Electric HLT questions

[wilfonzo]

I recently upgraded to a keggle from a turkey pot and wanted to make the most use out of the turkey pot and decided on an eHLT. Now I just want to make sure that I have it right. I plan on using a temp controller like THIS and wire it to a 1500watt heating element. Seems simple enough and now I can have perfect temps for mash and sparge water. Am I missing anything?

 

[shortyjacobs]

The A419 isn't the best option for this, as it will have trouble controlling the temp exactly, (large hysteresis compared with a PID, and it only works in "on-off" mode, which will mean larger temp swings of your pot even with a tight differential). The thermistor is also a bit slow to respond, compared with an RTD or thermocouple.

If you are any bit handy, you can buy an Auber PID, an SSR + heatsink, and an auber weldless RTD (http://goo.gl/HCHf2), and for less than the Johnson unit you'll have much better control.

(Note, many prefer the 2352 auber PID, which has more advanced features for +$10. Also, there are much cheaper RTD options out there on eBay, but you usually have to wait for chinese shipping...).

(edit: can't make that third link work....weird...)

 

[wilfonzo]

Perfect, thanks for the great info. For some reason I thought the A419 would be better, can't remember why anymore.
Anyone have an idea at how long it would take to heat 4 gallons to the 160's with this set-up?

 

[shortyjacobs]

Water has a specific heat of 4.18 J/gK. 160F is 71C. Assuming a starting temp of 45F, or 7C, and 4 gallons is 15142 mL of water, and water weighs 1 g / mL, that's 4.18 * (71-7) * 15142 = 4051000 Joules. A watt is a joule/second, so 4051000 joules is 4051000 watt*seconds. Assuming you have a 1500 watts element, and you lose 200 watts of heat to the atmosphere, that's 4051000/(1500-200) = 3100 seconds, or 51 minutes. My guess was originally "1 hour", so pretty close.

 

[wilfonzo]

Wow, and that is why I am not an engineer.

maybe i'll go with two elements

So, J/gK*(final temp - starting temp)*volume = joules
joules/watts=seconds

my water is usually 24C

4.18*(71-24)*15142=2975000
2975000/2600=19 minutes

can two elements run off one controller wired to a standard house outlet (with gfci,) would I have to have to power sources going to the controller?

On second thought I could just buy a second burner, good thermometer with thermowell, and get better at hitting my temps.

 

[shortyjacobs]

Each element will be pulling 1500W/120V = 12.5 amps. Standard household wiring handles 15-20 amps, depending on the wiring, so you need two separate circuits. You'd need to find two outlets in your house that run off of different circuit breakers, (not too tough). Then you run a cord from each one into your control box.

You will need a second SSR like I linked above, but still only one PID and RTD probe. The two SSRs will be wired to the PID control in parallel. One power source will go into "SSR 1", one power source into "SSR 2". The PID can be run off of either (not both) power sources. Each SSR will then control each 1500 watt element, but because they will be run off of the same PID, they will both switch on and off at the same time.

The rest of your math was right, except you assumed that your heat loss to atmosphere would double when you doubled the elements. In reality, it would stay the same, so you'd have (2 * 1500) - 200 = 2800 watts of effective heating power, not 2600 watts.

 

[wilfonzo]

Ok, that makes it easier. two outlets on different circuit breakers is not a problem.

I missed that, so it would then be 17 minutes which is probably close to how long it takes now (always forget to time it) and I would have perfect temperatures every time, making it easy to reproduce the same results in my brews.

This is starting to look like a fun project, Thanks again for the help.