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What type of panel would you use to charge a 12 V battery and why?
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10 stepsHere are the solutions to Activity 6.2:
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a) I would use a monocrystalline or polycrystalline solar panel. b) These types of panels are commonly available and efficient. For a 12 V battery, a panel designed for 12 V systems (typically having 36 cells) is suitable because its peak power voltage (around 17-18 V) is appropriate for charging a 12 V battery.
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A peak sun hour is a unit of measurement that indicates the intensity of sunlight in a specific location over a period of time. One peak sun hour is equivalent to one hour of full sun at an intensity of .
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The derate factor (or performance ratio) takes into consideration various losses that reduce the actual output of a solar panel compared to its rated power. These losses include temperature effects, shading, dust, wiring losses, inverter efficiency, and module mismatch.
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Solar irradiance is the power per unit area received from the Sun in the form of electromagnetic radiation. It is typically measured in watts per square meter ().
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Solar panel insolation (or solar irradiation) is typically measured in kilowatt-hours per square meter per day () or joules per square meter ().
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The formula for the efficiency of solar conversion () is:
where is the electrical power output, is the solar power input, is the area of the solar panel, and is the solar irradiance.
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Step 1: Identify the given values. Rated output of solar panel () = Solar irradiance (daily insolation, ) = (This is equivalent to 3.5 peak sun hours) Assume a derate factor () = 0.75 (based on the example provided in the image).
Use the formula for daily electricity output.
$$ E_{daily} = P_{rated} \times H \times PR $$Substitute the values and calculate.
$$ E_{daily} = 4 \text{ kW} \times 3.5 \text{ h} \times 0.75 $$ $$ E_{daily} = 14 \text{ kWh} \times 0.75 $$ $$ E_{daily} = \boxed{\text{10.5 kWh}} $$8. Step 1: Calculate the area of the single solar cell. Length () = Width () =
Calculate the electrical power output of the cell.
Voltage ($V$) = $0.6 \text{ V}$ Current ($I$) = $3 \text{ A}$ $$ P_{out} = V \times I = 0.6 \text{ V} \times 3 \text{ A} = 1.8 \text{ W} $$Calculate the solar power input to the cell.
Solar insolation ($I_{solar}$) = $900 \text{ W/m}^2$ $$ P_{in} = A_{cell} \times I_{solar} = 0.015625 \text{ m}^2 \times 900 \text{ W/m}^2 = 14.0625 \text{ W} $$Determine the efficiency of the solar cell.
$$ \eta = \frac{P_{out}}{P_{in}} \times 100\% $$ $$ \eta = \frac{1.8 \text{ W}}{14.0625 \text{ W}} \times 100\% $$ $$ \eta \approx 0.1280 \times 100\% $$ $$ \eta = \boxed{\text{12.80\%}} $$9. Step 1: Calculate the area of a single cell. Length () = Width () =
Calculate the power output of a single cell.
Insolation ($I_{solar}$) = $800 \text{ W/m}^2$ Efficiency ($\eta$) = $13\% = 0.13$ $$ P_{cell\_out} = \eta \times A_{cell} \times I_{solar} $$ $$ P_{cell\_out} = 0.13 \times 0.015625 \text{ m}^2 \times 800 \text{ W/m}^2 $$ $$ P_{cell\_out} = 1.625 \text{ W} $$Determine the current per cell, assuming a typical cell voltage.
A standard silicon solar cell typically produces about $0.6 \text{ V}$. $$ I_{cell} = \frac{P_{cell\_out}}{V_{cell}} = \frac{1.625 \text{ W}}{0.6 \text{ V}} \approx 2.7083 \text{ A} $$Calculate the number of cells required in series () to achieve the target voltage.
Target voltage ($V_{total}$) = $3.8 \text{ V}$ $$ N_s = \left\lceil \frac{V_{total}}{V_{cell}} \right\rceil = \left\lceil \frac{3.8 \text{ V}}{0.6 \text{ V}} \right\rceil = \lceil 6.333 \rceil = 7 \text{ cells} $$ (We round up to ensure the target voltage is met or exceeded.)Calculate the number of parallel strings () required to achieve the target current.
Target current ($I_{total}$) = $2 \text{ A}$ $$ N_p = \left\lceil \frac{I_{total}}{I_{cell}} \right\rceil = \left\lceil \frac{2 \text{ A}}{2.7083 \text{ A}} \right\rceil = \lceil 0.738 \rceil = 1 \text{ string} $$ (One string provides enough current.)Calculate the total number of cells.
$$ N_{total} = N_s \times N_p = 7 \text{ cells} \times 1 \text{ string} = 7 \text{ cells} $$ The 7 cells connected in series would produce $7 \times 0.6 \text{ V} = 4.2 \text{ V}$ and $2.7083 \text{ A}$, which meets the requirements of $3.8 \text{ V}$ and $2 \text{ A}$. The number of cells required is $\boxed{\text{7 cells}}$.
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