That is great question. So in these problems, we usually calculate for the gage pressure. The absolute pressure P = Pg (gage pressure) + Patm. If you substitute the absolute pressure equation for PA and P2 (at the open column), the Patm cancels and that is why Patm is usually considered zero when we have open column. I hope this helps and thank you for this comment, I will actually post a video clarifying this issue. Thank you for watching and good luck with your studying!
On page 102 of the FE handbook 9.5, the specific weight of Hg is 847 lbm/ft^3. Because it is lbm, I assumed I had to divide it by the gravitational constant to make lbf. Should we assume provided numbers are always lbf even if it doesn't specify? Using lbf, 2140 psf is the result which is not close to any of the answers, so in this case, I'd assume 847 is already lbf/ft^3
Unless otherwise specified, all problems take place in Earth's standard gravity, where 1 lbm is defined to weigh 1 lbf. This means the specific weight in lbf/ft^3 has the same numeric value as the density in lbm/ft^3. Where you have to work with the "gravitational constant" when using pounds, is adapting the unit of pounds to Newton's 2nd law, and anything derived from it. I use the quotes, because this is typically in situations that have nothing to do with gravity. What the gravitational constant is doing for you, is translating mass units to slugs, for consistency with force in pounds and acceleration in ft/s^2. The SI system takes the opposite approach. Units for force and mass that work elegantly in N's 2nd law by default, and a constant with physical meaning for finding weight.
Why didn't you multiply density and height by gravity? I understand you get the desired units if you don't multiply gravity. But isn't gravity affecting that point of the fluid hence Density x gravity x height.
When using specific weight (gamma), it includes both density and Earth's gravity by definition. As opposed to density (rho) which is just density, and is gravity-agnostic.
Hey students I am a civil engineering graduate from India , for those who need help with understanding their concept and help with home work can reach out to me 👍
@@franciscosanchez8572 The question isn't worded well enough, in my opinion to know for sure that they are asking you for gauge pressure. But had you calculated absolute pressure, you'd get 1102.7 psf, which isn't one of the choices. But, it is off by 14.7 psf from the choice we are given, so we can infer that it was gauge pressure they are asking for.
I love that saying! " A problem a day keeps the F away" I'm gonna use that one.
Good refresh video, I learned a lot.. thank you 🤗
Thank you for watching and please share with your friends who might find it helpful!
Excellent walkthrough of solutions...thank you
Thank you for watching Tomasi. Please share with your friends who might find it helpful 😊
Thank you for sharing. Awesome refresher
Thank you for watching 😊
thanks genie!
Ref book page 108: specific weight of water at standard conditions is 62.4 lbf/ft3. I get it
Awesome! Thank you for watching and good luck with your studying!
When exactly the Patm is considered? why Patm=2168psf is not taken in consideration for the calculations? Manuals and texts do consider Patm
That is great question. So in these problems, we usually calculate for the gage pressure. The absolute pressure P = Pg (gage pressure) + Patm. If you substitute the absolute pressure equation for PA and P2 (at the open column), the Patm cancels and that is why Patm is usually considered zero when we have open column. I hope this helps and thank you for this comment, I will actually post a video clarifying this issue. Thank you for watching and good luck with your studying!
On page 102 of the FE handbook 9.5, the specific weight of Hg is 847 lbm/ft^3. Because it is lbm, I assumed I had to divide it by the gravitational constant to make lbf. Should we assume provided numbers are always lbf even if it doesn't specify? Using lbf, 2140 psf is the result which is not close to any of the answers, so in this case, I'd assume 847 is already lbf/ft^3
Unless otherwise specified, all problems take place in Earth's standard gravity, where 1 lbm is defined to weigh 1 lbf. This means the specific weight in lbf/ft^3 has the same numeric value as the density in lbm/ft^3.
Where you have to work with the "gravitational constant" when using pounds, is adapting the unit of pounds to Newton's 2nd law, and anything derived from it. I use the quotes, because this is typically in situations that have nothing to do with gravity. What the gravitational constant is doing for you, is translating mass units to slugs, for consistency with force in pounds and acceleration in ft/s^2.
The SI system takes the opposite approach. Units for force and mass that work elegantly in N's 2nd law by default, and a constant with physical meaning for finding weight.
Thank you
how much is the value of SG?
Why didn't you multiply density and height by gravity? I understand you get the desired units if you don't multiply gravity. But isn't gravity affecting that point of the fluid hence Density x gravity x height.
When using specific weight (gamma), it includes both density and Earth's gravity by definition. As opposed to density (rho) which is just density, and is gravity-agnostic.
Why is there no atmospheric pressure, isn't it taken as 101kPa?
Because it is gage pressure.
enGENIEer what if you used absolute pressure?
Hey students
I am a civil engineering graduate from India , for those who need help with understanding their concept and help with home work can reach out to me 👍
@@franciscosanchez8572 The question isn't worded well enough, in my opinion to know for sure that they are asking you for gauge pressure. But had you calculated absolute pressure, you'd get 1102.7 psf, which isn't one of the choices. But, it is off by 14.7 psf from the choice we are given, so we can infer that it was gauge pressure they are asking for.
How can you get 62.46 lb/ft3?
Yea how did she get that?