Submersible Water Heater element load question

[Joe Silver]

Thanks in advance for anyone that knows the answer to this. If I have a 12 volt 600 watt dc Low Voltage Submersible Water Heater element that I am using with a 280 watt panel, will I burn the element out by supplying excess voltage to the heater? Or will I be ok, just as long as I do not exceed the 600 watts?

[Mike90250]

It's unlikely that a SUBMERGED 600w element can be damaged by connecting it to a single 280W panel. The panel, even if a 40V panel, will see a loading mis-match, and never be able to develop it's full voltage.
Your mileage may vary
read the operators manual
batteries not included

[Joe Silver]

It's unlikely that a SUBMERGED 600w element can be damaged by connecting it to a single 280W panel. The panel, even if a 40V panel, will see a loading mis-match, and never be able to develop it's full voltage.
Your mileage may vary
read the operators manual
batteries not included

thank you for the info!

[CraziFuzzy]

thank you for the info!

This will also cause it to absolutely never produce anywhere near 280W.

[Joe Silver]

This will also cause it to absolutely never produce anywhere near 280W.

Is there a formula that will give me an idea of what it will produce? If the panel is 40 volts x 7 amps.

What is the best way to power these heaters? Should I be using 600 watts but at 12 volts?

[Mike90250]

are you wanting to heat water, or have a diversion load for something ?

[Sunking]

Is there a formula that will give me an idea of what it will produce? If the panel is 40 volts x 7 amps.

Absolutely! The 600 wat element is a fixed passive resistor, to find the resistance: R = VV/P = .24 Ohms.

Thw panel current is 7 amps. 7 amps flowing through .24 Ohms = 7 x 7 x .24 = 11.76 watts. Voltage will be 7 x .24 = 1.68 volts from the panel. It really sucks.

What is the best way to power these heaters? Should I be using 600 watts but at 12 volts?

You could but 12 volt panels are 18 volts. So to actually get the full 600 watts would require 900 watts of 12 volt panels in parallel, and that would only be for about an hour or two out of the day. Still sucks, that's solar. Best way is to use a MPPT Charge Controller with dump load.

[CraziFuzzy]

I don't know the specific panel you are using, but here are some things we do know:

That heater element is a fixed resistor of 0.24Ω. R = E²/P = (12V)²/600W = 0.24Ω
A Solarworld 280w monocrystalline panel has an Isc of 9.71A.
That panel producing the most current it possibly could, 9.71A, flowing through that heater will produce 22.6W of heat. P = R x I² = 0.24Ω x (8.42A)² = 22.6W

You could eat a bunch of chocolate and sit in the water for a while and heat it faster.

This is not to say connecting a panel to a heating element directly won't work, but it does have to be matched well. That same panel connected to a 120V, 1000W element, for instance, could produce much more:

R = E²/P = (120V)²/1000W = 14.4Ω
P = R x I² = 14.4Ω x (9.71A)² = 1358W (theoretical maximum with the panel as a perfect current source - this is 140V though, so would be far over the panels Voc or 39.5V, which would limit it to:
P = E²/R = (39.5V)² / 14.4Ω = 108.4W

So a 14.4Ω is not ideal either, though it is far out the other end (where the panel is voltage limited)

The idea is to find a resistance that is closest to the panels published max power point, which for my solarworld example is: R = E/I = 31.2V/9.07A = 3.44Ω
Find a heater element with somewhere near that resistance, and a power rating to ensure it doesn't burn up at that power, and you'll be in the best situation for good sun (above numbers based on 1000W/m² solar radiation directly hitting the panel at 25°C.

You might look at the numbers for a 120V 4kW element.

[Mike90250]

Or, you can purchase some heat sink style load resistors

and wire them in parallel, till you get a good match for your panel. Bolt the resistors onto the tank you want heated.
(pic of a 50W resistor)50watt.jpg

Or for a huge boost in efficiency, you can use a solar water heater, with heat pipe tubes in it, close to 50% harvest efficiency.

[Joe Silver]

I don't know the specific panel you are using, but here are some things we do know:

That heater element is a fixed resistor of 0.24Ω. R = E²/P = (12V)²/600W = 0.24Ω
A Solarworld 280w monocrystalline panel has an Isc of 9.71A.
That panel producing the most current it possibly could, 9.71A, flowing through that heater will produce 22.6W of heat. P = R x I² = 0.24Ω x (8.42A)² = 22.6W

You could eat a bunch of chocolate and sit in the water for a while and heat it faster.

This is not to say connecting a panel to a heating element directly won't work, but it does have to be matched well. That same panel connected to a 120V, 1000W element, for instance, could produce much more:

R = E²/P = (120V)²/1000W = 14.4Ω
P = R x I² = 14.4Ω x (9.71A)² = 1358W (theoretical maximum with the panel as a perfect current source - this is 140V though, so would be far over the panels Voc or 39.5V, which would limit it to:
P = E²/R = (39.5V)² / 14.4Ω = 108.4W

So a 14.4Ω is not ideal either, though it is far out the other end (where the panel is voltage limited)

The idea is to find a resistance that is closest to the panels published max power point, which for my solarworld example is: R = E/I = 31.2V/9.07A = 3.44Ω
Find a heater element with somewhere near that resistance, and a power rating to ensure it doesn't burn up at that power, and you'll be in the best situation for good sun (above numbers based on 1000W/m² solar radiation directly hitting the panel at 25°C.

You might look at the numbers for a 120V 4kW element.

Thank you to everyone for the replies....I learned a thing or two