# Energy Required to Fill a Tank

• thatboi
In summary: No, that's not right. If you pass the hose over the edge then down into the tank the work would be the same as pumping from the bottom because you are lifting the water center of mass to a height of 5 meters.If, instead, the hose were to pass over the edge then down into the tank it could be done with the same work as in case (1), but the water would be at a lower pressure and therefore require more energy. In summary, it would require twice the work to pump the water over the top edge of the tank as it does to pump it from the bottom.

#### thatboi

Thread moved from the technical forums to the schoolwork forums
Hey all,
I am working through some practice problems and am stuck at the following one:
"A cylindrical tank 10m high has an internal diameter of 4m. How much work would be required to fill the tank with water if the water were pumped in i.) at the bottom and ii.) over the top edge?"

Part i is simple enough via applying potential energy formula but for part ii, the solution says that it takes twice as much work but I am having difficulty seeing why.
Any help would be appreciated!

Please post your solutions/attempts. They will serve as starting points to any help you might receive. Include part (a). Even though you may think you got it right, you may have made a mistake somewhere.

thatboi said:
Hey all,
I am working through some practice problems and am stuck at the following one:
"A cylindrical tank 10m high has an internal diameter of 4m. How much work would be required to fill the tank with water if the water were pumped in i.) at the bottom and ii.) over the top edge?"

Part i is simple enough via applying potential energy formula but for part ii, the solution says that it takes twice as much work but I am having difficulty seeing why.
Any help would be appreciated!

kuruman said:
Please post your solutions/attempts. They will serve as starting points to any help you might receive. Include part (a). Even though you may think you got it right, you may have made a mistake somewhere.
The solution to part i.) is ρgπr^2h*5 where ρ is the density of water and h is the height of the tank.
I believe the interpretation to part ii could be that the water is initially pumped from the ground (or where we define U=0) to top of the tank before letting the water fill the tank and thus it will take twice the energy to get the water to that height?

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thatboi said:
ρgπr^2 h*5
What do you mean by *5?

Orodruin said:
What do you mean by *5?
My logic was that ρgπr^2h is the weight of the water needed to fill the tank and when we filled it up from the bottom, it would be equivalent to moving the center of mass up 5 meters since the tank itself is 10 meters so to get the work I just multiplied the force by the distance.

FWIW the way you have written your formulas makes it hard on the people who are trying to help.
Please use latex. If you do not know "how to", please look on the left most column just below the very last post in the thread ---"Latex Guide".

thatboi said:
My logic was that ρgπr^2h is the weight of the water needed to fill the tank and when we filled it up from the bottom, it would be equivalent to moving the center of mass up 5 meters since the tank itself is 10 meters so to get the work I just multiplied the force by the distance.
Ok, it is a bit confusing when you mix symbolic expressions with numerical ones. I suggest just using symbolic expressions until the very end and only then put in numbers. Also, the number 5 by itself is not saying much, you really want to multiply by 5 m or you will have a dimensional error. Either way, the 5 should be replaced by a h/2 if you want to leave the height symbolic and therefore ##\rho g \pi r^2 h^2/2##.

So in the case of pumping from the bottom, you are lifting the water center of mass ##h/2## and multiplying by mass. How far up do you need to pump each bit of water in order for each bit of water to pass over the edge of the tank?

Yes sorry about that. Right, so then my interpretation in my first response is correct where I need to move the water to a height of ##h## and thus you get double the work?

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thatboi said:
Yes sorry about that. Right, so then my interpretation in my first response is correct where I need to move the water to a height of ##h## and thus you get double the work?

You should try to show how you get that result.

thatboi said:
Yes sorry about that. Right, so then my interpretation in my first response is correct where I need to move the water to a height of ##h## and thus you get double the work?
Yes… assuming the water is simply allowed to fall through the air from the top. If, instead, the hose were to pass over the edge then down into the tank it could be done with the same work as in case i.

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thatboi
haruspex said:
If, instead, the hose were to pass over the edge then down into the tank it could be done with the same work as in case i.
Though the lip of the tank is close to the edge of feasibility for siphon-like action to be fully effective.

Very close, in fact. One might have to make sure to use chilled water to avoid vaporization.

Seems ok to me for an idealization.

Comparing apples to apples, it would be more than double the work (I doubt this was the intended interpretation) but, in order to get it from the base of the tank to the top of the tank it must travel up a "pipe" with some area ##A_p##. There would be work done ##\frac{1}{2} \rho g A_p H^2## just to get it to the top of the tank in addition to the work done to fill the tank through the top ##\rho g A_{tank} H^2## . When you are ignoring viscosity its ok to make ##A_p## arbitrarily small...

jbriggs444 said:
Though the lip of the tank is close to the edge of feasibility for siphon-like action to be fully effective.

Very close, in fact. One might have to make sure to use chilled water to avoid vaporization.
A pump is still required to raise it 5m, so the siphon only has to manage the other 5, on average. If any risk of creating a vacuum, the pump could do a little more at the start.

jbriggs444
What are the "two cases"? We have
1. pump from bottom
2. pump over top
3. pump over top wit6h siphon maybe with cavitation
I'm lost.

bob012345
hutchphd said:
What are the "two cases"? We have
1. pump from bottom
2. pump over top
3. pump over top wit6h siphon maybe with cavitation
I'm lost.
I believe just #1 and #2 were part of the OP.

If you wanted to gravity feed
1. you slowly fift the feed tank tank as you fill. Not all the water needs to be lifted all the way
2. You need all the water in the feed tank to be lifted all the way

I hardly think the problem constructor had neither siphon or viscosity in mind for this problem.

erobz
Orodruin said:
I hardly think the problem constructor had neither siphon or viscosity in mind for this problem.
But the use of a pipe to fill the tank in part b is implicit, which is conceptually no different than part A itself. Comparing apples to apples, that's why I mentioned the work to fill the pipe, and ignoring viscosity, why we can neglect it.

Also implicit here is the use of a pump, and pumps add kinetic energy.

The problem statement should clarify that we are after the minimum possible work required to fill the tank in each scenario. IMO.

erobz said:
Also implicit here is the use of a pump, and pumps add kinetic energy.

The problem statement should clarify that we are after the minimum possible work required to fill the tank in each scenario. IMO.
In my opinion, that would be overkill. We have an original poster here who was concerned about where the kinetic energy loss was for water pouring over the rim and splashing onto the growing pool of water below. [No offense meant to the original poster. It was a reasonable concern].

But here we are worrying about siphons, cross-sectional areas of pipes and kinetic energy of flows. The next thing you know, we'll be looking at relative humidity and barometric pressure so that we can accurately account for the buoyancy of the water in air and asking about latitude so that we know how much higher the water is at the rim of the tank than at the center due to centrifugal force and the rotation of the earth.

bob012345 and erobz
jbriggs444 said:
In my opinion, that would be overkill. We have an original poster here who was concerned about where the kinetic energy loss was for water pouring over the rim and splashing onto the growing pool of water below. [No offense meant to the original poster. It was a reasonable concern].

But here we are worrying about siphons, cross-sectional areas of pipes and kinetic energy of flows. The next thing you know, we'll be looking at relative humidity and barometric pressure so that we can accurately account for the buoyancy of the water in air and asking about latitude so that we know how much higher the water is at the rim of the tank than at the center due to centrifugal force and the rotation of the earth.
Sure, but why stop there? I'd imagine when we consider everything it becomes impossible to solve this problem.

What I'm trying to say (perhaps ineffectively), is this could also set up the idea that this is the work done to fill a tank. Is it all that damaging, probably not, but it could cause a moment of "wait a second, I remember the work done to fill a tank being..." when they examine filling a tank under more realistic conditions later on.

erobz said:
Sure, but why stop there?
Because one has to stop somewhere. One wants to get at least one principle learned before going on to the next. Physics and engineering are not about ultimate truth. They are about approximations that work.

jbriggs444 said:
Because one has to stop somewhere. One wants to get at least one principle learned before going on to the next. Physics and engineering are not about ultimate truth. They are about approximations that work.
Yeah, I completely agree. Understanding the assumptions is important too. The fact that a pipe is used in part B to fill the tank, which is part A, shouldn't be completely ignored without just cause given the context of the problem.

Personally, if was sitting in class and that point was glossed over without concern, that would bother me.