Compound manometer example problem 

Thanks Robb =)

Thanks for letting me know, really glad to hear this helped!! I do plan on getting more fluids videos out later this year :)

Hey You're welcome! I try to keep the method as transparent as possible. Breaking it into part by part seems to be the best way to not get confused! =)

Thanks for letting me know that! Glad the videos are helping! :)

+humaid alkhateri Awesome glad to hear it, those +/- signs can be tricky!

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I will get to them one day!

Your welcome :) thanks for watching!

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abs .. i think coz its open

Yeah this whole video is done in absolute pressure. If you define Patm = 0 then take the difference between any P and Patm and you will have the gauge pressure instead (actually in this video, P2,P3,P4 will all be vacuum). This is a good video on the difference between absolute, gauge, and vacuum pressure: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-oOMzRzpNBEE.html

from what i have learnt , it is absolute

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+Ghufran Aldawood Thanks for the feedback! Glad to hear its helping :)

Thanks!

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Hey awesome, glad that it finally clicked for you =) =)

Glad it was helpful, thanks for watching! 😁

Glad it was helpful!!

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+RikkuCloud sometimes you just need to hear things said in a slightly different way. Glad it helped and thanks for commenting!

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Hey thanks! Appreciate the feedback :)

+Kevin Nguy Glad to hear it, thanks for reaching out!

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=) =) glad I can help

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Glad it helps, I noticed after the fact that I got a little lazy with saying/writing kPa vs Pa, but hopefully the little text box that pops up clears things up!

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Hey sorry not yet, but hopefully one day!

hahaha you're welcome :). Thanks for watching and sending the love!!

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Yasssss mission accomplished 🤜🤛

Happy to help! :)

Yes, and in this case 81 - 101 = -20 . A negative guage pressure is referred to as vacuum. -20 kPa guage = 20 kPa vacuum.

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I was thinking thins as well. There should be an extra pho*g*h for that pressure difference because pgas is higher

Because it’s a gas and not a liquid, you can ignore that height. Change in “depth” of gas is minuscule compared to that of liquid.

Because it’s a gas and not a liquid, you can ignore that height. Change in “depth” of gas is minuscule compared to that of liquid.

@@Engineer4Free good to know. Cheers

You're welcome!

The 81 kPa is absolute pressure

We consider pressure to increase in the positive direction when we move down through a fluid.

Welcome ☺️

Awesome glad to help. I only have a few videos about fluids right now, but the whole playlist is here if you are looking for it: ru-vid.com/group/PLOAuB8dR35oeOIPMOBH6hjwobuIJHPKSN

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Tanx for watching!

Daniel Lovik Hey where are you getting 0.1 from? We want need to use the vertical distance between 3 and 4 to find the pressure at point 4. The difference in pressure between any two points in a continuous body of a single fluid is ρgh where h is the difference in elevation between those points. Points 3 and 4 are both touching the same continuous body of water and are separated vertically by 0.2m. The shape of the body of water has nothing to do with the pressure as long as it's continuous. Once you know the pressure at the boundary between the oil and the water (point 3) write it down, then forget all about the oil and just consider the heights of the two ends of the water. This was the same logic used to find P2 and P3. Let me know if that helps or if you're still confused about it!

Engineer4Free Ya, still somewhat confused. I miswrote the last sentence in my previous comment meaning 0.1 not 0.2. As far as I can see the vertical distance from point three to point four is 10cm or 0.1m. Perhaps I just read it wrong and the distance between 3 and 4 is supposed to be 20cm not 10cm? Hopefully that's the case. Thanks

Yo, check out videos 6 - 10 here: Fluid Mechanics: ru-vid.com/group/PLOAuB8dR35oeOIPMOBH6hjwobuIJHPKSN they will clear your doubts with some easier examples 👌👌

You always want to calculate the difference in pressure between two different points in the same fluid. Typically, with a compound manometer problem like this, you want to find the difference between the high boundary and low boundary of each fluid. When doing that, use the density of the fluid that you are considering the top and bottom boundaries of.

@@Engineer4Free thx!

You need to consider all of the liquids/gases including the Atm. Press. My technique is just look for the first end of the gas/liquid and determine if the other end is above (negative) or below (positive).

Yes, P1 would be zero if we consider atmospheric pressure to be 0, which is the norm. This example was done in the frame of absolute pressure though, not gauge pressure. Pgas in this case would be -20kPa gauge (or also you could say 20 kPa vacuum. Either way, Pgas is 81 kPa absolute, which is 20kPa less that the surrounding atmospheric pressure.

As the density of air is negligible compared to the densities of mercury and water, point 2 and point 3 would have the same pressure

@@Orleezinc ok ok, that's what I thought, thanks for the reply!

+Bayan Btoush Thanks for the comment, glad to help!

Hey, it's 81kPa absolute, or -20kPa gauge

@@Engineer4Free omg thanks for replying but how do you calculate that?

Sorry I wrote that above reply wrong. Should be "-20 kPa gauge". Atmospheric pressure is given as Patm = 101kPa. Gauge pressure is how much pressure we are above reference (atmospheric) pressure. So Patm 101kPa Absolute = 0 kPa Gauge, because it's 0 kPa different from its self. 81 is 20 less than 101, so we are 20Kpa less than the reference, or - 20kPa gauge. Sometime you refer to negative gauge pressure as vacuum. So we could say "-20 kPa gauge" or "20 kPa vacuum" or just "81kPa absolute". Watch this video too: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-oOMzRzpNBEE.html

@@Engineer4Free ahh yes okay that makes sense. Thank you!

SG is density relative to water (if liquid) or relative to atm if a gas, density of H2O at 20deg is 998, so it's 0.8*998 to give actual density.

because it is relative , Specific gravity = Specific weight of oil/ specific weight of water and since density =specific weight /gravity density then is specific gravity multiplied to the density of the water.

P1 = Patm because it is in contact with the atmosphere, and the air on the outside of the manometer in this problem has pressure = Patm, including the air that is right on the boundary (in contact with) the mercury. If you go down into the mercury, the pressure in the fluid starts increasing with Patm+ρgh. When you go around the bend and start coming back up the other side, the pressure in the mercury drops by ρgh until you reach the same depth on the left hand side (which I mention at 1:31): that point also has a pressure of Patm. Moving even higher up pas that point, pressure continues to drop at a rate of ρgh, and once we pass point #2, the ρ changes to that of oil. I think you are seeing the difference in height of each side of the mercury and thinking that somehoe the Patm is lefting the other side, but for compound manometers like this, you can't look at them and come to a conclusion like that, you need to methodically go through each fluid one at a time and calculate as I did in this video. Hope that helps!

I teach all sorts of things, glad you found your way back :). What degree are you studying? I'm planning to release several more courses, hopefully you'll be able to use them!

Engineer4Free Chemical Engineering

Good luck!

Thanks!

i got about 2429...

Thank bru 🙌🙌

We talk about pressure from the "top down" because that's where the interface with the air is, and thus a known pressure. If you inverted the manometer in this video, the mercury on the furthest right part of the manometer would just fall out, just like turning a cup upside down. That's why you always see a boundary of liquid and gas with the liquid on the bottom, otherwise it wouldn't work. It doesn't mean that the first measurement has to be "going down" though. In this video, point 2 is above point 1, and although the column of fluid initially extends downward from 1, we don't care at all how far down it goes, we just care about the relative height difference between 1 and 2, and in this case 2 is above 1. Does that clear it up?

+Raeii Lobete Hey, I made 10 fluids videos so far, you can find the whole playlist here: ru-vid.com/group/PLOAuB8dR35oeOIPMOBH6hjwobuIJHPKSN if you have time, it's worth watching all 10. If you are in a rush, the last 3 videos in the playlist are directly about manometers. Basically, if you are looking at pressure in two different heights in a uniform colunm of liquid, they will have different pressures. Lets use the example from this video, where P1 and P2 are both connected by the single column of mercury, and P1 is lower in height than P2. P1 will have a greater pressure, because there is more fluid on top of it (15 cm more of mercury to be exact). So that means P1 is greater than P2. Think of how the bottom of the ocean is under a great amount of pressure, that's because the weight of all the water above it is creating the pressure. So if P1 > P2, then that means P1 = P2 + something. Conversely, if P2 < P1, then that means P2 = P1 - something. (where "something" is a positive value). "something" is always = ρgh. So in other words, if you know point is lower than another in a continuous fluid, then the pressure at the lower point = the pressure at the higher point + ρgh. If you know the pressure at the lower point, then simply subtract ρgh to get the pressure at the higher point. It seems a little backwards because you add when you go down and subtract when you go up, but remembering the example that the bottom of the ocean is under high pressure, that can help sort out which way adds pressure and which way subtracts pressure. Hope that helps!

Thank you so much sir

And it will be 101,000 - 19865.25?

Yeah I just rounded prematurely. You should keep more sig digs in calculation, but when I make these videos sometimes it's just super time consuming to write out the full value for such big numbers. But essentially 101 kPa - 19.9 kPa = 81.1 kPa.

حج صلي على النبي ...لما تنتقل من نقطه مرتفعه الى نقطه منخفضه نضع موجب للقانون (رو *جي*اتش) اما عندما ننتقل من نقطه منخفضه الى نقطه اعلى نضع سالب ل(رو*جي*اتش) 1

بالمثال نجد ان بي 1 اقل نخفاضا من بي 2 وهذا يعني ان بي 1 ضغطها اعلى لذلك عندما نريد ايجاد بي 2 نطرح (رو*جي*اتش)من بي 1 فنجد بي 2 والتي ضغطها اقل