1ya but it doesnt always remain at atm pressure does it...
or does air come into the tank while the water goes out [7]
a large closed vessel of height H contains a liquid. at the intial moment, the height of the liquid in the vessel is H/2, the space above being filled with air at atmospheric pressure. when the height falls to H/4, find the velocity of the outflow of the liquid.[atmospheric pressure is equal to the pressure due t a couloumn of H/4 of liquid.]
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1yeah, even i was in a dilemma while posting this statement. I deleted the statement thrice and finally dared to post it ;) , hence gave exclamation marks (!!!) as safe-side :P
Let forum beginner conclude this issue !!
33in general, the force applied by a given amount of mass will be same what ever it's density may be. same condition can be applied here !!!
You are taking weight of air...so what if g=0.. we have to take its pressure...not wt of air.
1@priyam
it's not given that the process is either adiabatic or isothermal, so we can't just assume it to be isothermal.
in general, the force applied by a given amount of mass will be same what ever it's density may be. same condition can be applied here !!!
1@philip, from ur point of view, the equation can be re-written as below
Given that atmospheric pressure will be equal to d pressure applied by H/4 of liquid coloumn.
Now, considering the liquid layer of height H/4 from bottom of the vessel, the pressure acting on it will be equal to 2Patm (due to air coloumn+due to remaining water coloumn of height H/4)
applying bernoulli's theorem, we get
(ÏgH/4+ÏgH/4) + 0 + (ÏgH/4)* = ÏgH/4 + 1/2.Ïv2 + ÏgH/4 ...... (*here height is taken to be H/4 as we are taking only H/4 water coloumn under consideration)
solving this equation also gives the same answer.
hence we can conclude that the pressure applied by air coloumn remains constant as the vessel is closed.
correct me if i am wrong !
33That one is wrong..(#9)
But if it is closed and volume has increased then it should be given whether it is adiabatic or isothermal..
I think isothermal...
so P when h=H/4 can be found out...
33Mujhe laga tha ki tough hoga ye but q me jab de hi diya hai ki it is being filled with air at atm pressure tab kya sochna.. it is behaving just like open container..
1maybe i am just making it complicated but
when the height falls to H/4, then also does the pressure remain constant.....?
wat u people have done wud also be the answer for an open container wouldn't it
1anything mentiond about the height of the hole or whatevr it is that the fluid is flowin out thru?
33where? above the fluid??
the space above being filled with air at atmospheric pressure
1Patm=ÏgH/4
Applying Bernoulli's theorem for d entire process, we get
ÏgH/4 + 0 + ÏgH/2 = ÏgH/4 + 1/2.ÏV2 + ÏgH/4
simplifying.... we get V = √(gh/2)
correct me if i am wrong !
1well..nothin mentioned about the radius or anything..so do we apply energy conservation here? or does Bernoulli want to step in here too?
1well..nothin mentioned about the radius or anything..so do we apply energy conservation here?
39well thats all the infrmation i have got. i mean thats the question. [267]
1I CANT FIND .............
since its a closed vessed.. lol.;-)
jus kiddin .. carry on..