I will post some shunts that I find. I will try looking for some immersion ones.

red leader, are you sure i can measure it with a clamp meter, and will it be as accurate as the shunt??

looking at the immersion heaters on mcmaster im quickly comign to the conclusion that they are not the way to go. making my own coper block that seals a normal heater cartridge from the water seems a more sensible way of going about things.

Ben, dont most of the DC psus go up to like 36volts max? if that is the case that would need almost 10 amps to get to 350watts of load..

check out this mcmaster part. at 120volts its 1200watts. so its 10watts per volt.. at 35volts thats 350watts... perfect for waht I need and most of the psu's on ebay are of the 35 volt variety...

4877K466
(Same as 3618K287)
Cartridge Heater With SS Lead Covering High Temp, 120 Volts, 3/4" Dia, 6" L, 1200 Watts

Ok.

1200W and 120 volts, that means 10 amps max.

You're shooting for 350 Watts.

Using:
P = V * I
and V = R * I

we get P = R * I * I (or P = R * I^2)

P=1200
I=10
so R is going to be 12 Ohm.

using:
P=350
and R=12
we get 5.4 amps.

The voltage would be 65 VDC.


If you could find a heater that would be less resistive, you could lower that voltage.

any suggestions on the specs i will be needing??

at this point im kinda thinking i should reduce my 350watt requirement to 250watts. It would be alot more feasable no?

using:
P=250
and R=12

I=4.56
and V=54

No big difference.

That heater above is fine. The only thing is that if you could find one with a lower resistance, you could pump more power into it, with a lower voltage. You might just want to stick with that one, and concentrate on finding the PSU.

I think I've got other heater cartridge manufacturers in the WBTA WebLinks, if you really want to check it out.

hmm... well i guess i will have to find one with lower resistance.. fun fun fun in the sun..

well i found one that is 2000watts at 120volts..
R=7.5Ohms..

not sure on the calculations that i will need otherwise.. can you show me the foruma to calcualte the amperate and volage needed.

Sure.

The equations are:
P = V * I
and
V = R * I

Where:
P = power (Watts)
V = voltage (volts)
R = resistance (Ohm)
I = amperage (amps)

With a substitution, you can get P=VI to be P=R*I^2.

What we know is:
P=2000 W
and
V=120

so I resolves to 16.7 amps.

then swapping V=RI to read as V / I = R, you get R = 7.2 Ohm.

Now for the target power. What we know is that R=7.2 Ohm, and that P should be 350 W.
P = 350 W
R = 7.2 Ohm

so using the substitution above, where P=R*I^2, and a little swapping around, we get I = sqrt (P / R). I resolves to 6.97 amps.

Then, we go back to V=R*I, and we get V = 50 volts.

So now your task is to find a DC power supply, capable of outputting at least 50 volts DC and at least 7 amps. tip: if the PSU is capable of more, it'll last you longer; there's no sense in taxing a power supply to its limit, all the time. Also, look out for voltage drifts as the unit warms up.

hmm thanks

btw I'm gonna ask my electrician friend if he can find an autotransformer for AC that willl stablalize the voltage so that it will give constant voltage.