero and the flow will stop.

The fix for this would seem to be using a bigger mass (probably much bigger than the mass of the fluid inside the syringe) but then the initial impact would produce a very high pressure that might burst the syringe rather than producing the flow you want.

Playing with a toy syringe will show that you need a steady (and fairly small) force for the whole time the syringe is emptying, not a big force for a short time.

If the "syringe" is really just a cylinder with two end caps but no nozzle, then the need to maintain the pressure goes away. In that case you could calculate the time to empty the fluid farly well using conservation of momentum - the same as school physics impact problems. The initial pressure pulse during the impact should be high enough to knock the cap off. You would have do design it so the "plate+mass" could travel down the syringe like a piston (without too much friction) to push the fluid out.

Well actually I was looking at this system as a "brake" for the weight, where the "syringe" would slow it down to a standstill. So if seen in that light, what you say is good I guess. So I want to calculate the time for the weight to come to a stop.

Why not use a piston damper in a viscous fluid?

AlephZero
Homework Helper
Unless ejecting the fluid is an important part of what the device does, then buying a ready made damper (shock absorber) would be a better plan than trying to design one from scratch.

For commercial dampers with a lnear response, the force is specified as proportional to the velocity. (There are other types available with more complicated response for applications that need them). You should be able to work out what force you need to stop the mass in the way you want and then find a damper the right "size" to do the job. Usually the distance needed to stop the mass is more critical than the time, but obviously time and distance are related by the initial velocity.

There are 3 sort of fluid mechanics calculations in my experience:

1 "school level", which are fine for understanding the theory of how fluid devices operate (Bernoulli's principle, etc, etc) but usually much too simplified to give realistic answers.

2 semi-empirical, i.e the key properties are taken from component data sheets which are probably based on test measurements. These do give realistic answers provided the parts are used as their designers intended. Basically you are designing a system by selecting from existing parts, not designing the parts themselves.

3 Accurate computer simulations from first principles. Commercial software packages are now becoming available, but it's still a serious undertaking to do this successfully and the results still need to be validated against experiments.

You don't say what size your device is, but here's one somebody designed already.... http://www.airpot.com/beta/html/velocity_control_falling.html [Broken]

Last edited by a moderator:
Thanks for the detailed replies AlephZero. Thing is, for my application, I cannot buy something off the shelf, I need to "replicate" the behavior of a damper, which I will have to construct myself. What I have to do is make some kind of collision protection mechanism for colliding trolleys. So I started thinking in this direction. I decided to phrase my initial question in a simplified way. May have been a mistake.

Q_Goest
Homework Helper
Gold Member
Hi Roger,
In situations like this, where you are trying to model a system, I'd always break it down into the most fundamental principals and build up a model from there.

I think we can safely assume the mass (M1) impacts the piston and stays with it as opposed to bouncing off. From this, we find there is a conservation of momentum between the before impact and after impact system. If the mass of the piston/liquid system (M2) is small compared to M1, we can neglect M2. If not, recalculate velocity of the system using conservation of momentum. This will be the velocity of the system just as the liquid is about to begin squirting out of your orifice.

Next, you need to model how this combined mass (Mc=M1+M2) is going to be slowed. Here, the simple F=ma (or F=Mc * a) equation is applicable. The force opposing the motion of Mc is due to pressure acting on the piston, so F = PA where P = pressure on piston and A = area of piston. But the question is, how can you find the pressure inside the cylinder at any time after impact, t?

With Mc moving at some initial velocity, V(0) (this represents velocity at time 0) you can find the velocity of the liquid shooting out of your orifice using the continuity equations for conservation of mass. This is the equation Fred was eluding to. The velocity is a function of the areas. You might also want to include some discharge coefficient for your orifice such as suggested on this web page. It should be just as simple as adding the coefficient (K on the web page) to the equation just as they show.
http://www.mcnallyinstitute.com/13-html/13-12.htm

Knowing the velocity of the liquid through the orifice, and the pressure on the orifice outlet, you can determine the pressure, P, acting on the piston.* That's the "head pressure" on the above web page. Note this equation is simply derived from Bernoulli's equation, so you should be able to derive this yourself. With pressure inside your cylinder, you can then determine the force exerted on the piston, F. With that, and knowing your mass, you can calculate the acceleration from F=ma as discussed previously.

I'd suggest doing this iteratively. You may need to make very small time increments at the beginning as that's when the largest forces will be acting on your mass Mc. Generally, I'd suggest creating a spread sheet using something like Excel. You'd create a section at the top of your sheet to put variables in such as M1, M2, Vi of mass M1, Piston diameter, Orifice diameter, Discharge coefficient, Liquid density, etc… Alternatively, you might be able to create a single equation that represents the velocity of the mass over time. Think about doing it both ways, and let us know what you come up with.

* Note, that for this rough calculation, I've made the assumption that the liquid is completely incompressible, which isn't exactly true. All liquids have some bulk modulus, just like solids. In reality, a pressure wave would be sent through the liquid at sonic velocity, but let's not consider this complicating factor for the moment. We could revisit this if need be, but I think if you're intent is to design a shock absorbing device, you are safe if you want to ignore this as velocities should be small compared to sonic velocity in the liquid.

Gokul43201
Staff Emeritus
Gold Member
Thanks for the detailed replies AlephZero. Thing is, for my application, I cannot buy something off the shelf, I need to "replicate" the behavior of a damper, which I will have to construct myself. What I have to do is make some kind of collision protection mechanism for colliding trolleys. So I started thinking in this direction. I decided to phrase my initial question in a simplified way. May have been a mistake.
This will not replicate any reasonable damper. It will be way overdamped. Also, as Aleph0 points out, you will not likely expel all the liquid from the nozzle, and if you do, you have the plunger hitting the forward wall of the syringe - not good. You can get rid of the liquid and leave the syringe air-filled and it will still be overdamped, but less badly compared to before.

What you're missing, to achieve anything close to critical damping, is a restoring force - you need a spring! With an essentially incompressible fluid in the chamber, you've got no restoring force, with air in an open syringe, you've got a little. The simplest way to do this correctly, with a purely pneumatic system is to build an properly sized enclosure on the outlet (nozzle) side of the syringe. In essence, you have a two-chambered air container with an orifice separating the two chambers. This is how most passive pneumatic isolators are built.

You need to give yourself a quick crash course on noise/vibration isolation before you jump into this project. A design project is useless if the design isn't based on understanding of the theory.

Last edited:
Q_Goest
Homework Helper
Gold Member
Hi Gokul.
This will not replicate any reasonable damper. It will be way overdamped.
Interesting you'd say that, and I can kinda understand why. The image of a syringe is one with a piston that's relatively enormous in comparison to the orifice. I think the OP unfortunately, sets up a bad image of how this thing will look, one which when you push on it, the response (reaction to load) will be a huge force. But hopefully a syringe won't actually be used. I would hope the project team will work through the analysis and determine piston and orifice sizes that will work for the application and be economical at the same time.

You can get rid of the liquid and leave the syringe air-filled and it will still be overdamped, but less badly compared to before.
I suppose there are some applications where a gas is used, but honestly I can't think of many. Industrial shock absorbers are used to control the motion of something at the end of its travel. They are generally hydraulic and can have some air or gas pocket to account for the oil displaced, but the gas isn't for damping. ThomasNet is a terrific resource for locating suppliers of industrial and hydraulic shock absorbers, of which there are many.
http://www.thomasnet.com/prodsearch.html?cov=NA&which=prod&what=shock+absorber&navsec=search

What you're missing, to achieve anything close to critical damping, is a restoring force - you need a spring!
One time shock absorbers using water are not uncommon. You'll find http://www.barriersystemsinc.com/dynamic/docs/filename_314.pdf" [Broken]driving down the highway, they're often used to absorb the impact of vehicles before hitting immovable barriers. A spring isn't necessary.

And I'd agree a better shock absorber would 'reset' itself. Perhaps in this application, that would work, but then again that's up to the folks working on the project. A one time shock absorber using a piston inside a water filled cylinder with a hole on the outlet will suffice. One could also put a valve on the outlet instead of an orifice, so it would be easy to adjust.

Last edited by a moderator:
AlephZero
Homework Helper
There are two possible uses for a spring here. One is "resetting". The other is in series with the damper (not in parallel as a resetting spring would be) to reduce the initial shock forces, both in the damper and to the trolley.

In other words, the first the spring absorbs the energy, then the damper dissipates it.

The benefits of adding the spring are an easier damper design (since it doesn't have to withstand the short term peak force of the initial impact force without failing) and less collateral damage to the trolley.

FredGarvin