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PostPosted: Dec 21st, '10, 22:48 
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Hi All, i'm just doing a bit of planning & using my base data based on BYAP layouts, i'm trying to work out ratio of ex clay to water in a GB... ie how much water is required to fill a 2200X1200X300 GB if u put 500lt of expand Clay in the GB - (i note that BYAP 2 bed system allows 1,000L of ex clay) i'm supposing that a GB is approx 800L in volume?? therefore 800L of GB less 500L of Ex Clay means approx 300L of water to fill in the flood stage???? can someone just check i'm on the right track here - my heads just rolling about with figures & info :oops: cheers ST


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PostPosted: Dec 21st, '10, 23:16 
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The exact amount of water needed to flood a bed will vary with time and plant roots etc. However, when figuring sump size and water volumes needed. I usually round up and just figure you need 50% of the water. So if it is a 500 liter grow bed then you need at least 250 liters of sump capacity to flood it. If the grow bed is 1000 liters, then you need 500 liters of sump or excess fish tank capacity to flood it. I know that is an over estimation but you always need some excess water beyond what is needed to flood the beds since most pumps can get the last couple centimeters of water off the bottom of a tank without overheating.

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PostPosted: Dec 22nd, '10, 00:31 
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thanxs TCL thats the practical v's mathmatical answer i needed. makes me realise now how important the size of Ft is!!!!! start pulling off 500l out of 1,000l tank gets to the half full half empty point....hmm iguess thats where the second sump tank comes in handy to stabailise FT levels - must have read that dozen times but not till i'm drawing plans & doing calculations that i see how many variables there can be in one basic principal .....back to the drawing board, don't count sheep to sleep anymore, just try working out AP systems :) :) cheers ST


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PostPosted: Dec 22nd, '10, 00:49 
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The sphere of radius r has volume (4/3)Pi(r^2)
The smallest cube that can contain that sphere has volume of (2r)^2 = 8(r^2)

The volume of water surround the sphere = 8(r^2) - (4/3)Pi(r^2) = [ 8 -4/3 Pi] r^2
about 3.81 r^2

So the ratio of water : space(GB) = 3.81 : 8
about 1:2

This confirm TCL info.


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PostPosted: Dec 22nd, '10, 11:03 
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Working on our growbeds, the beds are probably around 800L in total volume. We put in 500L of media in the bed. To flood the bed to about an inch below the surface takes roughly 160-180L of water.

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PostPosted: Dec 22nd, '10, 19:04 
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Ao+ wrote:
The sphere of radius r has volume (4/3)Pi(r^2)
The smallest cube that can contain that sphere has volume of (2r)^2 = 8(r^2)

The volume of water surround the sphere = 8(r^2) - (4/3)Pi(r^2) = [ 8 -4/3 Pi] r^2
about 3.81 r^2

So the ratio of water : space(GB) = 3.81 : 8
about 1:2

This confirm TCL info.

change the power
The sphere of radius r has volume (4/3)Pi(r^3)
The smallest cube that can contain that sphere has volume of (2r)^3 = 8(r^3)

The volume of water surround the sphere = 8(r^2) - (4/3)Pi(r^3) = [ 8 -4/3 Pi] r^3
about 3.81 r^3


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PostPosted: Dec 23rd, '10, 06:14 
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From my experience with a BYAP 500L growbed you only need about 140L of water to flood the growbed, I know this because I measured the volume of water on the drain cycle with a 20L bucket. My growbed is filled with media to 30mm from top and I flood to 30mm below this and I drain with a bell siphon, also you can't get all the water out of the growbed during a drain cycle.

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PostPosted: Dec 23rd, '10, 11:48 
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thanxs all, thats exactly what i needed to know - more calculations now, this damned bug has got me.......list of jobs for MrsST over xmas now looking very doubtful :lol: :lol: will have to build a few proto type syphons :laughing3: :laughing3:


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