### Solar Panel Rating

Our Sunpower panels are rated at a peak power of 425 watts with a voltage and current at 73 volts and 6 amp respectively. We needed to find the right combination of panels (number of panels per string and number of strings) with the right resistance (based on the specific heating element) that would place us at the Maximum Power Point, or MPP, on the characteristic IV curve.

Sunpower Ratings |

E20/435W SunPower Solar Panel IV curve |

The MPP occurs at the knee of the curve. In order to determine the resistance of the element we divide the power rating by the rated voltage to get the current, and then divide the voltage by the current. Adding panels in series causes the knee on the IV curve, and the voltage available from the solar panels, to increase by 73V. The same occurs to the current on the IV curve when connecting panels in parallel; the knee would raise, and the current available, would increase by 5A at an input of 800W/m^2.

We want to use a 4500W, 240V element to heat our water:4500W/240V = 18.75A240V/18.75 = 12.8 ohmsThe element resistance is around 13 ohms, so dividing our panel’s rated voltage by our element’s resistance gives us the current we need to draw to power the heater:73V/13 ohms = 5.6A, which is conveniently close to a irradiance input of 900W/m^2.By multiplying our rated voltage with our calculated current drawn, we get a peak power of 425W, which is not close enough to the 4500W needed to power our element; however, by having three panels in series, or about 210V, and then three strings of three panels, which increases our current to 18A, we would get a peak power of 3780W.

### Solar Panel and power needs-estimates

70 gallon system

From 50 °F to 150 °F

ΔT= 100°F

70gal(8.345lbs/1galH20)=584.15lbs

584.15lbs*100°F=58415.0 BTU

1BTU=.000293kWh

58415.0 BTU(.000293kWh/1BTU)=17.115kWh

For 17.115kWh/4.5kW element=3.8hrs

30gal(8.345lbs/1galH20)=250.2lbs

250.2lbs*100°F=25021.0 BTU

1BTU=.000293kWh

25021.0 BTU(.000293kWh/1BTU)=7.33kWh

For 7.33kWh/3.7kW element=1.98hrs

These calculations are with a 4.5kW element the will be powered by 210V.

The SunPower panels we are using are rated at 73V output with a load across.

Based on the I-V Curve for the panels we are using at solar noon the current with a load will be about 6amps, and the voltage is about 73V. The resistance of a standard water heating element rated at 240V, 4500W is 12.8ohms.

It just so happens that the resistance of 12.8ohms is matched very well for these panels. Because using Ohm’s Law we see that I=V/R, so I=73V/12.8ohms=5.7A. So with one panel we can directly hook up the element and get close to the maximum amount of power from that panel. In this case the power is P=IV, so P=5.7A*73V=416W. Because the element is rated for 240V we want to put as close to that amount of voltage through it. So 240V/(73V/panel)=3.3panels. So we want about 3 panels in series to get to the max power. But as we increase voltage the amount of power the element draws will not change unless the current changes as well. And because the slope of the line from the origin to the point of max power on the I-V curve is equal to the inverse of resistance and the resistance of 12.8ohms is ideal for this curve when we triple voltage we want to triple current as well. So ideally we have three panels in series as a string, and we have three strings hooked in parallel. So a total of 9 panels. With 220V and 17A, this makes a power output of 3750W at solar noon. This is a reasonable amount of power to work with to heat the water tank in a few hours.

Adjusting the initial calculations above:

To heat 30 gals 100°F.

7.33kWh/3.7kW element=2hrs to heat the water

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