Lets assume a world of 10 billion people with an average dem

Let’s assume a world of 10 billion people with an average demand for electricity of 1 kWe

per person (720 kWh/month) for providing lighting and running refrigerators, other appliances,

TVs, computers, etc. What surface area of polycrystalline silicon PV collectors would be

required to provide the electrical needs for 10 billion people? How much electrical storage

capacity would you estimate would be needed? If lead acid batteries were used to store the

energy, what mass of batteries would be required?

Solution

    The global annual average solar radiation that reaches the earth’s surface is 198 W/m² (this is the sum of energy absorbed plus energy reflected; for this problem, for simplicity, we will assume an average value of 200 W/m².

    Polycrystalline silicon cells with an efficiency of 8-9% as deployed and 18.2% for experimental models. We will assume an efficiency of 10% for this calculation.

    Now with a 10% efficiency and a 200 W/m² solar irradiation the electric energy available in one month (720 hours) is (10%)(200 W/m²)(720 h) = 14,400 Wh/m² = 14.4 kWh/m². The energy demand for 10 billion people in one month would be 720x10¹⁰ kWh. The area required then would be:

(720x10¹⁰ kWh) / (14.4 kWh/m²) = 5x10¹¹ m² = 500,000 km².

    How much energy should we store? This depends on the deployment of the solar cells. There are many factors to consider: the imbalance in the world’s population between the northern and southern hemispheres, storing energy overnight and during seasons with low solar irradiation, the possibility of having floating oceanic arrays that could be towed to regions of maximum solar irradiation, the need for transmission lines, etc. Since we already have the number that one month’s energy demand is 720x10¹⁰ kWh.

    Let’s estimate the battery mass required to store that much energy:

    The energy density of lead-acid batteries as ranging from 60 to 180 kJ/kg. Picking the high end of this range:

180 kJ/kg = 180 kW×s/kg = 0.05 kWh/kg.

    Means that we would need:

(720x10¹⁰ kWh)/( 0.05 kWh/kg) = 1.44x10¹⁴ kg of lead-acid batteries.