CHEMICAL ENGINEERING
THERMODYNAMICS I
KI 1221
Chemical Engineering Program
Department of Industrial and Process Technology
Rancangan Pembelajaran

Mata Kuliah
Kode / SKS
Tujuan

Pustaka


: Termodinamika Teknik Kimia I
: KI 1221 / 3 SKS
: Agar mahasiswa menguasai penyelesaian
problem pada unit – unit operasi dengan
mengaplikasikan hukum 1 & II termodinamika,
persamaan keadaan (EoS), panas reaksi, efek
panas, serta siklus rankine dan siklus carnot
: Smith, J. M., Van Ness, H. C., and Abbott, M.
M. (2001); Introduction to Chemical
Engineering Thermodynamics 6th edition;
McGraw-Hill Book Company
Institut Teknologi Kalimantan
Rancangan Pembelajaran

Materi Pokok
: - Hukum I dan II Termodinamika
- PVT Behavior
- Konsep gas ideal dan gas nyata
- Persamaan keadaan (EoS)
- Thermodynamics Networking
- Heat Effects
- Entropy Balance dan Lost of Work
- Siklus Carnot, Siklus Rankine, dll
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Rancangan Pembelajaran

Garis Besar Perkuliahan
Minggu ke-
Pokok Bahasan
Uraian
1
Hukum I
Termodinamika
• Pengantar Termodinamika
• Hukum I Termodinamika
• Neraca Energi pada Sistem Tertutup
• Phase Rule
• Proses Reversible
2
Idem
• Constant-V and Constant-P Process
• Kapasitas Panas
• Neraca Massa Energi pada Sistem Terbuka
3
Volumetric
Properties pada
Fluida Murni
• PVT Behavior
• Persamaan Keadaan Virial
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
Garis Besar Perkuliahan
Minggu ke-
Pokok Bahasan
Uraian
4
Idem
• Gas Ideal
• Persamaan Keadaan Kubik
5
Idem
• Generalized Correlations for Gases
• Generalized Correlations for Liquids
6
Heat Effects
• Efek Panas Sensibel
• Panas Laten pada Zat Murni
7
Idem
• Heat of Reaction
• Heat of Formation
• Heat of Combustion
8
Evaluasi Tengah
Semester
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Rancangan Pembelajaran

Garis Besar Perkuliahan
Minggu ke-
Pokok Bahasan
Uraian
9
Hukum 2
Termodinamika
• Hukum 2 Termodinamika
• Entropi
10
Idem
• Perubahan Entropi pada Gas Ideal
• Model Matematika Hukum 2 Termodinamika
11
Idem
• Neraca Entropi
• Perhitungan Ideal of Work
12
Properties
Termodinamika
pada Fluida
• Hubungan Properties pada Fase Homogen
• Residual Properties
• Residual Properties dari EoS
13
Idem
• Sistem Dua Fase
• Diagram Termodinamika
• Hubungan Properties Umum untuk Gas
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
Garis Besar Perkuliahan
Minggu ke-
Pokok Bahasan
Uraian
14
Aplikasi
Termodinamika
• Steam Power Plant
• Siklus Rankine
• Siklus Carnot
• Diesel Engine
• Gas-Turbine Engine
15
Idem
• Carnot Refrigerator
• Siklus Kompresi Uap
• Refrigerant
• Proses Liquefaction
16
Evaluasi Akhir
Semester
Institut Teknologi Kalimantan
Rancangan Pembelajaran

Penilaian
No.
Jenis Penilaian
Bobot Penilaian
1
Tugas Kelas
25 %
2
Evaluasi Tengah Semester
30 %
3
Evaluasi Akhir Semester
35 %
4
Kehadiran
10 %
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Scope of Thermodynamics
Thermodynamics Law
Physical and
Chemical Processes
Equilibrium
Properties Estimation
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Measures of Amount or Size

Spesific Volume
𝑉𝑡
𝑉≡
𝑚

𝑎𝑡𝑎𝑢
𝑉 𝑡 = 𝑚𝑉
Molar Volume
𝑉𝑡
𝑉≡
𝑛
𝑎𝑡𝑎𝑢
intensive property
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𝑉 𝑡 = 𝑛𝑉
extensive property
Dimensions of Science






Force
Temperature
Pressure
Work
Energy
Heat
Energy Conservation
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Exercise......


At what absolute temperature do the Celcius and Fahrenheit temperature
scales give the same numerical value? What is the value?
The variation of fluid pressure with height is described by
𝑑𝑃
= −𝜌𝑔
𝑑𝑧
Here, ρ is spesific density and g is the local acceleration of gravity. For an
ideal gas, ρ = MP/RT, where M is molar mass and R is the universal gas
constant. Modeling the atmosphere as an isothermal column of ideal gas at
10oC, estimate the ambient pressure in Denver, where z = 1 (mile) relative to
sea level. For air, take M = 29 g/mol; values of R are 82,06 cm3. atm/mol. K
Institut Teknologi Kalimantan
First Law of Thermodynamics
Although energy assumes many forms, the total quantity of
energy is constant, and when energy disappears in one form it
appears simultaneously in other forms
∆(Energy of the system) + ∆(Energy of surroundings) = 0
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Energy Balance for Closed Systems
W
∆(Energy of surroundings) = ± Q ± W
∆(Energy of surroundings) = - Q – W
Closed
Systems
∆(Energy of the systems) = Q + W
∆Ut = Q + W
Where, Ut = mU or Ut = nU
Q
Surroundings
∆(nU) = n ∆U = Q + W
Although Vt and Ut for a homogeneous system of arbitrary size
are extensive properties, spesific and molar volume V (or density)
and spesific and molar internal energy U are intensive
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Thermodynamic State and State Functions
𝑑𝑈 𝑡 = 𝑑𝑄 + 𝑑𝑊
Thermodynamic State
𝑉2
𝑃2
𝑑𝑃 = 𝑃2 − 𝑃1 = ∆𝑃
𝑃1
𝑑𝑉 = 𝑉2 − 𝑉1 = ∆𝑉
𝑉1
State Functions
𝑑𝑄 = 𝑄
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𝑑𝑊 = 𝑊
Process Functions
Example.....

a)
b)
When a system is taken from state a to state b along path acb, 100 J of heat
flows into the system and the system does 40 J of work.
How much heat flows into the system along path aeb if the work done by
the system is 20 J ??
The system returns from b to a along path bda. If the work done on the
system is 30 J, does the system absorb or liberate heat?? How much??
𝑡
∆𝑈𝑎𝑏
= 𝑄𝑎𝑐𝑏 + 𝑊𝑎𝑐𝑏 = 100 − 40 = 60 𝐽
For path aeb,
𝑡
∆𝑈𝑎𝑏
= 60 = 𝑄𝑎𝑒𝑏 + 𝑊𝑎𝑒𝑏 = 𝑄𝑎𝑒𝑏 − 20
𝑄𝑎𝑒𝑏 = 80 𝐽
For path bda,
𝑡
𝑡
∆𝑈𝑏𝑎
= −∆𝑈𝑎𝑏
= −60 = 𝑄𝑏𝑑𝑎 + 𝑊𝑏𝑑𝑎 = 𝑄𝑏𝑑𝑎 + 30
𝑄𝑏𝑑𝑎 = −60 − 30 = −90 𝐽
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The Phase Rule
𝐹 =2−𝜋+𝑁



π, the number of phases
N, the number of chemical species
F, the degrees of freedom of the system
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The Reversible Process
A process is reversible when its direction can be reversed at any
point by an infinitesimal change in external conditions

Reversible Expansion of Gas
𝑑𝑊 = −𝑃 𝑑𝑉 𝑡
Surrounding
𝑉2𝑡
𝑊=−
System
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𝑉1𝑡
𝑃 𝑑𝑉 𝑡
The Reversible Process
A process is reversible when its direction can be reversed at any
point by an infinitesimal change in external conditions

Reversible Chemical Reaction
𝐶𝑎𝐶𝑂3 (𝑠) ⇄ 𝐶𝑎𝑂(𝑠) + 𝐶𝑂2 (𝑔)
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Exercise.....


One mole of gas in a closed system undergoes a four-step thermodynamic
cycle. Use the data given in the following table to determine numerical
values for the missing quantities, i.e.,“fill in the blanks”.
Step
∆Ut
Q
W
12
-200
???
-6000
23
???
-3800
???
34
???
-800
300
41
4700
???
???
12341
???
???
-1400
A tank containing 20 kg of water at 20 oC is fitted with stirer that delivers
work to the water at the rate of 0,25 kW. How long does it take for the
temperature of the water to rise to 30 oC if no heat is lost from the water?
For water, Cp = 4,18 kJ/kg. oC
Institut Teknologi Kalimantan
Assignment I

Do it the problems at Smith Van Ness number 2.9; 2.13
!!!
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Constant-V and Constant-P Processes
𝑑 𝑛𝑈 = 𝑑𝑄 + 𝑑𝑊
𝑑𝑊 = −𝑃 𝑑(𝑛𝑉)
𝑑 𝑛𝑈 = 𝑑𝑄 − 𝑃 𝑑(𝑛𝑉)

Constant-Volume Process
𝑑𝑄 = 𝑑 𝑛𝑈
(const.V)
Integration yields,
𝑄 = 𝑛 ∆𝑈
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(const.V)
Constant-V and Constant-P Processes

Constant-Pressure Process
𝑑𝑄 = 𝑑 𝑛𝑈 + 𝑃 𝑑(𝑛𝑉)
𝑑𝑄 = 𝑑 𝑛𝑈 + 𝑑 𝑛𝑃𝑉 = 𝑑[𝑛 𝑈 + 𝑃𝑉 ]
𝐻 ≡ 𝑈 + 𝑃𝑉
𝑑𝑄 = 𝑑(𝑛𝐻)
Integration yields,
𝑄 = 𝑛 ∆𝐻
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(const. P)
Enthalpy
𝑑𝐻 = 𝑑𝑈 + 𝑑(𝑃𝑉)
𝑑𝐻 = 𝑑𝑈 + 𝑃 𝑑𝑉 + 𝑉 𝑑𝑃
Constant Volume
Surroundings
Constant Pressure
System
∆𝐻 = ∆𝑈 + ∆(𝑃𝑉)
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(U, P, dan V)
Heat Capacity
𝑑𝑄
𝐶≡
𝑑𝑇

Heat Capacity at Constant Volume
𝜕𝑈
𝐶𝑉 ≡
𝜕𝑇

𝑉
Heat Capacity at Constant Pressure
𝜕𝐻
𝐶𝑃 ≡
𝜕𝑇
𝑃
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Mass and Energy Balances for Open Systems

Mass Balance for Open Systems
𝑑𝑚𝑐𝑣
+∆ 𝑚
𝑑𝑡
𝑓𝑠
=0
Steady State
∆ 𝑚
𝑓𝑠
= 𝑚3 − 𝑚2 − 𝑚1
𝑢1 𝐴1 𝑢2 𝐴2 𝑢𝐴
𝑚=
=
=
𝑉1
𝑉2
𝑉
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Mass and Energy Balances for Open Systems

General Energy Balance
𝑑 𝑚𝑈
𝑑𝑡
𝑐𝑣
+∆
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1 2
𝐻 + 𝑢 + 𝑔𝑧 𝑚
2
=𝑄+𝑊
𝑓𝑠
Mass and Energy Balances for Open Systems

Energy Balances for Steady State Flow Processes
𝑑 𝑚𝑈
𝑑𝑡
𝑐𝑣
+∆
1 2
𝐻 + 𝑢 + 𝑔𝑧 𝑚
2
=𝑄+𝑊
𝑓𝑠
∆𝑢2
∆𝐻 +
+ 𝑔 ∆𝑧 = 𝑄 + 𝑊
2
∆𝐻 = 𝑄 + 𝑊
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Exercise......

Water flows through a horizontal coil heated from the
outside by high-temperature flue gases. As it passes
through the coil, the water changes state from liquid at
200 kPa and 80oC to vapor at 100 kPa and 125oC. Its
entering velocity is 3 m/s and its exit velocity is 200 m/s.
Determine the heat transferred through the coil per unit
mass of water. Enthalpies of the inlet and outlet streams
are 334,9 kJ/kg and 2726,5 kJ/kg
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Assignment II

Do it the problems at Smith Van Ness, number 2.19; 2.27;
2.31 !!!
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