Pressure in Parallel Circuits 

Each simulation is built from scratch, using our own specialized tools built on standard web technologies (SVG and JavaScript). Our animators draw out all the shapes, and then write custom code to animate them. I hope that answers your question!

Wow that's a super fantastic explanation. Thanks for sharing Lovro.

And thank you Carl (I guess) for this great animations. I've seen most of your videos, and although I understand the topic quite well already, I enjoy in this visualizations. As a electrical/controll engineer starting to work as a commissioning engineer for complex offshore cranes about a year ago, I didn't know anything about hydraulics and I had to learn a lot the hard way. If only I found this channel sooner :) However, there is is still one more (basic) thing that is bugging me, I found it important for understanding, not really for practice. I'm sure that the explanation is simple though. I would be very thankful if you could give me an answer, or even better, if you make a video about it. I will c/p the question I made some time ago on one forum. There was a huge interest in topic, I got a lot of people to think, but nobody could give me an actual answer! I'm sure many people here would be interested as well :) Here's the c/p "Hi everyone, I have some uncertainties in understanding the orifice pressure drop equation. First of all, I perfectly understand why do we have a certain pressure drop across the orifice, speed increases on the cost of the pressure drop according to formula (In ideal situations with no turbulence/viscosity factor Cd would be 1), but after the normal flow area is restored and velocity returns to original value, shouldn't the pressure increase (at the cost of kinematic energy) and end up restored by the very same formula again? (In ideal conditions, I understand there are certain losses). Does the pressure restore after it passes the orifice, and if not, why not? I'm sure I'm missing something here. I try to google for this but didn't find the info I need. Is the only pressure drop here related to losses? If so, I don't see the formula formulated in that way, it is based on pressure drop across the orifice because of gain in speed. Following that logic, after the speed is back to pre-orifice one, shouldn't the pressure be restored (minus the losses)? What would be the formula if the system is ideal (no losses)? p1=p2? Again, there must be some simple thing I'm missing here but if someone could give me an explanation of this, it would be great :) p.s. the source I found mentioning this pressure restoration, but then not including it or saying anything about it in their formulas! neutrium.net/fluid_flow/calculation-of-flow-through-nozzles-and-orifices/ Thanks in advance, L"

Well this is a deep subject. I'll just tackle a little bit of your question for now. After the orifice, one in a series of resistances is now behind the flowing hydraulic fluid molecules. The fluid is on it's way back to tank (atmospheric pressure usually, and also a low potential), and so maximum circuit pressure (potential energy) is highest at the pump outlet (for many/most systems, when functioning normally) and all other orifices are pressure drops, as fluid moves back towards the tank. This is very much the same as for voltage. Potential energy (voltage) in an electrical circuit is highest before electrons flow through a resistance on their way to the neutral/ground pole. Energy losses are permanent in a hydraulic system if heat was created/radiated outward, due to friction. The only time that I am aware of, that pressure is the same after an orifice as before, is when no flow is occurring (Pascal's law). I am speaking in general terms around your complex question. So I will give it more thought yet, especially where you are comparing only pressure to speed.

- Carl

Thanks again! I get the general idea and I understand the concept in case you have only a pump and an orifice. You need some pressure/force to squeze the fluid through the small opening but on the orifice outlet you have tank connection. It gets a bit fuzzier when you have more complex system, like flow-metered directional valve and an actuator/load after it. Of course, it's understandable how the pressure drops are pressure losses, which translates to energy losses in form of heat. The funny thing is that I understand intuitivly these pressure drops accross the orifice (it makes sense), but the way most sources are explaining why do we have pressure drops is Bernoulli equation, and Bernoulli equation also mandates that after passing through the narrower are, the pressure will rise once again, but it doesn't, it is converted into heat. Simple question that kinda gets you thinking :) We have 2 full time hydraulic designers in our office. I asked them both the same question, and surprisingly, nobody could give me an answer. They told me I got them thinking now. When they were learning about this in their study days, they simply excepted the common explanation, without thinking too much deeper. Would be great if you could make a video about it one day!

See here: www.lunchboxsessions.com/help/what-software-do-you-use-to-make-the-simulations

Mech E no because there is no flow through the gauge.

To your first question, no. The 600 would be in series with the check valves above so would add to the upstream pressure. Ie with both taps open, pressure at gauge would be 700psi. With both taps closed, gauge would show 900. Second question, yes that would mean all valves open and flow splits evenly between the 3 lines.

It will be 5 12 9 and 33

@@midwest9757 😆

That could be correct depending on the system, but generally no it's not correct unless your volumetric flow rate is increased by a larger amount than the ratio of your pressure to your flow in your equation... (not sure if I explained that well)

Hi Michael. We don't have any way to download these videos, but there are lots of "youtube downloader" websites. Good luck!

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