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
 
how big is your solar array? or rather, how much excess power do you normally generate?
 
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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