tme 221 2nd law
TRANSCRIPT
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Termodinamika Teknik
Hukum ke Dua Termodinamika
@2014 Harjadi Gunawan
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Todays main concepts:
Understand the need for and the usefulness of the 2ndlaw Be able to write and use the entropy balance
Be able to predict the maximum possible efficiency and COP of
power and refrigeration or heat pump cycles, respectively.
Be able to provide several different expressions which explain
the 2ndLaw of Thermodynamics.
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The First Law of Thermodynamics in an energy balance and tells us the
magnitudes energy flows.
3Sec 5.1: Introducing the Second Law
2
CV V
2CV CV
dEQ W m h gz
dt
It does not say anything about the spontaneousdirection.
Can we spontaneously cool the
refrigerator by removing heat
from the environment?
Q
out
Qin
Can we spontaneously heat the house by
removing heat from the environment?
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4Sec 5.1: Introducing the Second Law
The fridge will warm.
Qout
The house will cool.
We know intuitively, that the spontaneousdirection for heat flow is from
warm to cold.
Qin
If TA> TBTA= TB.
But, we have refrigeration and heating, so what is required to
change the direction of heat flow?
Work needs to be done to move in the non-spontaneousdirection
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5Sec 5.1: Introducing the Second Law
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6Sec 5.1: Introducing the Second Law
The Second Law of Thermodynamics answers the following
Which direction the process will move spontaneously?
What is equilibrium?
What is the maximum efficiency?
What parameters can be changed to move closer to
the maximum efficiency?
What is temperature?
How can we measure uand h?
Since the Second Law answers all these
questions, there is not a single statement
of the second law.
Popular forms of the Second Law include:
Everything degrades to chaos
No perpetual motion machine
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7Sec 5.2: Statements of the Second Law
Clausius Statement:
It is impossible for any system to operate in such a
way that the sole result would be an energytransfer by heat from a colder to a hotter body.
Kelvin-Planck StatementIt is impossible for any system to operate in
a thermodynamic cycle and deliver a net
amount of energy by work to its surroundings
while receiving heat transfer from a single
thermal reservoir. Q Thermal Reservoir:
A body where energy exchange as
heat does not effect the temperature.
Kelvin-Planck Statement tells us that
It is not possible to convert heat completely to work.
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8Sec 5.2: Statements of the Second Law
Entropy Statement:
It is impossible for any system to operate in a way that
entropy is destroyed.
entropywithin
system
Net entropy
transferred
into the
system
Entropy
generated
with system
[ ]= +
[ ] [ ]Note: Entropy can be generated (unlike mass)
Most processes do not operate at Ssys = 0, so generally, Ssys
sys genQST
( + 0 - ) ( + 0 - ) ( + or 0 )
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9
Energy Balance:
For a Closed Systems:
sys
sys sys
dEQ W
dt
sysdE
dt
Q W
Entropy Balance:
sys gen
QS
T
Entropy Rate Balance:
sys
gen
dS Q
dt T
sys sys sysE Q W
Energy Rate Balance:
sysdS
dt
Q
T
gen
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Process Entropy
Change
Entropy
Transfer
Entropy
Production
Possible,
Impossible, or
Indeterminate.
A > 0 0 if > 0 Possible
B < 0 if < 0 > 0 Possible
C 0 > 0 Impossible
D > 0 > 0 Possible
E 0 < 0 if > 0 Possible
F > 0 < 0 Impossible
G < 0 < 0 Possible
10
Problem 5: 3 Classify the following processes of a closed system as
possible, impossible, or indeterminate.
--------------------------------------------------------------------------------------------------------
SYS SYS
QS
T
Entropy
production
Entropy
transfer
Entropy
change
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11Sec 5.2: Statements of the Second Law
An alternate way of looking at Entropy:Entropy, S, is a measure of the Disorderwithin a system
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12Sec 5.3: Identifying Irreversibility
Reversible process:
Both the system and surroundings can bereturned to the initial state.
One grain of sand
is removed.
Gas
Sand
The grain of sandis replaced.
Irreversible process:
System and surroundings cannot bereturned to the original state.
Gas
Sand
Gas
Sand
Gas
Rock
Removerock
Irreversibility can be
internal to the system
external to the system
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14Sec 5.4: Interpreting the Kelvin-Planck Statement
Kelvin-Planck Statement
It is impossible for any system to operate in a thermodynamic
cycle and deliver a net amount of energy by work to itssurroundings while receiving heat transfer from a single
thermal reservoir.
0cycleW< 0 : Internal irreversibility present
= 0 : No internal irreversibility
For a single reservoir:
The inequality (
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15Sec 5.6: Second Law Aspects of Power Cycles
Efficiency of a Power Cycle interacting with two reservoirs.
H
C
Hot
Cold
in
out
in
outin
in
cycle
QQ
QQ
QQ
QQQ
Q
W
111
Second Law statements for power cycles(Carnot Corollaries):
The thermal efficiency of an irreversible
power cycle is always less than the thermal
efficiency of a reversible power cycle when
each operates between the same two
thermal reservoirs.
All reversible power cycles operating
between the same two thermal reservoirs
have the same maximum thermal efficiency.
If QC 0, then 1.0 = 100%
16S 5 7 S d L A f R f i i d H P
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16Sec 5.7: Second Law Aspects of Refrigeration and Heat Pumps
COP of refrigerationand heat pumps cycles interacting with two reservoirs
CH
H
cycle
in
CH
C
cycle
out
QQ
Q
W
Q
QQ
Q
W
Q
&
As WCycle0, then and
Second Law statements for refrigeration:
The coefficient of performancee of an
irreversible refrigeration cycle is always less
than the coefficient of performance of a
reversible refrigeration cycle when each
operates between the same two thermalreservoirs.
All reversible refrigeration cycles operating
between the same two thermal reservoirs
have the same maximum coefficient of
performance.
Such an arrangement would violate the Clausius Statement.
17S 5 8 T t S l
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17Sec 5.8: Temperature Scales
The Celsius Temperature Scale:
0Cfreezing point of water
(at 1 atm)100Cboiling point of water
(at 1 atm)
The Fahrenheit Temperature Scale:
0Ffreezing point of water &NaCl solution (at 1 atm)100Faverage normal human body
temperature (at 1 atm)
It is desirable to have a scale that is notdependent on a single substance.
Phase changes of many substancesallows extension of the scale based onproperties
18S 5 8 T t S l
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18
The driving force for heat transfer is a TThis causes a heat transfer QT
Sec 5.8: Temperature Scales
HCreversibleH
C TTQ
Q,
Choose
H
C
T
T
H
C
reversibleH
C
T
T
Q
Q
Thus
The ratio of the temperatures is equal to the ratio of the
heat rejected and the heat absorbed.
Used to define the Kelvin temperature scale.
reversibleTPQ
QT
16.273
19S 5 8 T t S l
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19
Consider three heat engines , operating reversibly
Sec 5.8: Temperature Scales
CHH
C TT
Q
Q,1
ICI
C TTfQ
Q,
1
2
3
1
2
3 IHI
H
TTfQ
Q,
2
3
IC
IH
IC
IH
TTf
TTf
QQ
QQ
,
,
C
H
C
H
C
H
T
T
Tf
Tf
Q
Q
Therefore
20Sec 5 9 : Ma im m Performance
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20
With
Sec 5.9 : Maximum Performance
H
C
T
T1max
H
C
reversibleH
C
T
T
Q
Q
The Carnot efficiency For a
reversible cycle
increase TH
decrease TC
Thus, ideally we want to
maximize T= (TH - TC)
TH limited by equipment costs
(highpand high T)
To increase
21Sec 5 9 : Maximum Performance
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21Sec 5.9 : Maximum Performance
Typically, the cold reservoir is the atmosphere and large bodies of water
and thus TC= 298 K, since it will cost too much to use a refrigerated
reservoir.
What is the maximum efficiency of a power cycle operating between
TH= 745 K
and TC = 298 K
298
1 1 60%745C
H
T
T
22Sec 5 9 : Maximum Performance
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22
Example: (5.19) A power cycle operating at steady state receives energy by heattransfer at a rate of QHat TH=1000 K and rejects energy by heat transfer to a coldreservoir at a rate QCat TC= 300 K.
For each of the following cases, determinewhether the cycle operates reversibly,
irreversibly, or is impossible.
Sec 5.9 : Maximum Performance
a) QH= 500 kW, QC=100 kW
2001 1 0.60
500
Cd
H
Q
Q
b) QH= 500 kW, Wcycle= 250 kW,
and QC= 200 kW
d) QH= 500 kW, QC= 200 kWc) Wcycle= 350 kW, QC= 150 kW
max3001 1 0.701000
C
H
T
T
100
1 1 0.80500
C
a
H
Q
Q cycle H C W Q Q
3500.70
350 150
cycle cycle
c
H cycle C
W W
Q W Q
250 500 200 300kW kW
?b
Impossible: a > max
Impossible
Reversible: c = max Irreversible: a < max
Max system efficiency:
23Sec 5 9 : Maximum Performance
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23
Example: (5.63) The refrigerator shown in the figure operates at
steady state with a coefficient of performance of 4.5 and a power
input of 0.8 kW. Energy is rejected from the refrigerator to the
surroundings at 20C by heat transfer from the metal coils whoseaverage surface temperature is 28C. Determine
Sec 5.9 : Maximum Performance
W=0.8kW
TH= 20C= 293 K
TC= ?
(a) The rate energy is rejected, in kW
(b) The lowest theoretical temperature
within the refrigerator, in K
(c) The maximum theoretical power, in
kW, that could be developed from a
power cycle operating between the
coils and surroundings.
=4.5
24Sec 5 9 : Maximum Performance
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Example: (5.63) Determine
Sec 5.9 : Maximum Performance
W=0.8kW
TH= 20C= 293 K
TC= ?
(a) The rate energy is rejected, in kW
=4.5max
cycle
in
W
Q
cycle H C W Q Q
max H cycle
cycle
Q W
W
COP of refrigeration cycle
Work Energy Balance:
C H cycleQ Q W
Combining:
max1 0.8 1 4.5 4.4H cycleQ W kW kW
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26Sec 5 9 : Maximum Performance
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Example: (5.63) Determine
Sec 5.9 : Maximum Performance
W=?
TH= 28C= 301 K
TH= 20C= 293 K
c) The maximum theoretical power, in kW, that
could be developed from a power cycle operating
between the coils and surroundings.
max 1 cycleC
H H
WT
T Q
kWQH 4.4
1 Ccycle H H
TW Q
T
Note: Such a power cycle wouldoperate between reservoir
temperatures of 20 oC and 28 oC.
Maximum Possible Power Cycle Efficiency:
Solve for possible Work done:
293
(4.4 ) 1 (4.4 ) 0.0266 0.117
301
kW kW kW
From part a) the rejected heat is