Thermodynamics - Test 1 Problem 1 - Multifluid manometer

Change in pressure with fluid depth.
Absolute vs. gage pressure Like and subscribe! And get the notes here:
Thermodynamics: drive.google.c...
Mechanics of Materials: drive.google.c...
Statics: drive.google.c...
Dynamics: drive.google.c...

[@darrenquaye7197]

This is hands-down the best video I have watched on manometer problems

[@zaraakkuly1861]

Wow, thank you so much! You have no idea how much of a life saver you are. I couldn't figure out why you sometimes minus and other times add pgh until I came across your video. You saved me hours of grief.

[@engineeringdeciphered]

Thanks for the kind words! Good luck!!

[@diskar-Tip3553]

Thank you your lesson are much easier to understand, your slow talking makes easily process the information

[@PonquecitoMortal]

Thank you!, I am so happy to find videos like this: completly clearly and don´t need to go around and around the problem to solve and to understand what to do. ^--^

[@engineeringdeciphered]

Thanks for the kind words! Good luck!!

[@MrAnon73]

Also thank you from NZ, I spent all day trying to get my head around this reading from a text book, fluid start to fluid end, end up=Pa(- )/end down=Pa(+) very helpful, I just could see it, can now thanks.

[@jakobrassi9816]

Hey thanks for this video, it is very consistent and affective. It's better than other videos. The last video I watched had 100k+ views or something but had a contradiction within it which made it confusing. I wish I came across this video first hopefully you got more views, I appreciate the content thank you.

[@H-cranky]

I have a test tomorrow and this is the exact question i have in my workbook but couldnt solve it. thank you man you came in clutch

[@buragayedrians.7233]

Thank you for this well explained steps in solving pressure. You helped a lot. Keep it up, sir.

[@shashikantdas6983]

Thanks sir. very nice concept, from back to first. I was doing always first to last. back to first is easy to understand in some problem

[@aqanni]

Thank you you are already a legend ❤️. Keep it up.

[@syedmuhammadqasim1494]

Thanks a lot for the clear and precise explanation!!!

[@lifefan1]

Your students are lucky 😪. Thank you for the video!

[@Darwin-bg1pm]

I just checked out your thermo google file and its pretty good. hopefully you put some more for thermo 2 (ch 9 and up). I also dont know how this problem is thermo, this is more like fluid mechanics problems xD but cool thanks!

[@engineeringdeciphered]

darwin ! I don’t have any Thermo 2 videos :( I’ve never taught that class.

[@MutammimChowdhury]

Thank you, I finally understand this topic!!!

[@manaralhamoud1941]

Thank you so so so much you made this SUPER EASY

[@kareliasilvestrini6671]

Loved this! I was having a major struggle with this guy!
By any chance, could you share the unit’s conversion sheet?

[@engineeringdeciphered]

drive.google.com/file/d/1Wa5SS1EmwBGj05YlKhfMdUNYWdbwGfYN/view?usp=sharing

[@engineeringdeciphered]

this is in the front (or back?) of the Cengel and Boles textbook.

[@kareliasilvestrini6671]

@@engineeringdeciphered I see... But thank you so much!

[@roycechisala8444]

You're the best 👏🏾👏🏾

[@aniruthdinesh9156]

Alas' the saviour shows up making the fog of the minus and plus lift up.

[@muhdafif8329]

Thanks bro helped me a lot

[@tisamusonda8799]

Thank you so much.. nicely explained

[@jim2395]

why is going up negative and going down positive? is there an explanation for this? great video btw!

[@sunmichoi6888]

Thank you so much!

[@jonnym610]

You are a god

[@kafee_Alqahtani]

thank you so much

[@Lovee_MRK]

Sir for my problem, at the end of the question they stated that the local atmospheric pressure to be 95.5 kPa. They still asked to determine the gage pressure though. So is this value just meant to throw the students off or am I to consider P2 as 95.5 kPa?
Edit: This was a past assignment and just stumbled upon it while reviewing. I do not know if I actually got it right as I considered P2 as 95.5 kPa.

[@engineeringdeciphered]

You were right to begin with. You don’t need that value to calculate the gage pressure. You would need it to calculate the absolute pressure.

[@Lovee_MRK]

@@engineeringdeciphered Thank you sir

[@TopNotchVideosYT]

thank you

[@HABF4]

thank you so much

[@fraciashiebadraper1967]

thank you

[@kritiko1795]

thanks bro!

[@fareskamalfares6076]

Omg you 😍 amazing 👏

[@OmarKhaled-mg2ok]

i have a question : to use or apply the rule that we all know that any point at the same horizontal plan has the same pressure ;; is the point has to be in the same material or liquid?? or it could be different???

[@engineeringdeciphered]

Only for the same liquid!

[@calozanhielchristianssaliv4537]

Thank you

[@adlymuhd4590]

thank you sir

[@kylecatman7738]

It's opened to the ATM, you can't discount it in this case.

[@engineeringdeciphered]

If we were calculating absolute pressure, you would be correct. But in this problem we are calculating gage pressure, so the atmospheric pressure is zero (gage).

[@leeroymajoni3451]

Can you please explain why you didn't subtract P(atm) from each term as you did for P2.

[@ninyomangitamurti3408]

Thanks for the explanation but may i know from which book you got this question?

[@engineeringdeciphered]

Don’t remember, but probably Cengel and Boles’ Thermodynamics textbook.

[@ninyomangitamurti3408]

@@engineeringdeciphered Okay, thank you very much. Sorry but may i request a conversion factor file?

[@engineeringdeciphered]

@@ninyomangitamurti3408 I’m having trouble copying and pasting from my phone. But it’s in a different comment on this video. Hopefully you can see it.

[@ninyomangitamurti3408]

@@engineeringdeciphered I found it in the comments. Thank you very much

[@Ghostahmad]

Isn't P(gage) =P(abs) - P(atm)?
Why p(atm)=0

[@diplomchee]

Thanks brother

[@ashokch1246]

Thank you sir

[@iantheodorejesalva8170]

sir is it still correct if i got the same answer but it is negative? I started finding out the gage in point 1.

[@jeffersonumara2371]

just before the test i came up to this, lucky me

[@engineeringdeciphered]

Good luck!!

[@jeffersonumara2371]

@@engineeringdeciphered thanks

[@danzzz3869]

Hi thanks for the tutorial. Can I have a copy of that conversion and derived units you have pleaseee. 😁

[@engineeringdeciphered]

It’s linked in another comment on here if you can see it

[@tac-anchristiang.5305]

good day, i just want to ask if can we have a copy of the pdf conversion file of yours? thank you

[@ahmedroshdy7270]

bro i need this conversion sheet so bad pls give it to me + why gage = zero

[@Shampoo2004]

I have the same exact equation in my lectures

[@elmerarts9124]

What book is this from sir?

[@emti69]

sir can you me the link of the conversion pdf

[@Darwin-bg1pm]

My problem specified that P2=Patm so Im good. but what if it didnt specify or didnt put gauge pressure instead, it says "determine air pressure" instead of gauge pressure, would P2 = Patm ?

[@engineeringdeciphered]

Correct, if it is open, the pressure at that open area would be the atmospheric pressure.

[@megathadif6892]

but in our version it gave us the atm pressure value. do we need to take into account?

[@engineeringdeciphered]

Yes, if it is asking for the absolute pressure. You would have that value where I have the 0. But if it is asking for gage pressure, you wouldn’t use that, even if it is given.

[@omarjallad2001]

This is not Thermo dynamic , it's fluids mechanic !!

[@lifefan1]

Yeah, but I learn this problem in Biothermodynamics course.

[@omarjallad2001]

@@lifefan1 ok .

[@naeem6583]

Wouldnt the water be 'down' too?

[@engineeringdeciphered]

No, I’m starting at P2 and the going to the left. If you were to start at P1 and go to the right then yes, water would be down.

[@naeem6583]

@@engineeringdeciphered I rewatch the vid and now I understand. I thought that you're referring to the position of the oil column when you say it's up but actually it's the h and where it's headed

[@loudnotes8409]

Patm= 101Kpa

[@raileehhh1039]

hey give me unit conversion

[@nnadonador3038]

Can u add arabic translation 🙂