Electric Heating Element Current Supply Concept

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Jared311

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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 :)
 
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.
 
jca said:
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.
 
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 ?
 
Jared311 said:
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).
 
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?
 
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.
 
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?
 
Yuri_Rage said:
An 11.5 amp average draw might be enough to trip even a 15A circuit, though, especially as the load changes with temperature and/or initial power up, and most certainly if you're using anything else on the same circuit.

11.5 amps shouldn't trip a 15A breaker and JFYI a breaker is designed to trip at 5X the rated current instantaneously. Oh and most heaters have a positive temperature coefficient.
 
I believe the newer electrical codes require 20A circuits to kitchen outlets, but even older codes required 15A.

If you're willing to spend all this money, why not just go with a steam system? Just get a pressure cooker and use it on your stovetop. I modified mine - drilled and tapped the lid for a needle valve, I run a copper steam manifold right off the valve. Works *very* well, and it's portable. It wasn't even expensive and it's quite safe compared to a (presumed) amateur playing with electricity and water. I use a 9L pressure cooker and even using it on one of my small (electric) burners, it heats up the mash very effectively - my last mash was 13.5lb with 3.75gal water. Very effective and you can mash to any temp you want.

Plus, you can cook potatoes really fast, and sterilize containers for washed yeast.
 
I recently got some 15.5 gallon kegs and I am going to turn one of them into an HLT. So I was looking into using a resistive heating element to heat up roughly 10 gallons of water to strike temperature.
 
Jared311 said:
I recently got some 15.5 gallon kegs and I am going to turn one of them into an HLT. So I was looking into using a resistive heating element to heat up roughly 10 gallons of water to strike temperature.
In that case, +1 to steam!
 
Ha, if we are going to talk about building regulations, this house was built in the 40's. It is probably grandfathered from most of the current regulations.
 
Well Jared311, is the electrical system in your apartment current? Do you have breakers or fuses? In either case I would not be worried about plugging in to the wall with a 11.5 amp load. Here let me ask you some more questions. Do you have a microwave or a coffee pot or even a toaster? These are high draw items. A microwave all by itself can draw 11.5 amps and the average coffee pot is 1500 watts, also a large load, but I would bet you do not think twice about plugging them in? If you live in an apartment I would be surprised if you did not have a GFCI in the kitchen or the bathroom. If you worried about safety, plug into a GFCI. If your apartment does not have one, you can pick one up at a hardware store: get one on an extension cord. Anyway that is my two cents.
 
Jared311 said:
Ha, if we are going to talk about building regulations, this house was built in the 40's. It is probably grandfathered from most of the current regulations.


Yeah, well, just remember, any time you do any kind of remodeling, which everyone does, your grandfather dies.
 
Bernie Brewer said:
Yeah, well, just remember, any time you do any kind of remodeling, which everyone does, your grandfather dies.

That isn't necessarily true since it is not nearly that cut and dry......it depends on what you are changing, your current system and ultimately the AHJ.
 
wihophead said:
That isn't necessarily true since it is not nearly that cut and dry......it depends on what you are changing, your current system and ultimately the AHJ.

Well, of course, but the point I was trying to make is that assuming your home is immune to current codes just because it was built 70 years ago is silly.
 
Just to comment on the original topic... What you're talking about is pretty much exactly what a step-up switching converter is. It's a common enough concept, and it can certainly be done. But don't forget, you are still limited by power - a 120V@20A outlet is only going to give you 2400W, so no matter how you convert it you'll still only be able to get that much power to a heating element from that outlet.

But as already mentioned, the cost would be huge. A custom-built switching step-up converter capable of 2400 watts would be big, heavy, and cost a small fortune.

If you absolutely had to make it work, you'd probably be better off getting an off-the-shelf 120V to 240V voltage converter like you'd use for running european appliances in the US. A 3000W unit would still cost you well over $100 and weigh a lot, and likely be inefficient and thus generate a lot of heat (ie - waste power), but it could work.

But the simple answer is to find a way to do it with just 120V, and/or install a 240V outlet.
 
The basic idea is interesting. Just look at your electric hot water heater for a guide. If there was a faster way to heat water these guys would use it. Yuri is correct you would need to be careful with your local building code. Good luck
 
To step up the voltage you would simply use a transformer nothing really complicated about that. It would work great if you had heaters rated for 480 volts and you wanted to get everything out of them you could but this would raise many other issues that are out of the scope of this thread.

The issue with the primary idea of raising the load voltage to establish a higher load current while at the same time maintaining the line current under 15A is impossible since you would be in essence creating free power and we all know there is no such thing but if you could get it to work you would be a very rich man...:mug:
 
I know a way to get free power < evil laugh >

All you have to do is wrap your neighbors power lines with a large inductor.
 
Hell tap ahead of the meter then brew on. No jail time and fines will get ya big time. I own 4 houses dating from 1905 to 1936 with 100 to 200 amp services each. The 1905 once had a 30 amp service now that was ok back in the teens.
 
Just a thought, havent checked the numbers,...

You may be able to use a stack of UPS'es ?

cheers,
walther
 
Interesting thought, I will have to look into that. I can buy a bunch of UPS'es for dirty cheap through work. I will check the specs tonight and see what I can get out of them and for how long.
 
Jared311 said:
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.

Ah, it's more power you need and not more current. You'd need a rack of car batteries to get that done. It would be a cheaper to go with two elements and run it off two 120 circuits. I think my next setup will be propane.
 
jca said:
Ah, it's more power you need and not more current. You'd need a rack of car batteries to get that done. It would be a cheaper to go with two elements and run it off two 120 circuits. I think my next setup will be propane.

Power (W) is the product of voltage (E or V) and current (I). If you raise the the voltage in a resistive circuit you inherently raise the current in turn producing more power.
 
I think with steam you will have alot of hot spot in your mash. Stirring will help but I think recirculating throw a 120vac heating element is a better method and won`t disturbed your grain bed.

Here a clip of a RIM setup that I'm getting Ideal from.

I don`t think it worth heating a HLT. But to kept your MLT temp consistant, it a excellent ideal.
 
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