It all started with a "bang"...
Literally - with a bang, when I was testing this long hydraulic cylinder equipped with a piloted check valve. The cylinder was "stretching" on the floor, hooked to our test bench, and I was cycling it to see if everything was in order. It was opening and closing just fine, but there was a sharp banging noise, happening each time the cylinder would switch direction after hitting an end of travel.
Of course, the pilot-operated check valve was the culprit. I would let the pressure reach the relief setting at the end of each stroke, which meant that a certain level of pilot pressure had to build to open the check valve when the cylinder was to switch, and this abrupt venting of a compressed volume was causing the banging sound.
There was nothing too special or unheard of about what was happening, but it did get me thinking about how such a cylinder + pilot-operated check valve combination would behave when subjected to trapped and/or load-induced pressures - and my head began to hurt immediately.
Let me explain why. A pilot-to-open check valve, unlike its "more sophisticated" over-center brother, is locked by a trapped or load-induced pressure - the higher the pressure that it's holding, the higher the pilot pressure required to open it. And then, to complicate things further, we have our classic double-acting hydraulic cylinder, which is a pressure amplifier in one direction, and a pressure divider in the other. So if you have a load that is "hanging", for example, on the rod side of such a cylinder, like so:
There's already a certain load-induced pressure in the rod end closing the piloted check valve, and then when you pressurize the tail end, you boost the pressure in the rod section by a factor, defined by the cylinder's area ratio, and this pressure is adding to the load-induced pressure. And so, as the tail end pressure rises, so does the pilot pressure necessary to open the valve, albeit not as fast because the pilot-to-open checks usually have a pilot ratio much higher than that of a typical double-acting cylinder, but still - thinking about the pressures adding up and working against each other, and how one would calculate at exactly what pressure the valve would open made my brain hurt. So, today's post is essentially me wrapping my head around the operation of a pilot-to-open check valve mounted on a hydraulic cylinder.
This is a typical construction of a pilot-operated (pilot-to-open) check valve:
Let's define/assume a few things about it:
In my opinion, the best way to look at the valve's behavior is through the definition of Equivalent Pressure Differential (EPD) across the check valve. If the EPD is higher than the cracking pressure - the valve is open, if it's lower - the valve is closed. Easy to grasp, right? Even though it is not a "real" pressure differential, but rather a force differential, caused by the larger pilot piston from the opening side, and the cylinder pressure actively closing the check valve from the other.
So, how would we calculate the EPD in our pilot-to-open valve + cylinder system?
Once again, it's quite easy - all we need to know is:
EPD = (P(c1) x R(valve)) - P(c2)
If you look at our system through the lens of the Equivalent Pressure Differential, the rest of the math becomes easy. There's only one other thing that you need to understand - the difference between the trapped pressure and the load-induced pressure in such a system.
The trapped pressure is the pressure "left" in our cylinder by the hydraulic system - something that can only happen at the end of its travel. For example, an outrigger, that was extended till it hit the end of stroke, and the operator kept pushing on the lever for a couple of seconds, leaving it pressurized at the system's relief setting.
The trapped pressure is just that - a volume of pressurized fluid trapped inside a container, like a compressed spring, or a gas-less accumulator, and therefore it can only be affected (increased) by the load (or the cylinder's pressure amplification) when the load-induced (or cylinder-amplified) pressure is greater than the trapped one.
Let's consider our cylinder system again. Say we closed it and "left" 200 bar in the rod side. And say the rod area is 10 cm² - i.e. 2 tons of force. If you pull the rod out with a force of 1000 kg, will the pressure in the rod side change? That's right, it will not! You'll need to pull with more than 2000 kg to change it. This means that the trapped pressure is not additive to the load or cylinder-induced pressures but rather a hard barrier you must "climb over" to make things move.
Now, the external load interaction is different. The external load, being what it is - an external force, is always "helping" our cylinder-based pressure amplifier/divider in its pressure-creating endeavors.
Let's turn to our system once again. Say we are pulling on the rod with 1000 kg of force, thus inducing 100 bar in the rod end. When we start injecting oil in the tail end, the pressure, amplified by the cylinder's area difference, will always add to the 100 bar created by the load. So if the cylinder has a ratio of 1.5, and we would pressurize the tail to 40 bar - the 60 bar (40 x 1.5) would add to the 100, and the pressure in the rod end would increase to 160 bar.
So, what does all this mean to us then? It means that when we have trapped pressure + some load situations, and we need to calculate the Equivalent Pressure Differential, we must consider the P(c2) to be the higher of the two values.
And now, finally, since we know which pressures to use, and we also know that our cylinder is a pressure multiplier in one direction and a pressure divider in the other, if we were to attribute R(cyl) as the area ratio of the cylinder, and P(cracking) as the cracking pressure of the check valve, we would calculate the opening pressure of our valve like so:
For the trapped pressure scenario it would be simple:
P(opening) = (P(cracking) + P(c2)) / R(valve)
And for the situations where there's no "pressurized cushion" in our cylinder, or the load-induced pressure is above it, we would do
For the rod-end-mounted valve:
P(opening) = (P(cracking) + P(c2)) / (R(valve) - R(cylinder))
For the tail-end-mounted valve:
P(opening) = (P(cracking) + P(c2)) / (R(valve) - (1 / R(cylinder)))
Now, don't worry, there's no need for you to do all this math to see how this works! I built an interactive graph, that does the math for you, and allows you to visualize how the P(c1) and P(c2) pressures plot out in relation to the Equivalent Pressure Differential.
Before I give you the link, I want to underline that the interactive graph that you are about to see clearly demonstrates that:
Now I can give you the link to the Cylinder Mounted Pilot-to-Open Check Valve Interactive Pressure Chart. Experiment at will!
And, finally, I also want to point out that we only looked into a simple valve design, and assumed that there's no back-pressure in the V2 port. Obviously, in real life, these valves come in many forms and flavors:
The cutaway view on the right shows a so-called balanced pilot-to-open check valve. In this design, the effect of the back pressure on the valve's pilot is greatly reduced.
I hope that I said enough for you to start wrapping your head around the operation of an actuator equipped with a pilot-to-open check valve.