Resistive heater - free electricity

nate

0
Jan 7, 2014
34
Sydney, Australia
yes i know... i'm VERY aware of the cost & efficiency of using resistive heating, power consumption, cable runs etc...

However...
Solar is installed & i need to use more of it (In Aus we get paid close to zero for generating), so thinking about setting up a resistive element to get *A FEW DEGREES* of increase, and maybe extend my season a few weeks.
Its a 40,000L pool, i've got a solar cover & do everything else for efficiency.

Thinking of doing a piddly 2.3kw heater, with a PWM solution to be able to control the load & ensure it ONLY uses excess solar.

Running at 2.3kw - what temp increase can i expect per hour, assuming to heat loss?
(i've done some looking & getting wildly different numbers)
 
2.3 kw is 7848 btu per hour. 40,000 liters is 88,183 pounds. 1 btu will raise 1 pound of water 1 degree Fahrenheit.
So, you would get 0.089 degrees Fahrenheit per hour or 2.1 degrees per day.
What about using a heat pump?
 
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I've got a 7kw air con unit there... but that'd mean fridgework, titanium exchanger, lots of plumbing - labor costs here are INSANE. Plus it'd be heating when i want cooling etc... i'd expect about $2-3k for that setup *easy*
Heat pump - $10k install.

I could DIY for $500 or so if there's value.
So realistically it's 0.8deg per day (8hrs electricity) which isnt too bad overall
 
its a 5kw setup, use 1/8th the capacity through the day, so would have 4kw to spare.
I'm going to setup some CT's on the feeds & connect it to an arduino, see if i can successfully log & measure the excess power.
Then use that to control an SCR for initially my water heater, and maybe this
 
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