Fluid Mechanics: Topic 2.1 - Pascal's Law

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CPPMechEngTutorials

CPPMechEngTutorials

Күн бұрын

Пікірлер: 58
@abbiemielczarek4807
@abbiemielczarek4807 3 жыл бұрын
couldn't have found a more useful video! i missed a lecture and when i looked at the powerpoint all i saw was numbers, but i actually understand it now!
@itthus9552
@itthus9552 4 жыл бұрын
great video, I watched the lecture first before I watched these videoes, and I had to keep pausing it to think about how he got the numbers. This is so much better having it in short video. Thanks again for the great explanation! I look foward to mastering this fluids mechanics class!
@beoptimistic5853
@beoptimistic5853 4 жыл бұрын
kzbin.info/www/bejne/joGmmHqKbqefqLM 💐💐💐
@numankashif6158
@numankashif6158 Жыл бұрын
Sir you are awesome . One of the finest videos on KZbin for fluid mechanics topic. And you have gain one more subscribers.
@CPPMechEngTutorials
@CPPMechEngTutorials Жыл бұрын
Hooray!
@김찬수-i9j
@김찬수-i9j 8 жыл бұрын
awesome! I'm the korean middle school student who has suspected about the Pascal's law for quite a long time. But because of your proof, I think i can sleep well tonight!
@CPPMechEngTutorials
@CPPMechEngTutorials 8 жыл бұрын
Sweet dreams!
@김찬수-i9j
@김찬수-i9j 8 жыл бұрын
I think I almost understood the whole proof but let me give you some questions that i hardly understood. By proving P1=P2=P3, can we say that pressure is independent of direction if no shearing forces exist...? + why can we suppose that dx,dy,dz go to 0...?
@김찬수-i9j
@김찬수-i9j 8 жыл бұрын
Sorry for bothering... but I was just curious about some parts of proof.
@CPPMechEngTutorials
@CPPMechEngTutorials 8 жыл бұрын
If we shrink the cube to a point in space, we find that the value of P is the same from all directions (we initially did not assume that). This implies that pressure has no preferred direction and we can treat it as a point function -- that is, pressure only depends on position in space. We say the pressure is XX Pascals at a given location, not that the pressure is XX Pascals to the left at a given location. Rather, the pressure force has a direction.
@김찬수-i9j
@김찬수-i9j 8 жыл бұрын
CPPMechEngTutorials Got it! Thanks a lot! :)
@wolframalpha8634
@wolframalpha8634 Ай бұрын
Sir, is there a particular reason as to why we assumed a wedge shaped fluid element to prove this? , can we prove the same using cuboidal or other eucledian elements?
@CPPMechEngTutorials
@CPPMechEngTutorials Ай бұрын
The shape doesn't matter.
@willofdodge1
@willofdodge1 6 жыл бұрын
These videos are extremely helpful, Thank-You!
@CPPMechEngTutorials
@CPPMechEngTutorials 6 жыл бұрын
You're welcome!
@ralphanthonymallillin2712
@ralphanthonymallillin2712 3 жыл бұрын
I don't understand what he mean at 3:15..I'm finding answers why the back and front pressure are always separated unlike the other 3 equal pressures..Because pascal said the pressure are all equal in all direction then why does the front and back pressure are separated from the others??
@mujeebkhan09
@mujeebkhan09 7 жыл бұрын
I have one doubt...... at the end of proof, our conclusion was dependent on limit i.e dx,dy or dz approaches zero, which is correct but brings ambiguity. It could be much simplified if static equilibrium equation was applied initially as our assumption was already for static fluids, then RHS would have been zero, and then effect of differential increment in pressure due to weight of differential element could be ignored. Then things would have been much simple and easy to understand. Inclusion of acceleration, then applying limit at end brought ambiguity to a considerable extent.
@hassanaleian3111
@hassanaleian3111 4 жыл бұрын
Yes this proof states that pressure in any direction at a point is equal but it doesnt explain why. However your suggestion is to my understanding incorrect. That is because their can be non-zero acceleration and this law still holds. As long as all fluid elements accelerate at same rate there will still be no shear stress as the video suggests.
@pstark4
@pstark4 5 жыл бұрын
Hi, the description of the surface forces acting on only three surfaces "for clarity" looks like a Cauchy stress tensor. Why is it only 3 faces, nine components, rather than 6 faces, 18 components? What about bending?
@ralphanthonymallillin2712
@ralphanthonymallillin2712 3 жыл бұрын
That's also my question because pascal said that pressures in all directions but they always omit it without further explanation!!!
@arthikalexander316
@arthikalexander316 8 жыл бұрын
why do you particularly take a wedge or prism shaped fluid element?
@maheshk3368
@maheshk3368 8 жыл бұрын
I think if we use cube shape, all pressures acting on opposite sides will cancel each other except weight of fluid and no conclusion can be drawn
@CPPMechEngTutorials
@CPPMechEngTutorials 8 жыл бұрын
We didn't have to. A cube or any other shape would have given us the same results. A prism allows us to examine an arbitrarily shaped fluid element (notice we did not specify theta) that is relatively simple in shape.
@victormd1100
@victormd1100 5 жыл бұрын
@@CPPMechEngTutorials How would prove that the pressure are equal in a fluid element shaped like a circle or an elipse (or any weird curve)? It seems that there would be many more unknowns (pressures) than equations due to all of the different directions normal to the surface
@sahalnijadv9237
@sahalnijadv9237 4 жыл бұрын
6:51 pressure is actually a tensor.
@beoptimistic5853
@beoptimistic5853 4 жыл бұрын
kzbin.info/www/bejne/joGmmHqKbqefqLM 👍💐💐
@krishnaprasadps7262
@krishnaprasadps7262 Жыл бұрын
Wow awesome session sir, it will be very helpful if you could share the presesentation slides.
@CPPMechEngTutorials
@CPPMechEngTutorials Жыл бұрын
Sorry... we only have the video to share.
@gangadharvenkataramanap7651
@gangadharvenkataramanap7651 7 жыл бұрын
I like you exaplaination, I understand the concept, but what should we do if we want to prove the pressure in x direction is also equal to remaining directions.
@CPPMechEngTutorials
@CPPMechEngTutorials 7 жыл бұрын
Use the same technique, but in the x-direction. Summation of Fx = dm*ax. The forces in the x-direction are pressure forces -- call the pressures P4 and P5 on the two faces and initially do not assume they are the same. You will find that P4 = P5 if the acceleration is 0. You would then have to use the idea discussed in Topic 2.2 that pressure only varies in the direction of gravity in a static fluid, so the pressure in the x-direction and y-direction should be the same since gravity is in the z-direction.
@bluefinsxmovies
@bluefinsxmovies Жыл бұрын
Please I cannot stop wondering these two things Where did the sin & cos disappear to? Why did; P2dxdscosΩ become P2dxdy all of a sudden? Please I await your response
@pranavkankal5604
@pranavkankal5604 4 жыл бұрын
Great work !!! Thank you very much 😊
@beoptimistic5853
@beoptimistic5853 4 жыл бұрын
kzbin.info/www/bejne/joGmmHqKbqefqLM 💐💐💐
@spikeoac4248
@spikeoac4248 4 жыл бұрын
If pressure doesn't vary with direction, then how is it that in the next video you prove that pressure varies linearly in the z direction? Doesn't that disprove what you just proved in this video?
@weeskaterdude
@weeskaterdude 4 жыл бұрын
In considering an infinitesimal fluid element, we are approximating the pressure at a point. By Pascals law we can show that the pressure is independent of direction. This does not imply that pressure cannot vary in space and therefore it is entirely possible that we can obtain a pressure differential. Consider an open tank of water. We know that a fluid element at the bottom of the tank experiences greater pressure than an element approaching the free surface. I.e there exists a pressure gradient. We can describe this gradient via a chosen coordinate system and the result obtained would be that the pressure varies linearly in the chosen coordinate. I hope this is clear.
@beoptimistic5853
@beoptimistic5853 4 жыл бұрын
kzbin.info/www/bejne/joGmmHqKbqefqLM 💐💐💐
@wyattblah
@wyattblah 8 жыл бұрын
This is a great explanation!
@CPPMechEngTutorials
@CPPMechEngTutorials 8 жыл бұрын
Great!
@siten1
@siten1 3 жыл бұрын
Why is everything infinitesimally small(dx, dm, dv, etc) except density?
@TheFuckel
@TheFuckel Жыл бұрын
It is constant
@pankajkumar-jo5sp
@pankajkumar-jo5sp 4 жыл бұрын
Extremely interested concept
@beoptimistic5853
@beoptimistic5853 4 жыл бұрын
kzbin.info/www/bejne/joGmmHqKbqefqLM 💐💐💐
@dbestacademybymentorgodslo2529
@dbestacademybymentorgodslo2529 Жыл бұрын
Please wat is the name of the app you are using?
@siten1
@siten1 3 жыл бұрын
Why is Pressure not dP?
@LailaWiduri
@LailaWiduri 4 жыл бұрын
All I can say is THANK YOUUUUU. This video is really helpful
@beoptimistic5853
@beoptimistic5853 4 жыл бұрын
kzbin.info/www/bejne/joGmmHqKbqefqLM 👍💐💐
@YesHerself
@YesHerself 8 жыл бұрын
nice! very well explained.thank you!
@CPPMechEngTutorials
@CPPMechEngTutorials 8 жыл бұрын
+teynerdy You're welcome!
@amh8303
@amh8303 6 жыл бұрын
What is the name of the reference that ur make video with? Abather the way ur good👌
@CPPMechEngTutorials
@CPPMechEngTutorials 6 жыл бұрын
We made these videos to be independent of any textbook. However, many textbooks have a similar derivation for Pascal's Law
@amh8303
@amh8303 6 жыл бұрын
@@CPPMechEngTutorials I meant the text book that ur follow . Because i think it is the same my lecturer going with! I had an exam this sunday .
@mujeebkhan09
@mujeebkhan09 7 жыл бұрын
amazing great job... thumbs up
@CPPMechEngTutorials
@CPPMechEngTutorials 7 жыл бұрын
Hooray!
@alnajdia6
@alnajdia6 3 жыл бұрын
You angel, you.
@thangcogang2739
@thangcogang2739 4 жыл бұрын
thankss :>
@beoptimistic5853
@beoptimistic5853 4 жыл бұрын
kzbin.info/www/bejne/joGmmHqKbqefqLM 💐💐💐
@rahulravi661
@rahulravi661 3 жыл бұрын
You just went through all that to put everything to zero? (P.S it's a troll comment , I'm an idiot 😏)
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