• This forum is the machine-generated translation of www.cad3d.it/forum1 - the Italian design community. Several terms are not translated correctly.

adjustment power resistances

  • Thread starter Thread starter Andrea Bellachioma
  • Start date Start date

Andrea Bellachioma

Guest
Good morning to all:)
I apologize if I didn't put the discussion in the exact section, but I found nothing else appropriate.
I need resistance with 2 values.
I explain better, I have a 1500w resistance and I need to use both 1500w and 750w.
the 1500w I need in the initial phase (I have to heat a container) and the 750w I need in the maintenance phase.
in the application where this resistance is to be used there is no room to be able to put 2 of different value and turn on or turn off depending on the case.
You know how I could do that?
thanks to all:)!
 
Andrea Bellachioma said:
Good morning to all:)
I apologize if I didn't put the discussion in the exact section, but I found nothing else appropriate.
I need resistance with 2 values.
I explain better, I have a 1500w resistance and I need to use both 1500w and 750w.
the 1500w I need in the initial phase (I have to heat a container) and the 750w I need in the maintenance phase.
in the application where this resistance is to be used there is no room to be able to put 2 of different value and turn on or turn off depending on the case.
You know how I could do that?
thanks to all:)!
Click to expand...
the second resistance can be inserted, in series, even outside the application. attention that you will have to dissipate the unused 750w, a beautiful "stufetta".
 
Thanks for the answer.
I can't use your solution, though.
I must have resistance inside the application and not outside.
mine is a consumption problem, i.e. I need a way to reduce the wattage of the resistance by also decreasing the current consumed (i.e. that taken from the enel system).
 
if with a transformer remove 25% of the current (increasing voltage), you will halved the dissipated power.

this in theory. If you tell me the tensions in the game, the electric regime, the type of resistance, etc. (i.e. you make me understand better than you are talking about), maybe I can help you even in practice.
 
continuous or alternating current work? in ca you just need a transformer that reduces tension and you will also have reduced power. in cc is a little more expensive because you would need a converter to generate the new voltage. give us some details about the application
 
Er Presidente said:
continuous or alternating current work? in ca you just need a transformer that reduces tension and you will also have reduced power. in cc is a little more expensive because you would need a converter to generate the new voltage. give us some details about the application
Click to expand...
that increases tension, not that reduces it. higher tensions dissipate lower energies.

However there are also "transformers" for cc. are static switching machines, even more expensive given the prices of today's copper.
 
try to feed the resistance of your water heater to 380v then see if you reduce the power and the current, but stay away! :-) the "transformers" in cc are called voltage converters. They cost a little more to equal power because they need a greater amount of "electronics".
 
use of 220v resistors in alternation. They are similar to those that can be found in furnaces.
In practice I need this resistance to give me 1500w for 20 min (the time I calculated to heat the container) and then it must provide me 750 w,necessaries to compensate for the losses of the application. .
 
Andrea Bellachioma said:
use of 220v resistors in alternation. They are similar to those that can be found in furnaces.
In practice I need this resistance to give me 1500w for 20 min (the time I calculated to heat the container) and then it must provide me 750 w,necessaries to compensate for the losses of the application. .
Click to expand...
but to manage a duty cycle instead of changing the power? with a timer (fuzzy eventually) then check the temperature at tenth grade without problems.
 
Could be an idea.
in the morning, I'll see how it comes.
I thought about a triac, what do you think?
 
because with the thermostat I do not decrease the absorption of the resistance from the network..when the signal starts from the feared and active the resistance, it consumes me 1500w from the plant enel... and it blows since with the other machines the maximum roof of kw...the first cycle I do it when the other machines are turned off. ..I think the only solution is to review the project and call the resistor supplier to see if there are smaller ones, so as to find the way to put it 2.
 
Er Presidente said:
try to feed the resistance of your water heater to 380v then see if you reduce the power and the current, but stay away! :-)
Click to expand...
If you try to turn the 230v starred voltage into 380v starry and feed your water heater (I don't, because I don't have it), I can assure you that it heats less.

If you pick up the 400v* chained instead of the starship, well, you better get out of the bathroom before you attack the switch.

(*) 380v has already been out of order, although still available in various areas of Italy
Er Presidente said:
the "transformers" in cc are called voltage converters. They cost a little more to equal power because they need a greater amount of "electronics".
Click to expand...
Right. I realized you wanted to put a converter before a traditional transformer, which makes no sense.
 
Fulvio Romano said:
If you try to turn the 230v starred voltage into 380v starry and feed your water heater (I don't, because I don't have it), I can assure you that it heats less.

If you pick up the 400v* chained instead of the starship, well, you better get out of the bathroom before you attack the switch.
Click to expand...
That's what I said. That's exactly the same thing. at 380v the water heater warms less than at 230v, regardless of how you generate the voltage.

First law of ohm.
 
I've always worked in my experience, but I've always seen the resistances dissipate more power to increase the voltage applied to the extremes of resistance.

in particular the current circulating in the resistance I obtained from i = v/r
while the power dissipated by the resistance was i2r

I do not understand how a resistance to whose garments is present a voltage of 380v can heat less than the same resistance with the applied garments 220v.


with regard to the initial problem if the counter is not too sensitive you can use a pwm (also called dimmer) similar to that of the halogen lamps from apartment.

They don't cost too much, they are of common use and they are easily found, the only limit is that they are usually for max 500w, so you should put 3 smaller resistors in place of the 1500 big one
 
In fact, for the preciseoon I would say that for your case the kit rs 362 " regulator of light for incandescent lamps from 4kw" fccia just to your case, surely it is very expensive, costs more than 30 euros ( therefore many times the cost of the resistance you have to control) but it allows you to have a fine adjustment and without particular limits

http://www.elsekit.com/pag.20.htmlor the kit
462 "power regulator for ovens and stoves electrhce up to 2kw"
http://www.elsekit.com/pag.25.html
.
 
I attached the conditioner to the "8000", consume less consumes.. .
Shuttle Launch.webp@Wink: :Wink:
 
Sorry I didn't answer, but I had a lot of work.
then, in the end the solution that has been adopted is the simplest.
In practice I used the smaller resistors connected to a timer.
the time necessary to the heater is increased but since they are connected to the timer,they turn on before and at 8 of the morning, they immediately warm it:d
 

Forum statistics

Threads
44,997
Messages
339,767
Members
4
Latest member
ibt

Members online

No members online now.
Total: 0 (members: 0, guests: 0)
Back
Top