Missing Potential Energy?

The average hite iz still haf a tank, not a quarter tank.

Quote: "The average hite iz still haf a tank, not a quarter tank."

No-ways! The two tanks are half filled, so the center of mass of the water is at quarter tank level!

The average height is still 1/2 of a tank:

1st case: 1 is full, 0 is empty, and 1/2 is half full, a variable added to the numerator could be used for the height of the tank in both cases

(1 + 0)/2 = 1/2 average height of water

2nd case:

(1/2 + 1/2)/2 = 1/2 average height of water

So no missing potential energy. Was this a trick question?

I think the problem is a little different.

Imagine two blocks stacked on top of each other sitting on a table. As a unit (both blocks glued together) their potential energy can be thought of as the height of the center of mass over some floor. As individual blocks, the top block has a higher potential energy than the lower block it sits on. If you take the top block and sit it on the table next to the lower block, both blocks are at the same potential energy. As a system (glue both blocks together side-to-side) the center of mass is lower than the initial state.

By moving the top block to a lower height you have changed some of the top block's potential energy into kinetic energy, but not all of it.

The same thing is true with the water, but instead of two blocks you have billions of water molecules, each with a specific potential energy at any instant in time. When you lower the water level relative to the ground (whatever that is in this experiment) you must lower the potential energy of the system relative to that ground. In the example cited by Jorrie, he did that by moving water. Obviously, moving mass implies kinetic energy and that energy must come from somewhere, which was a portion of the stored potential energy of the initial state of the water.

Anonymous Hero,

Thanks for the breakdown, as an engineer, when I saw the weight x height, I immediately started thinking in feet of head.

Let me see if I have this right. PE_final = PE_initial = PE_tank1 + PE_tank2

PE_initial = 1/2 height

PE_tank1 = 1/4 height

PE_tank2 = 1/4 height

PE_final = 1/2 height = 1/4 height + 1/4 height

Is that right?

Err, not quite. You physically move some of the water down in height, so you lower its PE.

Imagine moving a bucket of water from the top of a hill to some point half way down the hill. The potential energy is lowered because you reduced the height.

Jorrie essentially takes a tall bucket and makes it wider, so the water level falls and so does the center of the water's mass. That means in the final state the total water of the system is at a lower potential energy.

However, the act of moving water implies kinetic energy. Some of the water moved via gravity (not pumped) from a higher potential energy state (higer in altitude) to a lower state (lower in altitude) by flowing into another tank.

While the total volume of water remains the same, it is spread out more horizontally among two tanks so that its center of mass is lower to the earth or the ground than when it was in only one tank.

The center of mass is a good way to average the total mass's height and so calculate the potential energy (height X mass). The PE must change, but the conservation of energy says that we must account for the change. Since this is a physical process where water flows or moves, the "missing" energy is really just kinetic energy.

When you do the formula the final PE must equal the initial PE minus the KE expended.

Does that make sense?

Anonymous Hero:

You are doing the right things, bar one. The question refers to ". . . the water levels stabilize . . .", so what kinetic energy are you referring to?

Huh? Have you not moved water from one tank at a higher potential energy or PE to another tank where the water that has moved now has a lower PE? Even if you wait for the system to equalize and there is no further exchange of water between the two tanks you have an initial and final state and work was done to get to the final state, which has a lower PE.

That must imply kinetic energy. It is no different than a rock at the top of the hill. It has a specific PE with respect to the botom of the hill. If the rock starts rolling (assuming it gathers no moss) downhill we convert some of that PE into kinetic energy. If the rock stops halfway down the hill it will have half the PE it did before the roll. The delta between the two states in PE must equal the kinetic energy (ignoring heat, etc.).

The law of conservation of energy will be obeyed. Unlike cival law, there is no adjuducation phase for the laws of physics, you go straight to the penalty phase. ;-)

Even if you pump the water, the energy used to move the water will at least equal the difference in PE plus some inefficiency loss for the pump and motor.

Quoting Hero: "If the rock stops halfway down the hill it will have half the PE it did before the roll. The delta between the two states in PE must equal the kinetic energy (ignoring heat, etc.)."

But if the rock has stopped at the halfway point, it has no kinetic energy relative to the top or the bottom! So where is the other half of the original energy? Maybe it's time to call Sherlock and Watson!

Sherlock was an observationalist. Watson was a medical doctor. All we need is a physics professor that teaches 101.

The "missing" energy was expended in the form of kinetic energy to produce work.

Let's go back to the water example. You can look at it two ways and the equations are the same, just the sense would be different. In the example you cited originally half the water in tank 1 is moved to tank 2. That transfer requires energy. It doesn't matter if that energy is gravity, a water pump, or the transporter on the Star Ship Enterprise. The minimal amount of energy to do that transfer is the delta in potential energy between the system's initial and final states. Anything less would be a candidate for perpetual motion! We should know this as the conservation of energy.

In the case where gravity equalizes the water level in both tanks, there is a net work performed. Water flows from tank 1 to tank 2. The volume of water that transfers by gravity moves from a higher altitude referenced to the tank bottoms to a lower altitude. It doesn't matter which water molecules transfer, the end effect is the same. During the point in time that the water moves, the potential energy of the static water is converted to kinetic energy (like a rock rolling down a hill (or Jack)), which is the moving water. When the water and the two tanks equalize the kinetic energy has produced work. Now the work is done and it is Miller time (although I would prefer a good Stout), so the system is at rest and there is a potential energy that is in a lower state, which the delta between the initial PE and final PE of the system and that difference is very, very closely equivalent to the energy required to produce the work (moving the water minus any heat from friction).

Now, let's reverse the process. For the sake of simplicity I just received my Acme water pump from the UPS truck at lunch and this new pump is a humdinger because it is the first pump to be 100% efficient. Let's pump out all of the water from tank 2 into tank 1. What happened? Work was performed to pump the water, against gravity's better judgment, uphill, if you will. How much energy would be required to do that if we use our new Acme pump? Elementary! We only need to look at the difference in potential energy of the system before and after pumping the water. We need at least that much energy to vacate tank 2 and place it in tank1. In the real world our pumps are less than 100%, but thermodynamics is a 102 course.

In example one, we have a system before and after state that results in a net gain of kinetic energy (work) and a loss of potential energy. In example two we have the same thing, but the sign is reversed for the KE and PE. Curiously, both experiments expend and require the exact same amount of energy to change states from the initial state to the final state and back to the initial state. Additionally, how much work was done? Well the vector definition of work tells us it is none! That is okay because the equation balances out.

Speaking of work, that is just what I need to get back to!

Quoting Hero: "The "missing" energy was expended in the form of kinetic energy to produce work."

I think we just have a difference of opinion as to what is kinetic energy and what is work. I view kinetic energy purely as a mass moving at a certain speed relative to some reference frame (0.5mv²) and work as a force applied through a distance (fd).

According to my view, once an object is stationary in a reference system, it has no kinetic energy. Whatever work has been done to get it into that state, cannot be called kinetic energy.

OK, so where does this bring us? The water has half the original potential energy and zero kinetic energy. The other half of the energy must have gone into 'work', but where is it?

The result is the displacement of water. That is "where it is at".

I agree that the definition for KE is mass times velocity squared. During the water displacement phase where water is moving, the total energy (KE) of the moving water is the difference in PE between the initial and final states of the system.

A solid rocket engine could be considered a PE device. Once fired and exhausted it has a chemical PE of zero. However, during the transition from fully loaded to completely depleted it produced energy in the form of heat and combustion, which produced a net force and presumably some work, too!

I also view the problem you cited as a temporal one. It is not simply one state of existence then a final state of existence. Things happened from the initial state to the final state. It is a dynamic event. So, when the water begins to move, it has kinetic energy and the PE of the system correspondingly drops. I am not a math major, but if you were to sum the instantaneous KE for the duration of the transition from initial to final state (I think this is the integral or area under the curve?), that sum would be equal to the delta PE for the system.

So the bottom line is we know that there is no missing energy. The delta PE is simply the energy required to move half of tank 1's water from tank 1 to tank 2. This is not Bistro Mathematics. Somebody has to pay for the work done and that payment is withdrawn from the potential energy bank.

Sorry Hero, but I cannot quite agree with you when you say: "The delta PE is simply the energy required to move half of tank 1's water from tank 1 to tank 2."

There's a lot more to it, but let's wait and see what other people has to say.

The gravity G as the force acting over the 1st tank water volume is pushing the water UP in the 2nd tank until the gravity forces over the two water volumes are balanced pushing against each other with equal forces. Thereby the first G becoming 2 x 1/2 G but acting by two directions from now until a new unbalanced status will occure and proportionaly modifying the G's values in each tank. The said (the real) Missing Potential Energy in today's physics is still beyond our understanding anywise but not for long time.

OK Hero, maybe it is just semantics and the definition of 'work' that make us disagree. My take is that the lost potential energy was wasted in friction, noise and even a very slight heating of the water itself, which would soon have been lost anyway. Now you may call that 'work done', while I don't.

To me, 'work' was what was done when the water was lifted up into the originally full tank. If I then punched a hole into the full tank, the water going into the ground wastes all that 'work' again. But, I agree it may only be semantics!

Okay. I think the principle here is simply the conservation of energy in a closed system.

I also made an assumption that we will ignore secondary and other effects such as friction, air pressure, etc. because the nut of the question revolves around what principles are at work in your question. So, to that end, I assume a closed system

My take on it is simply a classical physics problem and the two main players are PE and KE. From Phys 101 we know that:

PE = hgm (Height * gravity (9.8 m/s^2) * mass)

KE = 1/2 * mV^2

When an object is held at a distance of h above the floor it will have a PE. If the object is dropped the PE is converted into KE. When the object strikes the floor (ignoring rebound) its velocity is at its maximum and its height is now zero as so is its PE.

The challenge question is essentially the same thing and must follow the laws of conservation of energy.

Hero, if you ignore friction, you are more or less at my post #23 above. Remember, what the OP asked is about the stabilized water situation.

If you do consider friction, but work with a larger, closed system, then there is no energy lost. The lost potential energy will go into added temperature of the closed system.

Your ball dropped from a height is not quite an equivalent problem to my stabilized water one. That is unless you wait until the ball has expended all its kinetic and potential energy and is lying stationary on the floor (a slightly heated ball in a slightly heated room!)

Yes, I had intended the system to completely stabilize. I thought that was where I was going! ;-)

Heat, sound, and other forms of energy radiation will be some of the byproducts dissipated in the challenge example. Everything still follows the law of conservation of energy, where energy is neither created nor destroyed, at least from a Newtonian perspective. In my mind that was the most important point I was trying to make.

Tell me more about that ACME water pump...I've got an idea......

Wiley Coyote

The lost PE was used in the KE to move the rock. Work=FxD(horiz and vertical)

Qutoing Shooter: "So no missing potential energy. Was this a trick question?"

No, it's not a "trick question" and I'm sorry, but you're wrong with your analysis. What happened to the height of the center of mass of the water?

There's no free lunch. The missing energy was manifest as the work required to lift half the water into the initially empty tank.

Bingo.

rcapper, you said: "The missing energy was manifest as the work required to lift half the water into the initially empty tank."

Consider the problem for a moment in a 'loss-less' scenario and in it's simplest form - say the two tanks are replaced with a U-tube, with all the water initially being held in the left arm of the tube and then released. Without losses, the water will oscillate between the two sides ad infinitum, just like a loss-less pendulum.

After half a cycle, all the water will be in the initially empty right hand arm, momentarily stationary. No work or energy has been expended to lift the water up there - potential energy has just been exchanged between the two arms, via kinetic energy. In practice, what makes the water eventually level out between the two arms and lose half of its potential energy? Friction losses, that's all I can think of!

"Ask an engineer the time, and he'll tell you how to build a watch."

Simple, you have moved half of the volume water to a lower state of potential energy (PE) by draining it into the second tank. Where did the "missing" PE go? It went to kinetic energy (KE).

PE_initial = PE_final + KE

It's like having a bag with two rocks hanging in a tree. Pull one rock out and set it on a lower limb. You lowered the effective PE of the system by converting some of the system's PE into KE.

This seems too elemental. What am I missing?

I agree with you, Anonymous Hero. I just had to put some numbers in for it to make sense. Like starting with 100 lbs of water at 10 ft, or 1000 ft. lbs, opening the valve which gives 100 lbs at 5 ft, or 500 ft. lbs.

Thanks for being patient with a hard headed engineer.

Wow, Bet you didn't expect all that Jorrie. Um: the water gets warmer! Randall

"Wow, Bet you didn't expect all that Jorrie. Um: the water gets warmer! Randall"

You're right Randall, I never thought for a one moment it would be this tricky! But the water heats up - and as Bill has said later on, a lot of BS has been stated as 'fact' in the meantime. Just one problem - his was part of it! Jorrie

I wonder if adding a small paddle wheel in the connecting pipe that turned a generator connected to a battery might help illustrate that work is being done by the transfer of half the water from the full tank to the empty one. The water flowing through the pipe would turn the wheel storing energy in the battery. Obviously the adds some resistance to the flow of water but it will still reach equilibrium and its obvious at that point that work was done. I don't think that the fact that we did not harness that energy some how doesn't mean energy wasn't used/wasted.

Shawn

Yep, Shawn, this way (generator and battery) one could recover some portion of the lost potential energy. This further complicates rcapper's solution, where the lost potential energy went into the lifting of the water in the originally empty tank - since we 'stole' some energy for the battery, there may not be enough left to lift that water!

You keep mentioning energy being "lost," but you really mean it was just "converted." IC engines "lose" energy, but that really means the energy is just converted to heat, which is useless to us. If raising the water level in the previously empty tank is a "useful" thing - maybe it allows a trapped boat to exit through a channel - then we might not consider the energy "wasted," but only "converted" to something useful.

Quoting HOKIEinSC: "You keep mentioning energy being "lost," but you really mean it was just "converted.""

I think you agree that if the energy was converted to heat and sound, it is 'lost' in the sense that "it is useless to us". Now this is exactly what happened to the lost potential energy of the water. The 'missing' potential energy was not used for "raising the water level in the previously empty tank". Check posts #23 and #29.

By the time that the water have lost that fraction of a degree increased temperature, the energy is really lost into the environment. It's time to get our heads around this one; otherwise this thread may run forever... And we already stand accused of 50+ posts on a 'nothing' problem!

of course its kinetic energy as you are gonna use it again if you had to pump it back into the first tank

Hello Jorrie

The energy difference is dissipated due to turbulence and friction caused by flow in the connecting pipe, and converted into heat, the majority of which goes into the water.

Quoting Codemaster: "The energy difference is dissipated due to turbulence and friction caused by flow in the connecting pipe, and converted into heat, the majority of which goes into the water."

Bingo!

I'm not sure about that.

If you add a turbine & generator, are you reducing the friction & turbulance? I dont think so.

I see the problem with the original post. You are refering to 2 different thingz:

1. The 2 tanks. ( 1 full + 1 empty = 2 haf full )

2. The mass uv water in tank #1, ignoring tank#2.

"Bingo" my foot! There's a serious lack of differences in energy, but there's no lack of BS. The average height of water is 1/2 h before and after the valve opens.

Hey Bill, I think you're wrong and Codemaster had it right.

So… are we going to redo the whole thing again? Not to be boring, maybe we can state it in terms of electricity (for electrical guys like myself).

Say we have one full battery and one identical 'flat' one, of the rechargeable type. We connect them in parallel and leave them. What electrical energy do we have left after a while? I think we all know the answer!!

Bill, when you say: "The average height of water is 1/2 h before and after the valve opens.", you might be right some microseconds after the valve opens, but wait for the water to stabilize - then think again!

You have a 1/2 mass x 1/2 height on each side. That gives 1/4 + 1/4 = 1/2 of the original total potential energy.

The potential energy is lost as heat or it raises the temperature of the water slightly if the tanks and piping are very well insulated.

Quoting Howetwo: "The potential energy is lost as heat or it raises the temperature of the water slightly if the tanks and piping are very well insulated."

Bingo2!

Jorrie, you say, "You have a 1/2 mass x 1/2 height on each side. That gives 1/4 + 1/4 = 1/2 of the original total potential energy."

The potential energy is not lost, and no significant amount of the potential energy is converted to heat. It is lost (used) to fill the other tank to a height equal to the source tank. That is, it is used to RAISE water from zero elevation in the second tank to 1/2 original height in the first tank.

If anyone else stated the same thing, I apologize. I didn't read everyones' posts thoroughly.

Hi Bill, read my post #23, where I 'prove' that the lost PE goes into friction, and see if it makes sense. Half the potential energy is lost just like ALL the potential energy of the water will be lost if you let the water run out onto the ground.

Jorrie

I'm no engineer, but I just don't understand why it seems when one tank is full you count it as having full PE(1). But when it's stabalized, and each are half full you count it at 1/4 PE, instead of 1/2 PE in each tank. Could you explain?

You lose 1/2 the mass and 1/2 the height of tank1, making PE, the product of mass and height, 1/4 of what it was. Tank2 eventually has the same PE as tank1, so the total PE is half of what it originally was.

You just have the same mass of water spread 'flatter' and thus lower, with less PE.

I'm kinda dissapointed it took 29 posts for that. Keep 'em comming Jorrie, I think we need the practice!

Wow! still going. I thought you were all done!

You've all still missed a rather large point. There is a rather large amount of work that has been done that no one has yet accounted for. First there is no such thing as an empty tank. If we assume the tanks are completely sealed at the starting positions one tank is full of water, the other is full of air. Opening the valve at the bottom results in some water transfer, a partial compression of the air in one tank, a partial vacuum in the other tank, and the syatem will stabilise but not at the same water column heights. The 'lost' potential energy of the water collumn is converted to the potential energy in the compressed and rarefied air. Assuming the tanks are open topped the situation does not change much. Instead of the 'lost' pe being converted to a different form of pe it is converted to the kinetic energy required to pump the displaced air from one tank to the other.

slo

The tanks could have an equalizer line between them for the air to travel or be open on top. The energy to lift the air up is about 850 times less than the stored energy lost as the water falls, because air is about 850 times less dense than water. If the tanks are 8 feet high and all the heat generated by the falling water level is spread evenly between the water in the two tanks, the water would be warmed by about .0026 degrees F.

Just to be pedantic I claim BINGO zero at post #20. Also I suspect that most (all?) of the additional heat would be in the "originally empty tank": so I make that 0.005144 °F warmer than the "originally full tank" (with an original head of 8 foot of water).

"Just to be pedantic I claim BINGO zero at post #20."

Hi Randall, sorry, I interpreted your statement wrong - I thought you were talking about the water getting warmer for ME!

Anyway, you can take half the honors - I think you would agree Codemaster's was a pretty good reply.

To complicate things further, europium (post #20) also issued a bingo, on a wrong answer from rcapper's post #17, if I recall correctly?

"0.005144 °F warmer"? Yep, I think you have it zipped up and that's good enough for the other half of the honors!

Regards, Jorrie

Quote: "If we assume the tanks are completely sealed at the starting positions one tank is full of water, the other is full of air."

I think it is pretty clear from the question and the stable state reached that the tanks are not sealed. The water would not have equalized in height in sealed tanks, as you have stated.

This is then a new problem. Maybe you can start a new thread on it?

I can't believe this question has attracted 50+ posts that quibble back and forth... Simple secondary school definitions provide the answer. 1) Conservation of Energy: Energy can not be created or destroyed it can only be converted to other forms. 2) Potential Energy: This is the energy an object has due to its position 3) Kinetic Energy: This is the energy an object has when it is moving 4) Work: Work is the force required to move an object through a distance, i.e. is this not moving 5) The liquid level in two vessels connected in parrallel will acheive the same height Obviously no energy was lost and in the end the sum of all the energies will be the same

From one Guest to another, who said: "I can't believe this question has attracted 50+ posts that quibble back and forth... Simple secondary school definitions provide the answer."

It apparently happens due to people who do not read the question properly - isn't that taught at secondary school as well? The OP simply stated/asked: "..so half of the original potential energy (weight x height) is now missing. How does this happen?"

It is not clear from your reply whether you agree with the OP's statement and have the correct answer, but if you don't, I suggest you consider going back to school!

Hi everyone. My apologies for dragging this out further but I am dissatisfied with all the answers so far. While I grant you the center of mass is now half in each tank I don't believe the PE is any less because while you originally had a PE of half the height of tank one and the water was released and stabilized to both tanks the original PE is still there albeit now stored in two tanks which cancel each other out as the original PE is now split into two equally opposing forces rather than one. If the two tanks are equal and the two half heights are equal then do they not both add up to the original half height of tank one?

Let's also take friction somewhat out of the equation. Instead of connecting the tanks together via a pipe let us slide them together. Let us imagine these tanks are connected now via a large door that when opened will allow the water from tank one to veritably rush unimpeded into tank two. Are we to believe that turbulence alone and friction of water molecules rubbing against each other will equal half of the original potential energy?

Even though there is no free lunch and each tank has now only a quarter of the potential energy of the first tank, only half of the first half was ever able to be used as the it was restored in tank two. What thinks you Jorrie? Can you set me straight here?

rollinshultz, you said: "Even though there is no free lunch and each tank has now only a quarter of the potential energy of the first tank, only half of the first half was ever able to be used as the it was restored in tank two."

I'm afraid this does not make sense. You seem to agree that "each tank has now only a quarter of the potential energy of the first tank", but then you don't accept that some potential energy was lost as heat and other forms of increase in entropy (relatively useless energy, like 0.005…. degrees F rise in water temperature).

This (somewhat debatably high) temperature increase should give us an indication of how little potential energy was actually converted or 'lost'. Real potential energy is measured from the center of the Earth. If we take the ½ or 1 meter, or whatever, of change in the average height of the water as a fraction of the radius of Earth, one would agree that the % is minuscule.

It is only when we, for convenience, take potential energy relative to some arbitrary datum like "the bottom of the tanks" or "mean sea level" that issues like "half of the potential energy is lost" surface!

If we would rephrase the original question as: "Originally, the water was at an average height of R + (half a tank), where R is the radius of the Earth. Now the same mass of water is at an average height of R + (quarter of a tank), so 0.00000001% (or whatever many zeros) of the original potential energy (weight x height) is now missing. How does this happen?"

Would that have been so problematic?

This 'missing energy' saga made me think of the Air Force Captain who wrote in the flight log: "No. 2 engine is missing" (meaning running uneven). The ground crew wrote in the log: "After a brief search, no. 2 engine was found fitted to the left wing".

... reminds me of a former (artist-type) partner, who insisted that a train jouney she'd been on was delayed for several hours because it had lost the gearbox. It had to be true, because the guard announced it over the PA. She reckons the time they were delayed was however long it took them to go back along the track to find it & then put it back in.

Nothing I can say will convince her of any alternative explaination for the 'train' of events.

Even more perplexing is to what gearbox could they have been referring? I know of no train that uses gears in its power train. There are gears driving the auxiliary alternator but those are fixed and not directly involved in the power train. Of course I am not intimately familiar with steam engine technology but maybe there was a period when diesels were geared. Even if that were to be the case what I would think of as loosing your gearbox would probably mean waiting for another engine to arrive to continue the journey. Any modern locomotive engine has electric motors, say six of them that pretty much fill the space between the pairs of wheels, and they are direct drive, no gearbox involved. How long ago was this train ride? Did the engineer loose the box he kept his gear in? Maybe it fell off the caboose? Maybe they did go back and find it.

Whatever the gearbox solution was, it looks like the 'lost potential energy' has now been found to every one's satisfaction - or shall we rather legalize it by: "above reasonable doubt"?

Re "lost" gearbox

Some locomotives used to have hydraulic drives that may have been referred to as gearboxes. A Swiss company, Voith was one maker. I believe they worked fine and you could run up against buffers with the engine flat out all day without any problems (apart from wasting a lot of diesel!). But they went out of use because they were less efficient than electric transmission.

Thanks!

Sorry, I worked on locomotive traction motors - they're definitely geared. There's no way you'd get the torque needed to start a mile-long freight train using direct drive motors.

So maybe my Ex was right all along!

I never knew that. Having only seen them from the outside, I suppose I made an erroneous assumption regarding what was inside. One of the things I love about this site is that almost without fail someone will point out wrong information. That's educational. Thanks.

Sorry Jorrie I am like a Ferret with a new toy, I just can't let go once I get a hold of it. While I agree with you on the PE being a function of radius of the earth plus height of tank where I disagree is on everything you do with the numbers afterwards.

I'll look at it two ways first: Before sharing the water.

PEtotal = PE-a or even PEtotal = PE-a + PE-b even though PE-b is zero.

After sharing PEtotal = PE-a + Pe-b where both are relatively equal. Where is the loss? Even a very small conversion of energy to heat during the equalization process is not actually a loss if you consider water density as a factor in the PE. The small heat change to the water carries with it an expansion of the said water and reduction of density and hence a slight probable difference in PE but be that as it may once the heat dissipated the water would condense to the original density and the PE would be unchanged.

So all I am saying is if we consider total PE there is no loss because it is always in the addition of the potential in both tanks. Would this fall under "conservation of energy"?

Hi rollinshultz. I'm not sure what you mean by PE-a and PE-b, but I think you still think incorrectly about PE.

If we choose the tanks' bottoms as a zero PE point and a tank's height is H, with the mass of water in a full tank M, then the original PE₁ = H/2 x M. After equalizing the water levels, the PE of the two sides combined is PE₂ = 2 x H/4 x M/2 = ½PE₁.

So in this reference frame, half of PE₁ is lost - how else do you get past the math?

Thanks guys,

I can honestly say that I spent a better part of an hour reading all the posts, And there is no telling how much energy be it potential or kinetic was wasted on that action.

Peace And Keep at it.

J.