a) If a household uses an average of 1500 kWh of electrical energy every 30 days
ID: 1775147 • Letter: A
Question
a) If a household uses an average of 1500 kWh of electrical energy every 30 days (month) and NREL estimates their house has a 3.1 kWh/m^2/day solar resource available, what's the minimum amount of roof area they would need to devote to (grid-tied) solar (photovoltaic) panels to meet their average electrical needs? (Assume they're buying standard residential-grade panels with an efficiency of 0.14.)
b) How many solar panels will the homewowner need to buy if they're buying 300 W panels that are 66 inches by 40 inches in size?
c) If the homeowner is able to buy panels and an inverter and do a DIY install at a cost of 0.79 dollars per Watt, what is the total cost of their system?
d) Based on the NREL estimate for solar resource and the panel efficiency quoted, what would be their system's annual enery production in kWh?
e) If the homeowner lives in a state with "net metering," the power company will buy the homeowner's excess production in the summer months. Under this assumption, after how many years will the homeowner recoup the costs of the install? Ignore tax incentives and assume a local electrical rate of 22 cents per kWh.
Explanation / Answer
A)Total Requirement=1500 kWh in 30 days
Per day generation capacity=3.1 kWh/m^2/day*0.14=0.434 kWh/m^2/day
30 days capacity=30*0.434 kWh/m^2/day=1.302 kWh/m^2
area required=1500 kWh/1.302 kWh/m^2=1152.07 m^2
B)300 W panels that are 66 inches by 40 inches in size so we can say 300/66*40=176.13 W/m^2
requirement=1.302 kWh/m^2
No of cells=1.302 kWh/m^2/176.13 W/m^2=7.32 units
C) Cost=0.79 dollars per Watt
Total cost=1500*0.79=1185 dollars
D) 1500 kWh*12=18000kWh
e) Need more that if any query ask just find how much excess enerygy is generated .You know the total cost.
Total cost/Your earning during summar
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.