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……..:|*
–
Raja Metronetizen
·
4 years, 3 months ago

@Raja Metronetizen
–
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
–
Saurabh Dubey
·
4 years, 3 months ago

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 \(\frac {A1}{A} \times \sqrt{\frac{2h}{g}}\)
–
Saurabh Dubey
·
4 years, 3 months ago

happie....now?
–
Riya Gupta
·
4 years, 3 months ago

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.
–
Edsel Salariosa
·
4 years, 3 months ago

Log in to reply

It is probably going to involve integrals. What have you tried so far?
–
Pranav Arora
·
4 years, 3 months ago

Log in to reply

assuming the upper height be 'h' and lower height be 'H-h' time = 2(H-h)/g whole under root
–
Mudit Tripathi
·
4 years, 3 months ago

## 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……..:|* – Raja Metronetizen · 4 years, 3 months agoLog in to reply

– Samuel Queen · 4 years, 3 months ago

That seems like the right answer using physical considerations.Log in to reply

– Saurabh Dubey · 4 years, 3 months ago

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 :PLog 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 \(\frac {A1}{A} \times \sqrt{\frac{2h}{g}}\) – Saurabh Dubey · 4 years, 3 months ago

Log in to reply

Log in to reply

Log in to reply

happie....now? – Riya Gupta · 4 years, 3 months ago

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. – Edsel Salariosa · 4 years, 3 months ago

Log in to reply

It is probably going to involve integrals. What have you tried so far? – Pranav Arora · 4 years, 3 months ago

Log in to reply

assuming the upper height be 'h' and lower height be 'H-h' time = 2(H-h)/g whole under root – Mudit Tripathi · 4 years, 3 months ago

Log in to reply