Sizing a Solar Electric System for Cathodic Protection |
Note: This applies to systems using our AERL maximum power point tracker type CP controls ONLY.
| Sizing the PV (photovoltaic) array and battery bank for CP systems is pretty straightforward. However, most people not familiar
with the "power point tracker" type controls and/or photovoltaic systems have trouble at first because the old standard of just specifying
a power supply of "X" Amps at a a fixed voltage gives you a much larger system than you need, and increases costs considerably. For example, a 6 panel (462 watt array) system does the same job, better, more efficiently and reliably, than the 10 panel (750 watt) systems that were previously being used, and costs about 35% less. |
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Concept |
Sizing a solar electric system for cathodic protection in past has been somewhat of an "iffy" proposition. This was mainly because the battery charge controls and CP controls were far less than perfect. It is difficult to size a system with several unknown variables, so many systems end up being under or oversized. Using maximum power point trackers (MPPT's) for charge regulation, and AERL or Power Tracker CP controls eliminates the main losses in PV systems, making design much easier. Using these two elements together can result in cost savings of 20% to 50%, and make design and system sizing much simpler.
The AERL Power Tracker CP controls are nearly 100% efficient. This is an important consideration when sizing a system. Its' major impact is that for all practical purposes you can ignore the losses usually associated with the standard CP controls and rectifiers. For example, if you previously specified 8 Amps at 24 volts, your power requirement was 192 Watts, regardless of the requirements of the ground bed/structure system. Your system resistance in this case is 3 ohms, even if the ground bed itself may be only 1 ohm. If you require 192 Watts, that is about 4600 Watt-hours per day. In the Southwest USA, this would require about 1000 Watts of solar panel.
With the 97% conversion efficiency of our CP controls, you only need to consider the actual requirements, and not take into account the losses in the rectifier. Since the ground bed is only 1 Ohm, you would only need 64 Watts, or about 1550 Watt-hours per day. Even adding in a 25% safety factor, your panel requirements would drop to about 380 Watts of solar panel. Approximate savings in the above example would be around $8,000. The system with the standard CP control would cost about $11,000, while the system with the our control would cost about $5,000.
With the our CP controls, we are interested only in two numbers: the ground bed (anode to structure) resistance, and the current (Amps) required. Our control does a direct DC to DC conversion at about 98% efficiency. You can figure everything on a "Watts to Watts" basis. In other words, if you need 100 Watts to your ground bed, then the batteries must supply 100 Watts. Your solar panels must be capable of supplying enough for the ground bed current, plus enough extra to take care of system and battery losses and panel inefficiencies. (These panel mismatch efficiencies are another major loss - see our section on MPPT's for battery charging go there).
1. If possible, get an actual ground bed resistance measurement. This is not always possible, as often the anodes and pipeline or wellhead are not even installed yet. In those cases, pick a "worst case" scenario. For most applications this will be on the order of 1/3 to 2 ohms, but may vary considerably in different locations. The lower your groundbed resistance is, the cheaper the solar power system.
2. Determine the current requirements in Amps.
3. Take the current requirement in Amps (I), and square it (I x I). Multiply this number by the ground bed resistance (R) in ohms. [ I x I x R, or "I squared R"]. This will give you the power needed in Watts. This is basic Ohms law for power into a resistance.
4. This power is continuous, so multiply it by 24 - this will give you Watt-Hours for one full day.
5. Determine the Winter Peak Sun Hours. For most of the Southwest and Mountain states, this ranges from 4.5 to 5.5, but can be as low as 1.8 in the Pacific Northwest. If you are not sure or don't know, give us a call. Note that this is NOT the same as the number of hours of sun, but is based on actual measurements of insolation, and gives the equivalent actual usable sun hours. You can take a look at our solar maps for an approximate guide (note: about 60K graphics file, takes a minute or so to load)
6. With solar electric systems, you have to produce all the power consumed in 24 hours during the Peak Sun Hours. Divide Watt-Hours {from 4} by the Peak Sun Hours - this gives you the power required per hour to break even (assuming no system losses).
7. Since PV arrays and batteries are not 100% efficient, divide that number by .66 (or multiply by 1.5). This takes care of all the other system and panel losses, and also adds in enough extra to charge up the batteries while still supplying enough power for the CP system. (To reduce these panel losses, you can use one of our Power Trackers.)
8. At this point, you should have a number that is equal to the power that the array must produce in one hour. To get the number of panels, divide this number by the power output of the PV panel - this will give you the number of panels needed. Since this is seldom an even number, you must usually go up to the next highest number of panels. For battery sizing, see the section down the page.
Short Form
I x I x R [GB] x 24
divide by Sun Hours
divide by .70.
This gives the required PV array watts per hour.
Array watts per hour divided by panel watts = number of panels required.
Example Calculation
From a wellhead site installed near Vernal, Utah, for Chevron USA
To figure how much battery you need, just take the watts required for how many days of backup you need. In the above example, you need 96 Watt-Hours per day. Most systems require 5 to 7 days. For 5 days, multiply 96 x 5 = 480 Watt-hours. If you have a 12-Volt system, divide the 480 by 12, which gives you 40 Amp-Hours. Since you want to cycle the batteries no more than 60% (to extend battery life), divide 40 AH by .6, which gives you about 67 Amp-Hours. In this case you would need 1 each of a 12-Volt 72 AH battery - something like the Concorde PVC-1272. In the Chevron site, the customer elected to go with two of the Concorde PVC-4D batteries, which gives a total of about 350 AH. Because the batteries are being cycled only about 10%, expected battery life is 15 to 20 years.
Note!:
You cannot just figure it directly using the amps to the ground bed to come up with amp-hours. The reason is that our controls down-convert
the battery voltage to exactly what is needed to push whatever current (amps) is set. You may have 5 amps at 5 volts going into the ground bed,
but into the controller you would have (for example) 1 amp at 25 volts. In both cases the POWER (in watts) is the same, which is why all calculations
are based on watts rather than amps or amp-hours.
To see what a real system will cost you, click me
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