32b- exam 3 review session
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ME200TRANSCRIPT
ME200- Thermodynamics I 1Jeff Engerer
ME200: Exam 3 Review Session
The Second Law
Jeff Engerer
Fall 2014
ME200- Thermodynamics I 2Jeff Engerer
• 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
ME200- Thermodynamics I 3Jeff Engerer
• 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
ME200- Thermodynamics I 4Jeff Engerer
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!
ME200- Thermodynamics I 5Jeff Engerer
• 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
ME200- Thermodynamics I 6Jeff Engerer
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
ME200- Thermodynamics I 7Jeff Engerer
• 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
ME200- Thermodynamics I 8Jeff Engerer
• 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
ME200- Thermodynamics I 9Jeff Engerer
• 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
ME200- Thermodynamics I 10Jeff Engerer
• 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
ME200- Thermodynamics I 11Jeff Engerer
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
ME200- Thermodynamics I 12Jeff Engerer
• 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
ME200- Thermodynamics I 13Jeff Engerer
• 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
ME200- Thermodynamics I 14Jeff Engerer
• 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
ME200- Thermodynamics I 15Jeff Engerer
• Possible Process:
– Electrical Resistance Heater
Possible vs Impossible: First Law
+
_
elecW
• Impossible Process:
– Electrical Resistance Heater
+
_
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!
ME200- Thermodynamics I 16Jeff Engerer
• 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
ME200- Thermodynamics I 18Jeff Engerer
• Possible Process:
– Electrical Resistance Heater
Possible vs Impossible: Second Law
• Impossible Process:
– Electrical Resistance Heater
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
ME200- Thermodynamics I 19Jeff Engerer
• 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)
ME200- Thermodynamics I 20Jeff Engerer
• 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
ME200- Thermodynamics I 21Jeff Engerer
• 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
ME200- Thermodynamics I 22Jeff Engerer
• 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?
ME200- Thermodynamics I 23Jeff Engerer
• 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!
ME200- Thermodynamics I 24Jeff Engerer
• Apply the second law to the reversible cycle
Entropy and Thermodynamic Cycles
Hot Reservoir
Cold Reservoir
TH
TC
QH
QC
Wcycle
0cv
THE Reversible Cycle
(NO Entropy
Generation)
jCVi i e e CV
j i ej
QdSm s m s
dt T
ME200- Thermodynamics I 25Jeff Engerer
• Apply the second law to the reversible cycle
Entropy and Thermodynamic Cycles
Hot Reservoir
Cold Reservoir
TH
TC
QH
QC
Wcycle
0cv
THE Reversible Cycle
(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
ME200- Thermodynamics I 26Jeff Engerer
• 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
ME200- Thermodynamics I 27Jeff Engerer
• The Carnot Cycle is a realization of these ideal thermodynamic
cycles:
Carnot Cycle
ME200- Thermodynamics I 28Jeff Engerer
• The Carnot Cycle can also take the form of a vapor-power
cycle:
Carnot Cycle
Recall how this
varies from the
Rankine Cycle
ME200- Thermodynamics I 29Jeff Engerer
• 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?
ME200- Thermodynamics I 30Jeff Engerer
• 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
ME200- Thermodynamics I 31Jeff Engerer
• 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
ME200- Thermodynamics I 32Jeff Engerer
• 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
ME200- Thermodynamics I 33Jeff Engerer
• 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
ME200- Thermodynamics I 34Jeff Engerer
• 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
ME200- Thermodynamics I 35Jeff Engerer
• An Isothermal Process:
Where Do We Put the Boundary?
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
ME200- Thermodynamics I 36Jeff Engerer
• A non-isothermal process:
Where Do We Put the Boundary?
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
ME200- Thermodynamics I 37Jeff Engerer
• 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
ME200- Thermodynamics I 38Jeff Engerer
• 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