1. Environmental Physics Chapter 3: Energy Conservation Copyright © 2008 by DBS

2. Introduction ‘Energy Intensity’ (Btu/GDP ratio) – shows conservation trend since 1980’s –High EI indicate high price of conversion of energy into GDP –Low EI indicates a lower price of conversion Shows that economy can grow without a commensurate increase in energy consumption Figure 1.16: United States energy use (Btu) compared to GDP over time, and their ratio.

3. Law of Conservation of Energy First law of thermodynamics - The total energy of a system can be increased by doing work on it or by adding heat W + Q = Δ(KE + PE + TE) When no work or heat is added (W + Q) = 0, we have a ‘closed system’ Δ(KE + PE + TE) = 0 The total amount of energy in an isolated system will always remain constant If our isolated system is the universe, then, law of conservation of energy follows: Total energy in the universe is a constant and will remain so In short, the law of conservation of energy states that energy can not be created or destroyed, it can only be changed from one form to another or transferred from one body to another, but the total amount of energy remains constant (the same).

4. Law of Conservation of Energy Note that the “law of conservation of energy”, a scientific law is very different to “energy conservation” – reducing energy use through the use of reduced activity or increased efficiency As an example of this law consider the “nosecracker” Total mechanical energy is conserved, no energy is transferred to the system by work or heat, Δ(KE + PE) = 0 Initial mechanical energy = final mechanical energy If the ball is released from ‘A’ to swing A-B-C- B-A, it will not go higher than A on its return Max PE G PE →KE Max KE PE = 0 In reality frictional forces at the pivot and air resistance result in thermal energy loss – stops the ball swinging

5. Law of Conservation of Energy Since energy in an isolated system is not destroyed or created or generated, one might wonder why we need to be so concerned since energy is a conserved quantity The final result of most energy transformations is waste heat Waste heat is not useful for doing work - Energy is said to be degraded

6. Energy Conversion Examples Passive solar home, Oswego, New York. Thirty-five percent of the heating needs are provided by solar energy. Passive solar technologies convert sunlight into usable heat, cause air-movement for ventilation or cooling, or store heat for future use, without the assistance of other energy sources

7. Energy Conversion Examples Change in energy of the system = Net energy added (energy in – energy out) ΔE = E in - E out Figure 3.2: A passive solar energy house. Energy in = energy out + energy stored. e.g. passive solar house, ΔE = 0, E in = E out Solar input = (energy loss through walls + energy stored in the materials of the house)

8. Energy Conversion Examples e.g. fossil-fueled steam power plant 1.Combustion of fuel in the boiler unit using air 2.Combustion generates heat that boils water into steam (water gains TE) along with combustion gases 3.High temperature, high pressure steam drives a turbine 4.Turbine drives a generator for producing electricity 5.Steam leaves the turbine and passes through condenser 6.Condenser takes cold water and passes it through a heat exchanger to condense the steam 7.Cold water becomes warmer water producing thermal pollution Figure 3.3: Block diagram of a fossil-fueled electric generating station. Energy in = energy out, because no energy is stored. E fuel + E air + E water in = E electricity + E water out + E combustion gases

9. Energy Conversion Efficiencies Although energy is conserved processes are not 100% efficient The efficiency (η) of an energy conversion process is defined as: η = E out /E in x 100 % The energy input that does not go into useful work goes into unusable energy forms (such as waste heat) e.g. Power plant: Only a fraction of the chemical energy in fuel is transformed into electricity… Efficiency = η = E electricity /E fuel x 100 % = 35 % …while essentially all of the fuel’s chemical energy is converted into heat during combustion, 65% of this heat is transferred to the water leaving the condenser and the gases leaving the stack η = eta

10. Energy Conversion Efficiencies e.g. light bulb is only 10% (η = 10/100 = 0.1) efficient E elec = E light + E heat and E heat = E elec – E light Where η = E light /E elec  E light = η E elec E heat = E elec – E light = E elec – η E elec = (1 - η)E elec If E elec = 60W, E heat = (1 – 0.1) E elec = 0.9 x 60W = 54W and E light = 60W – 54W = 6W

11. Energy Conversion Efficiencies Efficiencies range from 5 to 95 % The laws of thermodynamics place a limit on η

12. Energy Conversion Efficiencies Efficiencies can be multiplied for processes with several stages: e.g. coal → electricity = 35 % efficiency Transmsssion generator → home = 90% (loss 10 %) electricity → light = 5 % (loss 95%) Overall efficiency = η generation x η trans. x η conver. = 0.35 x 0.90 x 0.05 = 0.016  0.016 x 100% = 1.6 % Figure 3.4: Calculation of the overall efficiency for a multistep process involves multiplying the efficiencies of the individual steps.

13. Question 7 Gasoline ICE = 10 %, EV = 18 % Table 3.5 examines the conversion processes in an internal combustion engine and an electric vehicle. For both systems, calculate the overall efficiencies.

15. Energy use in Developing Countries Depends on muscle power! ¾ of the world’s population use ¼ of the energy Major disparity in energy use per capita US = 10 x France, 70 x Kenya Figure 3.5: An interesting method of pumping water in Burkina Faso, West Africa. The boy is doing work using his PE, gained by jumping in the air. Developing Countries Developed Countries

16. Energy use in Developing Countries Since 1960, developing countries have quadrupled their energy use and doubled their per capita use Caused massive environmental problems Figure 3.6: Growth in energy use, GDP, and population in developing countries: 1960–2000.

17. Energy use in Developing Countries A rapidly growing economy in China has led to increased use of coal combustion. About 75% of its electricity is generated with coal.

18. Huai River Basin http://www.nytimes.com/packages/khtml/2004/09/11/int ernational/20040912_CHINA_FEATURE.html Pollution in China and India

19. Energy use in Developing Countries Developing countries’ energy plans follow the example set by industrialized countries Using technologies fueled by fossil fuels China: coal (dirtiest fossil fuel) provides 70% of electricity (growing by 7% per yr)

20. Energy use in Developing Countries Industrialized countries are beginning to increase efficiency and move to renewable fuels, developing countries are burning more and more fossil fuels In the next 30 years global population is expect to grow 33% from 6 to 8 billion primarily in developing countries

21. Sustainable Development Building a house from handmade mud and straw bricks.

22. A Barrel, A Calorie, A Btu? Energy Equivalencies Food Calorie: 1 food calorie = 1kcal = 1000 calories = energy required to raise the temp. of 1 kg H 2 O by 1 °C British Thermal Units: 1 Btu = energy required to raise the temperature of 1 lb H 2 O by 1 °F 1 Btu = 252 calories = 1055 J Where 1 cal = energy required to raise the temperature of 1 g H 2 O by 1 °C

23. A Barrel, A Calorie, A Btu? Energy Equivalencies e.g. Barrel of crude oil (42 gallons) –Heat average home for 4 days to 20 °C when outside is 0 °C –Refined into gasoline and run a car for 1300 km (780 miles) Since different engines have different efficiencies far better to use ‘heating values’ Heating value is defined as the amount of heat the fuel could provide if completely burned

24. A Barrel, A Calorie, A Btu? Energy Equivalencies Heating value of coal, uranium, natural gas etc. can be equated to the heating value of so many barrels of oil MBPD = million barrels of oil per day e.g. burning 500 x 10 6 tons of coal provides same energy as burning 6 MBPD of oil for a year US consumption is 7 billion barrels per year or 20 MBPD

25. A Barrel, A Calorie, A Btu? Energy Equivalencies G

26. Question If one ton of bituminous coal is burned to generate electricity, how many kWh could be produced if the efficiency of this conversion is 35 %? One ton bituminous coal = 25 x 10 6 Btu (from Table 3.4) Process is 35 % efficient: η = E out / E in E out = 0.35 x 25 x 10 6 Btu = 8.9 x 10 6 Btu = 8.9 x 10 6 Btu x 1 kWh= 2560 kWh 3414 Btu

27. Summary Doing work on, or adding heat to an object increases an object’s total energy First law thermodynamics: W + Q = Δ(KE + PE + TE) Total energy of an isolated system is conserved Energy in = Energy out + Energy stored in the system (for an isolated system) Energy may be transformed from one form to another Efficiency is the ratio of useful energy output to the energy input

28. Problems 2. A ball of mass 0.5 kg is dropped from a height of 5 m. What is the greatest velocity it will have just before hitting the ground? PE = mgh = 0.5 kg x 9.8 m/s 2 x 5 m = 24.525 J = 25 J At ground PE = KE = ½ mv 2 ½ x 0.5 kg x v 2 = 25 J v 2 = (2 x 25 J)/0.5 kg = 100 m/s v = 10 m/s

29. Problems 3. How many pounds of coal has the same heating value as 20 gal gasoline? From Table 3.4: 1 gal gasoline = 1.24 x 10 5 Btu 20 gal gasoline = 20 x 1.24 x 10 5 Btu = 2.48 x 10 6 Btu From Table 3.4: 1 ton coal = 25 x 10 6 Btu 1 ton = 2000lb 2000 lb coal = 25 x 10 6 Btu 1 lb coal = 1.25 x 10 4 Btu 2.48 x 10 6 Btu= 198.4 lb coal 1.25 x 10 4 Btu / 1lb coal

30. Problems 4. A household furnace has an output of 100,000 Btu/hr. What size electrical heating unit (in kW) would be needed to replace this? 1 watt = 3.41 Btu/hr 100,000 Btu/hr x 1 Wx 1 kW= 29.3 kW 3.41 Btu/hr 1000 W

31. Problems 7. If 60 % of US petroleum use goes for transportation, and 1 % of this goes for buses, how much oil could be saved per year if all buses were electrified (assuming the electrical energy was generated from non-petroleum resources)? What is this in MBPD? 1 % of 60% = (60/100) x 100% = 0.006 x 100 = 0.6 % 0.6 % of all oil consumed per year is for buses From Appendix G (for 2003): US oil = 39.07 x 10 15 Btu x 1 bbl= 6.74 x 10 9 bbl / yr 5.8 x 10 6 Btu 6.74 x 10 9 bbl / yr x (0.6/100) = 0.040 x 10 9 bbl / yr 0.040 x 10 9 bbl / yr = 1.1 x 10 5 bbl / day 1.1 x 10 5 bbl / day x MBPD = 0.11 MBPD 10 6 bbl / day