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- Thread starter PNachtwey
- Start date
- Thread starter
- #21
Define sudden. Let's assume the actuator is moving at 10 inches per second and the cylinder is moving 12 inches per second. Let's also assume the valve is mounted on the cylinder so the minimum of volume of trap oil is 0.
Lets also assume the actuator still has 2 inches of stroke to go when it hits the obstruction.
Therefore in 1 millisecond the volume the volume changes 0.1% so the pressure will change 200 psi assuming the bulk modulus of oil is 200,000 psi.
How many milliseconds will it take the pressure to go from 1234 to 3131 PSI?. It appears that it will takes at least 9.5 milliseconds but in actuality it will take much longer because as the pressure increases the difference between the pressure at the source and the pressure in the cylinder goes down so the rate of flow into the cylinder drops which reduces the rate of increase in the pressure but the pressure still increases but at ever slower rates.
I like Mike's explanation of how a sudden increase in pressure occurs much better than opening a valve or spinning up a pump. Valves and pumps don't do anything instantly at the pressure rise would take even longer.
So Mike's answer is the speed is 0 assuming the actuator hits an obstruction or reaches the end of of cylinder.
What if the actuator hasn't reached an obstruction? Bud didn't mention an obstruction or reaching the end of the cylinder . The bang would be hard to miss.
stroupaloop
Mechanical
Hello all.
I'm fairly new to this forum but noticed this interesting post and thought I would supplement it with some of my thoughts being that some answers have still yet to be shown. I believe the last remaining answer is whether or not equations are capable of being generated to prove such topics and the answer is YES they are, but NO they are not simple. So I will do my best to explain them.
Pressure:
The following mathematical equation has been developed to best represent characteristics of fluid passing through an orifice, which I believe our original post master has stated somewhere else in these forums:
Q = Kv * (deltaP)^0.5
The Kv value represents what people may associate with as "resistivity." However, this simple equation is not easily used in all fluid power instances. Further explanation must be validated to provide meaningful definement to this equation.
The true form of this mysterious Kv value is as follows:
Kv = Cd * A * (2/rho)^0.5
where:
Cd = orifice flow discharge coefficient
A = orifice area
rho = fluid density
The area and density are obvious and easy finds for this equation model, but the definement of the orifice flow is much more difficult. For now, since I can not post charts, I will refrain from discussion about transitional flow.
The Cd value is dependent upon the defined Reynolds number of the fluid. As the Reynolds increases so does the Cd value until reaching turblent flow, whereby, for the most part, it levels out to a constant value. The relationship between the Cd value and the Nr (Rey No) during laminar flow is as follows:
Cd = 0.125 * (Nr)^0.5
This value becomes fairly constant in turblent flow, which stays at about 0.61.
With this in mind, this equation answers your question regarding pressure being a resistivity of flow. It can be correlated that as flow increases so does pressure across the orifice and vice versa. Its understanding the mathematical relationship to better define the actual characteristics and identifing the necessary test conditions.
For your second question of whether or not flow makes it go, the over all answer is YES. Everyone should be able to calculate the expected flow rate leaving a pump based upon volumetric displacement and rotational shaft speed and adjusting the value due to mechanical and volumetric efficiencies within the pump. So the answer there would be that pumps produce flow not prsesure.
Specifically for your "flow makes it go" statement, if you have a cylinder in your circuit, one would want velocity and force. To develop a force from your cylinder you need pressure due to the simple equation of F=P*A, likewise to "make it go" you need flow to create a velocity, whereby Q=V*A. Assuming that your area remains constant and that there are no resistive forces acting against your cylinder, your velocity would be proportional to the flow divided by the acting area. Additionally, your force would be the product of your pressure against the acting area.
I hope this answers your "loaded" questions and that the above equations are suitable for what you were looking for.
Personal Note:
My background stints from having the opportunity to work with Dr. Fitch, the founder of the Fluid Power Research Center and Dr. Hong, his protege, specializing in hyrdaulic system design, service assurance and contamination control.
Regards
Andrew
I'm fairly new to this forum but noticed this interesting post and thought I would supplement it with some of my thoughts being that some answers have still yet to be shown. I believe the last remaining answer is whether or not equations are capable of being generated to prove such topics and the answer is YES they are, but NO they are not simple. So I will do my best to explain them.
Pressure:
The following mathematical equation has been developed to best represent characteristics of fluid passing through an orifice, which I believe our original post master has stated somewhere else in these forums:
Q = Kv * (deltaP)^0.5
The Kv value represents what people may associate with as "resistivity." However, this simple equation is not easily used in all fluid power instances. Further explanation must be validated to provide meaningful definement to this equation.
The true form of this mysterious Kv value is as follows:
Kv = Cd * A * (2/rho)^0.5
where:
Cd = orifice flow discharge coefficient
A = orifice area
rho = fluid density
The area and density are obvious and easy finds for this equation model, but the definement of the orifice flow is much more difficult. For now, since I can not post charts, I will refrain from discussion about transitional flow.
The Cd value is dependent upon the defined Reynolds number of the fluid. As the Reynolds increases so does the Cd value until reaching turblent flow, whereby, for the most part, it levels out to a constant value. The relationship between the Cd value and the Nr (Rey No) during laminar flow is as follows:
Cd = 0.125 * (Nr)^0.5
This value becomes fairly constant in turblent flow, which stays at about 0.61.
With this in mind, this equation answers your question regarding pressure being a resistivity of flow. It can be correlated that as flow increases so does pressure across the orifice and vice versa. Its understanding the mathematical relationship to better define the actual characteristics and identifing the necessary test conditions.
For your second question of whether or not flow makes it go, the over all answer is YES. Everyone should be able to calculate the expected flow rate leaving a pump based upon volumetric displacement and rotational shaft speed and adjusting the value due to mechanical and volumetric efficiencies within the pump. So the answer there would be that pumps produce flow not prsesure.
Specifically for your "flow makes it go" statement, if you have a cylinder in your circuit, one would want velocity and force. To develop a force from your cylinder you need pressure due to the simple equation of F=P*A, likewise to "make it go" you need flow to create a velocity, whereby Q=V*A. Assuming that your area remains constant and that there are no resistive forces acting against your cylinder, your velocity would be proportional to the flow divided by the acting area. Additionally, your force would be the product of your pressure against the acting area.
I hope this answers your "loaded" questions and that the above equations are suitable for what you were looking for.
Personal Note:
My background stints from having the opportunity to work with Dr. Fitch, the founder of the Fluid Power Research Center and Dr. Hong, his protege, specializing in hyrdaulic system design, service assurance and contamination control.
Regards
Andrew
- Thread starter
- #23
Excellent.
One can see that deltaP changes every microsecond so Q is always changing.
dp/dt=B*Q(t)/V(t)
where:
dp/dt is the rate of pressure change
B is the bulk modulus of oil
Kv is the spool constant or orifice constant.
Q(t) is the flow at that instant.
V(t) is the volume of fluid at that instant.
Normally I compute Kv by using a valves rated flow and the rated pressure drop
RatedFlow=Kv*sqrt(RatedDeltaP/2) There are two lands or edges of the spool.
Kv=RatedFlow/sqrt(RatedDeltaP/2).
Note, Budt didn't include any of this data in his problem. If the actuator isn't dead headed there is no way of computing flows, accelerations, pressures or speeds.
This is the classic flow makes it go equation. It is wrong ( actually incomplete ) because it doesn't explain how the mass is accelerated to the velocity in Q=V*A. Not only does it assume there is no friction. It assume there is no mass to accelerate and it assume there is not opposing force cause by pushing the oil the port on the other side of the cylinder. That isn't very realistic. Those that use V=Q/A screw up their hydraulic designs. Then I/we get calls from customers that ask why the motion controller isn't making the actuator move as fast as they want.
Perhaps you should ask Dr Fitch and Dr Hong about 'flow makes it go'.
I am still going to the school of hard knocks and will never graduate.
stroupaloop
Mechanical
You are right. One can not solely rely on the Q=V*A equation when sizing or selecting a hydraulic cylinder. Much additional calculations must be included to accompany mechanical and volumetric efficiences as well as outside forces and cylinder design additions. To truly represent and model the hydraulic cylinder, non-linear differential equations are required to model dynamic performance to note pressure spikes caused by the necessary force to overcome the breakaway pressure and the inertial force of the flow impacting the piston.
But to briefly indulge into a further equation description, this is the general static equation I use, minus the bells and whistles for Force and Velocity
F(ext) = (Nm)*(P1*A1-P2*A2)
V(ext) = ((Nv)*Q1)/A1
where
Nm = mechanical efficiency
Nv = volumetric efficiency
Obviously much addition to the efficiencies are involved to properly characterize these values. Consideration of internal leakage, frictional forces due to seals at the piston and rod as well and the mass friticional force required. On top of that, one can go further by inducing vector calculations to accompany the mass and cylinder direction to better understand the external forces played at the end of the rod.
Beyond that one needs to look at external factors that do not encompass fluid power, such as wall thickness, material selection, and installation measures. To further complicate it, the issue of addressing cylinder cushioning into the equation greatly increases the complexity of the original static Q=V*A equation.
I'm glad to know that there is intelligent life out there, according to your forum post. Its remarkable how even government agencies that I have worked for who have dedicated hydraulic departments filled with engineers, can't answer these questions and find themselves designing systems that people rely their life on. Although I suppose we can chalk it up to the lack of opportunity for fluid poewr education. This is what we get when the engineering world considers an area "mastered," all the training goes away...
But to briefly indulge into a further equation description, this is the general static equation I use, minus the bells and whistles for Force and Velocity
F(ext) = (Nm)*(P1*A1-P2*A2)
V(ext) = ((Nv)*Q1)/A1
where
Nm = mechanical efficiency
Nv = volumetric efficiency
Obviously much addition to the efficiencies are involved to properly characterize these values. Consideration of internal leakage, frictional forces due to seals at the piston and rod as well and the mass friticional force required. On top of that, one can go further by inducing vector calculations to accompany the mass and cylinder direction to better understand the external forces played at the end of the rod.
Beyond that one needs to look at external factors that do not encompass fluid power, such as wall thickness, material selection, and installation measures. To further complicate it, the issue of addressing cylinder cushioning into the equation greatly increases the complexity of the original static Q=V*A equation.
I'm glad to know that there is intelligent life out there, according to your forum post. Its remarkable how even government agencies that I have worked for who have dedicated hydraulic departments filled with engineers, can't answer these questions and find themselves designing systems that people rely their life on. Although I suppose we can chalk it up to the lack of opportunity for fluid poewr education. This is what we get when the engineering world considers an area "mastered," all the training goes away...
stroupaloop wrote;
"Although I suppose we can chalk it up to the lack of opportunity for fluid poewr education. This is what we get when the engineering world considers an area "mastered," all the training goes away..."
My sentiments exactly, BUT, NO ONE SEEMS TO BE LISTENING NO MATTER HOW LOUD YOU SHOUT IT.
The Colege of Hard Knocks has been the main way to learn Fluid Power forever and from the BULK of the FEEDBACK feedback from this and other forums it will stay that way indefinitely.
However, I think Fluid Power is far from being mastered and will stay that way until it is recognized as a field of endeavor the same as Mechanical and Electrrical Engineering.
When someone who is mechanically adept and has 2 years of Industrial Technology from a Tech School can hire in as a Fluid Power salesman and almost immediately start designing Fluid Power circuits for Engineers and Mechanical Maintenance persons there is something wrong if that field of endeavor ever hopes to grow and advance.
Been asking for years, almost 21 to be exact when I wrote an article for H&P Magazine, why Fluid Power is not recognized as something besides an attachment to a Mechanical Engineers Title. Then I find out the Mechanical Engineers 4 years of College only includes about enough Fluid Power to learn how to spell Hydraulic and Pneumatic.
Had some good feedback from John Eleftherakis from down your way as well as others on the H&P article.
Maybe, Someday????
Bud Trinkel, Fluid Power Consultant
HYDRA-PNEU CONSULTING
"Although I suppose we can chalk it up to the lack of opportunity for fluid poewr education. This is what we get when the engineering world considers an area "mastered," all the training goes away..."
My sentiments exactly, BUT, NO ONE SEEMS TO BE LISTENING NO MATTER HOW LOUD YOU SHOUT IT.
The Colege of Hard Knocks has been the main way to learn Fluid Power forever and from the BULK of the FEEDBACK feedback from this and other forums it will stay that way indefinitely.
However, I think Fluid Power is far from being mastered and will stay that way until it is recognized as a field of endeavor the same as Mechanical and Electrrical Engineering.
When someone who is mechanically adept and has 2 years of Industrial Technology from a Tech School can hire in as a Fluid Power salesman and almost immediately start designing Fluid Power circuits for Engineers and Mechanical Maintenance persons there is something wrong if that field of endeavor ever hopes to grow and advance.
Been asking for years, almost 21 to be exact when I wrote an article for H&P Magazine, why Fluid Power is not recognized as something besides an attachment to a Mechanical Engineers Title. Then I find out the Mechanical Engineers 4 years of College only includes about enough Fluid Power to learn how to spell Hydraulic and Pneumatic.
Had some good feedback from John Eleftherakis from down your way as well as others on the H&P article.
Maybe, Someday????
Bud Trinkel, Fluid Power Consultant
HYDRA-PNEU CONSULTING
stroupaloop
Mechanical
Bud,
Your post make me laugh for multiple reasons and I'll explain as best I can. You are absolutely right regarding the fluid power education. As you probably know, the original hydraulics study began at MIT and then migrated to OSU as the FPRC. This began after WW2 and great advances were made. At the FPRC, students would attain a masters or doctorate degree in Fluid Power, where the founder was Dr. Fitch.
Once Dr. Fitch left the university the hydraulic training literally became non-existent for the most part and today, the only place to get somewhat formal training is at MSOE. Ironically, Dr. Fitch and Dr. Hong started a company named FES-BarDyne, Inc. where I came from and learned the majority of my hydraulic training from.
Ironically I work for John E. now who also came from the FPRC/FES-BarDyne days, which is why I laugh about your comment and literally how small of a world this fluid power industry is of compentent people.
I truly am honored to have had the opportunity to work with Dr. Hong and Dr. Fitch on very unique projects, but even then, these men of fluid power expertise will not be around forever. So the question remains, what will happen to the industry once the power houses of knowledge are no longer to be?...
Your post make me laugh for multiple reasons and I'll explain as best I can. You are absolutely right regarding the fluid power education. As you probably know, the original hydraulics study began at MIT and then migrated to OSU as the FPRC. This began after WW2 and great advances were made. At the FPRC, students would attain a masters or doctorate degree in Fluid Power, where the founder was Dr. Fitch.
Once Dr. Fitch left the university the hydraulic training literally became non-existent for the most part and today, the only place to get somewhat formal training is at MSOE. Ironically, Dr. Fitch and Dr. Hong started a company named FES-BarDyne, Inc. where I came from and learned the majority of my hydraulic training from.
Ironically I work for John E. now who also came from the FPRC/FES-BarDyne days, which is why I laugh about your comment and literally how small of a world this fluid power industry is of compentent people.
I truly am honored to have had the opportunity to work with Dr. Hong and Dr. Fitch on very unique projects, but even then, these men of fluid power expertise will not be around forever. So the question remains, what will happen to the industry once the power houses of knowledge are no longer to be?...
stroupaloop
Mechanical
Quick follow up.
I was looking up your posts and threads and I appreciate your willingness to educate people and try to spread the knowledge of the fluid power industry. You may be one of the few out there that will help the industry stay alive and knowledgable.
I was looking up your posts and threads and I appreciate your willingness to educate people and try to spread the knowledge of the fluid power industry. You may be one of the few out there that will help the industry stay alive and knowledgable.
- Thread starter
- #28
Andrew, your Nm and Nv terms are not valid.
I write the force equation like this
Force(t)=PCap(t)*ACap-PRod(t)*ARod+Friction(v)
You appear to use Nm as a substitute for friction. The friction function is a topic unto itself. The friction opposes the other forces and varies as a function of velocity. Using the velocity you can determine if the friction should be static or dynamic. This is essential for simulating the slick/stick that some actuators have. Vertical systems have and extra force term for the force due to gravity.
After computing F(t) one computes A(t)=F(t)/mass
Then A(t) is integrated to to get V(t).
V(t) is integrated to get the position, X(t).
It is calculating the net force that is the trick. The rest is easy.
To calculate the steady state velocity one needs to calculate the speed at which the net force is 0. V=Q/A does not work for reasons I stated above.
I write the force equation like this
Force(t)=PCap(t)*ACap-PRod(t)*ARod+Friction(v)
You appear to use Nm as a substitute for friction. The friction function is a topic unto itself. The friction opposes the other forces and varies as a function of velocity. Using the velocity you can determine if the friction should be static or dynamic. This is essential for simulating the slick/stick that some actuators have. Vertical systems have and extra force term for the force due to gravity.
After computing F(t) one computes A(t)=F(t)/mass
Then A(t) is integrated to to get V(t).
V(t) is integrated to get the position, X(t).
It is calculating the net force that is the trick. The rest is easy.
To calculate the steady state velocity one needs to calculate the speed at which the net force is 0. V=Q/A does not work for reasons I stated above.
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