saltynay
Patron
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- 5,149
- Location
- West Coast, Singapore
I am building a fluid bed filter which basically provides the maximum surface area for bacterial colonisation which is chemical filtration for an aquarium. The shape of this will be a reverse cone and the sand needs to be kept in a constant state of suspension to allow for the optimum flow of water and surface area of the sand. The water pump I will be using is 670-700 litres per hour.
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.
Can you please check through that this is correct as I have a gut feeling that it is wrong somewhere.
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.
Can you please check through that this is correct as I have a gut feeling that it is wrong somewhere.