#SolarMOOC Random Problems from 2009 NABCEP STUDY GUIDE

Random Solutions from the 2009 NABCEP Study

GuideSolPowerPeople, Inc.

Austin, TX

March 22, 2012

Problem 30

Problem 30

30. (b). The altitude for months between March and September is high enough to avoid shading.

Problem 30

Problem 32

Problem 32

Using the sun path chart provided, the minimum annual sun altitude between the hours of 9 a.m. and 3 p.m. sun time is closest toa. 10°b. 20°c. 30°d. 45°

Problem 35

Problem 35

35. If a proposed PV installation site has an unobstructed south-facing roof area of 60 m2, and if thin-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 module ratings) at STC will be closest to

a. 4500 wattsb. 3600 wattsc. 2250 wattsd. 1900 watts

Convert square meters to square feet. 1 m2 equals 10.76 ft2, 1 square meter of PV will produce 10.76 × 6 = 64.6 watts. The available roof area is 30 m2 which will produce 30 × 64.6 watts = 1937 watts (d is closest answer).

Problem 36

Problem 36

Which of the following devices does the NEC require to be a part of PV systems mounted 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

Problem 39

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?

A PV system is to be selected for operating a PV water pumping system. The pump will require 300 watts of PV modules for proper operation. A 12 -Vdc model and a 48-Vdc model 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 the 12-V pump, assuming the same percentage voltage drop in the wiring, may be

a. 1/16th as muchb. 1/4th as muchc. 4 times as muchd. 16 times as much

Problem 39

Step 1: determine current level for both pumps at 300W

12 volt pump: 300 watts / 12 volts = 25 amps48 volt pump: 300 watts / 48 volts = 6.25 amps

Step 2: choose a random voltage drop percentage and determine actual voltage drop for each system (let’s assume 2%)

12 volts x 0.02 = .24 volts48 volts x 0.02 = .96 volts

Step 3: plug both values into voltage drop formula

Rc = Vd/Imp

12V: Rc = .24 / 25 Rc = 0.0096

Step 4: find the ratio between the two resistance values

0.1536 / 0.0096 = 16 Answer “D” the resistance in the wire of the 48Vdc pump can be 16 times as great as the 12Vdc pump.

48V: Rc = .96 / 6.25 Rc = 0.1536

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

Problem 40

Problem 40

b. B is best because presumable the modules are on a sloped rooftop and the heat behind the modules would have to travel the shortest distance to escape. B is better than A because some of the heat can escape from both the sides and the middle as well.

Problem 41

Problem 41

281 lbs x 2.75 inches = 773 lbs

A ¼″x3½″ lag screw that has a 3″ thread is used to attach an L-bracket to an asphalt-shingle roof. If the combined thickness of the L-bracket, shingles, and roof membrane is ¾ inch, and if the screw penetrates directly into a roof truss made of Southern Yellow Pine, into a properly sized pilot hole, then the withdrawal resistance will be closest to

a. 632 poundsb. 773 poundsc. 843 poundsd. 984 pounds

2.75”

.75”

Problem 42

4 modules x 10 ft2 = 40 ft2

40 ft2 x 75 lbs/ft2 = 3000 lbs.

3000 lbs / 6 attachments = 500 lbs/attachment

500 lbs / 300 (lbs/in) = 1.67 inches

Answer (b)

Problem 42

Problem 44

Problem 44

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.

For the situations described, which would result in the most cost-effective use of a two-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

Problem 45

Problem 45

45. d. Article 314.28 is used when the size of any conductors entering the junction box is larger than 6 AWG.

Problem 45

Problem 46

Problem 46

Four current carrying conductors in conduitTHWN-2 has 90C rating

Temp °C

310.15(B)(2)(a)

310.15(B)(3)(a)

Step 1 (determine minimum): Isc x 1.25 x 1.25 7.2 x 1.25 x 1.25 = 11.25 A

Step 2 (calculate with conditions of use): Isc x 1.25 / derate for temperature / derate for conduit fill7.2 x 1.25 / .8 / .87 = 12.9 A

Step 3 compare both values and choose the largest: 12.9A is larger than 11.25.

Problem 52

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 dropb. battery typec. maximum input powerd. maximum input current

Problem 54

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.

Problem 56

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.

Problem 56

The purpose of a linear current booster is to

a. keep its output voltage the same as its input voltage and boost the output current to a value larger than the input 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 the input voltage

Problem 57

Problem 57

The purpose of an inverter is to

a. 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

Problem 58

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).

Problem 58

Harmonic 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, because

a. 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

Problem 59

Problem 59

A 2500-W inverter is used to supply a 120-V ac load of 1500 watts. This means that the ampacity of the wire at the inverter output must be at least

a. 12.5 Ab. 15.6 Ac. 20.8 Ad. 26.0 A

The inverter output current can be determined by dividing the inverter rated power by the rated output voltage. Then multiply by 1.25 (safety factor for conductor) to determine ampacity requirement:

2500÷120 = 20.8 A 1.25 x 20.8 = 26 A

Problem 60

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 the maximum inverter input current at full rating will be closest to

a. 129 Ab. 100Ac. 89 Ad. 69 A

Problem 61

Problem 61

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 be

a. 15 Ab. 20 Ac. 25 Ad. 30 A

The rated output current is 1500÷120 = 12.5 A

125% of 12.5 A is 15.6 A

The next highest circuit breaker (i.e., 20 A) should be used (from NEC 240.6)

Problem 64

Problem 64

NEC 705 (D) (2)

Problem 64

NEC 690.8(B)(1)

Problem 64

a. 20 Ab. 25 Ac. 56 Ad. 70 A

64. c. The bus rating is 225 A, but the main breaker is only 200 A. 120% of 225A is 270A. 270-200 = 70A. Therefore, the PV system may be connected using a 70A breaker so the sum of the main plus the PV breakers is 270A. However, because the PV breaker is rated at 125% of the inverter-output current, the inverter-output current can not exceed 56A.

Problem 68

Problem 68

From Table 250.122, if the current carrying conductor is protected with a 30 A overcurrent protection, use an AWG#10 equipment ground.

If the output of the inverter is connected to the optional standby system panel through a 30-A circuit breaker, the appropriate size of the equipment-grounding conductor between the inverter and the emergency panel is

a. 14 AWG copperb. 12 AWG copperc. 10 AWG copperd. 8 AWG copper

Table 250.122

Problem 69

Problem 69

A 5-kVA, 120-V generator has a rated output current of 42 A. It does not have a mechanism to limit its output current to the rated value. Assuming they are run in conduit, the output conductors should have an ampacity of no less than

a. 8 AWG THWNb. 8 AWG THHNc. 6 AWG THWNd. 6 AWG THHN

Article 445.13 indicates ampacity of the conductors needs to be 115% the current.

115% of 42 A is 48.3 A. 8 AWG THWN has an ampacity of 50 A.

Problem 70

Problem 70

Batteries 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

Problem 72

Problem 72

In a PV system, the equipment-grounding conductors should be

a. whiteb. blackc. redd. green

Problem 73

The 2011 NEC allows marking conductors with colored tape, provided that they are larger than

a. 10 AWGb. 8 AWGc. 6 AWGd. 4 AWG

Problem 73

The 2011 NEC allows marking conductors with colored tape, provided that they are larger than

a. 10 AWGb. 8 AWGc. 6 AWGd. 4 AWG

Grounding Conductors: 250.119(A) Grounded Conductors: 200.6(B)

Problem 74

Problem 74

NEC 110.26(A)(2)

The width of the working space in front of an inverter that is 24-inches wide must be at least

a. 24 inchesb. 30 inchesc. 36 inchesd. 42 inches

Problem 75

Problem 75

NEC 110.26 (A)(1)(b) Low Voltage.

The minimum depth of the working space in front of a charge controller for which the input voltage never exceeds 60 V dc is

a. 30 inchesb. 36 inchesc. 42 inchesd. negotiable

Problem 77

Problem 77 77. In order for a PV array to directly face the sun at 2:30 p.m. solar time on June 21 at 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

Problem 78

Problem 78

70% of 10/32 is 7/32

7 is 70% of 10.

Therefore, 7/32 is correct.

If 5/16-inch lag screws are used to fasten a charge controller to wooden studs, an appropriate pilot hole size would be closest to

a. 1/8 inchb. 19/64 inchc. 7/32 inchd. 1/4 inch

Problem 80

Problem 80

The first step in system checkout after completing the installation is

a. test open-circuit voltageb. visually check the entire systemc. install the source-circuit fusesd. close all disconnects

Problem 81

Problem 81

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).

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 outputb. install the source circuit fusesc. call the electrical inspectord. close all disconnects

Problem 82

Problem 82

-0.4%/°C x (50°C-25°C) = -10% loss due to temperature

Voltage = 68.4 × 0.9 ×0.984 = 60.6 V.

82. b. 60.6V

Derate due to temp Loss due to VD

Problem 83

Problem 83

4,000 W × .90 × .9 × 0.875 x .98 x .94 = 2,612 watts

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


Irradiance 900 w/m2

Module mismatch soiling etc. (not included in the problem statement but referred to in 2009 study guide)

Loss due to temperature:50-25=25 x .5% = 12.5% loss100% - 12.5% = 87.5%

Voltage Drop

Inverter Efficiency

Problem 84

Problem 84

Inverter Output = 4000W x 0.9 x 0.97 x 0.95 x (52 V/68.4 V) = 2522 W

A typical 4-kW crystalline silicon array is operating at STC in a utility-interactive system with battery backup. The STC maximum-power voltage rating of the PV array is 68.4 volts. The system uses a conventional charge controller that doesnot track maximum power. Wiring losses are 3% and inverter losses are 5%. If the batteries are at full charge at a voltage of 52 V, and if all PV output is delivered to the grid (assume that no power is being used to hold the batteries at 52 volts), the inverter output power will be closest to


Because there is no MPPT the STC array rating is reduced by the ratio of the battery voltage to the maximum-power voltage of the array.

Module mismatch soiling etc. (not included in the problem statement but referred to in 2009 study guide)

Voltage Drop

Inverter Eff.

Problem 85

Problem 85

4000 W x 0.95 (charge controller losses) x 0.97 (wire losses) x 0.95 (inverter losses) x 0.9 (mismatch) = 3152 W

A typical 4-kW crystalline silicon array is operating at STC in a utility-interactive system with battery backup. The STC maximum power voltage rating of the PV array is 68.4 volts. The system uses a MPT charge controller that has 5% losses. Wiring losses are 3% and inverter losses are 5%. If the batteries are at full charge at a voltage of 52 V, and if all PV output is delivered to the grid (assume that no power is being used to hold the batteries at 52 volts), the inverter output power will be closest to


*Key point here is that because the system uses MPT charge controller, there is no losses due to the voltage difference between Vmp of the array and the battery voltage like there was in problem 84.

Problem 87

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.

Problem 88

Problem 88

Answer: (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.

Problem 89

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

Problem 90

Problem 90

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

A utility-interactive PV system with no batteries consists of 15 100-W modules in series that feed a 1500-W inverter. The inverter output power is found to be 780 W when the modules are operating at 50°C with an irradiance level of 800 W/m2. If three 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

Max power

irradianceTemp losses

mismatch

Voltage drop

Assumed inverter efficiency

Problem 90

Since measured power was 780 watts and the expected performance was 697 watts, the system is likely working properly and exceeding expectations. Answer B.

#SolarMOOC Random Problems from 2009 NABCEP STUDY GUIDE

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