Can anyone post the answers to #SolarMOOC: Problem set #1 dated 3/3/12? Thanks.

1 reply

Hey John,

I copied this from the SOLARMOOC tab, near the bottom of March. It's a little hard to navigate because it's hard to see what was posted per day from the column on the far right of the page. On the right side column of the #SOLARMOOC tab- you can click on "March" - & you can see all the posts, etc from the month of March.

#SolarMOOC: PROBLEM SET #1 Solutions

Sunday 3/4/12

PROBLEM #1

A home is located in a town in the valley of a mountainous area at an elevation of 5,280 feet above sea level. To the south is a 20-mile wide cliff mountain ridge that runs east-to-west with an elevation of 11,500 above sea level. The latitude of the town is 40° N. At 4:00 p.m. on December 21st, the solar altitude angle is at 6°, and shading is just beginning to occur. How far away is the home from the base of the mountain? Choose the closest answer.

6,220 ft.

23,213 ft.

59,179 ft.

d. 43,978 ft.

First:

11,500-5,280=6220

tan 6°=6220/adjacent

adjacent=6220/0.1051042

adjacent=59,179 (shadow length)

Then:

Determine azimuth of the sun at 4:00PM Dec. 21st at 40°N Latitude

53° azimuth

cos 53°=adjacent/59,179

0.60181=adjacent/59,179

adjacent=59,179x0.60181

adjacent=35,614

Closest answer: 43,978 (Answer D)

Forgive the fact that the correct answer is so far away from the actual answer. Error in creating the answer choices caused this. However, since the problem asks for “closest answer”, it is still valid and accurate.

See my video for full explanation of this problem.

As promised in the video, here is another stab at the question under different circumstances, using different variables and considerations, but applying the same principles. Hint: this one requires an additional step.

Alternate Problem Set #1 Exercise #1

You are designing a ground mount PV array at 30°N Latitude with multiple rows that faces true south. The tilt of the modules are 20°. The width of the modules are 39 inches and they will be installed in landscape layout. What is the closest distance the rows of the modules can be and not cause any shade during the hours of 8AM and 4PM solar time?

PROBLEM #2:

A bipolar PV array has outputs of +244 V and -244 V under standard test conditions. The modules are crystalline silicon, and the manufactures coefficients have not been provided. Assuming the lowest expected temperature is -19° C, what is the expected maximum system voltage of the array? Choose the closest answer.

488 V

288 V

c. 576 V

600 V

|244V|+|-244V|=488V

488V x 1.18 = 575.84

Closest answer: 576 (Answer C)

In a bipolar array the voltage is considered to be the total voltage differential between the positive side and the negative side. Therefore, we add the absolute value of the voltages of both sub-arrays (positive and negative sides) to get the total voltage to get 488V under STC. Then, given the fact that this location has a lowest expected temperature of -19°C and the Voc temperature coefficient was NOT provided and these are crystalline PV modules, then we are obliged to utilize Table 690.7 of the NEC (page 51 Figure 56 of the NABCEP resource guide). The correction factor for -19°C is 1.18 and that is what we multiply the STC voltage by to get 575.84V. See also page 55 & 56 and Figure 66 of the NABCEP Resource Guide for more info on bipolar systems.

PROBLEM #3:

A carpenter and his wife live on a ranch in Ocala, FL. They have horses, and a shop where the carpenter makes wooden furniture. The property is located at 30° North Latitude. The couple attends church or volunteers at the Church Thrift Store in the mornings, but generally return to the ranch at lunchtime. They have recently decided to build a new stable on the North side of the pasture and want to install a 8 kW PV system on the barn.

Which orientation and roof pitch for the long-side roof would produce the most annual utility savings?

magnetic south, 40°pitch

true southwest, 20° pitch

true southeast, 30° pitch

d. true south, 30° pitch

True south, 30° pitch (Answer: D)

There is a lot of “noise” in this question. But given the available information, the rule of thumb says that true south, tilt equal to latitude will produce the most kwh over the course of the year. However, had there been information about predominate weather patterns of the area, then an optimal orientation might have been something other than true south. By the same logic, information about seasonal weather patterns (mostly cloudy in the winter, for example) could suggest something other than tilt equal to latitude would maximize annual production.

PROBLEM #4:

A PV Array is installed in a location that has an expected low temperature of -7° C. The installation includes a 5 kW Bimodal 240V Inverter. The location receives on average 5 Peak Sun Hours per day. What is the maximum current expected between the inverter and the AC Disconnect? Choose the closest answer.

a. 21 Amps

32.5 Amps

26 Amps

23 Amps

5000 Watts / 240 Volts = 20.833 Amps

Closest answer: 21 Amps (Answer A)

One of the most concise explanations comes from Bill Brooks and the SolarABC’s Expedited Permitting Process for PV Systems

“i) AC OUTPUT CURRENT

Explanation: The ac output current, or rated ac output current as stated in the NEC, at the point of connection is the maximum current of the inverter output at full power. When the rated current is not specifically called out in the specification sheets, it can be calculated by taking the maximum power of the inverter, at 40°C, and dividing that value by the nominal voltage of the inverter.

From the example in Appendix A:

Maximum Inverter Power = 7,000 watts

Nominal Voltage = 240 Volts

IRATED = 7,000 W/ 240 V = 29.2 amps”

PROBLEM #5

Two dissimilar crystalline silicon PV modules are connected in parallel. The STC rated Isc of module A is 2.3 Amps. The STC rated Isc of module B is 4.3 Amps. What would the measured current be under STC at short circuit with the modules in parallel?

4.6 A

b. 6.6 A

8.6 A

you cannot connect modules in parallel that have different currents.

4.3 A + 2.3 A = 6.6 Amps (Answer B)

Explanation: Modules in parallel add current regardless of whether they are similar or dissimilar (see figure 65 on page 56 of NABCEP Resource Guide).

Further information: Note that dissimilar modules in series are held down to the current of the lowest performing module. The voltage of dissimilar modules in series is additive without penalty. Regarding the voltage of modules in parallel, I have a point of contention. I have seen in multiple resources (the PV textbook, this NABCEP resource guide, and at least one SEI book), that the voltage between two dissimilar modules in parallel is the average between the two, in actuality, based on personal experience of having run dozens of modules labs, the voltage is actually held down to the voltage of the lower voltage modul

I copied this from the SOLARMOOC tab, near the bottom of March. It's a little hard to navigate because it's hard to see what was posted per day from the column on the far right of the page. On the right side column of the #SOLARMOOC tab- you can click on "March" - & you can see all the posts, etc from the month of March.

#SolarMOOC: PROBLEM SET #1 Solutions

Sunday 3/4/12

PROBLEM #1

A home is located in a town in the valley of a mountainous area at an elevation of 5,280 feet above sea level. To the south is a 20-mile wide cliff mountain ridge that runs east-to-west with an elevation of 11,500 above sea level. The latitude of the town is 40° N. At 4:00 p.m. on December 21st, the solar altitude angle is at 6°, and shading is just beginning to occur. How far away is the home from the base of the mountain? Choose the closest answer.

6,220 ft.

23,213 ft.

59,179 ft.

d. 43,978 ft.

First:

11,500-5,280=6220

tan 6°=6220/adjacent

adjacent=6220/0.1051042

adjacent=59,179 (shadow length)

Then:

Determine azimuth of the sun at 4:00PM Dec. 21st at 40°N Latitude

53° azimuth

cos 53°=adjacent/59,179

0.60181=adjacent/59,179

adjacent=59,179x0.60181

adjacent=35,614

Closest answer: 43,978 (Answer D)

Forgive the fact that the correct answer is so far away from the actual answer. Error in creating the answer choices caused this. However, since the problem asks for “closest answer”, it is still valid and accurate.

See my video for full explanation of this problem.

As promised in the video, here is another stab at the question under different circumstances, using different variables and considerations, but applying the same principles. Hint: this one requires an additional step.

Alternate Problem Set #1 Exercise #1

You are designing a ground mount PV array at 30°N Latitude with multiple rows that faces true south. The tilt of the modules are 20°. The width of the modules are 39 inches and they will be installed in landscape layout. What is the closest distance the rows of the modules can be and not cause any shade during the hours of 8AM and 4PM solar time?

PROBLEM #2:

A bipolar PV array has outputs of +244 V and -244 V under standard test conditions. The modules are crystalline silicon, and the manufactures coefficients have not been provided. Assuming the lowest expected temperature is -19° C, what is the expected maximum system voltage of the array? Choose the closest answer.

488 V

288 V

c. 576 V

600 V

|244V|+|-244V|=488V

488V x 1.18 = 575.84

Closest answer: 576 (Answer C)

In a bipolar array the voltage is considered to be the total voltage differential between the positive side and the negative side. Therefore, we add the absolute value of the voltages of both sub-arrays (positive and negative sides) to get the total voltage to get 488V under STC. Then, given the fact that this location has a lowest expected temperature of -19°C and the Voc temperature coefficient was NOT provided and these are crystalline PV modules, then we are obliged to utilize Table 690.7 of the NEC (page 51 Figure 56 of the NABCEP resource guide). The correction factor for -19°C is 1.18 and that is what we multiply the STC voltage by to get 575.84V. See also page 55 & 56 and Figure 66 of the NABCEP Resource Guide for more info on bipolar systems.

PROBLEM #3:

A carpenter and his wife live on a ranch in Ocala, FL. They have horses, and a shop where the carpenter makes wooden furniture. The property is located at 30° North Latitude. The couple attends church or volunteers at the Church Thrift Store in the mornings, but generally return to the ranch at lunchtime. They have recently decided to build a new stable on the North side of the pasture and want to install a 8 kW PV system on the barn.

Which orientation and roof pitch for the long-side roof would produce the most annual utility savings?

magnetic south, 40°pitch

true southwest, 20° pitch

true southeast, 30° pitch

d. true south, 30° pitch

True south, 30° pitch (Answer: D)

There is a lot of “noise” in this question. But given the available information, the rule of thumb says that true south, tilt equal to latitude will produce the most kwh over the course of the year. However, had there been information about predominate weather patterns of the area, then an optimal orientation might have been something other than true south. By the same logic, information about seasonal weather patterns (mostly cloudy in the winter, for example) could suggest something other than tilt equal to latitude would maximize annual production.

PROBLEM #4:

A PV Array is installed in a location that has an expected low temperature of -7° C. The installation includes a 5 kW Bimodal 240V Inverter. The location receives on average 5 Peak Sun Hours per day. What is the maximum current expected between the inverter and the AC Disconnect? Choose the closest answer.

a. 21 Amps

32.5 Amps

26 Amps

23 Amps

5000 Watts / 240 Volts = 20.833 Amps

Closest answer: 21 Amps (Answer A)

One of the most concise explanations comes from Bill Brooks and the SolarABC’s Expedited Permitting Process for PV Systems

“i) AC OUTPUT CURRENT

Explanation: The ac output current, or rated ac output current as stated in the NEC, at the point of connection is the maximum current of the inverter output at full power. When the rated current is not specifically called out in the specification sheets, it can be calculated by taking the maximum power of the inverter, at 40°C, and dividing that value by the nominal voltage of the inverter.

From the example in Appendix A:

Maximum Inverter Power = 7,000 watts

Nominal Voltage = 240 Volts

IRATED = 7,000 W/ 240 V = 29.2 amps”

PROBLEM #5

Two dissimilar crystalline silicon PV modules are connected in parallel. The STC rated Isc of module A is 2.3 Amps. The STC rated Isc of module B is 4.3 Amps. What would the measured current be under STC at short circuit with the modules in parallel?

4.6 A

b. 6.6 A

8.6 A

you cannot connect modules in parallel that have different currents.

4.3 A + 2.3 A = 6.6 Amps (Answer B)

Explanation: Modules in parallel add current regardless of whether they are similar or dissimilar (see figure 65 on page 56 of NABCEP Resource Guide).

Further information: Note that dissimilar modules in series are held down to the current of the lowest performing module. The voltage of dissimilar modules in series is additive without penalty. Regarding the voltage of modules in parallel, I have a point of contention. I have seen in multiple resources (the PV textbook, this NABCEP resource guide, and at least one SEI book), that the voltage between two dissimilar modules in parallel is the average between the two, in actuality, based on personal experience of having run dozens of modules labs, the voltage is actually held down to the voltage of the lower voltage modul