PV Student

Submitted: 04/23/2016

Training: NABCEP PV Technical Sales Exam Prep Discussion Group / Online / Anytime

How would you rate this course?
10/10

What did you like about the course?:
The pace for me was good, also questions with answers really helped. In the 5th week I got bombarded with work so was playing catch up.

How effective was the instructor's communication during the course?:
Very good, seems like my teammates asked the same things I was thinking.

What would you change about the course?:
Can't really think of anything.

If you enjoyed the course and are interested in providing a testimonial, please enter one here.:
Course was very good, and Learned so much. I really feel confident in meetings.

Comment:

Hey Sean, the exam went great! Youe prep course and material more than prepared me for the exam. I think you did a great job with us these last 6 weeks. There was not 1 question I was not able to handle.

Reply:

Thanks for the feedback!

Glad to be able to help!!

Sean

8-days until the exam!

My advice is now is the time to study a lot. No procrastinating. Do not worry too much about obscure things that will probably not be on the exam. Do not spend too much time trying to understand the one thing that you do not understand. It is a numbers game and nobody ever gets all of the questions right.

Plan to be done cramming 2 days before the exam, when it is time to get rest and review. Remember that a good state of mind and a clear head will be the best thing that you can accomplish the night before.

Now is time to start getting up early every morning if that is not your normal habit, so you will not be tired waking up earlier than usual on exam day.

Make sure you know where the exam is, where you will park and that you remember your ID.

Onward through the information!!

Sean White

Q:

Thanks Sean. Do you know which formula they would expect us to use? I've run across several in previous studies of varying accuracy levels:

Basic inter-row shading distance

(d) = 3H

H = height of module from back edge to surface

Better inter-row shading distance

(d) = H/tan AltA

H = height of module from back edge to surface

AltA = Altitude Angle on December 21st at 9:00 AM

Best inter-row shading distance:

(d) = [ (Sin TiltA) x M ] / tan AltA ] x cos AzAD

M = Module Length

TiltA = Desired Tilt Angle

AltA = Altitude Angle on December 21st at 9:00 AM

AzA = Azimuth Angle on December 21st at 9:00 AM

AzAD = Difference between AzA and due south

A:

I don't think that the NABCEP Exam is going to make you do the inter-row shading 3-part calculation.

I use the calculation when teaching trig often, because it covers sine, cosine and tangent all on one calculation. Very cool.

Usually nabcep will have a single stage trigonometry calculation on their exams. Often it will have something to do with shading from a tree.

It will probably be easier than any of the equations that you are promoting. Usually the trick that will get people on an inter-row shading calculation is the azimuth correction angle.

For your first example:

Basic inter-row shading distance

(d) = 3H

H = height of module from back edge to surface

This is very common in the industry, however for most places a 3:1 ratio is overkill. The California Solar Initiative used to require a 2:1 ratio. If you are closer to the equator, the 2:1 ratio is overkill and if you are closer to the arctic circle, 3:1 is good and moving south is better:) Not really, I loved living in Alaska.

For your second example:

Better inter-row shading distance

(d) = H/tan AltA

H = height of module from back edge to surface

AltA = Altitude Angle on December 21st at 9:00 AM

With this calculation, you are not taking into consideration the fact that at 9am, your sun is not yet facing the PV and you would be able to reduce your inter-row space with an azimuth correction. At 9am in the winter, your sun can be low and at an angle around 40 degrees from south in many places. 3 hours later at solar noon when your sun is south (and hopefully your array facing south), then the north to south distance between your rows will be less than that longer shadow that the 9am sunbeam will have to take at a 40 degree angle from south.

Here is something that I put together years back on my 1970s era website that has an inter-row spacing example with an image of the azimuth correction half way down the page:

http://www.pvstudent.com/Advanced-PV-Course-Notes-2.html

For your 3rd example:

Best inter-row shading distance:

(d) = [ (Sin TiltA) x M ] / tan AltA ] x cos AzAD

M = Module Length

TiltA = Desired Tilt Angle

AltA = Altitude Angle on December 21st at 9:00 AM

AzA = Azimuth Angle on December 21st at 9:00 AM

AzAD = Difference between AzA and due south

This is the correct industry way to do the calculation, however it is not an exact science for the following reasons:

1-Closer to the arctic circle the sun dies not rise until after 9am, so you would have to improvise.

2-At place with different air densities morning sun has different potentials (Arizona morning sun is a lot brighter than a humid or smoggy place).

3-If your array is not facing directly south, west facing for example, then 9am would be less important than 4pm.

4-If you have thin film PV where the solar cells are uniform stripes from the bottom to the top of the module, then the cells will all be shaded to the same proportion, so inter-row shading effects are not quite as severe as with typical crystalline PV modules.

5-Some places might work better to do the calculation for sun angles at 8:45am and others might be best at 9:15am, also it depends on the price of the system, 9am weather, inverter prices, etc, etc, etc...

Many racking manufacturers have pre determined inter-row spaces, such as ballasted rooftop mounting systems.

Horizontal axis trackers have the best option, they will back-track and go flat at sunrise and sunset to prevent any-inter-row shading.

For large solar farms, they will spend a lot of time using software simulations with software, such as PVSYST or HelioScope to determine the best inter-row shading distance.

2.5:1 is also a good distance used by many.

Remember, solar is related to weather and is not an exact science, unless you are in space.

Thanks,

Sean White

Q:

Looking at the joint task analysis for non-financial benefit analysis it says to know how calculate CO2 avoidance, tons of coal saved, water saved, miles not driven, etc. Is that really something they would ask us how to calculate? Tons of assumptions go into each of those and several vary by geographic location. Any particular rules of thumb to remember for them or is it more likely they would give us all the inputs/assumptions and just ask us to calculate them based on lifetime kWh production of the system?

A:

When I took the test these environmental questions were obvious and the answers are easy if you know how to do conversions. There are some examples in the book.

If they give you X pounds of CO2 per kWh, and tell you how many MWh were used, convert MWh to kWh and then multiply by X pounds of CO2.

They are not going to expect you to memorize how many pounds of CO2 are produced by a fossil fuel plant in Hawaii vs. a coal plant in North Dakota.

When I looked up the conversion for the book, I ended up at the EIA website:

https://www.eia.gov/tools/faqs/faq.cfm?id=74&t=11

It is interesting to know that it is in the range of 1.2 to 2.2 for fossil fuels according to EIA.

There are plenty of other websites that will give you different numbers. I'm sure you can find some fossil fuel funded foundation that will tell you PV pollutes more than clean coal. The funniest one I heard was from a Coursera course where the teacher said that wind energy leads to global warming because wind turbines slow down the wind which cools the earth. Very funny. He was very pro-nuke.

PV takes about a year to offset the energy that it took to make the PV system, longer if your crew drives stretch Hummers to work. This energy is getting to be less as the process is more efficient and less silicon is wasted.

Thanks,

Sean White

Q:

Oh and I was reminded today that mV/C(elsius) and mV/K(elvin) are the same thing but sometimes the spec sheets show it in mV/K which can be confusing.

A:

I think that the engineers making the datasheets that are using Kelvin instead of Celsius are just trying to impress people at how educated they are. They probably tell their family the weather forecast in Kelvin and have a big Kelvin thermometer on their front door. They probably have birthday parties for when their kids turn pi years old or the square root of 3.

Sean

Q:

I'm not super clear on the 10 ft. and 25ft. tap rules. Can you provide some concrete examples of how the conductor size is calculated for each of those?

 

A:

 

Let’s start first with the 10-foot tap rule and then infer the 20-foot tap rule from what we learned with the 20-foot rule.

 

Imagine a feeder going from a main panel to a sub panel. That feeder has a 200A breaker at the main panel.

 

Imagine that we want to connect a very small 8A inverter to the middle of that feeder.

 

The question is how large does the feeder tap conductor have to be. The feeder tap conductor is the conductor that is, in this case less than 10 feet.

 

The answer is it has to be at least 10% of the sum of the inverter feeder breaker plus 125% of the inverter current.

 

125% of the inverter current is 8A x 1.25 = 10A

 

The feeder breaker is 200A

 

200A + 10A = 210A

 

10% of 210A is 21A

 

We would need a feeder tap conductor for the 8A inverter that is at least 21A.

 

Now for the same numbers, except lets say that the feeder tap conductor between the inverter and the tap point is between 10 and 25-feet.

 

The answer is it has to be at least one-third of the sum of the inverter feeder breaker plus 125% of the inverter current.

 

125% of the inverter current is 8A x 1.25 = 10A

 

The feeder breaker is 200A

 

200A + 10A = 210A

 

One-third of 210A is 70A

 

The reason that the longer conductor has to be beefier is that there is more resistance on a longer conductor and if there is a fault, we want the breaker to open up before the wire heats up and starts a fire. A shorter wire has less resistance and a better chance at popping open the 200A breaker.

 

This is not common sense, because you would think that the conductor should be 200A to pop open the 200A breaker, but with a fault, you can have a better chance at opening a breaker than a mere overcurrent. A fault is hard-core! Like a short circuit!

 

Thanks,

Sean White

Q:

I've been reading recently about 2014 requirements for arc fault protection (690.11) and rapid shutdown (690.12). Both seem relevant for sales personnel, particularly the rapid shutdown requirements because of the potential for added equipment. Any good resources folks can recommend in this area?

 

A:

 

You are right, it is important for sales people to understand 690.11 dc arc fault protection and 690.12 rapid shutdown requirements.

 

690.11 became into play in the 2011 NEC for series arc faults. PV systems have to have dc arc fault protection and you would have a hard time finding an inverter that didn’t have dc arc fault protection.

A series arc fault is what happens if a PV connector started to come apart and there was a gap. The gap would start arcing and could start a fire if it didn’t burn itself out first.

 

If you have a large central inverter, then the inverter would have trouble detecting the arc with so many strings to keep track of and you would have to have the dc arc fault protection at the combiner.

 

690.12 Rapid Shutdown of PV Systems on or in Buildings came into play in the 2014 NEC and has been controversial. Many solar installers do not like having more rules that cost money and time. In defense of 690.12, firefighters are reluctant to climb on burning buildings that have electricity hidden in the walls when the power to the building is turned off. We hear them say “let it burn”. This is not good and Rapid Shutdown requirements are intended to be the cure.

 

The 2014 690.12 rules say that there has to be a way to make sure that there is less than 30V within 10 seconds of initiating shutdown on conductors within 10 feet of the array outside of the building or within 5 feet of the array inside of the building.

 

Rapid shutdown is accomplished 3 ways

1.    Inverter within 10 feet of the array will immediately shut down when ac power is no longer available to the inverter. This would include microinverters, ac modules or string inverters by the array.

2.    Power optimizers can perform rapid shutdown, because they are brains that can be controlled from down below. They need to be 690.12 compliant systems.

3.    A contactor is a remote disconnect that is powered from down below. Midnite solar and others make these switches that can be placed by the inverter shut-off switches and when the button is pushed, the array will be disconnected on the roof.

 

Warning: 2017 NEC Rapid Shutdown requirements will be stricter.

 

Thanks,

Sean White

Q:

One thing I remember from working on temperature co-efficients before is the importance of paying attention to whether the co-efficient is given as a percentage or in volts/degree. If it's in voltage/degree, which I've seen on some spec sheets, than you've got one less step to do.

A:

Good point! Usually it is given in percent per degrees C and it can confuse some people if all of the sudden the coefficient is given in volts or mV per degrees C.

 

You can also convert mV per degrees C to % per degrees C and vice versa.

 

Typical 60 cell module: 38Voc, Temp coef = -0.3%/C

Math:            -0.3%/C = -0.003/C

More Math: -0.003/C x 38V = -0.114V/C

 

And to go back

Math:            -0.114V/C divided by 38V = -0.003/C

                       -0.003/C x 100% = -0.3%/C

Thanks,

Sean

Q:

 

How much time is it worth to spend on the various types of cells and their chemical structures and construction methods? Or is it sufficient just to be aware in general of the different cell types (mono, poly, CIGS, CdTe, etc.) and their performance characteristics?

 

A:

Since you are studying for a Technical Sales Exam, I would recommend not worrying about getting into the different details about the chemistry if you have limited time. It is good to know, however there are many other things that are more likely to be on the exam. Yes it is a good idea to know that Monocrystalline is usually better than Polycrystalline and that CIGS and CdTe are thin film and have a history of being less efficient than crystalline.

 

That being said, there are interesting developments. It has been in the news recently that the world record efficient CdTe cell has just passes the world record Polycrystalline cell efficiency. The world record efficiencies are interesting, but do not translate to cost effective solutions.

 

Also, did you know that gallium arsenide is a thin film technology that is more efficient than monocrystalline. Gallium arsenide is also very very expensive and is that they typically use in space. Too expensive for terrestrial PV.

 

I also like to point out that I would rather have 16 percent efficient tier 1 polycrystalline PV than 15 percent monocrystalline. Mono is usually more efficient, but not always.

 

Another thing is that polycrystalline silicon often refers to the raw silicon that comes in chunks. Another term used for polycrystalline modules is multicrystalline.

 

Thanks,

Sean White