1. Unit B 3.3 Useful Energy and Efficiency

2. Useful Energy The purpose of a machine is convert input energy into types of energy needed to do work. Anything else is waste energy. The efficiency of a system can never be 100% (no perfect machines  ) as energy is always lost to the surroundings in energy transfer.

3. Useful Energy Initial energy source is called energy input The desired energy needed to do the work is called useful energy output The work the machine is supposed to do is the useful work output

4. Efficiency Efficiency is a measurement of how effectively a machine converts energy input into useful energy output Efficiency = useful work output total work output Efficiency is expressed as a percent Systems with moving parts always loose some energy as heat

5. 1. Conversion of Total Mechanical Energy to Useful Mechanical Energy Percent Efficiency = work output x 100 work input

6. Example A crane lifts a load of construction materials from the ground to the second floor of a building. In the process, the crane does 2.30 x 10 4 J of work or mechanical energy input, while doing 8.00 x 10 3 J of useful work or mechanical energy output lifting the load. What is the mechanical percent efficiency of the crane?

7. Percent Efficiency = useful mechanical energy x 100 total mechanical energy Percent Efficiency = 8.00 x 10 3 J x 100 2.30 x 10 4 J = 34.8% Practice Problem p. 216 #1

8. Example An internal combustion engine with an efficiency of 15.0% is used to do 3.20 x 10 4 J of useful work, or mechanical energy output. Calculate the mechanical energy input that had to be supplied by the combustion of fuel in the engine.

9. Percent Efficiency = useful mechanical energy x 100 total mechanical energy 15.0 = 3.20 x 10 4 J x 100 total mechanical energy Total mechanical energy = (3.20 x 10 4 J)(100) 15.0 = 2.13 x 10 5 J Practice Problem p. 217 #2

10. 2. Transfer of Total Thermal Energy to Useful Thermal Energy Percent efficiency = heat useful output x 100 heat total output

11. Example In heating a pot of water, 2.00 x 10 3 J of heat was supplied by the stove element. If only 5.00 x 10 2 J of heat was actually gained by the water, heat output, what was the percent efficiency of the stove element?

12. Percent efficiency = heat useful output x 100 heat total output Percent efficiency = 5.00 x 10 2 J x 100 2.00 x 10 3 J Percent efficiency = 25.0% Homework: read Science 10 p. 215 – 220 Check and Reflect 1-9 Practice Problem p. 217 #3