KONSEP DASAR
TERMODINAMIKA
AGUS HARYANTO
FEBRUARI 2010
THERMO vs. HEAT TRANSFER
• Thermodynamics stems from the Greek words
therme (heat) and dynamis (power or motion),
which is most descriptive of the early efforts to
convert heat into power. Today thermodynamics
is broadly interpreted to include all aspects of
energy and energy transformations, including
power generation, refrigeration, and
relationships among the properties of matter.
• Heat transfers the science that deals with the
determination of the rates of such energy
transfer.
THERMO vs. HEAT TRANSFER (cont)
• Thermodynamics membicarakan sistem keseimbangan
(equilibrium), bisa digunakan untuk menaksir besarnya
energi yang diperlukan untuk mengubah suatu sistem
keseimbangan, tetapi tidak dapat dipakai untuk
menaksir seberapa cepat (laju) perubahan itu terjadi
karena selama proses sistem tidak berada dalam
keseimbangan.
• Heat Transfer tidak hanya menerangkan bagaimana
energi itu dihantarkan, tetapi juga menaksir laju
penghantaran energi. Inilah yang membedakan Heat
Transfer dengan thermodinamika.
APLIKASI
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Tubuh manusia
Meniup kopi panas
Perkakas elektronik (sirip, heat sink)
Refrigerator (AC, Kulkas)
Mobil (siklus engine, sirip, radiator)
Pembangkit listrik (turbin, boiler)
Industri (penyulingan, pendinginan,
pengeringan, dll).
DIMENSI dan SATUAN
• Dimensi (M,L,T,θ)  homogen
• Satuan : SI Units (m, s, kg, K)
• Kesalahan umum:
1. Tidak paham homogenitas
2. Usaha minimal, kurang
latihan
3. Tidak terampil melakukan
konversi satuan
• Trik: perhitungan harus
menyertakan satuan
SECONDARY UNITS
• Secondary units can be formed by
combinations of primary units. Example:
• F = m.a
• P = F/A
m
N  kg 2
s
N
Pa  2
m
kg
Pa 
m.s 2
kg.m / s
Pa 
2
m
2
SISTEM vs. LINGKUNGAN
• A system is defined as a quantity of matter or
a region in space chosen for study.
• The mass or region outside the system is
called the surroundings.
• The real or imaginary surface
that
separates the system from
its surroundings is called the
boundary
CLOSSED vs. OPEN SYSTEMS
• Closed system (= control
mass): Mass can’t cross the
boundary, but energy can.
• Volume of a closed system
may change.
• Special case, if no energy
cross the boundary, that
system is called an isolated
system.
CLOSSED SYSTEM
A closed system
with a moving
boundary.
OPEN vs. CLOSSED SYSTEMS
• Open system (= control volume) is a properly
selected region in space. It usually encloses a
device that involves mass flow such as a
compressor, turbine, or nozzle.
• Both mass and energy can cross the boundary
of a control volume.
• The boundaries of a control volume are called
a control surface, and they can be real or
imaginary.
OPEN SYSTEM
OPEN SYSTEM
Open system (= control
volume) with one inlet
and one outlet (exit) and
a real boundary.
SIFAT-SIFAT SISTEM TERMODINAMIKA
• Any characteristic of a system is called a property.
• Some familiar properties are pressure P, temperature T,
volume V, and mass m. The list can be extended to include
less familiar ones such as viscosity, thermal conductivity,
modulus of elasticity, thermal expansion coefficient,
electric resistivity, and even velocity and elevation.
• Intensive properties are those that are independent of the
mass of a system, such as temperature, pressure, and
density.
• Extensive properties are those whose values depend on the
size—or extent—of the system.
• Extensive properties per unit mass are called specific
properties (specific volume (v = V/m), specific energy (e =
E/m).
SIFAT INTENSIF vs. EKSTENSIF
TUGAS (dikumpul Senin) :
Sebuah apel dibelah dua.
Buatlah daftar sifat
intensif dan ekstensifnya
Criterion to differentiate
intensive and extensive
properties.
SIFAT-SIFAT SISTEM PENTING
• Densitas atau massa jenis:
masa per satuan volume
• Volume spesifik, kebalikan dari
densitas: volume per satuan
masa (m3/kg)
• Densitas relatif atau specific
gravity: nisbah densitas suatu
substansi dengan densitas
substansi standar pada suhu
tertentu (biasanya air pada 4oC
di mana  = 1000 kg/m3)
ENERGY SISTEM TERMODINAMIKA
• BENTUK ENERGI:
1
2
KE

mV
1. Energi Kinetik (KE) 
2
2. Energi Potensial (PE)  PE = mgh
3. Energi dakhil atau Internal Energy (U)
• ENERGI TOTAL:
E = U + KE + PE
e = u + ke + pe
(per satuan massa)
POSTULAT KEADAAN
• All properties (can be measured or calculated)
completely describes the condition, or the state,
of the system. At a given state, all the properties
of a system have fixed values. If the value of even
one property changes, the state will change to a
different one.
• The number of properties required to fix the state
of a system is given by the state postulate:
The state of a simple compressible system is
completely specified by two independent,
intensive properties.
PROSES dan SIKLUS
• Any change that a
system undergoes from
one equilibrium state to
another is called a
process
• The series of states
through which a system
passes during a process
is called the path
(lintasan) of the process.
MACAM-MACAM PROSES
• Proses isotermal: proses pada suhu T konstan.
• Proses isobaris: proses pada tekanan P konstan.
• Proses isokhoris (isometris): proses pada
volume spesifik  konstan.
• Proses adiabatik: proses di mana tidak terjadi
pertukaran kalor dengan lingkungan.
• Proses isentropik: proses pada entropi S
konstan.
STEADY-FLOW PROCESS
• The terms steady and uniform are used
frequently in engineering, and thus it is
important to have a clear understanding of
their meanings.
• The term steady implies no change with time.
• The opposite of steady is unsteady, or
transient.
• The term uniform, however, implies no change
with location over a specified region.
PROSES dan SIKLUS
• A system undergoes a cycle if it returns to its
initial state at the end of the process.
Siklus dengan 2 lintasan
Siklus dengan 4 lintasan
TEKANAN
• Tekanan (P) : gaya (F) per satuan luas (A).
• Satuan tekanan adalah pascal (Pa) = N/m2.
• Untuk benda padat gaya per luas satuan tidak disebut
tekanan, tetapi tegangan (stress).
• Untuk fluida diam, tekanan adalah sama ke segala arah.
• Tekanan di dalam fluida meningkat sesuai dengan
kedalamannya akibat berat fluida (pengaruh gravitasi)
sehingga fluida pada bagian bawah menanggung beban
yang lebih besar daripada fluida di bagian atas.
• Tetapi tekanan tidak bervariasi pada arah horisontal.
• Tekanan gas di dalam tangki dapat dianggap seragam
karena berat gas terlalu kecil dan tidak mengakibatkan
pengaruh yang berarti.
TEKANAN: UKUR, ATM, VAKUM
• Tekanan aktual pada posisi tertentu disebut tekanan
absolut dan diukur secara relatif terhadap tekanan
vakum, yaitu tekanan nol mutlak.
• Kebanyakan pengukur tekanan dikalibrasi untuk
membaca nol di atmosfer (tekanan atmosfer lokal).
• Perbedaan tekanan absolut dan tekanan atmosfer
disebut tekanan ukur (pressure gage).
• Tekanan di bawah tekanan atmosfer disebut tekanan
vakum (vacuum pressure) dan diukur dengan
pengukur vakum yang menunjukkan perbedaan antara
tekanan atmosfer dan tekanan absolut.
• Pgage = Pabs – Patm
(untuk P > Patm)
• Pvac = Patm – Pabs
(untuk P < Patm)
TEKANAN UKUR,
TEKANAN ATMOSFER,
TEKANAN VAKUM
PENGUKUR TEKANAN
MANOMETER
PRESSURE GAGE
BAROMETER
PRINSIP MANOMETER
Perhatikan gambar:
• Seimbang F = 0
• P1 = P2
• A P1 = A Patm + W
di mana W = m g =
Vg=Ahg
• P1 = Patm +  h g
• P = P1 - Patm =  h g = Tekanan ukur di dalam tangki
EXAMPLE : Manometer
A manometer is used to
measure the pressure in a
tank. The fluid used has a
specific gravity of 0.85, and
the manometer column
height is 55 cm, as shown in
Figure. If the local
atmospheric pressure is 96
kPa, determine the absolute
pressure within the tank.
EXAMPLE: SOLUTION
EXAMPLE: MULTIFLUID MANOMETER
Water in a tank is pressurized by
air, and the pressure is measured
by a multifluid manometer (see
Figure). The tank is located on a
mountain at an altitude of 1400 m
where the atmospheric pressure is
85.6 kPa. Determine the air
pressure in the tank if h1 = 0.1 m,
h2 = 0.2 m, and h3 = 0.35 m. Take
the densities of water, oil, and
mercury to be 1000 kg/m3, 850
kg/m3, and 13,600 kg/m3,
respectively.
SOLUTION
APLIKASI MANOMETER
Measuring the
pressure drop across
a flow section or a
flow device by a
differential
manometer:
P1 + 1g(a + h) - 2gh - 1ga = P2
P1 - P2 = (2 - 1)gh
Untuk 2 >> 1 :
P1 - P2 ≈ 2 g h
BAROMETER Torricelli
Patm =  g h
EXAMPLE3: BAROMETER
• Determine the atmospheric pressure at a
location where the barometric reading is 740
mm Hg and the gravitational acceleration is g
9.81 m/s2. Assume the temperature of
mercury to be 10oC, at which its density is
13,570 kg/m3.
EXAMPLE3: SOLUTION
TEKANAN ATMOSFER
ELEVASI
(m)
0 (sea level)
1000
2000
5000
10,000
20,000
TEKANAN
(kPa)
101.325
89.88
79.50
54.05
26.5
5.53
TEKANAN
(mmHg)
760.00
674.15
596.30
405.41
198.77
41.48
Rule of thumb: naik 10 m, tekanan atmosfer turun 1 mmHg
EFEK KETINGGIAN
“….and
whomsoever He
wills to send astray,
He makes his
breast closed and
constricted, as if
he is climbing up
to the sky. Thus
Allah puts the
wrath on those who
believe not.” (QS,
6:125)
TEMPERATURE
• Thermodinamika  SUHU MUTLAK
• Satuan kelvin (K) untuk SI
• Satuan renkine (R) untuk USCS
Konversi:
CAUTION:
T(K) = T(oC) + 273.15
T(R) = T(oF) + 456.67
T(oC) = 1.8T(oC) + 32
T(R) = 1.8 T(K)
T(K) = T(oC)
T(R) = T(oF)
EXAMPLE4: TEMPERATURE
• During a heating process, the temperature of
a system rises by 10°C. Express this rise in
temperature in K, °F, and R.
PR:
• Soal No: 1-6C, 1-7C, 1-15C, 1-16C, 1-17C, 120C, 1-21C, 1-22C, 1-23C, 1-24C, 1-29, 1-31, 134C, 1-35C, 1-36C, 1-39C, 1-40, 1-42, 1-43,
1-44, 1-45, 1-48, 1-51, 1-53, 1-55, 1-57, 1-59,
1-61, 1-62, 1-63, 1-65, 1-66, 1-73, 1-85, 1-88,
1-101, 1-103, 1-105, 1-106, 1-108, 1-120, 1121, 1-122, 1-123, 1-125.
• 1 Mhs : 1 Concepts + 1 soal