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Accounting for Energy - Part 2

Accounting for Energy - Part 2. Chapter 22 b. 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. Recall.

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Accounting for Energy - Part 2

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  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 0 (Energy is conserved) State Energies • Kinetic • Potential • Internal (Independent of Path) Path Energies • Work • Heat (Depend on Path) 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. v1 F v2 F Initial state Dx Final state Kinetic Energy • A rigid mass accelerates from an initial velocity v1 to a final velocity v2 because of applied force, F. • Thus, an input of mechanical work causes the object to change its kinetic energy, Ek

  7. Kinetic Energy From the UAE Energyfinal - Energyinitial = Net Input Thus, for a constant force F: DEk = Net Mech. Work Input = FDx After applying Newton’s Laws we get DEk = ½ mv22 - ½ mv12 (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. Fup Fdn = mg Dx Gravitational Potential • Consider a rigid mass lifted vertically by a force F a distance Dx. • Thus, an input of mechanical work changes the object’s potential energy, Ep

  10. Gravitational Potential From the UAE: Energyfinal - Energyinitial = Net Input Thus, for a constant force F: DEp = Net Mech. Work Input = FDx The force acting on the mass is F = mg so DEp = mgDx

  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. Solution to Pairs Exercise #1 + M = 4000 kg 20 m + + Data

  13. Spring Potential Energy • Consider a force F compressing a spring a distance Dx. (See p. 587 Foundations of Engineering) • Thus, an input of mechanical work causes the spring’s potential energy to change. • From the UAE: Energyfinal - Energyinitial = Net Input DEp = 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. Solid Liquid Gas Plasma Review: States of Matter

  18. Pressure Liquid Pcritical Critical Point Plasma Solid Ptriple Triple Point Gas Vapor Ttriple Tcritical Temperature Review: Phase Diagram

  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, Ufinal - Uinitial = DU = Net Path Energy Input

  20. Heat Capacity for Constant- Volume Processes (Cv) • Heat is added to a substance of mass m in a fixed volume enclosure, which causes a change in internal energy, U. Thus, Q = U2 - U1 = DU = m CvDT The v subscript implies constant volume insulation DT Heat, Q added m m

  21. Dx DT Heat, Q added m m Heat Capacity for Constant- Pressure Processes (Cp) • 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.

  22. Cp Defined • Thus, Q = DU + PDV = DH = m CpDT The p subscript implies constant pressure • Note: H, enthalpy. is defined as U + PV, so dH = d(U+PV) = dU + VdP + PdV At constant pressure, dP = 0, so dH= dU + PdV For large changes at constant pressure DH = DU + PDV

  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): DEk + DEp + DU = Win - Wout + Qin - Qout • For an open system, with M defined as energy entering or leaving the system with the mass: DEk + DEp + DU = Win - Wout + Qin - Qout + Min - Mout

  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. Step 1 Step 2 Step 3 Sequential Energy Conversion E2 E3 E1 E4 h1 h2 h3 hoverall =h1h2h3

  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/s2)(X-Xtaut) • 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

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