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SwissFlow 800 settings?

jaydee116

New member
Hello,

I am setting up a test bench to test water blocks for PC's. I have the CF633 and for now I am using it for temp logging. I just got a couple SwissFlow 800's one for review and one for my test bench. Anyway I got the thing hooked up to one of the CF633's fan headers and it is reading just fine. The only problem I have is deciphering the reading?

The pulse/L is 6000. Here are the basic specs: http://www.swissflow.com/en/SF800/Flow_Meter_Specifications.

What should I set the PPR to and then how do I calculate a accurate flow rate from the reading?

Thanks for any suggestions.
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jaydee116

New member
I would just like to know how to do the math to manually calculate the flow rate for now. If you can implement that into software in the future all the better.

Right now I have the PPR set to 2 and it gives me a reading of 15500.

This is how the meter works:
6000 pulses = 1 L
so 100 Hz = 100 pulses/sec = 1 LPM , 200 Hz = 2 LPM, etc.

What is the 15500 value the meter is giving me?

How do I do the math to convert that number to liters per minute given the meters specs (not by software just on paper)?
 

jaydee116

New member
Looks as if I got it figured out. The PPR should be set to one and then divide the value by 6000 to get LPM.

So thats 31000/6000 = 5.16LPM.

This is good. I can now log the temps and the flow rate. Hopefully you guys can come up with a plug in or whatever to convert it into LPM and then log it to a .txt file some day. Then people can show flow rate on their LCD to.
 

BizyB

New member
I'm having a 635 with the SCAB and the sf800. With this combination i can easily change the fan RPM divider and fill in the "6" in CC2 for the correct readings in LPM . If you're use CC2 you should be able to change that too. But then again i don't have a 633 to test it. alternatly You can connect it to your motherboard and use for example "speedfan" for monitoring. But be aware, some motherboards use a standard divider of 2. So in addition you have to divide it to 3 to get the correct readings. Give a message if you got the solution !

Greetingkz
 

svtran

New member
I am testing the SwissFlow 800 on our dispenser. The output is connected to either a dual binary counter (16-bit) or a microcontroller counter. The flow rate is 5 oz/sec. As I change the dispensing time from 3 to 5 seconds, etc... I expect the count and the volume to change. However, the ratio should still be constant. It is not...I have a feeling that I forgot something in my equation.

6100 pulses = 1 L

I weighed my liquid in grams.
At 5 seconds, my count is around 1320.
At 4 seconds, my count is around 1055.

Even though my liquid is in grams, shouldn't my count per grams be constant?
 

jc634

Administrator
Well, here goes...

5 oz/sec=8.8L/min. Good. Middle of the range for the SF800.

8.8L/min=53680 Pulses/min (6100 pulses/L) = 894 Pulses/sec.

At a constant rate of 5oz/sec, you should have 3578 pulses in 4 secs.

At a constant rate of 5oz/sec, you should have 4473 pulses in 5 secs.

So... I'm not sure what your 1055 and 1320 counts represent, however, your total count will change over time. Your controller logic needs to convert pulses over a time span to a rate/volume/mass depending on what you need to display (simplified):

(6100 Pulses/min) * (time span) * (L to desired output conversion factor)

I don't think I've answered your question very well. I'm simply not sure of your question, so above is my simple logic to perform a calculation.

JC
 

svtran

New member
Well, here goes...

5 oz/sec=8.8L/min. Good. Middle of the range for the SF800.

8.8L/min=53680 Pulses/min (6100 pulses/L) = 894 Pulses/sec.

At a constant rate of 5oz/sec, you should have 3578 pulses in 4 secs.

At a constant rate of 5oz/sec, you should have 4473 pulses in 5 secs.

So... I'm not sure what your 1055 and 1320 counts represent, however, your total count will change over time. Your controller logic needs to convert pulses over a time span to a rate/volume/mass depending on what you need to display (simplified):

(6100 Pulses/min) * (time span) * (L to desired output conversion factor)

I don't think I've answered your question very well. I'm simply not sure of your question, so above is my simple logic to perform a calculation.

JC
JC,
That was very helpful. I don't think my flow count is correct either for the 5 seconds or 4 seconds pour. I need to check my calculation and flow counter. I will inform you of my finding.
Regards,
ST
 
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