Electric Motor Heater 7

[gurse]

I have a 7.5 kW motor with 110V, 40W anti condensation heater. But i have only 240 V supply for the heater, can I put a resistor in series with the heater to reduce the voltage please.

 

[rhatcher]

Pete,

Look past the SQRT(2) because it is not relevant to the theoretical question of whether a 1/2 wave rectifier puts out 1/2 power or 1/4 power. The SQRT(2) strawman can be knocked down later.

I will accept the premise of your equation. Remove the SQRT(2) from the denominator of my voltage equation and it matches yours. I assert that the current equation takes the same form as voltage and that power equals voltage times current.

P = V * I
For a half wave rectifier;
Vrms = Vp/2
Irms = Ip/2
Prms = Vrms * Irms = Vp/2 * Ip/2 = (Vp * Ip)/4

 

[electricpete]

P = V * I
For a half wave rectifier;
Vrms = Vp/2
Irms = Ip/2
Prms = Vrms * Irms = Vp/2 * Ip/2 = (Vp * Ip)/4

Then I think we are in violent agreement.

We haven't really defined Ip, but assuming we are hooked up to resistor R, I'll just call it Vp/R.
Then your result for average power output of hwr (call it Phwr)
Phwr = Vp^2 / (4*R)

What is the power if you hooked up the resistor directly to a sinusoid (call it Psinusoid)
Psinusoid = Vrms^2/R = (Vp/sqrt2)^2 / R = Vp^2/(2*R)

Compare the two results, and we see:
Phwr = Psinusoid /2
If direct connection to the sinusoid gave 192w, the hwr would give 192/2 = 96w.

1-ph bridge half wave rectifier - Vdc = 0.45 Vac

Vdc is not used to calculate power. Vrms is used to calculate power.



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(2B)+(2B)' ?

 

[edison123]

pete - I have posted theories. I have posted practicals. Both disprove your contentions.

QED.


Muthu

http://www.edison.co.in

 

[LionelHutz]

RMS is always used when calculating heating.


"It is only logical that the half wave RMS voltage output is 1/2 of the full wave RMS voltage input"

It might seem logical, but this statement is wrong. Anyone who wants to actually test the RMS difference mathematically can do this.
- Put a column of degrees from 0 to 360 into a spreadsheet.
- Take the sine of the angles in the next column. Keep it simple and make a 1V peak signal.
- Calculate the square of each of the voltages in the next column to get the squares.
- Sum the column of squares and divide the sum by 360 to get the mean.
- Finally take the square root of the mean which gives you the RMS voltage. You should get 0.707V.

Now, do the above again but make the negative half of the waveform all 0. You should get 0.5V.

The half wave RMS voltage is not half of the full wave RMS voltage. It seems wrong, but the math says it's right.

 

[electricpete]

pete - I have posted theories. I have posted practicals. Both disprove your contentions

But everything you posted and measured relies on using the average voltage.
The average value Vdc simply does not have relevance for computing average power.
It looks to me like your link is using average voltage for computing average power.

https://www.electronics-tutorials.ws/power/single-phase-rectification.html

Sorry, but if that's what it's saying, that link is wrong.

The instantaneous power of a voltage feeding a resistor is proportional to the square of the voltage. To find the average power you need to average the instantaneous power over time. You cannot do that if you immediatley average the voltage. The higher voltages should get more weighting in the average than the lower voltages (because of the square function), but if you average right off the bat then you don't get that - the voltage average counts all voltage equally. To find the time average of the instantaneous power (remembering instantaneous power is proportional to voltage squared), you need to average the voltage squared. That's what rms does.

Example: square wave voltage going between 0 and 10 volts with equal duty cycle, connected to a 1 ohm resistor. Find the power.

You (Muthu) want to say that we just use the average voltage Vdc = 5 for calculating average power. So average power would be Pavg = Vavg^2/R = 5^2/1 = 25 watts

I (Pete) want to say that we use the RMS voltage. Vrms = sqrt(50) = 7.07. So average power would be Vrms^2/R = 7.07^2/1 = 50 watts.

What is the real answer? Just look at the instantaneous power and take the average. The instantaneous power is 100 watts half the time and 0 watts half the time. So the average power over time is 50 watts. The rms approach gives the right answer, the average voltage approach does not.

This is turning into an odd marathon, on something I wouldn't think should be controversial. I guess I must be enjoying it, otherwise I wouldn't have spent so much time posting this weekend adding to the pile of posts on this subject.


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(2B)+(2B)' ?

 

[Nic.]

I find this rather odd - nobody has practically measured the actual output of a heater in these scenarios, you are still measuring voltage, and then inferring what that means practically. As the differences you are debating are huge- surely you could just rig up a heater and see if it gets hot twice as quickly? Then when that is settled maybe you could turn your amazing hive mind to my PSC question?

 

[itsmoked]

You double the voltage across a resistor you get 4 times the power dissipated. (Back to physics 2A if this is a problem.)

Add a diode which is classically defined as a simple switch. Closed half the cycle and open half the cycle.

That will reduce the power dissipated by half.

Half of 4 times normal power leaves you with 2 times normal power.

It's that simple. It requires no averages, no RMS, no current magic, no voltage magic, no integrals, no derivatives, no hardware, no tests.

Again, simply HALF of the FOUR TIMES normal power in this resistor heater application.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[itsmoked]

Sorry NIC. I don't actually have an answer for your screwy PSC problem. It unlike this subject is a head scratcher.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[electricpete]

For those still advocating that that HWR reduces power to 25%, I would invite you to respond to any of the following:

1 -crshears post 29 Jun 19 11:03 where he talks about cycling an ac heater on and off with a 50% duty cycle. If we flip the switch every month, every hour, every minute, or every second then I think we all agree the average power will be 50% of the power you’d have with the switch continuously on. Why should it be any different if we flip the switch every half cycle with a half wave rectifier?? (if necessary, note the symmetry of the positive/negative half cycles).

2 – Keith’s post 29 Jun 19 03:13 where he went to the trouble to do a computer analysis and share the results. I know it’s hard to view computer simulation results, but if there’s a factor of 2 error it should be obvious there is some difference between the plotted parameters and what we expect them to be. Where’s the difference/error? Tell us what you expect the instaneous power curve to look like and how is it different than what is shown? (imo simply thinking about the shape of the instantaneous power vs time plot - which is the same as for sinuousoid except zero every other half cycle - should reveal the basis for the 50%)

3 – my posts - please correct me on anything in my meandering posts you disagree with. I think my simplest argument illustrating the error of the average dc voltage approach is in my last post 30 Jun 19 04:56 about the square wave. Do you agree a 0/10 volt square wave with 50% duty cycle feeding 1 ohm resistor creates average power of 50 watts (half of the 100 watts that would result if we applied 10vdc continuously).? Do you agree using the average voltage value 5 volts in a power calculation would lead to a prediction of 25 watts which is not the correct average power? If it is not correct to use the average voltage value of a 0/10 volt square wave in a power calculation, then why would it be correct to use the average voltage value of a half wave rectified voltage in a power calculation?
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(2B)+(2B)' ?

 

[crshears]

You double the voltage across a resistor you get 4 times the power dissipated. (Back to physics 2A if this is a problem.) Add a diode which is classically defined as a simple switch. Closed half the cycle and open half the cycle. That will reduce the power dissipated by half. Half of 4 times normal power leaves you with 2 times normal power. It's that simple. It requires no averages, no RMS, no current magic, no voltage magic, no integrals, no derivatives, no hardware, no tests. Again, simply HALF of the FOUR TIMES normal power in this resistor heater application.

First principles! I love it! 4 LPSs!

Once this basic arithmetic part is accepted, the real-life adjustments for diode voltage drop etc. can be applied to tweak the calculations.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[itsmoked]

Hi crshears. Cracks me up the wrong answers get the LPS. Hopefully visitors don't take them to mean those were the correct answers.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[crshears]

I can only recognize what I see as the right way to go...anything else is out of my control.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[rhatcher]

Wow, I'm surprised that this question is still not correctly answered. It seems like there is agreement that a half wave rectifier only provides 1/2 voltage and it should be apparent that if only 1/2 of the voltage is present, only 1/2 of the current is drawn. Per Ohm's Law; 1/2V * 1/2R = 1/4P and, coincidentally, P = V^2/R. What am I missing?

 

[electricpete]

you're a smart guy, Ray. I appreciate your knowledge on motor repair.

No disrespect intended, but in this particular case, you seem to be stuck in a rut of thinking only one way, which happens to be incorrect (you're talking about 50% voltage, which is an average voltage, but average voltage has no relevance to power calculation...rms voltage is used for power calculations).

maybe read back over what the rest of us have said. I tried to summarize some key points in my post dated 30 Jun 19 15:59.


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[bacon4life]

rhathcher- There is no "agreement that a half wave rectifier only provides 1/2 voltage." A half wave rectifier provides voltage half of the time. Energy linearly correlates to time whereas energy correlates to the square of voltage. In equation form:
V^2*(t/2) does not equal (V/2)^2*t

As an example with a more physical representation, what happens to the area of a circle if the radius is cut in half?
pi*R^2/2 does not equal pi*(R/2)^2

 

[LionelHutz]

Wow, I'm surprised that this question is still not correctly answered. It seems like there is agreement that a half wave rectifier only provides 1/2 voltage and it should be apparent that if only 1/2 of the voltage is present, only 1/2 of the current is drawn. Per Ohm's Law; 1/2V * 1/2R = 1/4P and, coincidentally, P = V^2/R. What am I missing?

No, it's not agreed on. Try reading my last post. The diode reduces the RMS voltage to 71%, not 50%. Do the basic RMS calculation math on a sinewave signal as I suggested to prove it to yourself.

Just think what RMS means. The square root of the mean of the squares. So, work through the numbers.
The RMS voltage is 240V. So the mean of the squares is 57,600.
The diodes gets rid of half of the waveform, so it will cut the mean of the squares in half to 28,800.
Take the square root of that and you get 169.7V RMS. Definitely not half of the 240V RMS.


As I posted before, it's confusion between providing the voltage half the time and providing half the voltage.

 

[rhatcher]

I have read through the entire post again and I figured out 'what am I missing'. The prevailing analysis in this thread is based on AC RMS voltage. My understanding, education, and experience is that the output of a rectifier is considered to be DC. My equations from 29 Jun 20:00 were correct except for the 'form factor' of 0.9 which changes the single phase 1/2 wave voltage output from being 0.5 to 0.45 of the AC RMS input voltage. Edison123's post from 30 Jun 01:07 supports this. I do not have my textbooks at hand but I feel confident that this represents the prevailing theory for rectifier circuits; they are DC.

https://www.electronics-tutorials.ws/power/single-phase-rectification.html

Having said this, it is an interesting question that, although going against theory, is worth pondering. How would a heater, essentially a resistor, know the difference? Why wouldn't it respond to the AC RMS voltage instead of the average DC voltage. In all honesty, I don't know. My answers have been, as they always are, based on theory as I understand it and hands on experience. I've never measured a rectifier output with an AC RMS meter because my belief and training has been that there is no point, it is a DC circuit. The difficulty of answering this question is that there is no practical way to measure the actual heat output of a heater operating on a rectifier circuit, this value is always derived based on input voltage and current...in DC.

 

[crshears]

Lionel, I'm no mathematician, but I think I see where you're going...and I humbly submit that I might also see where you're going down a rabbit hole, viz., you are using all of the time in the cycle for your calculations whereas it seems to me to be generally agreed on that, viewed simplistically, and ignoring diode drop for the moment, for half the time there is no current. Consequently, it is IMHO incorrect to use all of the time to calculate what's going on.

Scanning the skies for incoming tomatoes...

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[rhatcher]

INCOMING!!....LOL, I'm with you crshears

 

[electricpete]

crshears - What Lionel posted is absolutely correct. The rms voltage of a half wave rectified voltage is 1/sqrt2 times the rms of the original sinusoid. This leads directly to output power of 50% when the voltage is applied to a Resistor.

But there are other ways to look at it. As you pointed out, turn an ac heater on and off with a 50% duty cycles and we can deduce the average power is 50% without invoking rms concepts.

There are many correct ways to solve the same problem. There is however only one correct answer, which you and Lionel agree on: the average power of resistor fed by half-wave rectified sin wave is 50% of what it would be if you applied the sinusoid directly to the resistor (diode voltage drops neglected as you mentioned).

rhatcher - I have mentioned that website is not a reliable source. You can't believe everything you read on the internet. Maybe you say that applies to eng-tips (myself, Lionel, crshears, itsmoked etc who are saying 50%). I guess so. Don't believe it because we say so. At some point you need to think carefully enough about the problem to convince yourself of which answer is right. I would suggest to draw graphs of v(t) and p(t)=(v(t))^2/R for the two situations. Then compare the average value of p(t) between the two situations of resistor fed by sinusoidal voltage and resistor fed by ideal HWR output. phwr(t) for HWR looks identical to psin(t) for sinusoid, except that phwr(t) is zero half the time. The inescapable conclusion looking at the graphs will be that the average value of phwr(t) over time is half the average power of psin(t). of note, nowhere during this entire process did we care about the dc value or average value of v(t). we care about the average value of p(t) which is proportional to the square of v(t). If we wanted a representative voltage to use for this power calculation, we would not use the dc value of the hwr output voltage, we would use the rms of the hwr output voltage (since rms squared = square(root(mean(square)) = mean(square) = the average of v(t)^2 which is what we care about for the power calculation)

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