Just wondering, your solution at 3:52, would that Pgas be absolute pressure because it's the total pressure, Patm + Pgas(gauge)? Would Pgas(gauge) be the difference between the two, ie. 39.877 kPa?
Change in diameter should have no effect so long as the diameter isn't so small that capillary action is in play. Otherwise, ignore and changes in diameter, and only focus on the vertical change within a continuous fluid column.
You pretty much need Rho. If you're not given that value, you will need to be given specific gravity or something instead that you can then convert into Rho. Take a look at my other fluids videos for talk about SG and stuff: ua-cam.com/play/PLOAuB8dR35oeOIPMOBH6hjwobuIJHPKSN.html
Pa = N/(m^2), but 1N = (1kg)*(1m/(s^2)) so sub in that for the N in the first expression to get Pa = [(1kg)*(1m/(s^2))]/(m^2) and rearrange to get Pa = 1kg/(m*(s^2)). Inspecting the term ρgh = (13,500kg/m^3)(9.81m/s^2)(0.3m) ----> the units reduce kg/(m(s^2)) which is Pa. Both terms on the right hand side are in units of Pa, so we can add them, and get the answer in Pa.
It's the density of Mercury at 1 atm pressure and 20 degrees celsius. It's a commonly used number that's often given in the problem or supplementary tables. Check out videos 1 and 2 here: ua-cam.com/play/PLOAuB8dR35oeOIPMOBH6hjwobuIJHPKSN.html
Pgas = absolute pressure of the gas. Patm = absolute pressure of the atmosphere. ρ = rho = density of mercury at 20 degrees celsius. g = acceleration due to gravity. h = change in height of fluid column. I really recommend taking about an hour and the first 8 videos that lead up to this one in the playlist here: ua-cam.com/play/PLOAuB8dR35oeOIPMOBH6hjwobuIJHPKSN.html 13,550 is a table value for the density of mercury at 20 degrees celsius which can be found in any density table in a fluids or thermo textbook, I use it several times in the videos leading up to this one so I don't always repeat where everything comes from in every single video.
Thank you!! I been putting off my fluids hw since i didn't understand this till now. Thank you!!
Just wondering, your solution at 3:52, would that Pgas be absolute pressure because it's the total pressure, Patm + Pgas(gauge)? Would Pgas(gauge) be the difference between the two, ie. 39.877 kPa?
really helped me with my assignment thanks man
Thanks for the upload! Do you think you could do video to show how to tackle manometer problems with a change in diameter?
Change in diameter should have no effect so long as the diameter isn't so small that capillary action is in play. Otherwise, ignore and changes in diameter, and only focus on the vertical change within a continuous fluid column.
@@Engineer4Free thank you :)
Thanks for the upload. :)
Super helpful, thank you good sir.
Thank you!
Pressure on right side of above mercury column will be 0 as we are measuring gauge pressure.
Yes, in both situations we would take Patm = 0 for our reference of gauge pressure.
Nice vid!
Is there a method of solving this without using Rho ?
You pretty much need Rho. If you're not given that value, you will need to be given specific gravity or something instead that you can then convert into Rho. Take a look at my other fluids videos for talk about SG and stuff: ua-cam.com/play/PLOAuB8dR35oeOIPMOBH6hjwobuIJHPKSN.html
Wait, very quickly, are we supposed to memorize rho of things?
21Greatest basic ones yea
Where'd the 9.8 come from
The value of g (gravity) is always 9.81 on planet Earth. It will only change if the given problem is related to space or other planets
thank you so much ive learn alot
Glad to hear it!
Dear, units are not matching P gas= Patm (Kpa ) +hdg( pa) , how come to add or subtract ?????
Pa = N/(m^2), but 1N = (1kg)*(1m/(s^2)) so sub in that for the N in the first expression to get Pa = [(1kg)*(1m/(s^2))]/(m^2) and rearrange to get Pa = 1kg/(m*(s^2)). Inspecting the term ρgh = (13,500kg/m^3)(9.81m/s^2)(0.3m) ----> the units reduce kg/(m(s^2)) which is Pa. Both terms on the right hand side are in units of Pa, so we can add them, and get the answer in Pa.
on the left side. u write, Pgas= Patm + Pmercury. why u didnt minus the Pgas? notice me pls i have final exam tomorrow
To get from P_atm to P_gas, you must go lower into the liquid. The lower you go, the higher the pressure. Therefore, P_gas = P_atm + ρgh
where is that 13,500 coming from??
It's the density of Mercury at 1 atm pressure and 20 degrees celsius. It's a commonly used number that's often given in the problem or supplementary tables. Check out videos 1 and 2 here: ua-cam.com/play/PLOAuB8dR35oeOIPMOBH6hjwobuIJHPKSN.html
Someone explain what all the variables in this equation are
and where tf did 13,550 come from
Pgas = absolute pressure of the gas. Patm = absolute pressure of the atmosphere. ρ = rho = density of mercury at 20 degrees celsius. g = acceleration due to gravity. h = change in height of fluid column. I really recommend taking about an hour and the first 8 videos that lead up to this one in the playlist here: ua-cam.com/play/PLOAuB8dR35oeOIPMOBH6hjwobuIJHPKSN.html 13,550 is a table value for the density of mercury at 20 degrees celsius which can be found in any density table in a fluids or thermo textbook, I use it several times in the videos leading up to this one so I don't always repeat where everything comes from in every single video.
What does g stand for? Speed?
+Nadine Joseph gravity
+Tristan Smith Yeah, acceleration due to gravity. Hence it is 9.81 m/s^2. If I was using imperial units then g would have been 32.2 ft/s^2!
Nadine Joseph gravity
wtf! he did it wrong
nice bait tryhard
cool
😊
Lora land