Hydraulic cylinder calculations

Hello experts,

I'm making my first steps in hydraulics and recently I have come across to a very old thread in which the following task was given, but never answered:
http://www.plctalk.net/qanda/showthread.php?t=42533&page=13
Piston: 17"
Rod: 15.5"
Piston weight: 2800 lbs.
Pressure: 1,000 PSI.
No flow controls, simple bang/bang valve, pressure compensated pump.
1. What is the max force that can applied on the blind end of the piston?
2. What is the max force that can be applied to the rod end of the piston?
3. What is the available pressure drop across the bang/bang valve when the piston is extending that dictates maximum flow through the valve?
4. What is the available pressure drop across the bang/bang valve when the piston is retracting?

The question was asked by a forum member TConnolly.
I have made some calculations and would like to verify if I understand the theory correctly.
I assumed given data for Piston and rod are diameter sizes.

1. Max force on blind end of the piston is Fb = 1000 psi*Area of blind end = 1009 kN. Area of blind end is Dp^2*pi/4.
2. Max force on the rod end is: Fr = P*(AreaP-AreadR) = 1000 psi*pi/4*(Dp^2-Dr^2) = 170 kN.
3. Piston is extending. This means pressure on blind end is 1000 psi. The force that corresponds to this pressure is F = p*AreaP = 1009 kN. This force on rod side creates higher pressure since the area is smaller. Prod = F/(AreaP-AreaR) = 2259 PSI. So the pressure drop between rod and blind side of a cylinder is 1259 PSI. Is this correct?
4. When piston is retracting, pressure on the rod side is 1000 psi. This corresponds to a force Fr = P*(AreaP-AreaR) = 446,6 kN. This force is moving the piston, so the pressure created on the blind side is Pb = Fr/AreaP = 442.5 psi. So the pressure drop is 557 psi

Am I on the right track?

Thank you

the force on the push side of the cylinder (not the rod side) is
push force = PI * R squared * Pressure
force = 3.1415926 * 8.5 * 8.5 * 1000 = 226,980 lbs force

the force on the retract (piston rod side) is
pull force = 3.1415926 * [ (R squared) - (piston rod squared)] * 1000
force = 3.1415926 8 [ (8.5 * 8.5) - (7.75 * 7.75)] * 1000
force = 3.1415926 * 12.19 * 1000 = 38,288 lbs force.

as far as the other questions, I left that up to my hydraulics designer.

james

James Mcquade said:

the force on the push side of the cylinder (not the rod side) is
push force = PI * R squared * Pressure
force = 3.1415926 * 8.5 * 8.5 * 1000 = 226,980 lbs force

the force on the retract (piston rod side) is
pull force = 3.1415926 * [ (R squared) - (piston rod squared)] * 1000
force = 3.1415926 8 [ (8.5 * 8.5) - (7.75 * 7.75)] * 1000
force = 3.1415926 * 12.19 * 1000 = 38,288 lbs force.

as far as the other questions, I left that up to my hydraulics designer.

james

Click to expand...

Ok, in short, first two tasks are correct (1 lbf = 4.448 N). Waiting for the remaining confirmation.

to answer questions 3 & 4 you will need the following information.

brand of valve.
size of valve
size of ports
size of hose / conduit connected to the valve.

google the valve and look at the specs, that might tell you.
Sorry, I do not have my hydraulic design books at the office.
james

James Mcquade said:

to answer questions 3 & 4 you will need the following information.

brand of valve.
size of valve
size of ports
size of hose / conduit connected to the valve.

google the valve and look at the specs, that might tell you.
Sorry, I do not have my hydraulic design books at the office.
james

Click to expand...

Hello James, I have what TConnolly specified in the thread posted above. It is ideal case I supposed.

I recommend that you to use the same unit system in all the calculations and best if it is the one used habitually in your country, that probably is SI units.

So it's best to start with cylinder sizes in milimeters and pressure in Pascal, if not your calculations can end like the Mars Climate Orbiter probe.

Valves are usually chosen knowing the working flow and pressure and applying these data in the tables of the manufacturer's valve catalog.

Yes, I used SI units when calculated final result. That is why Newtons are written. This was later confirmed by James, he calculated in lbf and when converted to Newtons, the results are the same.

you may also need the length of each line from the valve to the cylinder if its more than 5-10 ft. as I stated earlier, you need to answer the questions for # 3 and 4 to figure flow restriction and losses. generally speaking 5 ft is ok.
why is this needed, each valve has an associated flow and restriction. the bigger the valve, the more flow. a 1" port valve feeding a 1/2" hydraulic line can be figured as a 1" port valve, with a huge restriction.

james

Adenitz said:

Hello experts,

I'm making my first steps in hydraulics and recently I have come across to a very old thread in which the following task was given, but never answered:
http://www.plctalk.net/qanda/showthread.php?t=42533&page=13
Piston: 17"
Rod: 15.5"
Piston weight: 2800 lbs.
Pressure: 1,000 PSI.
No flow controls, simple bang/bang valve, pressure compensated pump.
1. What is the max force that can applied on the blind end of the piston?
2. What is the max force that can be applied to the rod end of the piston?
3. What is the available pressure drop across the bang/bang valve when the piston is extending that dictates maximum flow through the valve?
4. What is the available pressure drop across the bang/bang valve when the piston is retracting?

The question was asked by a forum member TConnolly.
I have made some calculations and would like to verify if I understand the theory correctly.
I assumed given data for Piston and rod are diameter sizes.

1. Max force on blind end of the piston is Fb = 1000 psi*Area of blind end = 1009 kN. Area of blind end is Dp^2*pi/4.
2. Max force on the rod end is: Fr = P*(AreaP-AreadR) = 1000 psi*pi/4*(Dp^2-Dr^2) = 170 kN.

Click to expand...

This is good as long as the cylinder is pushing against an obstruction.

3. Piston is extending. This means pressure on blind end is 1000 psi.

Click to expand...

You were doing OK up to here. The pressure on the blind or A side of the piston doesn't consider the pressure drop while extending

The force that corresponds to this pressure is F = p*AreaP = 1009 kN. This force on rod side creates higher pressure since the area is smaller. Prod = F/(AreaP-AreaR) = 2259 PSI. So the pressure drop between rod and blind side of a cylinder is 1259 PSI. Is this correct?

Click to expand...

No.

4. When piston is retracting, pressure on the rod side is 1000 psi. This corresponds to a force Fr = P*(AreaP-AreaR) = 446,6 kN. This force is moving the piston, so the pressure created on the blind side is Pb = Fr/AreaP = 442.5 psi. So the pressure drop is 557 psi

Am I on the right track?

Thank you

Click to expand...

No!

See this.
https://forum.deltamotion.com/t/the-vccm-equation/378?u=peter_nachtwey

There are two very important formulas you should know when designing hydraulic systems.
1 The VCCM equation.
2 How to calculate the Natural frequency of the cylinder and load.

Ok, I thought it was to easy to be true. Especially since I didn't use given data about mass and since James said the rest is left to designers.

Is it fair to say that given task was not fully specified in order to be solved with the given data?
BTW, this is how I imagined cylinder with bang-bang valve.

I really wonder how TConnolly managed to calculate the following (on page nr. 14):
The available pressure drop across the valve when extending is 1,000 PSI. Now if you actually have that much pressure drop across the valve the piston won't move, but that is the AVAILABLE pressure drop across the valve. The available pressure drop across the valve when retracting is 168 PSI. Actaul pressure drop across the valve will be less, but that is the maximum avaialble.

Hello Peter Nachtwey,

thank you for the information and the link. I have went through this example and I concluded that in any case I need to know Kvalve. Then I would calculate velocity in steady state and based on that I can calculate pressure drop across the valve: DeltaPa.
For the Fl (load force) I would use formula u*m*g (for friction assuming some coefficient for u, for example u = 0.2)

Since Kv is not known, it is not possible to estimate the pressure drop.

Please have a look at the attached picture. I went step by step based on your equations.

Adenitz said:

Please have a look at the attached picture. I went step by step based on your equations.

Click to expand...

This is very good.
What is the model of the valve. Can you provide a .pdf for the valve?
Most valves have a rated pressure drop for a rated flow.

Peter Nachtwey said:

Can you provide a .pdf for the valve?
Most valves have a rated pressure drop for a rated flow.

Click to expand...

Is this a question for me?
I have no idea. I was trying to answer to a couple of tasks that forum member TConnolly asked some other guy in a thread that is old more than 10 years. Please have a look at the link posted in my first post on this thread, page 13, at the end.

Adenitz said:

Is this a question for me?
I have no idea. I was trying to answer to a couple of tasks that forum member TConnolly asked some other guy in a thread that is old more than 10 years. Please have a look at the link posted in my first post on this thread, page 13, at the end.

Click to expand...

I thought you were designing a real system.
I write magazine articles for
https://www.hydraulicspneumatics.com/
The next article will be about how to calculate and use the valve flow constant, Kv.

Peter Nachtwey said:

I thought you were designing a real system.
The next article will be about how to calculate and use the valve flow constant, Kv.

Click to expand...

No no, I'm trying to solve these tasks: http://www.plctalk.net/qanda/showpost.php?p=297836&postcount=195

When it comes to hydraulics, I'm a complete newbie. Just trying to learn few things. I'm not even a mechanical engineer.

I wokd really like to know how TConnolly calculated 1000 psi and 168 psi. I assume there is something he forgot to mention like Kv.