Running a 60 Hz Heating Element at 50 Hz

It will get as hot as usual, however if there is a fan it will run slower. If this a compact heater that relies on the fan to move a lot of air then it may overheat. Fans are cube law devices so that running it at 83% of normal speed will reduce the air volume to 57% of normal.

It is a heating element only.

No motors involved.

Thanks

Berger316

Flow is proportional to speed, pressure to speed², and power to speed³.

I'm curious who voted this OT; clearly somebody doesn't know their affinity laws.

If it's a normal resistive heating element there'll be no difference between using 50Hz and 60Hz.

When energised, either:

• it will produce around 1800W of heat or
• it won't

depending on the overload protective devices on the circuit onto which it is connected.

How do you work that out?

It's a resistive load.

It's voltage dependant.

Think about it.

I think you need to specify the conditions to produce 1800 watts or you are in the OT zone. The overload device will be in one state or another, but not controlling the wattage over a range.

Tony has a point.

I would be surprised if you could tell any difference how this heating element worked using these two different power distribution frequencies. Now if this heating element is part of any appliance that is operated by an internal clock mechanism you might have things actually running 6/5 longer than the time you expect. Most clocks today run off of an internal time reference so not even this is likely.

The heating element will be at zero state(not producing any heat) 100 times a second, at 50 Hz, instead of 120 times a second at 60Hz.

So, my uneducated guess would be a very slight increase in heat produced.

Measurable? Don't know.

Notice that Ohm's law does not have frequency in it, only inductance and capacitance are affected by frequency, so for a purely resistive element frequency has no effect, even down to 0 Hz (DC).

Maybe I'm thinking about induction heaters. Now, I'm not going to sleep tonight until I figure this out. (that's not really true)

Let's see what Wiki says.

Oh, never mind, I defer to you.

I don't want to milk pigeons with boxing gloves. But considered the element is a wirewound coil element, it has some inductance too. Just to help out Mr. Lyn.

This is correct!

But then please explain your #20 after your #5 contributuion.

Thanks

If the windings for the oven are coiled, there could be a change in inductive reactance.

The element will also be at maximum state 100 times versus 120 times per second.

There should be no measurable diffference.

Don't confuse me with facts.

Point well taken.

The thermal mass of the element will even out these excursions.

Dear Mr. lyn,

The Heat produced in a Pure Ressistance Circuit is dependent on Two Factors.

1. Voltage. 2. Resistance.

Power in pure RESISTIVE CIRCUIT is governed by the equation,

P = V^2/R or I^2/R or V x I and in this Equation, NO FREQUENCY is INVOLVED, since it is a PURE RESISTANCE CIRCUIT.

DHAYANANDHAN.S

berger316-

A resistive heater element operating anywhere between the common (or not so common) power frequencies of DC to 400 Hz will not have any noticeable difference in heat output.

Also at 1,800 watts, 6,120 BTU's, it should be able to thaw out you derriere and even your hands at the same time.

Good Luck, Old Salt

Dear Mr.berger316,

For Heating Element, the Frequency Variation has nothing to do (OR) it can be termed as HEAT OUT PUT INDEPENDENT of Frequency. It will develop same Heat.

HEAT developed depends upon the VOLTAGE and RESISTANCE of the element.

The Heat produced in a Pure Ressistance Circuit is dependent on Two Factors.

1. Voltage. 2. Resistance.

Power in pure RESISTIVE CIRCUIT is governed by the equation,

P = V^2/R or I^2/R or V x I and in this Equation, NO FREQUENCY is INVOLVED, since it is a PURE RESISTANCE CIRCUIT.

DHYANANDHAN.S

Quote:HEAT developed depends upon the VOLTAGE and RESISTANCE of the element.

The Heat produced in a Pure Ressistance Circuit is dependent on Two Factors.

1. Voltage. 2. Resistance.

Actually, the current and resistance are the predominant factors and since current can be changed by the total impeadance the inductance and capacitance can change the output slightly, but not enough to be noticible with only a 20% change in frequency. Resistance will be predominant.

Formula 2 isn't correct:

P= U X I and (U = R.I ) so:

P = R.I.I or

R I square or U square/R

R I square or U square/R

Are I square or You square'er

Just a little comic relief

WALOOB

I am surprised to see so many inputs on a simple discussion.

Basically one needs to inform the member raising the enquiry that it is 120V RMS, that is Root Mean Square value or equivalent DC voltage that will produce the same effect and thus is independent of frequency.

If someone puts more difficult or genuine queries - which require expertise in Electrical Engineering and Experience, I generally see comments like: Ask manufacturer, refer to manufacturer's O&M manual, hire an expert, What the standard says, or search the web, do your homework first or polish the wire etc. etc. a sad.....

It will require adding a "flux capacitor" in order to avoid time warp interference.

The element will produce an amount of watts based upon and limited only by the impedance in/of the circuit and the voltage magnitude applied.

"the element is "ON" 50 % and "OFF" 50 %" - only if the supply is a unipolar square wave (i.e. chopped DC)!!!!!

How about, "It doesn't really matter".

<unsubscribe>

Look at Ohm's law and its derivative the Power triangle, where do you see AC, DC, or frequency mentioned? From a Basic Circuits site:

The answer wouldn't be any different if the question was stated as, "What's the power dissipated in a straight piece of nichrome wire that's powered by a power supply that outputs a DC offset triangle wave with a 30% duty factor" or any other such nonsense.

You simply calculate the RMS value of the waveform, square it, and divide by the hot resistance of the nichrome. That's what a "Hot-wire Wattmeter" does, integrates the heating effect of the combined voltage and current imposed on it, waveform doesn't matter.

Really?

Do you know what the actual peak-to-peak voltage is in a 120 VAC circuit ?

Voltage, heater resistance, current flow. None of these will change with frequency, hence wattage won't either.

Yes, RAMConsult and WJMFIRE it was an error in my thinking. I was concentrating on the increased time around the peak value being extended for much longer periods, as the change is much slower there. For any one cycle, it would always be the RMS value, with no change in average heating. But if timed correctly, a heater run at 27X10^-6 Hz, would provide more heat during a night's sleep, for instance, than the same heater run at 50Hz, during the same time period... unless I'm thinking poorly again. (Senility is looming on my horizon.)

Hz. Hz................it Hz me to respond to this...but here goes...

With any frequency, with a given RMS voltage, the current applied to a resistive load will always be the same. This applies to DC and AC. This is true for 0 HZ, 60 HZ, 400 HZ, or 1,000,000 HZ.

There is no exemption. Time (or even overnight) cannot alter this fact.

The RMS current, yes, but if the frequency is non-zero but so low that the period is best measured in hours, the temperature of a heating element would vary significantly during the cycle.

No Sir

If frequency is near zero or it tends to zero, then it is a DC voltage.

In case of DC Voltage Vpeak = Vrms

In case of sinusoidal, 50HZ, 60Hz, 400Hz or any frequency (ignoring skin effect on resistance of element)

Vpeak = 1.414 x Vrms

When we say 120V AC that means 120V RMS and not average or peak values.

Root mean squares means divide waveform in to nth number of small segments with time time 0 deg to ∏/2, each forming a small rectangle of base dθ and height Vθ. Multiply these two to get area of each imaginary rectangle, then square area of each rectangle, add all these squares and then take square root. That is RMS V.

Since Power of any circuit is proportional to square of voltage, hence the RMS voltage thus calculated is the voltage which produces power (in this case heating).

No where frequency comes in to play.

Think this small topic has engaged many as it is pure mathematics or imagination between zero and infinity.

Why not to close this topic - my sincere request to all.

Consider a supply with the frequency reduced to just one cycle per hour (while maintaining the sinusoidal nature of the waveform).

What will be the heating effect during the 10 minute period centered around zero crossing?

How will this compare with the heating effect during the 10 minute period centered around one of the peaks?

OK, the element may burn out as its power rating may be exceeded, but I think you must agree that it will tend to be hotter?

Once the period of supply variation is of the order of or longer than the thermal time constant of the element, the temperature will vary over time.

Do not close this topic until the correspondents get it right! To do so would perpetuate their ignorance.

You are making an argument while changing the paramaters, which is not what the OP is asking.

If the sine wave period is one hour and heat is measured for one hour as well, the heat dissipated is the same as a DC voltage, 60 HZ, etc.

If the sine wave period is 60 HZ and you measure heat dissipated for 1/100000 second, a different result will be measured.

By the same token, if you measure accumulated water in a bucket from a dripping roof, you will get a different rate of accumulation by using an hourly measurement versus a time period so short that the first drip hasn't reached the bucket.

You can call a supply which changes from 0 to 169.68V as a very very poorly regulated DC and not commercial AC.

And if this is a DC fluctuating supply - it should not have so poor regulation, that is, -100% to +41.4% in half an hour.

I am sure that you are an Engineer and must be aware that a good grid stays stable to the extent that variation at Consumer's point is with in +/- 5%.

Hence in this case peak shall be 126V maximum and you will agree that nothing shall happen to the heater element.

You also may be aware why 50HZ/60HZ frequency of power supply was standardized - to avoid flicker in incandescent lamps (specially the carbon coated filament lamps of earlier era) - hence please stay within Engineering parameters - even 1HZ or 0 Hz is not a problem but supply shall stay within limit. Please look again at my earlier explanation, on which you had raised the issue once again, where I said for DC supply Vrms = Vpeak = 120V.

I think this shall be my last comment on this subject.

"You are making an argument while changing the paramaters" - Passerby changed the parameters in post #39, in which he cited a frequency of 27X10^-6 Hz (period = 10.29 hours).

In your reply, you stated that "With any frequency, with a given RMS voltage, the current applied to a resistive load will always be the same". This is inaccurate, and would be better stated as "With any frequency, with a given RMS voltage, the RMS current taken over an integral number of cycles will always be the same".

"If the sine wave period is one hour and heat is measured for one hour as well, the heat dissipated is the same as a DC voltage, 60 HZ, etc." - this is only true of the total integrated heat, not the heat output at any given time.

Passerby was right (assuming that the element didn't burn out!).

Thank you JohnDG for picking up the slack that can happen to posts I make on Friday evenings. I rarely get on CF4 on weekends. And your answer was more efficient that mine would have been.

Not that it needs clarifying, but in the interest of a summation, the answer to the OP question is that for 50Hz the heating would be the same as 60Hz in practical experience. My post that said I would vote for a tad bit more heating at 50Hz was wrong.

Sometimes my convoluted, myopic thinking leads to mixing apples and oranges, without that being my intent. I'm sorry for that happening here and detracting from the discussion of the OP.

Pardon me if I restate material from other posts, but I am trying to bring this all together:

…between 50 and 60 Hz...? It would be hardly noticeable...but not immeasurable. Of note is that the heating element is cycling at 120x per second (60 Hz), or 100x (50 Hz). The frequency (Hertz) measures complete AC cycles. However, the resistive element is powered on both sweeps of the waveform.

Alternating current at a lower frequency, however, will cause production of less heat. In this case the lower the frequency, the lower the duty cycle.

A 10 Hz 120VAC RMS power source produces noticeable flicker in a tungsten filament incandescent lamp. (An incandescent lamp is more of a heat bulb than a light bulb with efficiency around 10% light-wise, so we can use it as a demonstration of the facts.) The 50 or 60 Hz waveform is pretty close to ideal to keep the tungsten filament at heat capacity, thus producing apparent continuous light (and heat) without noticeable flicker. The perception limit of the human eye is many times below this.

High speed footage, however, can catch this.

The heat capacity of the tungsten filament (or heating element material) must be taken into consideration to determine the rate of change.

Therefore, the frequency of the incoming power will have an effect on a resistive heating element. The effect is measurable with the proper equipment and can be calculated, but the effect is not noticeable by human only...unless your perceptions are outside of the norm.

"If the sine wave period is 60 HZ and you measure heat dissipated for 1/100000 second, a different result will be measured." However, 50 Hz is .020 seconds and 60 Hz is about .017 seconds. With this difference of roughly .003 seconds or 3/1000 second. But measurable to .00001 seconds due to ????? Is this an assumption? Perhaps your thinking is similar to the analogy of tank of water leaking at a fixed rate (heat output) yet never overflowing the top because the rate of water pouring in averages out to the same amount (120 V RMS)?

Pardon me if I restate material from other posts, but I am trying to bring this all together:

…between 50 and 60 Hz...? It would be hardly noticeable...but not immeasurable. Of note is that the heating element is cycling at 120x per second (60 Hz), or 100x (50 Hz). The frequency (Hertz) measures complete AC cycles. However, the resistive element is powered on both sweeps of the waveform.

Alternating current at a lower frequency, however, will cause production of less heat. In this case the lower the frequency, the lower the duty cycle.

A 10 Hz 120VAC RMS power source produces noticeable flicker in a tungsten filament incandescent lamp. (An incandescent lamp is more of a heat bulb than a light bulb with efficiency around 10% light-wise, so we can use it as a demonstration of the facts.) The 50 or 60 Hz waveform is pretty close to ideal to keep the tungsten filament at heat capacity, thus producing apparent continuous light (and heat) without noticeable flicker. The perception limit of the human eye is many times below this.

High speed footage, however, can catch this.

The heat capacity of the tungsten filament (or heating element material) must be taken into consideration to determine the rate of change.

Therefore, the frequency of the incoming power will have an effect on a resistive heating element. The effect is measurable with the proper equipment and can be calculated, but the effect is not noticeable by human only...unless your perceptions are outside of the norm.

The resistance of the conductor also depends on the frequency. The resistance of the conductor increase with an increase in frequency and gets decrease with decrease in frequency. This happens because of the SKIN EFFECT.

If 120 V 60HZ 1800 W heating element is supplied 120 V,50 Hz the resistance of the heating element will slightly decrease and the power will slightly increase.

P=V^2/R

Read More:Effect of frequency on Resistance

The key word being slightly. As the graph in your link shows, the difference in skin depth between 50 and 60Hz is minimal.

Will not be difference