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...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 focusCurrent focus

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)

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.

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

Initial stateF

v1

F

v2Dx

Final state

Kinetic EnergyFrom the UAE

Energyfinal - Energyinitial = Net Input

Thus, for a constant force F:DEk = Net Mech. Work Input = FDx

After applying Newton’s Laws we getDEk = ½ mv2

2 - ½ mv12

(see p. 584 Foundations of Engineering for derivation)

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)

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

Fup

Fdn = mgDx

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

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?

20 mM = 4000 kg

+

+

+ Data

Solution to Pairs Exercise #1

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

Spring Potential Energy

This time the force is not constant along x. By Hook’s Law, the relationship is

F=kxwhere k is the spring constant and x=0 is the uncompressed (relaxed) state.

)( 21

222

1

2

1

2

1

xxk

kxdx

FdxE

x

x

x

x

p

Internal Energy The energy stored inside the

medium. The energy associated with

translational, rotational, vibrational, and electronic potential energy of atoms and molecules.

Internal Energy

Translation

Rotation

Vibration

MolecularInteractions

Review: States of Matter

Solid Liquid

Gas Plasma

Review: Phase Diagram

Plasma

Gas

Vapor

Liquid

Solid

Ttriple Tcritical

Ptriple

Pcritical

Pressure

Temperature

Critical Point

TriplePoint

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

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 Cv DTThe v subscript implies constant volume

Heat, Qaddedm m

DTinsulation

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.

Heat, Qadded

DT

m m

Dx

Cp Defined

Thus,Q = DU + PDV = DH = m Cp DT

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

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

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

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.

ininoutoutp gzmgzmE 2

212

21

ininoutoutk vmvmE

m

Sequential Energy Conversion

Step 1 Step 2 Step 3E1

E2 E3E4

h1 h2 h3

hoverall =h1h2h3

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)

Individual Exercise #1 Solve the problem outlined on the

next slides Document your steps Solution will be turned in

separately

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...

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...

Bungee Jumping Exercise1. 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.

15 minutes

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

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