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Messages - Arvore

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General PV Discussion / Re: Example Question Corrections
« on: March 23, 2012, 07:42:35 PM »
Here's how I did it:

(6900kwh x .90/365) / 4.8 hours (latitude -15 tilt) x 1.25 = 4.43kw

The latitude -15 comes from the fact that the roof is at 26 degrees which is 15 degrees less than the latitude of 41.53 degrees. The 1.25 factor is the reciprocal of .8. Hope this helps.

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Section 9: Study Guide Review Questions / Re: Question #5 Error?
« on: March 23, 2012, 07:37:23 PM »
Disregard my previous comment. I must have read their answer incorrectly, because they did use the -15 tilt angle to arrive at the answer, oops.

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Page 64, second paragraph, header reads, "PV Power Source Maximum Circuit Current". Should be, "PV Power Output Maximum Circuit Current".

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Section 4: Installing Electrical Components / Include 690.35 Exception
« on: March 22, 2012, 04:03:24 PM »
On page 92, first paragraph, the requirement that "one conductor of a 2-wire system...must be grounded if the maximum PV system voltage is over 50V (NEC 690.41)." It might be good to include the exception (690.35) here, as well. In all fairness, this exception is mentioned in the following page; however, it might be beneficial to mention the exception in context with 690.41. I think it would make it less confusing.

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Section 8: Case Study Examples / Re: Voltage Drop Calculation
« on: March 22, 2012, 03:53:23 PM »
Derrick,

You're not mistaken. They based the nominal voltage on the inverter output. It was done correctly on the second case study, though.

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Section 9: Study Guide Review Questions / Question #5 Error?
« on: March 22, 2012, 03:43:30 PM »
From my perspective, this problem should have used "Latitude -15" in its calculation. If we go with the understanding that an "at latitude" tilt results in a 41.53 degree array tilt, then in this example, with the roof being at 26 degrees (which just happens to be 15 degrees less than latitude) we should use a solar irradiation of 4.8 hours based on the "latitude -15". Figure 14 on page 17 bears this out. Any dissenting view points?

7
Where did the 9A of continuous current value come from (paragraph 3). If we're still using the original source circuit example, the continuous current should be 10.5A (8.41 x 1.25).

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Section 2: Verify System Design / Re: Calculating Conductor Ampacity
« on: March 17, 2012, 02:11:55 PM »
Tony,

This does indeed look confusing at first glance. However, my thinking is that it failed because they opted for a 15A fuse instead of 14A (OCPD calculation resulted in 13.1A which should lead to a 14A fuse). According to NEC 690.9(C), second paragraph, OCPD <15A go in 1A increments. So, technically a 14A fuse should have been used and would have satisfied NEC 690.9(C). But because they went with the 15A fuse it seems to violate NEC 240.4(B), which states,"The next higher standard overcurrent device rating (above the ampacity of the conductors being protected) shall be permitted to be used..."

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Section 2: Verify System Design / Bypass Diodes Figure 67 (page 56)
« on: March 17, 2012, 01:57:15 PM »
The bypass diodes depicted are configured incorrectly, i.e. they're forward biased. During normal (unshaded) operation, bypass diodes are reversed biased (not conducting). During a shading event, the bypass diode corresponding to the shaded cell(s) becomes forward biased and conducts current from the unshaded cells.

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