1. Accounting for Energy - Part 2 Chapter 22 b

2. Objectives Know that energy is conserved Understand state energies Kinetic, Potential, Internal Understand flow work Understand sequential energy conversion Be able to do calculations involving accounting for energy

3. Recall... Accumulation = Net Input + Net Gen State Energies Kinetic Potential Internal (Independent of Path) Path Energies Work Heat (Depend on Path) 0 (Energy is conserved) Previous focus Current focus

4. State Energies State energies do not depend on path. Three kinds: Kinetic Potential Internal Often specified by other state quantities (e.g., velocity, height, temperature, pressure)

5. Kinetic Energy Energy associated with a moving mass. Often mechanical or shaft work is used to produce kinetic energy. Example: Shaft work from a car engine produces the car’s kinetic energy.

6. Kinetic Energy A rigid mass accelerates from an initial velocity v 1 to a final velocity v 2 because of applied force, F. Thus, an input of mechanical work causes the object to change its kinetic energy, E k Initial state F v1v1 F v2v2 xx Final state

7. Kinetic Energy From the UAE Energy final - Energy initial = Net Input Thus, for a constant force F:  E k = Net Mech. Work Input = F  x After applying Newton’s Laws we get  E k = ½ mv 2 2 - ½ mv 1 2 (see p. 584 Foundations of Engineering for derivation)

8. Potential Energy Potential energy is associated with the interactions with other bodies. It is always between two or more items Examples: Gravitational potential Spring potential Others: electrical (capacitors), magnetic (inductors), hydraulic (pumped storage)

9. Gravitational Potential Consider a rigid mass lifted vertically by a force F a distance  x. Thus, an input of mechanical work changes the object’s potential energy, E p F up F dn = mg xx

10. Gravitational Potential From the UAE: Energy final - Energy initial = Net Input Thus, for a constant force F:  E p = Net Mech. Work Input = F  x The force acting on the mass is F = mg so  E p = mg  x

11. Pairs Exercise #1 A 4000-kg elevator starts from rest, accelerates uniformly to a constant speed of 1.8 m/s, and decelerates uniformly to stop 20 m above its initial position. Neglecting friction and other losses, what work was done on the elevator?

12. 20 mM = 4000 kg + + + Data Solution to Pairs Exercise #1

13. Spring Potential Energy Consider a force F compressing a spring a distance  x. (See p. 587 Foundations of Engineering) Thus, an input of mechanical work causes the spring’s potential energy to change. From the UAE: Energy final - Energy initial = Net Input  E p = Net Mech. Work Input

14. Spring Potential Energy This time the force is not constant along x. By Hook’s Law, the relationship is F=kx where k is the spring constant and x=0 is the uncompressed (relaxed) state.

15. Internal Energy The energy stored inside the medium. The energy associated with translational, rotational, vibrational, and electronic potential energy of atoms and molecules.

16. Internal Energy Translation Rotation Vibration Molecular Interactions

17. Review: States of Matter Solid Liquid Gas Plasma

18. Review: Phase Diagram Plasma Gas Vapor Liquid Solid T triple T critical P triple P critical Pressure Temperature Critical Point Triple Point

19. No Phase Change (Sensible Energy) When path energy (heat, work) is added to a material, IF THERE IS NO PHASE CHANGE, temperature increases. This added energy changes the internal energy of the medium, U final - U initial =  U = Net Path Energy Input

20. Heat Capacity for Constant- Volume Processes (C v ) Heat is added to a substance of mass m in a fixed volume enclosure, which causes a change in internal energy, U. Thus, Q = U 2 - U 1 =  U = m C v  T The v subscript implies constant volume Heat, Q added m m TT insulation

21. Heat Capacity for Constant- Pressure Processes (C p ) Heat is added to a substance of mass m held at a fixed pressure, which causes a change in internal energy, U, AND some PV work. Heat, Q added TT m m xx

22. C p Defined Thus, Q =  U + P  V =  H = m C p  T The p subscript implies constant pressure Note: H, enthalpy. is defined as U + PV, so d H = d (U+PV) = d U + V d P + P d V At constant pressure, dP = 0, so dH= dU + PdV For large changes at constant pressure  H =  U + P  V

23. Phase Changes (Latent Energy) When path energy (heat, work) is added to a material, IF THERE IS A PHASE CHANGE AT CONSTANT PRESSURE, temperature stays constant. Examples… boiling water melting ice cubes

24. Total Energy Conservation For a closed system (no mass in or out):  E k +  E p +  U = W in - W out + Q in - Q out For an open system, with M defined as energy entering or leaving the system with the mass:  E k +  E p +  U = W in - W out + Q in - Q out + M in - M out

25. Flow Energies In open systems, we must consider the flow of energy across the system boundary due to mass flow. The mass flow rate is indicated by Potential: Kinetic: Etc.

26. Sequential Energy Conversion Step 1Step 2Step 3 E1E1 E2E2 E3E3 E4E4     overall =  1  2  3

27. Pair Exercise #2 An incandescent lamp is powered by electricity from a coal-fired electric plant. To produce 10 W of visible light, what is the required rate of heat release (W) from coal combustion? (Hint: See Figure 22.30)

28. Individual Exercise #1 Solve the problem outlined on the next slides Document your steps Solution will be turned in separately

29. Bungee Jumping Exercise George is going to bungee jump from a bridge that is 195 meters above a river on a cord with a taut length of 50.0 meters taut length = length of unstretched cord prior to reaching the taut length, the cord exerts no force and George is in freefall More...

30. Bungee Jumping Exercise Once the taut length is reached, use the following equation to determine the force on George due to the bungee cord: F=(15 kg/s 2 )(X-X taut ) You may neglect air drag in all tasks. You may assume this is a one-dimensional motion problem, i.e., you may assume that George falls straight down and on the rebound follows the same path upward. More...

31. Bungee Jumping Exercise 1. George has a mass of 75 kg. Determine his velocity at the instant the cord becomes taut. 2. Determine the maximum distance George will travel from the bridge and the minimum distance from the bridge once he bounces back. 3. Determine the maximum mass of person that can jump from bridge and NOT impact the river.

32. 15 minutes Be prepared to turn in your well-documented solution 15 minutes from now.

33. Team Exercise #1 (10 minutes) As a team.. compare answers resolve differences prepare a team solution Submit team solution original solutions from each team member