I was also taking a look at your blog about the relay board but if I understand correctly I would still have to replace the Triacs anyway, Is this correct?
Not necessarily. If the relay board is of ‘active high’ type, you just need to solder a wire from the gate of the triac (basically the shift register output) of each channel to the relay’s data pin. The blog post shows replacing triacs by transistors because the relay board I was using is of ‘active low’ type. So the transistors serve as inverters to reverse the logic.
It is also a little unclear for me how the relay board works as there seems to be more outputs then inputs.
Typically each relay channel has three pins on the input side: VCC (e.g. 5V), GND, and data pin (connected to microcontroller, or in this case, a shift register output pin); and three pins on the output side: NO (normally open), CO (change over), NC (normally closed). At resting position, CO and NC are connected; when relay is activated, CO and NO are connected. To put everything together: first, one wire from each solenoid should come together and go to the COM (common) terminal; then, the other wire of each solenoid can go into the relay NO pin. Finally, the CO pin of each relay is connected to GND. This way, when the relay is activated, it completes the circuit from 24V AC common wire -> solenoid -> ground (which is tired to the other wire of 24V AC). Hope this makes sense.