Answer :
Answer:
a) [tex]\eta = 13.455\%[/tex], b) [tex]E_{day} = 812.716\,kJ[/tex], c) [tex]C_{month. total} = 19.505\, USD[/tex], d) [tex]t = 40.588\,years[/tex]
Explanation:
a) The area of the solar panel is:
[tex]A = (20\,ft^{2})\cdot (\frac{0.3048\,m}{1\,ft} )^{2}[/tex]
[tex]A = 1.858\,m^{2}[/tex]
The energy potential is determined herein:
[tex]\dot E_{o} = (1000\,\frac{W}{m^{2}} )\cdot (1.858\,m^{2})[/tex]
[tex]\dot E_{o} = 1858\,W[/tex]
The efficiency of the solar panel is:
[tex]\eta = \frac{\dot E}{\dot E_{o}}\times 100\%[/tex]
[tex]\eta = \frac{250\,W}{1858\,W}\times 100\%[/tex]
[tex]\eta = 13.455\%[/tex]
b) The energy generated by the solar panel is presented below:
[tex]E_{day} = (0.135)\cdot (150\,\frac{W}{m^{2}} )\cdot (20\,ft^{2})\cdot \left(\frac{0.3048\,m}{1\,ft} \right)^{2}\cdot (6\,h)\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{1\,kJ}{1000\,J} )[/tex]
[tex]E_{day} = 812.716\,kJ[/tex]
c) The energy generated per month and per panel is:
[tex]E_{month} = 30\cdot E_{day}[/tex]
[tex]E_{month} = 30 \cdot (812.716\,kJ)\cdot \left(\frac{1\,kWh}{3600\,kJ} \right)[/tex]
[tex]E_{month} = 6.773\,kWh[/tex]
Monthly energy savings due to the use of 18 panels are:
[tex]C_{month, total} = 18\cdot E_{month}\cdot c[/tex]
[tex]C_{month, total} = 18\cdot (6.773\,kWh)\cdot (\frac{0.16\,USD}{1\,kWh} )[/tex]
[tex]C_{month. total} = 19.505\, USD[/tex]
d) The payback of the solar energy system is:
[tex]t = \frac{9500\,USD}{12\cdot (19.505\,USD)}[/tex]
[tex]t = 40.588\,years[/tex]
