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Voltage drop calcuations for PV systems.

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- 1. Voltage Drop Problems from2009 NABCEP Study Guide SolPowerPeople, Inc. Austin, TX March 19, 2012
- 2. Problem 47
- 3. Problem 47
- 4. Problem 47 Use Chapter 9, Table 8 of the NEC to determine wire resistance. Choose from the stranded, uncoated copper wire.
- 5. CHOOSE FROM UNCOATED COPPER IN ohm/kFTINDICATESSTRANDED
- 6. Problem 47 Distance* x Imp x wire resistance/1000 ft.VD% = System voltage*Distance is round-trip One way distance = 60 feet Im = 7 amps System voltage = 24 V 2 x 60ft x 7 amps x 0.491 ohms/1000 ft.VD% = = 0.017 24 V You can guess at the size conductor that will work and then plug in values for wire resistance from ch9 table 8. Here we have guessed that #6 will work, which it does.
- 7. Problem 47 Distance* x Imp x wire resistance/1000 ft. VD = System voltage One way distance = 60 feet Im = 7 amps*Distance is round-trip System voltage = 24 V 2 x 60ft x 7 amps x 0.778 ohms/1000 ft. VD = = 0.027 24 V While we have determined that #6 is less than 2% (1.7% actually), we still don’t know if that is the smallest conductor that will stay below 2%. We must plug in the resistance of the next smallest conductor, #8 at 0.778 ohms. Result is that #8 has too much resistance and puts us over 2% (at 2.7%) so #6 is the correct answer.
- 8. Problem 47 Another thing to consider with this problem is that the answer implies that we are to be using the nominal system voltage to calculate the %VD. Later, in problem 50, the voltage drop calculation is done with the actual Vmp of the modules instead of system voltage. This, presumably, is done because the question states “under maximum POWER conditions” instead of just when “Im is flowing” or “when current is the maximum power current”. I believe this distinction is made to illustrate the difference in voltage drop calculations between systems that have MPPT tracking versus those that do not (hence being held down closer to the nominal system voltage).
- 9. Problem 49
- 10. Problem 49
- 11. Problem 49
- 12. Problem 49 Calculating the necessary Ω/kFT, using V = 24 V, I = 7 A, %VD = 1, and d = 5 gives 3.42, which corresponds to 14 AWG copper, which has the next lowest value.Referring the Figure provided, if the length from the junction box to the circuit combiner is fivefeet, the smallest wire size needed to keep the voltage drop in this circuit less than 1% whenthe current in the circuit is the maximum power current, isa. 14 AWG copperb. 12 AWG copperc. 10 AWG copperd. 8 AWG copper
- 13. Problem 49 System Voltage 24 VDC 1% VD in our 24 VDC system is 0.01 x 24 = 0.24 VDC Imp = 7 amps One Way Distance = 5 feet VD = 2 x one way distance x Imp x wire resistance/1000 ft.Wire Resistance/1Kft (Rc) = VD x 1000 (2 x one way distance x Imp) 0.24Rc = = 0.24 x 1000 = 3.42 Ω/kFT 2 x 5ft x 7A 70 Therefore, conductor resistance needs to be < 3.42 ohm/1000 ft.
- 14. Problem 49Referring to the Figure, if the length from the junction box to the circuitcombiner is five feet, the smallest wire size needed to keep the voltage drop inthis circuit less than 1% when the current in the circuit is the maximum powercurrent, isa. 14 AWG copperb. 12 AWG copperc. 10 AWG copperd. 8 AWG copper
- 15. Problem 50
- 16. Problem 50
- 17. Problem 50If the distance from the junction box to the combiner box is 60 feet, to keep thevoltage drop between the module junction box and the source circuit combinerbox less than 2% under maximum power conditions at STC, the smallest wiresize that can be used for each source circuit in the system in the Figureprovided isa. 10 AWG copperb. 8 AWG copperc. 6 AWG copperd. 4 AWG copper In this case, the problem is asking you to use the maximum power voltage as the system voltage, likely because of the use of a MPPT charge controller.
- 18. Problem 50VD = 2 x one way distance x Imp x resistance/1000 ft. VDConductor Resistance = X 1000 (2 x one way distance x Imp)Voltage Drop of 2% of 34.2 V = 34.2 x 0.02 = 0.684 V 0.684 0.684 Rc = = x 1000 = 0.814 2 x 60 x 7 840Wire resistance < 0.814 Ω/kFT
- 19. Problem A What conductor should you use in the PV output circuit in a two string system with 12 modules per string, each module with a Vmp of 34.2V and Imp of 8.2A? The distance from the combiner box to the DC disco is 200 feet. Keep the voltage drop under 2%.
- 20. Problem A What conductor should you use in the PV output circuit in a two string system with 12 modules per string, each module with a Vmp of 34.2V and Imp of 8.2A? The distance from the combiner box to the DC disco is 200 feet. Keep the voltage drop under 2%. Distance x Imp x wire resistance/1000 ft.VD = System voltage 2(200)x 2(8.2)x 1.24/1000=8.1 34.2 x 12 = 410.4 8.1/410.4 = 1.9% 2(200) x 2(8.2) x 1.98 / 1000 = 12.98 12.98/410.4 = 3.1% Therefore #10 is the smallest conductor that will work.
- 21. Problem B What is the smallest size conductor you can have from an AC disconnect of a 2000W/120V Inverter to the main panel if it is 150 feet away? Assume maximum voltage drop of 1%.
- 22. Problem B What is the smallest size conductor you can have from an AC disconnect of a 2000W/120V Inverter to the main panel if it is 150 feet away? Assume maximum voltage drop of 1%. VDConductor Resistance = X 1000 (2 x one way distance x Imp) 2000/120=16.666 amps 2AWG with a resistance .01 x 120 = 1.2 V (voltage loss at 1%) of .194 (1.2 / (16.66 x 300)) x 1000 = .240 ohms/kFT
- 23. Problem C What AWG would you use to design for less that 2% voltage drop between a PV output circuit and DC disconnect, assuming the distance between is 14 feet and the system consists of 4 modules in series with an Imp. 8.2A and a Vmp of 27.6V?
- 24. Problem C What AWG would you use to design for less that 2% voltage drop between a PV output circuit and DC disconnect, assuming the distance between is 14 feet and the system consists of 4 modules in series with an Imp. 8.2A and a Vmp of 27.6V? VD Conductor Resistance = (2 x one way distance x Imp) 2% voltage drop = 0.02 x (4 x 27.6) = 2.2 V Imp = 8.2 amps Distance = 28 feet System Voltage = 110.4 (2.2/(28 x 8.2)) x 1000 = 9.58 ohms/kFT 18AWG at 7.95 is the smallest conductor in the table.
- 25. Problem D In a system consisting of 3 series strings of 14 modules, each with a Vmp of 29.5V and an Imp of 7.8A, what is the smallest conductor available to allow for less than 1.5% voltage drop between a PV output circuit and the DC disconnect that is 32 feet away?
- 26. Problem D In a system consisting of 3 series strings of 14 modules, each with a Vmp of 29.5V and an Imp of 7.8A, what is the smallest conductor available to allow for less than 1.5% voltage drop between a PV output circuit and the DC disconnect that is 32 feet away? VD Conductor Resistance = X 1000 (2 x one way distance x Imp) (6.195V/(64 x 23.4A)) x 1000 = 4.13 ohms/kft29.5 x 14 = 413 V orImp = 3 x 7.8A = 23.4 A 64 x 23.4 x 3.14 / 1000 / 413 = 1.13%Distance = 32 x 2 = 641.5% x 413 = 6.195V Answer: 14AWG with resistance of 3.14
- 27. Problem E What wire size would you use to design for less than 1% voltage drop over a 160 foot distance from a 3000W/240 V inverter to the main distribution panel?
- 28. Problem F Which size conductor would be required to design for less than 1% voltage drop from the 8A PV output circuit to the DC disconnect stand alone system with a 24V battery bank, assuming the distance is 22 ft?

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