Transcript Lecture 1: Rotation of Rigid Body

Chapter 19: The Second Law of
Thermodynamics
Directions of thermodynamic processes
 Irreversible
and reversible processes
• Thermodynamic processes that occur in nature are all irreversible
processes which proceed spontaneously in one direction but not
the other.
• In a reversible process the system must be capable of being returned
to its original state with no other change in the surroundings.
• A reversible process proceeds slowly through equilibrium states.
Heat engines
 Heat engine
• Any device that transforms heat partly into work or mechanical
energy is called a heat engine.
• A quantity of matter inside the engine undergoes inflow and outflow of
heat, expansion and compression, and sometimes change of phase.
• This matter inside the engine is called the working substance of the
engine.
• The simplest kind of engine to analyze is one in which the working
substance undergoes a cyclic process.
• All heat engines absorb heat from a source at a relatively high
temperature, perform some mechanical work, and discard or reject
some heat at a lower temperature.
U  U2  U1  0  Q  W  Q  W
For a cyclic process
Heat engines
 Energy flow diagram for a heat engine
TH
Temperature of hot reservoir
QH  0 Heat flow from hot reservoir
per cycle
TC
Temperature of cold reservoir
QC  0 Heat flow from cold reservoir
per cycle waste!
Net heat Q absorbed per cycle:
Q  QH  QC  QH  QC
Net work done W per cycle:
W  Q  QH  QC  QH  QC
Heat engines (cont’d)
 Energy flow diagram for a heat engine (cont’d)
Net heat Q absorbed per cycle:
Q  QH  QC  QH  QC
Net work done W per cycle:
W  Q  QH  QC  QH  QC
Thermal efficiency e:
e
W
Q
Q
 1 C  1 C
QH
QH
QH
Heat engines (cont’d)
 Steam engines (external combustion)
Internal-combustion engines
 Gasoline engine (a heat engine)
• First a mixture of air and gasoline vapor flows into a cylinder through an open intake
valve while the piston descends, increasing the volume of the cylinder from a
minimum of V to a maximum of rV (r: compression ratio 8-10).
• At the end of intake stroke, the intake valve closes and the mixture is compressed,
approximately adiabatically, back to volume V during the compression stroke.
• Then the mixture is ignited by the spark plug, and the heated gas expands,
approximately adiabatically, back to volume rV, pushing on the piston-power stroke.
Internal-combustion engines (cont’d)
 The Otto cycle
b  c : QH  nCV (Tc  Tb )  0
d  a : QC  nCV (Ta  Td )  0
e
QH  QC Tc  Tb  Ta  Td

QH
Tc  Tb
a  b : Ta (rV ) 1  TbV  1
c  d : Td (rV ) 1  TcV  1
e
e
Td r
r
 1
 1
 1
a  b : compression stroke
 1
 Ta r  Ta  Td (Td  Ta )(r  1)

 1
 1
Td r  Ta r
(Td  Ta )r 1
1
r 1
 1
1
r  1
thermal efficiency
in Otto cycle
r=8,=1.4->e=56%
(adiabatic compression)
b  c : ignite fuel (heating at
constant volume)
c  d : power stroke
(adiabatic expansion)
d  a : Reject heat to environment
(cooling at constant volume)
Internal-combustion engines (cont’d)
 The Otto cycle
Internal-combustion engines (cont’d)
 The Diesel cycle
a  b : Qab  0, U ab  nCV (Tb  Ta ),
Wab  U ab  nCV (Tb  Ta )
b  c : Qbc  nCp (Tc  Tb ), U ab  nCV (Tc  Tb ),
Wbc  Qbc  U bc  n(C p  CV )(Tc  Tb )
c  d : Qcd  0, U cd  nCV (Td  Tc ),
Wcd  U cd  nCV (Td  Tc )
d  a : Wda  0, U da  nCV (Ta  Td ),
a  b : compression stroke
Qda  U cd  Wda  nCV (Ta  Td )
(adiabatic compression)
b  c : ignite fuel (heating at
Wnet  Wab  Wbc  Wcd  Wda
constant pressure)
c  d : power stroke
 nCp (Tc  Tb )  nCV (Td  Ta )
(adiabatic expansion)
d  a : Reject heat to environment
Wnet
Td  Ta
Va
Va
e
 1
(  56%  15,  5)
(cooling at constant volume)
Q
 (T  T )
V
V
H
c
b
b
c
Internal-combustion engines (cont’d)
 The Diesel cycle
Refrigerators
 Refrigerator
A refrigerator works as a reversed heat engine.
Refrigerators
 Energy flow of a refrigerator
A refrigerator takes heat from low
temperature reservoir and gives it
off to high temperature reservoir.
W  W , QH  QH
From the first law of thermodynamics
QH  QC  W  0  QH  QC  W
QH  QC  W
 Coefficient of performance
QC
QC
TC
K

(
W
QH  QC
TH  TC
Carnot refrigerator)
Refrigerators
 Air conditioner
A air conditioner works on exactly
the same principle as a refrigerator.
For air conditioners, the quantities of
greatest practical importance are the
rate of heat removal (the heat current
H from the region being cooled) and
the power input P=W/t.
QC
QC
Ht H
K



W
QH  QC Pt P
H in Btu/h, P in watts
H/P (energy efficiency rating EER)
in (Btu/h)/W. Note 1 W=3.413 Btu/h.
EER~7-10.
The second law of thermodynamics
 The second law of thermodynamics
It is impossible for any system to undergo a process in which it absorbs
heat from a reservoir at a single temperature and converts the heat
completely into mechanical work, with the system ending in the same
state in which it began (Kelvin-Planck statement)
• The kinetic and potential (due to interactions between molecules)
energies associated with random motion constitute the internal energy.
• When a body sliding on a surface comes to rest as a result of friction,
the organized motion of the body is converted to random motion of
molecules in the body and in the surface.
• Since we can’t control the motions of individual molecules, we can’t
convert this random motion COMPLTELY back to organized motion.
It is impossible for any process to have as its sole result the transfer of
heat from a cooler to a hotter body (Clausius statement).
The second law of thermodynamics
(cont’d)
 Proof : Kelvin-Planck statement = Clausius statement
impossible
impossible
(b) An impossible 100% efficient engine
(a) A workless refrigerator violates
2nd law (K-P statement). If it existed, violates C statement. If it existed, it
could be used to make a workless
it could be used to make a 100%
refrigerator, which violates C stateefficient engine, which violate K-P
ment.
statement.
K-P statement
C-statement
The second law of thermodynamics (cont’d)
 The second law of thermodynamics
• The conversion of work to heat, as in friction or viscous fluid flow,
and heat flow from hot to cold across a finite temperature gradient,
are irreversible processes.
• Two equivalent statements of the 2nd law state that these processes
can be only partially reversed
• Gases, for example, always seep spontaneously through an opening
from a region of high pressure to a region of low pressure.
• The 2nd law is an expression of the inherent one-way aspect of this
and many other irreversible processes.
• To have the maximum efficiency of a heat engine, we must avoid
all irreversible processes.
Answer: Carnot cycle
• In any process in which the temperature of the working substance
of the engine is intermediate between TH and TC , there must be
NO heat transfer between the engine and either reservoir because
such heat transfer could not be reversible.
• Any process in which the temperature of the working substance
changes must be adiabatic.
• As heat flow through a finite temperature drop is an irreversible,
during heat transfer, there must be NO finite temperature difference.
The Carnot cycle
 The Carnot cycle
reversible expansion
adiabatic compression
adiabatic expansion
reversible compression
The Carnot cycle
 The Carnot cycle : ideal gas approximation
a  b : Q  QH  0, W  0
Vb
isothermal
Q

W

nRT

n
(
)
H
ab
H
expansion
Va
c  d : Q  QC  0, W  0
isothermal
Vd
Vc
Q

W

nRT

n
(
)


nRT

n
(
)  0 (Vc  Vd )
expansion
C
cd
C
C
Vc
Vd
 TC 
QC
TC n(Vc / Vd )
 ( )
  
QH
TH n(Vb / Va )
 TH 
Adiabatic processes:
TH Vb 1  TCVc 1 , TH Va 1  TCVa 1  (Vb / Va ) 1  (Vc / Vd ) 1  Vb / Va  Vc / Vd
QC
TC

QH
TH
e  1
TC TH  TC

TH
TH
thermal efficiency
of a Carnot engine
The Carnot cycle (cont’d)
 The Carnot refrigerator
A reversed Carnot engine = a Carnot refrigerator
QC
QC
QC / QH
K


,
W
QH  QC 1  QC / QH
TC
K
TH  TC
QC TC

QH TH
coefficient of performance of a Carnot refrigerator
When the temperature difference is small, K is much larger than
unity, in which case a lot of heat can be pumped from the lower
to the higher temperature with only a little expenditure of work.
The Carnot cycle (cont’d)
 The Carnot cycle and the second law
More efficient
than a Carnot
engine
impossible
A Carnot refrigerator
Since each step in the Carnot
cycle is reversible, the entire
cycle may be reversed. If you
run the entire backward, the
engine becomes a refrigerator.
The Carnot cycle (cont’d)
 The Carnot cycle and the second law (cont’d)
• The refrigerator does work W=-|W|,
takes in heat QC from the cold reservoir,
and expels hear |QH| to the hot reservoir.
• The superefficient heat engine expels heat
|QC|, but to do so, it takes in a greater
amount of heat QH+. Its work output is
then W+.
• The net effect of the two machines together
is to take a quantity of heat  and convert
it completely into work, which violates the
2nd law.
• No engine can be more efficient than a Carnot engine.
• All Carnot engines operating between the same two temperatures
have the same efficiency, irrespective of the nature of the working
substance.
Entropy
 Entropy and disorder
Entropy provides a quantitative measure of disorder.
Define the infinitesimal entropy change dS during an infinitesimal
reversible process at absolute temperature T as:
dQ
dS 
T
If a total amount of heat Q is added during a reversible isothermal
process at absolute temperature T, the total entropy change is given
by:
S  S2  S1 
Q
T
• Higher temperature means greater randomness of motion
• Adding heat Q causes a substantial fractional increase in molecular
motion and randomness
Entropy (cont’d)
 Entropy and disorder (cont’d)
Generalization of definition of entropy change is to include ANY
reversible process leading from one state to another, whether
it is isothermal or not:
Let us represent the process as a series of infinitesimal reversible
steps. During a step, an infinitesimal quantity of heat dQ is added
to the system at absolute temperature T. Then the change of
entropy for the entire process is:
S  
2
1
dQ
T
entropy change in a
reversible process
• The change in entropy does not depend on the path leading from
the initial to the final state but is the same for all possible processes.
• Since entropy is a function only of the current state of the system,
we can compute entropy change in irreversible processes using the
above formula. ( we need to invent a path connecting the given initial
and final state that consist entirely of reversible, equilibrium processes
and compute the total entropy change as for the actual path).
Entropy (cont’d)
 Entropy in cyclic processes
For any reversible process
(e.g. Carnot cycle):
b
a
dQ
dQ
a T  b T  0
I
II
b
a
b
dQ
dQ
dQ
a T   b T  a T
I
II
II
dQ

0
T
The entropy of a system in
a given state is independent
of the path taken to get there,
and is thus a state variable.
Entropy (cont’d)
 Entropy in irreversible processes
Consider a combined system with a Carnot engine and
a cyclic system.
dQR  (dWrev  dWsys )  dUC
heat reservoir TR
dQR
dQR dQ
dQ

 TR
 dQR Carnot
cycle
TR
T
T
dQ
dWC  TR
 dU C
T
WC  TR 
dQ
  dU C ,  dU C  0 ( cyclic )
T
combined cyclic system
dWC
Carnot
engine
dWrev
dQ
T
Cyclic
system
dWsys
Entropy (cont’d)
 Entropy in irreversible processes
Consider a combined system with a Carnot engine and
a cyclic system.
WC  TR 
dQ
  dU C ,  dU C  0 ( cyclic )
T
heat reservoir TR
dQ
T
If WC > 0, a cyclic device exchanging
heat with a single heat reservoir and
producing an equivalent amount of work.
- Violation of K-P statement of 2nd lawWC  TR 
WC  0  
dQ
0
T
combined cyclic system
dQR
-the 2nd lawThe equality
is true for a
reversible process.
Clausius inequality
Carnot
engine
dWrev
dQ
T
Cyclic
system
dWsys
Entropy (cont’d)
 Entropy in irreversible processes (cont’d)
Processes: aAb and aBb reversible
dQ
dQ
dQ
T

0


aAbBa T
aAb T bBa T  0
Process : aCb
irreversible
• A reversible heat transfer causes changes
in entropy of both the system and the
A
reservoir (at least one needed except for
adiabatic process)
C
• In an irreversible heat transfer process,
B
there is a finite source of energy instead
of an energy reservoir, and the temperature
difference during the heat transfer process is finite.
S
dQ
dQ
dQ

Tx
aAbCa
0
aAb
T

bCa
Tx
0
dQ
dQ
dQ
dQ

 0 ( 

 0 subtracted)
bBa T
bCa T
aAb T
bBa T
x
 
dQ
bCa T
x
 Sba  Sa  Sb  
 dS 
For an isolated system
dQ=0, so dS>=0.
dQ
(> for irreversible and = for reversible process)
T
Entropy (cont’d)
 Entropy and the second law
When all systems taking part in a process are included, the entropy
either remains constant or increases.
OR
No process is possible in which the total entropy decreases, when all
systems taking part in the process are included.
Exercises
Problem 1
A Carnot engine takes 2000 J of heat from a reservoir at 500 K, does
some work, and discards some heat to a reservoir at 350 K.
How much work does it do, how much heat is discarded, and what
is the efficiency?
Solution
The heat discarded by the engine is:
TC
350K
 (2000J )
 1400 J .
TH
500K
From the first law, the work done by the engine is:
QC  QH
W  QH  QC  2000J  (1400J )  600 J .
The thermal efficiency is:
TC
350K
e  1
 1
 0.30
TH
500K
Exercises
Problem 2
One kilogram of water at 0oC is heated to 100oC. Compute its
change in entropy.
Solution
In practice, the process described would be done irreversibly, perhaps
by setting a pan of water on an electric range whose cooking surface is
maintained at 100oC. But the entropy change of the water depends only
on the initial and final states of the system, and is the same whether
the process is reversible or not. Hence we can imagine that the temp.
of the water is increased reversibly in a series of infinitesimal steps,
in each of which the temp is raised by an infinitesimal amount dT.
dQ  m cdT
S  S2  S1  
2
1
T2
dQ
dT
T2
  mc
 m cn  1.31 103 J / K .
T1
T
T
T1
Example: Refrigerator
• A refrigerator pumps heat from the inside of the freezer (-5°C) to
the room (25°C). What is the maximum coefficient of
performance?
TL  5o C  273  268 K
TH  25 o C  273  298 K
Kideal
TC
268K


 8.9
TH  TC 298K  268K
• i.e. 8.9 Joules of heat would be pumped from the freezer for
every Joule of work done by the compressor. (typical K = 3-5)
Exercises
Problem 3
What is the entropy
change in a free
expansion process,
when the volume is
doubled.
Solution
The work done by n moles of ideal gas in an isothermal expansion
from V1 to V2 is:
W  nRTn(V2 / V1 )
Q  W  nRT n
2V
 nRT n 2.
V
Therefore the entropy change for n=1 is:
S 
Q
 nRn 2  (1mol )[ 8.314 J /(mol  K )]( n 2)  5.76 J / K .
T
Problem 4
Exercises
A physics student immerses one end of a copper rod in boiling water at
at 100oC and the other end in an ice-water mixture at 0oC. The sides of the
rod are insulated. After steady-state conditions have been achieved in the
rod, 0.160 kg of ice melts in a certain time interval. For this time interval find
(a) the entropy change of the boiling water; (b) the entropy change of the
ice-water mixture; (c) the entropy change of the copper rod; (d) the total
entropy change of the entire system.
Solution
(a) S=Q/T=-mLf/T=-(0.160 kg)(334x103 J/kg)/(373.15 K)=-143 J/K.
(b) S=Q/T=mLf/T=(0.160 kg)(334x103 J/kg)/(273.15 K)= 196 J/K.
(c) From the time equilibrium has been reached, there is no net heat
exchange between the rod and its surroundings, so the entropy
change of the copper rod is zero.
(d) 196 J/K-143 J/K=53 J/K.
Exercises
Problem 5
An experimental power plant at the Natural Energy Lab generates
electricity from the temperature gradient of the ocean. The surface and
deep-water temperatures are 27oC and 6oC, respectively. (a) What is
the maximum theoretical efficiency of this power plant? (b) If the power
plant is to produce 210 kW of power, at what rate must heat be extracted
from the warm water? At what rate must heat be absorbed by the cold
water? (c) The cold water that enters the plant leaves it at a temperature
of 10oC. What must be the flow rate of cold water through the system?
Solution
279 .15 K
 7.0%
(a) e  1 
300 .15 K
(b) Pout / e  210 kW / 0.070  3.0 MW , 3.0 MW  210 kW  (1/ e  1)(210 kW )  2.8 MW.
6
(c) dm  d Q / dt  (2.8  10 W )(3600s / h)  6  105 kg / h  6  105  / h.
dt
cT
[4190J /(kg  K )](4 K )