Resisting the force of a flow through a restricted area in a piston

[RichB2]

So I'm working on a method to allow control of a drop at different speeds. An idea my team had was to sit our test object on top of a piston, and drop the object by allowing the piston/plunger to drain out of a port at the bottom. We can't seem to wrap our heads around how the size of the port/valve at the bottom of the piston will effect the drop speed. We're currently playing around with Bernoulli equation and the Pascal principle, but haven't had any luck yet. Would love some insight.

 

[tnich]

So I'm working on a method to allow control of a drop at different speeds. An idea my team had was to sit our test object on top of a piston, and drop the object by allowing the piston/plunger to drain out of a port at the bottom. We can't seem to wrap our heads around how the size of the port/valve at the bottom of the piston will effect the drop speed. We're currently playing around with Bernoulli equation and the Pascal principle, but haven't had any luck yet. Would love some insight.

Bernoulli's principle should work for that. The water level sinks at some rate. That gives you the kinetic energy of the water in the piston. The potential energy plus pressure of the water in the piston is proportional to the depth d of the water. The pressure on the outside of the vent is zero, so all of the energy there is kinetic.
$\frac 1 2 v^2_{vent} + 0 = \frac 1 2 v^2_{piston} + gd$

 

[tnich]

Bernoulli's principle should work for that. The water level sinks at some rate. That gives you the kinetic energy of the water in the piston. The potential energy plus pressure of the water in the piston is proportional to the depth d of the water. The pressure on the outside of the vent is zero, so all of the energy there is kinetic.
$\frac 1 2 v^2_{vent} + 0 = \frac 1 2 v^2_{piston} + gd$

Oh, and maybe you are not thinking of the flow rate being constant:
$v_{vent}A_{vent} = v_{piston}A_{piston}$

 

[RichB2]

Right, so that's how our equation is set up right now, except we have a pressure term on the piston side, in terms of the weight of the test article and the area of the piston. Also got the constant flow rate portion, so our exit velocity (Vvent) is in terms of our dependent variable Avent. This is all well and good, but it doesn't help us understand the resistive force. We basically need to calculate the force that is pushing back up on the piston so that we know the acceleration of the piston downward.

 

[Chestermiller]

Could you please draw a diagram?

 

[RichB2]

Here's a quick diagram. Again, trying to figure out how A2 resists the flow. The goal is the get the resistive force in terms of A2 so that we can control acceleration. Going to make a freebody once we understand the force.

Edit: can't see the image showing up, so here's the imgur link: https://imgur.com/a/dCSTV

 

[Chestermiller]

What do you mean by "how A2 resists the flow?"

 

[RichB2]

What do you mean by "how A2 resists the flow?"

Because the area change limits the flow out of the piston, we assume there is a force pushing back. We're thinking about a syringe: the smaller the opening, the harder you will have to push the plunger to keep the same flow rate out. So, if the force is kept constant, as we plan to do, changing the exit area will change the acceleration of our test object downward.

 

[Chestermiller]

Because the area change limits the flow out of the piston, we assume there is a force pushing back. We're thinking about a syringe: the smaller the opening, the harder you will have to push the plunger to keep the same flow rate out. So, if the force is kept constant, as we plan to do, changing the exit area will change the acceleration of our test object downward.

In the case of a syringe, it is primarily viscous forces that resist the flow. When you apply the Bernoulli equation, you are assuming that the viscosity of the fluid is zero. The actual resistance encountered depends on the detailed geometry of the exit oriface/pipe and the properties of the liquid. The analysis can get more complicated if you include viscous forces; are you prepared to face up to this?

 

[RichB2]

Based on the weight we will be applying to the system, it will be rather large. We think it may be best to disregard the viscous forces. We really just want to know how much pushback we will get from decreasing the exit area.

 

[tnich]

Based on the weight we will be applying to the system, it will be rather large. We think it may be best to disregard the viscous forces. We really just want to know how much pushback we will get from decreasing the exit area.

Large weight implies large velocity in the vent and viscous forces will probably dominate. So you will probably not get a very good estimate of the acceleration by ignoring them. However, if you really want to do that, use the constant flow rate equation to eliminate vent velocity in the Bernoulli effect equation. Then you have a differential equation in piston height and rate of change of piston height. Solve that for piston height as a function of time, and you can get the acceleration of the piston.

 

[Chestermiller]

Based on the weight we will be applying to the system, it will be rather large. We think it may be best to disregard the viscous forces. We really just want to know how much pushback we will get from decreasing the exit area.

OK. Then let's forget about the analogy to the syringe.

Here is a thread that looks at this problem in much more detail, including calculation of the height of liquid in the tank as a function of time: https://www.physicsforums.com/threads/velocity-of-efflux.868030/#post-5456985

 

[zul8tr]

If I was still doing consulting engineering I would look for a rather simple practical approach that represents a good proxy for the behavior of the proposed system to achieve reasonable practical relative results within the client’s budget.

Assumptions and approximations (see attached sketch for layout):

1) The area of the round piston Ap is >> Ao (in the range Ao/Ap < 0.1) and Ao is a round orifice area and the cylinder is prismatic container
2) The internal pressure that the piston exerts on the fluid in the cylinder is constant over time and the weight added onto the piston is gently placed on it
3) The flow is 1D and quasi steady state
4) There is no leakage of fluid up along the perimeter sides of the piston. Friction is neglected for a seal on the piston against the cylinder wall
5) The fluid in the cylinder has unit weight w. Ex if water, w = 64.5 lbs/cu. ft and varies with temperature if temp is considered.
6) The control volume region of analysis has streamlines that are parallel to the vertical sides of the container. Therefore near the orifice Ao calculations should refrain from the piston getting to close to the Ao at the bottom where 3D effects start to predominate flow toward the orifice
7) The fluid in the container is incompressible and conservation of volume holds such that the Ap x dy(t) volume over time t equals the volume that leaves the container through Ao over the same time interval.
8) The discharge through the orifice with assumptions 1 and 3 at any time from the energy and continuity equations is approximated by:

Q = CAo (2gy)^ 0.5 where:

C = Cd / [(1 – n(Ao/Ap)^2]^0.5 where Cd is the discharge coefficient of the orifice with values that
range from about 0.6 for sharp edge to 0.98 for a rounded entrance. Refer to hydraulic text (or
Google) for other orifice conditions and Cd values. It is further assumed that Cd does not vary with
water depth and the orifice is in a thin plate with no additional piping as indicated in the poster’s
sketch. The value of n = 1 just using the energy and continuity equations and n = 2 for Ao/Ap < about
0.5 was derived by Chestermiller in a rather detailed analysis located in his post #12:

The system can reasonably be represented by a tank that drains down through an orifice with the addition of a weighted piston exerting additional internal force on the fluid to drain the cylinder faster than a no piston container with an open top and an same orifice.

The internal imposed pressure of the piston and weight is:

Pp = (W + Wp)/Ap where W = weight put on piston and Wp = piston weight

This can be converted to equivalent fluid head by dividing by fluid unit weight w

Hp = Pp/ w = (W + Wp)/(wAp) = constant uniform pressure head over time

A variable quasi steady draining container can be represented by:

Q = flow = - d(volume)/dt = approx. CAo(2gy)0.5 ; negative sign for time increases as y decreases, using d(volume) = Ap dy

(Ap dy)/dt = - CAo(2gy)^0.5

dt = - Ap dy / ( CAo(2gy)^ 0.5

Integrating from time t = 0 at (Hp + h) to a later time Tt at (Hp + y) where h is the starting height of the piston in the cylinder and y is the piston position at Tt and y < h:

Tt = from t = 0 to Tt = { (2Ap / {CAo(2g)^0.5) } x [(Hp + h)^ 0.5 – (Hp + y)^0.5]

If y = h, Tt = 0 at start position of piston, if Hp >> h, Tt approaches 0 not desired for Hp value

Select values for Ao, Ap, h, w and Cd, and a y < h and calculate Tt then:

Average Velocity of piston (Vt) over the drop distance (at time Tt):

Vt = (h - y) / Tt with piston velocity = 0 at time = 0

Average Acceleration of piston (At) over the drop distance (at time Tt):

At = Vt / Tt = (h –y) / (Tt)^2 with acceleration = 0 at time = 0.

Note assumption 4): friction of piston seal neglected so the calculated Tt will be greater than Tt. These equations should provide practical relative changes in piston drop times, piston velocity and acceleration for comparisons with different Ao, Ap, h, w, Cd and y < h.

Spread sheet it to play around

 

[russ_watters]

Based on the weight we will be applying to the system, it will be rather large. We think it may be best to disregard the viscous forces. We really just want to know how much pushback we will get from decreasing the exit area.

The assumption in the method you are using is that the force is always the same, regardless of the orifice size. This works for relatively small orifices.

Bernoulli tells you the velocity of the air out of the orifice(for relatively low pressures). The size times velocity is flow rate (which then tells you how fast the piston drops).

 

[JBA]

Unless the "fluid" in the vessel is a gas, the pressure at the orifice, which is a determining factor in the the flow rate a given orifice size will it will allow, the orifice flow is always going to be a function of the sum of the fluid height, and properties plus the piston weight, regardless of the orifice size.

The only way to eliminate that effect is to have a variable area nozzle actuated by sensing the pressure at the bottom of the vessel, the height of the piston, the changing weight of the total vessel and its contents, etc; or, adding fluid on top of the piston that maintains the original filled fluid height in in the vessel as the bottom fluid is discharged.

Other than that, the only way this effect can be minimized, but not eliminated is by using a fluid with very low weight relative to the piston weight or a very shallow vessel and and short piston travel; regardless of the total weight of the fluid in the vessel.

 

[zul8tr]

The assumption in the method you are using is that the force is always the same, regardless of the orifice size. This works for relatively small orifices.

Bernoulli tells you the velocity of the air out of the orifice(for relatively low pressures). The size times velocity is flow rate (which then tells you how fast the piston drops).

My assumptions state the orifice area Ao is << piston area Ap and the container uses a incompressible liquid. You never mentioned a gas in your problem statement ? If a gas is in the cylinder that is a different problem. Please state more specifically the problem when posting.

 

[russ_watters]

My assumptions state the orifice area Ao is << piston area Ap and the container uses a incompressible liquid. You never mentioned a gas in your problem statement ? If a gas is in the cylinder that is a different problem. Please state more specifically the problem when posting.

Apologies, I think you may be right, but this actually works better if it's a liquid under gravity than with pressurized air (which doesn't really work with Bernoulli's alone). Pressure is just a function of depth and Bernoulli's applies well. Either way, yeah, a clarification f er on the OP would help.