I have a 700lhr pump, 16/22 piping, 12/16 piping and hose clamps. I will need a clear cone approx 100mm in diameter 250mm in length, check valve, Sand (standardise the grain size by using a 1mm sieve (conviently its the same size as mossy proof window/door covers)), way to attach cone to pump, lubricant to seal rubber grommets.
Maths bit
Simple key to my algebra:
x = unknown value or multiply
r = radius
a = area
d = diameter
Work through of problem:
So I have found that I need 1.13m/min flow to lift 1mm sand which is standard medium sized blasting sand. I also know 2000 litres per hour lifts 0.033 cubic metres a minute of sand, so I converted this to 650L/h as an understatement of my pump to account for pipe resistance for the flow. So if 650 = 2000/x then 2000/650 = x, x = 3.0769230769231, so if I apply this to the original known value 2000/x = 0.033/x this gives me 0.010725 cubic metres a minute is lifted by my pump.
So combine this with the water velocity required to lift the sand we get 0.010725 x 1.13 = 0.01211925 so to suspend the sand my cone must have a cross sectional area at the bottom of less then 0.01211925 cubic metres.
So to convert this to a circle I used a=pi r^2, a/pi = 0.00385767709 then re transferred to the previous equation 0.00385767709^(1/2) = r, r = 0.0621102012, d = 2(0.0621102012), d = 0.124220402 x 100 (this value is in m so convert to cm), d = 12.4220402cm so the base of my cone must be smaller then this value for sufficient lift and wider then this value to allow for sufficient dropping of the particles so that they are not blown out of the top.
In the article I got the original data from for flow rates and cubic metres moved by 2000l/h pump, it states the filter built will have sufficient surface area to provide bacteria to run a 500-1000L tank. Following on from this 750L / (2000/650) = 243.75L filtration, bare in mind my tank is 60 Litres in total and I am already currently running 2 fluval 1+, interpet pf1 plus my diy filter/moss wall. That is one hell of a lot of filtration I might have to pump oxygen in to keep levels high enough for fish to breathe.
Maths bit
Simple key to my algebra:
x = unknown value or multiply
r = radius
a = area
d = diameter
Work through of problem:
So I have found that I need 1.13m/min flow to lift 1mm sand which is standard medium sized blasting sand. I also know 2000 litres per hour lifts 0.033 cubic metres a minute of sand, so I converted this to 650L/h as an understatement of my pump to account for pipe resistance for the flow. So if 650 = 2000/x then 2000/650 = x, x = 3.0769230769231, so if I apply this to the original known value 2000/x = 0.033/x this gives me 0.010725 cubic metres a minute is lifted by my pump.
So combine this with the water velocity required to lift the sand we get 0.010725 x 1.13 = 0.01211925 so to suspend the sand my cone must have a cross sectional area at the bottom of less then 0.01211925 cubic metres.
So to convert this to a circle I used a=pi r^2, a/pi = 0.00385767709 then re transferred to the previous equation 0.00385767709^(1/2) = r, r = 0.0621102012, d = 2(0.0621102012), d = 0.124220402 x 100 (this value is in m so convert to cm), d = 12.4220402cm so the base of my cone must be smaller then this value for sufficient lift and wider then this value to allow for sufficient dropping of the particles so that they are not blown out of the top.
In the article I got the original data from for flow rates and cubic metres moved by 2000l/h pump, it states the filter built will have sufficient surface area to provide bacteria to run a 500-1000L tank. Following on from this 750L / (2000/650) = 243.75L filtration, bare in mind my tank is 60 Litres in total and I am already currently running 2 fluval 1+, interpet pf1 plus my diy filter/moss wall. That is one hell of a lot of filtration I might have to pump oxygen in to keep levels high enough for fish to breathe.
