Dr. Morrison can you please do a mor complicated manometer where there's more than 2 arms and more than 2 liquids with different density an with both ends of the manometer closed. Thank you
For those of us who are familiar with tranposition and the fact that rho is the density of the substance, i.e. water = 1000, I dont think it was clearly explained what rho meant in your video? Presumably a certain amount of this you are expecting your viewers to understand, for instance g - gravity - 9.81 m/s^2 ?
thanks a bunch now i can go make my own manometer , before this though my calculations would be off by Pa = 1 atmosphere.No one could teach me how to tie a tie ,that is untell lady did.
thanks for the video lesson, very easy to understand. Fluids seems like the hardest course for me this semester. not looking forward to turbomachinery.
If I understand your question correctly, you have the b-side open to air (and the pressure at the top is Patm) and the a-side open to some other pressure Pa. If you do it my way, you get rho*g*h+Patm=Pa. In this case Pa is the absolute pressure. If you do it your tutor's way and say the pressure at the top is zero (gauge pressure), then you get rho*g*h=Pa but now Pa is in gauge pressure. In typical gauges, they read zero when open to atmosphere.
Hi.This will really help me in my test which is coming up in few days...but i have a question dat really confuses me
ok so we have P1=P2 wher the top pressure from each side is Pa and Pb respectivly.what if we have one side open to air.lets say Pb end is open to atmospher.therfore Pb wil be equal to atmospheric pressure right? But why does my tutor never take that atmsperic presure in consideration wen doing the calculation.he's P2 in this case wud simply be rho(of fluid g)*gravity*h1.how come?
Its a good basic explanation.. but if the manometer problem is really complicated.
Where you have multiple (Jump Across Points) P1 = P2
P3=P4 , P5=P6 , and the some "Jump Across Points" are on the top so the "column" is actually inverted.
In the end if you want the pressure difference between the entry and the exit.. how would you add them together using your method. I can't figure it out. It gets messy
Hi, for the complex case you cite, you need to choose "jump across points" that are in the same fluid and not at the top, but at the bottom. Always remember that the pressure is the same in the same continuous fluid at the same elevation. I will try to post a video on this. Dr. Morrison
Greetings Dr Morrison, I was just wondering if you made a tiny mistake or perhaps I'm wrong. Starting at 3:00 of the video, you say that the pressure at the bottom ( P1) is (PA)(rho)(g)(h1+h2), shouldn't it be PA + (rho)(g)(h1+h2)
This shows how they derive all the formulas for these kinds of problems, awesome. Too bad my teacher never showed this, or never explained these connnections to the formulae. This was takes longer but it makes a lot more sense to me then just random formulas.
Oh my goodness...You're left handed!
moviericky 1 week ago
shouldn't it be volume, not height?
frahs993 2 weeks ago
Dr. Morrison can you please do a mor complicated manometer where there's more than 2 arms and more than 2 liquids with different density an with both ends of the manometer closed. Thank you
itsd4real 4 months ago
good vid! the guitar was WAY too overbearing for my ears.
MattSpon 4 months ago
@MattSpon Haha! My bad... that was king crimson playing on my itunes! lol wow and I'm in the health sciences... look out.
MattSpon 4 months ago
so basically PA-PB=ρg*g*h1?
jimmertzis 7 months ago
This has been flagged as spam show
I believe that can be simplified to P1-P2=(rho)*g*h, or finding the absolute pressure applied at P1, P1=P2+(rho)*g*h
jbulldrummer 8 months ago
Comment removed
jbulldrummer 8 months ago
For those of us who are familiar with tranposition and the fact that rho is the density of the substance, i.e. water = 1000, I dont think it was clearly explained what rho meant in your video? Presumably a certain amount of this you are expecting your viewers to understand, for instance g - gravity - 9.81 m/s^2 ?
NuclearDispute 8 months ago
Brilliant but very dull. Life is short, go get a beer.
phoebelala 9 months ago
thankkkss a lot now i can sit my fluids exam!
zakariyya90 9 months ago
AWESOME.....
THANKS A LOT MA'M....U MADE IT SO SIMPLE FOR ME TO UNDERSTAND THE THING.... THANKS ALOT..
anks181190 9 months ago
Thank you so much for explaining the pitot static tube problem! Awesome!
jannarodriguez95348 11 months ago
thanks a bunch now i can go make my own manometer , before this though my calculations would be off by Pa = 1 atmosphere.No one could teach me how to tie a tie ,that is untell lady did.
LINEHACKER 1 year ago
Thank you
kungman190 1 year ago
Comment removed
kungman190 1 year ago
I love clever ladies, I bet your a really good looking lass
vibeflyer 1 year ago
Thanks a lot ! :)
KrishanBorwal 1 year ago
thanx. now i can get my chemical engineering degree. very helpful vid =)
strongwiller 1 year ago 5
Thank you!! That really helped!! Your a life saver!!
awana79 1 year ago
this stuff doesnt make any senseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee help meeeeeeeeeeeeee
NMLP1 1 year ago
Hello from Sydney!
thanks for the video lesson, very easy to understand. Fluids seems like the hardest course for me this semester. not looking forward to turbomachinery.
thanks for the video =D
Bahbohram 1 year ago
thanks.your explanation really helped me.
chellyash 1 year ago
that did not help me what-so-ever. the one day i don't pay attention in class we actually learn something i didn't already know how to do.
oh well, thanks for posting anyways.
the method in our bok is completely different (ideal gasses).
Sharod101 1 year ago
If I understand your question correctly, you have the b-side open to air (and the pressure at the top is Patm) and the a-side open to some other pressure Pa. If you do it my way, you get rho*g*h+Patm=Pa. In this case Pa is the absolute pressure. If you do it your tutor's way and say the pressure at the top is zero (gauge pressure), then you get rho*g*h=Pa but now Pa is in gauge pressure. In typical gauges, they read zero when open to atmosphere.
DrMorrisonMTU 1 year ago
Hi.This will really help me in my test which is coming up in few days...but i have a question dat really confuses me
ok so we have P1=P2 wher the top pressure from each side is Pa and Pb respectivly.what if we have one side open to air.lets say Pb end is open to atmospher.therfore Pb wil be equal to atmospheric pressure right? But why does my tutor never take that atmsperic presure in consideration wen doing the calculation.he's P2 in this case wud simply be rho(of fluid g)*gravity*h1.how come?
rallyx8 1 year ago
Its a good basic explanation.. but if the manometer problem is really complicated.
Where you have multiple (Jump Across Points) P1 = P2
P3=P4 , P5=P6 , and the some "Jump Across Points" are on the top so the "column" is actually inverted.
In the end if you want the pressure difference between the entry and the exit.. how would you add them together using your method. I can't figure it out. It gets messy
divinenuker 1 year ago
Hi, for the complex case you cite, you need to choose "jump across points" that are in the same fluid and not at the top, but at the bottom. Always remember that the pressure is the same in the same continuous fluid at the same elevation. I will try to post a video on this. Dr. Morrison
DrMorrisonMTU 1 year ago 2
Greetings Dr Morrison, I was just wondering if you made a tiny mistake or perhaps I'm wrong. Starting at 3:00 of the video, you say that the pressure at the bottom ( P1) is (PA)(rho)(g)(h1+h2), shouldn't it be PA + (rho)(g)(h1+h2)
divinenuker 1 year ago
Oh never mind, i shouldve watched the video further you corrected yourself. Sorry!
divinenuker 1 year ago
Great explanation.
This shows how they derive all the formulas for these kinds of problems, awesome. Too bad my teacher never showed this, or never explained these connnections to the formulae. This was takes longer but it makes a lot more sense to me then just random formulas.
luthmhor 2 years ago
thnx Dr.faith U R the best
mmaaeai 2 years ago
did u invent thermodynamics
TheSurya786 2 years ago
my fluids professor could take a lesson from you b/c that 5 min video made more sense than most of my semester. Good job.
c10seth 2 years ago 13
@c10seth two days before my first fluids test and those are my exact thoughts!!!!!!
Dinty51 3 months ago
thats what I was going to say!
phoebelala 2 years ago