Parallel/Series combination of PV modules

[DumpmeAdrenaline]

Homework Statement

You are to design a 24-volt, all-dc, stand-alone PV system to meet a 2.4 kWh/day demand for a small, isolated cabin. You want to size the PV array to meet the load in a month with average insolation equal to 5.0 kWh/m2-day. Your chosen PVs have their 1-sun maximum power point at VR = 18V and IR = 5A. Assume a 0.80 derate factor for dirt, wiring, module mismatch (i.e. 20% loss). You'll use 200-Ah, 12-V batteries with 100% Coulomb efficiency. How many PV modules are needed (you may need to round up or down)?

Relevant Equations

P=IV

The solar irradiance incident on the PV modules is 5 kwh/m^2-day.The current each module can provide is 5 A and the voltage due to the separation of charge carriers is 18 V under standard conditions (1 Sun=1000 W/m2) so we would have to connect two modules in series to exceed the voltage of the load. I am confused on the directions of current will current flow from the PV module flow to charge the battery or will it flow to the DC loads to meet the power demand of DC loads or both?

 

[scottdave]

What work have you done towards solving this problem, so far?

 

[DumpmeAdrenaline]

To charge the DC load, we would have to exceed 24 V. So, we connect 2 PV modules in series, thus producing 36 V. The power produced by the 2 PV modules is Power = IV * Derate factor = 36 * 5 * 0.8 = 144 W. Therefore, 144 * 5hr/day = 720 Wh/day = 0.72 kWh/day. The power demand is 2.4 kWh/day. So, we would need 6 modules (3 parallel * 2 in series arrangement) which would deliver a total of (36 * 15 * 5/1000) = 2.7 kWh/day. The extra 0.3 kWh/day would be used to charge the battery.

 

[scottdave]

Since it is a 24 volt system, you will need to limit it to 24 volt output, so the amount of current delivered should be multiplied by 24, correct?

 

[DumpmeAdrenaline]

But how can we obtain 24 V from the PV modules when each module produces 18 V?

 

[scottdave]

A voltage regulator circuit can drop to the desired output voltage.

The simplest example would be a resistor in series with the load. Take the difference between source (solar panels) output voltage and desired voltage. You will need to start off with more than 24 volts, so two of the 18 volt panels in series will give 36 volts, then subtract for the losses.

Interesting note - I did an online search about dirty solar panels. Many of the links that I checked out talk about decreasing power output, but they don't specify whether it is a voltage decrease or current (or a combination).

For the moment, suppose the panel has circuitry to deliver constant voltage at a variety of lighting conditions, so the max current is reduced. To get a 12 volt drop (36-24), calculate the necessary series resistance at the required current load.

If two in series are not enough to provide the required current, then you need to add more pairs in parallel.

 

[Tom.G]

Interesting note - I did an online search about dirty solar panels. Many of the links that I checked out talk about decreasing power output, but they don't specify whether it is a voltage decrease or current (or a combination).

For every 6 or so photons that hit the active material, 1 of those photons wil knock an electron loose. Current is a measure of how many of those electrons are flowing per second.

It follows then, that the fewer the number of photons that hit the active material (blocked by dirt), the lower the current.

Cheers,
Tom