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Author Topic: Connecting 24v DC to 3kw 230v AC  (Read 4337 times)
camillitech
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« on: April 08, 2009, 10:48:39 AM »

Morning chaps,

Having just successfully upgraded my mates hydro turbine to around 100m of head  Shocked (can't be specific cos the pressure gauge is off the scale  Grin) he now finds himself with two DC dumps constantly heating his shed  Roll Eyes he does have a 24v immersion element to replace the unused 3kw 230v in his thermal store but I'm not going to be able to fit this for him for a couple of weeks at least and of course the system will need drained down. Is there any reason why I can't just connect the DC dump to the original 230v element for now? We have the turbine putting out about 600w now but it's capable of much more. The original element will have to come out anyway as it has a slight leak but it seems a shame to be wasting all this power.

Cheers, Paul
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billt
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« Reply #1 on: April 08, 2009, 11:19:19 AM »

You can connect your 24v DC to a 230v immersion, but it won't dump much heat. Power dissipated is proportional to the square of the voltage, so 24v will only dissipate about 30w in a 3kw/230v heater.
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rob26440
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« Reply #2 on: April 08, 2009, 11:35:09 AM »

Was just about to post this.  And got beaten to it by billt!  And I agree with him.  However, here it is for what it is now worth.....

Here are my calcs on using 24v DC into a 240v 3kW immersion heater....

Ignoring the a.c. effect and just seeing the heater as a pure resistive load, using Ohms laws and DC. Then, using W/V = Amps, a 3kW immersion at 240v = 3000/240 = 12.5 Amps.  Resistance of the 3kW heater = V/I = 240/12.5 = 19.2 Ohms.  So at 24volts DC then your current = 24/19.2 = 1.25 Amps.  Watts = V x I = 24 x 1.25 = 30 Watts.  Hardly seems worth the bother of connecting it up.  (Can probably work it out using fewer of the formulae.)

Rob.
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camillitech
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« Reply #3 on: April 08, 2009, 11:47:34 AM »

You can connect your 24v DC to a 230v immersion, but it won't dump much heat. Power dissipated is proportional to the square of the voltage, so 24v will only dissipate about 30w in a 3kw/230v heater.

Thank's for that Bill, pardon my ignorance but does that mean it will only draw off 30w from the dump or it will draw the full amount but only produce 30w of heat  Huh

Cheers, Paul
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'Off grid' since 1985,  Proven 2.5kW, Proven 6kW direct heating, SMA SI6.OH, 800ah Rolls, 8kW PV ,4xTS45, Lister HR2 12kW, , Powerspout pelton, Stream Engine turgo, 60 x Navitron toobs and a 1500lt store. Outback VFX3048 and 950ah forklifts for backup,
dhaslam
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« Reply #4 on: April 08, 2009, 11:55:07 AM »

I would have thought the output should be  300 watts approx because the voltage is ten times less.   Power usage and heat produced should be the same. 
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Ivan
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« Reply #5 on: April 08, 2009, 12:06:50 PM »

You're assuming that the resistance of the wire is constant....which it isn't. When dumping 3kW, the resistance wire inside the immersion element will glow, and as it heats up, its resistance will increase, reducing the current. At low voltage, you'll certainly get a lot less....but as the wire isn't going to get very hot, its resistance wont change a lot. I think empirical measurement is the only way to find out.
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rob26440
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« Reply #6 on: April 08, 2009, 12:22:01 PM »

Quote
I would have thought the output should be  300 watts approx because the voltage is ten times less.   Power usage and heat produced should be the same.

To get 300 watts with 24v DC means a current of 12.5 Amps.  That requires a resistance V/R = 24/12.5 = 1.92 ohms.  The immersion heater is 19.2 ohms and that is a constant (for the purposes of this calculation).  At 24 v the current is 24/19.2 = 1.25 Amps.  30 watts is all you will get.
« Last Edit: April 08, 2009, 12:39:54 PM by rob26440 » Logged

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« Reply #7 on: April 08, 2009, 01:05:57 PM »

power increases by a factor of four as the voltage doubles into a fixed resistance.

power = (voltage x voltage) / Resistance

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rob26440
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« Reply #8 on: April 08, 2009, 01:31:22 PM »

Might be teaching Granny, etc.  But for ref here are the basics of Ohm's law for DC:

Volts = Amps X Resistance.  V=IR
Resistance = Volts / Amps.  R = V/I
Amps = Volts / Resistance.  I=V/R
Watts = Volts X Amps.  W=VI
Watts also = V2/R and I2R
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« Reply #9 on: April 08, 2009, 01:34:34 PM »

You're assuming that the resistance of the wire is constant....which it isn't.

That's true for something like a light bulb where the temperature change is large, or for a resistive material deliberately chosen for its non linearity, but for things like immersion heaters and kettle elements the resistance change is insignificant in normal use.
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billt
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« Reply #10 on: April 08, 2009, 01:37:12 PM »

Thank's for that Bill, pardon my ignorance but does that mean it will only draw off 30w from the dump or it will draw the full amount but only produce 30w of heat  Huh

Cheers, Paul

It will only draw 30W from the generator which will produce 30w of heat. In these things electrical power in is the same as the heat power dissipated.
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« Reply #11 on: April 08, 2009, 01:44:31 PM »

If power and heat are the same amounts, why does a tungsten lightbulb give 5% light and 95% heat?


opppps, just noticed the previous post.

tungsten is not the same resistance as an element?
but in that case it must be a lot higher for it to give so much heat
why wasnt something chosen to make more light and less heat?
« Last Edit: April 08, 2009, 01:46:39 PM by Hunny Bunny » Logged

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rob26440
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« Reply #12 on: April 08, 2009, 01:53:34 PM »

Once read a statement that said all energy ends up as heat.  Whatever method is used to generate it.  Movement, fire, friction, electricity, decomposition, etc.  I tend to believe it and I can't disprove or prove it.  Probably just as well.
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camillitech
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« Reply #13 on: April 08, 2009, 02:47:25 PM »

Thank's for that Bill, pardon my ignorance but does that mean it will only draw off 30w from the dump or it will draw the full amount but only produce 30w of heat  Huh

Cheers, Paul

It will only draw 30W from the generator which will produce 30w of heat. In these things electrical power in is the same as the heat power dissipated.

Thanks for that Bill, so basically it's not worth it and I'll just hang on a couple of weeks to fit the 1kw 24v DC element then  Wink

Cheers, Paul
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http://lifeattheendoftheroad.wordpress.com/

'Off grid' since 1985,  Proven 2.5kW, Proven 6kW direct heating, SMA SI6.OH, 800ah Rolls, 8kW PV ,4xTS45, Lister HR2 12kW, , Powerspout pelton, Stream Engine turgo, 60 x Navitron toobs and a 1500lt store. Outback VFX3048 and 950ah forklifts for backup,
rob26440
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« Reply #14 on: April 08, 2009, 04:30:31 PM »

Quote
why wasnt something chosen to make more light and less heat?

Back in the late 19th century, they tried a few things, including carbon as in the arc lamp.  Doesn't last long and needs the carbon ends winding in and out to get the best arc.  Think they also tried some kind of impregnated string - but that didn't last long either.  Then someone tried Tugsten - that did the trick - the filaments lasted and no-one seemed to bother about the heat.  Swan (Geordie) was the first person to have an electric lamp that lasted more than a few hours but Edison (Yank) made it work better and eventually they joined forces.

More about it here: http://www.rpsgb.org.uk/informationresources/museum/exhibitions/themotherofinvention/swanelec.html

About 100 yrs ago, the fluorescent lamp was invented by a French scientist and it has stayed much the same since.  Now we have CFLs that fluoresce and provide, so they say, more "home friendly" light than the "office" type lighting of the normal fluorescent tube.  Both still use mercury which is nasty stuff.  (Beats me why we walk about with it in our tooth fillings!  Maybe someone will tell me the amalgam is not real mercury or it has been superseded for general fillings by something less toxic.  I think they use ground glass in cosmetic filings)
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