Fluids (part 10)

[adil06944]

very good video but P2 is not zero unless the outside is a vacuum as well. P2 should be equal to atmospheric pressure. Your example where the pressures cancels out would work if the top was open and P1 was atmosphere as well. Then P1-P2=0.

[bfich123]

you cant divide the continuity equation by A1, you have to divide each vairable by A1 which leaves you with a mess... you cannot just exclude it because it appears on both sides

[xNanashiKun]

@bfich123 You can divide the continuity equation by A1 because it is essentially just one variable. V2A1/1000.

[AboveMeIsARetard]

you forgot atmospheric pressure... you also forgot it in your buoyancy force balance

[thekingsfc17]

@AboveMeIsARetard there is no atmospheric pressure because the area above the fluid is a vacuum

[adil06944]

@thekingsfc17 AboveMeIsARetard is correct about Atm pressure for point 2. This is a simple mistake to make.

[mlatti2]

@helllocutiepie i believe m/s just like it would normally be. if you do the derivations you could figure it out, too

[Helllocutiepie]

What are the units in the output velocity?

[Challsoer]

@Helllocutiepie Velocity is always m/s, if it was in other units it wouldn't be velocity.

[slmUSA]

Thank you!

[archiemedes42]

@everyone

He is assuming, like all physics problems, that everything is in a vacuum. If there was atmospheric pressure, water wouldn't even flow out and air would be sucked in and v2 wouldn't even be positive!!

[aayushkhanna]

@Froody1911 but even if there is no atmospheric pressure. shouldn't the pressure at a certain depth be denoted by the formaula rho*g*h ?? hence should'nt there be work done at the hole. ??

[NATHANPOULIN1]

if there is a vacuum, then p2 is not zero, if you assume one atm for barometric pressure. One atm for p2 would mean that h would have to be more than 10 meters high in order for the fluid to exit the punctured hole. If there is no vacuum at the top, then p1 and p2 are the same, they cancel out, and thus, have no effect on the velocity.

[onlyAerik]

indeed. If the jar is closed, then he'd also need to talk about a volume of air bubbling into the canister until the pressure is equalized, which would still come to when the water level dropped to dh below the hole.

A vacuum gap changes things. Ever held soda in your straw? Or had to open both sides of a can to get your refried beans to fall out? same thing.

[nikan4now]

I agree with the case wherein the container is surrounded by atmospheric pressure. But in this particular video (if we're using absolute values for pressure) P2 must NOT be zero. And if we're using relative pressure values then P2 must be zero and P1= -Patm [since P (REL)=P(abs)-P(Atm)]

[postsaccadic]

Very nice.

[laloksdm]

can P2 also be atm pressure?

[dst7ssa]

Ok I didn't get the external pressure too well. Did you assume Patm =o also if h=0, is "V" being calculated based on the top of the liquid or at the bottom of the liquid?

[nikan4now]

Here's a question: What if the top were not vacuum? We'd still get the same velocity right? I think with control volumes we can use relative pressure and in this case it would be -Patm for the inlet and zero for the outlet. Correct me if I'm wrong. Thanks.

[aayushkhanna]

@nikan4now if there would be no vacuum.

then u would have to consider atmospheric pressure also , which wud change the velocity of the fluid comming out of the hole.

[ayush9psycho]

i didnt kind of get dis lecture,he assumes a closed container with vaccum above it and he says that the fluid flowing from the smaller hole does no work ,i didnt get dat the fluid has has to work against atmospheric pressure, well anyways this man provides beautiful insites into the subject

[techgirls]

thank you =)

[ft8a]

thanks man, this is useful

[rick7917]

thx