# Calculation Confirmation of the Solar Panel Power in the DC Lighting System

7 Posts
Mar 25, 2011 09:41 am
Calculation Confirmation of the Solar Panel Power in the DC Lighting System

Dears,

Here is:
Off Grid
Sunshine Duration: 4hr per day
System: 24V DC, 48W
Operating hr = 10 hr

How much solar panel power needed?

Power needed in the system should be: = 48 X 10 = 480Wh
80% efficiency assume: =480/0.8 = 600
Power needed to be generated in the solar panel in one hr
= 600Wh/4
= 150Wh

Imply that I have to use 150W solar panel, right?

However I get another calcuation from someone, and I am not sure it is wrong or not:

Current is calculated first: 48W/24V = 2A;
Then go to the current consumption:  = 2A X 10 = 20Ah;
Then divide by sunshine hr = 20Ah/4 = 5A;
Power needed to be generated in the 34V panel  = 5 X 34
= 170W
80% efficiency assume = 170/0.8 = 212W

So which one is right? or both are wrong?

Thanks and Best Regards,
Eric Lee

99 Posts
Mar 27, 2011 12:35 am
Re: Calculation Confirmation of the Solar Panel Power in the DC Lighting System

The "other" calculation is certainly wrong.  You can't multiply 5A @ 24V by 34V to arrive at a sane wattage.  The way to do that part would be 5A * 24V = x * 34V, and so x = 5A * 24V / 34V = 3.529A.  3.529A * 34V = 120W, not 170W. And 120W / 0.8 = 150W, which is what you got.

But anyway, you had the right idea.  Don't convert to amps first, as that will throw you off completely.  You got it right the way you did it and it was obviously simpler and easier to understand.

48W x 10h = 480 Wh
480Wh / 0.8 = 600Wh
600Wh / 4h = 150W

That's all correct.  But that assumes that the combined efficiencies of both the load and the supply is 80%.  In my experience, that's not enough.  I would assume no more than 80% efficiency from the solar panels.  And also assume at least 10% inefficiency (90% efficiency) in the rated consumption of the load due to losses in the wiring, conversions, battery charging, etc.  You can of course directly measure the load and see if that's the case or not.  If 48W is what it actually consumes, even after all of the losses, then a 150W panel will be sufficient.  If it turns out that the load consumes 53W on average, then you should account for that.  To be safe, I would.

150W / 0.9 = 167W, which probably means a 170W panel

This will protect you in case you don't get exactly 4 hours of sunlight as well.  It's always better to oversize slightly.

BTW, where are you getting panels rated at 34V?  In my experience, they are rated at 12, 24, or maybe 48V.

7 Posts
Mar 27, 2011 10:06 am
Re: Calculation Confirmation of the Solar Panel Power in the DC Lighting System

Dear Mr. Thomas Anderson,

Best Regards,
Eric Lee

12 Posts
Mar 31, 2011 06:47 am
Re: Calculation Confirmation of the Solar Panel Power in the DC Lighting System

Example..

Sharp NE-170UC1 170 Watt 24 Volt Solar Panel

Type of Cell: Polycrystalline silicon
Cell Configuration: 72 in series
Maximum Power(Pmax): 170 W
Open Circuit Voltage: (Voc) 43.2 V
Maximum Power Voltage: (Vpm) *34.8 V (Round Down)
Short Circuit Current: (Isc) 5.47 A
Maximum Power Current: (Ipm) 4.90 A
Module Efficiency: 13.10%

Divide by 24 = 20 amps (needed in aHr.)
Divide by 4 = 5 amps (4 being the daily charge time)
Multiply 5 by *34 = 170 watts in panel

*34 being the Maximum Power Voltage or "Average Working Voltage" of a normal 24 volt panel. The sum of *17 works just as well for 12 volt systems.

99 Posts
Mar 31, 2011 03:36 pm
Re: Calculation Confirmation of the Solar Panel Power in the DC Lighting System

Ah, yes, that makes sense.  34V would be correct at maximum insolation.  I wonder how often you could expect it though.  My panels routinely put out only 80% or so of their rated power in full sun (1.2kW out of 1.5kW rated).  When they say "maximum" they really mean "maximum" as in an extreme outlier.

12 Posts
Mar 31, 2011 08:19 pm
Re: Calculation Confirmation of the Solar Panel Power in the DC Lighting System

We both know that maximum insolation will probably never happen but by using the Maximum Power Voltage (rounded down) as the mulitplier will include the fudge factor automatically with a slight error to the plus.

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