1. EGEE 102 – Energy Conservation And Environmental Protection Energy Efficiency

2. EGEE 102 - S. Pisupati2 Efficiency of Energy Conversion If we are more efficient with the energy we already have there will be less pollution, less reliance on foreign oil and increased domestic security.

3. EGEE 102 - S. Pisupati3 Energy Efficiency Energy Conversion Device Energy Input Useful Energy Output Energy Dissipated to the Surroundings

4. EGEE 102 - S. Pisupati4 Illustration An electric motor consumes 100 watts (a joule per second (J/s)) of power to obtain 90 watts of mechanical power. Determine its efficiency ? = 90 W x 100 = 90 % 100 W

5. EGEE 102 - S. Pisupati5 Device Electric Motor Home Oil Furnace Home Coal Furnace Steam Boiler (power plant) Power Plant (thermal) Automobile Engine Light Bulb-Fluorescent Light Bulb -Incandescent 90 65 55 89 36 25 20 5 Efficiency Efficiency of Some Common Devices

6. EGEE 102 - S. Pisupati6 25% Of the gasoline is used to propel a car, the rest is “lost” as heat. i.e an efficiency of 0.25 Source: Energy Sources/Applications/Alternatives Vehicle Efficiency – Gasoline Engine

7. EGEE 102 - S. Pisupati7 Heat Engine A heat engine is any device which converts heat energy into mechanical energy. Accounts for 50% of our energy conversion devices

8. EGEE 102 - S. Pisupati8 Temperature is in ° Kelvin !!!!!!! Carnot Efficiency Maximum efficiency that can be obtained for a heat engine

9. EGEE 102 - S. Pisupati9 Illustration For a coal-fired utility boiler, The temperature of high pressure steam would be about 540°C and T cold, the cooling tower water temperature would be about 20°C. Calculate the Carnot efficiency of the power plant ?

10. EGEE 102 - S. Pisupati10 540 ° C = 540 +273 ° K = 813 °K 20 °C = 20 + 273 = 293 °K

11. EGEE 102 - S. Pisupati11 Inference A maximum of 64% of the fuel energy can go to generation. To make the Carnot efficiency as high as possible, either T hot should be increased or T cold should be decreased.

12. EGEE 102 - S. Pisupati12 Schematic Diagram of a Power Plant

13. EGEE 102 - S. Pisupati13 Boiler Components

14. EGEE 102 - S. Pisupati14 Overall Efficiency Overall Eff = Electric Energy Output (BTU) x 100 Chemical Energy Input (BTU) = 35 BTU x 100 100 BTU = 35% Overall Efficiency of a series of devices = ( Thermal Energy ) x ( Mechanical Energy ) x ( Electrical Energy ) Chemical Energy Thermal Energy Mechanical Energy

15. EGEE 102 - S. Pisupati15 Overall Efficiency Cont.. ( Thermal Energy ) x ( Mechanical Energy ) x ( Electrical Energy ) Chemical Energy Thermal Energy Mechanical Energy = E boiler x E turbine x E generator = 0.88 x 0.41 x 0.97 = 0.35 or 35%

16. EGEE 102 - S. Pisupati16 System Efficiency The efficiency of a system is equal to the product of efficiencies of the individual devices (sub-systems)

17. EGEE 102 - S. Pisupati17 Extraction of Coal96%96% Transportation98%94% (0.96x0.98) Electricity Generation38%36 % (0.96x0.98x0.38) Transportation Elec91%33% Lighting Incandescent5%1.7% Fluorescent20%6.6% Step EfficiencyCumulative Efficiency Efficiency of a Light Bulb Step

18. EGEE 102 - S. Pisupati18 System Efficiency of an Automobile Production of Crude96%96% Refining87%84% Transportation 97%81% Thermal to Mech E25%20% Mechanical Efficiency- Transmission50%10% Rolling Efficiency20%6.6% Step Step Efficiency Cumulative Efficiency

19. EGEE 102 - S. Pisupati19 Efficiency of a Space Heater Electricity = 24% Fuel Oil = 53% Natural Gas= 70%

20. EGEE 102 - S. Pisupati20 Heat Mover Any device that moves heat "uphill", from a lower temperature to a higher temperature reservoir. Examples. Heat pump. Refrigerator.

21. EGEE 102 - S. Pisupati21 Heat Pump Heating Cycle Source: http://energyoutlet.com/res/heatpump/pumping.htmlhttp://energyoutlet.com/res/heatpump/pumping.html

22. EGEE 102 - S. Pisupati22 Heat Pump Cooling Cycle Source: http://energyoutlet.com/res/heatpump/pumping.html

23. EGEE 102 - S. Pisupati23 Coefficient of Performance (C.O.P) Effectiveness of a heat pump is expressed as coefficient of performance (C.O.P)

24. EGEE 102 - S. Pisupati24 Example Calculate the ideal coefficient of performance (C.O.P.) For an air-to-air heat pump used to maintain the temperature of a house at 70 °F when the outside temperature is 30 °F.

25. EGEE 102 - S. Pisupati25 Solution Cont. T hot = 70 °F = 21°C =21 + 273 =294K T cold = 30 °F = -1°C =-1 + 273 =272K C.O.P = 294 294 - 272 = 294 22 = 13.3

26. EGEE 102 - S. Pisupati26 Consequences For every watt of power used to drive this ideal heat pump, 13.3 W is delivered from the interior of the house and 12.3 from the outside. Theoretical maximum is never achieved in practice This example is not realistic. In practice, a C.O.P in the range of 2 - 6 is typical.

27. EGEE 102 - S. Pisupati27 More C.O.P.’s Compare the ideal coefficients of performance of the of the same heat pump installed in Miami and Buffalo. Miami: T hot = 70°F, T cold = 40°F Buffalo: T hot = 70°F, T cold = 15°F Miami: T hot = 294°K, T cold = 277°K Bufalo: T hot = 294°K, T cold = 263°K

28. EGEE 102 - S. Pisupati28 = 294 (294-277) =17.3 = 294 (294-263) =9.5 MiamiBuffalo