Yup i'd just pay the extra few dollars and get the 40A, it wont output much heat because it wont be getting taxed at all.A 240V 25A SSR would be the bare minimum you'd need for a 5500W element. For a 2500W element you could get away with a lot less. You'll only be pulling a little over 10A. But the price difference is pretty small and 40A SSRs are all over the place and frequently used by electric home brewers, so it might be just as easy and inexpensive to go that way. Doesn't hurt anything to have the extra capacity.
Just to elaborate on what has already been said: 2500 watts / 240 volts = 10.42 amps --> you need a SSR that can handle more than 10.42 amps. A 40 amp relay will do just fine and leave room for upgrades later.E-Brewers,
I can't remember what type of SSR I need, but I believe that to run a 240v 2500watt RIMS I need a 40AMP DC in/AC out using a SYL-2352 PID. Correct?
Not to raise a fuss, but is this a true statement? Seems to me that the heat produced by an SSR would be related to the current flowing through the relay, not the amperage rating of the relay.Yup i'd just pay the extra few dollars and get the 40A, it wont output much heat because it wont be getting taxed at all.
The heat dissipated by an SSR is equal to the current flowing thru it times the forward voltage drop within the SSR. A higher current rated SSR is usually going to have a slightly smaller forward voltage drop for the same current vs. a lower current rated SSR. The internal heat spreader and bonding may also be slightly better in order to be able to better dissipate the heat at high currents. Lastly, the junction cross section area may be slightly larger in order to keep the current density down at higher currents.Not to raise a fuss, but is this a true statement? Seems to me that the heat produced by an SSR would be related to the current flowing through the relay, not the amperage rating of the relay.