Is it current or voltage that is fixed in transformer output

[Umar Awan]

In a transformers output, is it the current that is fixed or the voltage?

I have been trying to resisitively heat some tungsten foil and I am stuck at trying to find out the amount of current I need in my transformers output. The problem I am facing is that I know that the tungsten resistivity will increase as it heats up, but I need to make sure my tungsten reaches the temperature of 1035 Celcius that I need at it. Specificly, I want to know if my power dessipated to tungsten foil, will Increase or decrease, as the resistance increases due to heating.
The reason is that I am confused between the two formula's: P=I^2R, P=V^2/R.
In P=I^2R, resistance increases, power increases(considering transformer gives constant ampere output)
In P=V^2/R, resistance increases, power decreases(considering transformer gives constant voltage output).
My question to you is:
1)Which is fixed in a high current transformer,(Current or Voltage)?
2)Is it possible that fall in current will be negated by increase in Resistance, so that power dessipated remains constant?

Umar Awan

 

[sophiecentaur]

Once you get your causes and effects the right way round and also define the required performance, then things become simpler.
The output Volts from the transformer are set (subject to practicalities) by the supply volts and the transformer ratio. The current that is supplied depends on the Resitance of the load with that value of V_s.
I_s = V_s/R
What's the load Resistance, R, going to be?
You need to decide what Power the tungsten foil needs to have supplied in order to maintain the temperature you want. Only you can tell that. It will depend entirely on its size and surroundings and it is the hardest thing to determine. You have the same problem that a light bulb manufacturer has and they use a lot of past experimental data to work out the dimensions of their filaments. You may have a choice of heater element size for a given R. (Long and fat or short and thin can both have the same R)
Do you already have the foil and is it in the form of tape of fixed dimensions? What do you know about the Power required?
You are right about the resistance changing with temperature and your transformer must be able to cope with an initially high value of current when the cool foil is first connected.
Can you find out your actual Power requirement first? (Experimentally)

 

[Umar Awan]

Once you get your causes and effects the right way round and also define the required performance, then things become simpler.
The output Volts from the transformer are set (subject to practicalities) by the supply volts and the transformer ratio. The current that is supplied depends on the Resitance of the load with that value of V_s.
I_s = V_s/R
What's the load Resistance, R, going to be?
You need to decide what Power the tungsten foil needs to have supplied in order to maintain the temperature you want. Only you can tell that. It will depend entirely on its size and surroundings and it is the hardest thing to determine. You have the same problem that a light bulb manufacturer has and they use a lot of past experimental data to work out the dimensions of their filaments. You may have a choice of heater element size for a given R. (Long and fat or short and thin can both have the same R)
Do you already have the foil and is it in the form of tape of fixed dimensions? What do you know about the Power required?
You are right about the resistance changing with temperature and your transformer must be able to cope with an initially high value of current when the cool foil is first connected.
Can you find out your actual Power requirement first? (Experimentally)

Thank you!*I mean it!*
My "mc delta t" 's and "p=j/s" 's give me a power requirement of 1030 watts...

My calculation of the resistance(Its a very thin foil) of the tungsten foil gives me a value of 3.2 ohms resistance.

I also need to heat this foil to about 1050 Celcius in about 3 minutes time

I am confident about my formulae though not very confident about efficiency.
It is a vacuum chamber, so no convection losses will be not be involved, and I have gone form some really HUGE thickness wires, so I think, apart from core losses from transformers, I am satisfied with my calculation's here.

I must confess, I am really in the dark about the amount of amperes I might need to make this work, and I see with all those post's of your's, I hope I can extort a fair bit of knowledge of you! :)*extra wide grin* aka. Please somebody help me out here!

 

[Umar Awan]

The output Volts from the transformer are set (subject to practicalities) by the supply volts and the transformer ratio. The current that is supplied depends on the Resitance of the load with that value of V_s.
I_s = V_s/R

Refering to my earlier reply to you. I just experimented with parameters set by my requirements and my afore mentioned requirements and I noticed that with 57.6 output voltage, I could get an output of 18 amperes. These many amperes would be sufficient to provide 1036.8 Watts (Taken formulate P=I^2R) Meaning I manage to procure a transformer providing atleast 1.5KVA(thus leaving room for efficiency). My Question to you now is? Are these the right equations I am using *sigh*? Do I need to leave more room for efficiency? Or should I just pick up my violin and forgot about this?

I'll be online quite a while. Now Is that day to get your milestone of a 1000 likes!

 

[sophiecentaur]

I don't think efficiency is a particular issue unless the transformer gets hot. But that wouldn't be a problem if the transformer you use is well up to spec. (VA rating).
I think you are going the right way.
1. How many Watts to do the job. (That's the hard bit and you seem to be there already)
2. Mechanical size of the element to get the heat where you want it.
3. Convenient / available output Volts from a typical off the shelf transformer. Gives the required Resistance (Power=V²/R)
That tells you the Current rating that you need - with some to spare
4. Length and CSA of element for 2. and 3.
5. One point I would make and that is 18A is a lot of current for convenient supply wires and connections. You may be better to choose a higher V_s to bring it down to more like half that. But look around at available transformers and prices. ...(not sure whether you means that you already had a beefy enough transformer)
Tungsten is probably expensive so you won't want to be trying many combinations of element dimensions.
PS will you have some thermostatic control? That could make life easier.
Jeez - 999. I can't remember when I last looked at that total.

 

[OCR]

Now Is that day to get your milestone of a 1000 likes!

sophiecentaur [COLOR=#black].[/COLOR] →[COLOR=#black]...[/COLOR]Likes Received: 1,000.

And, you're welcome...[COLOR=#black]..[/COLOR] [COLOR=#black]..[/COLOR]

 

[Tom.G]

This is just a 'Pay Attention To' note.
Ordinary incandescent light bulbs operate around 2400 Celsius (2700K), and Rule-of-Thumb says to design for inrush current 10 times operating current. One reference I found (CRC Handbook of Chemistry and Physics, 88th ed.) was at 627C (900K) the resistance of Tungsten was 4.5 times room temp resistance. Another reference (http://hyperphysics.phy-astr.gsu.edu/hbase/tables/rstiv.html) showed Temp Coeff. of Res. to be .0045 per Degree C at 20C. For short term or experimental use, the 1.5KVA transformer should be sufficient. If there will be rapid on/off cycling of ribbon heating (causing heat build up in the transformer), or long term production use, a 2KVA to 3KVA may be a better choice. You don't want to smoke that expensive (custom?) transformer!

 

[sophiecentaur]

You don't want to smoke that expensive (custom?) transformer!

Good point about the time for the element to warm up but having control over the R value, it should be possible to get hold of a low cost (110V?) off the shelf transformer with many kVA spec.

sophiecentaur [COLOR=#black].[/COLOR] →[COLOR=#black]...[/COLOR]Likes Received: 1,000.

And, you're welcome...[COLOR=#black]..[/COLOR] [COLOR=#black]..[/COLOR]

Cheers my boy.

 

[Umar Awan]

Guys Thank you! All of you.
This has probably been the most productive discussion of my life.
Now I have another question going on in my head: The reason we use a high current transformer is because plugging a low resistance heating element directly into a power supply( e.g. 1 ohm), will pull a current of 240 amperes (Power supply in Indian homes, 240V). The problem here is that the power supply is not capable of handling that current( my meter reads 240V, 10-20 ampere output). So we use a high current transformer so that it can handle the load of ampere output required. Does this mean that I can completely eliminate my transformer(save a few 100$'s), simply by increasing the resistance of my tungsten by reducing its thickness and increasing its length(I found a site that sells one of those tungsten fillaments that fit 30 inches of tungsten into 2 cm lengths[light bulb filaments]), and therefore plugging in my heating element directly from the power supply?Ofcoarse, I will have to fit in a Thick gauge wire from the mains directly, and watch out for electrical shock hazards.

Here is the specifications I think I might need to make this work without a transformer:
Resistance=16 ohms(So that I=V/R ---> I=240/16 ---> I= 15 amps)
Current=15 amps
At the same time, therefore increasing power output, P=I^2R, P=15*15*16, to 3600 watts.
Effectively reducing my heating time to about a minute

Do you think this will work? Is it only my equations that are problematic?

The transformer in my power supply is 3-phase(are they all 3 phase?), and I remember reading somewhere about some 3-phase transformer equation involving the "root of 3" value. May I know if this value is applicable in one of these equations.

 

[Umar Awan]

Also Tom. [O]G. Great stuff this temperature coefficient of resistance table. [Like emogy]

 

[sophiecentaur]

plugging a low resistance heating element directly into a power supply( e.g. 1 ohm), will pull a current of 240 amperes

Changing the secondary volts of a transformer will make no difference to the current demand on the mains. VI (=Power) is the same in the primary as the secondary, irrespective of the chosen secondary Volts and load Resistance. If your Power demand is, as you said, around 1kW, then that only represents less than 5A. The Resistance needed for an element for direct connection would have to be greater, of course, to keep the Power the same . (Proportional to V², aamof.) You would need to look at the cost of a longer / thinner element versus the extra cost (weight and volume) of a transformer.
All these points are worth considering before you actually part with money for this project.