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If the efficiency of the panel is 30%, determine the power input into the photovoltaic plate. Above the earth's atmosphere, the solar constant is 1.4 kWm⁻². However, only 40% of this power per unit a

Physics

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Here are the solutions to your questions:

i) If the efficiency of the panel is 30%, determine the power input into the photovoltaic plate.

  1. State the formula for efficiency.

    Efficiency(η)=Power output(Pout)Power input(Pin)\text{Efficiency} (\eta) = \frac{\text{Power output} (P_{out})}{\text{Power input} (P_{in})}
  2. Rearrange the formula to solve for power input.

    Pin=PoutηP_{in} = \frac{P_{out}}{\eta}
  3. Substitute the given values.

    Given: Pout=50.0 WP_{out} = 50.0 \text{ W}, η=30%=0.30\eta = 30\% = 0.30

    Pin=50.0 W0.30P_{in} = \frac{50.0 \text{ W}}{0.30}
  4. Calculate the power input.

    Pin=166.666... WP_{in} = 166.666... \text{ W}

    The power input into the photovoltaic plate is approximately 167 W\boxed{\text{167 W}}

    ii) Above the earth's atmosphere, the solar constant is 1.4 kWm−21.4 \text{ kWm}^{-2}. However, only 40% of this power per unit area reaches the earth. Estimate the surface area of the panel in figure 10.

  5. Calculate the power per unit area (intensity) that reaches the Earth's surface.

    Solar constant (S)=1.4 kWm−2=1400 Wm−2(S) = 1.4 \text{ kWm}^{-2} = 1400 \text{ Wm}^{-2}
    Intensity at Earth's surface (Iearth)=40% of S(I_{earth}) = 40\% \text{ of } S

    Iearth=0.40×1400 Wm−2I_{earth} = 0.40 \times 1400 \text{ Wm}^{-2} Iearth=560 Wm−2I_{earth} = 560 \text{ Wm}^{-2}
  6. State the relationship between power input, intensity, and surface area.

    Power input(Pin)=Intensity(Iearth)×Surface area(A)\text{Power input} (P_{in}) = \text{Intensity} (I_{earth}) \times \text{Surface area} (A)
  7. Rearrange the formula to solve for surface area.

    A=PinIearthA = \frac{P_{in}}{I_{earth}}
  8. Substitute the power input from part (i) and the calculated intensity.

    Using Pin≈166.67 WP_{in} \approx 166.67 \text{ W} from part (i).

    A=166.67 W560 Wm−2A = \frac{166.67 \text{ W}}{560 \text{ Wm}^{-2}}
  9. Calculate the surface area.

    A=0.297625 m2A = 0.297625 \text{ m}^2

    The estimated surface area of the panel is approximately 0.298 m2\boxed{\text{0.298 m}^2}

    iii) Explain why solar energy is referred to as renewable source energy.

    Solar energy is considered a renewable energy source because it originates from the sun, which provides a virtually inexhaustible supply of energy on a human timescale. The sun continuously emits vast amounts of energy, and its supply is not depleted by human use, unlike finite fossil fuels.

    i) What is the ozone layer?

    The ozone layer is a region within the Earth's stratosphere that contains a relatively high concentration of ozone (O₃) molecules. Its crucial function is to absorb most of the Sun's harmful ultraviolet (UV) radiation, preventing it from reaching the Earth's surface and protecting living organisms from its damaging effects.

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