Page 404SciencePower 9

Think about itIn every step between mining coal,

generating electricity , transmitting it to your home and lighting a lamp some energy is used or escapes in the form of heat in the process.

In this activity we will calculate what percentage of energy stored in the coal is actually converted into light in the light bulb.

Summary of steps1. Mining the coal: the process of

extracting coal from the ground is about 99% efficient.

Some energy is used by the machines in the mining process.

2. Transportation of the Coal: transporting the coal from the mine to the power plant is also quite efficient. Some energy is used to power the train. Efficiency about 97%

3. Generation of Electricity: Thermo – electric generating plants are not very efficient most of the energy from the coal escapes from the stacks or from coolant water as heat. Only about 33% of the energy from coal that is burned is converted into electricity.

4. Transmission of Electricity: When electrical energy travels through the power lines, some of the energy goes into heating the conductors. Electrical transmission is about 85% efficient.

5. Conversion to light: Incandescent light bulbs are very inefficient. Most of the electrical energy is converted into heat. At best an incandescent light bulb is about 15% efficient.

What to Do......Make a Table with the headings shown below.

STEP EFFICIENCYOF STEP

(%)

STORED ENERGY

REMAINING(out of original

1.0 X 106 J)

OVERALL EFFICIENCY

(%)

You will need enough space for the 5 steps

STEP EFFICIENCYOF STEP

(%)

STORED ENERGY

REMAINING(out of

original 1.0 X 106 J)

OVERALL EFFICIENCY

(%)

1.

Mining the coal

99%

1, 000 000J X 99% 100 = 990 000J

990,000J X 100 1,000,000 J

99%

2.

Transportation of the coal.

97%

990 000J X 97% 100

= 960300J

960300 X100 1,000 000J

96.03%

3. Generation of

electricity 33%

960300 X 33%100

= 316899 J

316899 X 1001,000 000J

31.69%

STEP EFFICIENCYOF STEP

(%)

STORED ENERGY

REMAINING(out of

original 1.0 X 106 J)

OVERALL EFFICIENCY

(%)

4.Transmission of Electricity

85%316899 J X

85%100%

269364.15 J

269364.15 X 100

1, 000, 000 J

26.94%

5. Conversion to

Light

15%

269364.15J X 15%

100%40,404 J

40404 X 1001,000,000J

4.04%

6. Fluorescent

Light90%

269364J X 90%100

= 242427.6 J

242427.6J X 100

1,000,000J24. 24%

Electrical Power & EfficiencyElectrical devices are labelled with a power

rating. Power is measured in units called Watts (W) after James Watt who discovered that steam could move things and then developed the steam engine.

Electrical Power Power is the rate at which energy is

transformed, or the rate at which work is done.

Electrical Power is the rate at which electrical energy is produced or consumed in a given time.

1 Watt = 1 Joule per second1W = 1J/s

Measuring Electrical UsageGenerating stations have power

ratings in Mega Watts (MW) = 1 million watts OR

Giga Watts (GW) = 1 billion wattsThe Kilowatt hour is the SI unit used to

measure energy usage An average Canadian family consumes

over 16, 000 kW-h of electrical energy in one year

Efficiency Not all appliances use energy efficiently. Efficiency is a measure of how much useful energy an electrical device produces compared to how much energy was supplied to the device.

Older devices tend to be less efficient than newer models.

This is an example of an Energuide label found on newer appliances. It helps consumers make informed choices when buying a new appliance

The energy star program was introduced in 1972 by the US Environmental Protection Agency and the US department of Energy.

Cost of ElectricityIn Ontario the cost of electricity is regulated at 5.6¢ for the first 1000kW-h during the winter and the first 600kW-h during summer. After 1000kW-h the cost increase to 6.5¢/kW-h

Cost to operate a Lap top Computer Cost to operate = power used X time X

cost of electricity

A laptop uses a 75W adapter when it is plugged in. Electricity costs 5.6¢/kW-h

Calculate the cost to operate the computer for 1 year for 24 hours a day.

Use the G.R.A.S.S. method for solving problems

Given: Power = 75W (convert to kW =75W X 1kW = 0.075kW 1000W Time = 24 hours per day for 365

days = 8760 hours Cost of Electricity = 5.6¢/kW-hRequired: cost to operateAnalysis: Cost to operate = power used X time X cost of

electricity

Solution: Cost to operate = 0.075kW X 8760h X

5.6¢ kW-h = 3679¢

Statement: It would cost 3679¢ or $36.79 to operate a laptop computer for 24 hours a day for 1 year.

PracticeCalculate the cost of operating a 1500W

hairdryer for 6 minutes per day for 3 days.The cost of electricity is 5.6¢/kW-hUse the G.R.A.S.S. method.

Given: Power = 1500W = 1.5kW

Time = 6mins X 3 days = 18 mins 18/60 = 0.3h Cost = 5.6¢/kW-hRequired: Cost to operate

Analysis: cost to operate = power X time X cost of electricity

Solution:

cost = 1.5kW X 0.3h X 5.6¢

= 2.52¢

Statement: the cost to operate the hairdryer for 6 mins for 3 days is 2.5 ¢

Calculate the difference between the operating cost of a 60W incandescent light bulb and a 13W CFL, each operating for 100 hours. The cost of electricity is

11¢ per kW-h

Given: Power = 60W X 1kW = 0.06kW

(incandescent) 1000W = 13W X 1kW = 0.013kW (CFL) 1000W Time = 100 hours

Cost = 11¢/kW-h

Required: Difference in cost between an incandescent

light bulb and a CFL

Analysis:

cost = power X time X cost of electricity

Solution: incandescent bulb cost = 0.06kW X 100h X 11¢ = 66 ¢ CFL cost = 0.013kW X 100h X 11 ¢ = 14.3 ¢Statement: Obviously the CFL is much cheaper to run than

the incandescent bulb by a saving of 51.7 ¢

Cost of operatingCalculate the cost of operating a refrigerator

power 750W for 1 year. Cost of electricity is 12 ¢/kW-h

Given: Power = 75W, time = 365 X 24 = 8760 hours

cost = 12 ¢/kW-h

Reguired: Cost to operate for one year

Analysis: Power X Time X cost

Substitute: 0.75KW X 8760 hours X 12 ¢/kW-h

Solution: $788.40

Calculating the Efficiency of a light bulb.

A light bulb uses 100J of electrical energy and produces 35J of light energy. Calculate the percentage efficiency of the light bulb.

Given: Energyout = 35J

Energyin = 100J

Required: Percent efficient (% efficiency)Analysis: % efficiency = Eout X 100

Ein

Solution: % efficiency = 35J X 100 100J = 0.35 X 100 = 35% efficientStatement: the light bulb is 35% efficient.

A toaster oven uses 1200J of energy to produce 850J of thermal energy. Calculate the percent efficiency of the oven.

Given:

Required:

Analysis:

Solution

Statement: