Energy Efficiency Calculations
Name: ________________________________________
A typical power plant in the United States operates at a capacity of 500 megawatts per hour.
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How many megawatts of electricity does the power plant generate per day (pay attention to the unit)?
500 x 24 = 12000 MW/day
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How many kilowatt-hours would the power plant generate per day?
12000 MW x 1000 kW/1 MW = 12000000 kW/day
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How many kilowatt-hours would the plant generate per year if the plant operated 365 days per year?
12000000 kW/days x 365 days/ 1 year = 438000000 kw/year
However, plants do not run 365 days per year. The fraction of time a plant is operating is known as capacity
factor. Most thermal power plants have capacity factors of 0.9.
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Calculate the capacity of this power plant if the capacity factor is 0.9.
438000000 kW/year x 0.9 = 394200000 kW/year
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Think of some reasons why a power plant would not be able to operate 365 days per year.
Practice
A typical power plant in the United States operates at a capacity of 350 megawatts per hour.
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How many kilowatts of electricity does the power plant generate per hour (pay attention to the unit)?
350 MW/hr x 1000 kW/1 MW = 350,000 kW/hr or 3.5 x 105 kW/hr
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How many kilowatt-hours would the power plant generate per day?
3.5 x 105 kW/hr x 24 hr/day = 8,400,000 kW/day or 8.4 x 106 kW/day
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How many kilowatt-hours would the plant generate per year if the plant operated 365 days per year?
8.4 x 106 kW/day x 365 days/yr = 3,066,000,000 kW/yr or 3.066 x 109 kW/yr
However, plants do not run 365 days per year. The fraction of time a plant is operating is known as capacity
factor. Most thermal power plants have capacity factors of 0.85.
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Calculate the capacity of this power plant if the capacity factor is 0.85.
3.066 x 109 kW/yr x 0.85 = 2,606,100,000 kW/yr or 2.6061 x 109 kW/yr
A few things to know…
1. Power: rate at which energy is used
2. Kilowatt-hours: measure of energy
3. Energy Equation: Energy (kWh) = power (kW) x time (hour)
4. 1 kilowatt (KW) = __1000_ Watts (W)
5. Calculating consumption example 1:
Your electric company charges 14 cents per kilowatt-hour. A coffee maker has a power rating of 1,050 watts.
a. How many kilowatts does the coffee maker use?
1050 watts x 1 kW/1000 watts = 1.050 kW
b. How much energy does the coffee maker use to run for one hour?
1.050 kW x 1 hour = 1.050 kWh
c. How much does it cost to use the coffee maker one hour?
1.050 kWh x $.14 = $0.15
d. If you used that coffee maker 1 hour every day, how much would it cost you for the month of January?
$0.15 x 31 days = $4.65
e. How much would it cost you for an entire year?
$4.65 x 12 months = $55.80
6. Calculating consumption example 2:
Your electric company charges 12 cents per kilowatt-hour. A 60 Watt light bulb has a power rating of 60 watts.
a. How many kilowatts does the light bulb use?
60 W x 1 kW/1000 W = 0.060 kW
b. How much energy does the light bulb use if turned on for 3 hours?
0.060 kW x 3 hrs = 0.18 kWh
c. How much does it cost to use the light bulb for 3 hours?
0.18 kWh x $0.12 = $0.02
d. If you used that light bulb for 3 hours every day, how much would it cost for the month of January?
$0.02 x 31 days = $0.62
e. How much would it cost to use it 3 hours a day for an entire year?
$0.02 x 365 days = $7.30
Calculating Price of Common Appliances
Object
Power Rating
Iron
1100 W x 1 kW/1000W =
1.1 kW
CFL bulb
15 W x 1 kW/1000 W =
0.015 kW
0.4 W x 1 kW/1000 W =
0.0004 kW
Clock
Electric
Stove
5000 W x 1 kW/1000 W =
5 kW
Time
used
1.1 kW x
0.5 hr =
0.55 kWh
Cost to use for given
time
Cost per month
0.55 kWh x $0.12 =
$0.07
$0.07 x 31 days =
$2.17
$0.07 x 365
days/year =
$25.55
0.015kW
x 8 hrs =
0.12 kWh
0.12 kWh x $0.12 =
$0.01
$0.01 x 31 days =
$0.31
$0.01 x 365
days/year = $3.65
0.0004
kW x 24
hrs =
0.0096
kW
0.0096 kWh x $0.12 =
0.0012
$0.0012 x 31
days = $0.04
$0.0012 x 365
days/year = $0.44
5 kW x 1
hr = 5
kWh
5 kWh x $0.12 = $0.60
$0.60 x 31 days =
$18.6
Cost per year
$0.60 x 365
days/year =
$219.0
1. Do you notice any patterns? What types of things seem to cost more and why do you think that happens?
Answers will vary.
2. What types of things seem to cost less to use?
Answers will vary.
3. Which of the appliances surprises you the most? Why?
Answers will vary.
Do The Math
According to the U.S. Department of Energy, a typical home in the United States uses approximately 900 kWh
of electricity per month. On an annual basis, this is…..
How many kilowatt-hours per year will one home need?
900 kWh/month x 12 months/year = 10800 kWh/year
How many homes can a 500 MW per hour power plant with a 0.9 capacity factor support?
500 MW/hr x 1000 kW/1 MW = 500000 kW/hr x 0.9 = 4500000 kW/hr
Start by determining how much electricity the plant can provide per month:
4.5 x 106 kW/hr x 24 hr/day x 31 days/month = 3,348,000,000 kW/month or 3.348 x 109 kW/month
Divide the amount of energy in kWh of one power plant by the amount of energy one home needs for one
month.