IIT JEE , AIEEE Forum - Physics , Mechanics

kislay

Two cylindrical tanks having cross sectional area A and 2A are kept on horizontal floor. First tank is filled with water to a height H and the other tank is empty.If two tanks are connected by a pipe of cross sectional area a(a << A)at the bottom at t = 0, calculate time t when level of water in two tanks becomes same.

rajat

is the answer sqroot(H/12g)

kislay

answer is (2A/3a)*

(2H/g)

edison

Apply Bernoulli's theorm at the two points of fluid flow.

kislay

SIR I COULDN'T GET BY DOING THAT,PLS HELP

frenied

i got tht answer.ill give u the solution.gimme ur email address ill need to send the diagram.and promise me a salute if i give u the soln.

kislay

yes friend i will give u a salute.
my email address is kislay_1990@yahoo.co.in

frenied

ok ill send u the solution.and if u dont gimme the salute well....that'd be really mean...but ill send the solution.

frenied

ok ill send u the solution.and if u dont gimme the salute well....that'd be really mean...but ill send the solution.or ill

frenied

ill post the thoery of my working here for i need to use the formula editor and the diagram ill mail.

frenied

_{^{so u have 2 tanks with ht. y" in 2nd tank and y' in 1st tank.so 2A.y"+A.y'=A.H.this gives us y'=H-2y"----i .now picture the two tanks filled with liquid to these hts. and connected by the pipe of area a.so applying the bernoullis th. for the two ends of connecting pipe}}

^{_{y'.d.g+1/2dv'}2_+0=ydg+1/2dv2₊₀}

_{^{cancelling d and multiplying by 2 we get}}

^{_2y'g+v'2_=2yg+v2 -------ii}

^{now dy/dt=v" and dy'/dt=v' being the rates of increase and decrease in hts.of water respectively in 2nd and 1st tank.and v is also the velocity of efflux from the pipe.so from eqn of continuity}

^{A.v'=a.v thus v=A/a.v'}

^{subs this value in ii to get}

^{_v2_-v'2_=2g(y'-y)}

^{_(A/a)2_v'2_-v'2_=2g(y'-y)}

_^thus

^_v'=

2g(y'-y)/((A/a)²-1

differntiate i wrt time to get

-2v"=v' - indicating y' decreases as y increases.

thus

2v"=

2g(y'-y)/((A/a)²-1

now put 2v"=2dy/dt and y'=H-2y.integrate with time limits as 0-t and y limits as 0-H/3 becuase if uve realised that the hts. will be H/3 at which water level is equal.if u didnt then heres that also

A.h+2A.h=A.H find h which is the assumed ht at which level is same.besides neglect he subtraction of 1 from the denominator as a<<A so A/a>>1.u will get the answer.refer to the diagram and u'll understand it better.

kislay

^{_{EXPERTS PLEASE SEE

here is ur soln. friend i have highlighted the doubtful steps

ok so u have 2 tanks one of area A and the other of 2A.now initially the total volume of water is A.H this remains conserved.now let us consider a time when there is water to ht. y in 2nd tank and y' in 1st tank.so 2A.y+A.y'=A.H.this gives us y'=H-2y----i .now picture the two tanks filled with liquid to these hts. and connected by the pipe of area a.so applying the bernoullis th. for the two ends of connecting pipe}}

^{_{y'.d.g+1/2dv'}2_+0=ydg+1/2dv2₊₀}

_{^{why the pressure energy at 2 pt.s =0

cancelling d and multiplying by 2 we get}}

^{_2y'g+v'2_=2yg+v2 -------ii}

^{now dy/dt=v and dy'/dt=v' being the rates of increase and decrease in hts.of water respectively in 2nd and 1st tank.and v is also the velocity of efflux from the pipe.so from eqn of continuity}

^{A.v'=a.v thus v=A/a.v'}

^{yqu have taken v as the velocity with which water rises in tank with area 2A,not the velocity with which it comes out of pipe so how can u apply continuity eqn. to pipe with velocity v?
subs this value in ii to get}

^{_v2_-v'2_=2g(y'-y)}

^{_(A/a)2_v'2_-v'2_=2g(y'-y)}

_^thus

^_v'=

2g(y'-y)/((A/a)²-1

differntiate i wrt time to get

-2v=v' - indicating y' decreases as y increases.

thus

2v=

2g(y'-y)/((A/a)²-1

now put 2v=2dy/dt and y'=H-2y.integrate with time limits as 0-t and y limits as 0-H/3 becuase if uve realised that the hts. will be H/3 at which water level is equal.if u didnt then heres that also

A.h+2A.h=A.H find h which is the assumed ht at which level is same.besides neglect he subtraction of 1 from the denominator as a<<A so A/a>>1.u will get the answer.refer to the diagram and u'll understand it better.

please explain properly

frenied

ok i think i made a mistake in my explanation.actually v is just the velocity of efflux from the pipe and so the bernoulli's equation holds good as also the continuity equation.so using these we find v' which we can write 2dy/dt as i have already said.i hope it is clear now.u see i had applied the bernoulli's eqn.for the 2 ends of the pipe so v is just the efflux velocity.i wrote it wrongly as the velocity of rise of water.i hope that makes it clear now.just delete that part of my explanation.and as for ur first doubt its not the pressure enrgy its the PE that is zero for both.i have taken acct.of pressure enrgies in the first terms on both sides.

frenied

and take v as only the velocity of efflux and no other velocity.i think then my solution would be correct.if u still have doubts do feel free to contact.as it is i have not used the fact that velocities of rise of liq.in 2nd tank and velocity of efflux are equal at any time.ill repost the solution with corrections.

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