WE ARE HERE INTERESTED IN FINDING THE TIME REQUIRED TO EMPTY A TANK,IF A HOLE IS MADE AT THE BOTTOM OF THE TANK..........:::::::::
you have given that tank is filled with water....so the density of water is 1 gm/cc.......we would take this during our calculation.......ok………...now think that the cross sectional area of the hole is "a"........... area of the cross section of the tank is surely pi(r^2)=A (let us consider)........let at some instant of the time the level of liquid in the tank is “y”……..velocity of efflux at this instant of time would be……..v=sqrt(2gy)…………::::::::::
now , at this instant volume of liquid coming out of the hole per second is (dV/dt)=av…………..volume of liquid coming down in the tank per second is also (dV/dt)=A(-dy/dt)……….i think you have understood till now………….
Now, av =A(-dy/dt)………..asqrt(2gy)=A(-dy/dt)……….now integrate both sides taking same quantities in the same side….you would get…….integration [0 to t] dt= (-A/{asqrt(2g)})* integration[H to 0] y^(-.5) dy…...…………….so we get,........... t=(A/a)sqrt(2H/g) where A=pir^2....(as you gave)……:)(:……..PLEASE LET ME KNOW IF ANYTHING IS WRONG……..:|*

This is a hydraulics problem in orifice.To determine the time to empty the tank.Use Bernoulli's theorem v1^2/2g+p1/y+h1=v2^2+p2/y+h2.Since the container is open.The pressure is zero and to the nozzle because of atmospheric pressure.Where h1=h and h2=0 reference to the datum of orifice.neglect the velocity in v1 ,it will not effect your computation because v1 are to small velocity.You will get h=v1^2/2g equation 1.The discharges of water in the orifice looks like a projectile.It will occupy range or distance from the orifice .V1=X/t equation 2.substitute equation 1 to 2 then you will get the time to empty the tank.This is theoretical not including the head loss.

## Comments

Sort by:

TopNewestWE ARE HERE INTERESTED IN FINDING THE TIME REQUIRED TO EMPTY A TANK,IF A HOLE IS MADE AT THE BOTTOM OF THE TANK..........::::::::: you have given that tank is filled with water....so the density of water is 1 gm/cc.......we would take this during our calculation.......ok………...now think that the cross sectional area of the hole is "a"........... area of the cross section of the tank is surely pi

(r^2)=A (let us consider)........let at some instant of the time the level of liquid in the tank is “y”……..velocity of efflux at this instant of time would be……..v=sqrt(2gy)…………:::::::::: now , at this instant volume of liquid coming out of the hole per second is (dV/dt)=av…………..volume of liquid coming down in the tank per second is also (dV/dt)=A(-dy/dt)……….i think you have understood till now…………. Now, av =A(-dy/dt)………..asqrt(2gy)=A(-dy/dt)……….now integrate both sides taking same quantities in the same side….you would get…….integration [0 to t] dt= (-A/{asqrt(2g)})* integration[H to 0] y^(-.5) dy…...…………….so we get,........... t=(A/a)sqrt(2H/g) where A=pir^2....(as you gave)……:)(:……..PLEASE LET ME KNOW IF ANYTHING IS WRONG……..:|*Log in to reply

That seems like the right answer using physical considerations.

Log in to reply

Ya that's how it's derived. I was a bit lazy(I gave the answer late night) so just gave the answer without the derivation :P

Log in to reply

I'am assuming that the area of cross section of the cylinder is A1 and A<<<<A1. Then the answer should be \(\frac {A1}{A} \times \sqrt{\frac{2h}{g}}\)

Log in to reply

Comment deleted Apr 29, 2013

Log in to reply

Comment deleted May 07, 2013

Log in to reply

happie....now?

Log in to reply

This is a hydraulics problem in orifice.To determine the time to empty the tank.Use Bernoulli's theorem v1^2/2g+p1/y+h1=v2^2+p2/y+h2.Since the container is open.The pressure is zero and to the nozzle because of atmospheric pressure.Where h1=h and h2=0 reference to the datum of orifice.neglect the velocity in v1 ,it will not effect your computation because v1 are to small velocity.You will get h=v1^2/2g equation 1.The discharges of water in the orifice looks like a projectile.It will occupy range or distance from the orifice .V1=X/t equation 2.substitute equation 1 to 2 then you will get the time to empty the tank.This is theoretical not including the head loss.

Log in to reply

It is probably going to involve integrals. What have you tried so far?

Log in to reply

assuming the upper height be 'h' and lower height be 'H-h' time = 2(H-h)/g whole under root

Log in to reply