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]

Not sure where are you going with this, Keith.

It's an insult to the planet. It will consume an additional waste of 438kWhrs a year. That's 1.13 x 438 = 500 pounds of completely pointless carbon dioxide.

Muthu

http://www.edison.co.in

 

[crshears]

Hang on there, che; I didn't propose anything at all! Looks to me like you were right the first time; with all due respect to zlatkodo, I agree with you and Keith that his proposal is not the way to go.

CR

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

 

[edison123]

For a single phase, half wave rectifier (what zlatkodo suggests), isn't Vdc supposed to be around 45% of Vrms, which in this case would 0.45x240 = 108 Vdc?

Muthu

http://www.edison.co.in

 

[waross]

We don't want to increase the voltage to the point that a core saturates.
For a given coil/core there is a maximum safe voltage just below the saturation knee of the flux curve.
But the inductive reactance increases with increased frequency.
We could say that the safe voltage for a 60 Hz coil to avoid saturation is 120 Volts.
If we express that as 120V/60 Hz or V/Hz = 2, then we can use any frequency and voltage combination that the insulation will withstand. eg: 60 Volts @ 30 Hertz, 240 Volts @ 120 Hertz.
Originally the V/Hz was used to determine safe working voltages for transformers and motors being swapped from 50 Hz to 60 Hz and vice versa.
Now V/Hz is much used in conjunction with VFDs.
It is a maximum and the most efficient voltage.
eg: Running a transformer with a V/Hz-2 at 1/2 voltage will not harm the transformer but will cut the KVA rating in half.
eg: 10 KVA, 240 Volt transformer used at 120 Volts. No harm to the transformer but the safe load is now 5 KVA.
Specifically, a 380 Volt transformer used on 240 Volts will work fine but the new KVA will be 240V/380V x the rated KVA.
Or; If the transformer current is not listed, use Voltage and KVA to calculate the maximum allowable current.
That current times the applied voltage will give you the new VA rating.
I hope that this makes it better, not worse.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[crshears]

It all helps, Bill.

As to Keith's "insult to the planet," it makes sense if the electricity is generated with fossil fuels; waste of energy will create needless greenhouse gases.

CR

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

 

[itsmoked]

What crshears sez Muthu. Using energy derived by any manner to do something pointless is lame. Using a resistor to drop voltage is, of course, just that! It also will likely create a hand burning hazard and if not "properly installed" also an electrocution hazard. The offered transformer solution covers the enclosure and wiring connections protection issues simply.

As for the V/Hz issue it's everything the transformer maestro (Bill) said and the one I linked was a 250VA version so it has way more than enough capacity to run a 20W/40W heater. It's bigger than I'd normally suggest except it had all the taps and the ability to directly run conduit to it that no control transformer will provide.

If there already is space available in some control panel then any old 100VA control transformer will work just fine. If it's a 380V to 120V version 50VA would even suffice.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[gurse]

Hi Keith and Waross
For a half wave rectifier Vdc = 0.45Vrms = 240 * 0.45 = 108 V DC
Effectively, the voltage across the heater will be the same as applying 110V AC so the power will be almost the same (110V 40W) whether you apply 110V AC directly or 240V AC through a diode.

 

[itsmoked]

Hi gurse.

Remember the power goes up with the square of the voltage. Hence the 40W@110V heater would put out 190W at 240V.

P = V[sup]2[/sup]/R
40W = 110[sup]2[/sup]/R

R = 110[sup]2[/sup]/40W
= 302Ω

Likewise:
P = (240-0.7)[sup]2[/sup]/302Ω (0.7V ~ diode drop)
= 190W

The diode is going to block one direction which cuts the power by 1/2 or 190W/2 = 95W.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[gurse]

Hi Itsmoked,

The way I look at it was as follows,
Vdc across the heater coil is 240V rms * 0.45 = 108 through a diode (half wave rectification).
Current across the heater coil is V/R = 108/302 = 0.36A
and power = I^2 * R = 38.6 W.

Therefore, 110V AC directly or 240V AC through a diode will produce the same heat (40W) on the heater.

 

[crshears]

No, it doesn't work like that.

Keith is right; with a diode in series you are not effectively applying 110VAC; you are in actual fact applying 240VAC half the time.

Think it through.

CR

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

 

[edison123]

I did some tests with a single diode today.

No-load test:
AC Input / DC Output
100V / 60V
200V / 123V
240V / 148V

Load test with a 220 V, 1000 W heater:
AC Input / DC Output
100V, 0.92A / 45V, 0.95A
200V, 1.80A / 89V, 1.7A
240V, 2.15A / 106V, 2.0A

LPS for zlatkodo for this thinking outside the box.

Keith - Thanks, I wasn't sure if you were jesting about power wasting.


Muthu

http://www.edison.co.in

 

[gurse]

Calculations rule out using a diode so I will go ahead and use a resistor in series with the heater coil as planned originally and waste some energy in the process.

 

[LionelHutz]

I would agree with zlatkodo and edison123. It also works in theory which backs up the testing proving it works.

For a full wave rectifier feeding a resistor, the equivalent DC voltage is the average voltage of the AC waveform. For a sinewave, the average voltage is 90% of the RMS voltage. For 240VAC the average DC voltage would be 216VDC. For a half-wave rectifier it would then theoretically be 108VDC since the voltage is only applied half the time. Applying 108VDC would give a very similar heating as 110VAC would.

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.

Look at Edisons numbers. At 240V there appears to be an input power of 516W and an output power of 212W. We know the diode isn't dissipating 304W so that means the input wattage can't be correct. I've seen similar input vs output power arguments when metering a phase controller - people trying to figure out where the power is going except it's not being consumed in the first place.

 

[che12345]

Dear Mr gurse (Electrical)(OP)28 Jun 19 14:50

1. "Calculations rule out using a diode so I will go ahead and use a resistor in
series with the heater coil as planned originally and waste some energy in
the process". I would strongly urge you to reconsider the use of [ step down transformer
proposal].

2. From all the learned 32 replies, [NOT] a single person supports the use of " ...a
resistor in series with the heater ..." This 33rd reply also objects your insistence
of using a resistor in series...

4. As a responsible engineer, please do [NOT waste] the energy/money unnecessarily.
Save the planet for [us], if you do not care! The choice is yours but be considerate
for others as well.

Che Kuan Yau (Singapore)

 

[bacon4life]

zlatkodo, Per IEEE 519 Recommended Practices and Requirements for Harmonic Control in Electrical Power Systems 7.3.2:

Half-wave rectification produces even harmonics that have a dc component that saturates transformers. These are to be avoided.

Muthu, In your testing, what do you mean by DC output? The output of a half wave rectifier is a mix of DC and even harmonics. The heater responds to the total RMS, not just the DC component. Many meters only measure average values. In order to display an "RMS" value to the user, the meter uses a multiplier for either DC or sinusoidal AC (or perhaps even for a triangle or square wave).

Lionel, With double the voltage plus a diode, the current is doubled for half the time. Since average power is I^2*R *t/T, this results in ~80W.

 

[itsmoked]

Thanks bacon.

To check for myself I decided to simulate it on everyone's favorite simulator LT SPICE. I used a 301Ω resistor (a standard size).

I used a 60Hz sine wave and since this all shows instantaneous you'll note the 338V peaks from 240Vac.

Here's no diode: The Average power shown in the "waveform box" is 187W without a diode.

Here's with a diode the average shows 91W:

Keith Cress
kcress -

http://www.flaminsystems.com

 

[edison123]

bacon
AC voltage measured across diode. DC voltage measured across heater.
AC Current before the diode. DC current after the diode.

I double checked today with all analog meters (as compared to the above with all digital meters) and the readings are almost the same.

Keith - So, the DC voltage is 48% of AC input?

Muthu

http://www.edison.co.in

 

[electricpete]

I'm with Keith, Bill, bacon, and crshears.

The average dc value of the half wave rectified voltage has no relevance for computing watts of a connected resistor.

That's why we have rms (computing watts). the rms of a half wave rectified 240v rms ac wave is 240/sqrt2. The heating in a resistor connected to that is
(240/sqrt2)^2/R = [240^2/R]/2
that is half what it would be if you connected it directly to 240 (ie in excess of 80 watts).


=====================================
(2B)+(2B)' ?

 

[edison123]

pete

The testing numbers don't lie. Using a single rectifier did reduce the output DC voltage to about 45% of AC input, the current for 1 KW heater did fall proportionately along with the power (voltage squared). So, using zlotkodo's idea should meet OP's requirement.

Muthu

http://www.edison.co.in

 

[electricpete]

Using a single rectifier did reduce the output DC voltage to about 45% of AC input,

the dc / average is irrelevant to power computations.

along with the power

power computed by multiplying average / dc current times average / dc voltage would be meaningless

=====================================
(2B)+(2B)' ?