Originally Posted by defenestrate
that was going to be my next question... if i could keep the pwm for bk control and heating off the mash water in the bk, then use a PID in a cooler for the HLT.
You can do it one of two ways...
(Disclaimer: It is my preference to use one SSR for each leg of a 240 volt load)
You can use 2 SSRs for one element hookup, and use a switch to connect either
the PWM or
the PID to the SSR triggers. This way you flip the switch to "Manual" or "Auto" And plug the RIMs/HERMs element in for the mash, and then plug in the BK element for the boil.
Or you can use 4 SSRs, and hook the PWM to one pair to drive the BK element, and the other pair to the PID to drive a RIMs or HERMs. Essentially, the cost difference is in the extra parts. To double the outputs, you need to double the SSRs, wire, and 240v plugs.
Both are great options, and both are used by many here. Just be aware of your load - If you go with option 2 you can never use both at the same time on a 30 amp circuit. If you want to use both, you'll need to move up to a 50 amp service - Which requires heavier wire and components.
Volts X Amps = Watts
Watts / Volts = Amps
So 5500 / 240 = 22.9 Amps - Well inside your 30 amp service
5500(BK) + 2000(RIMs) = 7500 /240 = 31.25 amps - 30 amps isn't enough
2 X 5500 = 11000 / 240 = 45.83 amps - Time for a bigger breaker, 6
gauge wire instead of 10, yadda yadda yadda
The other thing to remember (In option 2, with a 30 amp service) is that you can't even run both elements at the same time at 50% power. Because of the way pulse width modulation works...
A 50% power setting at the PWM means 100% power to the element, 50% of the time. It's the same for a PID as well.
So if both the PID and the PWM are set for 50%, and they both fire (100%) at the same time, it's a 200% load on your system.
Edit: that's probably in the top 2 most used elements here - You'll be in good company.