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RE: Aquatic Plants Digest V5 #51
- To: "APD" <aquatic-plants at actwin_com>
- Subject: RE: Aquatic Plants Digest V5 #51
- From: "Charles Kuehnl" <ckuehnl at mmcable_com>
- Date: Thu, 25 Apr 2002 14:18:31 -0500
- Disposition-notification-to: "Charles Kuehnl" <ckuehnl at mmcable_com>
- Importance: Normal
- In-reply-to: <200204241948.g3OJm1608062 at acme_actwin.com>
> Date: Wed, 24 Apr 2002 11:13:18 -0500
> From: Dan Dixon <dandixon at mac_com>
> Subject: spray bar
> > Hi list members, I don't post often.I read the list every day and
> > learned a lot from this list. Now I would like to add a spray bar to
> > planted tank. I have a125 gal tank with a 20 gal sump powered by a
> > driven 250 gal per hour pump. The piping is 11/4 in. I think that I
> > the bar to be 5 ft. it will fit between the return and the overflow
> > other end of the tank. The return is about 5 in below the water
> What I
> > need to know is should I make the holes in the spray bar different
> > all the size? I will have to decide what size holes that I will
> > there any formulas for this? Thanks for any help.
> > Blendia
> You can make them all the same size. Hydraulic pressure will force
> less the same volume through all the holes. Same way a garden
> hose works. With air it's a different matter, since air is more
> compressible, but water behaves hydraulically.
> I don't know of any hole size formulas, but basically you'll want to
> the total hole size of your spray bar (area of all the holes added
> together) at least the size of your return pipe's opening area.
I am not sure it works this way. I would think though that if the holes
are pretty small (a lot of holes). I would like to qualify this with it
has been over ten years and I am a EE type not a mechanics or physics
type although I used to work a lot with compressible flow and a little
with incompressible flow. I probably should look this up or call my
college roommate (a design ME) before saying anything. If no one else
posts and anyone really wants to know I would be glad to check.
> Assuming a return pipe diameter of 11/4ths inches:
> 1.25/2 = 0.625 (radius of pipe)
> 0.625 squared = 0.390625
> 0.390625 x pi (3.1415) = 1.227148437 square inches
Dan, I do not see where you got 1.25 from 11/4ths.
11/4" = 2.75"
Following your method, the radius would then be 11/4 * 1/2 = 11/8 or
Area = 1/2*pi*r^2 = 2.97 in2
This is a mighty big pipe. Blendia, are you sure about the pipe
diameter or was it perhaps 11/4ths circumference?
> If you drill one hole every 6 inches (10 holes total) you need holes
> an area of at least 0.1227148437 square inches each. Reversing the
> area formula we get a hole radius of 0.1976423537, or a diameter of
about > 0.4 of an inch. Use a 7/16 inch drill bit.
> If you drill one hole every 3 inches (4 per foot, 20 total) you need
> with a diameter of about 0.14 inch each, or about a 3/16 inch drill
> Etc, etc.
> Just always err on the high side (round up) and you won't restrict
> Dan Dixon
> PS. Someone be sure to check my math!!! :)
If my math is right, the rest of Dan's method looks correct but would
obviously need to be redone with the larger cross sectional area of the
pipe. Running the numbers in my head I think a rough estimate would be
about 5/16" for 20 holes to get the same total flow area. I would drill
more and follow Dan's method of calculation to determine the size (make