Solar Panels and Solar Radiation Flux Density Help - Very Confused

In summary: the calculated value of the solar radiation flux density in part c b is an underestimate", the student really isn't expected to do much (or anything, really) in the way of independent research in order to answer the problem.
  • #1
AN630078
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Homework Statement
Hello, I have a question which I am really struggling with attached below. I have tried to answer all sections, and areas in which there is an absence of solutions I have tried to suggest possible methods or approaches I could take. I am very uncertain though and would thus be very grateful of any help. I know there are quite a few questions here but they are related to one another, therefore, where I am uncertain about aspects this may be because I have made a mistake earlier, so if anyone could offer possible suggestions to help I would be very grateful. I am most stuck on question 1 b and c, and the same for 2 b and c which is essentially repeating the method but for the winter reading. This is supposedly an A-Level question I am completing for revision purposes yet I have never come across solar radiation flux density, which is causing me great confusion.

A company invests in 14 solar panels to generate electricity in the north-east of England. Each panel has an area of 1.38 m^2.
The average energy produced per day during the summer is 16.1 kWh and the average energy produced per day during the winter is 2.1 kWh. The panels are all oriented South and the roof is inclined at 35° (which gives an angle of 55° between the panel and the radiation).

Question 1:

a) Calculate the power generated in the summer
b) The solar panels are 100 % efficient, find the solar radiation flux density? Consider the angle of inclination in your calculation.
c) It is found that the solar panels are only 20.8 % efficient, meaning the calculated value of the solar radiation flux density in part c is an underestimate. Find the actual solar radiation flux density in north-eastern England during the summer?
Question 2: Repeat for the winter reading.
Question 3: It is often very cold and cloudy in the north-east of England, considering this is the textbook’s value of 300 Wm^-2 a reasonable example?
Relevant Equations
P=E/t
Intensity (Wm^-2)= Power (W)/ Area (m^2)
Question 1;

a) P=E/t
E=5.796*10^7 J energy produced per day during the summer
However, I am not certain how to calculate the time period, since although this concerns the energy produced per day, the sun does not shine for the entire duration of this 24 hour period. Also, I am unsure of the average length of daylight hours in the north-east of England in the summer, and I do not want to insert such a value as it has not been specified by the question. Consequently, I have proceeded taking the time as one day but I do not think this calculation will produce the correct result for the power generated in the day during the summer.

t= 24 hours * 60 minutes * 60 seconds = 86400 s

P=5.796*10^7/86400
P=4025/6=640.8333… ~ 641 W to three significant figures

b) This is where I am having significant difficulty as I am utterly unfamiliar with the concept of solar radiation flux density. I have researched the term further but have had little fortune in finding anything helpful. I do not have anyone else to ask therefore I would be very grateful for any help or governance to assist my understanding.
The concepts I have come across neglect the angle of inclination also which is why I do not think I have arrived at the correct conclusion.
I did however come across the equation for the Solar Flux Density (Sd), which is the amount of solar energy per unit area on a sphere centred at the Sun with a distance d:
Sd =L/(4πd^2) W/m^2

Or perhaps the formula for the Solar Constant (S), which is defined as the solar energy density at the mean distance of Earth from the sun (1.5 x 1011 m):

S = L / (4 π d^2)
S=(3.9*10^26 W)/[4*3.14*(1.5 *10^11 m)^2] = 1370 W/m^2

I do not know whether either of these are applicable here?

c) Again, I am not certain how to calculate the solar radiation flux density, regardless of the efficiency of the solar panels, which is why I have not answered this section. 


Question 2;

a) P=E/t
E=7.56*10^6 J per day
t=? Again, I am uncertain of the time period, as the sun does not shine for the entire duration of a 24 hour period, and the average length of daylight hours in the north-east of England in the winter is shorter than the summer. However, taking the energy reading as per day I shall also take the time period as one day;
t= 86400 s
P=7.56*10^6/86400
P=175/2=87.5 W

b) and c) I do not know how to find the solar radiation flux density of the solar panels in the winter when the solar panels are 100% efficient or 20.8% efficient.

Question 3;
I am very confused here, firstly I have no idea which textbook the question is referring to. Secondly, I assume the value of 300 Wm^-2 is a measure of intensity calculated by; Intensity (Wm^-2)= Power (W)/ Area (m^2)

Consequently, to evaluate the reading of 300 Wm^-2 would I input the values for the power in the summer and for the winter, to see whether this is a reasonable example? I do not think my values found for the power are correct but I will proceed nonetheless to show how I may continue.

In the summer;
Intensity= 641/ 1.38
Intensity = 464.4927.. ~ 464 Wm^-2 to three significant figures

In the winter;
Intensity= 87.5/ 1.38
Intensity=63.40579… ~ 63.4 Wm^-2 to three significant figures

I do not think my calculated values for power in the summer or winter are correct, however, if they were evaluating the calculations for the intensity I would assume the textbook’s reading of 300 Wm^-2 is not a reasonable example, as this figure is too low for the summer and too high for the winter.

Thank you to anyone who replies 👍
 
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  • #2
Congratulations, you seemed to have overthunk the entire problem... well, mostly.

I find it reasonable to assume that, since the author has felt the need to include reasonably obvious statements, like "consider the angle of inclination in your calculation" and "...meaning the calculated value of the solar radiation flux density in part c b is an underestimate", the student really isn't expected to do much (or anything, really) in the way of independent research in order to answer the problem.

(except maybe part c, where it looks like they want you to either do a calculation or look something up, elsewhere)

As far as "solar radiation flux density" is concerned, all the words are pretty self-explanatory : what do you think it means within the context of the problem ? What units would be appropriate to use ?
 
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  • #3
Hello, upon further reearch I have found that the terms intensity and radiation flux can be used interchangeably,
hmmm27 said:
Congratulations, you seemed to have overthunk the entire problem.

I find it reasonable to assume that, since the author has felt the need to include reasonably obvious statements, like "consider the angle of inclination in your calculation" and "...meaning the calculated value of the solar radiation flux density in part c b is an underestimate", you really aren't being expected to do much(or anything, really) in the way of independent research in order to answer the problem.

As far as "solar radiation flux density" is concerned, all the words are pretty self-explanatory : what do you think it means within the context of the problem ? What units would be appropriate to use ?
@hmmm27 Thank you for your reply. Would "solar radiation flux density" be another term for intensity, which has units of Wm^-2?

So in calculating the solar radiation flux density could I use the equation for intensity?
Intensity (Wm^-2)= Power (W)/ Area (m^2)

But I do not know how to include the angle of inclination here?
 
  • #4
AN630078 said:
Hello, upon further reearch I have found that the terms intensity and radiation flux can be used interchangeably,

@hmmm27 Thank you for your reply. Would "solar radiation flux density" be another term for intensity, which has units of Wm^-2?
It's an actual term, which I found out by typing the term into an Internet search engine.

But I do not know how to include the angle of inclination here?
How do you think the relationship between a) the incoming sunlight, and b) the angle of the panel with respect to that, affects the amount of sunlight received by the panel.
 
  • #5
hmmm27 said:
It's an actual term, which I found out by typing the term into an Internet search engine.How do you think the relationship between a) the incoming sunlight, and b) the angle of the panel with respect to that, affects the amount of sunlight received by the panel.
Thank you for your reply. Really? I have not found the term specifically related to solar radiation flux density and I have searched online and in my textbooks, which show nothing to do with it.
I think the amount of incoming sunlgiht, and thus the count of photons, is going to increase more rapidly when the solar panel is facing the sun (at 90 degrees) and will be lowest when it is edge on (at 0 degrees) to the sun. However, I do not know how to relate this angle within the equation I=P/A?
 
  • #6
google.
use ""s around the term ...
"solar radiation flux density"

re: the latter, have you done any trigonometry ? sines, cosines, etc.

If there's a flat piece of paper between you and something you're trying to look at, how does the angle of the paper determine how much of your vision is being blocked ? Obviously if it's flat-on to your line of sight, it's the area of the paper, and if it's edge-on, it doesn't block at all. But, what if it's at an angle ? How much does it block, then ? If you shine a flashlight at it, how does the angle determine how much light hits the paper ?
 
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  • #7
hmmm27 said:
google.
use ""s around the term ...
"solar radiation flux density"

re: the latter, have you done any trigonometry ? sines, cosines, etc.

If there's a flat piece of paper between you and something you're trying to look at, how does the angle of the paper determine how much of your vision is being blocked ? Obviously if it's flat-on to your line of sight, it's the area of the paper, and if it's edge-on, it doesn't block at all. But, what if it's at an angle ? How much does it block, then ? If you shine a flashlight at it, how does the angle determine how much light hits the paper ?
Thank you for your reply. I tried googling "solar radiation flux density" in quotation marks as you suggested but I have either just come across academic research or companies selling solar panels.

Yes, I have done trigonometry. Also, in reference to your analogy as the angle of the paer increases from being edge-on to flat-on it blocks more of my line of sight. Similarly, as the angle increases if light is shone onto the paper the amount of light hitting the paper increases and more is absorbed.
 
  • #8
AN630078 said:
Thank you for your reply. I tried googling "solar radiation flux density" in quotation marks as you suggested but I have either just come across academic research or companies selling solar panels.

<sigh> got to love "targeted search results" (...not). Sorry about that : my first few hits concern a device called a "pyranometer", which measures the aforementioned "solar radiance flux density" in ##kW/m^2## . Actual photon counts not required, nor relevant.

Yes, I have done trigonometry. Also, in reference to your analogy as the angle of the paer increases from being edge-on to flat-on it blocks more of my line of sight. Similarly, as the angle increases if light is shone onto the paper the amount of light hitting the paper increases and more is absorbed.

Okay, so given that you now know you're dealing with units in square - not cubic - metres, you should be good with the calculation, yes ?
 
  • #9
hmmm27 said:
<sigh> got to love "targeted search results" (...not). Sorry about that : my first few hits concern a device called a "pyranometer", which measures the aforementioned "solar radiance flux density" in ##kW/m^2## . Actual photon counts not required, nor relevant.
Okay, so given that you now know you're dealing with units in square - not cubic - metres, you should be good with the calculation, yes ?
Thank you for your reply. Oh ok I will instead seach for a "pyranometer"

I am still not sure of the calculation. Would I use Intensity (Wms^-2)=Power (W)/ area(m^2)
Power in the summer was calculated to be 671 W
Then take the total area = area of panel * sin θ
Total area=1.38 m^2 * sin 55
Total area = 1.1304... ~ 1.13m^2
Radiation flux(or intensity)=671/1.13
Solar radiation flux = 593.5795.. ~ 594 Wm^-2 to 3.s.f

I am still rather confused here, also would my calaulated value for the summer be correct?
 
  • #10
The problem statement has several doubtful parts.

As you note, the total output over 24 hours is not directly useful in finding the solar flux. For one thing, the angle changes, so the stated angle of incidence only applies at one instant.

Assuming that is midday, we really want the peak output. Now, we could assume the output follows a sine curve from dawn to dusk (different intervals in summer and winter, of course) and deduce the peak kW from the total kWh.

Then again, the given 55 degrees is curious. This is at a latitude of around 55N. If the panels are tilted up at 35 degrees then at midday in midsummer the sun's angle to the normal will be 55-23.5-35 degrees, almost square-on. In midwinter, 55+23.5-35 degrees, around 45 degrees.
 
  • #11
haruspex said:
The problem statement has several doubtful parts.

As you note, the total output over 24 hours is not directly useful in finding the solar flux. For one thing, the angle changes, so the stated angle of incidence only applies at one instant.

Assuming that is midday, we really want the peak output. Now, we could assume the output follows a sine curve from dawn to dusk (different intervals in summer and winter, of course) and deduce the peak kW from the total kWh.

Then again, the given 55 degrees is curious. This is at a latitude of around 55N. If the panels are tilted up at 35 degrees then at midday in midsummer the sun's angle to the normal will be 55-23.5-35 degrees, almost square-on. In midwinter, 55+23.5-35 degrees, around 45 degrees.
Thank you for your reply. Yes I agree with you and was thinking surely the angle of incidence applies at one instant.
How would I find the actaul power output then without overcomplicating matters and using the information in the question?
How could I find the peak kW from the total kWh also?

Thank you for explaining that 55 degrees is an unsual angle also, how this would be almost square-on in the summer. I think the entire question is a little peculiar.
 
  • #12
AN630078 said:
How could I find the peak kW from the total kWh also?
AS I wrote, you can model it as a sine curve from 0 to pi.
If the peak power is A then at time t it is A sin(t). The output over time T is ##\int_0^T A\sin(\omega t).dt=[-\frac A{\omega}\cos(\omega t)]_0^T=\frac{2A}{\omega}##, where ##\omega=\frac{\pi }{12h}##.
So, given this is 16.1kWh we deduce ##A=\frac{\pi 16.1}{24}## kW, or about 2kW.
 
  • #13
haruspex said:
AS I wrote, you can model it as a sine curve from 0 to pi.
If the peak power is A then at time t it is A sin(t). The output over time T is ##\int_0^T A\sin(\omega t).dt=[-\frac A{\omega}\cos(\omega t)]_0^T=\frac{2A}{\omega}##, where ##\omega=\frac{\pi }{12h}##.
So, given this is 16.1kWh we deduce ##A=\frac{\pi 16.1}{24}## kW, or about 2kW.
Thank you for your reply, I truly appreciate it, although I do not understand it fully and think it offers greater depth than required by the question. Would the peak kW then be used to find the power output using
I have been thinking how to find the solar radiation flux density. If the power generated in the summer is 641W then for question 1 b could I proceed as follows;
solar flux = 1/sin(θ) *(power output of the panels / area of panels)
solar flux = 1/sin(55) *(671 W / 1.38 m^2)
solar flux = 593.57952 ~ 594 Wm^-2

1.c When it is found that the solar panels are only 20.8 % efficient, would I calculate the actual solar radiation flux density using the formula:
efficiency= useful power output/ total power input * 100%
useful power output=efficiency/100%*total power input
useful power output=0.208 * 671 W
useful power output=139.568 ~ 140 W to 3.s.f

Question 2; Repeating the previous method for the winter:
P=E/t
E=7.56*10^6 J energy produced per day during the summer

To find the solar radiation flux density in the winter;
solar flux = 1/sin(θ) *(power output of the panels / area of panels)
solar flux = 1/sin(55) *(87.5 W / 1.38 m^2)
solar flux = 77.40418... ~ 77.4 Wm^-2If it is found the solar panels are only 20.8% efficient;
useful power output=efficiency/100%*total power input
useful power output=0.208 * 87.5 W
useful power output= 18.2 W to 3.s.f

Question 3;is 300 Wm^-2 a reasonable example?
In hindsight I believe that the values I have calculated for the power generated are correct, though I am still a little uncertain of them. Therefore, I do not think the example of 300 Wm^-2 is reasonable since this is 294 Wm^-2 too low for the summer and 222.6 Wm^-2 too high for the winter. Even though it is often overcast and cloudy in northern England, the value of 300 Wm^-2 is too general and midrange for the contrast between the summer and winter months.
 
  • #14
AN630078 said:
solar flux = 1/sin(55) *(671 W / 1.38 m^2)
It doesn’t make sense to use an average power over 24 hours (671 or 641?) and a fixed angle of 55 degrees. That's why you should be using the peak output.
No need to do the calculus, just argue that most of the energy will be generated over 6 to 8 hours, so multiply the average power by 3 or 4 to get peak power.
But you are forgetting this is the total over 14 panels, each with the given area.
AN630078 said:
efficiency= useful power output/ total power input * 100%
useful power output=efficiency/100%*total power input
Think that through again.
And bear in mind that it ought to be giving you a bigger answer for the solar flux.
AN630078 said:
Question 3;is 300 Wm^-2 a reasonable example?
Reasonable for what? Solar flux? Irradiation of the panels? Output from the panels? Peak or average over 24h?
Btw, doesn't matter if it is cold in the NE, that actually improves efficiency. Just need the sunshine.
 
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  • #15
@haruspex Thank you for your reply. I am still really confused here.

a) So in the summer, to find the power generated multiply the average power per day by 4 (since the peak power is generated in 8 hours for example). Also, you mention that I forgot that this is the total over 14 panels, each with the given area.
5.796*10^7 J * 4=2.32*10^8 J to 3.s.f
Area = 1.38*14=19.32 m^2
P=2.32*10^8/19.32=11904761.9 ~ 1.19*10^7 W
And for the winter the peak power would be;
7.56*10^6 J*4=3.02*10^7 J to 3.s.f
Then P=3.02*10^7/19.32=22878787.88 ~ 2.30* 10^7 W

b) Could I still use (1/sin θ) *(power output of the panels / area of panels) to find the solar radiation flux?c) You said to rethink the formula for efficiency, so would my answer have been wrong ? How can I correct this?

Question 3; I have no idea what "reasonable" is here referring to. I think by the units of 300 Wm^-2 I would assume the intensity?
 
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  • #16
AN630078 said:
a) So in the summer, to find the power generated multiply the average power per day by 4 (since the peak power is generated in 8 hours for example)
I see you have tagged this "a)". Does that mean you are using this to answer part a)?
I didn't pay much attention to that part before... not sure what it is asking, but my best guess is that they just want the average power over 24 hours, which is a simple matter of dividing the daily energy output by the length of a day.

You do need it for part b).
Suppose the peak output power is Ppeak. If most of the energy is generated over h hours each day then the total for the day will be about hPpeak, so the average power will be hPpeak/24.
In the present case, you are (unusually) given the average and want to find the peak, so we reverse this to get Ppeak=Pavg*24/h.
In Sydney, the standard is that the daily total is only equal to about 4 hours of peak output, so h=4, and you would multiply by 6 to get peak from average. But if you try to derive it from simply assuming a sine curve you get h=24/π, and a multiplier of only π. This discrepancy is partly because in the early morning and late afternoon it is not just that the angle of incidence is low; there is also the issue of all that extra atmosphere the light has to pass through. So the sine curve overestimates the power at those times. Another reason is that the peak is only achievable when the sun is square on to the panel, and with typical installations that only happens at the equinoxes.

Go with a multiplier of four.
AN630078 said:
b) Could I still use (1/sin θ) *(power output of the panels / area of panels) to find the solar radiation flux?
Yes, if that means peak output, theta is measured as angle to the panel (not angle to the normal), and we are assuming 100% efficiency here. The difficulty is knowing what the question setter intends to be used for theta. If we assume the panels are oriented to maximise output at the equinoxes, theta would be 23.5 degrees at the solstices.
AN630078 said:
You said to rethink the formula for efficiency
I think the lack of parentheses led me to misinterpret your formula.
AN630078 said:
Question 3; I have no idea what "reasonable" is here referring to. I think by the units of 300 Wm^-2 I would assume the intensity?
The units would also be appropriate for power output per unit area of panel, but given the context of the immediately preceding question I would think it means solar irradiance per unit area of Earth's surface in NE England. Question is, when?

For square-on irradiance, it is about 1000W/m2, so 300 would be with the sun at only 18 degrees above the horizon. It is 35 degrees above midday at an equinox, 12 above in midwinter. So I don't have a good guess as to what the 300 represents.
 
  • #17
haruspex said:
I see you have tagged this "a)". Does that mean you are using this to answer part a)?
I didn't pay much attention to that part before... not sure what it is asking, but my best guess is that they just want the average power over 24 hours, which is a simple matter of dividing the daily energy output by the length of a day.

You do need it for part b).
Suppose the peak output power is Ppeak. If most of the energy is generated over h hours each day then the total for the day will be about hPpeak, so the average power will be hPpeak/24.
In the present case, you are (unusually) given the average and want to find the peak, so we reverse this to get Ppeak=Pavg*24/h.
In Sydney, the standard is that the daily total is only equal to about 4 hours of peak output, so h=4, and you would multiply by 6 to get peak from average. But if you try to derive it from simply assuming a sine curve you get h=24/π, and a multiplier of only π. This discrepancy is partly because in the early morning and late afternoon it is not just that the angle of incidence is low; there is also the issue of all that extra atmosphere the light has to pass through. So the sine curve overestimates the power at those times. Another reason is that the peak is only achievable when the sun is square on to the panel, and with typical installations that only happens at the equinoxes.

Go with a multiplier of four.

Yes, if that means peak output, theta is measured as angle to the panel (not angle to the normal), and we are assuming 100% efficiency here. The difficulty is knowing what the question setter intends to be used for theta. If we assume the panels are oriented to maximise output at the equinoxes, theta would be 23.5 degrees at the solstices.

I think the lack of parentheses led me to misinterpret your formula.

The units would also be appropriate for power output per unit area of panel, but given the context of the immediately preceding question I would think it means solar irradiance per unit area of Earth's surface in NE England. Question is, when?

For square-on irradiance, it is about 1000W/m2, so 300 would be with the sun at only 18 degrees above the horizon. It is 35 degrees above midday at an equinox, 12 above in midwinter. So I don't have a good guess as to what the 300 represents.
Thank you for your reply. I know that you stated it would not make sense to use average power over 24 hours and that I should use the peak output, I am just unsure about doing so as no specific information has been given regarding the peak daylight hours? I think it is negligible on the part of the way the question is written but I do think it is implying that one should find the average over 24 hours, since this can be found with the information given. Even the questions like "calculate the power generated in the summer" are a little ambiguous, is this per day, or the average across the whole of the summer, or the total adding the power each day in summer?

a) Yes, I was trying to approach the problem in order. So is this asking for the average power generated across the whole of the summer? i.e if summer is approximately 91 days?
Ok, so if the peak power occurs over 4 hours;
t=4*60=240 s
Average Power=(240*5.796*10^7)/(24*3600)
Average Power=161,000 W

b) I think the angle I am supposed to use is 55 degrees between the solar panels and the radiation.
Solar radiation flux= (1/sin θ) *(power output of the panels / area of panels)
Solar radiation flux= (1/sin 55)*(161,000/(1.38*14))
Solar radiation flux=10173.12157 ~ 102000 Wm^-2 to 3.s.f.

c) If the panels are 20.8% efficient
useful power output=(efficiency/100%)*total power input
useful power output=0.208*161,000
actual solar radiation flux = 33488 W

Question 2; For the winter reading:
iI the peak power occurs over 4 hours;
t=4*60=240 s
Average Power=(240*7.56*10^6)/(24*3600)
Average Power=21000 W
Solar radiation flux= (1/sin θ) *(power output of the panels / area of panels)
Solar radiation flux= (1/sin 55)*(21000/(1.38*14))
Solar radiation flux=1326.92890~ 1327 Wm^-2 to 3.s.f
If the panels are 20.8% efficient:
useful power output=(efficiency/100%)*total power input
useful power output=0.208*21000 W
actual solar radiation flux = 4368 W

Would this be correct? The readings are far greater than my previous calculations though.

Question 3;

If 300 Wm^-2 is the sun at only 18 degrees above the horizon, would this mean that this occurs only at a specific time of day at this angle, therefore it would not be a reasonable example as this is limited to a specific point of solar radiation flux output, compared to the averages previously calculated. Honestly I do not know and I am grasping at straws here 😂
 
  • #18
AN630078 said:
is this asking for the average power generated across the whole of the summer? i.e if summer is approximately 91 days?
Ok, so if the peak power occurs over 4 hours;
t=4*60=240 s
Average Power=(240*5.796*10^7)/(24*3600)
Average Power=161,000 W
The question confuses me. It asks for the "power generated in the summer". If it asked for the energy generated it would be a matter of multiplying 16.1 kWh x however many days you think summer is.
But it asks for power... is that average power over a summer's day? If so, 16.1 kWh/24h.
Anyway, I cannot make sense of your calculation. You cannot multiply by 91 days and end up with a number of Watts. Joules, possibly.

AN630078 said:
b) I think the angle I am supposed to use is 55 degrees between the solar panels and the radiation.
Solar radiation flux= (1/sin θ) *(power output of the panels / area of panels)
Solar radiation flux= (1/sin 55)*(161,000/(1.38*14))
Solar radiation flux=10173.12157 ~ 102000 Wm^-2 to 3.s.f.
The peak combined power output from the panels cannot be more than 4kW at most (16.1kWh/4h). That solar flux is 100 times reality.

AN630078 said:
c) If the panels are 20.8% efficient
useful power output=(efficiency/100%)*total power input
useful power output=0.208*161,000
actual solar radiation flux = 33488 W
You are confused between input and output. Actual solar radiation would be input, so = output *100%/efficiency.
AN630078 said:
Question 3;
I cannot figure out what this 300W/m2 refers to, so I cannot help with this one.
 
  • #19
haruspex said:
.
I cannot figure out what this 300W/m2 refers to, so I cannot help with this one.
Hi. Can I chip in? The question is poorly written. I believe all references to "solar radiation flux density"
(intensity) are intended to be *average value over 24 hours*.

If I'm correct, then 300W/m2 is meant to be the generally accepted average intensity over a full day. (Obviously during the daytime, the intensity is much larger, and at night time it is effectively zero.)

However, it is not entirely clear if 300W/m2 is meant to be the summer or winter value.
 
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  • #20
Steve4Physics said:
Hi. Can I chip in? The question is poorly written. I believe all references to "solar radiation flux density"
(intensity) are intended to be *average value over 24 hours*.

If I'm correct, then 300W/m2 is meant to be the generally accepted average intensity over a full day. (Obviously during the daytime, the intensity is much larger, and at night time it is effectively zero.)

However, it is not entirely clear if 300W/m2 is meant to be the summer or winter value.
Yes, that looks reasonable. So a) is asking for the average power over 24h on summer's day.
Since it asks the same questions re summer and winter, the 300W/m2 is probably the average of the two.

Any similarly brilliant insight re the 35 degree and 55 degree info?
 
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  • #21
haruspex said:
… a) is asking for the average power over 24h on summer's day.
Since it asks the same questions re summer and winter, the 300W/m2 is probably the average of the two.
1a) and 2a) are the average output powers of the panels in summer and winter.
1b) and 2b) are a the average intensities of the sunlight in summer and winter.
If 300W/m2 is meant to be an annual average, then it should broadly match the average of the answers to 1b)and 2b). But that’s just a guess.

[Any similarly brilliant insight re the 35 degree and 55 degree info

I hope you're not being sarcastic! Yes, the wording in the question is poor. But I think
“an angle of 55° between the panel and the radiation”
means that the angle of incidence (to thepanel's normal) is 90° – 55° = 35°. So if the intensity is I, the useful component of intensity is Icos(35°). But again, that’s just a guess.
 
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  • #22
@Steve4Physics Thank you for your reply.
So to find 1a) The average power output in the panels in the summer;
P=5.796*10^7/86400
P=670.8333… ~ 671 W to three significant figures

b) Would the total area be per panel or per 14 panels? i.e. 1.38 m^-2 or (1.38*14)=19.32 m^2
Also, should the angle used here be 35 degrees? If so why would it not be 55 degrees?
Solar flux = 1/sin(θ) *(power output of the panels / area of panels)
solar flux = 1/sin(35) *(671 W / 19.32 m^2)
solar radiation flux = 60.5513... ~ 60.6 Wm^-2

Or would it be with the area per panel: 1/sin(35) *(671 W / 1.38 m^2)=847.719~848 Wm^-2

c)Have I used the efficiency formula incorrectly here? Would it be efficiency = (useful power output/total power input)*100%
Therefore, rearranging in terms of useful power output (actual solar radiation flux);
useful power output = (efficiency/100%)*total power input
useful power output = 0.208 *671
useful power output =139.568 ~ 140 W to 3.s.f.

2. a) The average power output in the panels in the winter;
E=7.56*10^6 J per day
P=E/t
P=7.56*10^6/86400 = 87.5 W

b) Again, this depends on the area and angle so I am still uncertain.
Solar flux = 1/sin(θ) *(power output of the panels / area of panels)
solar flux = 1/sin(35) *(87.5 W / 19.32 m^2)=7.8960~7.89 Wm^-2
or
solar flux =1/sin(35) *(87.5 W / 1.38m^2)=110.5446 ~ 111 Wm^-2

I think the are must be per panel because otherwise the value for the intensity appears far too low for the area of the 14 panels.

c)useful power output = 0.208 *87.5
useful power output =18.2 W to 3.s.f.

3. If 300 Wm^-2 is supposed to be the average of the summer and winter reading, would this be 111 Wm^-2+848 Wm^-2 =479.5 Wm^-2
This far surpasses the value of 300 Wm^-2, meaning the textbook's example is an underestimate of the average intensity.

Would my calculations be correct, and if not how and where can I amend them? Thank you very much for your help. 😁👍
 
Last edited:
  • #23
AN630078 said:
So to find 1a) The average power output in the panels in the summer;
P=5.796*10^7/86400
P=670.8333… ~ 671 W to three significant figures
Agreed. But if you understand what power means and how to use units you can much more simply:
P = 16.1kWh/day = 16100W h/(24h) = 16100/24 W = 671W (the answer to 1a))

b) Would the total area be per panel or per 14 panels? i.e. 1.38 m^-2 or (1.38*14)=19.32 m^2
The total area A = 1.38*14=19.32 2 m2 How else could you interpret the word 'total'?!

Also, should the angle used here be 35 degrees? If so why would it not be 55 degrees?
The original question is so badly written we can't be certain what angle to use. I believe the angle of incidence (relative to the normal) is 35º for the reasons I already explained in post #21. Note that cos(35º) = sin(55º) so you can use either. But your guess is as good as mine.

Solar flux = 1/sin(θ) *(power output of the panels / area of panels)
solar flux = 1/sin(35) *(671 W / 19.32 m^2)
solar radiation flux = 60.5513... ~ 60.6 Wm^-2
I don't like the term solar flux. Intensity (I) is the correct term. I would write:
I = Power/effective area = 670.8/(19.32cos(35º)} = 42.4W/m²
So that's the answer to 1b).

Or would it be with the area per panel: 1/sin(35) *(671 W / 1.38 m^2)=847.719~848 Wm^-2
No.

c)Have I used the efficiency formula incorrectly here? Would it be efficiency = (useful power output/total power input)*100%
Therefore, rearranging in terms of useful power output (actual solar radiation flux);
No! Useful power output is NOT actual solar radiation flux. Solar radiation flux is what is going IN, Useful power is what is coming OUT. And there is a simpler way to do the calculation if you can think mathematically.

Useful power output = 671W (as calculated in part 1a). That hasn't changed. But now we realize the efficiency was much less than 100%. This means the intensity must have been much higher than 42.4W/m². Since the efficicency was 20.8%, by proportion intensity must have actually been 42.4 * 100/20.8 = 203.8 = 204W/m² (so that's the answer to 1c)

If you check, you will see I've actually done 1a), 1b) and 1c) for you! 2a), 2b) and 2c) are exactly the same calculation but using 2.1kWh rather than 16.1kWh. If you think carefully, the answerS to 2a, 2b) and 2c) are the same as for 1a), 1b) and 1c) but multipled by 2.1/16.1.

So you should be abe to complete Q2.

For Q3, 300W/m² is well above the answer to 1c) and it will be even more the answer to 2c). This suggests 300W/m² is much too high.
 
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  • #24
Steve4Physics said:
Agreed. But if you understand what power means and how to use units you can much more simply:
P = 16.1kWh/day = 16100W h/(24h) = 16100/24 W = 671W (the answer to 1a))The total area A = 1.38*14=19.32 2 m2 How else could you interpret the word 'total'?!The original question is so badly written we can't be certain what angle to use. I believe the angle of incidence (relative to the normal) is 35º for the reasons I already explained in post #21. Note that cos(35º) = sin(55º) so you can use either. But your guess is as good as mine.


I don't like the term solar flux. Intensity (I) is the correct term. I would write:
I = Power/effective area = 670.8/(19.32cos(35º)} = 42.4W/m²
So that's the answer to 1b).No.


No! Useful power output is NOT actual solar radiation flux. Solar radiation flux is what is going IN, Useful power is what is coming OUT. And there is a simpler way to do the calculation if you can think mathematically.

Useful power output = 671W (as calculated in part 1a). That hasn't changed. But now we realize the efficiency was much less than 100%. This means the intensity must have been much higher than 42.4W/m². Since the efficiency was 20.8%, by proportion intensity must have actually been 42.4 * 100/20.8 = 203.8 = 204W/m² (so that's the answer to 1c)

If you check, you will see I've actually done 1a), 1b) and 1c) for you! 2a), 2b) and 2c) are exactly the same calculation but using 2.1kWh rather than 16.1kWh. If you think carefully, the answerS to 2a, 2b) and 2c) are the same as for 1a), 1b) and 1c) but multiple by 2.1/16.1.

So you should be abe to complete Q2.

For Q3, 300W/m² is well above the answer to 1c) and it will be even more the answer to 2c). This suggests 300W/m² is much too high.
Thank you for your reply I truly appreciate your help and explanation.
Repeating the procedure you have exhibited in question 1;

2. a) P=2.1 kWh/day = 2100 h/(24h) = 87.5W
b) No I agree, I had not actually come across the term "solar radiation flux density" prior to this question and am far more familiar with using intensity.
I = Power/effective area = 87.5/(19.32cos(35º)} = 5.5288 ~ 5.53W/m²
c) Oh ok so would 5.53W/m² be the intensity of the solar panel at 20.8% efficiency?
5.53*100/20.8=26.5811 Wm^-2 (the actual solar radiation flux)

Question 3) This is not a reasonable example as 300Wm^-2 is a tremendous overestimate of the average intensity per day in both the summer and winter. Even if this is assumed to be the average intensity over the year (summer and winter), that would be equal to (42.4+5.53)/2=13.29055 Wm^-2 which is still far lower than the example given.

Thank you again, do I need to amned anything here? 👍
 
  • #25
Steve4Physics said:
I hope you're not being sarcastic!
I worried you might think that, but I assure you not. Hence the Like.
Steve4Physics said:
But I think “an angle of 55° between the panel and the radiation”
means that the angle of incidence (to thepanel's normal) is 90° – 55° = 35°.
Yes, it feels like it is trying to say something like that, but it says:
"the roof is inclined at 35° (which gives an angle of 55° between the panel and the radiation)."
No, that gives an angle of 55° to the vertical. The angle to the radiation depends also on the angle of the rays to the vertical.
As I noted, the location is around 55°N. Angling the panel at 35° to the horizontal gives an average angle between the normal and the midday sun of 20°. At the solstices we can add and subtract 23.5°, so nearly square on in midsummer but around 45° in midwinter.
This matters because the student has to make use of the angle in the calculation.

I tried looking up UK websites advising on mounting angle to see what is done in practice. The results are worrying. More than one says to elevate the panel at 35 to 40, on the basis of 90-latitude (!). http://www.reuk.co.uk/wordpress/solar/solar-panel-mounting-angle/ says latitude+10 to 15 degrees, but offers no explanation. My guess is so as to favour wintertime output, but it will cut total output over the year.
 
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  • #26
Update.. maybe the UK advice of 35 to 40 elevation maximises the annual output. Trouble is, peak demand will be winter.
 
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  • #27
Steve4Physics said:
I don't like the term solar flux. Intensity (I) is the correct term

Intensity is another word for energy flux density. But in common (maybe sloppy) parlance, people like to drop the word "density". So you just have to keep your wits about you as to whether someone is talking about the evaluated surface integral (formally flux), or the integrand (formally flux density)!
 
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  • #28
haruspex said:
I worried you might think that, but I assure you not. Hence the Like.

Yes, it feels like it is trying to say something like that, but it says:
"the roof is inclined at 35° (which gives an angle of 55° between the panel and the radiation)."
No, that gives an angle of 55° to the vertical. The angle to the radiation depends also on the angle of the rays to the vertical.
As I noted, the location is around 55°N. Angling the panel at 35° to the horizontal gives an average angle between the normal and the midday sun of 20°. At the solstices we can add and subtract 23.5°, so nearly square on in midsummer but around 45° in midwinter.
This matters because the student has to make use of the angle in the calculation.

I tried looking up UK websites advising on mounting angle to see what is done in practice. The results are worrying. More than one says to elevate the panel at 35 to 40, on the basis of 90-latitude (!). http://www.reuk.co.uk/wordpress/solar/solar-panel-mounting-angle/ says latitude+10 to 15 degrees, but offers no explanation. My guess is so as to favour wintertime output, but it will cut total output over the year.
Thank you very much for your reply and for pursuing this further. My apologies but I am a little confused what the correct angle to use would be when finding the intensity?
 
  • #29
haruspex said:
Update.. maybe the UK advice of 35 to 40 elevation maximises the annual output. Trouble is, peak demand will be winter.
I do not know whether this will help but I found this regarding radiation flux (intensity) in a textbook online 👍
Here is a link to it also, https://www.pearsonschoolsandfecolleges.co.uk/AssetsLibrary/SECTORS/Secondary/PDFs/Science/EdexcelScience/ALevelRevisionGuides/EdexcelASPhysicsRG_9781846905957_pg70-78_web.pdf
It also mentions efficiency, as required by part c.
Screenshot 2020-09-18 at 23.19.51.png
 
  • #30
AN630078 said:
2a) P=2.1 kWh/day = 2100 h/(24h) = 87.5W
2b) I = Power/effective area = 87.5/(19.32cos(35º)} = 5.5288 ~ 5.53W/m²
Agreed.

2c) Oh ok so would 5.53W/m² be the intensity of the solar panel at 20.8% efficiency?
No. It is meaningless to refer to the "intensity of the solar panel". The solar panel does not have an intensity. The intensity is the amount of power from the sun per square metre. 5.53W/m² is the intensity *calculated assuming the solar panels are 100% efficient*. So 5.53W/m² is not an accurate value of intensity (because efficiency is not 100%).

But we know the panels are only 20.8% efficient. So intensity must be (roughly 5 times) bigger than 5.53W/m².

5.53*100/20.8=26.5811 Wm^-2 (the actual solar radiation flux)
Yes but round to 3 sig. figs.

Question 3) This is not a reasonable example as 300Wm^-2 is a tremendous overestimate of the average intensity per day in both the summer and winter. Even if this is assumed to be the average intensity over the year (summer and winter), that would be equal to (42.4+5.53)/2=13.29055 Wm^-2 which is still far lower than the example given.
No. First, 'example' is the wrong word. And we physicists don't use words like 'tremendous' in this context! Just say (for example) "300W/m² is an overestimate ...
'per day' is wrong. You = mean averaged over a day.
13.29055 is too many significant figures.

The most *accurate* values of intensity are the ones calculated allowing for efficiency. So you should be averaging the answers from 1c) and 2c), not from 1b) and 2b).
(204 + 26.6)/2 = 115Wm/s²
So the 300W/m² (assumed to be an annual average), compared to 115W/m², is roughly between two and three times too big.
 
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  • #31
Steve4Physics said:
Agreed.No. It is meaningless to refer to the "intensity of the solar panel". The solar panel does not have an intensity. The intensity is the amount of power from the sun per square metre. 5.53W/m² is the intensity *calculated assuming the solar panels are 100% efficient*. So 5.53W/m² is not an accurate value of intensity (because efficiency is not 100%).

But we know the panels are only 20.8% efficient. So intensity must be (roughly 5 times) bigger than 5.53W/m².Yes but round to 3 sig. figs.No. First, 'example' is the wrong word. And we physicists don't use words like 'tremendous' in this context! Just say (for example) "300W/m² is an overestimate ...
'per day' is wrong. You = mean averaged over a day.
13.29055 is too many significant figures.

The most *accurate* values of intensity are the ones calculated allowing for efficiency. So you should be averaging the answers from 1c) and 2c), not from 1b) and 2b).
(204 + 26.6)/2 = 115Wm/s²
So the 300W/m² (assumed to be an annual average), compared to 115W/m², is roughly between two and three times too big.

Thank you for your reply and your continued insight.

2. c) Sorry my mistake, clearly I misused intensity here.

Question 3) "example" was the wording from the original question. Oh right, I will correct my average value also, thank you for remarking on that.

Can I verify that it would be correct to use an angle of 35 degrees when finding the intensity, as I have become a little uncertain?
 
  • #32
etotheipi said:
Intensity is another word for energy flux density. But in common (maybe sloppy) parlance, people like to drop the word "density". So you just have to keep your wits about you as to whether someone is talking about the evaluated surface integral (formally flux), or the integrand (formally flux density)!
Yes indeed. In general, we have to be prepared to interpret loosely used terminology in context. It's very difficult for the younger students who have had little exposure to different terminologies.
 
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  • #33
AN630078 said:
Can I verify that it would be correct to use an angle of 35 degrees when finding the intensity, as I have become a little uncertain?
I can't verify that. The wording in the question is unclear. 35º is is simply my best guess using the available information, as I explained in message #21. Is this a written piece of work you have to hand-in? If so precede it with a statement such as:
"The direction of solar radiation relative to the panel is unclear. I have assumed an angle of incidence of 35º".
You will not be penalised if the angle is not 35º because the question is so obviously poorly-worded.
 
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  • #34
Steve4Physics said:
I can't verify that. The wording in the question is unclear. 35º is is simply my best guess using the available information, as I explained in message #21. Is this a written piece of work you have to hand-in? If so precede it with a statement such as:
"The direction of solar radiation relative to the panel is unclear. I have assumed an angle of incidence of 35º".
You will not be penalised if the angle is not 35º because the question is so obviously poorly-worded.
Thank you very much for your reply and for your advice, I admit I will probably use your statement. It is a question my physics tutor sent to me in a quantum physics revision package of questions, but I had not come across a homogenous problem to this before, at least not within the context. If you like, once I have feeback from them I can inform you of the angle demanded by the question. Thank you again 👍😁
 
  • #35
haruspex said:
I worried you might think that, but I assure you not. Hence the Like.
Thankyou!
 
<h2>1. What are solar panels and how do they work?</h2><p>Solar panels are devices made up of photovoltaic cells that convert sunlight into electricity. The cells are made of silicon and other materials that create an electric field when exposed to sunlight. When light particles (photons) from the sun hit the cells, they knock electrons loose from the atoms, generating an electric current.</p><h2>2. How much solar radiation flux density is needed for solar panels to work?</h2><p>The amount of solar radiation flux density, or the amount of sunlight, needed for solar panels to work depends on various factors such as the type of solar panel, location, and weather conditions. Generally, a minimum of 1000 watts per square meter of solar radiation is needed for efficient functioning of solar panels.</p><h2>3. What is the difference between direct and diffuse solar radiation flux density?</h2><p>Direct solar radiation flux density refers to the sunlight that reaches the Earth's surface without being scattered or reflected. This type of radiation is strongest on clear, sunny days. On the other hand, diffuse solar radiation flux density is sunlight that has been scattered or reflected by particles in the atmosphere. This type of radiation is present on cloudy or overcast days.</p><h2>4. How do I calculate the solar radiation flux density at a specific location?</h2><p>The solar radiation flux density at a specific location can be calculated using the solar radiation data available for that location. This data includes the average daily solar radiation, which can be obtained from various sources such as the National Solar Radiation Database. The calculation involves using the average daily solar radiation and the latitude and longitude of the location to determine the solar radiation flux density.</p><h2>5. Can solar panels work without direct sunlight?</h2><p>Yes, solar panels can still work without direct sunlight. While direct sunlight is ideal for maximum efficiency, solar panels can still generate electricity on cloudy or overcast days. This is because solar panels can still absorb the diffuse solar radiation present in the atmosphere. However, the electricity production may be lower compared to clear, sunny days.</p>

1. What are solar panels and how do they work?

Solar panels are devices made up of photovoltaic cells that convert sunlight into electricity. The cells are made of silicon and other materials that create an electric field when exposed to sunlight. When light particles (photons) from the sun hit the cells, they knock electrons loose from the atoms, generating an electric current.

2. How much solar radiation flux density is needed for solar panels to work?

The amount of solar radiation flux density, or the amount of sunlight, needed for solar panels to work depends on various factors such as the type of solar panel, location, and weather conditions. Generally, a minimum of 1000 watts per square meter of solar radiation is needed for efficient functioning of solar panels.

3. What is the difference between direct and diffuse solar radiation flux density?

Direct solar radiation flux density refers to the sunlight that reaches the Earth's surface without being scattered or reflected. This type of radiation is strongest on clear, sunny days. On the other hand, diffuse solar radiation flux density is sunlight that has been scattered or reflected by particles in the atmosphere. This type of radiation is present on cloudy or overcast days.

4. How do I calculate the solar radiation flux density at a specific location?

The solar radiation flux density at a specific location can be calculated using the solar radiation data available for that location. This data includes the average daily solar radiation, which can be obtained from various sources such as the National Solar Radiation Database. The calculation involves using the average daily solar radiation and the latitude and longitude of the location to determine the solar radiation flux density.

5. Can solar panels work without direct sunlight?

Yes, solar panels can still work without direct sunlight. While direct sunlight is ideal for maximum efficiency, solar panels can still generate electricity on cloudy or overcast days. This is because solar panels can still absorb the diffuse solar radiation present in the atmosphere. However, the electricity production may be lower compared to clear, sunny days.

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