Entropy - Energi Masa Depan Weblog

flinkexistenceMechanics

Oct 27, 2013 (3 years and 9 months ago)

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AH


Define entropy to quantify the 2nd
-
law
effects.


The increase of entropy principle
.


Calculate the entropy changes


Examine
isentropic processes


Reversible steady
-
flow work relations.


Isentropic efficiencies for steady
-
flow
devices.

3


When you open a bottle of perfume you
can smell the aroma as the molecules
leave the bottle and reach your nose.


Why don’t they spontaneously go back
into the bottle?


It would not violate the first law of
thermodynamics.


But
they most probably will not.


The explanation is
entropy
.

4

Entropy

is:


a
measure of the
disorder

of
a system.


a
measure of the energy in a system or process
that is unavailable to do work. In a reversible
thermodynamic process, entropy is expressed
as the heat absorbed or emitted divided by the
absolute temperature.



dS

=
(
dQ
/T
)
int

rev


Entropy
is a thermodynamic property (
state
function
)


Pertidaksamaan

Clausius
:



Engineers are usually concerned with the
changes in entropy.


Entropy change of a system during a process:





The relation between
Q
and
T
during a process
is often not available
.



Most cases we rely on tabulated data for
S
.


Entropy is a property


It has fixed values at
fixed states.


Entropy change

S
between two specified

states is the same
(whether reversible or
irreversible) during a
process.

7


Entropy,
S
, can be used to define the state of a
system, along with
P, T, V, U,
and

n.


We can calculate entropy only for reversible
processes.


To calculate entropy for an irreversible
process,
find a reversible process that takes the system
between the same two states and calculate the
entropy.


It will be the same as for the irreversible process
because it depends only on the two states.


Isothermal heat transfer processes are
internally reversible.


The entropy change of a system during an
internally reversible isothermal process :





or


A piston

cylinder device
contains a liquid

vapor mixture
of water at 300 K. During a
constant
-
temperature process,
750 kJ of heat is transferred to
the water. As a result, part of
the liquid in the cylinder
vaporizes. Determine the
entropy change of the water
during this process


From the
Clausius

inequality,























For an isolated system (adiabatic closed
system, Q = 0)



S
isolated

≥ 0


Total entropy change:



S
gen

=


S
total

=

S
syst

+

S
env

≥ 0


3
posibilities
:





The 2
nd

law: the entropy of an isolated system
never decreases. It either stays constant
(reversible process) or increases (irreversible).


Entropy in one part of the universe may decrease
in any process, the entropy of some other part of
the universe always increases by a greater
amount, so the total entropy always increases.


A heat source at 800
K loses 2000 kJ of
heat to a sink at (
a
)
500 K and (
b
)

750 K.
Determine which
heat transfer process
is more irreversible.

(a)
For the heat transfer process to a sink at 500 K:







(b)
Similarly, for process at 750 K:


Process (b) is less irreversible since it involves a
smaller
T

difference (smaller irreversibility).


In the saturated mixture region:


s

=
s
f

+
x
.
s
gf



(kJ/kg)


In the Compressed liquid region:



For a specified mass
m
:




Entropy is commonly used as a
coordinate on diagrams such as the
T
-
s
and

h
-
s
diagrams
.


A rigid tank contains 5 kg of refrigerant
-
134a initially at 20
°
C and 140
kPa
. The
refrigerant is now cooled while being
stirred until its pressure drops to 100
kPa
.
Determine the entropy change of the
refrigerant during this process.


The volume of the tank is constant,
V
2

= V
1
.


The properties of the refrigerant are:







At final state refrigerant is a saturated mixture
since
v
f

< v
2

< v
g

at 100 kPa, therefore:


Thus,

s
2

= s
f

+ x
2
.s
fg







=
0.07188 + 10.8592(10.879952)






= 0.8278 kJ/kg.K


The entropy change of the refrigerant





The
negative

sign indicates the entropy of the
system is decreasing. This is not a violation of
the second law, however, since it is
S
tot

that
cannot be negative



A process during which the entropy remains
constant is called an
isentropic process
:



s = 0

or


s
2

= s
1



(
kJ/kg.K)


Isentropic process can serve as an appropriate
model for actual processes.


A
reversible adiabatic
process is isentropic (
s
2

=
s
1
),

but an isentropic process is not necessarily
a reversible adiabatic. (The entropy increase of
a substance during a process as a result of
irreversibilities

may be offset by a decrease in
entropy due to heat losses)

Steam enters an
adiabatic turbine at
5
MPa

and 450
°
C
and leaves at a
pressure of 1.4
MPa
.
Determine the work
output of the turbine
per unit mass of
steam if the process
is reversible.


Take the
turbine
as the system (
control
volume
, since mass crosses the system
boundary during the process). There is
one inlet and one exit, thus:


The power output of the turbine is
determined from energy balance,








area under the curve =
Q
int

rev

Isentropic
proces

on T
-
s
diagram


commonly used in
engineering


valuable in the analysis
of steady
-
flow devices
such as turbines,
compressors, and
nozzles

For adiabatic steady
-
flow devices,
the

h
on an
h
-
s
diagram
is
a measure of work, and
the

s
is a measure
of
irreversibilities
.


Show the Carnot cycle on a
T
-
S diagram
and indicate the areas that represent the
heat supplied
Q
H
,
heat rejected
Q
L
,
and
the net work output
W
net,out

on this
diagram.

Carnot cycle is made
up of two reversible
isothermal (
T
constant)
processes
and two isentropic
(s
constant) processes

Area
A12B
represents
Q
H
, the area A43B
represents
Q
L
, and
the area in color

is
the net work since:


The entropy of a system can be considered a
measure of the disorder of the system
.


Then the second law of thermodynamics can be
stated as:

“Natural processes tend to move toward
a state of
greater disorder
.”


Jika

KIAMAT
adalah

puncak

kekacauan

alam

semesta



周Trmo獵灰ort

29

The
second law of thermodynamics

can be
stated in several equivalent ways:

(a)
Heat flows spontaneously from a hot object
to a cold one, but not the reverse.

(b)
There cannot be a 100
% efficient
heat
engine (one
that can change a given
amount of heat completely into
work).

(c)
Natural processes tend to move toward a
state of greater disorder or greater
entropy.

30


In any natural process, some energy becomes
unavailable to do useful work.


As time goes on, energy is degraded, in a sense; it
goes from more orderly forms (such as
mechanical) eventually to the least orderly form,
internal or thermal energy.


The amount of energy that becomes unavailable
to do work is proportional to the change in
entropy during any process.


A natural consequence of this is that over time,
the universe will approach a state of maximum
disorder
.

Heat Death
! (KIAMAT!)


Bahan
:
Bab

1, 2, 3, 4, 6, 7


Jumlah

soal
: 10 (
konsep

dan

esay
)


Tidak

boleh

menghidupkan

HP


Waktu

55
menit


Mulai

jam 8
tet

(no
ganjil
)


Mulai

jam 9 (no
genap
)


Tanggal

17 April