Problem 1 a) Calculate the space required in Istanbul to meet the time-averaged

Question

Problem 1

a) Calculate the space required in Istanbul to meet the time-averaged power requirements
of a typical home. Assume a 15% efficient PV system, and that the energy burn rate of an average home is around 2kW.

b) How do these values change, if peak demand must be met (i.e. there is no energy storage battery available)? You can assume that peak energy is roughly double the average load so 4kW

c) Now think about you have a Tesla Model S. How many solar panels are required to charge Model S?
(Assume average car drives 12k miles a year and you need to find Tesla model S maximum motor power from official Tesla web-side)

Explanation / Answer

a)

note that 2 kW = 2 kWh/h = 48 kWh/day

land required = (power needed)/(solar resource * efficiency)

= (48 kWh/day)/(4.2 kWh/m2 /day * 0.15) = 76.19 m2

b)

Twice the power needed means twice the area requirements.

Land required = 152.38 m2

Better Solution:

Assume that peak power is needed in the middle of the day, i.e. when the sun is highest in the sky and the irradiance is around 1000 W/m2 .

Land required = (power needed)/(solar resource * efficiency)

= (4000 W)/(1000 W/m2 * 0.15) = 26.7 m2

  how this area is actually SMALLER than the area in part (a)

c)

C)         Clearly there is not enough roof space for solar panels even to power the car using storage and meeting the car’s average power. Homes, which have a much larger roof area, are clearly more suitable for PV deployment.

engine is rated at 215 kW . its efficiency is 21.7 kWh per 100 miles. The average car drives around 12,000 miles a year (source: epa.gov), which means the average person drives around 33 miles a day and would consume 7.1 kWh per day in their car if they drove the Tesla roadster.

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