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Old 04-10-2008, 09:38 PM   #1
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Default Electric Heating Element Current Supply Concept

Alright, I have been thinking about a whole new concept for supplying the current for the electric heating element. If you have the money, you could design a switching capacitor bank that charged up using a smaller current draw. Then you could fire the capacitors with a large discharge creating the necessary current. I am literally sitting in class right now thinking of this, so I haven't really investigated the concept. The whole idea will rely on the calculations of how fast you can charge and discharge a capacitor. I figured I would start the thread now and add more info when I am not in class

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Old 04-10-2008, 10:32 PM   #2
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What's the goal with that setup? What's the power source? If it's AC, you can just use a transformer to lower the voltage and increase the current. If it's DC, you can buy a power inverter off the shelf which is using capacitors to get the job done.

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Old 04-11-2008, 12:15 AM   #3
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Quote:
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What's the goal with that setup? What's the power source? If it's AC, you can just use a transformer to lower the voltage and increase the current.
I am planning on using 120VAC as the power source for a resistive heating element. That won't work because the heating element will have a set resistance that doesn't depend on either the voltage or current. The power can be calculated using P=V^2/R which shows that with a constant resistance, reducing the voltage will result in also reducing the power delivered.
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Old 04-11-2008, 12:49 AM   #4
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The size capacitor you'd need to make it even remotely useful would be about 10x the cost of running a 220v line.

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Old 04-11-2008, 01:06 AM   #5
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I'm not entirely sure the idea has any merit at all. Your goal is to emulate a constant high voltage/current source using a low voltage/current source to charge capacitors. The problem is that the time it takes to charge the capacitors is going to be detrimental to the system. Sure, you'll get bursts of high power, but the heating element will cool while "waiting" for another burst of power. I'm almost positive that the net result will be that the element operates at an output of slightly less than what it would if you simply plugged it into the low voltage source (due to losses in the system). You can store energy, and you can use energy, but you can't pull it out of thin air.

We can help in finding a solution, but I think you need to define the problem a little more clearly. What power source is available? What heating elements are you considering? Most importantly, what are you trying to heat with the system ?

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Old 04-11-2008, 01:17 AM   #6
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Quote:
Originally Posted by Jared311
The power can be calculated using P=V^2/R which shows that with a constant resistance, reducing the voltage will result in also reducing the power delivered.
That equation is for instantaneous power. You need to calculate energy as a function of power and time to prove your concept (E = P x t).
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Old 04-11-2008, 01:19 AM   #7
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Yeah, Yuri's right. Your mass of water will already be acting like an energy store. Given the right kind of insulation and enough time, you can get up to just about any temp. It's not like you can charge a cap to the point where you can run a 5500watt element for an hour at full power. Again, why not run some 10/3 romex back to the panel?

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Old 04-11-2008, 01:29 AM   #8
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Yeah, this was all just a brain fart that came to me while zoning out in class tonight. I was going to see if maybe I had enough capacitors if I could charge and discharge while maintaining an increase in temperature. The idea doesn't seem plausible after a little bit of research.

Heres the problem I am facing. I am currently living in a rented house and can't mess with the circuit breaker at all. Its an old home and so it was wired extremely poorly. I don't think I can get away with running a 1375W heating element at 120VAC because it will draw around 11.5A. If it was my own place then I would just wire in a 30A breaker dedicated for just my brewing station.

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Old 04-11-2008, 01:37 AM   #9
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IIRC, most building codes require 20A circuits for the kitchen. If you have a garage, check those circuits for a slightly higher rating as well. A large current draw like that could be enough to trip even 20A circuit if you're using anything else on the same circuit.

What are you trying to heat? A mash tun? Boil kettle? What capacity?

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Old 04-11-2008, 01:39 AM   #10
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Quote:
Originally Posted by Yuri_Rage
most building codes require 15A circuits for the kitchen.
I am guessing this was a typo....
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