What a nice video. It explains perfectly what happens in those systems and makes lots of examples. I think this is one of those problems that just are tricky but in reality they're really simple, as you made it look at the end
I think the problem has deviated from the original question, which is: what is the result pressure in the combined pipe considering its end closed? So the system is isolated from any external effects. Can you answer this question please?
@@ELHAOUARIAbdelhalim If there are no valves, nothing is preventing the pressurized tanks from depressurisation. so it's safe to assume that both of them have a tank on their pipeline. the whole segment afterwards, up until the closed (in this case) pipe, is full with air at a certain pressure. what is that pressure? it's the same exact problem as before, just that now that pressure is not automatically 1 bar since it does not go in the atmosphere but could be anything depending on how you filled this pipe. If you only filled it with air from the top tank, it would be the top pressure. if you filled it with atmospheric air, it would be still 1 bar. if you vacuumed it, would less.
If you take the original problem as stated (similar lines, no valves), then the outlet pressure is 15 bar (minus pressure losses) and there is no flow from source B. If fact, some or all the flow from source A would go into B. The key here is to distinguish between Rating an existing system versus Designing a new one. If I can add valves wherever I want to, I am designing. If I need to calculate pressures for a fixed system, I am Rating.
This is what I really don't get in this problem. If you want to stay "inside the box" of this odd though-experiment, then A is simply going to back-flow into B as well as exit on the open end. After all, there is no pressure drop in the pipes and the source ALWAYS provides 15 bar, those 15 bar are there all the way until the exit. B can only be at 15 bar too without any additional devices in-between.
@@leocurious9919that's what I thought to. Although I thought a little less than 15 bar since you will drop some pressure with the flow going to the 10 bar side.
@@Jonas_Aa Just like a source that always provides 15 bar, I would assume that there is no pressure loss. The diameter etc. is not given, so it could really be anything from 0 to 15 bar. It could be a capillary tube for all I know.
Actually weirdness happens - I worked on a 30 bar pure nitrogen system where we detected a leak because atmospheric oxygen was contaminating the nitrogen - yes, one atmosphere air was leaking back into a 31 bar nitrogen system… And tightening the leaking tube fitting eliminated the ppm level contamination.
To think how many students (and engineers alike) have never truly understood this concept. Extremely clear and entertaining explanation. If only more lecturers had your gift of unwrapping a problem and coaxing out the answer in such a logical and comprehensible form.
@@ProcesswithPat I do remember the moment fairly clearly. But I think that "joke" was erased from my memory haha. Maybe I was too stressed at the time to appreciate cabbage based humour
I’ve seen similar problems in electrical systems. It’s very easy to get trolled if you don’t consider the full context or spot what’s yet to be defined. Great video, thanks Pat!
Is it valid to reason about systems like the one described above that unless you know "the load," you cannot know the behavior of the whole system? As in electrical engineering, you don't speak of current until there's a force pushing it (i.e. voltage difference), and even adding a seemingly minor element affects the characteristics of the whole system.
It is the same in electrical engineering. The impedance of a line determines the voltage at a given current flow... I love it when seemingly independent fields of engineering go hand in hand. Just as how Navier Stokes and Maxwell equations have a lot of symmetries.
@@philipoakley5498 I'm not that deep in electrical engineering. I'm more the mechanical guy that has kept an interest for the electronics... rare in my branch, but helps a lot.
@@jackmclane1826 for fluid flows you can get sidewall effects and turbulent flow so there are some differences between plumbing flows and electrical flows. Electric just depends on potential (pressure) difference. It's all about how the flow 'valves' are constrained adjusted;-)
@@philipoakley5498 But you also get eddy currents that have a similar effect to turbulent flows in fluids. They can pretty much push the flow of current more to the inside of the conductor, see 'skin effect' for more info. This can pretty much work as an additional impedance/resistance, creating voltage loss across the wire, just like you can get pressure loss from transfering fluids over ditances. Although I'm not saying the two are the same. It's interesting to see that similiar effects arise in both cases.
@@vojtechsmetana5261 Do watch out for some of these analogies, as they fall prey to the fallacy of pushing analogies too far. Things like skin effect have a strong frequency effect. And electrical flow has to be circular, which isn't the case in fluid design. You also don't consider EMC effects with fluids, so caution where caution is due. ;-)
Good explanation. As an electronics/maintenance person it really describes a conversation I had with a chemical engineer. I was not successful in my attempts to explain it
Interesting - I do underground process water systems in mining and deal with this constantly. These are generally closed systems (no atmospheric dumps unless we’re filling reservoirs etc and even then we’d use orifice plates with PRV or sustaining)so we deal with frictional losses as back pressure in miles of pipe, a lot of it horizontal. Joining two streams isn’t really ideal underground, the higher pressure from one would just push back against the other until the pressure drop was big enough to cause mixing. We do force this sometimes using sustaining valves though. Anyway, great video and great explanation on how the entire system needs to be understood.
I actually think there's a bigger point that I notice some people oversee. You nicely demonstrate that anything that happens downstream has a big effect on what's upstream, which is also the case with external flow. And that is what I think people sometimes forget/don't know.
The problem is that the more restricting the mixing pipe is and the following equipment the more likely source A is going to flow into source B. That is why back flow preventers often get installed.
Also important that the piping after the junction has the combined flow of the two. So in a proper mixing situation, more pressure drop per meter in that piping than upstream.
@@mikefochtman7164 true, but you can't guarantee that blockages won't happen. You need to have back flow preventers any time that you bring two different chemical lines together. You also forgot to mention the angle of entry into the larger pipe being important as well. Ideally you want both lines to meet at a Y instead of a T to insure that the flow is toward the mixing line and not toward each other.
Hi, how about let's say during design, I have 2 pumps for mixing. One bigger pump A (200m3/hr) is designed to generally operate at 10 barg, while the other smaller pump B (50m3/hr) is at 6 barg. However, I need the outlet to be at 10 barg (more than 200m3/hr), would the pump B be able to provide the additional flow? For example, to achieve 10 barg, the flow is smaller in the pump curve.
I’m happy to learn that I was correct to become confused when you drew a T connection but were saying “combined streams.” My gut reaction was “how they are they being combined?”
Great explanation. One thing to add is that geomtery plays a big role in the way fluids mix and pressure change occurs. For example ejectors and venturis are practical devices for mixing fluids and allowing a low pressure stream to increase in pressure past the mixing point. Whereas two pipes coming to a T can cause backflow unless pressure is reduced in the high pressure stream.
Guess... You get to control the flow, by selecting the output pressure... and have to attach the 10bar at a point in the pipe - where the pressure has already dropped below 10... they can't be connected in "parallel".
Sometimes it’s easier to change a flow problem, to an equivalent electrical problem. If you think about preassure as voltage, pipes and valves as resistors, it might be easier to see how to solve this problem, or actually that it can’t be solved without additional information.
Having 10 bar in the line A, and 5 bar in the line B, wont you have reverse flow from A to b as well (from higher pressure medium to lower pressure medium)? Or we are assuming that the mixing point geometry is not allowing such flow or something.
Hey pat the way you simplify and explain things is just so helpful ...why did you stop making these videos man, we want our passionate process Engineer back man
Love your videos. Your ability to simplify these complex concepts into simple explanations is great. I have a bachelor degree in chemical engineering and have worked as a process engineer in numerous plants, and your videos are super helpful and applicable. Thank you!
Coming from electrical engineering, I see the problem as: given two voltaga sources of 15V and 10V, what is the voltage if we connect them together? It's immediately apparent that the problem hasn't been setup properly since 1) there's no reference to ground and 2) if there was you are shorting the sources and will cause current to flow in the smaller voltage. We need a load (the reactor) with reference to ground (1 bar) and appropriate resistors (valves) to match the voltage drops!
Nicely explained,each sentence is well framed.yes it is true pressure mixing is rare and most engineer are confused about it because it is counter intuitive.
Holy moly, this makes me ask so many questions as I’m trying to devise my pool pump location, supply nd return piping . Parallel nd singular pipe options. Nd you gave me all the answers I need. Bravo , very happy you nd your channel popped up in my feed , while looking into flow , diameter nd pumps. Thank you again
Another analogy that helped me is to transpose it to circuit theory and use the node laws. Voltages into a node have to be the same, and sum of currents into and out of a node always sum to zero. I tackled this years ago with a mixing issue on fluids. Worked back from the outlet pressure and wrote an iterative spreadsheet to solve for equal remaindered pressures on both lines. Centrifugal pumps involved so I baked in the pump curves on each side and baked each line down to an equivalent Cv to make things quick to calculate.
In lymphatic vessel surgery, we must bridge a lymphatic vessel with a fluid having a lot of pressure, with a capillary (a vein) having little pressure, so all the lymphatic fluid can dissolve in the capillary ( vein).
what would happen if you do not install control valves to balance the pressure? would the 15 bar line flow slightly into the 10bar stream while the majority would flow into the outlet stream?
Lovely video. Another way to design the process and perhaps even a preferred way for safety and management in most cases would be to have the two pipes combine in the reactor vessel then the feed lines have a constant back pressure (as you choose) and are not affected by pressure or feed rate changes in the other reactant. I can think of no compelling reason to combine the two feeds inside the plumbing, a mechanically coupled two gang feed pump would be preferred even if the excuse of saving on a pressurised feed pump was the excuse. Saving on a reactor fitting would be a rare reason to justify the early mixing. Someone on Facebook recommended your video, said you were worth watching. I will keep an eye out for more.
Vastly depends on the process. Most solid-liquid injection processes for scrubbers can be done in piping to optimize against pluggage and make sure the slurry is homogenized ie soda ash for scrubbing. Lots of applications for static pipeline mixers as well
I also graduated in chemical/process engineering and I am delighted to find a channel dedicated to this subject. Unfortunately, my career took another turn.
As electrical engineer this problem seems easy if you try to imagine it with voltage instead of pressure. In a schematic all connections between components are considered equal voltage. If you want to model the conductors realistically, you'd have add resistors etc. into the schematic, even though they aren't physical components, but models for the properties of the conductor. If you connected two different voltages with a straight connection, you'd get a short circuit with infinite amps towards the lower voltage. So the schematic wouldn't be sufficient this way, you need to model the properties of the conductors/pipes and of whatever comes after the combination point. My bet is that if the resistance after the combination point is big enough, the fluid in the lower pressure source would actually flow backwards.
You mentioned that water flows from a high pressure location to a low pressure location. A hose connected to a pump that is going down hill would have higher pressure due to head gain at the nozzle compared to at the pump outlet, but the water still flows down hill. Can you explain what's going on in that situation?
@Process with Pat I think the best way to get an indication that we divide mass by pressure value and this value could indicate the final value of the mixture pressure and more precisely the Specific energry multiplied by the total mass of course the first case you mentioned pipe bleeding air to atmosphere no change in pressure happens coz we have in finite valve of air mass having atmospheric (pressure the absolute pressure value indicates the flow direction only)
Thanks for the enlightenment. For the longest time i treated pipe pressures like electrical laws and it kept confusing me: if A=15 and B=10 meet, shouldn't the point C be either a sum, a difference or the average? Turns out line losses are huge and the result is determined by if the outlet is open or closed
or you can use the high pressure fluid to create a type of jet pump. the low pressure fluid gets sucked into the suction side of the jet pump. in steam systems you get something called thermocompressors that do this, but obviously they are designed for steam service. im not sure what the fluid alternative is to that
The result of the video is clear, but I notice it's completely independent from the mixing problem. I can establish pressure on the outlet, for 1, 2, 3 o more tubes converging ok. But this does not tell much about the 2 flows of the example interact which each other. What is the flow from B and from A to outlet? I can't motivate Flow-B being outgoing (positive), even with a "simple valve" (a one-way valve I mean). I think Flow-B could only be ingoing (negative) without a valve, or zero with a valve. The reason is obvious: pA>pB. Using pressure regulator valves to equalize pressure is a solution to a different problem, not what I would like to understand. I hope a complete explanation comes out :p Anyway good job, I really appreciated your approach to the problem.
Hmm~ interesting, I guessed wrong thinking the pressures would average. That makes sense though that the pressure downstream of the mixing point is dependent on the backpressure downstream.
This is exactly the analogue to an electrical engineering problem. Where the stream pressures are like voltages and the lines are like interconnects and valves being resistances.
Excellence presentation. Everyone has the basic understanding that things flow from a higher pressure to a lower pressure (not trying to be sarcastic here). The actual downstream pressure needs to be defined. And yes, depending upon the operating pressure, A will flow more than B, unless a valve is introduced (which will create a pressure drop). Fun problem.
This isn't a question, the higher pressure will always dominate, if you put regulators in line that's a different question, if the end of the system is open to atmosphere you can use a Ventura type valve that will allow different pressures or substances to mix.
Being in electronics, my first question was what are the load and source impedances, since they were not given I called it a trick question in under 20ms.
Ok i completely misunderstood the assignment, but i also only looked at the thumbnail and title before thinking about it, i thought we needed to keep the flowrate and diameter or the pipe the same, so i thought we need to install a compressor at the end of both streams before they mix and compress them down in order to keep the flowrate and pipe diameter the same, because more stuff in the same volume at the same speed = higher pressure than before, i thought we would simply need to add up the different pressures
Pat, thanks for this video. Could you do a video of pump head calculation in the situation where on the discharge side, several lines connect to the main discharge line ? For the pressure drops should we consider the total flow rate for the section of the main pipe after connection of the other pipes ? And/or should we consider also the head of the pumps from other lines than connect to the main one ? I realized that everybody knows how to calculate pump head in the easiest scenario (2 tanks, 1 pipe, no other connections) but no one explains how to deal with connections. Have a nice day !
i have electronics knlowlege, and tried to solve this by comparing it with the water analogy of electricity. so pressure = voltage, and volume of liquid moved = current. the analogy is far from perfect, but i still came to the conclusion that there is not enough information. so yeah i tried to solve this using ohms laws lol
Nice video! Hypothetical question: If I had a stream of water at 10 bar colliding with a stream of water at 15 bar and there was a single pipe making a T with the intersection point of the two streams (so basically your drawing), would water from the 10 bar stream be able to mix with the 15 bar stream, or would the 15 bar stream just shut out the 10 bar and exit via the single pipe?
It hypothetically possible. The only time the 15 bar stream would flow back into the 10 bar stream is if it was easier for it to do that rather than for both streams to simply flow out of the combined pipe. This would mean that either: 1. the "outlet" (combined, T-piece) actually discharges into a pressure between 10 and 15 bar, say a pressurised tank (but that's weird because why then did we form our analysis from the point of view of the 10 and 15); or 2. The capacity from the 15 bar stream is so high that it is able to put enough flow through the combined pipe such that the pressure drop in that pipe becomes greater than 10 bar, making it possible to send some of the 15 bar stream into the 10 bar stream.
@@ProcesswithPatif it behaves anything like electric current then there will always be a flow from higher level to lower level in a parallel system. More will flow in the route of less resistance but there will be a tiny flow in the higher resistance route proportional to the resistance. Current cannot choose to not take the route of more resistance completely as long as the voltage is high enough to overcome the resistance. Speaking in fluid dynamics I'd say the 15 bar stream will split up in a certain manner, one part flowing "out" and one flowing in the direction of the 10 bar source.
I am confused. At 7:42 you have drawn an output @process line A as 10 bar, while line B is 5 bar. Is this just an oversight and both should be equal, or am I missing something?
As a random bloke that just landed on this video, i'm lost at the atmosphere part... only got worse after that. Why does it leave at atmosphere? I turn my hose on, water comes out at high or low pressure. I'm thinking I don't have a certain understanding to help with this. Usually im pretty good with physics but never ever looked at plumbing stuff before. Fells like when i looked at circuits the first time.
Basically at any point, there is only one pressure. So at the point they meet, they always have the same pressure. If you don’t have a resistance in front of the mixing part. They must have equal pressure and will never have different pressures. If p(A)>p(B) -> p(B)=p(C) with negative flow.
Our professor put it like this: Pressure is always produced by the acuators, never by the pump. The only thing a pump produces, is flow. Same with shower heads. To get higher pressure for showering, you need to replace the shower head, not the pump.
For some reason i always thought If one flow A is 15, and the other one B is 10, wouldn't the flow with higher pressure not "allow" the lower pressure to go inside C? How are they still able to mix? Am i making a mistake of treating liquids as having perfect flow?
Like in electrical systems, the only reason the system exists in the first place is to provide energy to the consumer. You start with them and then you go up stream, making sure they satisfy the requirements. Great video!
Thanks for the video. I've been trying to find a solution for mixing water at different pressures AND with varying pressures for A and B. If A and B are 2 different lines controlled by pressure switches (i.e. for well water pumping), then the varying pressures mean any mixing valves cannot be static. And since you cannot perfectly match any pressure switches, either A or B will eventually go to the minimum switch activation setpoint more frequently. It's an interesting problem for which I have not found a good, automated solution.
Looks like a resistor bridge problem. You can make parallel with electronics. Admitting both pipes have same section and there's no leak in the system, you should say Da = Db = Dtot/2 = 2,5bars and output closing manometer should say 12,5 bars. Admitting pipe sections are enough small to prevent current overflow for source and sink.
I assume that adding pressures works similarly to adding resistance in electricity, so I think it'd follow the inverse of a sum of inverses. 1/10+1/15=1/x 15/150+10/150=1/x 25/150=1/x 1/6=1/x therefore x=6 bar
Good Video. Unregulated the 15 bar line product flow will flow back into the 10 bar pressure line ,if connected one with each other and no exhaust to the atmosphere. Once connected together and provided an escape to a the atmosphere or other processes via pipeline the potential combinations possible are dictated by many variables. That's the Engineers job.
im havign exactly this same problem but in a reformer. m,y fuel and air reaacting are mixing at different temperatures and pressures. dealing with the temperature part is simpler, one can use enthalpy balance, but the pressure one is mind boogling. and papers really dont say much about this
I’m glad to hear it was helpful. Yeah this one really messed with my head early in my career and it’s not something that is explained very well, that’s why I thought I’d do it! Good luck with your reformer!
At the mixing point the 15 bar would flow out as well as down the 10 bar line because it is 5 bar higher pressure. The B component would not flow at all because it is at a lower pressure. Even with a backflow preventer the 10bar pipe cannot over come the 15 bar at the mixing point. If I have this wrong please explain why? Thanks.
I m facing a problem that I don t understand and the thing is that i am facing 2 inlets at different pressures and pipe sized and I am guessing why the fluid at the highest pressure will not block the fluid at the lowest pressure because fluid pass from high pressure to low pressure. In ur case you installed valves and made the pressure at each inlet the same so you would know your outlet pressure. What if we don t have these valves what would happen? will the 2 flows mix up and go to the outlet or if the pressure in one inlet is higher than the other one will it block it or go throw it even? Thx :)
I have a video that’s a follow-on from this one on reverse flow. Go check it out and see if it helps. About the valves - they are in the example but even if they weren’t it wouldn’t matter. Valves cause pressure drop. So does piping. The flows of each stream regulate themselves so that the pressure drops of each stream are such, that the streams combine at the same pressure.
Thank you, I appreciate you posting these videos. I’m currently a field power engineer in the oil and gas sector. Can you suggest a layout program where I can drop in fittings and pipe that will automatically do the flow calculations for me? I’m trying to re design a glycol heating system, but with my own money so price is an issue. Thank you.
I think the problem has deviated from the original question, which is what is the result pressure in the combined pipe considering its end closed? So the system is isolated from any external effects. Can you answer this question please?
If it's open like a water tap it would simply run 1 bar or atmosphere pressure. Otherwise the flow would expand until it reaches it. Simple. It's more difficult to mix different sources at different pressures and combine them in a fixed amount of substance ratio. Something like an ammonia synthesizer (N2+3H2). You want the mixture at a fixed pressure and temperature (compressing and decomps heats or cools like it's in a heat pump) and all you got was the two sources and a few pumps or valves (and you want to use the least of them because of the electricity bill...) Tough job that's not what I can do.
wow, I'm really impressed. I come from mechanical engineering and I'm currently at the final level. but I got the same case for my final assignment, I want to know what this pressure phenomenon is called?
Umm isnt it like this: If the system is somewhat closed, the pressure will be 15 bar, simply because thats the highest pressure provided by one of the two lines. If it isnt closed and the flow is free to move the pressure sets according to the resistance that just happens to be in the system with a maximum of 15 bar.
A problem that baffled me on many occasions, and particularly when those two different pressures were fed into a mixing valve. Even more so when one had a constant cv/kvs value within the valve. Curiously, the mixing valve always achieved the desired result.
The practicality when using two valves (pressure regulating devices) to control a determined outlet pressure from independent pressure supplies is idealistic and will not work to provide mixing with any certainty.
I think after 5bar mixing tank to satisfy 2 bar criterion you need to install Pressure Regulating Valve at downstream to maintain 2bar, rather increasing 2 bar at mixing Junction Please correct me if I am wrong
once I made a water gun using 10liter canister, drilled a hole in bottom and inserted car windshield motor and water is sprayed via silicon tube. But then I installed additional same size windshield motor and but also now each motor had an aquarium one way valve and connected to silicon tube via T connector - to my surprise running two motors did not increase the water flow of the water gun :D
Because windscreen wiper pumps are positive displacement pumps and water is mostly incompressible (380bar pressure produces a 2% compression on water - Titan oops pressure). Compressible fluids and centrifugal pumps give different results (but pump sizing is important too).
Very interesting video. @3.16 states that pressure will be 0 gauge at the point of leaving to the atmosphere. So, in that condition what about pressure at Points A and B. Is it also 0 gauge (plus friction loss in the pipe )?. I am just confused about how upstream pressure will be maintained when an outlet is open to atmospheric ?.
A problem with this analysis is if you put in a valve that has a pressure drop that equals the pressure difference between the reactor and source you have zero flow. Of course, a simple valve has a resistance roughly proportional to flow, so as flow rate drops pressure drop decreases and you get flow. It would simplify your design concept if you just looked at pressures between points to calculate flows. Look at each pipe segment as a resistor. You also have the reactor. As it fills the pressure will increase. If the reactor has a pop-off valve you will have a constant pressure. If there is a chemical reaction you will have pressure/temperature effects. The whole analysis would be simplified if you had a flow meter on each of the feedstocks and adjusted the valves to get the feed rates you needed. After looking at feed rates you could calculate prsssures since the currents would be knowns. A problem in your atmospheric analysis is it does not take into account fluid momentum. Point that hose upwards and look how high the stream goes. Pipes have a pressure drop per unit length which is flow dependent. Therefore the exit pressure is not the atmospheric pressure but rather the pressure that explains the height of the fluid column ejected into the container. The analysis differs in flow limited systems, which might be defined as a system where pressure drop over the common pipe is negligible. In that case the pressure at the mixing point is equal to the reactor pressure and you’ll need two post docs to control the valves and get correct volume delivery for each reactant.
Hi i am i bit confuse .. so you have 2 cylinders high pressure and Low pressurecand in middle you have a Pressure regulator .. so when i open the Pressure regulator what will happen if i open the Pressure regulator??? Cna help me what isbthe answer thanks
Great video but it's a little bit confusing because with an engineer approach you consider the lost in conduit, but just with this diagram we can just think you are connecting three pressure level with a perfect conduit system, and it doesn't make any sense. As an electrical engineer it's like drawing two different voltages sources in parallel without considering any resistance, which is an heresy.
What a nice video. It explains perfectly what happens in those systems and makes lots of examples. I think this is one of those problems that just are tricky but in reality they're really simple, as you made it look at the end
I think the problem has deviated from the original question, which is: what is the result pressure in the combined pipe considering its end closed? So the system is isolated from any external effects. Can you answer this question please?
@@ELHAOUARIAbdelhalim If there are no valves, nothing is preventing the pressurized tanks from depressurisation. so it's safe to assume that both of them have a tank on their pipeline. the whole segment afterwards, up until the closed (in this case) pipe, is full with air at a certain pressure. what is that pressure? it's the same exact problem as before, just that now that pressure is not automatically 1 bar since it does not go in the atmosphere but could be anything depending on how you filled this pipe. If you only filled it with air from the top tank, it would be the top pressure. if you filled it with atmospheric air, it would be still 1 bar. if you vacuumed it, would less.
If you take the original problem as stated (similar lines, no valves), then the outlet pressure is 15 bar (minus pressure losses) and there is no flow from source B. If fact, some or all the flow from source A would go into B. The key here is to distinguish between Rating an existing system versus Designing a new one. If I can add valves wherever I want to, I am designing. If I need to calculate pressures for a fixed system, I am Rating.
This is what I really don't get in this problem. If you want to stay "inside the box" of this odd though-experiment, then A is simply going to back-flow into B as well as exit on the open end. After all, there is no pressure drop in the pipes and the source ALWAYS provides 15 bar, those 15 bar are there all the way until the exit. B can only be at 15 bar too without any additional devices in-between.
@@leocurious9919that's what I thought to. Although I thought a little less than 15 bar since you will drop some pressure with the flow going to the 10 bar side.
@@Jonas_Aa Just like a source that always provides 15 bar, I would assume that there is no pressure loss. The diameter etc. is not given, so it could really be anything from 0 to 15 bar. It could be a capillary tube for all I know.
@@leocurious9919 The pipes are actually carbon nanotubes.
Actually weirdness happens - I worked on a 30 bar pure nitrogen system where we detected a leak because atmospheric oxygen was contaminating the nitrogen - yes, one atmosphere air was leaking back into a 31 bar nitrogen system…
And tightening the leaking tube fitting eliminated the ppm level contamination.
To think how many students (and engineers alike) have never truly understood this concept. Extremely clear and entertaining explanation. If only more lecturers had your gift of unwrapping a problem and coaxing out the answer in such a logical and comprehensible form.
Do you remember asking this? Boyle's law, Charles' Law, coleslaw - that was the answer...
@@ProcesswithPat I do remember the moment fairly clearly. But I think that "joke" was erased from my memory haha. Maybe I was too stressed at the time to appreciate cabbage based humour
@@ProcesswithPat ah, cole's law
@@ProcesswithPat2
I’ve seen similar problems in electrical systems. It’s very easy to get trolled if you don’t consider the full context or spot what’s yet to be defined. Great video, thanks Pat!
Is it valid to reason about systems like the one described above that unless you know "the load," you cannot know the behavior of the whole system? As in electrical engineering, you don't speak of current until there's a force pushing it (i.e. voltage difference), and even adding a seemingly minor element affects the characteristics of the whole system.
It is the same in electrical engineering. The impedance of a line determines the voltage at a given current flow...
I love it when seemingly independent fields of engineering go hand in hand.
Just as how Navier Stokes and Maxwell equations have a lot of symmetries.
And Reynolds ;-) Though we've yet to see that properly in electrical engineering!
@@philipoakley5498 I'm not that deep in electrical engineering. I'm more the mechanical guy that has kept an interest for the electronics... rare in my branch, but helps a lot.
@@jackmclane1826 for fluid flows you can get sidewall effects and turbulent flow so there are some differences between plumbing flows and electrical flows. Electric just depends on potential (pressure) difference.
It's all about how the flow 'valves' are constrained adjusted;-)
@@philipoakley5498 But you also get eddy currents that have a similar effect to turbulent flows in fluids. They can pretty much push the flow of current more to the inside of the conductor, see 'skin effect' for more info. This can pretty much work as an additional impedance/resistance, creating voltage loss across the wire, just like you can get pressure loss from transfering fluids over ditances. Although I'm not saying the two are the same. It's interesting to see that similiar effects arise in both cases.
@@vojtechsmetana5261 Do watch out for some of these analogies, as they fall prey to the fallacy of pushing analogies too far. Things like skin effect have a strong frequency effect.
And electrical flow has to be circular, which isn't the case in fluid design.
You also don't consider EMC effects with fluids, so caution where caution is due. ;-)
I always find it amazing that intuitive understanding of pressures is such a bad guide to reality. Great video!
Similarly, temperature is a construct...
Good explanation. As an electronics/maintenance person it really describes a conversation I had with a chemical engineer. I was not successful in my attempts to explain it
Interesting - I do underground process water systems in mining and deal with this constantly. These are generally closed systems (no atmospheric dumps unless we’re filling reservoirs etc and even then we’d use orifice plates with PRV or sustaining)so we deal with frictional losses as back pressure in miles of pipe, a lot of it horizontal. Joining two streams isn’t really ideal underground, the higher pressure from one would just push back against the other until the pressure drop was big enough to cause mixing. We do force this sometimes using sustaining valves though. Anyway, great video and great explanation on how the entire system needs to be understood.
I actually think there's a bigger point that I notice some people oversee. You nicely demonstrate that anything that happens downstream has a big effect on what's upstream, which is also the case with external flow. And that is what I think people sometimes forget/don't know.
The problem is that the more restricting the mixing pipe is and the following equipment the more likely source A is going to flow into source B. That is why back flow preventers often get installed.
Also important that the piping after the junction has the combined flow of the two. So in a proper mixing situation, more pressure drop per meter in that piping than upstream.
@@mikefochtman7164 true, but you can't guarantee that blockages won't happen. You need to have back flow preventers any time that you bring two different chemical lines together. You also forgot to mention the angle of entry into the larger pipe being important as well. Ideally you want both lines to meet at a Y instead of a T to insure that the flow is toward the mixing line and not toward each other.
Hi, how about let's say during design, I have 2 pumps for mixing. One bigger pump A (200m3/hr) is designed to generally operate at 10 barg, while the other smaller pump B (50m3/hr) is at 6 barg. However, I need the outlet to be at 10 barg (more than 200m3/hr), would the pump B be able to provide the additional flow? For example, to achieve 10 barg, the flow is smaller in the pump curve.
I’m happy to learn that I was correct to become confused when you drew a T connection but were saying “combined streams.” My gut reaction was “how they are they being combined?”
Great explanation. One thing to add is that geomtery plays a big role in the way fluids mix and pressure change occurs. For example ejectors and venturis are practical devices for mixing fluids and allowing a low pressure stream to increase in pressure past the mixing point. Whereas two pipes coming to a T can cause backflow unless pressure is reduced in the high pressure stream.
Guess... You get to control the flow, by selecting the output pressure... and have to attach the 10bar at a point in the pipe - where the pressure has already dropped below 10... they can't be connected in "parallel".
Sometimes it’s easier to change a flow problem, to an equivalent electrical problem. If you think about preassure as voltage, pipes and valves as resistors, it might be easier to see how to solve this problem, or actually that it can’t be solved without additional information.
Having 10 bar in the line A, and 5 bar in the line B, wont you have reverse flow from A to b as well (from higher pressure medium to lower pressure medium)? Or we are assuming that the mixing point geometry is not allowing such flow or something.
The 15 bar will flow into the T junction and the 10 bar will be equalised and effectively not flow.
Hey pat the way you simplify and explain things is just so helpful ...why did you stop making these videos man, we want our passionate process Engineer back man
1:45 Depends on the volume of the outgoing pipe. Make it to small and you have backflow in the 10 Bar pipe😜
This is such a well paced video, something I’ve thought about a lot.
Love your videos. Your ability to simplify these complex concepts into simple explanations is great. I have a bachelor degree in chemical engineering and have worked as a process engineer in numerous plants, and your videos are super helpful and applicable. Thank you!
Thanks for the really nice comment!
Coming from electrical engineering, I see the problem as: given two voltaga sources of 15V and 10V, what is the voltage if we connect them together? It's immediately apparent that the problem hasn't been setup properly since 1) there's no reference to ground and 2) if there was you are shorting the sources and will cause current to flow in the smaller voltage. We need a load (the reactor) with reference to ground (1 bar) and appropriate resistors (valves) to match the voltage drops!
Nicely explained,each sentence is well framed.yes it is true pressure mixing is rare and most engineer are confused about it because it is counter intuitive.
It's a delight to see you speak again. Guess I'm gonna be a chemical engineer now
Holy moly, this makes me ask so many questions as I’m trying to devise my pool pump location, supply nd return piping . Parallel nd singular pipe options.
Nd you gave me all the answers I need.
Bravo , very happy you nd your channel popped up in my feed , while looking into flow , diameter nd pumps.
Thank you again
What way to explain this simple problem. Thanks Pat.
Really good, clear video Pat! Fair to say it was informative and concise- great job, keep it up 👏 👍
I really appreciate it!
No worries, fully deserved!
Another analogy that helped me is to transpose it to circuit theory and use the node laws. Voltages into a node have to be the same, and sum of currents into and out of a node always sum to zero.
I tackled this years ago with a mixing issue on fluids. Worked back from the outlet pressure and wrote an iterative spreadsheet to solve for equal remaindered pressures on both lines. Centrifugal pumps involved so I baked in the pump curves on each side and baked each line down to an equivalent Cv to make things quick to calculate.
Delighted to see this!
0:25 my answer 12.5
In lymphatic vessel surgery, we must bridge a lymphatic vessel with a fluid having a lot of pressure, with a capillary (a vein) having little pressure, so all the lymphatic fluid can dissolve in the capillary ( vein).
what would happen if you do not install control valves to balance the pressure? would the 15 bar line flow slightly into the 10bar stream while the majority would flow into the outlet stream?
I did a video in reverse flow using the exact same setup. Check it out!
4:10?
Why would the ocean water flow back into the pipe if the pressure
Lovely video. Another way to design the process and perhaps even a preferred way for safety and management in most cases would be to have the two pipes combine in the reactor vessel then the feed lines have a constant back pressure (as you choose) and are not affected by pressure or feed rate changes in the other reactant.
I can think of no compelling reason to combine the two feeds inside the plumbing, a mechanically coupled two gang feed pump would be preferred even if the excuse of saving on a pressurised feed pump was the excuse. Saving on a reactor fitting would be a rare reason to justify the early mixing.
Someone on Facebook recommended your video, said you were worth watching. I will keep an eye out for more.
Vastly depends on the process. Most solid-liquid injection processes for scrubbers can be done in piping to optimize against pluggage and make sure the slurry is homogenized ie soda ash for scrubbing.
Lots of applications for static pipeline mixers as well
You still give the same value even after I failed year 2.2. Remain sublime
I also graduated in chemical/process engineering and I am delighted to find a channel dedicated to this subject. Unfortunately, my career took another turn.
As electrical engineer this problem seems easy if you try to imagine it with voltage instead of pressure. In a schematic all connections between components are considered equal voltage. If you want to model the conductors realistically, you'd have add resistors etc. into the schematic, even though they aren't physical components, but models for the properties of the conductor. If you connected two different voltages with a straight connection, you'd get a short circuit with infinite amps towards the lower voltage. So the schematic wouldn't be sufficient this way, you need to model the properties of the conductors/pipes and of whatever comes after the combination point. My bet is that if the resistance after the combination point is big enough, the fluid in the lower pressure source would actually flow backwards.
You mentioned that water flows from a high pressure location to a low pressure location. A hose connected to a pump that is going down hill would have higher pressure due to head gain at the nozzle compared to at the pump outlet, but the water still flows down hill. Can you explain what's going on in that situation?
@Process with Pat I think the best way to get an indication that we divide mass by pressure value and this value could indicate the final value of the mixture pressure and more precisely the Specific energry multiplied by the total mass of course the first case you mentioned pipe bleeding air to atmosphere no change in pressure happens coz we have in finite valve of air mass having atmospheric (pressure the absolute pressure value indicates the flow direction only)
Thanks for the enlightenment. For the longest time i treated pipe pressures like electrical laws and it kept confusing me: if A=15 and B=10 meet, shouldn't the point C be either a sum, a difference or the average? Turns out line losses are huge and the result is determined by if the outlet is open or closed
or you can use the high pressure fluid to create a type of jet pump. the low pressure fluid gets sucked into the suction side of the jet pump.
in steam systems you get something called thermocompressors that do this, but obviously they are designed for steam service. im not sure what the fluid alternative is to that
At what altitude the vacuum of space starts sucking?
The result of the video is clear, but I notice it's completely independent from the mixing problem. I can establish pressure on the outlet, for 1, 2, 3 o more tubes converging ok. But this does not tell much about the 2 flows of the example interact which each other.
What is the flow from B and from A to outlet? I can't motivate Flow-B being outgoing (positive), even with a "simple valve" (a one-way valve I mean).
I think Flow-B could only be ingoing (negative) without a valve, or zero with a valve. The reason is obvious: pA>pB.
Using pressure regulator valves to equalize pressure is a solution to a different problem, not what I would like to understand.
I hope a complete explanation comes out :p
Anyway good job, I really appreciated your approach to the problem.
Hmm~ interesting, I guessed wrong thinking the pressures would average. That makes sense though that the pressure downstream of the mixing point is dependent on the backpressure downstream.
Thank you for addressing very basic but highly confusing problem. 👏
This is exactly the analogue to an electrical engineering problem. Where the stream pressures are like voltages and the lines are like interconnects and valves being resistances.
Excellence presentation. Everyone has the basic understanding that things flow from a higher pressure to a lower pressure (not trying to be sarcastic here). The actual downstream pressure needs to be defined. And yes, depending upon the operating pressure, A will flow more than B, unless a valve is introduced (which will create a pressure drop). Fun problem.
This isn't a question, the higher pressure will always dominate, if you put regulators in line that's a different question, if the end of the system is open to atmosphere you can use a Ventura type valve that will allow different pressures or substances to mix.
Understanding science . . . fascinating! Nicely done.
That was fascinating. Thank you for the presentation. I learned a lot from this. I also like the piccy on the wall behind you; very powerful!
Being in electronics, my first question was what are the load and source impedances, since they were not given I called it a trick question in under 20ms.
Ok i completely misunderstood the assignment, but i also only looked at the thumbnail and title before thinking about it, i thought we needed to keep the flowrate and diameter or the pipe the same, so i thought we need to install a compressor at the end of both streams before they mix and compress them down in order to keep the flowrate and pipe diameter the same, because more stuff in the same volume at the same speed = higher pressure than before, i thought we would simply need to add up the different pressures
Pat, thanks for this video.
Could you do a video of pump head calculation in the situation where on the discharge side, several lines connect to the main discharge line ? For the pressure drops should we consider the total flow rate for the section of the main pipe after connection of the other pipes ? And/or should we consider also the head of the pumps from other lines than connect to the main one ?
I realized that everybody knows how to calculate pump head in the easiest scenario (2 tanks, 1 pipe, no other connections) but no one explains how to deal with connections.
Have a nice day !
i have electronics knlowlege, and tried to solve this by comparing it with the water analogy of electricity. so pressure = voltage, and volume of liquid moved = current. the analogy is far from perfect, but i still came to the conclusion that there is not enough information.
so yeah i tried to solve this using ohms laws lol
It seems like true.
thank you! great video and clear! looking forward to more of your videos on the process design!!
Nice video! Hypothetical question: If I had a stream of water at 10 bar colliding with a stream of water at 15 bar and there was a single pipe making a T with the intersection point of the two streams (so basically your drawing), would water from the 10 bar stream be able to mix with the 15 bar stream, or would the 15 bar stream just shut out the 10 bar and exit via the single pipe?
It hypothetically possible. The only time the 15 bar stream would flow back into the 10 bar stream is if it was easier for it to do that rather than for both streams to simply flow out of the combined pipe. This would mean that either:
1. the "outlet" (combined, T-piece) actually discharges into a pressure between 10 and 15 bar, say a pressurised tank (but that's weird because why then did we form our analysis from the point of view of the 10 and 15); or
2. The capacity from the 15 bar stream is so high that it is able to put enough flow through the combined pipe such that the pressure drop in that pipe becomes greater than 10 bar, making it possible to send some of the 15 bar stream into the 10 bar stream.
@@ProcesswithPatif it behaves anything like electric current then there will always be a flow from higher level to lower level in a parallel system. More will flow in the route of less resistance but there will be a tiny flow in the higher resistance route proportional to the resistance. Current cannot choose to not take the route of more resistance completely as long as the voltage is high enough to overcome the resistance.
Speaking in fluid dynamics I'd say the 15 bar stream will split up in a certain manner, one part flowing "out" and one flowing in the direction of the 10 bar source.
I am confused. At 7:42 you have drawn an output @process line A as 10 bar, while line B is 5 bar. Is this just an oversight and both should be equal, or am I missing something?
As a random bloke that just landed on this video, i'm lost at the atmosphere part... only got worse after that. Why does it leave at atmosphere? I turn my hose on, water comes out at high or low pressure. I'm thinking I don't have a certain understanding to help with this. Usually im pretty good with physics but never ever looked at plumbing stuff before. Fells like when i looked at circuits the first time.
Basically at any point, there is only one pressure. So at the point they meet, they always have the same pressure. If you don’t have a resistance in front of the mixing part. They must have equal pressure and will never have different pressures.
If p(A)>p(B) -> p(B)=p(C) with negative flow.
I guessed 12.5 because I had no idea of what I was doing, but your explantion made it all clear to me! Thanks!
Our professor put it like this:
Pressure is always produced by the acuators, never by the pump. The only thing a pump produces, is flow.
Same with shower heads. To get higher pressure for showering, you need to replace the shower head, not the pump.
For some reason i always thought If one flow A is 15, and the other one B is 10, wouldn't the flow with higher pressure not "allow" the lower pressure to go inside C? How are they still able to mix? Am i making a mistake of treating liquids as having perfect flow?
Like in electrical systems, the only reason the system exists in the first place is to provide energy to the consumer. You start with them and then you go up stream, making sure they satisfy the requirements. Great video!
Thanks for the video. I've been trying to find a solution for mixing water at different pressures AND with varying pressures for A and B. If A and B are 2 different lines controlled by pressure switches (i.e. for well water pumping), then the varying pressures mean any mixing valves cannot be static. And since you cannot perfectly match any pressure switches, either A or B will eventually go to the minimum switch activation setpoint more frequently. It's an interesting problem for which I have not found a good, automated solution.
This reminds me of some of the discussions I've had at work. Being in the weeds vs 10,000 foot view, verses a 50,000 foot view.
As a physicist my immeadiate reaction was: "Well you're not giving me enough information".
Looks like a resistor bridge problem. You can make parallel with electronics. Admitting both pipes have same section and there's no leak in the system, you should say Da = Db = Dtot/2 = 2,5bars and output closing manometer should say 12,5 bars. Admitting pipe sections are enough small to prevent current overflow for source and sink.
A question I used to come across quite frequently and to be honest I never knew the correct answer. I do now, so many thanks.
I assume that adding pressures works similarly to adding resistance in electricity, so I think it'd follow the inverse of a sum of inverses.
1/10+1/15=1/x
15/150+10/150=1/x
25/150=1/x
1/6=1/x
therefore
x=6 bar
Good Video. Unregulated the 15 bar line product flow will flow back into the 10 bar pressure line ,if connected one with each other and no exhaust to the atmosphere.
Once connected together and provided an escape to a the atmosphere or other processes via pipeline the potential combinations possible are dictated by many variables.
That's the Engineers job.
Not answered in the video? Is the flow from the second stream blocked or not? Does mixing even occur?
Great video. For some people it may be easier to understand this using circuit theory with voltages and currents. It's basically the same deal.
What will you do if the process requires temperature constraints
im havign exactly this same problem but in a reformer. m,y fuel and air reaacting are mixing at different temperatures and pressures. dealing with the temperature part is simpler, one can use enthalpy balance, but the pressure one is mind boogling. and papers really dont say much about this
I’m glad to hear it was helpful. Yeah this one really messed with my head early in my career and it’s not something that is explained very well, that’s why I thought I’d do it! Good luck with your reformer!
At the mixing point the 15 bar would flow out as well as down the 10 bar line because it is 5 bar higher pressure. The B component would not flow at all because it is at a lower pressure. Even with a backflow preventer the 10bar pipe cannot over come the 15 bar at the mixing point. If I have this wrong please explain why? Thanks.
wouldn't the higher pressure just push into the 10bar line and the output line?
I m facing a problem that I don t understand and the thing is that i am facing 2 inlets at different pressures and pipe sized and I am guessing why the fluid at the highest pressure will not block the fluid at the lowest pressure because fluid pass from high pressure to low pressure. In ur case you installed valves and made the pressure at each inlet the same so you would know your outlet pressure. What if we don t have these valves what would happen? will the 2 flows mix up and go to the outlet or if the pressure in one inlet is higher than the other one will it block it or go throw it even?
Thx :)
I have a video that’s a follow-on from this one on reverse flow. Go check it out and see if it helps.
About the valves - they are in the example but even if they weren’t it wouldn’t matter. Valves cause pressure drop. So does piping. The flows of each stream regulate themselves so that the pressure drops of each stream are such, that the streams combine at the same pressure.
@@ProcesswithPat thank you sooooo soooooooooooooooooooooooooooooo much :)
You bet!
And why 15 bar stream doesn't simply flow into 10 bar pipe?
Thank you, I appreciate you posting these videos. I’m currently a field power engineer in the oil and gas sector. Can you suggest a layout program where I can drop in fittings and pipe that will automatically do the flow calculations for me? I’m trying to re design a glycol heating system, but with my own money so price is an issue. Thank you.
Awesome videos, keep them coming Pat!
You bet!
I think the problem has deviated from the original question, which is what is the result pressure in the combined pipe considering its end closed? So the system is isolated from any external effects. Can you answer this question please?
What about the flow of both products? You might want to control the flow because you might need mol of product A and less mol of product B.
If it's open like a water tap it would simply run 1 bar or atmosphere pressure. Otherwise the flow would expand until it reaches it. Simple.
It's more difficult to mix different sources at different pressures and combine them in a fixed amount of substance ratio. Something like an ammonia synthesizer (N2+3H2). You want the mixture at a fixed pressure and temperature (compressing and decomps heats or cools like it's in a heat pump) and all you got was the two sources and a few pumps or valves (and you want to use the least of them because of the electricity bill...)
Tough job that's not what I can do.
wow, I'm really impressed. I come from mechanical engineering and I'm currently at the final level. but I got the same case for my final assignment, I want to know what this pressure phenomenon is called?
Thank you Pat very interesting.
Umm isnt it like this: If the system is somewhat closed, the pressure will be 15 bar, simply because thats the highest pressure provided by one of the two lines. If it isnt closed and the flow is free to move the pressure sets according to the resistance that just happens to be in the system with a maximum of 15 bar.
A problem that baffled me on many occasions, and particularly when those two different pressures were fed into a mixing valve. Even more so when one had a constant cv/kvs value within the valve. Curiously, the mixing valve always achieved the desired result.
The practicality when using two valves (pressure regulating devices) to control a determined outlet pressure from independent pressure supplies is idealistic and will not work to provide mixing with any certainty.
I think after 5bar mixing tank to satisfy 2 bar criterion you need to install Pressure Regulating Valve at downstream to maintain 2bar, rather increasing 2 bar at mixing Junction
Please correct me if I am wrong
once I made a water gun using 10liter canister, drilled a hole in bottom and inserted car windshield motor and water is sprayed via silicon tube. But then I installed additional same size windshield motor and but also now each motor had an aquarium one way valve and connected to silicon tube via T connector - to my surprise running two motors did not increase the water flow of the water gun :D
Because windscreen wiper pumps are positive displacement pumps and water is mostly incompressible (380bar pressure produces a 2% compression on water - Titan oops pressure).
Compressible fluids and centrifugal pumps give different results (but pump sizing is important too).
Very interesting video. @3.16 states that pressure will be 0 gauge at the point of leaving to the atmosphere. So, in that condition what about pressure at Points A and B. Is it also 0 gauge (plus friction loss in the pipe )?. I am just confused about how upstream pressure will be maintained when an outlet is open to atmospheric ?.
I didn't need the concept, but I enjoyed myself learning it. Excellent explanation, thanks
A problem with this analysis is if you put in a valve that has a pressure drop that equals the pressure difference between the reactor and source you have zero flow. Of course, a simple valve has a resistance roughly proportional to flow, so as flow rate drops pressure drop decreases and you get flow. It would simplify your design concept if you just looked at pressures between points to calculate flows. Look at each pipe segment as a resistor. You also have the reactor. As it fills the pressure will increase. If the reactor has a pop-off valve you will have a constant pressure. If there is a chemical reaction you will have pressure/temperature effects. The whole analysis would be simplified if you had a flow meter on each of the feedstocks and adjusted the valves to get the feed rates you needed. After looking at feed rates you could calculate prsssures since the currents would be knowns.
A problem in your atmospheric analysis is it does not take into account fluid momentum. Point that hose upwards and look how high the stream goes. Pipes have a pressure drop per unit length which is flow dependent. Therefore the exit pressure is not the atmospheric pressure but rather the pressure that explains the height of the fluid column ejected into the container.
The analysis differs in flow limited systems, which might be defined as a system where pressure drop over the common pipe is negligible. In that case the pressure at the mixing point is equal to the reactor pressure and you’ll need two post docs to control the valves and get correct volume delivery for each reactant.
Hi i am i bit confuse .. so you have 2 cylinders high pressure and Low pressurecand in middle you have a Pressure regulator .. so when i open the Pressure regulator what will happen if i open the Pressure regulator??? Cna help me what isbthe answer thanks
Good vibes learn a lot. Side question what the name of that art piece in the background??
Great video but it's a little bit confusing because with an engineer approach you consider the lost in conduit, but just with this diagram we can just think you are connecting three pressure level with a perfect conduit system, and it doesn't make any sense. As an electrical engineer it's like drawing two different voltages sources in parallel without considering any resistance, which is an heresy.
Excellent - similar conclusion one could find in economy and money stream transfers.
What is the meaning of the ( reactor) ?
Anyone can help, please.
is it a component such as valves, or is it a pressure generator?
This was well presented to help understand the solution to the problem stated.
The title was enough to make me watch the entire video