Collector Size of solar panel

In summary, if your conversion efficiency is 0.28, you would need input energy of 2000kWhr to get the same output energy as if your conversion efficiency was 28%.
  • #1
deenuh20
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Homework Statement



Your house requires 2000 kWh of energy per month and you want to power it using sunlight which has an average daylight intensity of 1 kW/m2. Assuming that sunlight is available 8 hours per day, 25 days per month (only 5 – 6 cloudy days per months!) and that you have a way to store energy when the sun isn’t shining, calculate the smallest collector size you would need if its conversion efficiency were R=28 % ___m^2

Homework Equations



Iave=Pave/Area

The Attempt at a Solution



First I found how many seconds that light was present in the month by dimensional analysis of the 25 days/month and the 8 hours light/day. I got 720,000 sec. Then, I converted 2000kWh to W/s and got 555.556 W/s. Then I multiplied 720,000s by 555.556 W/s and got 4*10^8 W as the power. Seeing that it was 28% conversion efficiency, I multiplied the power by .28 and got 1.12*10^8 W. Then, to find area, I used the above equation:

1kW/m^2=1.12*10^8/Area, and solved for area, which was 112000m^2, but it was the wrong answer.
 
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  • #2
There is a problem with your units, and thus the calculation. You cannot convert kWhr to W/s, that makes no sense. kWhr is a measure of energy, and W is a measure of Energy/time.

I like to carry units along in my calculations to be sure that I don't do the wrong multiplication or division. Try doing the calcs again, and carry the units along just like they were variables (I like to carry them along in square brackets [] to show that they are units).

What do you get if you're careful to keep the units consistent for each term and on both sides of each equation?
 
  • #3
Keeping units in check, here's what i did:

1. Covert 2000kWh to kWs = 7.2x10^9 Joules
2. Because of efficiency, 7.2x10^9 J * .28 = 2.016x10^9 J
3. Converted days to seconds: (25d)(8h/d)(3600s/h)=72000s
4. Power= 2.016x10^9 J/72000s= 2800J/s= 2800 W
5. Pave/Iave= Area so 2800 W/(10^3 W/m^2)=2.8m^2

However, this isn't the correct answer. Did I approach it correctly?
 
  • #4
*bump for viewing/replies*
 
  • #5
I think maybe step #2 was wrong. If your conversion efficiency is 0.28, and you want 2000kWhr output energy, how much input energy do you need?
 

1. What is the ideal collector size for a solar panel?

The ideal collector size for a solar panel depends on several factors, such as the location, climate, and energy needs of the user. Generally, a larger collector size will result in higher energy production, but it is important to consider the available space and budget as well.

2. How does the collector size affect the efficiency of a solar panel?

The collector size directly affects the efficiency of a solar panel. A larger collector size means a larger surface area to absorb sunlight, resulting in higher energy production. However, too large of a collector size can also lead to overheating and reduced efficiency, so it is important to find the right balance.

3. Can I increase the collector size of my existing solar panel system?

In most cases, it is possible to increase the collector size of an existing solar panel system. However, this may require additional equipment and adjustments to the system, so it is best to consult with a professional before making any changes.

4. How do I determine the appropriate collector size for my solar panel system?

The appropriate collector size for a solar panel system can be determined by considering the average daily energy consumption, the location and climate, and the available space and budget. Consulting with a solar panel expert can also help to determine the most suitable collector size for your specific needs.

5. Are there any disadvantages to having a larger collector size for a solar panel?

While a larger collector size can result in higher energy production, it can also come with some disadvantages. These can include higher costs for equipment and installation, as well as potential issues with overheating and reduced efficiency. It is important to carefully consider all factors before deciding on the collector size for a solar panel system.

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