G'day brewers! I'm in the planning stages for a ground-up custom-built E-BIAB system. The stainless steel pot I'm looking at is 180qt (45 US gallons) or roughly 170L. I'm anticipating boil sizes to run around the 150L/160L area. My question is, would a single 240v/5500w element be enough to bring this volume to a complete boil? I'm also potentially looking at a 240v/6500w element as well, in case that step up would make the difference. Any thoughts or insight would be appreciated. Cheers!

Take a look at some of the commercial electric brew-house units. For a 1 barrel system, they require somewhere in the 50-60 amp range at 240, around 12,000 watts. Just curious as to what type of bag you will use and how you will lift it. A bag of wet grain in a 5 gallon BIAB system starts to get heavy.

I have a 30 gallon pot and I'm anticipating that a single 30A/240v/5500w heater element will just barely do my boils which will start in the 25-27 gallon range. You'll need a lot more to get your 45 gallon pot going.

There has got to be some maths/formula for this. You can easily get 30lt to the boil with a standard 10A 240v element so if your brewing at 5X that batch size you'd need to scale up by that amount which gives that 50A 240v (3 Phase ) element Bubba was talking about plausible. There's gotta be a liquid to amp / whatt equation? All I know is it's better to have more than enough I've got two 10a 2400 watt elements at my disposal for when I want to move water temp quickly in my small rig like 1c/minute I'm gathering you'd need similar heating abilities if you want a pleasurable micro.brewing expierienced. Interested in how you get on.

I have a 25 gallon and one 5500 watt element and it's fine but takes 20 to 30 minutes to boil, for 45 gallon you'll need 2

Yes, there is. To heat 170 liters from 55 C to 100 C in 30 minutes, you need 13.8 kw (58 amps at 240V). This assumes an exceptionally well insulated system. I would estimate roughly 20% heat loss with decent insulation, so you probably need closer to 75 amps. This is based on the heat capacity of water. I knew that 4 years of engineering school would come in handy.

I'm just impressed with "I'm in the planning stages for a ground-up custom-built E-BIAB system. The stainless steel pot I'm looking at is 180qt (45 US gallons) or roughly 170L. I'm anticipating boil sizes to run around the 150L/160L area. My question is, would a single 240v/5500w element be enough to bring this volume to a complete boil?"

Your going to need a winch to lift the bag and a custom made bag and as I noted I don't think it will boil very vigorously at all and probably take 45 minutes to get there

That's why I love this forum always someone with a bit of expierence in all kinds if worldly helping things like engineering. Bloody Beauty Bubba maths def isn't my specialy but knowing there has got to be an equation for this or someone that can help me is! I think we've brain stormed up some action for the OP 1st contact electrical supplier who can provide two 240v/5500watt elements a control panel to run the Pid Pump solenoid valves ect ect. Then find yourself a one Tun lifting capacity winch.

I think we have discovered why there aren't many (any!?) commercial breweries doing BIAB. That being said......if the bag was in a cage and you did have an (high) engine winch, it might actually work. A quick google search indicates 2x 5500W elements + assorted shouldn't be a problem.

Constructing a customized basket for the system as well. Hoisted with a winch. All part of the build. Thanks for the pointer.

It's definitely doable ... check out Brewha Equipment (https://brewhaequipment.com/products/brewha-biac-all-in-one-for-nano-and-micro-breweries)

Pt = (4.2 × V × dT ) ÷ 3600. Pt is the energy in kWH required to heat the water. V is volume in liters. dT is the temperature rise (in Celsius). Take the Pt and divide by the size of the heating element and you'll get the time to heat the water. Example: V=100 liters, dT = 35 (mash temp to boiling). Pt = 4.08 kWH. If you have a 5000 W heating element, it will take 49 minutes to heat from 65 to 100 C. However, you also need to take into account thermal losses. I would add about 20% as a rough estimate. In this case, it would take about 58 minutes. You can calculate the thermal losses, but that is system specific and a bit more difficult. Might even need a differential equation.