Background Information, Data, and Assumptions As of 2011 the population of the w

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Background Information, Data, and Assumptions As of 2011 the population of the world is estimated to be about 7 billion people. The estimates vary widely but the medium UN prediction of world population is about 9 billion in 2050 Complete energy use (transportation, electricity, heating, etc.) of an American is estimated to be about 10,000 W per person around the year 2000. The land area of the earth is about 1.5 times 10^14 m^2. The average solar energy input from the sun is 100 W/m^2. The wholesale cost to generate electricity from coal is about $0.05/kWh. Biofuel Information The efficiency of photosynthesis is estimated to be less than 1%. Photovoltaic Solar Panel Information Assume about 80% of the time is sunny (limited cloudy cover). Assume about 50% of the land use for solar power is covered in PV cells (you have to allow space for trucks to drive between solar panels for maintenance and the structure below the panels). The efficiency of solar cells (photovoltaic cells) is estimated to be about 15%. A crude estimation of the cost per Watt of solar panels is less than $5 per peak Watt and a Watt is we can extract is roughly 1/4 a peak Watt. The lifetime of solar panels is estimated to be 25 years. Hydroelectric Information Assume about 1 m of rain per year falls on the land and the average elevation of land is 1000 m above sea level. The density of water is about 1,000 kg/m^3. Information The density of air is about 1 kg/m^3. The spacing between wind turbines needs to be 5-10 turbine diameters. A high estimate of wind speed is 10 m/s and assume the diameter of a typical turbine would be 10 m. Understanding the "solar" alternative energy sources, let's evaluate the practicality of each type to see if it would provide enough energy to meet our energy demands. Calculate the W/m^2 we could get from biofuels (photosynthesis) in an ideal situation. Calculate the W/m^2 we could get from photovoltaic cells. Calculate the W/m^2 we could get from hydroelectric power (using PE/yr = mgh/yr). Calculate the W/m^2 we could get from wind power. First we calculate the energy output of a windmill Kinetic Energy = 3/2 mv^2 Mass of air/time = (density of air) times (area of blades) times (speed of wind) Since required spacing is 5-10 diameters between windmills, let's assume 8, spacing needs to be 160 m times 160 m = 25,600 m^2. What is the W/m^2 for windmills?

Explanation / Answer

4. in ideal case scenario
   Average solar input, I = 100 W/m^2
   efficiency of photosynthesis is about 1 percent
   so, power per unit area produced by plants = 0.01I = 1 W/m^2
5. efficiency of photovoltaic cells = 15 percent
   sunny time per day = 80 percent of the day
   land use is 50 percent
   so power per unit area generated by the photovoltaic cell = 0.15*0.8*0.5*100 = 6 W/m^2
6. asssuming 1 m rainfall at 1000 m
   so Potential energy of this water = mgh
   here h = 1000 + 0.5 m [ taking average of the total rainfall height and 0]
   g = acceleration due to gravity
   m = rho*A*1 [ where A is the cross section cpvered by water]
   now PE/ year = rho*A*g(1000.5)/365*24*60*60 W = 0.311228 A W
   so Power per unit area = 0.311228 W/m^2

7. energy output of a windmill = KE = 0.5mv^2
   but mass flow per unit time m/t = rho*A*v [ A is area of blades, rho is density of air, v is wind velocity]
   so power generated = 0.5*rho*A*v^3
   now rho = 1 kg/m^3
   A = pi*(5)^2 [ as diameter is 10 m]
   now, this means every windmill covers an area of pi*(10*8)^2
   v = 10 m/s
   so, power per unit area = 0.5*1*pi*(5^2)*10^3/pi*(10*8)^2 = 1.95 W/m^2

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