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warmmilk
09-25-2013, 08:57 AM
So I have a theory on how to control pump speed on my next build, let me know if this will cause any issues.

Get this pump:
http://www.performance-pcs.com/catalog/index.php?main_page=product_info&cPath=59_201&products_id=21689

use this molex to 3pin adaptor:
http://www.performance-pcs.com/catalog/index.php?main_page=product_info&products_id=608

to plug it into this fan controller (30W per channel):
http://www.frozencpu.com/products/11435/bus-230/Lamptron_30W_-_4_Channel_Aluminum_Rheobus_w_Multi_Color_Backlit_ LED_-_Version_2_-_Black_FC-5V2_w_Brighter_Display.html?tl=g47c17s239


and for bonus points, I can add the rpm monitoring pin to the adaptor...

will this work? will it cause any issues? is 30W enough? any issues from tiny 3 pin connector instead of full size molex?

I'm planning on filling the other 3 channels with 8 more fans (3 Noctua NF-A14 FLX, 3 NB eloop, 2 Noctua NF-F12 respectively) but all running really slow

ying
09-25-2013, 01:28 PM
I think this will work :D

SwiFt_wayno
10-22-2013, 11:01 PM
You can get pump controllers specifically for this if you wanted, here is an example ----> http://www.pccasegear.com/index.php?main_page=product_info&products_id=18293&cPath=1285

other than that your idea should work just fine :)
EDIT: just noticed it was the 450 it needs 35w to start up

Makymaco
10-24-2013, 10:55 AM
Be carefull, some pumps have a hard time starting up with standard fan controler.

I don't know how this one will react.

I had the same idea as you but it didn't work well with ddc.

You may have not the possibility to lower the volts much

PWM pump are better in this aspect.

Or use a pump integrating directly this feature.

Or use a fan controller wich can deliver more current for starting up the pump

Makymaco
10-24-2013, 11:00 AM
In the description : Startup Voltage: 9 to 13.2 VDC

You won't be able to do 24V with your fan controller, only 12V

ryan92084
10-24-2013, 01:28 PM
the http://www.lamptron.com/2013/02/05/introducing-the-lamptron-cw611/ would likely be the better (but more expensive) choice as it is designed to do just that.