immersion heater for sous vide

[la_benji]

HI guys,
I live in Japan and I've just started to get into electronics with an arduino. But I have a goal to make a sous vide in the next month.
Problem is that I can't find reasonably priced immersion heating elements for the correct voltage in Japan (100V 60Hz). I can find 220-240 V 2000W elements for cheap but I'm not sure what would happen if I tried to use them.
I'm going to be running a 40A relay controlled by a temperature pid. My question is, is it ok to run a 220V 2000W immersion heater element with the 40A set up, or will I destroy my apartment?
My thinking was that it would work but be under powered.
Any help is appreciated.
Thanks for reading.

 

[(*steve*)]

At 100V, a 220V 2000W element (with a resistance close to 24 ohms) would have a dissipation of a little over 400W.

It will draw 4A and the 40W relay (a solid state relay I assume) is fine, since 40A is the maximum it can switch and you want to switch less than a quarter of that.

 

[la_benji]

Hi Steve
Thanks for the help. Would it be possible for you to show me how you got those numbers? I've been trying to figure it out with P=IV and V=IR but I can seem to do it.
Thanks again
Ben

 

[(*steve*)]

Sure.

If you assume that the element dissipates 2000W at 220V, then we can use P = IV to determine the current.

Rearranging, we get I = P/V = 2000/220 = 9.1A

Now we can use V = IR to determine the resistance.

Rearranging we get R = V/I = 220/9.1 = 24.2 ohms

Now we can calculate the current at a lower voltage, so rearranging V = IR we get

I = V/R = 100/24.2 = 4.13A

Now using P = VI directly we vet P = VI = 100*4.13 = 413W

However, we can do it even faster if we combine P = VI and V = IR. We just want something in terms of resistance and voltage. So we rearrange V = IR to be I = V/R and then substitute V/R for I in P= VI. This gives us P = V^2/R

So knowing that The power is 2000W and voltage is 220V, we rearrange to get R = V^2/P - 220^2/2000 = 48400/2000 = 24.2

Then using the formula again for 100V, we get P = V^2/R = 100^2/24.2 = 10000/24.2 = 413W

But... We can do it even faster.

With the knowledge of the previous equations we know that power is related to the square of the voltage. So, the new power is (100/220)^2 * 2000 = (0.455)^2 * 2000 = 0.207 * 2000 = 413W

BUT - w can do it even faster! We can quickly say that because 100V is a bit less than half of 220V that the power will be a bit less than 1/4 of 2000W, so under 500W -- you might guess between 400W and 450W, which would probably be close enough

 

[la_benji]

Haha great! Thanks so much Steve, very very helpful. I was skipping steps and using the wrong voltages etc.
Thanks again.
Ben

 

[la_benji]

Sorry one last question!
If I put 2 elements in series then the resistance will be double (~50Ohms) so the ampage will be half (~2A) so the power will be half too (~200W).
If I do them in parallel then the the resistance of them will be 1/(1/25+1/25)=12.5Ohms, so the current will double (~8A) and the power too will double (~800W).
Is that correct?
Thanks again

 

[(*steve*)]

That is correct.