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Formulas Review forNABCEP PVI Exam Part2Sarah RaymerDirector of Education and Training ServicesSolPowerPeople, Inc.
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Notice to viewers:Photovoltaic (PV) Installer Resource Guide(Revised August 2012, version 5.3) Please click hereto download a copy.The PV Resource Guide is an excellent resource forthose planning to take the NABCEP Solar PV InstallerCertification Exam.While it remains a useful study resource it, as always,is NOT the definitive guide to the exam. Also, wewould like to remind readers that unlike the examquestions in the NABCEP Certified PV InstallerExam, sample questions in this guide were notwritten under the supervision of a professionalpsychometrician.
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Notice to viewers: The newly revised NABCEP Installer Resource Guide Version 5.3 includes the Solar Installer Job Task Analysis WITH links and suggested readings on where to locate information to help you learn and understand those concepts!Verify System Design, ConfirmString Sizing Calculations [Sect. 5]
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http://www.solarpaneltalk.com/showthread.php?727-Any-NABCEP-certified-members-aroundPost:I wonder if this certification is as tough toacquire as some folks gossip it to be...Comment:Why, yes, yes it is.I have been in the industry for a few years as afull time professional & also teach the NABCEPCert of Knowledge and PV Advanced courses ata local community college & I had troublepassing it.
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Voltage DropThe formula: CM = [25 x I x L] ÷ VdropCM – Circular millsI – single phase current [amps]L – one-way lengthVdrop – % voltage drop
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Voltage Drop CM = [25 x I x L] ÷ VdropCM is the circular mills diameter of the conductor [seeNEC Chapter 9, Table 8]I is the Imp/amps [Are you looking at 1 string, 2strings, max power conditions or STC?]Length is ONE-WAY distanceVdrop is the % of the Voltage of the system that islost, so: If 240V, < 1%: 240V x .01 = < 2.4V If 48, < 2%: 48V x .02 = < .96V
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Voltage DropFormula approach: What do we know? What do we need to know? CM = [25 x I x L] ÷ Vdrop Look at the problem What is given What is asked Clarify Refer to diagrams and codebook
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Voltage Drop Example 1- # 47 of the NABCEP™ 2009 Study Guide:• D = 60 Feet• System is a 24 V system• Vdrop = < 2% [2% of 24V = 24V x .02 = .48]• I =7A Imp – one string “when Im is flowing”• What wire size [CM] could we use to stay under 2%?
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24 volt system, 2 strings of 2 modules inseries
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Voltage Drop WHAT DO WE KNOW? • D = 60 Feet • System is a 24 V system • Vdrop = < 2% [2% of 24V = 24V x .02 = .48] • A = Question involves junction box to combiner box circuit: 2 modules in series, 7 amps each, “when Im is flowing” • We are looking at one string! • SVA, PAA- series volts add, parallel amps add CM = [25 x I x L] ÷ Vdrop WHAT DO WE NEED TO KNOW?• What wire size [CM] could we use to stay under 2% [<.48V] CM = [25 x 7x 60] ÷[24V x .02]
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Voltage Drop CM = [25 x 7x 60] ÷[24V x .02]Solve: 25 x 7 x 60 = 10,500 10,500 ÷ .48 = 21,875 Now- what wire size is at least 21,875 CM? #8 is only 15,510, not big enough #6 is 26,240- bigger than 21,875! Answer is C, #6!
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Voltage Drop CM = [25 x I x L] ÷ Vdrop Example 2- # 49 of the NABCEP 2009 Study Guide:We know:I = 7AL = 5ftV = 24VVdrop = < 1%
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Voltage Drop CM = [25 x I x L] ÷ VdropWe know: #16 is onlyI = 7A 2580, not bigL = 5ft enoughV = 24VVdrop = < 1% #14 is 4110- bigger than 3,646- OK!Solve:CM = [25 x 7 x 5] ÷ [.01 x 24]CM = 875 ÷ .24 A, #14!CM = 3645.8 or 3,646
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Voltage Drop CM = [25 x I x L] ÷ Vdrop Example 3- # 50 of the NABCEP 2009 Study Guide:Note- this question is stating “under maximum power conditions atSTC”. The Max power voltage for this circuit which is 2 modules inseries would be 34.2V [Vmp x 2]. We have to be careful about what is being asked- this is tricky!
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• SVA- series volts add!• MAX power conditions!
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Voltage Drop CM = [25 x I x L] ÷ VdropWe Know:I = 7AL = 60ftV = 34.2VVdrop = < 2%So- Vdrop is 34.2V x .02 =.684Solve:CM = [25 x 7 x 60] ÷ [.684]CM = 10500÷ .684 How does it work? Forums are great!CM =15351 http://www.physicsforums.com/showthread.phSo B, # 8 would be it! p?t=577708
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Voltage Drop CM = [25 x I x L] ÷ VdropWhat size wire would carry a load of40A at 240V a distance of 500 feet with2% or less voltage drop?CM = [25 x 40 x 500] ÷ [240 x .02]CM = 500,000 ÷ 4.8CM = 104,167What Size AWG?Where do we find that?
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Voltage Drop CM = [25 x I x L] ÷ VdropWhat size wire would carry a load of40A at 240V a distance of 500 feet with2% or less voltage drop?CM = [25 x 40 x 500] ÷ [240 x .02]CM = 500,000 ÷ 4.8CM = 104,167What Size AWG? #1/0Where do we find that?
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Voltage Drop CM = [25 x I x L] ÷ VdropWhat size wire would carry a load of40A at 240V a distance of 500 feet with2% or less voltage drop?CM = [25 x 40 x 500] ÷ [240 x .02]CM = 500,000 ÷ 4.8CM = 104,167What Size AWG? #1/0Where do we find that?
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Voltage Drop CM = [25 x I x L] ÷ Vdrop What size wire would carry a load of 40A at 240V a distance of 500 feet with 2% or less voltage drop? CM = [25 x 40 x 500] ÷ [240 x .02] CM = 500,000 ÷ 4.8 CM = 104,167 What Size AWG? #1/0 Where do we find that?Chapter 9, Table 8 2011 NEC Codebook- the onlyreference you’re allowed during the NABCP PVI Exam
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Voltage DropWhat voltage drop would arise from carrying 50Aa distance of 250 feet with #8AWG? CM = [25 x I x L] ÷ Vdrop16510CM = [25 x 50A x 250] ÷ Vdrop16510CM= 312500 ÷ Vdrop16510CM x Vdrop = 312500Vdrop÷ 16510CM = 312500 ÷ 16510CMVdrop = 18. 92V• Remember: Vdrop is the % of the Voltage of the system that is lost, so: If 240V, < 1%: 240V x .01 = < 2.4V If 48, < 2%: 48V x .02 = < .96V
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Last one…What wire size would you use to design for lessthan 1% voltage drop in a circuit carried over a160 ft. distance from a 3000W/240V inverter tothe main distribution panel? D = 160 V = 240, V drop is < 1% of 240 = 2.4V I = 3000 ÷ 240 = 12.5A CM = [25 x 12.5 x 160] ÷ [2.4] CM = 50,000 ÷ 2.4 CM = 20,833 So - # 6 AWG at 26,240CM – is big enough
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Max Circuit Current, Conductor and OCPD Sizing 2011 NEC CodeFor this review, we are primarily dealing with: Article 240 - Overcurrent Protection 240.4 - OCPD/AWG imitations 240.6 - OCPD sizes Article 310 - Conductors for General Wiring Table 310.15[B[2][a] - ambient temp. correction Table 310.15[B][3][c] - # of current carrying conductors in conduit adjustment Table 310.15[B][3][c] – Rooftop temp. adjustment Article 690 – Solar PV Systems 690.8 – Circuit Sizing and Current
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Chapter 3 of the NEC 2011 Codebook SOME PV ReferencesDownload complete file on our webpage:http://solpowerpeople.com/2011-nec-2011-study-guide-database/
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Max Circuit Current, Conductor and OCPD SizingMAX Circuit Current: Use the Isc Multiply for 1.25 for Irradiance spikes If more than one, multiply by # strings: PAA [parallel - amps add]Minimum Conductor Ampacity and OCPD: Use the Max. Circuit Current and multiply by another 1.25 to calculate continuous load ampacity. Use the Max Circuit Current and apply corrections and adjustments for conditions of use. USE THE HIGHER of these 2 calculations for ampacity.
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Step 1: Max. Circuit Current690.8[A] Max. Circuit Current: 690.8[A][1] - Isc x 1.25 – to account for high irradiance fluctuations 690.8[A][2] - Multiply by # strings if more than one in parallel PAA [parallel- amps add] 690.8[A][3] – Inverter output circuit current = continuous inverter output rating [on spec sheet- or calculated] 2000W ÷ 240V = 8.3A
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Step 1: Max. Circuit Current Inverter Max. Cir. Current is the Max Output Current from the spec. sheet. Make sure you look at the right one for your nominal system voltage and inverter size.2000 ÷ 240 = 8.3A
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Step 1: Max. Circuit CurrentModule Isc is 12.5A, what is the max. circuitcurrent? 12.5 x 1.25 = 15.63AIsc of 12.5A, 3 series strings in parallel. What isthe PV output max circuit current [same asInverter input circuit]? 12.5 x 3 x 1.25 = 46.88AInverter’s max circuit current is the Max Cont.Output Rating [spec sheet] 8.3A [2000 ÷ 240 = 8.3A]
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Step 2: OCPD and MINIMUM Ampacity 690.8[B][1]- Overcurrent Devices690.8[B][1][a] – OCPD must carry at least 125%of max circuit current Isc x 1.25 x 1.25 or Isc x 1.56690.8[B][1][b] – must be installed in accordanceto listing/labeling {110.3[B]}690.8[B][1][c] – consider temp. rating of terminal.You must use the 75ºC column for sizingampacity of conductor to match that allowed bythe terminal. You can use the 90ºC column forampacity adj. and corr. with a 90ºC ratedconductor {110.14[C]}
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Step 2: OCPD and MINIMUM Ampacity 690.8[B][2]- Conductor Ampacity690.8[B][2] – conductors must be the larger ofthe two 690.8[B][2][a] – Must be 125% of 690.8[A] without corrections for conditions of use Isc x 1.25 x 1.25 690.8[B][2][b] – Must be max. circuit current calculated in 690.8[A] after conditions of use corrections have been applied690.8[B[[2][c] – The conductor must still beprotected by the OCPD
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Step 2: OCPD and MINIMUM Ampacity690.8[B[[2][c]OCPD must protect the wire that comes into it.The wire coming into the terminal must have anampacity HIGHER than the OCPD rating.Think- BAD: If the conductor could only carry 18A before failing, and your OCPD was a 20A breaker- it wouldn’t trip and prevent a failure in the conductor. GOOD: If you have a conductor that could carry up to18A under conditions of use before failing and your OCPD is a 15A fuse, the conductor will never get up to failure point of 18A because the fuse will trip and open the circuit at 15A- before the conductor fails.
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Step 2: OCPD and MINIMUM AmpacityMinimum Conductor Ampacity and OCPD: Use the Max. Circuit Current and multiply by another 1.25 to calculate continuous load ampacity. 690.8[B][1][a] – OCPD must carry at least 125% of max circuit current 690.8[B][1][b] – consider temp. rating of terminal. For example, if you are using a 75ºC rated terminal, you must use the 75ºC column for sizing ampacity of conductor to match that allowed by the terminal. You can use the 90ºC column for ampacity adj. and corr. with a 90ºC rated conductor {110.14[C]}Isc x 1.25 x 1.25 or Isc x 1.56Use the 75ºC column for this conductor calculationbecause it must be designed to not exceed the terminaltemp rating
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Step 2: OCPD and MINIMUM Ampacity What size conductor would be required for the PV output circuit from the combiner box to the disconnect in a system with 3 strings of 225 Watt modules with an Isc of 6.7A? (3) strings of Scheuten 225W Modules Isc = 6.7A Step 1- Determine Maximum Current- Isc x 1.25 x 3 6.7 A x 1.25 x 3 = 25.125 A 690.8[B][2][a] Step 2- Determine MINIMUM Conductor Ampacity and OCPD- 25.125 A x 1.25 = 31.4 A minimum
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Step 2: OCPD and MINIMUM Ampacity6.7 A x 1.25 x 3 = 25.125 A 690.8[B][2][a]Step 2- Determine MINIMUM Conductor Ampacity and OCPD- 25.125 A x 1.25 = 31.4 A minimumThe next size OCPD is A 35A OCPD [NEC 240.6]
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OCPD sizes- NEC 240.6 Minimum Ampacity = 31.4 A minimum OCPD = 35A (next standard size – NEC 240.6)
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Step 2: Ampacity andMinimum AmpacityOCPD = 31.4 AOCPD = 35A (next standard size – NEC 240.6)Step 3: Choose Conductor-Choose from Table 310.15(B)(16) from 75°C column (becauseof terminal rating) = #10 AWG (max 35A) {690.8 [B}[1][b]} 690.8[B][1][b] – consider temp. rating of terminal. You must use the 75ºC column for sizing ampacity of conductor to match that allowed by the terminal. Later, you can use the 90ºC column for ampacity adj. and corr. with a 90ºC rated conductor {110.14[C]}However, due to NEC 240.4(D)** which states the max. OCPDfor #10 AWG is 30A, you must go up to #8 AWG which canhandle up to 50A, and is not limited by the OCPD as per240.4[D]**.
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**240.4(D)At the bottom of Table 310.15(B)(16), there is a notation that must beconsidered for wire sizes #10 through #18 (marked with 2 asterisks **).
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Step 3: Conditions of Use 690.8[B][2][b] Minimum Ampacity = 31.4 A minimumOCPD = 35A (next standard size – NEC 240.6) #8 AWG which is not limited by 240.4D as#10AWG is- to be protected by no less than a 30A OCPD. Step 4: Based on 90°C column for ampacity, apply your derates based on conditions of use. 8 AWG allows for 55 AMPS.
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Conductor Sizing - Conductor Ampacity Under Conditions of Use 8 AWG allows for 55 AMPS. Ambient Temperature is 40.5°C Conduit is 4” off roof 3 strings have come to 1 in the PV output circuit, so we have 2 current carrying conductors
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Conductor Sizing Under Conditions of UseIn conduit 4” above the rooftop [Table310.15(B)(3)(c)] add 17°C to the 40.5° toget to 57.5°C , which derates to .71 (Table310.15(B)(2)(a)).There are only 2 conductors in conduit ,so there is no deration multiplier for howmany conductors in conduit (Table310.15(B)(3)(a)).
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“Conditions of Use”NEC Table 310.15(B)(2)(c) (2008)NEC Table 310.15(B)(3)(c) (2011)Temperature Adder is added to record hightemperature then derated using Table310.15(B)(3)(c) Roof to Distance From Bottom of Conduit Temperature Adder 0 to ½ inch 33°C ½ to 3 ½ inches 22°C 3 ½ to 12 inches 17°C 12 to 36 inches 14°C
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“Conditions of Use”Correct for Number of Conductors in Raceway NEC Table 310.15(B)(2)(a) (2008) NEC Table 310.15(B)(3)(a) (2011) Number of Current- Carrying Conductors Correction Factor 4 to 6 0.80 7 to 9 0.70 10 to 20 0.50 21 to 30 0.45 31 to 40 0.40 Over 40 0.35
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Conductor Sizing Under Condition of Use8 AWG allows for 55 AMPS (90° column) We are allowed to use 90ºC column as per 110.14C.55 A x .71 = 39.05 ABecause 39.05 A is greater than max current of31.4A, then #8 AWG is GOOD!If you did the calculation with 90ºC column and a#10AWG wire 40A x .71 = 28.4 Ampacity after COU. This is not OK for the expected Max current of 31.4A.Conductor can handle 39.05A, and so is protected bythe 35A OCPD.
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