Voltage Regulator Help

[Newcastle1996]

Hi, all.

I’m currently using the t500 and have bought a voltage regulator for when I use the pot still head so I can control the voltage better when i’m only putting 5 litres in the 25L still for example. So the still is 2kw so when I have 5L In I’d essentially like to only have to use 1/5th of the power once it has heated up and get some better quality theoretically. Now I’m aware lowering the voltage will lower the power but how do I calculate how much I am reducing the power? The controller can be adjusted from 0 to 220 volts and can be adjusted by 10 volts each turn. I just want to be able to calculate the power output in correlation to voltage. This is my controller: https://www.ebay.co.uk/itm/AC-220V-4000 ... 2cbcd6749d" onclick="window.open(this.href);return false;" rel="nofollow

Advice on how to use this + power correlation would be great!!

[rubelstrudel]

Measure the resistance over the heating element.
It will be somewhere close to 24 Ohm

Now if you know the voltage you're putting into your heater, the wattage is given by this formula:

Watt = (Voltage_in ^2) / Resistance_heater

So say you're giving it 100v, then your output watt = 100V *100V / 24 Ohm = 416 Watt

You get to this formula by fiddling with Ohms Law
U (voltage) =R (resistance) * I (current)
and
P (watt) = U (voltage) * I (current)


When you're measuring a phase switched AC, make sure you have a true RMS (root mean square) voltage meter or you can get almost any old measure.

[Yummyrum]

Agree
Shame you didn’t get the Auberins DSPR1 controller . It has 0-100% power so you know accurately what percentage you are reducing it by . ( fluctuations in mains supply voltage excluded )
https://www.auberins.com/index.php?main ... cts_id=444" onclick="window.open(this.href);return false;" rel="nofollow

[rubelstrudel]

BTW:

I am using this controller:
https://www.ebay.com/itm/220V-10KW-SCR- ... ctupt=true" onclick="window.open(this.href);return false;" rel="nofollow

It is rated for 10KW, more than double my heating element. It is recommended to have these rated for double your peak draw. I am very happy with this controller. It has a built in voltage meter that is accurate, so you always know exactly what you're putting into your heater. I have a slightly different version with a manual potmeter adjustment, not this digital version, but the main body of the thing is the same. Its been in heavy use for a year now without problems.

[Newcastle1996]

[quote="rubelstrudel"]Measure the resistance over the heating element.
It will be somewhere close to 24 Ohm


So say you're giving it 100v, then your output watt = 100V *100V / 24 Ohm = 416 Watt

Thanks for that! Just to make sure I’m doing it correctly...

If i set my device to 180 volts that is:

180 x 180 divided by 24 equals 1.35Kw of power going into my boiler?

Thanks!

[still_stirrin]

...So say you're giving it 100v, then your output watt = 100V *100V / 24 Ohm = 416 Watt

Just to make sure I’m doing it correctly...If i set my device to 180 volts that is: 180 x 180 divided by 24 equals 1.35Kw of power going into my boiler?

Yep. That's how you do it.

But realize that a little (very little, however) of the voltage (power) is consumed by the controller as well. That's why you need a heatsink for the power supply/controller...because it does consume some energy. So, the 1.35kW is the power input from your source. But, it's close enough for "assumption's purpose".
ss

[rubelstrudel]

But realize that a little (very little, however) of the voltage (power) is consumed by the controller as well. That's why you need a heatsink for the power supply/controller...because it does consume some energy. So, the 1.35kW is the power input from your source.

You can't fool Ohms Law. If you put 220V over a 24Ohm resistor, you will create 2016 Watt of power output in your resistor. So if the voltage you measure is over your resistor/heating element, the power output is what it is.

In addition to that the controller will generate some heat, but that is beyond the power you use in your heater. So your TOTAL system of heater + controller will use slightly more power, but we're talking about a few tens of watts - statistically insignificant. The amount of power that goes in to the boiler is the only thing we really care about - let the power utility company care about measuring your total consumption and send a bill.

[rubelstrudel]

If i set my device to 180 volts that is:

180 x 180 divided by 24 equals 1.35Kw of power going into my boiler?

Thanks!

If your heating element has a resistance of 24 Ohm, then yes that is correct. You will have to measure the resistance in your heating element.
I.E. If your heating element has a resistance of 26 Ohm and you put in 180V, then your heating element will consume/put into the boiler (180²/26)=1.24KW.

[shadylane]

Here's to Ohms law

Don't forget the Voltage Regulator output
Is a hard to measure sawtooth waveform

The big resister "heater" probably has at least a +/- 10% value

Personally, I gave up on accurately measuring Watts
Now there's a grease pencil mark on a analog voltmeter

[rubelstrudel]

So, to not complicate matter we say:

Trust the voltage scale on your dial. Whatever it says is probably close enough for practical purposes.
To find the approximate power output of your heater, use the formula P=V^2/R

[slowpoke]

I don't need to know a lot of electric clap trap to know how to use my drill to drill a hole in a piece of wood. Why would I want to know the same to turn a nob on a voltage controller. Would seem enuff to see what's coming out the spout. My controller increases the power when I turn the nob to the right and decreases when I turn it to the left. If the likker aint coming out the way I want it to, I turn the knob to change it. And who the hell is this Ohms feller anyway? I dont no and dont give a rats furry hind end.

[Newcastle1996]

I don't need to know a lot of electric clap trap to know how to use my drill to drill a hole in a piece of wood. Why would I want to know the same to turn a nob on a voltage controller. Would seem enuff to see what's coming out the spout. My controller increases the power when I turn the nob to the right and decreases when I turn it to the left. If the likker aint coming out the way I want it to, I turn the knob to change it. And who the hell is this Ohms feller anyway? I dont no and dont give a rats furry hind end.

I’m a believer in theoretical knowledge. I guess people are different. This is just something I wanted to find out. Plus i’m fairly new to this and just wanted to figure out how to quarter the power when I have quarter the amount in. With me being a beginner I guess I don’t yet have the knowledge to just see what is coming out of the still and measure it this way and feel I could waste a lot of spirit or have a poor quality. At least know roughly what the power is I will have some sort of control I guess.

No harm in having this information.

[rubelstrudel]

And who the hell is this Ohms feller anyway? I dont no and dont give a rats furry hind end.

According to legend, George Simon Ohm was a bugger for the bottle. He formulated his law when he observed, as a young poor student that the level of drunken stupor he would manage to achieve was the product the potency of his likker, and the speed with which he consumed it.

[Tennessee_Spirits]

The best way to accurately deliver a known amount of power to a heater element is to have a square wave generator that is made to display the time duration that the square wave is in the positive direction.

. This device on amazon will generate that calibrated squarewave. It needs about 20 ma of 9-12volt dc and can drive a solid state relay. The case and cords are more expensive than the components and if a fan is used can switch 40 amps AC for your element at 120 to 240 volts AC.

It has a rotary encoder knob and you can set it to any percent power you want confidently.

[clearspirit]

Or you could just buy one of these...

https://5stardistilling.com/product/pow ... 00w-10amp/

[shadylane]

Those don't survive very long.