Electric Element Voltage Specifications?

[mcharbs55]

Hello,

I have a question about electric elements and their corresponding voltages. I am looking at powering a 1500W 120VAC element out of an ordinary kettle with a DC voltage supplied by solar panels.

So, from what I understand, 120VAC represents the RMS voltage meaning that it would require 120VDC to operate correctly...but why? Is this the minimum rating or maximum rating? Should the element not operate correctly on all voltages; at 24vdc it would suck 62.5A instead of the usual 12.5A at 120VAC. As long as there is enough wattage provided, should the voltage not be irrelevant?

Just trying to figure this out...It should work with my array since it will be 155VDC and 1375W. Other sources say that the voltage must be higher than the rated voltage and the wattage of the panels should never increase the wattage of the element, this is why these values were chosen for the array.

Thanks in advance for any help!

 

[Henryk]

Hi,
A heater element of a kettle is just a piece of a resistive wire. The relationship between the current and the voltage for a resistor is given by Ohm's law V = I * R. The power dissipated is V* I and that equals to V²/R = I²R.

The kettle was designed to provide 1500 W of power at 110 V supply voltage (rms if AC or actual DC voltage). What it means is that the resistance of the heater element was made to be (110 V)²/ 1500 W = 8.066 Ohm.

To answer your questions: no, the voltage is important.
The kettle will actually work at any voltage applied, it is just that the power will be dependent on the voltage.
If you apply 24 Volts, the power will be 24²/8.066 = 71.4 W

Now, to your solar panel. You state it provides 1375 Watts at 155 Volt. I suppose you can use it to power the kettle. Applying 155 Volts to a 8.066 ohm resistor will dissipate close to 3000 Watts of power which is double that of what your solar power specs are saying. For a panel, if you try to take more current that it can provide, it will simply reduce the output voltage. So the power you will get out of your solar panel to the kettle will be something in between 3000 and 1500 Watts.
yup, you will be able to make your coffee using solar panel to power your kettle.
You can find out the actual power if you have the full specs for the panel. It will depend on the intensity of the sunlight.

 

[sophiecentaur]

1500W 120VAC element out of an ordinary kettle with a DC voltage supplied by solar panels.

Basically, you need an array of panels, covering several metres squared to supply 1500W. That's a major installation and, with the voltages and currents involved, requires professional experience or knowledge to guarantee safety.
If you want to use a more modes installation then you would need to match your load to the supply you are giving it. That element will have only about 10Ω resistance. That will be almost like a short circuit applied across the solar panel. You would need around 120V DC's worth of panels (i.e. 10 connected in series) and they would need to produce more than 10A. If you connect a 10Ω load to a single panel, the output voltage would be very low (a volt or less, say) and there would be no power delivered, worth having)
There are immersion heaters sold for use in cars, which work on 12V and take a very few Amps. They will heat a cup of coffee, eventually, and are quite popular. Also, a system of one (appropriately rated) panel and one such heater is quite safe to use.
I could point out that a cheaper way of providing that sort of heating effect would be a parabolic (not too high spec needed) reflector. More power would actually be available for a given area of reflector than the same area of PV cells and it would be very robust.

 

[Asymptotic]

What Henryk said, except instead of 110V used in his calculations it should be the 120V rating. This works out to an element resistance of 9.6 ohms rather than 8.066 ohms. One thing to keep in mind - manufacturers often specify wattage rating at operating temperature. Heater element wire has a positive temperature coefficient (often abbreviated as 'tempco') which means resistance increases as temperature increases, so don't be alarmed if an ohmmeter measurement at room temperature yields a value somewhat lower than the calculated 9.6 ohms.

>> 120 VAC ... Is this the minimum rating or maximum rating?
120 volts is a nominal voltage rating. Usually, heater manufacturers bracket the nominal rating by +/-10%, so this element was designed to operate from 108 to 132 volts.

powering a 1500W 120VAC element out of an ordinary kettle

Watt density plays a role, too. Watt density is the amount of power (watts) divided by area (US manufacturers still often specify in square inches) and typically ranges from 50 to 90 W/SqIn for immersion heaters. Ultimately, the limiting factor is sheath surface temperature. Immersion heaters are wattage rated with the expectation they'll be completely covered by fluid, and will fail prematurely (sometimes, the element sheath gets hot enough to melt) if operated in open air.

 

[sophiecentaur]

at 24vdc it would suck 62.5A

I just read this. it is totally wrong. The Current is given by
I = V/R
You reduce the volts and the current is reduced accordingly through a given value of resistor. You are confusing the fact that a (purpose built) 24V heater would need to draw 62A. The heater would need to be less than half an ohm resistance!

So, from what I understand, 120VAC represents the RMS voltage meaning that it would require 120VDC to operate correctly...but why?

That's just basic theory. The RMS (Root Mean Square) voltage of a varying voltage tells you the average power that's delivered. What it does is to add the various different powers for the values of the varying voltage over the cycle and find the average of those values. That allows you to say what the power for a equivalent single (DC Voltage) supply would be. You have to use RMS because the mean Voltage for a sinusoidal AC waveform is Zero - which would not be what you want.

But you are still not describing your actual setup in enough detail. Are you really using such a big PV array, operating at a domestic supply voltage? Are you just planning to use it as a heating source, rather than for an electrical supply. That could be very inefficient use of your available area. But if you already have the panels, you would obviously want to use them. Are you planning to store the energy in the form of hot water in a tank? Are you planning to make this system properly safe? You have a fire risk and a shock risk which need to be catered for.

 

[CWatters]

In cases it's of interest you can buy a box of tricks that works out how much electricity your house is consuming and diverts any excess generation from PV to an immersion heater.

 

[mcharbs55]

Hi,
A heater element of a kettle is just a piece of a resistive wire. The relationship between the current and the voltage for a resistor is given by Ohm's law V = I * R. The power dissipated is V* I and that equals to V²/R = I²R.

The kettle was designed to provide 1500 W of power at 110 V supply voltage (rms if AC or actual DC voltage). What it means is that the resistance of the heater element was made to be (110 V)²/ 1500 W = 8.066 Ohm.

To answer your questions: no, the voltage is important.
The kettle will actually work at any voltage applied, it is just that the power will be dependent on the voltage.
If you apply 24 Volts, the power will be 24²/8.066 = 71.4 W

Now, to your solar panel. You state it provides 1375 Watts at 155 Volt. I suppose you can use it to power the kettle. Applying 155 Volts to a 8.066 ohm resistor will dissipate close to 3000 Watts of power which is double that of what your solar power specs are saying. For a panel, if you try to take more current that it can provide, it will simply reduce the output voltage. So the power you will get out of your solar panel to the kettle will be something in between 3000 and 1500 Watts.
yup, you will be able to make your coffee using solar panel to power your kettle.
You can find out the actual power if you have the full specs for the panel. It will depend on the intensity of the sunlight.

Thank you for your reply,

I was thinking of it completely backwards. This cleared things up for me quite a bit and was very easy to follow.

I did the calculations with a 120v rating and got 9.6ohms for the element.
I then calculated the actual power of the panels with that ohmic value using v^2/R and my panel voltage of 155V giving me 2502.6W
I then calulated the required current to supply this wattage using p=I^2R and got 16.15A
Since the panels are series connected I only have a current of 8.8ish amps rated.
I did the voltage drop two ways... 8.8/16.15= 0.545... (.545)(155)= 84.48V
OR vdrop= IR= (8.8)(9.6)= 84.48V
So around 84.5V.
I then used that voltage to recalculate the actual wattage produced...p=v^2/R= 84.48^2/9.6= 743.42Watts! Just under half the rated wattage of the Element.
It was all the same thing as doing I^2R with the actual current of 8.8A...I just couldn't believe how small the output wattage was from 5 panels so I double checked! Am I doing something incorrect here? You mentioned that it should be between 1500 and 3000 watts...how did you get that value?

Thank you again for your help.

 

[mcharbs55]

I just read this. it is totally wrong. The Current is given by
I = V/R
You reduce the volts and the current is reduced accordingly through a given value of resistor. You are confusing the fact that a (purpose built) 24V heater would need to draw 62A. The heater would need to be less than half an ohm resistance!

That's just basic theory. The RMS (Root Mean Square) voltage of a varying voltage tells you the average power that's delivered. What it does is to add the various different powers for the values of the varying voltage over the cycle and find the average of those values. That allows you to say what the power for a equivalent single (DC Voltage) supply would be. You have to use RMS because the mean Voltage for a sinusoidal AC waveform is Zero - which would not be what you want.

But you are still not describing your actual setup in enough detail. Are you really using such a big PV array, operating at a domestic supply voltage? Are you just planning to use it as a heating source, rather than for an electrical supply. That could be very inefficient use of your available area. But if you already have the panels, you would obviously want to use them. Are you planning to store the energy in the form of hot water in a tank? Are you planning to make this system properly safe? You have a fire risk and a shock risk which need to be catered for.

Thank you for your replies. I was thinking of it completely wrong as you stated. I am boiling water with solar panels for a school project and am just trying to figure out how many panels I need so my current setup is nothing! It is looking like I will need far more panels than I anticipated with my most recent calculations, still in the design phase and will be seeking professional help when the time comes to ensure that it is a safe installation.

Thank you for your help.

 

[mcharbs55]

What Henryk said, except instead of 110V used in his calculations it should be the 120V rating. This works out to an element resistance of 9.6 ohms rather than 8.066 ohms. One thing to keep in mind - manufacturers often specify wattage rating at operating temperature. Heater element wire has a positive temperature coefficient (often abbreviated as 'tempco') which means resistance increases as temperature increases, so don't be alarmed if an ohmmeter measurement at room temperature yields a value somewhat lower than the calculated 9.6 ohms.

>> 120 VAC ... Is this the minimum rating or maximum rating?
120 volts is a nominal voltage rating. Usually, heater manufacturers bracket the nominal rating by +/-10%, so this element was designed to operate from 108 to 132 volts.

Watt density plays a role, too. Watt density is the amount of power (watts) divided by area (US manufacturers still often specify in square inches) and typically ranges from 50 to 90 W/SqIn for immersion heaters. Ultimately, the limiting factor is sheath surface temperature. Immersion heaters are wattage rated with the expectation they'll be completely covered by fluid, and will fail prematurely (sometimes, the element sheath gets hot enough to melt) if operated in open air.

Thank you for helping me clear this up, I appreciate the help.