Q1.
An interesting estimation problem that helps us to appreciate the enormous amount of energy that we get from the Sun is to estimate how much of the Earth's surface would be needed to be covered with solar panels to supply all the electric power that our country uses today (Fig. 13.4). To make this estimate, you can find these numbers on the internet, assume some insolation on the Earth's surface and consider that some fraction of this energy is converted into electricity. You will probably find that even a fraction of the area of the Thar desert, if covered with solar panels, could supply India's electricity needs.
Answer
Working it out below, about 4500 km2 of panels would do it — roughly a square 67 km × 67 km, which is only about 2 per cent of the Thar desert. Here is the full estimate with every unit carried through.
Step 1 — how much electric power does India use?
India's annual electricity consumption ≈ 1.6 × 1012 kW h (about 1600 billion units)
Number of hours in a year = 365 × 24 h = 8760 h
Average power, P = energy ÷ time
P = (1.6 × 1012 kW h) ÷ (8760 h)
P = 1.83 × 108 kW = 1.83 × 1011 W (about 180 GW)
Number of hours in a year = 365 × 24 h = 8760 h
Average power, P = energy ÷ time
P = (1.6 × 1012 kW h) ÷ (8760 h)
P = 1.83 × 108 kW = 1.83 × 1011 W (about 180 GW)
Step 2 — how much electric power does one square metre of panel give?
Peak insolation at the surface ≈ 1 kW m–2 = 1000 W m–2 (clear sky, Sun high)
But averaged over day and night, cloud and low Sun, India receives about
5 kW h m–2 per day → 5000 W h m–2 ÷ 24 h ≈ 200 W m–2 as a round-the-clock average
Efficiency of a solar panel, η ≈ 20% = 0.20
Electric power per square metre = 0.20 × 200 W m–2 = 40 W m–2
But averaged over day and night, cloud and low Sun, India receives about
5 kW h m–2 per day → 5000 W h m–2 ÷ 24 h ≈ 200 W m–2 as a round-the-clock average
Efficiency of a solar panel, η ≈ 20% = 0.20
Electric power per square metre = 0.20 × 200 W m–2 = 40 W m–2
Step 3 — divide.
Area required, A = power needed ÷ power per square metre
A = (1.83 × 1011 W) ÷ (40 W m–2)
A = 4.6 × 109 m2
1 km2 = 106 m2, so A = 4.6 × 109 ÷ 106 = ≈ 4600 km2
That is a square of side √4600 km2 ≈ 68 km
A = (1.83 × 1011 W) ÷ (40 W m–2)
A = 4.6 × 109 m2
1 km2 = 106 m2, so A = 4.6 × 109 ÷ 106 = ≈ 4600 km2
That is a square of side √4600 km2 ≈ 68 km
Step 4 — compare with the Thar desert.
Area of the Thar desert ≈ 2 × 105 km2
Fraction needed = 4600 km2 ÷ 2 × 105 km2 = 0.023 = about 2.3%
Fraction needed = 4600 km2 ÷ 2 × 105 km2 = 0.023 = about 2.3%
Why the answer is so small: the Sun delivers energy at an enormous rate — 1400 J every second on every square metre at the top of the atmosphere. India's entire electrical demand, spread over a whole year, is tiny compared with what falls on even a small patch of desert. The limit on solar power is therefore not the supply of sunlight; it is the cost of panels, the land, and above all storage, because the Sun does not shine at night or through the monsoon.
Check it yourself: your numbers will differ from these — use the latest consumption figure and your own assumption for efficiency and average insolation. As long as you write the units at every step and they cancel correctly (W ÷ W m–2 = m2), your estimate is a good one. The purpose of an estimation problem is the order of magnitude, not the last digit.