A family has installed a wind turbine. The problem provides the rated power output, product life, and installation cost of the wind turbine. We need to calculate the electrical energy generated in 2014, the cost of buying the same amount of energy from an electricity supplier, and the total savings over the product life of the wind turbine.

Applied MathematicsEnergy CalculationCost AnalysisFinancial ModelingUnits Conversion
2025/7/2

1. Problem Description

A family has installed a wind turbine. The problem provides the rated power output, product life, and installation cost of the wind turbine. We need to calculate the electrical energy generated in 2014, the cost of buying the same amount of energy from an electricity supplier, and the total savings over the product life of the wind turbine.

2. Solution Steps

a) Calculate the electrical energy generated in kilowatt-hours (kWh) during
2
0
1

4. The wind turbine generated energy for 2000 hours, and its rated power output is 1.5 kW. The energy generated is given by the formula:

Energy=Power×TimeEnergy = Power \times Time
Energy=1.5kW×2000hoursEnergy = 1.5 \, kW \times 2000 \, hours
b) Calculate the cost of buying the same quantity of electrical energy from an electricity supplier.
The electricity supplier charges 15 pence per kilowatt-hour (kWh). The cost is given by the formula:
Cost=Energy×PriceperkWhCost = Energy \times Price \, per \, kWh
Since the price is given in pence, we need to convert it to dollars. Assuming 100 pence = 1 dollar. The price is $0.15 per kWh.
Cost=Energy×0.15Cost = Energy \times 0.15
c) Calculate the total savings over the 20-year product life of the wind turbine.
First, calculate the total cost of buying electricity for 20 years.
Then, subtract the installation cost of the wind turbine ($2100) from the total cost of buying electricity.
Step-by-step calculation:
a) Energy generated in 2014:
Energy=1.5kW×2000hours=3000kWhEnergy = 1.5 \, kW \times 2000 \, hours = 3000 \, kWh
b) Cost of buying electricity in 2014:
Cost=3000kWh×Cost = 3000 \, kWh \times 0.15/kWh = 450450
c) Total savings over 20 years:
Total cost of buying electricity for 20 years:
Totalcost=20years×Total \, cost = 20 \, years \times 450/year = 90009000
Total savings = Total cost of buying electricity - Installation cost
Totalsavings=Total \, savings = 9000 - 2100=2100 = 6900$

3. Final Answer

a) The electrical energy generated in 2014 is 3000 kWh.
b) The cost of buying the same quantity of electrical energy in 2014 is $
4
5

0. c) The total money the householder will save over 20 years is $6900.

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