Compound manometer example problem

[@harvinderpal4662]

7 years on and this video hasn’t lost its relevance...very well explained, thank you so much👍

[@robbiegolds1234]

Don't apologize for redundancy. This was perfect. Perfect reps = Perfect steps.

[@Engineer4Free]

Thanks Robb =)

[@EE-pq8yn]

I just waned to take a minute to thank you for teaching me your way.same exact problem i got on my test and your explanition was golden! fluid mechanics could be a big tricky game but you made it easy to digest! Please make more videos on more topics in fluid mechanics. Thanks!

[@Engineer4Free]

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

[@a.h8606]

This concept is the one thing I fail to remember everytime i prep for exams and each time this video delivers. Thanks man.

[@Akki420ish]

This video is gold. Cleared my concept after 4 years. Thanks a ton! One thing l'd like to ask: The pressure you took at P1 = P(atm) is a gauge pressure or absolute pressure? I have seen in some books that P(atm) is taken zero sometimes. Can you explain?

[@fareedhaddad8240]

abs .. i think coz its open

[@Engineer4Free]

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: kzbin.info/www/bejne/pYCwq4WwpbN1e6c

[@wafa98_67]

from what i have learnt , it is absolute

[@lisagarrett3069]

Thank you for the video! No need to say sorry for the repetition. I actually didn't catch the reasoning the first time, so the second time made it click and the third time solidified the knowledge. :) Great video!

[@Engineer4Free]

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

[@shensley011]

Always be redundant! Thanks for the help (again) you're a true gem. Question, what is it wasn't open to the atmosphere and was just a fluid?

[@humaidalkhateri3095]

I HAVE BEEN IN REAL STRUGGLE with the plus or mins sign . I have a midterm tomorrow in fluid mech. and this's been a great help!!

[@Engineer4Free]

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

[@darkforestf]

Thank you so much! This vid has divided all parts of the eqn for a better understanding unlike the textbook always come up with a compound eqn

[@Engineer4Free]

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! =)

[@exogendesign4582]

this helps me a lot, I was trying to figure it out on my own, and had the signs flip out coz I didn't consider the action of the fluid after watching this video it cleared me up, thanks for this man.

[@Engineer4Free]

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

[@nchls5730]

Thank you so much for this video. This helps me prepare my fluid mechanics exam. Subscribed! :)

[@Freakybananayo]

great video. i have a question though, see how the gas inside the bulb is higher up than point 4 wouldn't there be a difference in pressure? thanks.

[@adaugoukaegbu4934]

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

[@Engineer4Free]

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]

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.

[@Freakybananayo]

@@Engineer4Free good to know. Cheers

[@wafa98_67]

thank you so much! i watched many vids and none could help! but u explained it so simply! =)

[@Engineer4Free]

Thanks for the feedback!

[@junior214223]

Do you have any videos dealing with pumps and the pressure at those pumps?

[@Engineer4Free]

Hey sorry not yet, but hopefully one day!

[@tombuxton6774]

This man is singlehandedly saving my degree at the moment.... I could kiss you

[@Engineer4Free]

I gochu 🤜🤛

[@comfortmankge7035]

THANKS SIR IT WAS REALLY HELPFUL.......

[@kiriyamarei7358]

You are a life saver

[@Engineer4Free]

=) =) glad I can help

[@oharahetuka3252]

Thank you Sir. Your explanation on why we should add or subtract "ρgh" helped me. I always stumbled when figuring this out. Thanks a lot again!

[@Engineer4Free]

Awesome! Glad to hear it :D

[@nicoleng4181]

is the final answer gauge pressure or absolute pressure ?

[@Engineer4Free]

The 81 kPa is absolute pressure

[@andyzh2950]

I'm SO GRATEFUL for your explanation, I finally understand when to use + or - in that equation, subscribed.

[@Engineer4Free]

Yasssss mission accomplished 🤜🤛

[@RikkuCloud]

Thank you sir, had some problem getting over the "unsidedown oil" part in a exercise problem I've been having, but when you said "as you go down deeper into the fluid P3 is going to be greater than P2" then my mind finally got it hahaa, thanks.

[@Engineer4Free]

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

[@omarahmed621]

What a legend, Thank you so much

[@Engineer4Free]

Thanks!!

[@danielgross8948]

Great video man!

[@Engineer4Free]

Thanks!! 😊

[@deathcomplex-t2z]

this was very useful thank you!!

[@Engineer4Free]

Glad it was helpful, thanks for watching! 😁

[@naq8405]

Hi, I'm confused as to which liquid density im supposed to choose everytime i calculate the pressure.

[@Engineer4Free]

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.

[@naq8405]

@@Engineer4Free thx!

[@drcommondrate12]

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).

[@qsdfcvgyjmkl]

Thank you for this video. It really made the explanation much easier to understand! Please continue your wonderful work!

[@sphanie]

Hi, is the pressure inside the tank the gauge pressure or the absolute pressure? I really hope you answer this🥺

[@Engineer4Free]

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

[@sphanie]

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

[@Engineer4Free]

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: kzbin.info/www/bejne/pYCwq4WwpbN1e6c

[@sphanie]

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

[@Gartaxo42]

helped me out fr fr. bless you

[@sreejithp2740]

fantastic work...cleared the concept..tnku so much

[@Engineer4Free]

Glad to hear it :)

[@hijxckk]

Hi I was just curious how pressure would be impacted if we had air in the tube in-between p2 and p3 instead of another fluid

[@Orleezinc]

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

[@hijxckk]

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

[@jallmo]

Thank you so much you saved me (:

[@ashencool8495]

very easy to understand thnk u

[@redalert1730]

thank you so much... i finally understand the measurement and depth in pressure.. hoping you will upload hydrostatic ,boyles law, buoyancy. orifice, and bernoullis theorem thank you soo muchhh

[@Engineer4Free]

I will get to them one day!

[@found2273]

Thanks sir / bro from India

[@piratenu1]

thanks a lot dude. Today Im taking an exam and was confused with a question of this type. Your video was much helpful to solve it. :)

[@Lykenyte]

Hello, I would just like to know if the final answer of the gas were in gage pressure, would the answer be P4 - Patm? Thank you!

[@Engineer4Free]

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

[@iValkyrie46]

thanks sir. this really helps me

[@Engineer4Free]

Great, glad to hear it!!

[@jacobstahlmann6958]

How can the two P1s have the same pressure if they have different fluids above them

[@PoproqzSwag]

Thank you. I was struggling with simple concepts and my teacher would not explain it thoroughly.

[@Engineer4Free]

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: kzbin.info/aero/PLOAuB8dR35oeOIPMOBH6hjwobuIJHPKSN

[@josiah8006]

Hiii i have a question, if the pressure at a certain point is lower than the current then does it have to be positive? I mean add

[@Engineer4Free]

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

[@dfunctual5]

Thanks for sharing. Shouldn't the equation to find P4 incorporate a height of 0.1m instead of 0.2m? As it's drawn 0.2m is the distance from point 3 to 4.

[@Engineer4Free]

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!

[@dfunctual5]

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

[@elhadeth_DZ]

how you convert .....pgh of mercury to pas ??????

[@teshd7108]

Very clear video, thanks!

[@cqaaliqcismaancllahi1487]

Thank you

[@Engineer4Free]

You're welcome!

[@AnilKumar-yv4iu]

I think P1=0 as gauge pressure in atmosphere is zero. In the manometer we are actually measuring gauge pressure.

[@Engineer4Free]

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.

[@narunstorm]

❤thank you❤

[@tbarak.shakir.b5003]

You r amazing thank you so much 🤩❤❤

[@Engineer4Free]

You’re welcome 😊!!

[@angelamargaritagallegomedr4274]

so helpful, thank you!

[@Engineer4Free]

Glad it was helpful!!

[@forrrest6609]

​ @Engineer4Free Help! How is p3 less than p1?? I'm really confused?

[@Engineer4Free]

Yo, check out videos 6 - 10 here: Fluid Mechanics: kzbin.info/aero/PLOAuB8dR35oeOIPMOBH6hjwobuIJHPKSN they will clear your doubts with some easier examples 👌👌

[@gelomelo3426]

Thank you so much🌹🌹🌹❤❤❤

[@muhittinselcukgoksu1327]

Thank you soooo much .Your explanations and drawings were wonderful, regards.

[@Engineer4Free]

Thanks for the kind words!

[@usmanshahid1913]

Thanks man for the video!

[@Engineer4Free]

Hey you're welcome Usman!

[@Toptop06]

Very helpful! Thank you ❤️

[@Engineer4Free]

You're welcome! =)

[@anilsharma-ev2my]

Work done by an aneroid barometer in joules. ? Please show actual data in energy terms ?

[@rko9887]

Thanks!!!! Really helped a lot,... Although u need to chk ur calculation otherwise ok....

[@Engineer4Free]

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!

[@diegofung95]

Thanks for posting this!

[@kustupsk1]

Thank you for your hard work!

[@Engineer4Free]

Your welcome :) thanks for watching!

[@alexwalton2132]

thanks sir, very helpful, much love

[@Engineer4Free]

Thanks for watching 🙂

[@popedope4842]

Great work man!

[@Engineer4Free]

Hey thanks! Appreciate the feedback :)

[@yiiyatschan1395]

Thank you for this video.

[@CBFD667]

awesome, definitely helped me out

[@jameslondon05]

13,580 * 9.81 * 0.15 =/= 19995 as you suggest in your first calculation. it is 19982.97, thus your final answer is a little off. great work through and still useful.

[@alhassanjd6670]

Thank you so much ..

[@Engineer4Free]

You're welcome Hassan!!

[@zaazazza6555]

Thanks!

[@Engineer4Free]

You're welcome!

[@worldpark90]

Many Thanks !!!

[@Engineer4Free]

You're welcome =) =)

[@madhukumars458]

thank you sir

[@Engineer4Free]

You're welcome Madhu =)

[@Aprst25]

Why is specific gravity of oil being multiplied with the density of water

[@jameslondon05]

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.

[@exogendesign4582]

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.

[@knowledgezone8748]

great work

[@Engineer4Free]

Thanks!

[@lyrauken]

Thanks man.

[@Engineer4Free]

You're welcome Berkay

[@montana4able]

great video .. great man .. thx!!

[@namnguy5024]

very helpful. thank you

[@Engineer4Free]

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

[@shamitfatin3516]

Thanks!!!!!!

[@Engineer4Free]

You're welcome!!!

[@Diya.Dasari]

THANK YOU......

[@GoatedMofo]

THANK YOUUU

[@Engineer4Free]

YOU'RE WELCOMEEE

[@lameesk.a162]

You are the best, thank you verrrrrry much, you've saved me💜💛😘😘

[@PoproqzSwag]

I also need help edith a differential manometer with 2 fluids. I'm going to check your videos for that.

[@lilonarry9693]

THANK YOU THANK YOUUU THANK YOUU ALL THE LOVE XXXXXX

[@Engineer4Free]

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

[@abdullahhouran96]

God bless this legend

[@Engineer4Free]

Thank bru 🙌🙌

[@mro2038]

Thanks alot

[@Engineer4Free]

Happy to help! :)

[@helper2004]

Thanks

[@Engineer4Free]

Welcome ☺️

[@YazzaY1]

THANK YOU!

[@gelomelo3426]

Great👏👏👏👏👏

[@bhgam3r769]

Thanks 🌹🌹

[@beboshi69]

Huh, I was doing your C++ tutorials and now I started my degree I'm back here

[@Engineer4Free]

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!

[@beboshi69]

Engineer4Free Chemical Engineering

[@Engineer4Free]

Good luck!

[@beboshi69]

Thanks!

[@asjalahmad4632]

Thanks a great deal

[@yugs6970]

Tanx a lot

[@Engineer4Free]

Tanx for watching!

[@baderk326]

i think p1 = p(atm)+pgH cuz the pressure that applied via p(atm) is higher than the oil

[@Engineer4Free]

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!

[@vuongngo2181]

ty

[@huhhhhh4497]

Isn’t 13500*9.81*0.15 is equal 19865.25?

[@huhhhhh4497]

And it will be 101,000 - 19865.25?

[@Engineer4Free]

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.

[@bayanbtoush2786]

thaanx a million

[@Engineer4Free]

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

[@fareedhaddad8240]

thanks ♥

[@Engineer4Free]

♥

[@رعدالوائلي-ض6ب]

على أي أساس أضع سالب أو موجب. ممكن أحد يفهمني.

[@johnorrange1113]

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

[@johnorrange1113]

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

[@lukmanali6426]

is the exact answer 81380pa?

[@Cleonhide2]

i got about 2429...

[@boukhadrachakib2108]

There is only one thing that is killing my mind we always talk about the pression from the top what about the pression ftom the bottom!?? :/

[@Engineer4Free]

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?

[@raeiilobete4117]

I dunno why there is a subtraction or addition since my teacher doesn't teach that kind of method. Please help me, do you have some lesson similar to this. My teacher said that he will include a lot of this in our exam Huhuhu

[@Engineer4Free]

+Raeii Lobete Hey, I made 10 fluids videos so far, you can find the whole playlist here: kzbin.info/aero/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!

[@raeiilobete4117]

Thank you so much sir

[@francoispasmore7896]

Hey .. i think u have sumn wrong ... Monometers measure gauge pressure ... meaning at 1 atm the gauge would read '0'

[@Engineer4Free]

+Francois Pasmore Good observation. I did this problem using the absolute pressure measurements. In this case where I find the pressure of the gas to be 81 kPA absolute, that could also be expressed as 20 kPa gauge (101 kPa - 81 kPa = 20kPa). Perhaps I should have been more clear that I was using absolute pressures!

[@francoispasmore7896]

+Engineer4Free ok that's for clearing that up

[@Short-Cut-bb6ud]

My answer is coming out as 81.4Kpa 😬

[@JoshCabatuando]

THANKSSSSSSSSSSSSSSSSSSs

[@Mehdiranjb]

🌹🌹

[@jxmsxi]

❤