Backyard Aquaponics
A place to discuss aquaponics

Hiya, my name's Jason and I'm new to this aquaponics forum. I've got a small system set up right now.. a 55 gallon barrel with two gold fish in there, the water is pumped out into a couple of pvp pipes that then run back into the barrel... very simple. Now that I have this small system running I want to upgrade just a bit. I'm about to start building a 300 gallon tank and I was going to hook up 150 gallons worth of growing medium to filter the water. I know it's nota 4x2x1 ratio but I figure if I go for lower stocking density it shouldnt be a problem... right? I was thinking I'd be alright up to .25lbs per 2 gallons of water instead of the 1lb per 2 gallons, sound reasonable?

Next question... I'm pumping the water out of the 300 gallon tank with a 500 gph pump.. I think it should be running at 450 gph at the head I'm going to have it pump up to. I'm pumping it out of the tank into 6 grow beds that are made from 55 gallons barrels cut sideways, so about 150 gallons of grow bed. I was going to have these all set up as seperate eb-flow beds using an autosiphon on each one. So... here's my main question... the water comes out of the pump through a 1" pipe and then I want to divert it into 6 seperate lines/hoses/whatever to feed into the grow beds... what sized pipe/hose should I use to make sure that whatever gets 'equal' distribution to all the grow beds? I was going to have the pump pump up to the 3' head I need then go into a 2" pipe to use as a manafold and then have all 6 hoses/pipe branch off at approximately the same spot... but what sized pipe/hose should I use?

Thanks for the read.

Jason

should be a divide math problem right?

forget the manifold to a degree you will need to have a valve on each to regulate the flow to each GB

Hi Jason, welcome on board. Good to see you are already planning an upgrade

Never one to turn down a math problem....

A(1in) = 1/4*pi in^2

So you need to divide that by 6. So each outlet pipe should have an area of 1/24*pi in^2. 1/24 = r^2; r ~= 0.2; diameter ~= .38. Closest English measurement would be 3/8" diameter.

You'll probably still need valves due to imperfections in exit height and turbulence in the T's and such.

not to sound to dense here but could you explain the equation?

A(1in) = 1/4*pi in ^2

i know finding the area of a circle is pi*radius^2, I'm not sure what you're doing with the 1/4...

Thanks

Welcome Jason..... don't know about the maths....

EB overcame a similar design problem by utilising a header tank/manifold as below......

Then as suggested you can utilise ball valve taps on each distribution line to equalize the flow...

This way you can also use cheap 19mm (3/4") irrigation tubing....

i'd go with thamos solution, makes sense to me. If you have issues then insert valves into each of the six legs and tweak.

or Area = pi*diameter ^2 /4 (the pipe sizes are the diameter measurements)

You've still got it Les!

Ah ... so... area of a 1" pipe would be .785" .... 3.14 * (1/2)^2 = 0.785 square inches

0.785/6 = 0.138 square inches.

so...

0.138 = pi*r^2
pi * .0138 = r^2
0.0417 = r^2
.2 = r
.4 = diameter

thanks for the math lesson

No problem