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Hopefully there is intelligent life here.

Unfortunately there isn't an option for a poll. 1. How many of you think pressure is resistance to flow? 2. How many of you think flow makes it go? Do you have an equation to prove either of the two statements? 

Me thinks this is a setup....
1. Obviously pressure can be caused by resistance to flow, but it can also be caused by trying to compress a trapped fluid. So is this taking "resistance to flow" to the extreme? 2. Flow makes what go? Work is a combination of pressure and flow.
ISZ 

Yes ISZ, this is a setup. You get brownie point for recognizing this.
I/my company, make hydraulic motion controllers. I/we know all about the math and physics of hydraulics. The problem we have is that ignorance isn't bliss. It is just plain ignorance and some how these problem become our problems.
So does any body believe in the the two statement? So does any body have a formula to justify either?
Be careful and think about your answer.


Hi PNachtwey
Well I'll say no with the exception that ISZ stated, firstly pressure is an analogous to voltage in an electrical circuit and secondly flow is analogous to current in the same sense, therefore likening it to Ohms Law the resistance would be (pressure/flow). Again with ISZ I don't understand the "flow makes it go" perhaps now you might explain further. No I have no formula to prove your above statements either.
regards
desertfox 

Peter...
I too struggle to understand the phrase "flow makes it go".
I feel that your request for an equation to quantify the pressure to flow ratio is a little simplistic.
I can give you the pressure loss coefficient, if you tell me the density of the fluid.
There is of course Bernoulli's equation.
I have a copy of the Handbook of Hydraulic Resistance, tell me what it is you want to know and I will see what I can do.
I spent a good 16 years in the the hydraulics "business". I could no longer stand the ignorance of the people who run it. The advent of electronic control was somewhat of a renaissance. However, the uptake of proportional control seemed to be very slow in the UK. That is with the exception of mobile hydraulics,which tends to be the complete opposite to industrial hydraulics.
I dont wish to labour the point, but there is saying that goes..."Power is nothing without control".
I have seen a few exceptionally good industrial hydraulic machines, but most people just seem happy to continue using big clunking hydraulic machinery as long as it keeps on bashing metal.
Adrian 

zdas04 (Mechanical) 
28 Jul 08 9:50 
These are pretty easy terms, but the loaded questions are simply stupid.
Pressure is an indication of the potential energy at a specific point within a fluid.Fluids will flow from a high pressure to a low pressure.The rate of fluid flow is a function of both the magnitude of the pressure differential over a measured distance and the total resistance to flow over the measured distance.
To say that "Pressure is resistance to flow" is nonsense. As everyone above has said, the second question/statement makes no sense.
David 

After giving this a little thought I must admit that your statement "pressure is resistance to flow" is actually true from a certain perspective. In a situation were a downstream valve can be used to constrain the flow path causing a pressure build up that restricts flow.
That said, "without pressure there is no flow" is also a true statement. Flow is defined by the pressure differential between two points. Without differential pressure, there simply is no flow.
So pressure is both the cause and the restriction to flow. Definable by Q = Cv(Dp/Sg)^0.5
As for "flow makes it go" seems to speak to the concept that flow can do work which is clearly true. However, the relationship between flow and horsepower (work) is entirely dependent on the efficiency of the mechanism doing the conversion.


Good there is intelligent life here. I have heard those two sayings for years and they drive me nuts.
1. Pressure is just force / area. 2. Force makes things go. See Newton's laws of motion flow or oil wasn't mentioned.


It's going to be a long week. 

budt (Industrial) 
30 Jul 08 16:42 
hydromech wrote; "I spent a good 16 years in the the hydraulics "business". I could no longer stand the ignorance of the people who run it. The advent of electronic control was somewhat of a renaissance. However, the uptake of proportional control seemed to be very slow in the UK. That is with the exception of mobile hydraulics,which tends to be the complete opposite to industrial hydraulics. I dont wish to labour the point, but there is saying that goes..."Power is nothing without control". I have seen a few exceptionally good industrial hydraulic machines, but most people just seem happy to continue using big clunking hydraulic machinery as long as it keeps on bashing metal." Where were you when all the post's were up about how I thouht there was a need for TRAINED/DEDICATED Fluid Power Engineers and Maintenance persons on the Fluid Power Forums??????? Sure could have used such great input. Bud Trinkel, Fluid Power Consultant HYDRAPNEU CONSULTING 

budt (Industrial) 
30 Jul 08 23:30 
I am the one that got Peter started on this subject. In all my hydraulic classes and in my Fluid Power traning books http://www.hydraulicspneumatics.com/200/eBooks/ I point out to the students that "Pumps don't make Pressure , only Flow" and that "Pressure is Resistance to Flow." It is my simplified way to get across the fact that a Pressure Gauge on a Hydraulic Circuit that is showing no pressure does not mean the pump is bad and conversely if the Pressure Gauge is showing full system prssure that the pump is good. No prssure may only mean the Pump Flow is all returining to tank with little resistance and a Flow Meter in that circuit could be showing 10 GPM Flow from a 10 GPM pump. OTH a 2,000 PSI pressure reading could be obtained from a pump producing 25% flow if there was enough resistance but the system would be operating slower than designed and a Flow Meter would show that reduced flow. Unfortuanately few Hydraulic Circuits have a Flow Meter installed so many pumps are changed unnecessarily when a Pressure gaauge shows less than the schematics set pressure and no adjusting of the pressure controls will bring pressure back to normal. At least that is the way I learned Hydraulics in the College of hard Knocks and the information has served me well. This subject has come up seveal times on other Forums over the past 10 years but still has not been resolved. It actually started with my asking, Should There be TRAINED/DEDICATED Fluid Power Engineers and Maintenance persons. Bud Trinkel, Fluid Power Consultant HYDRAPNEU CONSULTING 

budt (Industrial) 
30 Jul 08 23:39 
On "Flow Makes it Go" Peter has never answered my question on how fast a cylinder is moving when a Pressure Gauge at the port that is receiving oil that is moving the cylinder is reading 1,243 PSI and how much faster will it move if pressure suddenly jumps to 3,131 PSI. That would seem an easy one to answer if "Force Makes it GO" and if you knew the Area of the cylinder. Bud Trinkel, Fluid Power Consultant HYDRAPNEU CONSULTING 

Anybody? I have answered this before but Budt doesn't like the my answer.


I like to think of it like this...
A pump is simply a mechanical device for transfering energy from an electric motor and imparting the energy into the fluid.
In line with the the basics of Bernoulli's theory, there is a relationship between the velocity and the pressure of the fluid.
Let the the fluid move at high speed and the pressure will be low. Think of a pressure washer. High pressure fluid allowed to escape to low pressure will convert its stored energy into kinetic energy.
In a hydraulic system, the energy in the oil is contained by the pipework and the oils "energy" will carry the oil in the direction of the least pressure.
Thinking of cylinder moving with no load, the oil will obviously move towards the cylinder because that is where the pressure is lowest. When the cylinder reaches a resistant load its speed and therefore the speed of the oil inside will slow. The drop in speed will cause the pressure to rise and the pressure load across the area of the piston will create a force.
The ability of fluid to evenly distribute pressure means that when the oil is static the pressure is evenly distributed through the system.
Heat, vibration and noise all steal energy from the fluid and the net result at the cylinder is less power.
Flow makes the oil go because flow is kinetic energy. Resist the kinetic energy and flow will turn into pressure.
It is energy that make it go, energy in the form of pressure or flow.
All the of above is ignoring the changes caused by viscosity and reynolds numbers and laminar and turbulent flow and HOSES that seem to make a mockery of the basic laws of physics!
There...thats what I think. 

Answer? Answer what? All I see is a seemingly never ending FLOW of trick questions, catch phrases and exercises in semantics. Quite franky I'm way too busy for this crap.
And for the record, I've been in hydraulics, pneumatics and flow control since 1983. And while I may not agree with everything I've seen over the past 25 years, I am still in awe at the sheer volume of advancement in the past 150 years of fluid power that has made this country the manufacturing giant it has always been and has put true meaning in the words "Git R Done". 

Wayne... You are not that busy if you can post a reply.. What country are you in? I think the 80/20 law can be aplied to your statement about 150 years of advancement. 80% of the advancement was done in the first 20 years. The remaining 20% was completed during the following 130 years. As I said in a previous post, it is really the advent of electrohydraulic control that really pushed the boundries of hydraulic applications. Adrian 

I was hoping that some one would point out that Budt's question doesn't make any sense because there isn't enough information. Things do not accelerate instantly to a steady speed. Budt wants to know the speed when? 1 micro second after increasing the pressure? Pressure can not be changed instantly from 1234 to 3131 PSI. Hydromech, I thought you were doing well until Quote: It is energy that make it go, energy in the form of pressure or flow.
Newton didn't mention any of those terms in his three laws of motion. Hydromech, how does one calculate an acceleration from pressure or energy? Force makes it go. F=m*a or a=F/m. Then you integrate acceleration to get velocity and integrate velocity to get position. However, the acceleration will change from instant to instant. Budt, didn't supply an area to push against so we don't know the force. He didn't supply a mass so there was no way to calculate an acceleration. Also, Budt supplied no info about the pressure supply. Even if the first instant the pressure is 3131 PSI it will not be the next because as soon as the load moved the supply pressure would drop. No information was supplied as to how this pressure would drop as a function of flow so we have no idea what the delta p across the valve supplying the valve is. [quote] Answer? Answer what? [quote] Yes, it is a poor question. Quote: All I see is a seemingly never ending FLOW of trick questions, catch phrases and exercises in semantics.
There isn't anything tricky about them. The first two statements in the first post are 'jingles' that hear hydraulic people repeat over and over without and basis in math and physics. Budt's question is incomplete. It is typical of many questions I see on forums are the person asking the question doesn't know what information is required to arrive at the answer. Quote: Quite franky I'm way too busy for this crap.
Yes, I agree but as long as people are building crappy hydraulics I am afraid it is necessary. No one has provided an equation for calculating the gain of the hydraulic servo actuator system that calculates the the steady state speed. If you can't do that then how do you know the system will go as fast as required? There is nothing tricky about that. Here is a hint. An object will accelerate until the sum of forces acting on it 0. When an object has a 0 net force acting on it will be at a constant speed. 

budt (Industrial) 
31 Jul 08 14:47 
Peter answered this post: Quite franky I'm way too busy for this crap. With: Yes, I agree but as long as people are building crappy hydraulics I am afraid it is necessary. Peter, As long as the UNTRAINED are designing Hydraulic circuits the situation will not change. Just imagine where the Mechanical and Electrical field would be if they had been handled in a similar manner to how the Fluid Power field has been and is. Just a thought. Bud Trinkel, Fluid Power Consultant HYDRAPNEU CONSULTING 

Consider this training. The home work is to find the formula to compute the steady state velocity when the valve is open. In other words find a general solution to your question. You have seen posts where I have posted the solution.


A sudden jump in pressure at the cylinder's fill port suggests that the cylinder has stopped moving because it's hit a hard stop, the fill flow has gone to zero, and so has the pressure difference across the fill plumbing. Such a large jump in pressure suggests that, instead, the cylinder has not necessarily hit a hard stop, but that a closed center directional valve with essentially zero internal leakage has stopped it, the stored strain energy in the cylinder has caused the cylinder to act as an intensifier, and we have been looking at the rod end, where the pressure has now risen considerably above the pump pressure, in proportion to the cylinder's area ratio. Mike Halloran Pembroke Pines, FL, USA 

Quote: A sudden jump in pressure at the cylinder's fill port suggests that the cylinder has stopped moving because it's hit a hard stop, the fill flow has gone to zero, and so has the pressure difference across the fill plumbing.
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. 

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 

Quote: Cd = 0.125 * (Nr)^0.5 Kv = Cd * A * (2/rho)^0.5 Q = Kv * (deltaP)^0.5
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. Quote: Q=V*A.
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. Quote: 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.
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. 

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, nonlinear 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*A1P2*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... 

budt (Industrial) 
1 Aug 08 15:09 
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 HYDRAPNEU CONSULTING 

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 nonexistent 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 FESBarDyne, 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/FESBarDyne 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?... 

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. 

Andrew, your Nm and Nv terms are not valid. I write the force equation like this Force(t)=PCap(t)*ACapPRod(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.




