Transcript Chapter 22-1 - UCF Physics

Chapter 22
Heat Engines, Entropy and the
Second Law of Thermodynamics
First Law of Thermodynamics –
Review
• The first law states that a change in internal
energy in a system can occur as a result of
energy transfer by heat, by work, or by both.
• The law makes no distinction between the
results of heat and the results of work.
First Law of Thermodynamics –
Review
• In according with first law of thermodynamics the
energy is always conserved
• What does it mean to conserve energy if the
total amount of energy in the universe does not
change regardless of what we do?
• The first law of thermodynamic does not tell the
whole story: Energy is always conserved, but
some forms of energy are more useful than
others.
First Law – Missing Pieces
• There is an important distinction between
heat and work that is not evident from the
first law
• The first law makes no distinction between
processes that occur spontaneously and
those that do not
– An example is that it is impossible to design a
device that takes in energy and converts it all
to work
The Second Law of Thermodynamics
• The possibility or impossibility of putting energy to
use is the subject of the Second Law of
Thermodynamics.
• For example, it is easy to convert work into thermal
energy, but it is impossible to remove energy as
heat from a single reservoir and convert it internally
into work with no other changes.
• This experimental fact is one statement of the
second law of thermodynamics.
The Second Law of Thermodynamics
• Kelvin’s statement: No system can take energy
as heat from a single reservoir and convert it
entirely into work without additional net
changes in the system or its surroundings.
• Clausus statement: A process whose only net
result is to transfer energy as heat from a
cooler object to a hotter one is impossible.
The Second Law of Thermodynamics
• A common example of conversion of work into heat
is movement with friction.
• Suppose you spend two minutes pushing a block
along a tabletop in a closed path, leaving the block
in its initial position. Also, suppose that the blocktable system is initially in thermal equilibrium with
the surroundings.
The Second Law of Thermodynamics
• The work you do on the system is converted into
internal energy of the system and the block-table
system becomes warmer → the system is no
longer in thermal equilibrium with its
surroundings → the system will transfer energy
as heat to its surroundings until it returns to the
thermal equilibrium
The Second Law of Thermodynamics
• Because the final and initial states of the system
are the same, the 1-st Law of TD dictates that the
energy transferred to the environment as heat
equals to the work done on the system.
• The reverse process never occur – a block and
table that are warm will never spontaneously cool
by converting their internal energy into the work
that causes the block to push your hand around the
table!
The Second Law of Thermodynamics
• Such an amazing occurrence would not violate the
first law of thermodynamics or any other physical
law, it does, however, violate the second law of
thermodynamics.
• There is a lack of symmetry in the roles played by
heat and work that is not evident from the first law.
This lack of symmetry is related to the fact that
some processes are irreversible.
The Second Law of Thermodynamics
• Establishes which processes do and which do
not occur
• Some processes can occur in either direction
according to the first law
• They are observed to occur only in one direction
according to the second law.
Irreversible Processes
• An irreversible process is one that occurs
naturally in one direction only
• No irreversible process has been observed to
run backwards
• An important engineering implication is the
limited efficiency of heat engines
Heat Engine
• A heat engine is a device that takes in energy by
heat and, operating in a cyclic process, expels a
fraction of that energy by means of work
• A heat engine carries some working substance
through a cyclical process
Heat Engine.
• The working substance
absorbs energy by heat from
a high temperature energy
reservoir (Qh)
• Work is done by the engine
(Weng)
• Energy is expelled as heat to
a lower temperature reservoir
(Qc)
Heat Engine.
• Since it is a cyclical process,
ΔEint = 0
– Its initial and final internal
energies are the same
• Therefore, Qnet = Weng
• The work done by the engine
equals the net energy
absorbed by the engine
• The work is equal to the area
enclosed by the curve of the
PV diagram
– The working substance is a
gas
Thermal Efficiency of a Heat Engine
• Thermal efficiency is defined as the
ratio of the net work done by the engine
during one cycle to the energy input at
the higher temperature
e
Weng
Qh
Qh  Qc
Qc

 1
Qh
Qh
• We can think of the efficiency as the
ratio of what you gain to what you give
More About Efficiency
• In practice, all heat engines expel only a
fraction of the input energy by mechanical
work
• Therefore, their efficiency is always less
than 100%
– To have e = 100%, energy expelled as heat to
a lower temperature reservoir, QC, must be 0
Perfect Heat Engine
• No energy is expelled to
•
•
•
the cold reservoir
It takes in some amount
of energy and does an
equal amount of work
e = 100%
It is an impossible
engine
During each cycle a heat engine absorbs 200 J of heat
from a hot reservoir, does work, and exhausts 160 J to a
cold reservoir. What is the efficiency of the engine?
During each cycle a heat engine absorbs 200 J of heat
from a hot reservoir, does work, and exhausts 160 J to a
cold reservoir. What is the efficiency of the engine?
W

Qh
Qh= 200J
W = Qh – Qc = 200J – 160J = 40J
W
40 J


 0.20  20%
Qh 200 J
Heat Pumps and Refrigerators
• Heat engines can run in reverse
– This is not a natural direction of energy transfer
– Must put some energy into a device to do this
– Devices that do this are called heat pumps or
refrigerators
• Examples
– A refrigerator is a common type of heat pump
– An air conditioner is another example of a heat
pump
Heat Pump Process
• Energy is extracted from
•
•
the cold reservoir, QC
Energy is transferred to
the hot reservoir, Qh
Work must be done on
the engine, W
Coefficient of Performance
• The effectiveness of a heat pump is
described by a number called the
coefficient of performance (COP)
• In heating mode, the COP is the ratio of
the heat transferred in to the work required
energy transferred at high temp Qh
COP =

work done by heat pump
W
COP, Heating Mode
• COP is similar to efficiency
• Qh is typically higher than W
– Values of COP are generally greater than 1
– It is possible for them to be less than 1
• We would like the COP to be as high as
possible
COP, Cooling Mode
• In cooling mode, you “gain” energy from
a cold temperature reservoir
Qc
COP 
W
• A good refrigerator should have a high
COP
– Typical values are 5 or 6
A refrigerator has a coefficient of performance
equal to 5.00. The refrigerator takes in 120 J of
energy from a cold reservoir in each cycle. Find (a)
the work required in each cycle and (b) the energy
expelled to the hot reservoir.
A refrigerator has a coefficient of performance
equal to 5.00. The refrigerator takes in 120 J of
energy from a cold reservoir in each cycle. Find (a)
the work required in each cycle and (b) the energy
expelled to the hot reservoir.
Qc
C O P  refrigerator 
W
(a) Q c  120 J C O P  5.00
(b)
W  24.0 J
Heat expelled = Heat removed + Work done.
Q h  Q c  W  120 J 24 J 144 J
Carnot Engine
• A theoretical engine developed by Sadi
Carnot
• A heat engine operating in an ideal,
reversible cycle (now called a Carnot
cycle) between two reservoirs is the most
efficient engine possible
– This sets an upper limit on the efficiencies of
all other engines
Carnot’s Theorem
• No real heat engine operating between
two energy reservoirs can be more
efficient than a Carnot engine operating
between the same two reservoirs
– All real engines are less efficient than a
Carnot engine because they do not operate
through a reversible cycle
– The efficiency of a real engine is further
reduced by friction, energy losses through
conduction, etc.
Carnot Cycle
Overview of
the processes
in a Carnot
cycle
Carnot Cycle, A to B
• A → B is an isothermal expansion
• The gas is placed in contact with
•
•
the high temperature reservoir, Th
The gas absorbs heat |Qh|
The gas does work WAB in raising
the piston
Carnot Cycle, B to C
• B → C is an adiabatic expansion
• The base of the cylinder is
•
•
replaced by a thermally insulated
wall
No heat enters or leaves the
system
The temperature falls from Th to
Tc . The gas does work WBC
Carnot Cycle, C to D
• The gas is placed in contact with
•
•
•
the cold temperature reservoir
C → D is an isothermal
compression
The gas expels energy Qc
Work WCD is done on the gas
Carnot Cycle, D to A
• D → A is an adiabatic
•
compression
The gas is again placed against
a thermally nonconducting wall
– So no heat is exchanged with the
surroundings
• The temperature of the gas
increases from Tc to Th
• The work done on the gas is WDA
Carnot Cycle, PV Diagram
 The work done by the


engine is shown by
the area enclosed by
the curve, Weng
The net work is equal
to |Qh| – |Qc|
DEint = 0 for the entire
cycle
Efficiency of a Carnot Engine
• Carnot showed that the efficiency of the engine
depends on the temperatures of the reservoirs
QC
TC

Qh Th
TC
and eC  1 
Th
• Temperatures must be in Kelvins
• All Carnot engines operating between the same
two temperatures will have the same efficiency
Notes About Carnot Efficiency
• Efficiency is 0 if Th = Tc
• Efficiency is 100% only if Tc = 0 K
– Such reservoirs are not available
– Efficiency is always less than 100%
• The efficiency increases as Tc is lowered
and as Th is raised
• In most practical cases, Tc is near room
temperature, 300 K
– So generally Th is raised to increase efficiency
Carnot Heat Pump COPs
• In heating mode: COP 
• In cooling mode:
Qh
Th

W
Th  Tc
Qc
Tc
COP 

W
Th  Tc
A Carnot heat engine uses a steam boiler at 100C as the
high-temperature reservoir. The low-temperature
reservoir is the outside environment at 20.0C. Energy is
exhausted to the low-temperature reservoir at the rate of
15.4 W. (a) Determine the useful power output of the heat
engine. (b) How much steam will it cause to condense in
the high-temperature reservoir in 1.00 h?
A Carnot heat engine uses a steam boiler at 100C as the
high-temperature reservoir. The low-temperature
reservoir is the outside environment at 20.0C. Energy is
exhausted to the low-temperature reservoir at the rate of
15.4 W. (a) Determine the useful power output of the heat
engine. (b) How much steam will it cause to condense in
the high-temperature reservoir in 1.00 h?
Qc
Qc
Tc
Tc
The efficiency is: ec  1
 1
 Dt
Qh
Th
Qh
Th
Dt
Q c Th
 273  100 K

 15.4 W
 19.6 W
Dt
Dt Tc
 273  20 K
Qh
A Carnot heat engine uses a steam boiler at 100C as the hightemperature reservoir. The low-temperature reservoir is the
outside environment at 20.0C. Energy is exhausted to the lowtemperature reservoir at the rate of 15.4 W. (a) Determine the
useful power output of the heat engine. (b) How much steam
will it cause to condense in the high-temperature reservoir in
1.00 h?
(a)
Q h  W eng  Q c
The useful power output is:
W eng
Dt

Qh
(b) Q h
Dt

Qc
Dt
 19.6 W  15.4 W  4.20 W
 Qh 

Dt m LV

 Dt 


Q h Dt
3600 s
2
m
 19.6 J s 

3.
12

10
kg

6
Dt LV
 2.26  10 J kg 
Gasoline Engine
• In a gasoline engine, six processes occur
during each cycle
• For a given cycle, the piston moves up
and down twice
• This represents a four-stroke cycle
• The processes in the cycle can be
approximated by the Otto cycle
Otto Cycle
• The PV diagram of an
•
Otto cycle is shown at
right
The Otto cycle
approximates the
processes occurring in
an internal combustion
engine
The Conventional Gasoline Engine
Gasoline Engine –
Intake Stroke
• During the intake stroke,
•
•
•
the piston moves
downward
A gaseous mixture of air
and fuel is drawn into the
cylinder
Energy enters the system
as potential energy in the
fuel
O → A in the Otto cycle
Gasoline Engine –
Compression Stroke
• The piston moves upward
• The air-fuel mixture is
•
•
•
compressed adiabatically
The temperature increases
The work done on the gas
is positive and equal to the
negative area under the
curve
A → B in the Otto cycle
Gasoline Engine
– Spark
• Combustion occurs when
•
•
•
•
the spark plug fires
This is not one of the strokes
of the engine
It occurs very quickly while
the piston is at its highest
position
Conversion from potential
energy of the fuel to internal
energy
B → C in the Otto cycle
Gasoline Engine
– Power Stroke
• In the power stroke, the
•
•
•
•
gas expands adiabatically
This causes a temperature
drop
Work is done by the gas
The work is equal to the
area under the curve
C → D in the Otto cycle
Gasoline Engine
– Valve Opens
• This is process D → A in the
•
•
•
•
Otto cycle
An exhaust valve opens as
the piston reaches its bottom
position
The pressure drops suddenly
The volume is approximately
constant
– So no work is done
Energy begins to be expelled
from the interior of the
cylinder
Gasoline Engine –
Exhaust Stroke
• In the exhaust stroke,
•
•
•
the piston moves
upward while the
exhaust valve remains
open
Residual gases are
expelled to the
atmosphere
The volume decreases
A → O in the Otto
cycle
Otto Cycle Efficiency
• If the air-fuel mixture is assumed to be an
ideal gas, then the efficiency of the Otto
cycle is
e  1
1
V1
V2 
g 1
g is the ratio of the molar specific heats
• V1/V2 is called the compression ratio
Otto Cycle Efficiency
• Typical values:
─ Compression ratio of 8
─ g = 1.4
─ e = 56%
• Efficiencies of real engines are 15% to
20%
– Mainly due to friction, energy transfer by
conduction, incomplete combustion of the airfuel mixture
Diesel Engines
• Operate on a cycle similar to the Otto cycle
•
•
•
without a spark plug
The compression ratio is much greater and so
the cylinder temperature at the end of the
compression stroke is much higher
Fuel is injected and the temperature is high
enough for the mixture to ignite without the spark
plug
Diesel engines are more efficient than gasoline
engines
A 1.60-L gasoline engine with a compression ratio
of 6.20 has a useful power output of 102 hp.
Assuming the engine operates in an idealized Otto
cycle, find the energy taken in and the energy
exhausted each second. Assume the fuel-air
mixture behaves like an ideal gas with γ= 1.40.
A 1.60-L gasoline engine with a compression ratio of 6.20
has a useful power output of 102 hp. Assuming the engine
operates in an idealized Otto cycle, find the energy taken in
and the energy exhausted each second. Assume the fuel-air
mixture behaves like an ideal gas with γ= 1.40.
eO tto  1
1
V1 V2 
g 1
1
1
 1
 1
0.400
7
5

1


 6.20
 6.20
eO tto  0.518
We have assumed the fuel-air mixture to behave like a
diatomic gas.
Now
e
W eng
Qh

W eng t
Qh t
A 1.60-L gasoline engine with a compression ratio of 6.20
has a useful power output of 102 hp. Assuming the engine
operates in an idealized Otto cycle, find the energy taken in
and the energy exhausted each second. Assume the fuel-air
mixture behaves like an ideal gas with γ= 1.40.
746 W 1 hp
Q h W eng t

 102 hp
t
e
0.518
Qh
 146 kW
t
Q h  W eng  Q c
Q h W eng


t
t
t
Qc
 746 W 
3
 146  10 W  102 hp 
 70.8 kW

t
 1 hp 
Qc
Irreversibility and Disorder
 There are many irreversible processes that can
not be described by the heat-engine or
refrigerator statements of the second law, such
as a glass falling to the floor and breaking or a
balloon popping.
 However all irreversible processes have one
thing in common – the system and its
surroundings moves towards a less ordered
state.
 Suppose a box containing a gas of mass M at temperature
T is moving along a frictionless table with a velocity vcm.
The total kinetic energy of the gas has two components:
that associated with the movement of the center of the
mass ½Mvcm2, and the energy of the motion of its
molecules relative to its center of the mass
 The center of the mass energy ½Mv2cm is ordered
mechanical energy that could be converted entirely
into work.
 For example, if a weight were attached to a moving
box by a string passing over a pulley, this energy
could be used to lift the weight.

The relative energy of the molecules is the
internal thermal energy of the gas, which is related
to its temperature T. It is random, non-ordered
energy that can not be converted entirely into work.
 Now, suppose that the block hits a fixed wall and stops. This
inelastic collision is clearly an irreversible process. The ordered
mechanical energy of the gas is converted into random internal
energy and the temperature of the gas rises. The gas still has
the same total energy, but now all of the energy is associated
with the random motion of the molecules about of center of the
mass of the gas, which is now at rest.
Entropy
 Thus, the gas become less ordered (more disordered), and
gas lost some of its ability to do work.
 There is a thermodynamic function called entropy S that is a
measure of the disorder of the system.
 Entropy S, like pressure P, volume V, temperature T, and
internal energy Eint, is a function of the state of the system.
 The change in entropy dS of the system as it goes from one
state to another is defined as
dQrev
dS 
T
where dQrev is the energy that must be transferred to the
system as heat in the reversible process that bring the
system from the initial state to the final state.
Entropy of an Ideal Gas
 Consider an arbitrary reversible quasi-static process in
which a system consisting of an ideal gas adsorbs an
amount of heat dQrev.
 According to the first law, dQrev is related to dEint and W by
dEint  dQrev  dW  dQrev  PdV
or
dV
Cv dT  dQrev  nRT
V
If we will divide each term by T:
dT dQrev
dV
Cv

 nR
T
T
V
Entropy of an Ideal Gas
dT dQrev
dV
Cv

 nR
T
T
V
dQrev
dT
dV
dS 
 Cv
 nR
T
T
V
We will assume Cv to be constant and by integrating we will find
the entropy change of an ideal gas that undergoes a reversible
expansion from an initial state of volume V1 and temperature T1
to a final state of volume V2 and temperature T2
DS 

dQ
T2
V2
 Cv ln  nR ln
T
T1
V1
Entropy and the Second Law
• Entropy is a measure of disorder
• The entropy of the Universe increases
in all real processes
– This is another statement of the second law of
thermodynamics
Entropy and Heat
• The original formulation of entropy deals
with the transfer of energy by heat in a
reversible process
• If dQrev is the amount of energy transferred
by heat when a system follows a reversible
path then the change in entropy, DS is
dQr
DS 
T
Entropy and Heat
• The change in entropy depends only on
the endpoints and is independent of the
path followed
• The entropy change for an irreversible
process can be determined by calculating
the change in entropy for a reversible
process that connects the same initial and
final points
More About Change in Entropy
• dQrev is measured along a reversible path,
even if the system may have followed an
irreversible path
• The meaningful quantity is the change in
entropy and not the entropy itself
• For a finite process,
f
DS   dS  
i
i
f
dQr
T
Change in Entropy
• The change in entropy of a system going
from one state to another has the same
value for all paths connecting the two
states
• The finite change in entropy depends only
on the properties of the initial and final
equilibrium states
DS for a Reversible Cycle
• S = 0 for any reversible cycle
• In general,
dQrev
0
T

– This integral symbol indicates the integral is over a
closed path
Entropy Changes in Irreversible
Processes
• To calculate the change in entropy in a
real system, remember that entropy
depends only on the state of the system
• Do not use Q, the actual energy transfer in
the process
– Distinguish this from Qrev , the amount of
energy that would have been transferred by
heat along a reversible path
– Qrev is the correct value to use for DS
Entropy Changes in Irreversible
Processes
• In general, the total entropy and therefore
the total disorder always increases in an
irreversible process
• The total entropy of an isolated system
undergoes a change that cannot
decrease
– This is another statement of the second law of
thermodynamics
Entropy Changes in Irreversible
Processes
• If the process is irreversible, then the total
entropy of an isolated system always
increases
– In a reversible process, the total entropy of an
isolated system remains constant
• The change in entropy of the Universe
must be greater than zero for an
irreversible process and equal to zero for a
reversible process
Heat Death of the Universe
• Ultimately, the entropy of the Universe should
•
•
reach a maximum value
At this value, the Universe will be in a state of
uniform temperature and density
All physical, chemical, and biological processes
will cease
– The state of perfect disorder implies that no energy is
available for doing work
– This state is called the heat death of the Universe
DS in Thermal Conduction
• The cold reservoir absorbs Q and its entropy
•
•
•
changes by Q/Tc
At the same time, the hot reservoir loses Q and
its entropy changes by -Q/Th
Since Th > Tc , the increase in entropy in the cold
reservoir is greater than the decrease in entropy
in the hot reservoir
Therefore, DSU > 0
– For the system and the Universe
DS in a Free Expansion
• Consider an adiabatic free expansion
• Q = 0 and cannot be used since that is for
an irreversible process
DS  
i
f
dQr 1

T
T

i
f
dQr
DS in Free Expansion
• For an isothermal process, this becomes
DS  nR ln
Vf
Vi
• Since Vf > Vi , DS is positive
• This indicates that both the entropy and the
disorder of the gas increase as a result of the
irreversible adiabatic expansion
DS in Calorimetric Processes
• The process is irreversible because the system
•
goes through a series of nonequilibrium states
Assuming the specific heats remain constant
and no mixing takes place:
DS  m1c1 ln
Tf
Tc
 m2 c2 ln
Tf
Th
– If mixing takes place, this result applies only to
identical substances
DS will be positive and the entropy of the Universe
increases
Entropy on a Microscopic Scale
• We can treat entropy from a microscopic
•
viewpoint through statistical analysis of
molecular motions
A connection between entropy and the number
of microstates (W) for a given macrostate is
S = kB ln W
– The more microstates that correspond to a given
macrostate, the greater the entropy of that macrostate
• This shows that entropy is a measure of disorder
Entropy, Molecule Example
• One molecule in a two-sided container has a 1-in-2
•
•
chance of being on the left side
Two molecules have a 1-in-4 chance of being on the
left side at the same time
Three molecules have a 1-in-8 chance of being on
the left side at the same time
Entropy, Molecule Example Extended
• Consider 100 molecules in the container
• The probability of separating 50 fast
molecules on one side and 50 slow
molecules on the other side is (½)100
• If we have one mole of gas, this is found to
be extremely improbable
Entropy, Marble Example
• Suppose you have a bag with 50 red
marbles and 50 green marbles
• You draw a marble, record its color, return
it to the bag, and draw another
• Continue until four marbles have been
drawn
• What are possible macrostates and what
are their probabilities?
Entropy, Marble Example, Results
• The most ordered are the least likely
• The most disorder is the most likely
What change in entropy occurs when a 27.9-g
ice cube at –12C is transformed into steam at
115C?
What change in entropy occurs when a 27.9-g
ice cube at –12C is transformed into steam at
115C?
f
273 K
dQ 273 K m cicedT
273 K
DS  
 
 m cice  T 1dT  m cice ln T 261 K
T
T
i
261 K
261 K
  273 
DS  0.027 0 kg  2 090 J kg C   ln 273 K  ln 261 K   0.027 0 kg  2 090 J kg C   ln 

  261 
DS  2.54 J K
As the ice melts its entropy change is
Q m Lf
DS  

T
T

  32.9 J K
0.027 0 kg 3.33  105 J kg
273 K
What change in entropy occurs when a 27.9-g
ice cube at –12C is transformed into steam at
115C?
As liquid water warms from 273 K to 373 K,
f
DS  
i
 Tf 
 373
 m cliquid ln    0.027 0 kg  4 186 J kg C  ln 
  35.3 J K

273
 Ti 
m cliquid dT
T
As the water boils and the steam warms,
 Tf 
m Lv
DS 
 m csteam ln  
T
 Ti 
DS 

0.027 0 kg 2.26  106 J kg
373 K
  0.027 0 kg  2 010 J kg C  ln  388  164 J K  2.14 J K


373
The total entropy change is
 2.54  32.9 35.3 164  2.14
J K  236 J K