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Old 04-01-2009, 03:34 AM   #11
The Pol
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Quote:
Originally Posted by wilserbrewer View Post
Thanks Pol,

While I admire your work and am a bit jealous, not all of us are cut out for such high tech rigs.
Several months ago, I swore that I could not build this thing either. I learned how to wire the whole thing up via the interweb!

It isnt for everyone, but you have options.



 
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Old 04-01-2009, 03:40 AM   #12
DNisich
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I've been thinking about doing this for a long time as well. I've always wondered why a simple ordinary light dimmer switch couldn't be used to control an SSR which in turn controls the heater element? At the end of the day, if you use the SSR then don't you just need a simple way to control duty cycle in order to regulate the boil? Doesn't a simple wall dimmer do just that?

Does anybody know why this wouldn't work? How 'bout it Pol? Any input on this?



 
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Old 04-01-2009, 03:44 AM   #13
The Pol
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Quote:
Originally Posted by DNisich View Post
I've been thinking about doing this for a long time as well. I've always wondered why a simple ordinary light dimmer switch couldn't be used to control an SSR which in turn controls the heater element? At the end of the day, if you use the SSR then don't you just need a simple way to control duty cycle in order to regulate the boil? Doesn't a simple wall dimmer do just that?

Does anybody know why this wouldn't work? How 'bout it Pol? Any input on this?

It wont work...

The SSR is merely a switch, on and off. Its purpose is to switch large loads that normal switches cannot, and do it quickly, without mechanical relays.

A dimmer is a resistance switch that has an infinitely variable output. The SSR needs a ON and OFF input, not a variable constant input. Dimmers dont control duty cycle, if they did, your lights would turn on and off constantly, that is duty cycle. Dimmers are always ON, or off... they have no duty cycle.

The input voltage for most SSRs is about 3-10 volts DC... this is what causes the SSR to switch on and off.

Dimmer + SSR will not work, they are not compatible.

 
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Old 04-01-2009, 03:54 AM   #14
ApolloSpeed
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maybe even go more ghetto.... use two 3500watt elements, and just turn one off when ya get up to boiling.


....btw, I don't understand SSR, PID. Links? Where do I get this stuff?
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Old 04-01-2009, 03:58 AM   #15
The Pol
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Auberins.com

PID is a temp controller that switches the SSR ON and OFF. The SSR is just a fancy non mechanical relay capabe of switching power many times per second!

The PID is 120VAC, has a temp probe in the vessel. The power to the heating element is routed through the SSR, it is he switch. The PID controls the duty cycle.

65% equals 65% on, 35% off...

It is simple once you see it and understand that it is just a temp controller connected to a simple relay that has an input and an output

 
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Old 04-01-2009, 04:00 AM   #16
DNisich
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Quote:
Originally Posted by The Pol View Post
It wont work...

The SSR is merely a switch, on and off. Its purpose is to switch large loads that normal switches cannot, and do it quickly, without mechanical relays.

A dimmer is a resistance switch that has an infinitely variable output. The SSR needs a ON and OFF input, not a variable constant input. Dimmers dont control duty cycle, if they did, your lights would turn on and off constantly, that is duty cycle. Dimmers are always ON, or off... they have no duty cycle.

The input voltage for most SSRs is about 3-10 volts DC... this is what causes the SSR to switch on and off.

Dimmer + SSR will not work, they are not compatible.
I'm not yet convinced. I read at How Stuff Works that the old way that dimmers worked was with a variable resistor but the modern dimmers use a triac to vary the on-off time (duty cycle). Check out the article here:

HowStuffWorks "How Dimmer Switches Work"

Additionally, I've found a slew of A/C input SSRs at Grainger.com. So, just pull from one of the hot wires to power the dimmer and I would think you'd be in business.

Admittedly, I've haven't yet tried this. But the plan was to buy a dimmer, SSR and heatsink. Then use it with a simple light socket to see if I can control a light bulb. If it works for the bulb, I'm pretty sure it will work for the heater element too.

Not as sexy as PID but man, I still think this would work.

 
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Old 04-01-2009, 04:29 AM   #17
Bobby_M
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The point is, an SSR (relay) is an on/off state device that doesn't know what the hell a variable voltage (analog) input means. They have a voltage threshold at which the device goes from off to on.

I think one cheap way was already mentioned. Run two elements to heat quickly then kill one to maintain. Another way would be to run a 5500w element on 240v to ramp up quickly then cut it to 120v which would run about 1300 watts. I don't know if 1300 is enough to maintain boil though.
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Old 04-01-2009, 01:00 PM   #18
trimpy
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I am also a SSR user. You can pick up a 40A ssr on ebay for less than 15 bucks. My setup uses an AVR microprocessor (tiny2313) hooked to my laptop's serial port. I type a number into a hyperterminal window and it adjusts the PWM that controls the SSR. All my parts are < 10$ and it works great. I use serial because I plan to do more automation and logging, otherwise id use a 10$ lcd and some buttons to make a self contained unit.

If you don't have much microcontroller experience, i recommend an arduino board. They are about 35 bucks and are very easy to work with.

 
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Old 04-01-2009, 02:03 PM   #19
ChemE
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Calculating how much energy input is required to maintain a 6.5 gallon boil isn't too hard depending on the accuracy required. Energy input = conductive heat loss + boil off.

My boil off rate in my somewhat squat boiler is 0.97 gallons/hour or 1.02 mL/second. It takes 2,260 Joules/gram to vaporize water. A mL of water at 212F weighs 0.9584 grams (don't use the common approximation of 1 mL = 1 gram since we are far from 4C in a boiler) so the power required to maintain the boil rate is P = (1.02mL/s)(0.9584g/mL)(2,260J/g) = 2,209J/s = 2,209 watts.

To this we need to add the heat lost through coduction and radiation which is somewhat less trivial and requires a few assumptions. If anyone were seriously interested in this additional step I could probably swing the bat but it wouldn't be much more than a first-pass approximation of the true number given the number of assumed unknowns.
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Old 04-01-2009, 02:18 PM   #20
The Pol
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Quote:
Originally Posted by ChemE View Post
Calculating how much energy input is required to maintain a 6.5 gallon boil isn't too hard depending on the accuracy required. Energy input = conductive heat loss + boil off.

My boil off rate in my somewhat squat boiler is 0.97 gallons/hour or 1.02 mL/second. It takes 2,260 Joules/gram to vaporize water. A mL of water at 212F weighs 0.9584 grams (don't use the common approximation of 1 mL = 1 gram since we are far from 4C in a boiler) so the power required to maintain the boil rate is P = (1.02mL/s)(0.9584g/mL)(2,260J/g) = 2,209J/s = 2,209 watts.

To this we need to add the heat lost through coduction and radiation which is somewhat less trivial and requires a few assumptions. If anyone were seriously interested in this additional step I could probably swing the bat but it wouldn't be much more than a first-pass approximation of the true number given the number of assumed unknowns.
Cool calcualtions, but from practical testing in my garage on my system, 2,200 watts is going to give you a very anemic boil, if it boils at all. I require 3,500 watts in my keggle in a 60F ambient environment. If you have a well insulated vessel, this may change significantly, but that is my $.02



 
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