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Electric Motor Heater 7
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LionelHutz
Electrical
No, once you have RMS voltage you simply divide by the resistance to get the RMS current.
A period of the voltage waveform is the same amount of time for both the AC and 1/2 wave rectified voltage. So, it would be wrong to only use part of the period when calculating the RMS voltage or current for the 1/2 wave version.
A period of the voltage waveform is the same amount of time for both the AC and 1/2 wave rectified voltage. So, it would be wrong to only use part of the period when calculating the RMS voltage or current for the 1/2 wave version.
Am I following this correctly.
The heater is rated 110 Volts and the supply is 240 Volts. The heater is rated at 40 Watts.
At 240 Volts the heater will develop (240/110)^2 = 4.76 times 40 Watts or 190 Watts.
If the diode cuts that in half it will produce 95 Watts and will burn out.
If Lionel is correct and the RMS with the diode is 169.7 Volts the heater will produce (169.7/110)^2 x 40 Watts = 95 Watts and the heater will burn out.
RMS. A value of AC current that will produce the same heating as a DC current.
Average DC current.
Once upon a time almost all test meters used d'Arsonval movements.
The d'Arsonval movement responds to average DC current.
When an AC current (or an AC current derived from a voltage dropped across a resistor for voltage measurements) the form factor had to be applied to convert from RMS AC to average DC.
The form factor of 1.1 was only valid for a sign wave form.
A couple of times I have seen obvious errors due to distorted wave forms measured with d'Arsonval movements.
Who knows how many times there were errors that were not obvious.
Hint; If you measure and compare the line to line voltage with the line to neutral voltage on an unloaded wye transformer bank, the ratio will not be 1.73. The difference is enough to be obvious.
I have encountered this a couple of times.
Bill
--------------------
"Why not the best?"
Jimmy Carter
The heater is rated 110 Volts and the supply is 240 Volts. The heater is rated at 40 Watts.
At 240 Volts the heater will develop (240/110)^2 = 4.76 times 40 Watts or 190 Watts.
If the diode cuts that in half it will produce 95 Watts and will burn out.
If Lionel is correct and the RMS with the diode is 169.7 Volts the heater will produce (169.7/110)^2 x 40 Watts = 95 Watts and the heater will burn out.
RMS. A value of AC current that will produce the same heating as a DC current.
Average DC current.
Once upon a time almost all test meters used d'Arsonval movements.
The d'Arsonval movement responds to average DC current.
When an AC current (or an AC current derived from a voltage dropped across a resistor for voltage measurements) the form factor had to be applied to convert from RMS AC to average DC.
The form factor of 1.1 was only valid for a sign wave form.
A couple of times I have seen obvious errors due to distorted wave forms measured with d'Arsonval movements.
Who knows how many times there were errors that were not obvious.
Hint; If you measure and compare the line to line voltage with the line to neutral voltage on an unloaded wye transformer bank, the ratio will not be 1.73. The difference is enough to be obvious.
I have encountered this a couple of times.
Bill
--------------------
"Why not the best?"
Jimmy Carter
electricpete
Electrical
You are right on Bill.
I think a lot of the confusion here arises from an electronics site that has incorrect info.
Since this thread has wandered so far from the op, I posted in another forum asking folks if they agree that site is wrong:
thread248-455914
=====================================
(2B)+(2B)' ?
I think a lot of the confusion here arises from an electronics site that has incorrect info.
Since this thread has wandered so far from the op, I posted in another forum asking folks if they agree that site is wrong:
thread248-455914
=====================================
(2B)+(2B)' ?
LionelHutz
Electrical
Bill - why would 95W burn out the resistor not as quickly as 95W?
rhatcher - A half wave rectified voltage is a periodic waveform, not DC. The output of the diode is not a constant DC voltage, it is a periodic voltage which equals the AC input 1/2 of the time and 0V the other half.
Pavg = Vrms^2/R - the correct calculation to use for any periodic voltage waveform.
Pavg = Vdc/R - this is only valid when the voltage is a constant unchanging DC.
crshears - WTF, not happy unless you're disagreeing with someone????
rhatcher - A half wave rectified voltage is a periodic waveform, not DC. The output of the diode is not a constant DC voltage, it is a periodic voltage which equals the AC input 1/2 of the time and 0V the other half.
Pavg = Vrms^2/R - the correct calculation to use for any periodic voltage waveform.
Pavg = Vdc/R - this is only valid when the voltage is a constant unchanging DC.
crshears - WTF, not happy unless you're disagreeing with someone????
Well, doubling the voltage increases the power four times.
Using a diode cuts the wattage in half.
To cut the wattage in half, we must reduce the voltage by root 2.
Gosh. No wonder that the equations for wattage and for voltage are not the same.
And the point is that the heater will probably fail no matter if we cut the power in half or if we reduce the voltage by a factor of root 2.
Bill
--------------------
"Why not the best?"
Jimmy Carter
Using a diode cuts the wattage in half.
To cut the wattage in half, we must reduce the voltage by root 2.
Gosh. No wonder that the equations for wattage and for voltage are not the same.
And the point is that the heater will probably fail no matter if we cut the power in half or if we reduce the voltage by a factor of root 2.
Bill
--------------------
"Why not the best?"
Jimmy Carter
LionelHutz
Electrical
LOL, you got it Bill.
The visual way to show the power change with the diode is to draw the instantaneous power curves. The diode will remove half of the instantaneous AC power curve, hence it must reduce the heater power by half. Keith kind of showed the curves, but not very clearly.
The visual way to show the power change with the diode is to draw the instantaneous power curves. The diode will remove half of the instantaneous AC power curve, hence it must reduce the heater power by half. Keith kind of showed the curves, but not very clearly.
As an advocate toward proper execution of work, the "band-aid" approach of adding anything
to correct one's laziness in solving a problem is... it gets complicated.
The correct answer to the Op's question is to install the properly rated heater in the motor.
It was solved way back on June 27th when Keith Cress (itsmoked) suggested the above.
Any number of creative ways and methods can be derived to power anything.
In this instance, we're talking about a 10 horsepower motor found in an installation
that originally called for a heater driven by 110 volts.
The Op stated there was no 110 volt source to power the heater.
That means, the motor is not part of any original engineered installation.
So, technically... we don't even know if the motor even calls for the heater.
It's being retrofitted to best meet ones needs based on one's budget and means to an end.
To answer the OP's original question of, "... can I put a resistor in series with the heater to reduce the voltage please."
The answer is YES!
You can do what ever you want.
John
to correct one's laziness in solving a problem is... it gets complicated.
The correct answer to the Op's question is to install the properly rated heater in the motor.
It was solved way back on June 27th when Keith Cress (itsmoked) suggested the above.
Any number of creative ways and methods can be derived to power anything.
In this instance, we're talking about a 10 horsepower motor found in an installation
that originally called for a heater driven by 110 volts.
The Op stated there was no 110 volt source to power the heater.
That means, the motor is not part of any original engineered installation.
So, technically... we don't even know if the motor even calls for the heater.
It's being retrofitted to best meet ones needs based on one's budget and means to an end.
To answer the OP's original question of, "... can I put a resistor in series with the heater to reduce the voltage please."
The answer is YES!
You can do what ever you want.
John
LOL Hey Lionel, I disagree with your contention!
![[bigsmile]](/data/assets/smilies/bigsmile.gif "[bigsmile] [bigsmile]")
Mea maxima culpa; upon a more careful scrutiny, I now realize it was me went down the rabbit hole by mis-reading what you had written. Sorry about that!
Your graph shows in exquisite detail what I was trying to get at; nice work.
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
This thread has been challenging. As previously stated, I have replied, as always, from experience and that experience is that in an industrial setting a rectifier's output is considered to be DC and is measured with instruments on the DC setting. I cannot, from experience, vouch for the academic viewpoint that the output is RMS DC (?) as is espoused here nor, as stated, have I ever attempted to measure the output of a rectifier with meters set to AC. For those that wonder what this statement means, yes, I am an engineer but no, I was not taught to analyze rectifier circuits using RMS DC.
Having said that, I have also conceded that the question of considering the rectifier output to be RMS DC is an interesting one to consider, especially in the case of a resistor. However, the fact that I am seeing very few other posters conceding that the DC viewpoint is valid in an industrial setting leads me to believe that there is not a lot of direct experience with rectifiers reflected in this thread. I am in agreement with Edison123, the guy who posted on 28 Jun 19 12:01 that he had actually built a circuit and tested it.
This being said, I am done. To those that read this thread, as always, caveat emptor.
Having said that, I have also conceded that the question of considering the rectifier output to be RMS DC is an interesting one to consider, especially in the case of a resistor. However, the fact that I am seeing very few other posters conceding that the DC viewpoint is valid in an industrial setting leads me to believe that there is not a lot of direct experience with rectifiers reflected in this thread. I am in agreement with Edison123, the guy who posted on 28 Jun 19 12:01 that he had actually built a circuit and tested it.
This being said, I am done. To those that read this thread, as always, caveat emptor.
davidbeach
Electrical
The output of a half wave rectifier is far from being DC. Fed directly into the heater, unfiltered, it is simply every other half cycle of the source AC. It’s zero half the time, that’s not DC.
To begin to talk DC with some ripple you need at least a capacitor. And for a half wave rectifier it needs to be a pretty significant capacitor. I think this is the first post in this thread that has mentioned a capacitor or other filter.
So it’s twice rated voltage half the time, 2 X rated power; not half the voltage all the time, 1 X rated power.
To begin to talk DC with some ripple you need at least a capacitor. And for a half wave rectifier it needs to be a pretty significant capacitor. I think this is the first post in this thread that has mentioned a capacitor or other filter.
So it’s twice rated voltage half the time, 2 X rated power; not half the voltage all the time, 1 X rated power.
electricpete
Electrical
The “DC viewpoint” is apparently that the power dissipated in a resistor fed from a sinusoidal source is reduced to 25% or less by inserting a half wave rectifier between the sinusoidal source and the resistor. This is based on an assumption that the “dc” or "average" value of the voltage waveform (50% or less) should be used to calculate power (when we square 50% or less voltage we get 25% or less power).
Everyone else says that inserting a half wave rectifier between a sinusoidal source and resistor decreases the power to 50% of what it would be if the resistor was fed directly from the sinusoidal source. This is based on either simple examination of the instantaneous power curve (see Lionel’s graphs a few posts above this) or rms concepts (which tell us to use the rms value of voltages for power calculations, NOT average/dc values). The people in the 50% "camp" (*) are:
[ul]
[li]Waross[/li]
[li]LionelHutz[/li]
[li]Davidbeach[/li]
[li]Itsmoked[/li]
[li]IRStuff (from the linked thread in other forum)[/li]
[li]electricpete[/li]
[li]Bacon4life[/li]
[li]Crshears[/li]
[li]Gurse[/li]
[/ul]
That 50% camp is a pretty impressive group (even if I bring the credibility level down). Note also Lionel and Gurse started in the 25% camp and moved over to the 50% camp (people don’t often change their mind from right to wrong after thinking about something).
* It’s actually not something that is subject to to the need for opinions, experiments, or experience. I would venture to say it is covered in the first one or two years of most EE programs (the stuff which sometimes gets forgotten 20 years later). But if experiments float your boat, measuring dc proves nothing, you have to measure the true rms for power calculations.
=====================================
(2B)+(2B)' ?
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- #94
LionelHutz
Electrical
The answer becomes much simpler once you accept this.
As for RMS - there is no such thing as DC RMS or AC RMS. It's just RMS. You can calculate the RMS voltage for ANY periodic waveform.
When all else fails, don't go by your gut, don't go by some canned formula, don't go by a meter reading when you're not sure if the meter is correct. Go right back to basics. In this case, every engineer was first taught to calculate the RMS voltage of a waveform to determine the average power it will produce in a resistor. So, do that with this waveform.
As for RMS - there is no such thing as DC RMS or AC RMS. It's just RMS. You can calculate the RMS voltage for ANY periodic waveform.
When all else fails, don't go by your gut, don't go by some canned formula, don't go by a meter reading when you're not sure if the meter is correct. Go right back to basics. In this case, every engineer was first taught to calculate the RMS voltage of a waveform to determine the average power it will produce in a resistor. So, do that with this waveform.
aussiejohn2
Electrical
Hi rhatcher,
My experience in industrial, mining, oil and gas, and utility is different. Rectifier output is only ever measured with a *true RMS* meter on AC range. True rms hea^X^X^X meters measure the rms value (regardless of waveform), either by physically integrating and averaging in a resistor, or sampling the waveform and doing the rms calculation in software.
However there is no need to rely on anecdote or long-forgotten EE 101. It is not difficult or expensive to build a scale version of the circuit you are proposing (diode and resistive heater) and measure the voltages with an oscilloscope instead of a multimeter. You could do it for under $50 and some time...or with the sound card input of your PC, $1 and some time. If you do, you will be able to measure the voltage and current waveforms directly. You will find that they are "half sinewaves" as the esteemed group suggest. If you multiply the instantaneous voltage by instantaneous current to find the instantaneous power (A more expensive oscilloscope can do this for you, otherwise you can just do it by hand) you will find it follows a sin^2 curve. If you average that power over a period you will find the average power will be half that without the diode, will be equal to the true rms current times the true rms voltage, and will not be equal to the power calculated from an "AC" or "DC" measurement from a meter that is not true rms.
John.
My experience in industrial, mining, oil and gas, and utility is different. Rectifier output is only ever measured with a *true RMS* meter on AC range. True rms hea^X^X^X meters measure the rms value (regardless of waveform), either by physically integrating and averaging in a resistor, or sampling the waveform and doing the rms calculation in software.
However there is no need to rely on anecdote or long-forgotten EE 101. It is not difficult or expensive to build a scale version of the circuit you are proposing (diode and resistive heater) and measure the voltages with an oscilloscope instead of a multimeter. You could do it for under $50 and some time...or with the sound card input of your PC, $1 and some time. If you do, you will be able to measure the voltage and current waveforms directly. You will find that they are "half sinewaves" as the esteemed group suggest. If you multiply the instantaneous voltage by instantaneous current to find the instantaneous power (A more expensive oscilloscope can do this for you, otherwise you can just do it by hand) you will find it follows a sin^2 curve. If you average that power over a period you will find the average power will be half that without the diode, will be equal to the true rms current times the true rms voltage, and will not be equal to the power calculated from an "AC" or "DC" measurement from a meter that is not true rms.
John.
![[blush]](/data/assets/smilies/blush.gif "[blush] [blush]")
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