Verifying System Design

This Newsletter is packed full of great material! Great presentation from Jeff Gilbert can be viewed below, along with practice problems, additional resources, and the upcoming #SolarMOOC lecture topics.

But first, here is a great opportunity from Jeff Gilbert…

Success-In-Solar

Have you been thinking about starting your own solar contracting business, but don’t know how to get started?  Maybe you are already a solar contractor, but frustrated with your pace of growth?

Getting from where you are now to where you want to be in business may not have to be such a struggle. Jeff Gilbert of Azimuth Solar Training has developed a new program designed to make you highly successful.  The program is rightly called Success-In-Solar. Through the program Jeff will personally guide you and share his secrets to success from his 18 years of experience as a solar expert. In 2000 Jeff started Chesapeake Solar. He went through all the struggles and pains that are unique to a solar contractor and he learned tricks of the trade and strategies for success.  In 2007 his company was listed on Inc. Magazines 500 fasted growing small companies and in 2008 he sold his company to a large national company for a healthy price. He now wants to share what he has learned with you.

This program is your opportunity to pick Jeff’s brain, to learn how to avoid the common pitfalls and to gain the unfair advantage that will set you ahead of your competition. This program is not yet being advertised or offered to the public. Jeff is seeking a select few to initiate his new program.  To be considered for this unique training program send an Jeff an email at jeff.gilbert@azimuthsolar.net with “Success-In-Solar” in the subject box. Let him know you are interested in learning more and he will send you information about a special conference call where he will share the details of the program and allow you to ask questions. There’s really nothing to lose. Send your email now.

Verifying System Design: NABCEP IRG

Review of Case Study Example 8.1 with Jeff Gilbert

 

Upcoming Lectures:

September 6th, 2012

Ryan Light Owner at IRIS LLC

“Stand Alone System Design for NABCEP PVI Exam Preparation”

Ryan will walk through an example of a stand-alone/off-grid system design, providing a solid overvew of the ins and outs of:

  • Load analysis
  • Battery bank sizing and selection
  • Charge controller selection
  • Inverter sizing and selection

September 13th, 2012

Sarah Raymer

“PV Formulas Review for NABCEP PVI Exam Prep. Part 2”

September 17th, 2012

Richard Stovall

Troubleshooting and Maintenenance

 

Additional Resources to review:

NABCEP has issued a new NABCEP PV Installer Resource GuideTM! It has a Job Task Analysis with resource guidance to help you understand all the necessary subtopics within! This is incredible!

Download it here:

http://www.nabcep.org/wp-content/uploads/2012/08/NABCEP-PV-Installer-Resource-Guide-August-2012-v.5.3.pdf

 

This link is from: http://www.nabcep.org/resources

  

Home Power Magazine Code Corner by John Wiles: http://www.nmsu.edu/~tdi/Photovoltaics/Codes-Stds/codecorner.html

 

FAQs from Code Corner: http://www.nmsu.edu/~tdi/Photovoltaics/Codes-Stds/FAQcodes.html


Here’s a pocket index study card for the day. Write these down and carry them around with you- memorize them! These are a few of the specifics in safety and general practice that you may be required to know for the NABCEP PV Installer ExamTM.

verifying system design

 

Study Questions to go with Solar PV Design Formulas for NABCEP Preparation #SolarMOOC Webinar from August 23, 2012

Please see below for answers. Don’t cheat and read ahead! 😉

Other problems that should be reviewed  to test understanding of this webinar are in the NABCEP 2009 Study Guide:

Problem #s:  41, 42, 48, and 78

Please review and post your answers in the forum.

Download the Study Guide here: http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&ved=0CDYQFjAC&url=http%3A%2F%2Fwww.cabrillo.edu%2F~swadsworth%2Ffiles%2Fnabcepstudyguidev4-2april2009.pdf&ei=9RpBUPyiOKSi2wW_noHQDA&usg=AFQjCNFQSqjq74rlatZRuGJi2J9LDL6pig

This is a great study resource!!

1. If you have a budget of $23,000 and the cost of the PV system installed is $6.25/watt what  size system could you have installed?

 

2. What would the production of that system be with standard derations for DC to AC conversion?

 

3. What is the efficiency of a module that is 1.8m2 and 230 Watts at standard test conditions?

 

4. At standard test conditions, what would the efficient be of a module producing 175w/m2?

 

5. The usable portion of a roof with no shade is 12’ x 24’. Using a module that measures 62.2” x 31.4”, what is the maximum number of modules that can be installed?

 

6.To meet 40% of the energy demand for a home using 1340kWh/month, what size system DC would be required in a location with 5.3 peak sun hours/day?

 

7. How many attachment points will there need to be for an installation with the following circumstances: 7 Sharp 235 Watt modules arranged in portrait mode, each with a dimension of  39.1” x 64.6”. There can be no less more than 24” at each end of the rails unsupported, and no more than 48” in between attachments.

 

8. Determine the length of the lag screw required for the installation below:

Location has a wind load of 55psi, the roof trusses are southern yellow pine, and the width of the lag screw is 5/16”. There will be 14 modules arranged in 2 rows of 7, each with a dimension of 39.1” x 64.6”.  There will be 24 attachments and the roof is 1.5” thick.

 

9. What size hole should be drilled for a 7/16” inch lag screw in southern yellow pine?

 

[Questions below are from the NABCEP Study Guide 2009 Version 4.2] available online: http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&ved=0CDYQFjAC&url=http%3A%2F%2Fwww.cabrillo.edu%2F~swadsworth%2Ffiles%2Fnabcepstudyguidev4-2april2009.pdf&ei=9RpBUPyiOKSi2wW_noHQDA&usg=AFQjCNFQSqjq74rlatZRuGJi2J9LDL6pig

 

24. If an open circuit voltage of a crystalline silicone PV array is 315V at 25ºC, then according to the NEC, if the array is operating at -20ºC, the maximum system voltage must be corrected to:

a. 267V

b. 315V

c. 372V

d. 378V

 

26. If the maximum power voltage of a crystalline silicone module is 17.1V at standard test conditions, then at 60ºC [module temperature] and at 1000w/m2 incident on the module, the maximum power voltage will be closest to:

a. 20.1V

b. 17.1V

c. 14.1V

d. 12.0V

 

38. A crystalline silicone array that has bi-polar outputs of +252 and -252 with a common grounded conductor under STC is selected for a large single family dwelling. The lowest ambient temperature in which the system can be installed at a single family dwelling is:

a. -10ºC

b. 5ºC

c. -20ºC

d. 25ºC

 

37. A crystalline silicone array that has bipolar outputs of -252 and +252 with a common grounded conductor under STC is selected for a large single family residence. The lowest expected ambient temperature at the installation site is -25ºC. For this system, the max system voltage is:

a. 252V

b. 315V

c. 504V

d. 605V

 

——————————————————————————————————-

 

Answers:

1. If you have a budget of $23,000 and the cost of the PV system installed is $6.25/watt what  size system could you have installed?

1. $23,000 ÷ 6.25 = 3,680W DC

 

2. What would the production of that system be with standard derations for DC to AC conversion?

3.68kW DC x .77 = 2.83kW AC

 

3. What is the efficiency of a module that is 1.8m2 and 230 Watts at standard test conditions?

.230 ÷ 1.8 = .1277 or 12.8%

 

4. At standard test conditions, what would the efficient be of a module producing 175w/m2?

STC are 1000w/m2, 25ºC, an 1.5 Air Mass.

175/1000 = .175 or 17.5%

 

5. The usable portion of a roof with no shade is 12’ x 24’. Using a module that measures 62.2” x 31.4”, what is the maximum number of modules that can be installed?

12’ x 24’ = 144” x 288”

 

Portrait mode:

144” ÷ 62.2” = 2.32 or 2

288” ÷ 31.3” = 9.17 or 9

18 total

 

Landscape Mode:

144” ÷ 31.4” = 4.58 or 4

288” ÷ 62.2” = 4.63 or 4

16 total

 

But also consider the room required for end and mid clamps:

 

With 2 rows of :

2 x 62.2” = 124.4” , 144” – 124.4” = 19.6”

9 x 31.4” = 282.6” , 288” – 282.6” = 5.4” – this is not enough for the clamps that would be required to attach the modules to the rails. So this takes it down to 2 rows of 8. This is also 16 total modules. In fact, this installation would be easier install wise, because you would only need 4 rails rather than the 8 that you would need with 4 rows of 4.

Answer is- 16.

 

6.To meet 40% of the energy demand for a home using 1340kWh/month, what size system DC would be required in a location with 5.3 peak sun hours/day?

1340kWh x .40 = 536kWh AC

536kWh AC ÷ .77 = 696kWh DC

 

696kWh needed:

696kWh ÷ 30 days/month = 23.3kW/day

23.3kW/day ÷ 5.3 psh = 4.377kW or 4.4kW system

 

7. How many attachment points will there need to be for an installation with the following circumstances: 7 Sharp 235 Watt modules arranged in portrait mode, each with a dimension of  39.1” x 64.6”. There can be no less more than 24” at each end of the rails unsupported, and no more than 48” in between attachments.

39.1” x 7 = 273.7”

273.7” + 6” for midclamps + 4” for endclamps = 283.7”

283.7” – 48”  to allow for 24” at each end = 235.7”

 

235.7” ÷ 4spaces which is 5 attachments = 59”- NO

235.7” ÷ 5 spaces which is 6 attachments = 47.4” – YES!

With 2 rails each with 6 attachments, this is a total of 12 attachments

8. Determine the length of the lag screw required for the installation below:

Location has a wind load of 55psi, the roof trusses are southern yellow pine, and the width of the lag screw is 5/16”. There will be 14 modules arranged in 2 rows of 7, each with a dimension of 39.1” x 64.6”.  There will be 24 attachments and the roof is 1.5” thick.

39.1” x 64.6” = 2525.86”

2525.86” ÷ 144 = 17.54ft2

 

[17.54ft2 x 14 modules x 33psi wind load]÷ 24 attachments = 337.65lb each

337.65 lb. ÷ 332lb = 1.02

 

1.02 + 1.5” = 2.5 or 2.5” lag screw

 

9. What size hole should be drilled for a 7/16” inch lag screw in southern yellow pine?

 

Method 1:

7 x .75 = 5.25

5.25/16 or 5/16”

 

method 2:

7 ÷ 16 = .4375

.4375 x .75 = .33 or 33/100, or 1/3” , closest drill bit to 1/3 is 5/16”

 

[Questions below are from the NABCEP Study Guide 2009 Version 4.2] available online: http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&ved=0CDYQFjAC&url=http%3A%2F%2Fwww.cabrillo.edu%2F~swadsworth%2Ffiles%2Fnabcepstudyguidev4-2april2009.pdf&ei=9RpBUPyiOKSi2wW_noHQDA&usg=AFQjCNFQSqjq74rlatZRuGJi2J9LDL6pig

 

24. If an open circuit voltage of a crystalline silicone PV array is 315V at 25ºC, then according to the NEC, if the array is operating at -20ºC, the maximum system voltage must be corrected to:

a. 267V

b. 315V

c. 372V

d. 378V

 

315V x 1.18 [from NEC table 690.7] = 371.7

 

26. If the maximum power voltage of a crystalline silicone module is 17.1V at standard test conditions, then at 60ºC [module temperature] and at 1000w/m2 incident on the module, the maximum power voltage will be closest to:

a. 20.1V

b. 17.1V

c. 14.1V

d. 12.0V

 

60º -25º [STC] = 35º difference

35º x .5% change per degree standard = .175 or 17.5% decrease in power

.175 x 17.1V = 2.99V

17.1V – 2.99V = 14.11V

 

38. A crystalline silicone array that has bi-polar outputs of +252 and -252 with a common grounded conductor under STC is selected for a large single family dwelling. The lowest ambient temperature in which the system can be installed at a single family dwelling is:

a. -10ºC

b. 5ºC

c. -20ºC

d. 25ºC

 

252 + 252 = 504V

504V x [CF from NEC table 690.7] = 600V or less

 

504V x 1.18 [CF for -20ºC] = 595V

This is the lowest, and it is OK.

 

37. A crystalline silicone array that has bipolar outputs of -252 and +252 with a common  grounded conductor under STC is selected for a large single family residence. The lowest expected ambient temperature at the installation site is -25ºC. For this system, the max system voltage is:

a. 252V

b. 315V

c. 504V

d. 605V

 

252V + 252V = 504V

-25 CF from NEC table 690.7 is 1.2

504V x 1.2 = 604.8V, closest to 605V

verifying system design

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