any pv experts willing to help me,doing an assignment at college and cant get my head round some of these questions
A 4.3kwp photovoltaic array system is to be installed on a roof with a pitch of 30o and facing due south. The Voc measured on the day of commissioning was 464 .4 volts at a temperature of 45oc and a irradiation of 800watts.
(i)Calculate the number of modules required and method of connection.
(ii) Typical value for Isc under measured conditions.
(iii) KW of inverter required if under sized by 0.81%
(iv) DC module string cable rating, Array isolator rating
(v) String/array isolator voltage rating.
(vi) Size of area in M2 covered by PV array.
(vii) Does the system pass or fail commissioning from the above information, giving reasons for your answer.
Pmax 215 w
Vpm 42 v
Ipm 5.13 a
Voc 51.6 v
Isc 5.61 a
i came up with these answers
(i) 20 panels and not sure how to connect
i think i know the basics that in series voltage adds ,amps remain constant,so if i have 2 strings of 10 panels in prallel i get the same voltage ? but more amps as opposed to 1 string of 20 panels
(ii) to (v) i can get
(vi) dont know
not sure what the requirements are with regards to passing the commissioning and cant calculate how he got 464.4 Voc on the day (not sure if that really matters anyway)
any help would be greatly appreciated
Well I will start you off -
You mean Insolation not irradiation
This is measured in watt per m2 not in watts.
45C is somewhere quite hot?? Is that ambient or panel temperature?
You need to know the temperature coefficient of the panels to work out the VOC at 45c which would normally be set at 25c and 1000w/m2.
The area of the array depends on the area of the panels which does not relate to the output necessarily.
Do more research and you will understand why the question is inaccurate.
thanks freddy i get the solar radiation bit 800w/m2 and i didnt put the co-efficient in cause i understand that bit,
yes 45 degrees is the panel temp
somehow i dont think the question is wrong, possibly badly worded
as i said its q (i) and (vi) i cant grasp but thanks again
A single panel has an open circuit (oc) voltage of 51.6 V, and the installed system reads 464.4 V, that is, exactly 9 times a single panel. So we are sure of one first thing: the PV arrays consists of at least one string of 9 panels connected in series (which add their voltage, and share the same current).
How much power a string of 9 modules would yield at STC (Standard Testing Conditions)? Nine times a single panel, so 9 x 215 W = 1935 W, but this would be at STC (irradiation 1000 W/sq.m., instead of the measured 800 W/sq.m.). Lacking any graphics or tables for the modules to check, one can assume the same panel string is going to generate about 80% of the energy under the measured conditions, so to get the desing power of 4.3 KWp we have to take that into account. So the nine panel string would produce, under 800 W/sq.m., only 1548 W, that compared to the desing power, is just about a third.
So, I would say that the system must consist of three parallel strings of nine panels in series in each strings. Under STC, each string will output at most Isc amps, and will generate at most 9 x Voc. The power generated would be, without taking into account losses, 27 x 215 x 0.80 = 4.6 Kw. Peak power at STC would be 5.8 Kwp
All strings add their current to the input of the charge controller, so to get the Isc you would have to check the graphics for Isc according to irradiance, and multiply the Isc of a single panel by three (current in a single string is the lowest of the currents of the panels in the string).
(iii) So I understand there is no storage in batteries and energy goes directly to the inverter. I don't really understand the "if undersized", because if the inverted gets on its input more power than designed to handle it can stop working or just burn. But power entering the inverter is the one generated in the panel array minus losses in the cabling, that I'm not sure they have been specified in the problem.
This is country dependant, every electrical legislation says a different thing. Just check the voltage and current cables are going to withstand at maximum, apply all the safety factors legislation asks for (i.e. to take into account unusual sunny and bright, but cold summer noons), take into account the max amount of losses in the cabling, and check the appropiate tables for cable diameters and shielding.
I think this is a electrical circuit breaker for DC currents (so hardware is very different to, a much more expensive than that for AC) used when maintenance to the installation is needed, and you want to decouple the panel array from the rest of the installation. You should check the portfolios of this kind of hardware manufactures after checking legislation to see if there are any safety factors to apply to the array DC voltage.
(vi) Just simple mathematics. An array of 3 by 9 panels, you may have to decide on the panel orientation (portrait or landscape) depending on the roof space available.
No idea what to say :-)
But, please remember I'm just a PV afficionado, not a certified professional installer, expert or anything like that, so take my advice with great care and don't trust my calculations very much.
try this link to estimate solar PV install
place location with flag on the map
and wattage and orientation of panels to the side and press calculate
If the electricity wholesale rate is 5c per KWhr how much worth of electricity would this unit produce during an average winters day in Donegal ?
A 2.66kw system has done 80 kWh in west Cork since 10th January. On a sunny day it will output say 2kw peak for maximum 4 hours. On an average dull day it will drop back to 2-300w. It is not really viable to take one month as any sort of guide. Average out over 25 years and you get a fair idea of total average production.
Using SAP2009 the Irradiation in Donegal is about the same as Yorkshire so I would estimate total kWh for January to be 108kWh for a 4.3kw system.
So comparing the two the numbers do stack up pretty much spot on.