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Setting three way with outdoor sensor
Big Ed_4
Member Posts: 3,017
Number setting. Depends on the range of the mixing valve used . There should be a range listed on the valve. But you should just set it to disire temp from a temperture reading. That's why they just put numbers. For what you have now just run the boiler higher then what you want to set the low side heating and take a reading of the low supply pipe and set it for the max desired temperture.Some installers like to add a thermometer to the supply for convience.
Still need to set boiler for the highest load. Which is the high temperture radiation.
You can get fancy and install a second ODR and control a motorized three way valve let's say for large cast iron radiation.
With radiant I would install floor sensors to radiation.
Still need to set boiler for the highest load. Which is the high temperture radiation.
You can get fancy and install a second ODR and control a motorized three way valve let's say for large cast iron radiation.
With radiant I would install floor sensors to radiation.
There was an error rendering this rich post.
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Comments
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Three way
I'm using a mod -con with ODR, with one high temp and one low temp application. I'm am using a non-thermostatic, non motorized three way (ESBE) .
There are 1-10 settings, My linear thinking says setting the valve to 5 would give me 90* @ 180* boiler supply temperature? But that cant be right, how can I calculate and set a three way with boiler outdoor sensor.
Thanks0 -
Right Math, Wrong Temp
The manual 3 way valve will do as you say, 5 will give you approximately 50% of the flow. If on start up, the boiler puts out 180 F and your return before the valve is 50 F, your return to the boiler would be 115 F (180-50=130/2=65+50= 115 F.
After running for a bit, if your return temperature before the valve rises to 100 F, the return to the boiler would be 140 F (180-100=80/2=40=140.
If your return before the valve remains 100, but your boiler output drops to 140, the return after the valve would be 120 F. (140-100=40/2=20+100=120 F)0
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