Club 80-90 Forums

Hi,
As part of my ou I've been looking at solar power. Struggling to get my head around how to compose the formula to get power output.
I know, solar incident per sq.m.
Area covered by solar panels
And, efficiency.
So if solar incident is 1kJ over an area of 1m^2 and 3m^2 is covered with 20% efficient panels, then I worked out:
(1kJ*3m^2)*0.20=0.6kJm^2.
If, watt=J/t(s), then, 0.6kjm^2/1s = 600watts?

I hope this makes sense to someone and they're able to help me!
Thanks in advance!
Josh.

Sounds right to me, I remember the power of the sun equals roughly 1kw per square meter.

Almost 40 years ago doing this sort of stuff at engineering college,

The formula you have used is correct:
Panel Area (metres squared) X Solar Incidence(joules/metre squared) X 20/100 = Output in Joules
Therefore 3X1000X0.2=600
And with a Watt being I joule per second the answer is indeed 600 watts
But you need to watch your units, the answer to the first part is just 600 Jules, there should be no reference to metres squared

You can always run a units check on any formula to see if you have the right units as an answer.
In this case you have
(mXm) X (J / (mXm)) on the left hand side in which the (mXm) on the top and bottom of the formula cancel out, times the 20/100 number which is without units and so the answer is just Joules

Units (time), running out of the latter

Although this is all academic in the UK where one minute you get a bit of sun then three days of gloom and rain.

The maths is great....a blunt instrument that states definite's but in reality delivers only paper theories.
I find the talk interesting but at the same time slightly irrelevant. There are just so many compromises of the implementation and unknowns as to the function. The decisions on the hardware are rarely based on the maths and more often governed by the how and where.
Maths in this instance gives us nothing more than theoretical ideals, perhaps a starting point.

Martin

Thanks for the responses, I'm terrible at moving/cancelling units. I believe my next module is maths so hopefully that'll improve!
Yes, it is very much an academic exercise. I do appreciate there is a wide range of factors that can further influence the actual solar panel output. The example in the assignment is based on a solar car in the Australian desert (with exact figures altered).
I'm not relating it back to my van; that's just going to have a 150W flexi panel and a `poor man's sterling`.

Thanks again,
Josh

Okay, so the next bit I am struggling with...
If a solar powered fan with 2.4v battery draws 30mA how much power is required to run it for 10hrs?
I have... P(req.) = P * t and P = I *V so P(req.) = (i*v)*t so P(req.)=(2.4v*0.03A)*(10*3600) = 0.072 * 36000 = 2592 W s or Joules.
Does this sound correct?