Electric keggle heater calculations

[bull8042]

I am an Electrical Engineer, NO degree in thermal dynamics so I am looking for someone to confirm my calculations.
I am converting a 15.5 gal keg into a brew kettle and want to get away from the propane burner so I can brew indoors.
I am not taking into account heat loss of the SS kettle and plan to up-size the heater slightly if needed.
One BTU is required to raise one pound of water one degree.
Water is 8.34 lbs per gallon. So 8.34 BTUs per gallon per degree.
Water supply is 70* average, so I would have to raise the temp of the water by 142* to boil.
I want the capability to boil 10 gallons in case I ever decide to brew a double.
8.34 BTU * 10gal = 83.4 BTU per degree
83.4 BTU * 142* = 11842.8 BTU total
11842.8 BTU = 3.468 Kilowatt Hours
To deliver 3.468 Kilowatt Hours in 30 minutes requires 6937 Watts
That being said, it appears as if I should be able to boil 10 gal of water in 30 minutes with a 7000W heater.....
Can anyone verify that i haven't bastardized these calcs?

[baltobrewer]

Hey, im no electrical engineer either, but here's my take on your calcs: my 14 gallon HLT with a 7000W heater on 240v gets pretty hot, pretty quick.

There you have it. So yes, your mathematics are entirely correct, as proven by the above equation. er, formula....er........proof. whatever.

This post was unhelpful. I can see that. But seat of the pants, if you use 7000W on 240, my guess is you should be fine.

[Anthony]

I am an Electrical Engineer, NO degree in thermal dynamics so I am looking for someone to confirm my calculations.
I am converting a 15.5 gal keg into a brew kettle and want to get away from the propane burner so I can brew indoors.
I am not taking into account heat loss of the SS kettle and plan to up-size the heater slightly if needed.
One BTU is required to raise one pound of water one degree.
Water is 8.34 lbs per gallon. So 8.34 BTUs per gallon per degree.
Water supply is 70* average, so I would have to raise the temp of the water by 142* to boil.
I want the capability to boil 10 gallons in case I ever decide to brew a double.
8.34 BTU * 10gal = 83.4 BTU per degree
83.4 BTU * 142* = 11842.8 BTU total
11842.8 BTU = 3.468 Kilowatt Hours
To deliver 3.468 Kilowatt Hours in 30 minutes requires 6937 Watts
That being said, it appears as if I should be able to boil 10 gal of water in 30 minutes with a 7000W heater.....
Can anyone verify that i haven't bastardized these calcs?

Those calculations are about right. Many moons ago, a guy with an electric brew rig posted an Excel spreadsheet on the Green Board that really simplifies the various calculations. I've mirrored it on my website here. Hope that helps... also keep in mind, unless you're brewing extract, you won't ever really be going from water supply temperature to boiling, instead you'll be going from mash temperature to boiling.

[bull8042]

Those calculations are about right. Many moons ago, a guy with an electric brew rig posted an Excel spreadsheet on the Green Board that really simplifies the various calculations. I've mirrored it on my website here. Hope that helps... also keep in mind, unless you're brewing extract, you won't ever really be going from water supply temperature to boiling, instead you'll be going from mash temperature to boiling.

That spreadsheet pretty much mirrors my results, but is a lot "purtier". Thanks for the link. And you are exactly right about my starting temp. I absolutely forgot about that since I have been pretty much an extract brewer.... until you guys corrupted me and have given me a "burning desire" to go all-grain..... or maybe it was a "burning sensation"!
This is the first step in converting (on paper) the "Wort-O-Matic" setup to use the whirlpool chiller I fabbed last weekend instead of a CF chiller.

And to baltobrewer, your post was helpful. Real-world results usually seem to trump calculations every time.

[codewritinfool]

I get 3.48 kWH from 70F to 212F, so close enough. Double it to get 30 minutes, or about 6.96 KW.

All of that assumes, of course, that you have no thermal losses, which you will.

Oh, and from 155F to 212F is 1.4kWH.

[bull8042]

Hey, im no electrical engineer either, but here's my take on your calcs: my 14 gallon HLT with a 7000W heater on 240v gets pretty hot, pretty quick.

There you have it. So yes, your mathematics are entirely correct, as proven by the above equation. er, formula....er........proof. whatever.

This post was unhelpful. I can see that. But seat of the pants, if you use 7000W on 240, my guess is you should be fine.

HOLY DOO DOO Jay! I just realized that I am trying to emulate YOUR system, but with the whirlpool instead of the CFC! Dang small world.
I just haven't figured out how to eliminate moving the chiller between the BK and the HLT......
Thanks for the great pics and information...
Bull

[baltobrewer]

I think the CFC is a nice option because it serves to raise as well as lower temps. It's a heat exchanger, right?

Post some pics!

[bull8042]

I think the CFC is a nice option because it serves to raise as well as lower temps. It's a heat exchanger, right?

Post some pics!

You are exactly right, and I was going to purchase one. But I stumbled on some information touting the benefits of whirlpool chilling, found out I could get a 50' roll of 5/8" copper for $60, and it became a no-brainer after that. Did throw a wrench into my plans to emulate your system though.
However, I have since decided to stick with batch-sparging until I get more comfortable, so the complexity of your system is really overkill at the moment.