# Question: information background biofuel information the efficiency of photosynthesis is estimated...

###### Question details

**Information background**

**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 ¼ 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}.

**Wind 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.

Questions

1. Calculate the W/m^{2} we could get from biofuels
(photosynthesis) in an ideal situation.

2. Calculate the W/m^{2} we could get from photovoltaic
cells.

3. Calculate the W/m^{2} we could get from hydroelectric
power (using PE/yr = mgh/yr).

4. Calculate the W/m^{2} we could get from wind power.
First we calculate the energy output of a windmill

Kinetic Energy = ½mv^{2}

Mass of air / time = (density of air) × (area of blades) × (speed of wind)

5. Since required spacing is 5 – 10 diameters between windmills,
let’s assume 8, spacing needs to be 160 m × 160 m = 25,600
m^{2}. What is the W/m^{2} for windmills?