Hi Adam, I've corrected the answer in the ppt. Thanks for the heads up on this, and thanks also for your correction of the 110.26 mistake we made in newsletter 13. Please let us know if you find any other mistakes.
Adam.. good catch. I mistakenly circled D, but you are absolutely correct. Even the answer key from NABCEP had it correct. My mistake. I'll correct the slide and upload and repost. Thanks
The answer to problem 73 is incorrect. The answer should be c. 6 AWG. 250.119(A) refers to 'Conductors Larger Than 6 AWG.' 200.6(B) refers to 'Sizes 4 AWG or Larger.' which effectively means the same thing. So to restate the problem, answered properly it should read: The 2011 NES allows marking conductors with colored tape, provided that they are larger than 6 AWG.
Not sure how to do this without disconnecting modules.
Transcript
1. Random Solutions from the2009 NABCEP Study Guide SolPowerPeople, Inc. Austin, TX March 22, 2012
2. Problem 30
3. Problem 30 30. (b). The altitude for months between March and September is high enough to avoid shading.
4. Problem 30
5. Problem 32
6. Problem 32Using the sun path chart provided, the minimum annual sun altitudebetween the hours of 9 a.m. and 3 p.m. sun time is closest toa. 10°b. 20°c. 30°d. 45°
7. Problem 32Using the sun path chart provided, the minimum annual sun altitude between the hours of 9 a.m. and3 p.m. sun time is closest toa. 10°b. 20°c. 30°d. 45°
8. Problem 35
9. Problem 3535. If a proposed PV installation site has an unobstructed south-facing roof area of 60 m2, and ifthin-film modules with six watts-per-square-foot power output at STC are to be installed on 50%of the roof, then the maximum available PV array output power (based on the sum of moduleratings) at STC will be closest toa. 4500 wattsb. 3600 wattsc. 2250 wattsd. 1900 wattsConvert square meters to square feet. 1 m2 equals 10.76 ft2, 1 square meter ofPV will produce 10.76 × 6 = 64.6 watts. The available roof area is 30 m2 whichwill produce 30 × 64.6 watts = 1937 watts (d is closest answer).
10. Problem 36
11. Problem 36Which of the following devices does the NEC require to be a part of PV systemsmounted on residential dwellings?a. A stand-off mount for the PV modulesb. A utility interconnectionc. A ground-fault protection deviced. An accessible source circuit combiner box
12. Problem 39A PV system is to be selected for operating a PV water pumping system. The pump willrequire 300 watts of PV modules for proper operation. A 12 -Vdc model and a 48-Vdcmodel are available. If both pumps operate at the same power level, theresistance of the wire to the 48-V pump, compared to the resistance of the wire to the12-V pump, assuming the same percentage voltage drop in the wiring, may bea. 1/16th as muchb. 1/4th as muchc. 4 times as muchd. 16 times as much Problem stated differently:  How much higher can the resistance of the wire be if a 12 Vdc pump is replaced with a 48 Vdc pump?
13. Problem 39Step 1: determine current level for both pumps at 300W12 volt pump: 300 watts / 12 volts = 25 amps48 volt pump: 300 watts / 48 volts = 6.25 ampsStep 2: choose a random voltage drop percentage and determine actual voltage dropfor each system (let’s assume 2%)12 volts x 0.02 = .24 volts48 volts x 0.02 = .96 voltsStep 3: plug both values into voltage drop formulaRc = Vd/Imp12V: Rc = .24 / 25 48V: Rc = .96 / 6.25 Rc = 0.0096 Rc = 0.1536Step 4: find the ratio between the two resistance values0.1536 / 0.0096 = 16 Answer “D” the resistance in the wire of the 48Vdc pump can be16 times as great as the 12Vdc pump.
14. Problem 39 39. D Another way to solve the problem would be to take the ratio between the two voltages and then square them.  48 / 12 = 4  42 = 16  Answer 16  This works with any voltage
15. Problem 40
16. Problem 40b. B is best because presumable the modules are on a sloped rooftop and the heatbehind the modules would have to travel the shortest distance to escape. B isbetter than A because some of the heat can escape from both the sides and themiddle as well.
17. Problem 41
18. Problem 41 .75” 2.75” 281 lbs x 2.75 inches = 773 lbsA ¼″x3½″ lag screw that has a 3″ thread is used to attach an L-bracket to an asphalt-shingle roof. If the combinedthickness of the L-bracket, shingles, and roof membrane is ¾ inch, and if the screw penetrates directly into a rooftruss made of Southern Yellow Pine, into a properly sized pilot hole, then the withdrawal resistance will be closesttoa. 632 poundsb. 773 poundsc. 843 poundsd. 984 pounds
22. Problem 44For the situations described, which would result in the most cost-effective use of atwo-axis tracking mount?a. In areas of low wind, latitude less than 30°, and moderate daytime summer cloud coverb. In areas of low wind, latitude greater than 30°, and minimal daytime summer cloud coverc. In areas of moderate wind, latitude greater than 30°, and moderate year-around cloud coverd. In areas of moderate wind, latitude less than 30°, and minimal year-around cloud cover  Low wind is preferable to reduce the loads on the mechanical structure.  Latitude greater than 30° since the sun moves in both axes more in these latitudes.  Minimal daytime cloud cover - tracking systems don’t work as well in cloudy areas.  Answer (b) Higher latitudes and minimal summer cloud cover result in the highest performance gain for tracking mounts.
23. Problem 45
24. Problem 45
25. Problem 45 45. d. Article 314.28 is used when the size of any conductors entering the junction box is larger than 6 AWG.
26. Problem 45
27. Problem 45
28. Problem 46
29. Problem 46
30. Problem 46 310.15(B)(2)(a) 310.15(B)(3)(a) Temp °CFour current carrying conductors in conduit THWN-2 has 90°C rating
31. Step 1 (determine minimum): Isc x 1.25 x 1.25 7.2 x 1.25 x 1.25 = 11.25 AStep 2 (calculate with conditions of use): Isc x 1.25 / derate for temperature / derate for conduit fill 7.2 x 1.25 / .8 / .87 = 12.9 AStep 3 compare both values and choose the largest: 12.9A is larger than 11.25.
32. Problem 52
33. Problem 52
34. Problem 52 Different batteries have different charge control set points. Answer (b) The charge controller connected to the PV output circuit of Figure 4 normally requires adjustment for a. voltage drop b. battery type c. maximum input power d. maximum input current
35. Problem 54
36. Problem 54 54. d. None of the proposed connections would work. The first one (a) would result in overcharging of the 120 Ah batteries. The second (b) would result in even more serious overcharging of the 120 Ah batteries, and the third (c) would also result in serious overcharging of the 120 Ah batteries.
37. Problem 56
38. Problem 56 Some water pumps require higher currents to get them started. An LCB will pull back the voltage and boost the current so the pump can start.
39. Problem 56The purpose of a linear current booster is toa. keep its output voltage the same as its input voltage and boost the output current to a value larger than theinput currentb. convert a high input voltage and low input current to a lower output voltage and a higher output currentc. convert a low input voltage and high input current to a higher input voltage and a lower input currentd. keep its output current the same as its input current and boost the output voltage to a value larger than theinput voltage
40. Problem 57
41. Problem 57The purpose of an inverter is toa. convert dc at one voltage to ac at the same or another voltageb. convert ac at one voltage to dc at the same or another voltagec. convert dc at one voltage to dc at another voltaged. convert ac at one voltage to dc at another voltage
42. Problem 58
43. Problem 59 NEC 690.8 (A) (3) Inverter Output Circuit Current  The maximum current shall be the inverter continuous output current rating. NEC 690.8 (B) (1) Sizing Conductors  The circuit conductors and overcurrent devices shall be sized to carry not less than 125 percent of the maximum currents as computed in 690.8(A).
44. Problem 58Harmonic distortion relates to power quality. UL Standard 1741 establishes an upper limit of 5% for harmonic distortion.Sine wave inverters are required for connection to utility lines, becausea. they are more efficient than other types of invertersb. they are the only inverters that have low enough harmonic distortionc. only sine wave inverters can be designed to disconnect from the utility when utility power is lostd. non-sine wave inverters cannot develop adequate power for utility interconnection
45. Problem 59
46. Problem 59The inverter output current can be determined by dividing the inverter rated powerby the rated output voltage. Then multiply by 1.25 (safety factor for conductor) todetermine ampacity requirement:2500÷120 = 20.8 A1.25 x 20.8 = 26 AA 2500-W inverter is used to supply a 120-V ac load of 1500 watts. This means that theampacity of the wire at the inverter output must be at leasta. 12.5 Ab. 15.6 Ac. 20.8 Ad. 26.0 A
47. Problem 60
48. Problem 60 NEC 690.8 (A) (4) Stand-Alone Inverter Input Circuit Current. – The maximum current shall be the stand-alone continuous inverter input current rating when the inverter is producing rated power at the lowest input voltage. The maximum input current is the rated power divided by the lowest input voltage divided by the efficiency, which gives 2500 ÷ 22 ÷ 0.88 = 129 A.A 2500-W inverter with an input-voltage range of 22 V to 32 V has an efficiency of 88% at full output. This means themaximum inverter input current at full rating will be closest toa. 129 Ab. 100Ac. 89 Ad. 69 A
49. Problem 61
50. Problem 61The rated output current is 1500÷120 = 12.5 A125% of 12.5 A is 15.6 AThe next highest circuit breaker (i.e., 20 A) should be used (from NEC 240.6)If the maximum ac output rating of an inverter with 120-V ac output is 1500 W,the rating of the circuit breaker at the point-of-utility connection should bea. 15 Ab. 20 Ac. 25 Ad. 30 A
51. Problem 64
52. Problem 64 NEC 705 (D) (2)
53. Problem 64 NEC 690.8(B)(1)
54. Problem 6464. c. The bus rating is 225 A, but the main breaker is only 200 A. 120% of 225A is270A. 270-200 = 70A. Therefore, the PV system may be connected using a 70Abreaker so the sum of the main plus the PV breakers is 270A. However, because thePV breaker is rated at 125% of the inverter-output current, the inverter-outputcurrent can not exceed 56A.a. 20 Ab. 25 Ac. 56 Ad. 70 A
55. Problem 68
56. Problem 68If the output of the inverter is connected to the optional standby system panel through a30-A circuit breaker, the appropriate size of the equipment-grounding conductorbetween the inverter and the emergency panel isa. 14 AWG copperb. 12 AWG copper Table 250.122c. 10 AWG copperd. 8 AWG copper From Table 250.122, if the current carrying conductor is protected with a 30 A overcurrent protection, use an AWG#10 equipment ground.
57. Problem 69
58. Problem 69 Article 445.13 indicates ampacity of the conductors needs to be 115% thecurrent.115% of 42 A is 48.3 A. 8 AWG THWN has an ampacity of 50 A.A 5-kVA, 120-V generator has a rated output current of 42 A. It does not have amechanism to limit its output current to the rated value. Assuming they are run inconduit, the output conductors should have an ampacity of no less thana. 8 AWG THWNb. 8 AWG THHNc. 6 AWG THWNd. 6 AWG THHN
59. Problem 70
60. Problem 70Batteries discharge to 80% of their capacity every two days.The generator at C/10 rate will charge 10% per hour.It takes 8 hours to charge back up from 80% discharge.8 hours divided by 2 days is 4 hours per day average.The generator burns 1 gallon of fuel per hour.Therefore, the fuel usage is an average of 4 gallons per day.Answer: “A” 4 gallons per day
61. Problem 72
62. Problem 72In a PV system, the equipment-grounding conductors should bea. whiteb. blackc. redd. green
63. Problem 73The 2011 NEC allows marking conductors with colored tape, provided thatthey are larger thana. 10 AWGb. 8 AWGc. 6 AWGd. 4 AWG
64. Problem 73The 2011 NEC allows marking conductors with colored tape, provided thatthey are larger thana. 10 AWGb. 8 AWGc. 6 AWGd. 4 AWGGrounding Conductors: 250.119(A) Grounded Conductors: 200.6(B)
65. Problem 74
66. Problem 74The width of the working space in front of an inverter that is 24-inches widemust be at leasta. 24 inchesb. 30 inchesc. 36 inchesd. 42 inches NEC 110.26(A)(2)
67. Problem 75
68. Problem 75The minimum depth of the working space in front of a charge controller forwhich the input voltage never exceeds 60 V dc isa. 30 inchesb. 36 inchesc. 42 inchesd. negotiable NEC 110.26 (A)(1)(b) Low Voltage.
69. Problem 77
70. Problem 7777. In order for a PV array to directly face the sun at 2:30 p.m. solar time on June 21at 30° N latitude (see Figure 2), which array orientation is correct?a. 60° W of S with a tilt of 40° with respect to the horizontalb. directly west with a tilt of 60° with respect to the horizontalc. directly west with a tilt of 30° with respect to the horizontald. 45° W of S with a tilt of 60° with respect to the horizontal
71. Problem 78
72. Problem 78If 5/16-inch lag screws are used to fasten a charge controller to woodenstuds, an appropriate pilot hole size would be closest toa. 1/8 inchb. 19/64 inchc. 7/32 inchd. 1/4 inch70% of 10/32 is 7/327 is 70% of 10.Therefore, 7/32 is correct.
73. Problem 80
74. Problem 80The first step in system checkout after completing the installation isa. test open-circuit voltageb. visually check the entire systemc. install the source-circuit fusesd. close all disconnects
75. Problem 81
76. Problem 81 Before applying PV power to either an inverter, a charge controller, batteries or a load, one should first a. check the polarity of the PV output b. install the source circuit fuses c. call the electrical inspector d. close all disconnects Equipment can be destroyed by reversing the positive and negative legs of the array. Incorrect polarity can severely damage the electronics in a power conditioning unit (i.e. a charge controller or an inverter).
77. Problem 82
78. Problem 82 -0.4%/°C x (50°C-25°C) = -10% loss due to temperatureVoltage = 68.4 × 0.9 ×0.984 = 60.6 V. Derate due to temp Loss due to VD82. b. 60.6V
79. Problem 83
80. Problem 83 A 4-kWSTC crystalline silicon PV array is operated in a utility-interactive mode with no battery backup. The inverter tracks maximum power, and the array is operating at 50°C with 900 W/m2 incident on the array. There is a 2% power loss in the wiring and the inverter is 94% efficient. On a typical PV system, the inverter output power will be closest to a. 3316 watts b. 2985 watts Loss due to temperature: c. 2612 watts 50-25=25 x .5% = 12.5% loss d. 1492 watts 100% - 12.5% = 87.5% Inverter Efficiency 4,000 W × .90 × .9 × 0.875 x .98 x .94 = 2,612 watts Irradiance Voltage 900 w/m2 Drop Module mismatch soiling etc. (not included in the problem statement but referred to in 2009 study guide)
81. Problem 84
82. Problem 84A typical 4-kW crystalline silicon array is operating at STC in a utility-interactive systemwith battery backup. The STC maximum-power voltage rating of the PV array is 68.4volts. The system uses a conventional charge controller that doesnot track maximum power. Wiring losses are 3% and inverter losses are 5%. If thebatteries are at full charge at a voltage of 52 V, and if all PV output is delivered to thegrid (assume that no power is being used to hold the batteries at 52 volts), the inverteroutput power will be closest to Because there is no MPPT the STCa. 1261 watts Module mismatch soiling array rating is reduced by the ratio of the battery voltage to the maximum-b. 2207 watts etc. (not included in the power voltage of the array.c. 2522 watts problem statement but referred to in 2009 study Voltage Dropd. 3152 watts guide) Inverter Output = 4000W x 0.9 x 0.97 x 0.95 x (52 V/68.4 V) = 2522 W Inverter Eff.
83. Problem 85
84. Problem 85A typical 4-kW crystalline silicon array is operating at STC in a utility-interactive system with battery backup. The STC maximum power voltagerating of the PV array is 68.4 volts. The system uses a MPT charge controllerthat has 5% losses. Wiring losses are 3% and inverter losses are 5%. If thebatteries are at full charge at a voltage of 52 V, and if all PV output isdelivered to the grid (assume that no power is being used to hold the batteriesat 52 volts), the inverter output power will be closest toa. 1261 wattsb. 2207 watts *Key point here is that because the system usesc. 2522 watts MPT charge controller, there is no losses due to thed. 3152 watts voltage difference between Vmp of the array and the battery voltage like there was in problem 84. 4000 W x 0.95 (charge controller losses) x 0.97 (wire losses) x 0.95 (inverter losses) x 0.9 (mismatch) = 3152 W
85. Problem 87
86. Problem 87 87. c. The other answers are also good practices. The point here is that presumably there is a problem. This means that it is possible that a switch is bad, so that even if it is in the off position, maybe it is not off. So after turning off all relevant switches to remove power from points where wires are to be attached or removed, the voltages and currents should be measured to be sure there are no surprises. Even then, gloves and a facemask may be appropriate (especially on high voltage systems and battery systems) when making the connections.
87. Problem 88
88. Problem 88Answer: (d) measure the individual module voltages in this source circuit Tricky question. It assumes that you have access the junction boxes on the backs of the modules so you can get a voltage measurement.
89. Problem 89
90. Problem 89 From 3.8.2 of the Study Guide: “The lengths of battery cables also need to be checked to ensure that the batteries have not been connected with different lengths of cable that will result in uneven charge and discharge of batteries.” Answer: b. Uneven charge and discharge current
91. Problem 90
92. Problem 90A utility-interactive PV system with no batteries consists of 15 100-W modules inseries that feed a 1500-W inverter. The inverter output power is found to be 780 Wwhen the modules are operating at 50°C with an irradiance level of 800 W/m2. Ifthree modules are observed to be shaded, which conclusion is most likely?a. The inverter input current is probably too lowb. The system is probably functioning properlyc. The inverter is probably not tracking maximum powerd. The modules probably do not have bypass diodes Assume shaded modules aren’t operating (1200 W) Power loss due to modules = (50 - 25) x 0.005 = 0.125, 100% - 12.5% = 87.5% or .875 Inverter Output = 1200 x .8 x 0.875 x 0.9 x 0.97 x 0.95 = 697 W mismatch Assumed inverter efficiency Max power Temp losses Voltage drop irradiance
93. Problem 90Since measured power was 780 watts and the expectedperformance was 697 watts, the system is likely workingproperly and exceeding expectations. Answer B.
Best Regards,
Richard Stovall
P.S. good luck on the test!!
250.119(A) refers to 'Conductors Larger Than 6 AWG.' 200.6(B) refers to 'Sizes 4 AWG or Larger.' which effectively means the same thing.
So to restate the problem, answered properly it should read:
The 2011 NES allows marking conductors with colored tape, provided that they are larger than 6 AWG.