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Retirement and forum shutdown (17 Jan 2022)

Hi,

John Howell who has managed the forum for years is getting on and wishes to retire from the role of managing it.
Over the years, he has managed the forum through good days and bad days and he has always been fair.
He has managed to bring his passion for fish keeping to the forum and keep it going for so long.

I wish to thank John for his hard work in keeping the forum going.

With John wishing to "retire" from the role of managing the forum and the forum receiving very little traffic, I think we must agree that forum has come to a natural conclusion and it's time to put it to rest.

I am proposing that the forum be made read-only from March 2022 onwards and that no new users or content be created. The website is still registered for several more years, so the content will still be accessible but no new topics or replies will be allowed.

If there is interest from the ITFS or other fish keeping clubs, we may redirect traffic to them or to a Facebook group but will not actively manage it.

I'd like to thank everyone over the years who helped with forum, posted a reply, started a new topic, ask a question and helped a newbie in fish keeping. And thank you to the sponsors who helped us along the away. Hopefully it made the hobby stronger.

I'd especially like to thank John Howell and Valerie Rousseau for all of their contributions, without them the forum would have never been has successful.

Thank you
Darragh Sherwin

math geek required..calculate change in water temp

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19 Oct 2010 20:57 - 19 Oct 2010 21:03 #1 by Damian_Ireland (Damian_Ireland)
any math geeks on here. trying to work out an equation to find out the following
i have a 900l tank which is at 29oC. I have a water container with 200L of water at 11oC.
I will empty 200L out of my tank so it will then have 700L.
If I then put in 200l of water at 11oC what will the drop in my tank water temp be..
I am thinking i will be about 5oC
difference in temp of water divided by the percentage difference in volume
Last edit: 19 Oct 2010 21:03 by Damian_Ireland (Damian_Ireland).

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19 Oct 2010 21:31 #2 by murph (Tony Murphy)
Drop will be 2.57 degrees. However, the heat generated by pumps, lights and heater may reduce this. Depends whether you chuck it all in or add it slowly. Also, rock and sand/gravel has a higher heat capacity than water. these will contribute to a lower than expected temp drop if present.

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19 Oct 2010 21:40 - 19 Oct 2010 21:47 #3 by Damian_Ireland (Damian_Ireland)
equation please murph. dont think thats correct. 200l is 28% of 700L and the difference in the temp between the 2 is 18oC.
So would the change not be equal to 28% of 18 ?
Last edit: 19 Oct 2010 21:47 by Damian_Ireland (Damian_Ireland).

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19 Oct 2010 21:51 #4 by murph (Tony Murphy)
Ignoring the fact that the heat capacity of water is temperature dependent (and definitely not linear), we will assume it is here.
You have 700l @ 29^ (too lazy to find kbd degree symbol!).
You are adding water @ 11^.
It will take 700l @ 11^ to drop the water 9^.
Therefore 200 l will drop the temp by (9/700)*200.

Or, for the pedantic, ((deltaT/2)/vol tank)*(vol added).

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19 Oct 2010 22:37 - 19 Oct 2010 22:41 #5 by Ma (mm mm)
Factor time too in there, the slower you add the lower the temp loss. You can experiment with litres instead of hundreds, 9 litres in a bucket, replace two with cooler water, to get an idea I guess.


Having te answer might help you qork it out?


Look up Newton, Laws of cooling, might help






Mark

Location D.11
Last edit: 19 Oct 2010 22:41 by Ma (mm mm).

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19 Oct 2010 23:33 #6 by igmillichip (ian millichip)
12:30 am; too many variables.......and energies are mass based, so we'd also need to consider not the volume of the water but the mass of the water a 11 and 29 C. Plus the other things mentioned eg kinetic energy of dropping one water into the other etc etc

at this time of night, I'd simply guess it to be roughly a drop of between 4 and 2.5 C (assuming that the tank glass does not contribute to the temp....and that will be difficult to not allow its involvment)

It would actually be quicker to do the addition and take the temperature.

@Mark....you're not going to drag me into this scientific question either. So there. I've got my database designer hat on tonight. :)

Irish Tropical Fish Society (ITFS) Member.

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20 Oct 2010 08:40 #7 by stretnik (stretnik)
Replied by stretnik (stretnik) on topic Re:math geek required..calculate change in water temp
In my opinion, impossible to achieve, while you are adding the Water both bodies are losing temperature.

Kev.

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