How would the problem change if you have multiple openings along the height of the tank? In which case the flowrate would also reduce as the water level falls below the openings. What sort of calculations and integrations would be required?
Oh - that's a great problem. I made a note to make a video about that in the future. Won't be anytime soon, but I like it. First, I would make a simplifying assumption that each hole out the side is either full or empty. It becomes WAY too complicated to consider the case where the water level is midway through the hole. So, water would flow out of each hole until it immediately stopped. You would need to solve for mass/volumetric flow rate out of each hole based on height. And then change in height would be a function of the sum of each hole. The interesting thing is that every hole would be on a Bernoulli streamline with the top of the tank. So - for example, the initial velocity out of the bottom hole is not affected at all by the other holes. It's still just a regular bernoulli equation. Each hole is independent of the others, at any given moment in time. They only relate to each other in the sense that the change in height will be faster since they all contribute to it. This may need to be solved somewhat discontinuous, solving for the amount of time to drain down to the bottom each of each hole, then for the next calculation, omit that hole that is now above the waterline when calculating time to the next hole. However, I think in the end if you plot height against time, this would still be a smooth curve, without corners or jumps, because flow out of each hole would approach zero as height approached its level. Good problem.
185 seconds was completely arbitrary. I think I actually went backwards. Note that the final answer is exactly .500 m. That's too neat of a number to happen by chance. I probably solved for how much time it would take to get that 0.5m height ... and it turned out to be 185 seconds, so then I reversed the problem and asked what height would be after that time, now that I knew it would work out to a nice round final number. Nothing special about 185 seconds at all.
Imagine we have two tank, one with the hole on the bottom, one has the pipe throughth down to hole with same area and the tank is on the same height but pipe exit below morenthan the hole. Does the tank drain faster if we add the pipe throughth down? It is sound wrong but equations seems that way.
A little counterintuitive, but yes, it would drain faster with a pipe extension to make the exit lower. Same premise works for siphoning as well. If you want to empty out something like a fish tank using a siphon, don't have the exit of the siphon right below the tank, get a longer pipe so the exit can be all the way down on the ground, lower exit is faster exit velocity since you convert more potential energy to kinetic with greater change in height.
thank you, this was well explained! Just one question, can you explain the units of flow rate for Mdot=Min-Mout? is that kg/s? I guiess im just overall confused on the control volume thanks!
You are correct. Mass is in kg, and Mdot is (d/dt)M, the time derivative of mass. Same way that time derivative of position (meters) becomes velocity (meters/second), and then another time derivative acceleration (meters/second-squared), a time derivative puts seconds in the denominator of units. Time derivative of mass is then kilogram/second. Amount of mass, per unit time. Mass flow rate.
40 years ago I worked in a lab, which also had a special fluid dynamic section. There an engineer conducted a test series in which a cylindrical tank filled with water was emptied through a round hole in the bottom (exactly as described in your video). Before emptying the tank the water was given a long time to come "to rest", making sure there would not be any water motion. In a next test the water was put into a swirl motion before emptying the tank. As far as I remember the water in swirl motion took less time to empty the tank. Any idea how to calculate this?
That's a cool experiment. Modeling this - my first thought would still be with the Bernoulli Equation. Normally, we approximate the velocity at the top of the tank as zero. This isn't entirely correct because the water level is decreasing. What if you accounted for the swirl, as giving the water at the surface of the tank extra kinetic energy, maybe by calculating an "average" velocity of the water at the top of the tank. The direction is different, the streamline path for the bernoulli equation is down at the top of the tank, and this velocity would not be in the direction of the streamline, but, eventually the water does turn and exits the tank horizontally, so accounting for an initial horizontal velocity at the top of the tank doesn't seem completely unreasonable. I don't think this answer is perfect - but I think it might be good enough to illustrate the concept, and it would confirm the direction of the experimental results you saw, if there's extra kinetic energy at the top, then there should be extra energy at the bottom, which means faster time to empty. You've really got me thinking though and I think this might make for a really good classroom demo. It can be very easily duplicated with grocery store plastic water bottles, side by side. I think I'll test this myself next semester and see what the students think. Thanks so much, this is really getting my brain moving.
Thank you for your answer! Glad that you are considering going into details and thinking about further experiments. Exciting stuff! Would like to be in your class. Creative students and smart teachers make our future.
You're right - that is definitely the right way to do it. In practice I usually try to ensure all my units are consistent at the start so that I know the answer will work out to be the correct units, which allows me to get away with being lazy - but I fully concede that it is risky and lazy. I'm making a note to do better next time. Good luck this semester.
Can't believe such a brilliantly produced video has near to no views. Helped me tons. Thanks!!
Maybe later it will get a ton of views, and you'll always get to say you were a fan before it was cool.
Watching this video, I actually had fun learning, two things that hardly go together. This is a masterpiece, A+!
Awesome. Unfortunate that it's rare, but glad I can be the exception. Have a great semester!
How would the problem change if you have multiple openings along the height of the tank? In which case the flowrate would also reduce as the water level falls below the openings. What sort of calculations and integrations would be required?
Oh - that's a great problem. I made a note to make a video about that in the future. Won't be anytime soon, but I like it. First, I would make a simplifying assumption that each hole out the side is either full or empty. It becomes WAY too complicated to consider the case where the water level is midway through the hole. So, water would flow out of each hole until it immediately stopped. You would need to solve for mass/volumetric flow rate out of each hole based on height. And then change in height would be a function of the sum of each hole. The interesting thing is that every hole would be on a Bernoulli streamline with the top of the tank. So - for example, the initial velocity out of the bottom hole is not affected at all by the other holes. It's still just a regular bernoulli equation. Each hole is independent of the others, at any given moment in time. They only relate to each other in the sense that the change in height will be faster since they all contribute to it. This may need to be solved somewhat discontinuous, solving for the amount of time to drain down to the bottom each of each hole, then for the next calculation, omit that hole that is now above the waterline when calculating time to the next hole. However, I think in the end if you plot height against time, this would still be a smooth curve, without corners or jumps, because flow out of each hole would approach zero as height approached its level. Good problem.
very well explained! thank you very much.. 1 thing I'd like to ask, why used 185 seconds
185 seconds was completely arbitrary. I think I actually went backwards. Note that the final answer is exactly .500 m. That's too neat of a number to happen by chance. I probably solved for how much time it would take to get that 0.5m height ... and it turned out to be 185 seconds, so then I reversed the problem and asked what height would be after that time, now that I knew it would work out to a nice round final number. Nothing special about 185 seconds at all.
Imagine we have two tank, one with the hole on the bottom, one has the pipe throughth down to hole with same area and the tank is on the same height but pipe exit below morenthan the hole. Does the tank drain faster if we add the pipe throughth down? It is sound wrong but equations seems that way.
A little counterintuitive, but yes, it would drain faster with a pipe extension to make the exit lower. Same premise works for siphoning as well. If you want to empty out something like a fish tank using a siphon, don't have the exit of the siphon right below the tank, get a longer pipe so the exit can be all the way down on the ground, lower exit is faster exit velocity since you convert more potential energy to kinetic with greater change in height.
thank you, this was well explained! Just one question, can you explain the units of flow rate for Mdot=Min-Mout? is that kg/s? I guiess im just overall confused on the control volume
thanks!
You are correct. Mass is in kg, and Mdot is (d/dt)M, the time derivative of mass. Same way that time derivative of position (meters) becomes velocity (meters/second), and then another time derivative acceleration (meters/second-squared), a time derivative puts seconds in the denominator of units. Time derivative of mass is then kilogram/second. Amount of mass, per unit time. Mass flow rate.
40 years ago I worked in a lab, which also had a special fluid dynamic section. There an engineer conducted a test series in which a cylindrical tank filled with water was emptied through a round hole in the bottom (exactly as described in your video). Before emptying the tank the water was given a long time to come "to rest", making sure there would not be any water motion. In a next test the water was put into a swirl motion before emptying the tank. As far as I remember the water in swirl motion took less time to empty the tank. Any idea how to calculate this?
That's a cool experiment. Modeling this - my first thought would still be with the Bernoulli Equation. Normally, we approximate the velocity at the top of the tank as zero. This isn't entirely correct because the water level is decreasing. What if you accounted for the swirl, as giving the water at the surface of the tank extra kinetic energy, maybe by calculating an "average" velocity of the water at the top of the tank. The direction is different, the streamline path for the bernoulli equation is down at the top of the tank, and this velocity would not be in the direction of the streamline, but, eventually the water does turn and exits the tank horizontally, so accounting for an initial horizontal velocity at the top of the tank doesn't seem completely unreasonable. I don't think this answer is perfect - but I think it might be good enough to illustrate the concept, and it would confirm the direction of the experimental results you saw, if there's extra kinetic energy at the top, then there should be extra energy at the bottom, which means faster time to empty. You've really got me thinking though and I think this might make for a really good classroom demo. It can be very easily duplicated with grocery store plastic water bottles, side by side. I think I'll test this myself next semester and see what the students think. Thanks so much, this is really getting my brain moving.
Thank you for your answer! Glad that you are considering going into details and thinking about further experiments. Exciting stuff! Would like to be in your class. Creative students and smart teachers make our future.
Please add units, when you plug in numbers!
You're right - that is definitely the right way to do it. In practice I usually try to ensure all my units are consistent at the start so that I know the answer will work out to be the correct units, which allows me to get away with being lazy - but I fully concede that it is risky and lazy. I'm making a note to do better next time. Good luck this semester.
@@BrianBernardEngineering thanks! My semesters are long gone though! I am here for the joys of fluid dynamics!