# Termodinamika Teknik Kimia I

Urban and Civil

Nov 29, 2013 (4 years and 5 months ago)

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DICKY

DERMAWAN

www.di c kyder ma wan.ne t7 8.ne t

di ckyder mawan@gma i l.com

ITK
-
233

Termodinamika Teknik Kimia I

3
SKS

4
-

The Second Law: Entropy

Process Direction

No apparatus can operate in such a way that its
only

effect
(in system & surroundings) is to convert heat absorbed by a
system
completely

into work done by the system

It is impossible by a cyclic process to convert the heat
absorbed by a system

completely
into work done by the
system.

No process is possible which consist
solely

in the transfer of
heat from one temperature level to a higher one.

Partial

conversion of heat to into work is the basis for nearly all
commercial production of power.

Heat Engine

Heat engines produce work from heat in a cyclic
process

Essential to all heat
-
engine cycles are:

Absorption of heat into the system at a high temperature

Rejection of heat to surroundings at a lower temperature

Production of work

Conversion of heat to work always accompanied by
heat release to surroundings

Heat Engine

C
H
out
in
Q
Q
Q
Q
W

Thermal efficiency:

absorbed
heat
output
work
nett

H
C
H
C
H
H
Q
Q
1
Q
Q
Q
Q
W

For

o⁢eun楴yⰠQ
C

must be zero. No engine has ever been built
for which this is true; some heat is always rejected.

Carnot’s Theorem & Carnot Engine

Carnot engine operates
between two heat
reservoirs

in such a way that all heat
absorbed is absorbed at the
constant temperature of
the hot reservoir

and all heat
rejected

at the
constant temperature of
the cold reservoir

For two given heat reservoir, no engine can have a thermal
efficiency higher than that of a Carnot engine.

Carnot Cycle

Step1: A system at T
C

undergoes a
reversible

temperature rise to T
H

Step 2: The system maintains contact with
the hot reservoir T
H
, and undergoes a
reversible

isothermal process during which
heat Q
H

is absorbed from the hot reservoir

Step 3: The system undergoes a
reversible

adiabatic process in the opposite direction
of step 1 that brings its temperature back to
that of the cold reservoir at T
C
.

Step 4: The system maintains contact with the reservoir at T
C
, and undergoes
reversible

isothermal process in the opposite direction of the step 2 that returns
it to its initial state with heat rejection of Q
C

to the cold reservoir

Carnot Engine
-

Reversibility

Any reversible engine operating between 2 heat
reservoirs is a Carnot engine.

An engine operating on a different cycle must
necessarily transfer heat across finite temperature
difference and therefore cannot be reversible.

Carnot Cycle with an Ideal Gas as Working Fluid

Step1: a
-

b

Reversible

temperature rise from T
C

to T
H

Step 2: b

c

Reversible
isothermal expansion to arbitrary
point c with heat absorption of Q
H

Step 3: c

d

Reversible

temperature decrease T
C
.

Step 4: d

a

Reversible

isothermal compression to the initial state with heat rejection of Q
C

Entropy

Our analysis showed that

or

The equation suggests the existence of a property whose changes are given by
the quantities Q/T

When the
isotermal

steps are infinitesimal, the heat quantities become
dQ
:

or

Thus the quantities
dQ
rev
/T sum to zero for the arbitrary cycle, exhibiting the
characteristic of a property. We call this property as entropy, S.

0
T
Q
T
Q
C
C
H
H

C
H
C
H
T
T
Q
Q

0
T
dQ
T
dQ
C
C
H
H

0
T
dQ
rev

T
dQ
dS
rev

dS
T
dQ
rev

There exists a property called entropy S, which is an intrinsic property of a
system, functionally related to the measurable coordinates which
characterize the system

Characteristic of Entropy

Entropy owes its existence to the second law, from which it
arises in much the same way as internal energy does from the
first law.

The change in entropy of any system undergoing a finite
reversible

process is

When a system undergoes an
irreversible

process between
two states, the entropy change of the system is evaluated to an
arbitrary chosen reversible
process that accomplishes the
same change of state as the actual process. The integration is
not

carried out for the irreversible path.

T
dQ
S
rev

Characteristic of Entropy

Since entropy is a state function, the entropy changes of the
irreversible and reversible processes are identical.

The entropy change of a system caused by the transfer of heat
can always be calculated by

dQ
/T whether the heat transfer
is accomplished reversibly or irreversibly.

When a process is irreversible on account of finite differences
in other driving forces, such as pressure, the entropy changes
is not caused solely by heat transfer, and for its calculations
one must devise a reversible means of accomplishing the same
change of state.

Calculation of Entropy

For one mole of fluid undergoing a mechanically reversible
process in a closed system, using the first law, the defining
equation for enthalpy one find:

For an ideal gas:

Although derived for a mechanically reversible process, this
equation is a general equation for the calculation of entropy
changes, since it relates properties only and independent of
process causing the change of state.

T
dP
V
T
dH
T
dQ
dS
rev

0
T
T
ig
p
P
P
ln
T
dT
R
C
R
S
0

Example

Calculate
Δ
S for each step of the cycle shown below.
Assume ideal gas with constant C
p
.

Problem 5.6

A quantity of an ideal gas, C
p

= 7/2 R at 20
o
C & 1 bar
and having a volume of 70 m
3

is heated at constant
pressure to 25
o
C by the transfer of heat from a heat
reservoir at 40
o
C. Calculate the heat transfer to the
gas, the entropy change of the heat reservoir, the
entropy change of the gas, and

S
total
.

Problem 5.7

A rigid vessel of 0.05 m
3

volume contains an ideal gas,
C
v

= 5/2 R, at 500 K and 1 bar.

a.
If heat in the amount of 12000 J is transferred to
the gas, determine its entropy change

b.
If the vessel is fitted with a stirrer that is rotated by
a shaft so that work on the amount of 12000 J is
done on the gas, what is the entropy change of the
gas if the process is adiabatic? What is

S
total
?

Example 5.3

Methane gas at 550 K & 5 bar undergoes a reversible
adiabatic expansion to 1 bar. Assuming methane to
be an ideal gas at these conditions, determine its
final temperature.

Heat Engine

C
H
out
in
Q
Q
Q
Q
W

Thermal efficiency:

absorbed
heat
output
work
nett

H
C
H
C
H
H
Q
Q
1
Q
Q
Q
Q
W

For

o⁢eun楴yⰠQ
C

must be zero. No engine has ever been built
for which this is true; some heat is always rejected.

Mathematical Statement of The Second Law

0
S
total

This mathematical statement of the second law affirms
that every process proceeds in such a direction that the
total entropy change associated with it is positive.

The limiting value of zero being attained only by a
reversible process.

No process is possible for which the total entropy
decreases.

Thermodynamic Efficiency

In a process producing work, there is an absolute maximum
work attainable which is accomplished by completely
reversible process. For irreversible process,

W
actual

produced

<
W
ideal

In a process requiring work, there is an absolute minimum
amount of work required which is accomplished by
completely reversible process. For irreversible process,
W
ractual

required

>
W
ideal

t
ideal
s
W
W

ideal
t
s
W
W

Problem 5.12

An ideal gas, C
p

= 7/2 R, undergoes a cycle consisting of the
following mechanically reversible steps:

a.
1
, V
1
, T
1

to P
2
, V
2
, T
2

b.
An isobaric expansion from P
2
, V
2
, T
2
to P
3

= P
2
, V
3
, T
3

c.
3
, V
3
, T
3
to P
4
, V
4
, T
4

d.
A constant
-
volume process from P
4
, V
4
, T
4

to P
1
, V
1

= V
4
, T
1

Sketch this cycle on a PV diagram and determine its thermal
efficiency if T
1

= 500 K, T
2

= 800 K, T
3

= 2000 K, and T
4

=
1000 K

Problem 5.13

A reversible cycle executed by 1 mol of an ideal gas for which
C
p

= 5/2 R consist of the following processes:

a.
Starting at 600 K & 2 bar, the gas is cooled at constant
pressure to 300 K

b.
From 300 K & 2 bar, the gas is compressed isothermally
to 4 bar

c.
The gas returns to its initial state along a path for which
the product PT is constant.

What is the thermal efficiency of the cycle?

Example 5.4

A 40
-
kg steel casting (Cp = 0.5 kJ/(
kg.K
) at a
temperature of 450oC is quenched in 150 kg of oil
(Cp = 2.5 kJ/(
kg.K
) at 25oC. If there are no heat
losses, what is the change of entropy of:

a.
The casting

b.
The oil

c.
Both considered together

Problem 5.11

A piston/cylinder device contains 5 mol of an ideal gas,
C
p

= 5/2 R, at 20
o
C & 1 bar. The gas is compressed
reversibly and adiabatically to 10 bar, where the
piston is locked in position. The cylinder is then
brought into thermal contact with a heat reservoir at
20
o
C, and heat transfer continues until the gas also
reaches this temperature. Determine the entropy
changes of the gas, the reservoir, and

S
total

Irreversibility

One mole of an ideal gas, Cp = 7/2 R is compressed
adiabatically in a piston/cylinder device from 2 bar &
25
o
C to 7 bar. The process is irreversible and requires
35% more work than a reversible adiabatic
compression from the same initial state to the same
final pressure. What is the entropy change of the
gas?

Classical Lost Work & Process Analysis

The ideal work is the maximum amount of work which
can be done by the process by operating reversibly
within the system and by transferring heat between
the system and the surroundings reversibly.

The lost work:

For processes containing several units, lost
-
work
calculations can be made for each unit and summed
to determine the overall value.

actual
ideal
lost
W
W
W

Q
S
T
W
s
lost

Example 4.7

Steam enters a turbine at 1.5
MPa

& 500
o
C and
exhausts at 0.1
MPa
. The turbine delivers 85% of the
shaft work of a reversible
-
it is neither reversible nor adiabatic. Heat losses to
the surroundings at 20
o
C are 9 kJ/kg steam.

Determine temperature, entropy change of the steam
leaving the turbine, & the lost work

Example 4.8

Assume that 5000 kg/h of oil with a heat capacity of
3.2 kJ/
kg.K

is to be cooled from 220 to 40
o
C, using a
large quantity of water which can be assumed to be
at a constant temperature of 30
o
C. Determine the
lost work in the process and the thermodynamic
efficiency of the process.

Example 4.9

Assume that 100 kg of methane gas/h is adiabatically
compressed from 0.5
MPa

& 300 K to 3.0
MPa

&
500 K after which it is cooled
isobarically

to 300 K
by a large amount of water available at 290 K.
Determine the efficiency of the compressor.

If the surroundings are assumed to be at 290 K, do a
thermodynamic analysis of the process.

Assume ideal gas.

Cp = 35.58 J/
mol.K

Cv

= 27.27 J/
mol.K

8

Steam
dengan

tekanan

12 bar
dan

temperatur

200
o
C

diekpansi

satu

tahap

dalam

sebuah

turbin

sehingga

tekanannya

1,5 bar

kondisi

jenuhnya
.
Turbin

bekerja

secara

dengan

efisiensi

85%.
Perkirakan

(
dalam

sistem

satuan

SI):

Kualitas

uap

keluar

turbin

Kebutuhan

uap

air agar
dihasilkan

daya

sebesar

50
MWatt

Kerja

ideal,
kerja

musnah

dan

perubahan

entropi

total,
jika

temperatur

lingkungan

25
o
C

Problem 5.8

An ideal gas, C
p

= 7/2 R, is heated in a steady
-
flow heat
exchanger from 20
o
C to 100
o
C by another stream of
the same ideal gas which enters at 180
o
C. The flow
rates of the two streams are the same, and heat
losses from the exchanger are negligible.

a.
Calculate the molar entropy changes of the two gas
streams for both parallel and counter
-
current flow
in the exchanger

b.
What is

S
total

in each case?

Carnot Cycle

Suatu mesin Carnot menyerap panas sebesar 400 kJ
dari sumber panas pada 525 oC dan melepas panas
ke penerima panas pada 50 oC. Perkirakan kerja
yang dihasilkan, perubahan entropy sumber panas,
penerima panas dan total!

2
nd

law

Satu mol gas ideal dikompresi secara isotermal
-
dalam sebuah alat silinder berpiston. Kerja yang
dibutuhkan untuk kompresi ini 30% lebih besar
dari kerja untuk proses reversibelnya. Selama
kompresi, sejumlah panas dibuang ke lingkungan
yang memiliki temperatur konstan sebesar 25 oC.
Hitung perubahan entropi dari gas, penerima
panas dan total!

2
nd

law

Uap jenuh dengan tekanan 5 bar dan temperatur 151,87 oC
berada dalam silinder yang volumnya 0,750 m3.
Silinder tersebut dilengkapi dengan piston yang dapat
bergerak bebas. Uap air tersebut kemudian dikompresi
secara reversibel sampai diperoleh uap air yang
memiliki tekanan 12 bar dan temperatur 187,99 oC.
pembuangan panas ke lingkungan sejumlah 2000 kJ.
Temperatur lingkungan diketahui sebesar 27 oC.
Perkirakanlah:

a.

Prosentase uap air yang mengembun

b.

Perubahan entalpi dan energi dalam (dalam
kJ/kg)

Kerja yang dibutuhkan untuk kompresi tersebut (dalam kJ)

Perubahan entropy uap air, lingkungan dan total

2
nd

law

Gas nitrogen dengan laju 2 ton/jam mula
-
mula memiliki
temperatur
-
168 oC dan tekanan 1 Mpa. Gas nitrogen
tersebut kemudian dikompresi secara adiabatik dalam
sebuah kompresor sehingga temperatur dan tekanannya
masing
-

63 oC dan 6 MPa. Dari
deskripsi proses tersebut:

Dengan mengasumsikan gas nitrogen sebagai gas ideal,
perkirakan daya yang dibutuhkan (dalam kWatt),
efisiensi mekanik (dalam %) dan perubahan entropi gas
nitrogen pada proses kompresi tersebut! Ambil

= 1,4.

Dengan menggunakan diagram P
-
H gas nitrogen terlampir,
jawablah kembali pertanyaan soal a!

7

Karbon tetrakhlorida cair sebanyak 5 kg mengalami perubahan secara
reversibel pada tekanan konstan sebesar 1 bar dalam sistem tertutup.
Akibat perlakuan tersebut, temperatur fluida mengalami perubahan
dari 0 oC menjadi 56 oC. Sifat
-
sifat dari CCl4 cair pada 1 bar dan 0 oC
diasumsikan tidak dipengaruhi temperatur:

= 1,2 x 10
-
3 K
-
1, CP =
0,84 kJ kg
-
1 K
-
1, dan

= 1,590 kg/m3. Tentukanlah:

a.

Perubahan entropi dari karbon tetrakhlorida cair

b.

Temperatur lingkungan pada saat proses perubahan tersebut
berlangsung (dalam oC)

Jika proses berlangsung secara ireversibel, perkirakan efisiensi mekanik
alat apabila kerja yang mampu dihasilkan sebesar 11,67 kJ!

Berkaitan dengan soal c, hitunglah perubahan entropi total! Apa
kesimpulan yang dapat diambil!

Ambil nilai temperatur lingkungan sama dengan hasil perhitungan soal
b!