If my average monthly electric consumption is 200 units (1 unit 1 kWh), how many

Question

If my average monthly electric consumption is 200 units (1 unit 1 kWh), how many solar panels each of 300 W capacity will be needed to meet this demand if:

a. the panels worked at 10% efficiency.

b. You will get output equal to only 4 hours of peak power in a day.

c. The slant of your roof is not optimal so you get only 90% of the peak power.

d. 20% of the power is lost because of shade, transmission loss etc.

e. If you need to generate the same power with a wind turbine, what should be the diameter of the blades considering wind speed 15 meters per second and air density 1.3 kg/ m^3 and the turbine extracts 50% of the kinetic energy of the wind and the efficiency of mechanical to electrical conversion is 80%

Explanation / Answer

300*24*3600*30*0.1*n = 200*1000*3600

a)

n = 200*1000/(300*24*30*0.1)= 9.25 =10

b)

n = 200*1000/(300*4*30) = 5.55 =6

c)

n = 200*1000/(300*24*30*0.9) = 1.03 = 2

d)

n = 200*1000/(300*24*30*0.8)= 1.17 = 2

e)

0.5*1.3*15^2*0.5*0.8*24*3600*30*x = 200*1000*3600

x = 200*1000*3600/(0.5*1.3*15^2*0.5*0.8*24*3600*30) = 4.47 m

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.