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D. Electrical Calculations <br />1. PV Circuit current <br />PV circuit nominal current <br />Conrinuous current: adjustment factor <br />PV circuit conlinuous Guitent rating <br />15.3 amps <br />125% <br />19.125 amps <br />2013 CEC Article 705.60(Bb <br />2. Overcurrent protection device rating <br />PV circuit continuous current rating <br />Next standard size fuse/breaker to protect conductots <br />19. I 25 :itilps <br />20 amp breaker <br />Use 20 amp AC rated fuse or breaker <br />3. Conductor conditions of use adjustment (conductor ampacity derate) <br />a. Temperature adder <br />Average high temperature 40.9 °C (105.62 °13 <br />Conduit is installed 1" above the roof surface Add 22 °C to ombient <br />Adiust.cd maxinium ambicn[ temperature 62.9 °C (145.22°F) <br />2013 ¢ T.C Table 31 415(B)(2)(a) <br />b. PV Circuit current adjustment for new ambient temperature <br />Donte factor for 62.9 °C (145·22°ID 58% <br />Adjusted PV circuit continuous current 3229 amps <br />2013 CEC Table 310.16 (bottom of tabi¢) <br />c. PV Circuit current adjustment for conduit fill <br />Number of current-carfying conductors <br />Conduit fill derate factor <br />Final Adjusted PV circuit continuous currerit <br />3 conductors <br />100% <br />32.9 amps <br />2013 CEC Table 310.15(13)(2)(a) <br />Total derated ampacity for PV circuit 32.9 amps <br />Conductors (tag2 on 1-line) must be; rated for a minimum of 32.9 amps <br />THWN-2 (90°C) #10AWG conductoris rated for 40 amps (Use #10AWG or larger)2013 CECHTable 310.16 <br />4. Voltage drop (keep below 3% total) <br />2 parts: <br />1. Voltage drop across longest PV circuit micro-inverters (from modules to j-box) <br />2. Voltage drop across AC conductors (from i-box ro poinr. u f interconnection) <br />1. Mirco-inverter voltrage drop: <br />The largest number of micro-inverters in a row in the entire array is 9 inCircuit 1. According to <br />manufacturer': specifications this equals a voltage drop of 0.39 %. <br />0.39% <br />2. AC conductor volruge drop: <br />= IxRKD (+ 240 x 100 to €onvert to percent) <br />= (Nominal current of largest circuit) x (Resistance of #10.AWG copper) x (rotal wire run) <br />= (Circuit 1 norninal currtnt is 15.3 amps) x (0.00126@ x (1'50) + (240 volts) x (100)1.2% <br />Total system voltage drop:1.59% <br />vivint 30 3% r