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I am currently designing a custom tank build and I have a simple question that requires a good bit of background to explain.
I am aiming for a turnover rate of 30-40 for a 75gal frag tank that I plan to make as my display.
Dimensions: LxWxH 38”x38”x12”
I am going to drill holes in the center of the bottom panel for a true bean-animal overflow tower in the center. (Some call it a tower overflow or a center overflow, but it is basically a square weir in the center)
I want to make this overflow as small as possible so that it does not distract from my display, while leaving enough room to install 3 bulkheads for the bean animal design.
Essentially, I used the inner diameter value of schedule 80 PVC for the following nominal pipe sizes: 3/4”, 1”, 1 1/4”, 1 1/2”, and 2”.
The formula I used to calculate flow rate:
V= CdA(2gh)^.5
where Cd was calculated using (Cc-Cv)
Cc as contraction coefficient over a sharp edge aperture (straight edge) (.62)
Cv as the velocity Coefficient of water (0.97)
These calculations are much easier to do using metric values, so I converted all of the PVC pipe ID square inch areas into square meters.
The result of volumetric flow for a single aperture of each nominal pipe size are as follows, considering I used 11” as my height value (because I want my display tank filled to 1” below the rim)
3/4”= 148.9 gal/h
1”= 241.2 gal/h
1 1/4”= 419.76 gal/h
1 1/2”= 787.32 gal/h
2”= 939.2 gal/h
FINALLY, my question. I am not really sure how the bean animal piping is setup; I know that 1 of them should have a full siphon, so I can assume the full amount of my calculated volumetric flow rate. One pipe must have a partial siphon, right? And the last is an unused emergency drain, right? If anyone can relate my rambling to the bean animal drain setup and explain how much flow rate I can assume from the pipes that are only partially siphoned, I’d really appreciate it.
Thanks for reading this, even if you aren’t a math geek! I have put a lot of time into this because it’s still a ways out of reach, but it’s my dream build and I want to make sure it’s perfect. Happy reefing!
I am aiming for a turnover rate of 30-40 for a 75gal frag tank that I plan to make as my display.
Dimensions: LxWxH 38”x38”x12”
I am going to drill holes in the center of the bottom panel for a true bean-animal overflow tower in the center. (Some call it a tower overflow or a center overflow, but it is basically a square weir in the center)
I want to make this overflow as small as possible so that it does not distract from my display, while leaving enough room to install 3 bulkheads for the bean animal design.
Essentially, I used the inner diameter value of schedule 80 PVC for the following nominal pipe sizes: 3/4”, 1”, 1 1/4”, 1 1/2”, and 2”.
The formula I used to calculate flow rate:
V= CdA(2gh)^.5
where Cd was calculated using (Cc-Cv)
Cc as contraction coefficient over a sharp edge aperture (straight edge) (.62)
Cv as the velocity Coefficient of water (0.97)
These calculations are much easier to do using metric values, so I converted all of the PVC pipe ID square inch areas into square meters.
The result of volumetric flow for a single aperture of each nominal pipe size are as follows, considering I used 11” as my height value (because I want my display tank filled to 1” below the rim)
3/4”= 148.9 gal/h
1”= 241.2 gal/h
1 1/4”= 419.76 gal/h
1 1/2”= 787.32 gal/h
2”= 939.2 gal/h
FINALLY, my question. I am not really sure how the bean animal piping is setup; I know that 1 of them should have a full siphon, so I can assume the full amount of my calculated volumetric flow rate. One pipe must have a partial siphon, right? And the last is an unused emergency drain, right? If anyone can relate my rambling to the bean animal drain setup and explain how much flow rate I can assume from the pipes that are only partially siphoned, I’d really appreciate it.
Thanks for reading this, even if you aren’t a math geek! I have put a lot of time into this because it’s still a ways out of reach, but it’s my dream build and I want to make sure it’s perfect. Happy reefing!