Help- PV direct to immersion heater with no MPP

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  • COsimplegreenhouse
    Junior Member
    • Nov 2023
    • 3

    Help- PV direct to immersion heater with no MPP

    I want to see if I'm looking at this the right way and what I'm missing. I've looked through a lot about generating heat directly from panels to a heating element (no MPP). From research I see an MPP would improve output but I got the panels for free and all I want to do is heat one of my three barrels in a small greenhouse to increase the amount of radiant heat available at night to prevent the electric heater on grid power from kicking on for 4-6 hours on cold nights, I'll still have the heater available for colder nights. I'm in Colorado. My temperatures aren’t getting too high in the daytime so I have some room to allow the increase in daytime heat I’ll get from the heating barrel. I also want to start to learn about these panels to maybe use them for a different usage in the future and have more knowledge for setting up a system when I add a battery.

    I’ve found found David Pop’s ‘matching heating element to load resistance’ vidoeo, I found an article ’56 issue 152 renew magazinerenew.org.au, DIY PV water heating’ that agrees with this resistance matching too. I don’t really understand how impedance affects this but I know it’s a factor.

    3 panels = 840W
    Pmpp=280W
    Voc=38.97
    Vmp=31.67
    Imp=8.84
    I was told the panels are about halfway through their advertised life.

    Wired in parallel Vmpp=31.67V, Impp=26.52A, r= 1.194 (David Poz calculator, Vmpp/Vipp) 1.536-1.194=0.342
    David Poz excel recommends 48V 1500W 1.536 Ohm heater (1500/24=31.25amps)

    Wired in series Vmpp=95.01V, Impp=8.84A, r= 10.7477 (David Poz calculator, divide Vmpp/Vipp) 12.8-8.84=3.96
    David Poz excel recommends 240V 4500W 12.8 Ohm element (4500/240=18.75amps)

    From what I’ve read the panels are generating current and sending it to the heating element and the voltage and in turn wattage will drop off significantly if the resistance values do not match. It will be most efficient at max power and also drop off a lot around that, but I will have pretty steady direct sun on the panels for most of the day.
    Does it matter that the parallel configuration will be able to send more of the current the element would draw, or will the higher voltage balance that out in the series connection? Is the smaller difference between the resistances in the parallel system going to make it a better choice?

    How much power could the heating element generate?

    I know this voltage won’t be Vmpp but how will series vs. parallel affect my results below?
    Parallel: When connected to a 31.67 volt power supply (31.67*31.67)/1.536=668.66W
    Series: (95.01V*95.01V)/12.8 Ohms=705.23W

    To find out how much will this raise water temperature I found a chart that says I need 617.5W to heat 50gallons 5 degrees in an hour. If I have sun all day but maybe 4 hours when it’s striking the panels most directly that’s 20F in a day, which seems pretty good.
    I didn’t find the right equation for raising temperature using power in watts and time so hopefully I can rely on this chart until I do.

    I could also heat two barrels and I think running the heaters in parallel means I assume half the optimal load to each is the max?
    I don’t see an element that’s available in the xcel that’s close to half the resistance for the series setup (6.4ohms) but for the parallel setup half the resistance would be 0.768ohms and there is a 24V 600W heater that’s 0.96ohms (0.96-0.768=0.19)
    If the voltage in parallel is 31.67Vmpp half of that would be fine for a 24V heater.
    Would each element ideally generate (15.84X15.84)/0.96=261.4W? Though that would still be only 522.8W total, less than the 1500W element above.
    Doubling the surface area would double the rate of heat transfer so maybe two barrels radiating heat would be better than one hotter barrel.
    If I have 261W per barrel this would heat a barrel about 2degrees F per hour, so if I’m using 4 hours it’s 8 degrees each and is less heat so would only be better if doubling the rate of heat exchange to the air was worth the loss.

    If I use the proper size wires and ground the system is either setup safer, and what are advantages and disadvantages to each system?

    I could also run two sets of panels in series (not connected) to two barrels/heaters with two 120V/2000W heaters, each with a resistance of 7.2 Ohms, only a 0.035 difference from the system resistance (Vmpp/Impp) of 7.165.
    (63.34*63.34)/7.2 Ohms=557.22W*2=1114.43W
    I didn't try two independent sets of two panels in series or parallel when I was looking at combinations before so if I'm looking at this right then this is the best combination.

    Thanks to anyone who sees this and knows about these kind of systems or has worked on solar setups like this.
  • bcroe
    Solar Fanatic
    • Jan 2012
    • 5203

    #2
    Of course power = Volts X Amps. Current available from panels
    is directly proportional to sun intensity, if you use too low a load
    resistance the voltage will be too low, the rest of the potential
    voltage (and hence power) being burned up in the panels themselves.

    Getting V X A optimized is the job of MPPT. Best to follow DPs advice.
    With direct connection (no MPPT) I would set it up for slightly higher
    voltage at peak sun, that way as the sun moves and drops peak
    current, you will for a time move toward MPPT instead of away from
    it. With clouds power here drops to around 12% or even less, but
    that is using MPPT. Without MPPT the power transferred might be
    more like 1% of peak power capability. good luck, Bruce Roe

    Comment

    • COsimplegreenhouse
      Junior Member
      • Nov 2023
      • 3

      #3
      Thanks Bruce!
      How much higher would you adjust for the voltage at peak? I can run some numbers and see if/how it changes the recommended heating units.
      Also in terms of mppts, is there a low cost way to implement something like this? I haven’t gotten too deep into things like buck converters and power controllers and everything because based on what I read it didn’t really apply here, or maybe just applied less than the math and concept that I was trying to figure out. I’d love to be able to track the power that I’m getting too, and be able to adjust and compare from there (without a huge price tag) but I’m not sure if that’s possible in a simple system like this, besides just manually measuring the water temp and panel output.
      Jake

      Comment

      • bcroe
        Solar Fanatic
        • Jan 2012
        • 5203

        #4
        Originally posted by COsimplegreenhouse
        Thanks Bruce!
        How much higher would you adjust for the voltage at peak? I can run some numbers and see if/how it changes the recommended heating units.
        Also in terms of mppts, is there a low cost way to implement something like this? I haven’t gotten too deep into things like buck converters and power controllers and everything because based on what I read it didn’t really apply here, or maybe just applied less than the math and concept that I was trying to figure out. I’d love to be able to track the power that I’m getting too, and be able to adjust and compare from there (without a huge price tag) but I’m not sure if that’s possible in a simple system like this, besides just manually measuring the water temp and panel output.
        Jake
        There was a water heater control that managed MPPT into a wide
        range of loads, but it put some heavy RF noise into the wiring that
        would scare me and perhaps be unstable. Most MPPTs are for a
        battery of known voltage, which would then require only approximate
        load matching.

        Simple buck converters, etc do not work in this situation since they
        assume constant voltages and unlimited source current. MTTP has
        an entirely different feedback arrangement.

        As a starting point, you could figure load resistance (R = V max power
        divided by I max power) at best sun. I would then increase that resistance
        perhaps 15% which will reduce power transfered a smaller %, but will
        have operation first moving toward best operation at somewhat reduced
        sun intensity. By the time the sun gets to a pretty bad spot, there is not
        much power to worry about any more. good luck, Bruce Roe

        Comment

        • PNPmacnab
          Solar Fanatic
          • Nov 2016
          • 425

          #5
          David is entertaining and a start, but he lacks any real education. Ideal resistance isn't ideal. Studies have shown that using twice the ideal resistance will produce more average daily heat. Trouble is when you need the heat, it is usually hardest to harvest. Heating is a function of the square of the current. A little drop in current has a dramatic drop in power. I've been designing efficient PV resistive heating controls for years. This is very simple technology that nobody in the solar community can figure out. Google the following for a good article A Cost-Effective and Efficient Electronic Design for Photovoltaic Systems for Solar Hot Water Production

          Comment

          • COsimplegreenhouse
            Junior Member
            • Nov 2023
            • 3

            #6
            Thank you, I was wishing David's videos had a little more math and reasoning behind the resistance claims, and more data about the system later, but it got me on my way so I'm thankful for that. Thanks for your article, I was going to ask for a reference before I got to the end of the post.

            Comment

            • bcroe
              Solar Fanatic
              • Jan 2012
              • 5203

              #7
              I did find DPs video somewhat entertaining, as he demonstrated
              just how badly guesswork can go. He eventually did get to the
              right idea of matching the load to available sources, I take that
              as educational. I was once again entertained that a spreadsheet
              was created, when all that was needed were a couple very basic
              power load calculations such as we used to do on slide rules.

              So studies say a 200% load resistance out performs 115%, I can
              believe that. Your results will vary. Bruce Roe

              Comment

              • bcroe
                Solar Fanatic
                • Jan 2012
                • 5203

                #8
                Originally posted by PNPmacnab
                Google the following for a good article
                A Cost-Effective and Efficient Electronic Design for Photovoltaic Systems for Solar Hot Water Production
                Might be good if some efficient, legal, and safe PV to water heater
                controls were marketed.

                Yes that article does describe an approach, but is hardly a complete
                solution. Bruce Roe

                Comment

                • PNPmacnab
                  Solar Fanatic
                  • Nov 2016
                  • 425

                  #9
                  Solar people can't build anything, they are a Buy It Now group. Here is a guy in Lithuania that built one with no prior electronic experience. A little overkill but he gets about 5KWH in the winter from an array less than 1.5KW. https://www.youtube.com/watch?v=3TY_u6m9JXM

                  Here is the chart of resistance, 28.5 ohms was the calculated ideal resistance.
                  IDEAL28ohm.JPG



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