32b- Exam 3 Review Session

ME200- Thermodynamics I 1Jeff Engerer

ME200: Exam 3 Review Session

The Second Law

Jeff Engerer

Fall 2014


• Three Key Points from Lectures 1-21

• Entropy Balance Equation

• Motivation for the Second Law

• Ideal Thermodynamic Cycles (and Carnot)

• State-Simplifying Assumptions for Entropy Calculations

• Examples

Overview


• Point 1: Identify the system, process, and states

– Terms in equations defined by boundary location

• Example: Heat Exchangers

– Process and States

• Point 2: Mass & Energy Balance and Simplifying Assumptions

• Point 3: Properties and Simplifying Assumptions

• Application: Solving First-Law Problems

What We Learned in Lectures 1-21


Identify the System, Process, and States

ProcessState 1 State 2

Tcold

Plow

Thot

Phighheating

Initial Properties Final Properties

Cold

Air In Hot Air

Out

Heating

Process occurring

over time(typical of closed

systems)

Process occurring

over space(typical of open systems)

In many cases (especially

thermodynamic cycles &

heat exchangers) this is

non-trivial!


• Learned to simplify the equations by the following assumptions:

– Closed System

– Steady State

– Adiabatic/Insulated

– No Work

– Change in Kinetic and/or Potential Energy Negligible

• Also, learned to time-integrate equations when necessary (common

for closed systems)

Apply Equation-Simplifying Assumptions

2 2- - / 2 / 2cvcv cv e ie i

e i

dEQ W m h V gz m h V gz

dt

CVi e

i e

dmm m

dt


Apply State-Simplifying Assumptions

• Liquid/Solid Property Approximation:

– v and u are independent of pressure:

– Can be used to find properties

• Incompressible Assumption:

– Above assumption, plus v is constant

• Ideal Gas Assumption

– High Temperature, Low Pressure

• Constant Specific Heat:

– Can be applied to any of the above (and more)

𝒗 = 𝒗(𝑻, 𝒑) ≈ 𝒗(𝑻)

𝒖 = 𝒖(𝑻, 𝒑) ≈ 𝒖(𝑻)

𝒉 𝑻, 𝒑 ≈ 𝒖𝒇(𝑻𝟏) + 𝒑𝟏𝒗𝒇(𝑻𝟏)

𝒗 = 𝒗𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒄𝒗 = 𝒄𝒑 = 𝒄

Pv RT

𝑢 = 𝑢(𝑇, 𝑝) ≈ 𝑢(𝑇)

( ) ( )h T u T RT

∆𝒖 = 𝒄𝒗∆𝑻

∆𝒉 = 𝒄𝒑∆𝑻


• Connecting the dots– Apply first law to processes

between states

• Find state from process variables (work/heat)

• Find process variables from the states (beginning and end)

– Use this technique to:

• Solve a problem

• Find information needed to solve a problem

– Remember how we used this in exam 2

How We Applied These Ideas

Heat Out(To Hot Reservoir)

Heat In(From Cold Reservoir)

Work

In

Expansion

(Throttling)

Valve

Condenser

Evaporator

Compressor

4

32

1

Applying state principle tells us if

a state is known or unknown


• When discussing entropy in class, we took a ‘bottom-up’

approach

– Motivation for the Second Law

– Irreversibilities

– Reversible Processes and Cycles

– Carnot Cycles

– Calculating/Looking up Entropy Values

– Entropy Balance Equation

• To help tie this information together, today we’ll look at this

from a ‘top-down’ approach

Entropy and The Second Law


• Shown Schematically:

The Entropy Balance Equation

Inlets Outlets

Via Heat

Transfer

Storage

jCVi i e e CV

j i ej

QdSm s m s

dt T

Generation

Review Slide


• Our second-law analysis will tie into our problem-solving

philosophy from the first weeks of the course

– Point 1: Identify the system, process, and states

• Terms in equations defined by boundary location

– Example: Interaction with a reservoir

• Process and States

– Point 2: Entropy Balance and Simplifying Assumptions

– Point 3: Properties and Simplifying Assumptions

Applying the Entropy Balance Equation


Simplifying Assumptions for Entropy

Balance

• The entropy balance can also be simplified, using the following

assumptions:

– Closed System

– Steady State

– Adiabatic/Insulated

– Reversible

• Also, learned to time-integrate this equation when necessary

(common for closed systems)

jCVi i e e CV

j i ej

QdSm s m s

dt T


• Recall how to use the tables:

– Subcooled Liquid Data (water only, limited pressure data)

– Saturated (Two-Phase) Data

– Superheated Vapor Data

– Ideal Gas Data

• Involves using so values (discuss in more detail later)

• Patterns in these tables reveal two general trends:

– Entropy Tends to Increases as the Temperature of the Substance

Increases

– Entropy Tends to Increase as the Specific Volume of the Substance

Increases

Entropy as a Property


• Next, we will look at second-law concepts with the entropy

balance equation in mind:

– Motivation for the Second Law

– Ideal Thermodynamic Cycles (and Carnot)

– State-Simplifying Assumptions for Entropy Calculations

– Examples

Review of Second Law Concepts


• Reason for the second law is very similar to the reason we use

the first law

• What does the first law do?

– Uses a fundamental property

• Energy

– Tells us what processes will and will not occur

• Energy is conserved (neither generated nor destroyed)

– Describes this mathematically for application to complex processes

Motivation for the Second Law

2 2- - / 2 / 2cvcv cv e ie i

e i

dEQ W m h V gz m h V gz

dt


• Possible Process:

– Electrical Resistance Heater

Possible vs Impossible: First Law

+

_

elecW

• Impossible Process:


+

_

elecW

CV CV CVE Q W

outQ *

outQ

out elecQ W

CV CV CVE Q W

*

out elecQ W

*

out elecQ WThe violation of the first law (and our

intuition) tells us this is impossible!


• Reason for the second law is very similar to the reason we use

the first law

• What does the second law do?

– Uses a fundamental property

• Entropy

– Tells us what processes will and will not occur

• Entropy cannot be destroyed (but can be generated)

– Describes this mathematically for application to complex processes

Motivation for the Second Law

jCVi i e e CV

j i ej

QdSm s m s

dt T


• Possible Process:


Possible vs Impossible: Second Law

• Impossible Process:


CV CV CVE Q W

+

_

elecW outQ

out elecQ W

CV CV CVE Q W

in elecQ W

First law is NOT violated, suggesting that

this process is possible (but our intuition

tells us that it isn’t!)

+

_

elecWinQ


• From previous slide, electrical generation by adding heat to a

resistor is impossible:

– Requires 2nd Law to prove this

• Second law shows that this is impossible for ANY device (not

just an electrical resistor)

– We will develop this idea into the

more formal Kelvin-Planck Statement

Possible vs Impossible: Second Law

+

_

elecWinQ

+

_

elecWinQ

Additional Requirement: The properties in the

control volume are not changing with time (i.e.

work is not coming from a storage term)


• There are three different ways of viewing the second law (all of

which are equivalent)

– Clausius Statement:

– Kelvin-Planck Statement:

– Entropy Statement:

Three Statements of the Second Law

0 Entropy cannot

be destroyed


• Recall these are processes that occur ‘on their own’ in one

direction, but not the other

• Some commonly encountered examples (in ME200):

– Depressurization

• Gas flowing through a throttling valve

– Transfer of Heat across a gradient

• Heat transfer from a hot reservoir to a cold reservoir

– Direct conversion of work into heat

• Electrical resistance heater

• Paddle wheel

Irreversible/Spontaneous Processes


• Reversible, Irreversible, and Impossible: Turbine Example

Applying to Irreversibilities:

Depressurization

jCVi i e e CV

j i ej

QdSm s m s

dt T

SS Adiabatic

Turbine

Work

Pressure

Source

Pressure

Sink

Turbine

Reversible Turbine

0CV

No

Work

Pressure

Source

Pressure

Sink

1

2

Irreversible Case

0CV

Turbine

Work

(too

large)

Pressure

Source

Pressure

Sink

Turbine

Impossible Case

0CV

For which,

we solved:What if it was a

compressor?


• This implies that there is some optimal thermodynamic cycle

for which the entropy production is zero

Entropy and the Heat Engine

Hot Reservoir

Cold Reservoir

TH

TC

QH

QC

QC=QH

0cv

An Irreversible Cycle

(Entropy

Generation)

Hot Reservoir

Cold Reservoir

TH

TC

QH

QC=0

Wcycle=QH

Hot Reservoir

Cold Reservoir

TH

TC

QH

QC

Wcycle

0cv

THE Reversible Cycle

(NO Entropy

Generation)0cv

An Impossible Cycle

(Entropy

Destruction)

WORST BEST Too

good to

be true!


• Apply the second law to the reversible cycle

Entropy and Thermodynamic Cycles

Hot Reservoir

Cold Reservoir

TH

TC

QH

QC

Wcycle

0cv


(NO Entropy

Generation)

jCVi i e e CV

j i ej

QdSm s m s

dt T


• Apply the second law to the reversible cycle

Entropy and Thermodynamic Cycles

Hot Reservoir

Cold Reservoir

TH

TC

QH

QC

Wcycle

0cv


(NO Entropy

Generation)

jCVi i e e CV

j i ej

QdSm s m s

dt T

SS ReversibleClosed

0 H C

H C

Q Q

T T

C C

revH H

cycle

Q T

Q T


• This relationship is the one we used to

establish the maximum thermal efficiency and

coefficients of performance for thermodynamic

cycles:

– Heat Engine

– Heat Pump

– Refrigerator

Maximum Performance

th,max 1cycle C

H H

W T

Q T

,C C

R rev

cycle H C

Q TCOP

W T T

,H H

HP rev

cycle H C

Q TCOP

W T T

C C

revH H

cycle

Q T

Q T

Maximum

Performance,

Maximum Work

Maximum

Performance,

Minimum Work


• The Carnot Cycle is a realization of these ideal thermodynamic

cycles:

Carnot Cycle


• The Carnot Cycle can also take the form of a vapor-power

cycle:

Carnot Cycle

Recall how this

varies from the

Rankine Cycle


• For the Carnot Cycle:

– Heat addition/removal occurs isothermally

– Compression/expansion occurs isentropically

– And, of course, all of which are reversible

Carnot Cycle on the T-S Diagram

Gas Piston

‘version’

isothermal

isentropic

What is the

significance of

the area under

the curves?


• As was shown in class, the T-ds Relations are:

• For each state simplifying assumption, we can apply this

relationship to find the change in entropy for:

– Incompressible Substance with constant specific heat

– Ideal Gas Equation

• With variable specific heat

• With constant specific heat

Entropy and State-Simplifying Assumptions

Tds du Pdv Tds dh vdP


• So for an incompressible substance:

• And for entropy:

– If specific heat is variable:

• Don’t use this integral in class

– If specific heat is constant:

• DO use this relationship in ME200

Incompressible Substance

𝒖 = 𝒖(𝑻, 𝒑) ≈ 𝒖(𝑻) 𝒉 𝑻, 𝒑 ≈ 𝒖𝒇(𝑻𝟏) + 𝒑𝟏𝒗𝒄𝒐𝒏𝒔𝒕

𝒗 = 𝒗𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒄𝒗 = 𝒄𝒑 = 𝒄

2

1

2 1

( )T

T

c Ts s dT

T

22 1

1

lnT

s s cT


• For an Ideal Gas:

• And for entropy:

– For variable specific heat:

– For constant specific heat:

Ideal Gas

22 1 2 1

1

( ) ( ) lnP

s s s T s T RP

Temperature dependent

portion (from tables)

Pressure dependent

portion (plug-in values)

2 22 1

1 1

ln lnp

T Ps s c R

T P 2 2

2 1

1 1

ln lnv

T vs s c R

T v OR

Pv RT

𝑢 = 𝑢(𝑇, 𝑝) ≈ 𝑢(𝑇)

( ) ( )h T u T RT


• Each term in the entropy balance equation depends on where

the boundary for the control volume is placed

• There is a special consideration for the entropy balance

equation:

– What temperature is used to describe the heat transfer term if the

temperature of the heat source/sink is different from the temperature of

the system?

Applying the Second Law to Problem

jCVi i e e CV

j i ej

QdSm s m s

dt T


• If the temperature of the system and the surroundings (i.e.

reservoir) are different, which temperature do we use?

– In other words, where do we place the boundary?

• Recall our discussion on internal irreversibilities:

– Where we place the boundary determines:

• The temperature of heat transfer term

• The magnitude of the entropy generation term

Where Do We Put the Boundary?

Tsys

Tsurr

TCV 1 CV 2

CV 1: External Irreversibility

• excludes

CV 2: Internal Irreversibility

• includes

T

T CV

CV T

Q

sys

Q

T

or

surr

Q

T

Review Slide


• An Isothermal Process:


Isothermal System

2 1 system gen

sat

QS S S

T

sat

Q

T

Thermal resv at TR

Q

Thermal resv at TR

System temperature is

constant (no integral) and

entropy generation is

excluded (process is

internally reversible)

Review Slide


• A non-isothermal process:


Non-isothermal System

2

2 11

j

system gen gen

j j R

Q QS S S

T T

Thermal reservoir at TR

Q, 0

reservoir

R

gen reservoir

QS

T

System temperature is not

constant, so to avoid

having to do an integral,

use the reservoir

temperature and include

the irreversibility

Review Slide


• Electrical work is applied to a resistor in a gas-cylinder. The

air is heated from 300 K to 400 K as the piston expands from

an initial volume of 0.3 m3 at a constant pressure of 5 MPa.

– Find the entropy generation in the cylinder

• Assume that the cylinder is well-insulated

• Use Variable Specific Heat

Second Law Analysis of a Gas-Piston

Heating Process

Air

Welect


• A vapor power cycle consists of the following devices/processes:– 1-2: Adiabatic Reversible Turbine

– 2-3: Reversible Condenser

– 3-4: Adiabatic Irreversible Pump

– 4-1: Irreversible Boiler

• The following is known:

– State 1: T1=356.2 °C, P1=0.70 MPa

– State 2: T2=25 °C (two-phase)

– State 3: T3=25 °C (sat. liquid)

– State 4: T4=30 °C, P4=0.70 MPa

– Mass flow rate: 10 kg/s

• Find:– Entropy Generation and Work for Pump

– Entropy Generation in Boiler

– Work produced by turbine

– QC by applying first law to entire cycle

– Show that condenser is reversible

– Compare thermal efficiency to Carnot efficiency

Second Law Analysis of a Vapor Power

Cycle

Heat In(From Hot Reservoir)

Heat Out(To Cold Reservoir)

Cycle

Work

Pump

Work

Boiler

Condenser

Pump

Turbine

For each process, assume kinetic and

potential energy are negligible.

Liquid in pump is incompressible with

constant specific heat

1

2

3

4

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32b- Exam 3 Review Session

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