310 likes | 447 Views
Efficiency of Electric Lighting From Coal. Page 404 SciencePower 9. Think about it. In 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.
E N D
Efficiency of Electric Lighting From Coal Page 404 SciencePower 9
Think about it • In 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 steps • 1. 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. You will need enough space for the 5 steps
Electrical Power & Efficiency • Electrical 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 second • 1W = 1J/s
Measuring Electrical Usage • Generating stations have power ratings in Mega Watts (MW) = 1 million watts OR • Giga Watts (GW) = 1 billion watts • The 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 Electricity • In 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-h • Required: cost to operate • Analysis: 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.
Practice • Calculate the cost of operating a 1500W hairdryer for 6 minutes per day for 3 days. • The cost of electricity is 5.6¢/kW-h • Use 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-h • Required: 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 operating • Calculate 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 = EoutX 100 Ein
Solution: % efficiency = 35J X 100 100J = 0.35 X 100 = 35% efficient Statement: 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: