Camco 5500 watt elements

I am looking to go all electric on my next HLT which will be 55 gallon SS. Does anyone have an formula for calculating the watts needed to heat this volume of water? I will not be boiling just heating to mash and sparge temp. Based on what I have read on the forum, it looks as though I will need 2-5500 watt Camco Low density elements but this may make the automation of the system more difficult? Any ideas?

 

Timeframes are required to calculate this.

Here's some ideas tho. I used 55 gallons FULL, going from 100*F to 155*F, with a timeframe of 30 minutes. The following calculations take into account NO losses from water to air, kettle to air, ect. This is obviously not possible, but it gives you a starting point.

We need to calulate the number of joules required by the following formula.
Q=cm(dT)
where c = 4.186 J/g (Pure Water)
m = mass (in grams)
dT = Change in Temp (in degrees C)

In your case,
Q=(4.187 J/g)*(207,900g)*(30*C)
Q= 26,108,082 J

Then since 1 watt = 1 J/second,

Watts = 26,108,082 / (30minutes*60Seconds)
Watts = 14,504


I entered the formulas into Excel, so if you want to change the variables, just let me know.

 

You can control multiple elements from one PID. You can either use one SSR rated for the total amp draw or use one SSR per element. One other thing to consider is that two 5500 watt elements max out a 60 amp circuit.

 

This is what you want. This isn't mine but I got it from some thread on here a long time ago that I can no longer find.

calculate heating time spreadsheet

 

I also use that. It will tell you how long from temp to boil, temp to temp, wire size, breaker size, etc.