All PVIP nabcep test help needed

[ExtraordiNary]

I am reaching out. I have been installing solar for five years and hope to own my own business soon however I have failed the test three times all with a score of 68, yes I am a horrible test taker but need help, I am so close. So I would love to go over questions with people like:

How many microinverters can you put on a 30a breaker with 250w panels and 240v system
Is this 250÷240=1.04 30÷1.04=28 or do I debate the breaker 30/1.25=24 24/1.04 23??

 

[Carultch]

I am reaching out. I have been installing solar for five years and hope to own my own business soon however I have failed the test three times all with a score of 68, yes I am a horrible test taker but need help, I am so close. So I would love to go over questions with people like:

How many microinverters can you put on a 30a breaker with 250w panels and 240v system
Is this 250÷240=1.04 30÷1.04=28 or do I debate the breaker 30/1.25=24 24/1.04 23??

The short answer is that the 1.25 continuous load factor applies, for sizing ordinary breakers of all inverter output circuits. Your formula eventually becomes n = BKR/(1.25*P/V), and when rounding down to the nearest whole number, you get n=23.

The long answer is that you've had to make a couple assumptions, and the following would also matter in reality:
1. What device is actually rated for 250W? Because the inverter output circuit is always sized according to the inverter's own specifications, regardless of what module is feeding it. Only the microinverter's rating matters, for the circuit on the AC side.
2. Not every inverter's datasheet matches what you get, by estimating current from P=I*V. If there is a discrepancy, the ampere rating on the datasheet is what should be applied, instead of WattsAC/Volts or WattsAC/Volts/sqrt(3). Some manufacturers have their reasons for making this number larger than you'd expect.
3. In theory, a breaker could be built in a dedicated/listed assembly and listed for continuous duty, in which case the 1.25 factor would not apply. In practice, this is a rare situation, since ordinary breakers are only listed for noncontinuous loads at their full rating, and require the 1.25 factor for sizing them with continuous loads.

 

[ExtraordiNary]

Thank you for your input. As an installer I am not allowed to do the AC side so I have been slowly learning. This question comes from the test and as most of their questions there is more information that is needed.

 

[ExtraordiNary]

The short answer is that the 1.25 continuous load factor applies, for sizing ordinary breakers of all inverter output circuits. Your formula eventually becomes n = BKR/(1.25*P/V), and when rounding down to the nearest whole number, you get n=23.

The long answer is that you've had to make a couple assumptions, and the following would also matter in reality:
1. What device is actually rated for 250W? Because the inverter output circuit is always sized according to the inverter's own specifications, regardless of what module is feeding it. Only the microinverter's rating matters, for the circuit on the AC side.
2. Not every inverter's datasheet matches what you get, by estimating current from P=I*V. If there is a discrepancy, the ampere rating on the datasheet is what should be applied, instead of WattsAC/Volts or WattsAC/Volts/sqrt(3). Some manufacturers have their reasons for making this number larger than you'd expect.
3. In theory, a breaker could be built in a dedicated/listed assembly and listed for continuous duty, in which case the 1.25 factor would not apply. In practice, this is a rare situation, since ordinary breakers are only listed for noncontinuous loads at their full rating, and require the 1.25 factor for sizing them with continuous loads.

Thank you very much. Can you guide me to where in the NEC that it says for rmc conduit the at least one end needs a grounding bushing...also to secure emt to a metal box wouldn't you use a grounding bushing or is just a locknut washer sufficient.

 

[Carultch]

Thank you very much. Can you guide me to where in the NEC that it says for rmc conduit the at least one end needs a grounding bushing...also to secure emt to a metal box wouldn't you use a grounding bushing or is just a locknut washer sufficient.

In general, it doesn't. Metal conduit usually can get its required continuity to metal enclosures through standard locknuts and connectors. Where standard parts are insufficient, and bonding bushings are required in addition to the parts you use mechanically, is situations with missing or impaired continuity. Such as plastic enclosures, or ring knockouts remaining over 250V to ground.

It is metal service raceways specifically, that require at least one bonding bushing. I.e. raceways with circuits prior to the main service disconnect. Another specific case, is steel raceways with a GEC inside them, that require it on both ends.

 

[ExtraordiNary]

In general, it doesn't. Metal conduit usually can get its required continuity to metal enclosures through standard locknuts and connectors. Where standard parts are insufficient, and bonding bushings are required in addition to the parts you use mechanically, is situations with missing or impaired continuity. Such as plastic enclosures, or ring knockouts remaining over 250V to ground.

It is metal service raceways specifically, that require at least one bonding bushing. I.e. raceways with circuits prior to the main service disconnect. Another specific case, is steel raceways with a GEC inside them, that require it on both ends.

Ok thanks so concentric knockouts would require grounding bushing noooo locknut is sufficient man I get confused with this stuff...thanks

 

[Carultch]

Ok thanks so concentric knockouts would require grounding bushing noooo locknut is sufficient man I get confused with this stuff...thanks

Here's the explanation of why. The small sections of metal that hold each ring onto the rest of the enclosure, create a choke-point in how fault current will flow from the enclosure to the conduit. This is what impaired continuity means. A 277/480V source will generate over 5 times as much heat on that chokepoint, as a 120/208V source. Unless the enclosure is listed otherwise, the default is that these are only listed for faults driven by 250V or less. To avoid this problem, the code requires an EGC-sized bonding jumper between the conduit fittings and the enclosure's ground bar, to bypass the choke-point. Some enclosures have more heavy-duty punch-outs, that are manufactured and listed for higher voltage applications.

Where a bonding bushing is required, here are your options for what to put on the interior:
1. A bonding bushing electrically, and a standard locknut mechanically.
2. A bonding bushing serving both purposes. A bonding bushing can be the nut that secures the conduit mechanically, if it can tighten snug against the enclosure interior.
3. A bonding locknut/jumper mechanically and electrically, with a plastic bushing to protect wiring.
4. A bonding wedge/jumper electrically, a standard locknut mechanically, and the plastic bushing. This is what you do to recover, after you forgot it.

 

[ExtraordiNary]

Here's the explanation of why. The small sections of metal that hold each ring onto the rest of the enclosure, create a choke-point in how fault current will flow from the enclosure to the conduit. This is what impaired continuity means. A 277/480V source will generate over 5 times as much heat on that chokepoint, as a 120/208V source. Unless the enclosure is listed otherwise, the default is that these are only listed for faults driven by 250V or less. To avoid this problem, the code requires an EGC-sized bonding jumper between the conduit fittings and the enclosure's ground bar, to bypass the choke-point. Some enclosures have more heavy-duty punch-outs, that are manufactured and listed for higher voltage applications.

Where a bonding bushing is required, here are your options for what to put on the interior:
1. A bonding bushing electrically, and a standard locknut mechanically.
2. A bonding bushing serving both purposes. A bonding bushing can be the nut that secures the conduit mechanically, if it can tighten snug against the enclosure interior.
3. A bonding locknut/jumper mechanically and electrically, with a plastic bushing to protect wiring.
4. A bonding wedge electrically, a standard locknut mechanically, and the plastic bushing. This is what you do to recover, after you forgot it.

Thank you for the input.

 

[ggunn]

I am reaching out. I have been installing solar for five years and hope to own my own business soon however I have failed the test three times all with a score of 68, yes I am a horrible test taker but need help, I am so close. So I would love to go over questions with people like:

How many microinverters can you put on a 30a breaker with 250w panels and 240v system
Is this 250÷240=1.04 30÷1.04=28 or do I debate the breaker 30/1.25=24 24/1.04 23??

The 250W and 240V numbers are red herrings - very common in NABCEP questions. You need to know what the maximum output current of the microinverter is at the service voltage, multiply it by 1.25, and divide 30A by that number. Round down to the nearest whole number.

It could mean that the microinverters are rated at 250W maximum at 240V (although the specs for them are not normally written that way and a "250W panel" usually refers to the DC STC rating of the panel, not the rating of the microinverter), in which case Imax = W/V =250W/240V = 1.04A, (1.04A)(1.25) = 1.30A, and 30A/1.30A = 23.04. The max inverters on a 30A branch is 23.

 

[ExtraordiNary]

The 250W and 240V numbers are red herrings - very common in NABCEP questions. You need to know what the maximum output current of the microinverter is at the service voltage, multiply it by 1.25, and divide 30A by that number. Round down to the nearest whole number.

It could mean that the microinverters are rated at 250W maximum at 240V (although the specs for them are not normally written that way and a "250W panel" usually refers to the DC STC rating of the panel, not the rating of the microinverter), in which case Imax = W/V =250W/240V = 1.04A, (1.04A)(1.25) = 1.30A, and 30A/1.30A = 23.04. The max inverters on a 30A branch is 23.

Ok thanks that was great information. You are obviously familiar with nabcep. Can I go over questions with you. I am giving it another try on Sat. I guess the only way you really fail is to quit.
Thanks

 

[ExtraordiNary]

Ok thanks that was great information. You are obviously familiar with nabcep. Can I go over questions with you. I am giving it another try on Sat. I guess the only way you really fail is to quit.
Thanks

Accidently ungrounding the grounded conductor results in what safety risk?

 

[ggunn]

Accidently ungrounding the grounded conductor results in what safety risk?

These questions will be multiple choice; what are the options for this one? Floating the neutral makes the line to neutral voltages dependent on the magnitude of the line to neutral loads.

 

[ExtraordiNary]

These questions will be multiple choice; what are the options for this one?

Yeah don't quite remember...like shock, fuses and breakers won't work

 

[ggunn]

Yeah don't quite remember...like shock, fuses and breakers won't work

For most systems grounded = neutral. Floating the neutral makes the line to neutral voltages dependent on the magnitude of the line to neutral loads. It turns the system into a voltage divider; line to line voltage will stay the same but line to neutral voltages can be anywhere from nearly zero to nearly the line to line voltage although they will still sum to the line to line voltage. Use that knowledge to pick an answer.

 

[ExtraordiNary]

For most systems grounded = neutral. Floating the neutral makes the line to neutral voltages dependent on the magnitude of the line to neutral loads. It turns the system into a voltage divider; line to line voltage will stay the same but line to neutral voltages can be anywhere from nearly zero to nearly the line to line voltage although they will still sum to the line to line voltage. Use that knowledge to pick an answer.

Thanks

How about uplift force is 46lbs/ft2 using 3 inch lags there are three panels over four rafts label a,b,c,d. A is the inside b is inside what is the percent uplift force on lag b than lag a.
50, 100, 200, 400...very vague question lag a has one panel versus b has two panels on it so a guess is 200%

Any thoughts to consider on interrupt ratings. There was a question but it was so long all I remember is the answer had a breaker size and either 5000 or 10000 interrupt rating.

 

[ggunn]

From my experience with the NABCEP exam in 2009, any answer that mentions a painted wooden ladder is right out! There could be OSHA questions; that one got me.

 

[ExtraordiNary]

From my experience with the NABCEP exam in 2009, any answer that mentions a painted wooden ladder is right out! There could be OSHA questions; that one got me.

Thanks, just frustrating our utility makes us have the nabcep to install yet about 25 questions relate to AC and I am not allowed to do AC work. Thanks for your help

 

[ExtraordiNary]

Thanks, just frustrating our utility makes us have the nabcep to install yet about 25 questions relate to AC and I am not allowed to do AC work. Thanks for your help

Where do you live?? Colorado here

 

[ggunn]

Where do you live?? Colorado here

Me? Texas.

 

[ExtraordiNary]

Me? Texas.

How is solar down there these days. We installed two 40 panel jobs in Waco area three years ago