[RESOLVED] Draining the Pool - Serious

This is a serious question so please don't add the normal chit chat.

The water in my pool needs refershing. Rather than doing it all at once which can result in problems with the ground under the pool rising, I plan on draining 25% on each of 4 consecutive months. I know that the result will be less than a 100% replacement because the 2nd through 4th times some of the "new" water will be taken out. I'm having difficulty calculating how much "old" water will be left when I'm done. Can you help me figure it out?

Quote:

---

Rather than doing it all at once which can result in problems with the ground under the pool rising

---

Could you explain that a bit? I've never heard about that and it sounds interesting.

As for the water, by some rough calculations you will end up with 31% of 'old' water.

EDIT: After a fifth iteration of replacing 25% you will end up with about 25% of old water. Sixth will leave you with about 19% of old. On the other hand, if you do two iterations, removing 50% of water in each, you would end up with 25% of old water. So, the expense in doing it by draining a quarter (5 iterations) is a cost of 25% of water than if you were to drain half and half, to produce the same result of having 25% of old water at the end.

Well a US gallon of water weighs 8.35 pounds, and assuming I have 20,000 gallons in the pool, that's 167,000 pounds bearing down on the underlying soil. I believe that if that weight were removed all at once there would be some rebound which might push the pool shell up.

BTW as they say in US schools could you please "show your work"? In otherwords what did your calculation look like?

Well my method is pretty simple but I think fairly accurate. I drew four squares and added four circles/dots to each. Squares represent the water and dots represent old water. When you remove one square and add a clean one, you are left with 3 squares with 4 dots each and one empty. Difusion will redistribute the 12 dots evenly so you will have 3 dots in each of four squares. And I repeat that. First iteration is cleanly distributed, after that you get a bit of fractions. But after four iterations you end up with around 5 dots in total. You had 16 at the beginning so that's about 31% of the total.

That's an awesome solution. I don't think I would have come up with that in a week.

The issue of rebound can be very serious, though I haven't heard about it in that context. If you have an in-ground pool with a very shallow water table, as some of our hatchery raceways are, emptying the pool can cause the pool itself to float right up out of the ground.

Another way of approaching this problem would be to assume that you have 20,000 units of gunk in the pool. It doesn't matter what the gunk is, nor does it matter what the units are, and I used 20,000 so that some of the math starts out easier. Therefore, each gallon has 20,000/20,000 units per gallon, or 1 unit per gallon. When you take out 5,000 gallons, you are taking out 5,000 * (20,000/20,000) units of gunk, or 5,000 units of gunk. That leaves 15,000 units of gunk in 20,000 units of water. You can repeat the same equation over and over for each serial dilution that you perform. Of course, this is the same as what the badger was doing more visually.

Ok this is way out there but this is what I would do for totally clean water:

1: Get a huge roll (many rolls) of waterproof bin liners.
2: Fill a few with fresh water.
3: Get a pump.
4: Have the pump mount higher than the natural level of the water.
5: Start putting water filled bags in the pool.
6: Displacement raises the level to the pump.
7: Keep adding bags of water until the pool is empty (or to an acceptable level).
8: Pop the bags and take them out.

Granted I am a hygiene freak but it would be the easiest way to get rid of the maximum amount of bad water.

Quote:

---

Originally Posted by baja_yu

Well my method is pretty simple but I think fairly accurate. I drew four squares and added four circles/dots to each. Squares represent the water and dots represent old water. When you remove one square and add a clean one, you are left with 3 squares with 4 dots each and one empty. Difusion will redistribute the 12 dots evenly so you will have 3 dots in each of four squares. And I repeat that. First iteration is cleanly distributed, after that you get a bit of fractions. But after four iterations you end up with around 5 dots in total. You had 16 at the beginning so that's about 31% of the total.

---

Great solution, thanks. I owe you a rep.

This is a stupid question, but do you not have a sewer available to you? Maybe I was just lucky but I had access to a drain behind my house. When we changed the earth in the pump or drained some water out of the pool to close it down, we always ran a hose down the sewer and were done with it.

I miss the luxury of having a pool, maintenance nightmares however I am glad to be rid of.

Draining that much chlorine into a sewer is not a great solution, depending on where that sewer water goes. You might consider looking into using something to neutralize the chlorine, such as sodium thiosulfate.

Quote:

---

Originally Posted by Shaggy Hiker

Draining that much chlorine into a sewer is not a great solution, depending on where that sewer water goes. You might consider looking into using something to neutralize the chlorine, such as sodium thiosulfate.

---

I have a "pool guy" and he's going to help me with the chlorine problem.

Just to chip in, these two methods are exactly equivalent:

1) Take out some old water, put in new. Mix thoroughly. Repeat.

2) Take out 25% of old water from pool and 25% of new water from separate pool. Put replacement amount of new water back in separate pool. Repeat. Tip all the new water in at the end.


Basically, every time you take out 25% of the water, you're taking out 25% of the existing dirty water and 25% of the existing clean water, assuming they are completely mixed together, and that's the same as just taking 25% out each time and not mixing them at all.

Summary, each time you are left with 75% of the previous amount of dirty water.

And... 0.75 x 0.75 x 0.75 x 0.75 = 0.75 ^ 4 = .......... 31.64%

Quote:

---

Originally Posted by zaza

Just to chip in, these two methods are exactly equivalent:

1) Take out some old water, put in new. Mix thoroughly. Repeat.

2) Take out 25% of old water from pool and 25% of new water from separate pool. Put replacement amount of new water back in separate pool. Repeat. Tip all the new water in at the end.


Basically, every time you take out 25% of the water, you're taking out 25% of the existing dirty water and 25% of the existing clean water, assuming they are completely mixed together, and that's the same as just taking 25% out each time and not mixing them at all.

Summary, each time you are left with 75% of the previous amount of dirty water.

And... 0.75 x 0.75 x 0.75 x 0.75 = 0.75 ^ 4 = .......... 31.64%

---

Wow, I was making it way to complicated.