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.

 

[edison123]

pete, again, it's not theoretical calculations, it's actual measurements. Done twice. With two different sets of metering.

I am done.

Muthu

http://www.edison.co.in

 

[itsmoked]

Muthu; While I believe my logic is correct I am having a bit of trouble understanding why the alternative view offered by yourself is not lining up with mine.

I welcome Pete's input in the hopes it clarifies exactly what the issue is.

I was tempted to wire-it-up at my office then realized I had a Spice simulator that should work just as well. The Spice simulation seems to agree with me.

Perhaps Average verse RMS is the fly-in-the-oinment?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[electricpete]

Perhaps Average verse RMS is the fly-in-the-oinment?

yes. average is not relevant to power calculations. rms is.

muthu - if you have a digital instrument that measures "true rms", please measure your voltage with that. I'll bet the output of the half bridge rectifier is almost 70% of the input.

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(2B)+(2B)' ?

 

[edison123]

pete

The first test was with Fluke digital true RMS meters, hall effect clamp tester for DC load current and regular clamp tester for AC current.

The second test was with good old moving iron voltmeters and ammeters class 1.0 for both AC & DC.

I could actually feel the low heat effect of the heater on the back of my hand.

Keith - Can your simulator provide the voltage and current (both AC & DC) other than the power? That way, we can compare the numbers.

Muthu

http://www.edison.co.in

 

[electricpete]

The first test was with Fluke digital true RMS meters,

please look at the true rms of the input and output voltages (all I saw for output was dc, which again is irrelevant for power calculations)

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(2B)+(2B)' ?

 

[edison123]

How would an ac voltmeter measure DC volts (true rms or not)?

Muthu

http://www.edison.co.in

 

[electricpete]

the dc is not the only thing of interest.

as bacon said, the half wave rectifier output has dc plus even harmonics of line frequency.

you are looking at the dc component and ignoring the harmonics. that will give an artificially low result. a true rms meter combines all the components. I don't know the capabilities of your particular meter.

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(2B)+(2B)' ?

 

[edison123]

pete - May be you can do the test and post the results here? I am basically done.

Muthu

http://www.edison.co.in

 

[zlatkodo]

May be you can do the test and post the results here

I agree.

 

[itsmoked]

Here's the problem.

For a half wave rectifier Vdc = 0.45Vrms = 240 * 0.45 = 108 V DC

This is not correct. You want exactly the RMS value of the voltage and not some fraction of it to calculate the resistor's power. That's the entire point of RMS to render-down a crazy waveform into the equivalent DC heating value.

The RMS value of a half-rectified sinusoidal waveform is correctly given by RMS = V[sub]peak[/sub]/2.

The peak voltage of a 240V sine-wave is not 240V it's 338V.

V[sub]RMS[/sub] = 338V/2
V[sub]RMS[/sub] = 169V

NOT 108V

Keith Cress
kcress -

http://www.flaminsystems.com

 

[itsmoked]

gurse; Make sure you size the dropping resistor to be about 4 times it's power rating. This will limit it's temperature to something that won't instantly burn a human and will keep it from oxidizing into uselessness in a year or less.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[crshears]

Using a simplistic analysis. 40W @ 110V does become 190W @ 240VAC. But, you're not applying the power 1/2 the time, you are applying the voltage 1/2 of the time which leads to applying the current 1/2 of the time which means the average power gets reduced to 1/4.

Aha! Here's where the train goes off the rails.

Let's do this a little differently, ignoring harmonics for the moment...

Run any heater on any AC voltage continuously and you will get X units of heat.

Run that same heater for half a minute, then stop it for half a minute, then on 1/2 a mo, then off 1/2 a mo, and the output of the heater will average one-half of its output with the power on continuously.

IOW you are in fact applying both the power and the voltage half the time, but this does NOT reduce the heater's output to 1/4.

Agreed?

CR

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

 

[gurse]

Having re-considered the options suggested, i will use a diode and a resistor in series with the heating element.

Heater 110V 40 W
I = 40 / 110 = 0.36 A
Heating element resistance = 110 / 0.36 = 305 R

Diode will reduce the 240 rms to 240 * 1.414 / 2 = 170 V rms
A 300 R resistor in series with the heating element will give a current of 170 / 605 = 0.28 A
The heat generated across the resistor and heating element will be 0.28 ^2 * 300 = 24 W.
Reduced wattage will ensure lower failure rate and less energy will be wasted on the resistor.

 

[electricpete]

the half wave rectifier cuts the 192 watts by a factor of 2.

doubling the circuit resistance with series external resistor cuts current by a factor of 2, cuts the space heater watts by a factor of 4.

factor of 2x4=8 reduction in space heater watts from 192 to 24... math checks out.




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(2B)+(2B)' ?

 

[itsmoked]

OK gurse. Use a 100W dropping resistor in your exploits.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[rhatcher]

I gave Zlatkodo the second LPS yesterday and I stand by it.

Sine Wave AC or Full Wave Rectified DC
Vrms= Vp/SQRT(2)
Irms = Ip/SQRT(2)
Prms = (Vp*Ip)/2

Half Wave Rectified DC
Vrms = Vp/(2*SQRT(2))
Irms = Ip/(2*SQRT(2))
Prms = (Vp*Ip)/8

P(half wave) = P(full wave)/4

Keith, add the SQRT(2) to the denominator of your equation and you will get the correct half wave output voltage of 120V. It is only logical that the half wave RMS voltage output is 1/2 of the full wave RMS voltage input.

 

[electricpete]

Half Wave Rectified DC
Vrms = Vp/(2*SQRT(2))

This does not correspond to the rms value of the half wave rectifier output.

I'd think the discussion of switching on a heater for half the time would be enough to satisfy most people intuitively that the average power is 50%.

But here's my attempt at a proof using the rms.

start by examining rms of sinuousoid Vp*sin... (I'm leaving out the argument of the sin function for brevity)

Use the definition of rms (Root Mean Squared):
rms (Vp*sin)= root(mean(Vp*sin)^2)

Move the RHS above to LHS below and equate it to the well known solution for rms of a sinusoid:
root(mean(Vp*sin)^2)= Vp/sqrt2

square both sides:
mean((Vp*sin)^2)= Vp^2/2 [Equation 1]

let hwr represent the output of the half wave rectifier.

the left side of equation 1 is obviously twice as much as mean(hwr^2)....since hwr^2 is the same as (Vp*sin)^2 except for the fact that hwr^2 is zero half the time, so there is a factor of 2 difference in the "mean" (mean = average value over time).

so swap the lhs of equation 1 from: mean((Vp*sin)^2) to: 2*mean(hwr^2) :
2*mean(hwr^2)=Vp^2/2

divide each side by 2:
mean(hwr^2)=Vp^2/4

take sqrt each side:
root(mean(hwr^2)=Vp/2

the left side above you'll recognize as rms. the rms of the hwr output is Vp/2

so the rms of the hwr output is 1/sqrt2 =70.7% of the rms of the input. the heating in a resistor connected to the output will be 50% of the heating in the same resistor connected to the input

P(half wave) = P(full wave)/2

diode voltage drop neglected

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(2B)+(2B)' ?

 

[LionelHutz]

This half the time thing has created a conundrum with thoughts of applying voltage half the time being the same as applying half the voltage.

electricpete pointed out that RMS voltage and RMS current causes the heating and I do agree that to be true.

It made me ponder what RMS fundamentally means and the effect on the RMS voltage with half a sinewave. I now see a rather simple explanation.

Start with a 1V peak sinewave. The RMS of that sinewave is 0.707V.

The mean of the squares must be 0.5.

Removing 1/2 of the sinewave removes 1/2 of the values that were squared and meaned to produce 0.5. This gives a new mean of the squares that equals 0.25.

Root 0.25 gives a new RMS voltage of 0.5V.

Using a 1 ohm resistor.

RMS values for full wave rectified are 0.707V and 0.707A.

RMS values for half wave rectified are 0.5V and 0.5A.

Power goes from 0.5W to 0.25W, or reduces by half.

I hope I got the theory right this time.

 

[electricpete]

that looks correct and succinct to me.

with thoughts of applying voltage half the time being the same as applying half the voltage.

I think you put your finger on the intuitively appealing trap that people might fall into. The mean of the hwr signal is half the mean absolute value of the sinusoid. But of course none of the steps for computing the rms of a signal involves the mean of that signal (it's only the square of the signal whose mean is computed)

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(2B)+(2B)' ?

 

[edison123]

rhatcher

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

For me, it has always been

3-ph bridge full wave rectifier - Vdc = 1.35 Vac
1-ph bridge full wave rectifier - Vdc = 0.90 Vac
1-ph bridge half wave rectifier - Vdc = 0.45 Vac

For some theory,

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

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

Muthu

http://www.edison.co.in