I want your help to know how i can calculate the power in KW of a 3 phase heater when connected in star and delta. Each arm has a total resistance of 17 ohms and the coil is of Nichrome. So there are three arms in total. The supply voltage is 415 volts Ac. How do i know the value of inductance in the coil? I don't have a LCR meter.

please help. with thanx in advance

Anshuman
Elmex Power Controls
[email protected]

 

Then use ammeter, voltmeter and DC source. Or go to a fortune-teller (or better - back to physics lectures).

Jacek Dobrowolski, Ms. Sc. E. E.
Software Engineer

 

Hello Anshuman

You don't need an LCR meter. You can assume your load is purely resistive. I'm guessing that you measured the resistance with the heater cold. When the nichrome is hot it will have a higher resistance.
However, with 17 ohms:
The voltage from each phase to the star point will be 415/sqrt(3) = 239.6V
For a resistive load use Ohm's Law V=IR
I=V/R = 239.6/17 = 14.1 Amps
For a resistive load, Power P=VI
P=VI = 239.6 x 14.1 = 3377 Watts
Multiply by 3 for each phase
Total power is 10131 Watts or 10.1kW
But this won't be accurate because the resistance
will change when the heater gets hot.

A better way to calculate the power would be to run the heater and measure the current in one of the phases using a clamp tester.

Jamie Downs
[email protected]

 

1) In delta the voltage to each arm is 415 V.
The power in each arm is 415*415/17=10kW (total 30kW)

2) In star the voltage to each arm is 415/sqrt3
The power in each arm is (415/sqrt3)^2 / 17=3.3kW.

The inductance (L) is not interesting because only the resistance (R) will produce heat.

The resistance can depend on the temperature of the arm so the best is to measure the current per arm and the voltage per arm and just multiply.

 

Use Ohm's law to give you current per element.

415v / 17ohm = 24.4 amps = 10.126KW per element

240v / 17ohm = 14.1 amps = 3.384KW per element

 

Use Ohm's law. Inductance is negligible for electric resistance heaters.

For star, Power = 415^2/17 = 10.13 KW

For delta, Power = 3 x 415^2/17 = 30.39 KW

 

Hi Anshuman,

You could try calculating the inductance of the coil from first principles (number of turns, diameter and length) and then calculate reactance at 50 or 60 Hz. This will at least tell you if reactance is a significant factor in the total impedance of the heater. My suspicion is that the reactance of the coil will be negligible and you can get a good idea of the coil current from DC resistance and AC voltage and then calculate power from this. Coils that have significant reactance at power frequency usually need a ferrous core of some type.

Regards

Peter Whalley
Magenta Communications Pty Ltd
Melbourne, VIC, Australia
e-mail: peter*no-spam*@magentacomm.com.au
delete *no-spam* before sending

 

From the nature of the original question it is obvious that the person is lacking some technical expertise in a given subject. (Aren't we ALL??... none of us are experts in EVERYTHING!) Mr. Anshuman exposed his lack of experience in an attempt to seek answers and improve his knowledge. Most of the posts in response were very helpful and provided the needed information. However, the post from Mr. Dobrowolski served only to belittle Mr. Anshuman and did not offer any useful information. I don't believe that responses such as these deserve publication. I firmly believe that all posts should be of a helpful nature, for the benefit of the general audience. If Mr. Dobrowolski finds it too frustrating to respond in a helpful manner to someone asking a basic question, he should perhaps abstain from reading the A-List, because the A-List is by nature a group of people needing help at various levels. We all need help from time to time, and we are not ALL at the top of the technical pile.

Thanks for your consideration,

Gerald Beaudoin

 

Gerald,

Well stated! You are so correct, heck if we were all at the top of the pile why would we need a A-list or forum such as this? I believe the fault lies with the moderator as well for letting such a post to be presented in this forum.

matt

 

> From the nature of the original question it is obvious that the person is lacking some technical expertise in a given subject. (Aren't we ALL??... none of us are experts in EVERYTHING!) Mr. Anshuman exposed his lack of experience in an attempt to seek answers and improve his knowledge. <

Gerald, while I will admit that Mr. Dobrowolski's statements were harsh and perhaps even rude, I cannot totally agree with your statements. As a former college professor I recognize some of the questions posed in this forum as coming straight out of the text book. These students do not seek answers to improve their knowledge. Rather they seek answers which they can plagiarize for their homework assignments.

> Most of the posts in response were very helpful and provided the needed information. <

I feel there are situations where the needed information should be withheld. If someone asked for information on how to make nitroglycerin I don't think this forum would not reply but when a question about current transformers was asked there were endless replies. I can argue from personal experience that the volatility of an improperly handled current transformer is on par with small amounts of nitroglycerin. Sometimes there is a point where the line between being helpful and complicit in aiding abetting a crime must be drawn. When a question is so poorly posed, one must wonder if this is the proper time to be helpful.

Fred Townsend

 

Fred,

I must disagree with you on the point that you seem to be saying no is allowed to ask a stupid question. I for one as a rescue trainer, tell all of my students to ask all of the stupid questions they want, in fact Albert Einstein once said "the only stupid question is the one not asked" (something to that effect). Some questions on this forum do not get answered for any number of reasons, I don't believe that it is because those of us who respond think the person is stupid or asking a stupid question, I believe that they ask becuase they are stuck, require more information or are curious as to the general thoughts and accepted practices in the industry in which we all work. Based on your response I think you should find a new line of work, I for one would not want a professor such has you teaching my children, because to stifle the nature of a individual who is wanting to gain knowledge and clarify for themselves their understanding of the knowledge they have acquired is truly a horrible thing to hear from a former professor. By the way, information can be provided in such has way to lead a person to seek out more information and gain more knowldege about a subject, this can be done through mentoring and leading questions, but to flat out withhold information is wrong and does not serve any purpose, we are not the holders of the keys, we are the ones who should be preparing to pass the keys along so that others may follow in our footsteps, I am glad that I don't withhold that which I know, I just find other ways to share so that others may learn as well. If I have offend you, I am sorry, unfortunately you have struck a cord with me and probably others as well.

Matt Hyatt

 

On July 3, 2003 15:27, matt hyatt wrote: <clip>
> Well stated! You are so correct, heck if we were all at the top of the
> pile why would we need a A-list or forum such as this? I believe the fault
> lies with the moderator as well for letting such a post to be presented in
> this forum.
<clip>

I believe the moderator is doing a fine job, and if there is any blame it should be laid at the feet of whatever person makes statements you object to. The moderator is there to keep the list free of spam and profanity and has been doing so quite successfully. Keeping anyone from making an ass of themself is each person's own responsibility.

Blame the questioner for stupid questions, blame the answerer for stupid answers, but please don't blame the moderator for stupid list members.

--

************************
Michael Griffin
London, Ont. Canada
************************

 

I understand Mr. Beaudoin's point of view and find myself guilty of not being helpful to people who sign their posts with something like "any_name Power Controls" under their name and they do not posses even basic knowledge about that power. I also confess that the way I did it, as Mr. Townsend pointed it out, was rude or maybe even boorish but in this particular case I let myself to express my condemnation to such people. If it was too much I'm sorry. As the benefit of the general audience is concerned - well, I think now the audience knows that a heater from "any_name Power Controls" is a dangerous device because they're not sure how much heat it can generate.

Regards,

Jacek Dobrowolski, Ms. Sc. E. E. Software Engineer

 

A heater current CAN NOT be measured with the formlue as people in this forum said. A cold resistance value should not be used in such calculations.
Also....
All heaters are designed to operate at certain temparature. The resistance of the heater will INCREASE with the operating temp.
So if you can have a 1KW heater with 1000 or 800 or even 500 C as operating temp. and all these cold resistance will be different.
There are two ways to measure the current.
One to use an ammeter and measure the currents in Star and Delta SEPARATELY.
Second is if you know the operating temp and the Nicrome wire thichness and the length or resistance of the wire, refer the mfr book and calculate the current.
regards,
Sekar

 

Michael,

You are correct, it was wrong of me to lay blame at the feet of the moderator, I was responding while a little angry, because of the rudness expressed by an other responder. I do stand by my last response on this matter, though, those of us who respond to help others are providing a servcie based upon on our years of knowledge and we should not judge some else'e questions or them.

We can either respond or not, it is that simple.

Matt Hyatt

 

Sekar,

Don't be so critical. The "cold" resistance is a good start for a calculation. And, I believe most heater users know that the rated kw is given for "hot" resistance!

Regards,
Phil Corso, PE
Boca Raton, FL
[[email protected]] ([email protected]) {[email protected]}

 

Thank you for your concern,

The person who asked the question does not seem to know and it is wrong to misguide him. The cold resistance will not give the actual current or even any thing near it.
You can calculate the inrush current for the first few cycles when we know the cold resistance.

regards,
Sekar