Thermodynamics - Pressure example 2 manometer

[@i3enjaminmejia66]

this video just proves how absurdly confusing the imperial system is

[@5thHeelToe2nd]

We really appreciate you, I am getting ready for a test and your videos have helped immensely. Thank you for you devotion to teaching / instructing. Subscribed and will follow closely.

[@kylecatman7738]

Where does the gc = 32.2 get stickier?? I can't remember off hand where it'll mess you up most? Because it always cancel out g... but maybe in the B.E. somewhere, idk.

[@shanebriangalgo8147]

What if you should've convert it inches to meters? and instead lbm/in^3, you could've used the kg/m^3. Much easier than the way you solve, however different answer I will got.

[@kennydalid9438]

can you post a downloadable soft copy of this convertion table

[@engineeringdeciphered]

www.dropbox.com/s/gb23smh0c5glzyc/Conversion%20factors.pdf?dl=0

[@kamranhafeez4269]

@@engineeringdeciphered Thank you so much

[@mothug978]

Hello professor, awesome lessons, clear explanations. One question: since you add as you go down, but if you were to go up (from left to right), would you put a negative sign instead? Thank you!

[@engineeringdeciphered]

Yes, Anytime you go down - add. anytime you move up - subtract.

[@alvinlin8140]

@@engineeringdecipheredbut you went up when you went from the water to the Mercury so shouldn’t you be subtracting in that part? I can elaborate more if you want

[@engineeringdeciphered]

@@alvinlin8140 I think you’re talking about the part where I was going through the air. In that part, yes, going through the air you would technically subtract the rho g h of air, but we can neglect any gas because the rho of gas is much much smaller than the rho of the liquids. So I just skipped subtracting the rho g h of air. (you could do it, but it would be a very small number)

[@alvinlin8140]

@@engineeringdecipheredok gotcha thanks, my prof hasn’t gone over examples so this is helpful! One more thing, the pressure of a gas is negligible but in the question we are finding the pressure of the natural gas, can you explain that?

[@engineeringdeciphered]

@@alvinlin8140 well, it’s the change in pressure (due to height change of a gas) that is negligible.

[@jooniversalstudio5757]

I've calculate it just like in the video but I ignore the 14.2 lbf/in^2 and then I got 126.332288 is this KPa? because if it's KPa then I add the 14.2 lbf/in^2 I got 140.5332288 KPa. And then, if I convert 140.5332288 into Psia, I got 20.38 Psia not 18.1 Psia. Did I might be missed something?

[@Parekotv-s8o]

Absolute pressure is gauge pressure + atmospheric pressure right??

[@engineeringdeciphered]

Yes!

[@nehemiascaban2461]

What app do u use on tis video? Btw, nice problem!!!

[@engineeringdeciphered]

NotesPlus

[@deiji6893]

if i started from the natural gas i will subtract the pressure of both water and mercury, right? what about patm? should i add it or subtract?

[@engineeringdeciphered]

If you started from the natural gas, you would subtract both water and mercury and then you would “equal” it to Patm. The Patm would be on the right side of the equation

[@futurefundedd]

Watching just after paper and same problem came and paper warr gaya

[@azam8031]

Thank you very much 🙌

[@PrincyGaming14]

how do you put in the calculator, mind was to far to your answer but I put the same input? thank you

[@tonyatilano7057]

let me know if you find out have my exam tomorrow

[@choppsticcs]

at 4:40 why did you use 32.2 ft/s^2 instead of 9.81m/s^2

[@audufaisal728]

its not in S.I units

[@willowdesk]

just a thought, but the question is asking about pressure IN the pipeline right? not on the natural gas itself? so why wouldn't you also subtract the natural gas in the tube touching the mercury?

[@angrycyclone4353]

Because its a gas and gases have such a low density and are negligible

[@beshoymtsu2143]

Thank you!

[@carlosalzhrani4564]

What if we replace air with water what will be the height of it.

[@engineeringdeciphered]

You see in the problem that I neglected the part with air because it’s density is so small (compared to the others). If the air was water, then you would do the (rho)(g)(h) of that column as well, instead of neglecting it.

[@carlosalzhrani4564]

@@engineeringdeciphered I got the part but what I ment is that what will be the height if we replace air with oli for example.

[@engineeringdeciphered]

So for that type of problem you would have to be given the pressure of the natural gas. If you know that pressure and you don’t know the height of the right column - it’s still the same process and similar equation. Now h is the only unknown in the equation and you can solve for it.

[@carlosalzhrani4564]

@@engineeringdeciphered Thank you 👍

[@Miistik]

how do you solve it using ft instead of in?

[@alexisperry6010]

I’d convert it to inches

[@rizzmechanics]

Thank you

[@عليالعراقي-خ9ظ4ه]

Can you send me conversion factors please 🙏 I need him

[@engineeringdeciphered]

www.dropbox.com/s/gb23smh0c5glzyc/Conversion%20factors.pdf?dl=0

[@عليالعراقي-خ9ظ4ه]

@@engineeringdeciphered thanks 👍

[@gump7237]

letsss go