So - I've been digging into the cooling side of things. Specifically, because my argon is still having thermal shutdown issues when running above 20mw.
I currently have a 6" and an 8" inline duct booster fan in series. The 6" pulls 250cfm in free air, and the 8" pulls 500cfm in free air.
The spec for the argon states 140CFm in free air.... BUT it also says "with a pressure head of 0.95" of H2O"... uhhhmmm.... yeah... Hey JDSU - 0.95" of water does not equal free air...
To get a blower that will pull 140cfm at 0.95" of water takes quite a substantial motor. The closest I have found so far is a Dayton 1TD2 model that pulls around 500cfm in free air and around 150 at 1" of water (interpolated form a cruddy spec sheet graph). That model runs about $150 or there abouts. There is a local place that sells a function equivalent blower for $110. It is made by ROTOM and is model R7-RB445.
I'll dig in a bit more and refresh myself on my engineering fluid theory (its been about 15 years since I had fluids) and I'll see if I cant come up with some real numbers for what will work. HVAC ducts, which are easily adaptable to this use, come in 4", 6" and 8" sizes... so I want to see what the cfm / pressure equivalent in the 6" duct are to the 4" (the laser head tube is about 4" diameter".
Stay tuned...
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I forgot to ask my question!!
What specific models of fans are you guys using and how long can you run at max output?
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Further update...
There are a few ways to calculate the relative pressure / CFM in different ducts. Using the Bernoulli equation you can assume that if you keep one item constant, then the other variables can be calculated.
In our case, JDS specifies 0.95" of h20 as the pressure head. So, we will hold that constant and then do some calculations:
For a Dayton 1TDT2 fan, the output area is approximately 4.2" x 5", so its area is approx 21 in^2.
Lets call P1 the pressure at that output, and A1 and V1 also correspond to that output on the Dayton fan.
Lets then call P2, A1 and V2 the corresponding values at the argon laser output.
Keeping "P" constant and assuming the continuity equation is in effect (i.e. there is no density change or height change in the ductwork);
A1(V1) = A2(V2)
Assuming we are going to use the 4" exit at the argon head (A2 = 12.56 in^2), what is the CFM that this fan will give us for a 0.95" h2o pressure?
The specifications for the fan show that at approx 1" h20, the CFM is approx. 180CFM.
Therefore:
V2=((A1)(V1)/A2)
V2 = ((21)(180))/12.56
V2 = 300.95CFM
So - we are approximately double the JDSU specification for required cooling!
Now - I know what your saying - "What about the friction loss in the ductwork?"
I did some digging on that as well... assuming the roughest ductwork imaginable (i.e. made of rough concrete), and assuming your duct is 3 feet long, your pressure drop is only 2.229 Pascals, or 0.0089" H2O.... Negligible...
Please feel free to correct me if any of this is wrong... its only been about 15-20 years since I did any fluids and I had to look most of this up!
Citations:
http://www.electricmotorwarehouse.com/Dayton/blower_specs/ITDT2.pdf
https://en.wikipedia.org/wiki/Bernoulli's_principle
http://www.cibsejournal.com/cpd/2011-08/
http://www.princeton.edu/~asmits/Bicycle_web/continuity.html
http://www.engineeringtoolbox.com/colebrook-equation-d_1031.html
https://en.wikipedia.org/wiki/Colebrook_equation#Turbulent_flow_in_smooth_conduits
http://www.engineeringtoolbox.com/darcy-weisbach-equation-d_646.html
http://www.engineeringtoolbox.com/air-temperature-pressure-density-d_771.html
http://www.convertunits.com/from/in+H2O/to/pascal
I currently have a 6" and an 8" inline duct booster fan in series. The 6" pulls 250cfm in free air, and the 8" pulls 500cfm in free air.
The spec for the argon states 140CFm in free air.... BUT it also says "with a pressure head of 0.95" of H2O"... uhhhmmm.... yeah... Hey JDSU - 0.95" of water does not equal free air...
To get a blower that will pull 140cfm at 0.95" of water takes quite a substantial motor. The closest I have found so far is a Dayton 1TD2 model that pulls around 500cfm in free air and around 150 at 1" of water (interpolated form a cruddy spec sheet graph). That model runs about $150 or there abouts. There is a local place that sells a function equivalent blower for $110. It is made by ROTOM and is model R7-RB445.
I'll dig in a bit more and refresh myself on my engineering fluid theory (its been about 15 years since I had fluids) and I'll see if I cant come up with some real numbers for what will work. HVAC ducts, which are easily adaptable to this use, come in 4", 6" and 8" sizes... so I want to see what the cfm / pressure equivalent in the 6" duct are to the 4" (the laser head tube is about 4" diameter".
Stay tuned...
------------
I forgot to ask my question!!
What specific models of fans are you guys using and how long can you run at max output?
-------------
Further update...
There are a few ways to calculate the relative pressure / CFM in different ducts. Using the Bernoulli equation you can assume that if you keep one item constant, then the other variables can be calculated.
In our case, JDS specifies 0.95" of h20 as the pressure head. So, we will hold that constant and then do some calculations:
For a Dayton 1TDT2 fan, the output area is approximately 4.2" x 5", so its area is approx 21 in^2.
Lets call P1 the pressure at that output, and A1 and V1 also correspond to that output on the Dayton fan.
Lets then call P2, A1 and V2 the corresponding values at the argon laser output.
Keeping "P" constant and assuming the continuity equation is in effect (i.e. there is no density change or height change in the ductwork);
A1(V1) = A2(V2)
Assuming we are going to use the 4" exit at the argon head (A2 = 12.56 in^2), what is the CFM that this fan will give us for a 0.95" h2o pressure?
The specifications for the fan show that at approx 1" h20, the CFM is approx. 180CFM.
Therefore:
V2=((A1)(V1)/A2)
V2 = ((21)(180))/12.56
V2 = 300.95CFM
So - we are approximately double the JDSU specification for required cooling!
Now - I know what your saying - "What about the friction loss in the ductwork?"
I did some digging on that as well... assuming the roughest ductwork imaginable (i.e. made of rough concrete), and assuming your duct is 3 feet long, your pressure drop is only 2.229 Pascals, or 0.0089" H2O.... Negligible...
Please feel free to correct me if any of this is wrong... its only been about 15-20 years since I did any fluids and I had to look most of this up!
Citations:
http://www.electricmotorwarehouse.com/Dayton/blower_specs/ITDT2.pdf
https://en.wikipedia.org/wiki/Bernoulli's_principle
http://www.cibsejournal.com/cpd/2011-08/
http://www.princeton.edu/~asmits/Bicycle_web/continuity.html
http://www.engineeringtoolbox.com/colebrook-equation-d_1031.html
https://en.wikipedia.org/wiki/Colebrook_equation#Turbulent_flow_in_smooth_conduits
http://www.engineeringtoolbox.com/darcy-weisbach-equation-d_646.html
http://www.engineeringtoolbox.com/air-temperature-pressure-density-d_771.html
http://www.convertunits.com/from/in+H2O/to/pascal
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