If aiming for a particular power output \(^{109}\) from a PV array, describe explicitly/quantitatively how PV panel efficiency interacts with the physical size (area) of the array. For instance, what happens if the efficiency doubles or is cut in half, while keeping the same target output?

Short Answer

Expert verified
Answer: The area of a PV array is inversely proportional to its efficiency when maintaining a constant power output. If the efficiency doubles, the required area is halved, whereas if the efficiency is cut in half, the required area doubles.

Step by step solution

01

1. Understand the equation for power output

The power output of a PV array can be calculated using the following formula: Power output = Efficiency * Area * Solar radiation Where efficiency is the percentage of sunlight converted into electricity, the area is the surface area of the PV array, and solar radiation refers to the energy received from the sun.
02

2. Keeping power output constant

Our goal is to analyze what happens to the area of a PV array when efficiency changes under the condition that the desired power output remains constant. To do this, first, rewrite the power output formula as follows: Area = \(\frac{Power\, output}{Efficiency \times Solar\, radiation}\)
03

3. Compare changes in efficiency

When the efficiency of a PV array changes, we can compare how its area changes by finding the ratio of the new area to the original area. Let's denote the original efficiency as \(Eff_{1}\) and the new efficiency as \(Eff_{2}\).
04

4. Doubling efficiency

If the efficiency doubles, we have \(Eff_{2} = 2 \times Eff_{1}\). To find the new area (Area\(_{2}\)) required for the same power output, we can use the formula from step 2. \(\frac{Area_{2}}{Area_{1}} = \frac{Power\, output}{2 \times Eff_{1} \times Solar\, radiation} \cdot \frac{Eff_{1} \times Solar\, radiation}{Power\, output} = \frac{1}{2}\) This means that when the efficiency doubles, the required area for the same power output is halved.
05

5. Halving efficiency

If the efficiency is cut in half, we have \(Eff_{2} = \frac{1}{2} \times Eff_{1}\). To find the new area (Area\(_{2}\)) required for the same power output, we can again use the formula from step 2. \(\frac{Area_{2}}{Area_{1}} = \frac{Power\, output}{\frac{1}{2} \times Eff_{1} \times Solar\, radiation} \cdot \frac{Eff_{1} \times Solar\, radiation}{Power\, output} = 2\) This means that when the efficiency is cut in half, the required area for the same power output doubles. In conclusion, the area of a PV array is inversely proportional to its efficiency when aiming for a constant power output. If the efficiency doubles, the area required is halved, whereas if the efficiency is cut in half, the area required doubles.

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