File

10/3/2011
Chemical Engineering Thermodynamics
Prepared by:
by: Dr.
Dr NINIEK Fajar Puspita
Puspita,, M.Eng
M Eng August
August, 20
2011
11
2011Gs_IV_The
2011Gs_IV_
The First Law of Thermodynamics
1
Lesson 4
Lesson
Topics
Descriptions
Lesson 4A
Work
Mempresentasikan diskusi tentang difinisi-difinisi kerja.
M
Mempertimbangkan
ti b gk secara
s c
ringkas
i gk s bentuk-bentuk
b t k b t k kerja
k j
berikut: work, boundary, shaft, gravitational, and spring
Lesson 4B
Heat
Mempresentasikan diskusi tentang difinisi-difinisi heat.
Mempertimbangkan secara ringkas 3 mekanisme untuk
heat transfer: conduction, convection, and radiation.
Lesson 4C
FLT_1st Law
of
Thermodyna
mics
Mempelajari Hk I termodinamika dan dikenal sebagai
prinsip-prinsip
p
pp
p konservasi energi.
g
Mempelajari penerapan Hk I pada sistem adiabatis
tertutup dan kemudian sistem non-adiabatis tertutup
Lesson 4D
Problem
Solving
Procedure
Mempelajari penggunaan prosedur yang akan membantu
menyelesaikan persoalan secara sistematis. Ini untuk
mengingat mata kuliah thermo.
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Lesson 4
Lesson
Topics
Descriptions
Lesson 4E
Isobaric and
Isochoric
Processes
Mempelajari bagaimana menganalisa kasus-kasus khusus
yang terjadi
t j di pada
d proses-proses yang berada
b d pada
d
kondisi isobaric dan isochoric.
Lesson 4F
Thermodyna Memperkenalkan siklus-siklus power, refrigeration, heat
pump.
mic Cycles
3
Apakah yang disebut dengan Kerja?
Kerja?
`
Work (Kerja) adalah gaya F yang bekerja
sepanjang pemindahan jarak /displacement x,
searah dengan gaya.
gaya
`
`
`
Karena F = PA, kerja dapat diekspesikan dengan
tekanan dan luas area:
`
`
Persamaan ini didasarkan pada difinisi kerja
mekanik. Bagaimanapun kami perlu difinisi kerja
yang lebih luas yang mengijinkan kami untuk
memahami bentuk-bentuk lain dari kerja.
Compresi atau
eskpansi Gas (atau
Cairan)
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Difinisi Termod
Termodiinamik
namik
Difinisi kerja menurut
termodinamika :
Kerja dilakukan sistem pada
sekelilingnya jika pengaruh
tunggal pada eksternal apapun ke
sistem dapat menaikkan berat.
Garis putus-putus
merepresentasikan batasan sistem
(permukaan imajiner yang
memisahkan sistem yang distudi
dari lingkungannya). Daerah
did l
didalam
ruang d
darii garis
i putusputus merupakan sistem yang
tertutup dan yang lain
dipertimbangkan sebagai
lingkungan.
`
`
`
5
Arah mana kerja dipindahkan ?
Kami
menggambarkan
batasan dari 2
sistem, A and B.
Kerja melintasi
batas setiap
sistem.
Dapatkah anda
katakan dari arah
yang mana kerja
bergerak?
Kedalam atau
keluar dari setiap
sistem?
`
`
`
systems, A
systems, B
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Apakah yang disebut Konvensi tanda (Sign Convention
Convention))
?
Pertimbangkan System A: kincir
angin mentrasfer kerja melintasi
g
batas kedalam gas.
Ini kerja yang dilakukan oleh
sekeliling (kincir angin, motor dan
batere) pada sistem (gas).
Tidak ada tanda untuk kerja karena
tergantung pada perspektif anda.
Bagaimanapun sekali anda memilih
konvensi tanda (sign convention)
untuk suatu persoalan, hal ini
penting menjadi konsisten.
`
`
`
Konvensi disini:
WA < 0: kerja dilakukan pada
sistem
`
`
S t
System
A
Sekeliling (Surroundings)
7
Apakah yang disebut Konvensi tanda ?
Pertimbangkan System B:
generator mentransfer
kerja melintasi batasan
sistem. Ini adalah kerja
yang dilakukan oleh sistem
ke sekeliling.
Dan lagi tidak ada
konvensi tanda untuk kerja
karena ini tergantung pada
perspektif anda.
Bagaimanapun, sekali anda
memilih konvensi tanda
untuk persoalan, hal ini
penting untuk ditaati.
`
`
Surroundings
System B
B > 0: kerja dilakukan
pada sekeliling
`
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Kerja,, Daya dan satuanKerja
satuan-satuannya
`
`
Diagram ini menggambarkan difinisi
kerja yang melibatkan kenaikan berat.
Kerja mempunyai satuan Joules (J)
dalam satuan SI.
Æ1 J = 1 N*m
`
Ide baru: apakah kecepatan yang
mana kerja sedang lakukan?
`
`
W menunjukkan kecepatan .
Dalam satuan SI kecepatan adalah
Joules per second atau Watts (W): 1 W
=1J/s
Watt adalah satuan tenaga (Power).
`
American Engineering System:
Power [=] Btu/s or ft-lbf /s or hP
(horsepower)
9
Differentials of Properties are Exact
`
`
`
The differential of every
property is exact.
Thi is
This
i an inexact
i
differential
diff
i l
. So far we have only dealt with
exact differentials
State 2
`
`
Exact Differentials :Volume,
temperature pressure,
temperature,
pressure internal
energy, dan enthalpy
merupakan fungsi-fungsi
keadaan (mereka hanya
tergantung pada keadaan dan
tidak detail dari proses.
The differential of every
property is exact.
State 1
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Work is not a Property: It is Inexact
`
`
Kerja adalah bukan
sifat sistem atau
sekelilingnya Kerja
sekelilingnya.
adalah variabel path
dan variabel path
mempuntai
perbedaan yang tidak
tepat (inexact
differentials)
State 2
Untuk memahami ini
lebih baik ,
pertimbangkan alur
proses dari State 1 to
2:
State 1
Biarkan proses ini menjadi ekspansi
dari gas yang terisi didalam piston –
dan – peralatan silinder.
11
Boundary Work and its Process
`
Kami sedang mengamati
hubungan diantara kerja dan
alur proses kami.
kami
`
Ingat bahwa kerja adalah
produk dari daya gaya yang
mendorong ekspansi dan
memindahkan arah gaya.
`
Jika piston tidak pernah
bergerak cepat.
Initial State:
Final State:
Dimana:
12
P1
P2
P1>P2
V1
V2
V1<v2
Alur proses ini dapat divisualisasikan pada diagram P-V.
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Positive and Negative Boundary
`
`
`
Disini adalah diagram P-V
yang menunjukkan
keadaan initial dan final
dari sistem.
Disini kami melihat proses
kami dari State 1 (tekanan
tinggi, volume kecil) ke
State 2 (tekanan rendah,
volume besar).
Pikirkan tentang tanda
(sign convention) dari
kerja.
P > 0, saat dV > 0, gas
ekspansi dan W > 0.
Saat dV < 0, gas
terkompresi dan W < 0.
X
13
Expansion and Compression
`
Kerja yang dilakukan oleh
gas pada sekelilingnya
selama ekspansi
direpresentasikan oleh area
dibawah P versus V.
14
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Exact vs. Inexact Differentials
`
`
`
`
`
`
Exact
Ingat dari bab sebelumnya, bahwa entalpi
adalah
d l h sifat
if t atau
t variabel
i b l keadaan
k d
dan
d
turunannya (its differential) adalah tepat
(exact): dH.
Alur proses yang menghubungkan states 1
dan 2 tidak ada jalan mempengaruhi
perubahan entalpi dari state 1 to state 2.
Inexact
Pada bab ini, kami telah mempelajari
bahwa ke
kerja
ja adalah va
variabel
iabel path
ath dan
turunannya adalah tidak tepat (inexact)
Kami tidak mengevaluasi kerja tanpa
mengetahui secara tepat bagaimana P dan
V berubah selama proses sehingga kami
dapat mengevaluasi area dibawah alur
proses pada Diagram P-V.
15
Menentukan batas aktual kerja secara eksperimental
`
Kami dapat menggunakan sensor untuk mengukur P dan V.
Data dari sifat ini ditunjukkan dalam Diagram P-V, dibawah
ini
`
Mengintegrasikan data secara numerik untuk mencapai
suatu estimasi W12.
Bedanya berapa lama proses ini terjadi untuk melengkapi?
`
The Actual
expansion of gas
(or liquid)
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Non--QuasiNon
Quasi-Equilibrium Boundary
y
y
y
y
Mengapa ini beda, berapa cepat
piston berberak?
Key Reasons:
Kerja tergantung pada resistive force
dan bukan gaya yang diterapkan.
Saat piston bergerak cepat, gas tidak
tetap dalam keadaan
kesetimbangan. Sebagai hasil,
resistive force lebih besar dari pada
P A.
Untuk alasan yang sama, jumlah aktual kerja yang dilakukan saat gas
didalam silinder tiba-tiba dan secara cepat ekspansi yaitu kurang dari pada
integral dari P dV
Karena resistive force yaitu lebih besar dari pada P A, jumlah aktual kerja yang
diinginkan untuk mengkompres gas yaitu lebih besar dari pada integral dari P dV
17
Quasi--Equilibrium Boundary
Quasi
`
`
`
Persamaan ini hanya benar untuk proses
yang tidak pernah menyimpang dari
keseimbangan.
P
Persamaan
ini
i i adalah
d l h suatu perkiraan
ki
yang baik untuk quasi-equilibrium
processes
Satu contoh dari quasi-equilibrium
process adalah alat piston-silinder yang
mana isi silinder dikmpresi secara pelan.
Sebagaimana piston bergerak lambat sepanjang alur proses, tekanan
didalam gas tidak pernah menyimpang secara signifikan dari
keseimbangan. Oleh karena,
1. Sifat-sifat intensif seperti T dan P adalah uniform keseluruh sistem
pada semua waktu selama proses.
2. Integral dari P dV adalah cara yang akurat untuk menghitung kerja.
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Quasi--Equilibrium Boundary
Quasi
`
`
`
`
Contoh yang lebih baik dari quasiequilibrium process ditunjukkan disini.
Gas mula-mula pada keseimbangan.
Kemudian, suatu waktu berat yang
sedikit diambil dari belakang piston.
Sebagaimana setiap berat diambil, gas
terekspansi secara xpands sedikit dan
sistem diijinkan untuk mencapai
keadaan kesetimbangan baru.
Sistem tidak pernah berangkat jauh
dari keseimbangan.
Menggunakan berat yang lebih kecil,
seperti butiran pasir, uch as grains of
sand akan lebi jauh mengurangi
sand,
penyimpangan dari kesetimbangan.
• Dalam batas yang mana berat sangat kecil, sistem melewati deretan
keadaan kesetimbangan.
• Dalam batas ini, gas secara aktual mengalami proses ekspansi quasiequilibrium.
19
What is a Process Path?
`
Secara matematik, suatu
proses dapat diekspresikan
sebagai suatu persamaan
untuk P sebagai fungsi V.
• Kami dapat menyumbat persamaan alur ini menjadi integral diatas agar
supaya untuk menentukan kerja yang dilakukan dalam proses quasiequilibrium.
•Dua
Dua keadaa
keadaan apapu
apapun dapat d
dihubungkan
ubu gka dengan
de ga ju
jumlah
a aalur
u p
proses
oses
yang berbeda secara tidak terbatas (infinite).
•Suatu tipe alur proses adalah biasa. Yang lain digunakan sebagai kasuskasus benchmark untuk perbandingan dengan proses-proses nyata. Kami
akan pertimbangkan proses-proses isotermal dan proces-proces polytropic
pada halaman berikut.
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Isothermal Process for an Ideal Gas
`
Ideal Gas EOS/equation of
state: (Persamaan keadaan
gas ideal)
`
Batasan kerja untuk proses
isotermal pada gas ideal:
karena
21
Boundary Work for a Polytropic
`
Proses polytropic adalah proses dimana:
…
Dimana:
C & δ are constants
`
`
Boundary work untuk proses polytropic:
`
Untuk gas ideal (PV = nRT), persamaan
boundary work disederhanakan menjadi:
22
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What Other Forms of Work
`
Ada banyak bentuk kerja yang lain, yang terkait dengan gaya yang
bertindak sepanjang suatu jarak.
`
Bentuk lain dari kerja mekanik yang penting yaitu termasuk:
1.
2.
3.
4.
Shaft Work
Gravitational Work
Acceleration Work
Spring Work
23
What is Shaft Work?
`
`
`
Shaft Work:
Kerja yang terkait dengan
transmisi energi melalui poros
yang berputar biasanya
diperhitungkan dalam banyak
persoalan-persoalan teknis.
Silahkan mengisolasi shaft dan
pulley untuk menentukan
hubungan diantara kopling yang
diterapkan, jumlah revolusi
melalui yang mana shaft berputar
dan shaft bekerja
24
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Shaft Work and Torque
`
`
Shaft Work :
Anda mungkin mengingat kembali
dari pelajaran fisika anda nahwa
gaya (F) yang memutar shaft
melalui tangan momen (r)
menghasilkan torque (τ):
T
r
n
or
25
Determining the Amount of Shaft Work
`
T
Gaya ini bertindak melalui jarak (s), yang terkait
pada jumlah revolusi-revolusi dari perimeter,
g persamaan
p
berikut:
2πr, dengan
r
n
Daman N adalah jumlah revolusi yang
mana the shaft berputar.
Ingat kembali bahwa kerja adalah gaya
yang bertindak melalui suatu jarak.
K
Karena
it
itu
or
Hitung kecepatan yang mana shaft berputar, mengharap dalam
revolutions per minute (RPM), kami mendapatkan power tranmsi
melalui shaft yang diekspresikan seperti dibawah ini:
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What is Gravitational Work
`
`
`
`
Gravitational Work:
Kerja yang dilakukan oleh atau melawan gaya gravitasi.
Gaya gravitasi yang bekerja pada ‘body’
body , F,
F adalah:
Dimana:
m = masa ‘body’
g= percepatan gravitasi
gc = konstanta gravitasi
(yang tergantung pada sistem satuan yang anda gunakan)
` Tkerja yang diinginkan untuk
menaikkan berat dari z1 ke z2 adalah:
z2
z1
27
What is Acceleration Work
`
`
`
Acceleration Work:
Suatu gaya harus diterapkan untuk mempercepat suatu ‘body’.
Gaya yang diterapkan ini dikalikan dengan jarak yang
ditempuh.
Accelerational Work
•Initial state 10 km/h
•Final state 50 km/h
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Determining the Amount of Acceleration Work
`
`
Acceleration Work:
Menggunakan Newton's Second Law of
Motion, kami dapat menetapkan
acceleration work yang diinginkan untuk
mempercepat dari satu keadaan ke keadaan
yang lain.
Accelerational Work
•Initial state 10 km/h
•Final state 50 km/h
ds dihubungkan dengan kecepatan oleh :
Substitusi F dan ds kedalam persamaan kerja:
29
What is Spring Work?
`
`
`
Spring Work:
Saat gaya diterapkan pada spring, spring merentang
pada panjang yang baru
baru. Kerja spring ditentukan
dengan mengenali hubungan diantara gaya dan
panjang spring.
Untuk spring elastik yang linier, rentangan adalah
proporsional pada gaya yang diterapkan:
dimana:
ksp = spring constant (kN/m)
x = spring
p g extension from rest ((m))
Konstanta spring dapat ditentukan jika perpanjangan
spring diketahui untuk satu gaya yang diterapkan.
30
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Determining the Amount of Spring Work
Spring Work
x diukur dari
tumpuan
x = 0 cm, F = 0 N
x1 = 1 cm
F = 200 N
x2 = 2 cm
F = 400 N
Note:
Spring work also occurs
when a spring is
compressed to a length
shorter than its resting
length.
31 displacement, x of a linear spring doubles when the force is doubled
The
Example #1
4A-1 :
y
y
y
y
Work for a Cycle Carried Out in a Closed System
Air undergoes a three-process cycle. Find the net work
done for 2 kg of air if the processes are
Process 1-2: constant pressure expansion
Process 2-3: constant volume
Process 3-1: constant temperature compression
Sketch the cycle path on a PV Diagram.
32
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`
`
`
Data : T1=100oC
Given: T1=100oC
Find:
`
`
`
T2=600oC
T2=600oC
P1=200kPa
P1=200kPa m2.0kg
Sketch cycle on a PV Diagram.
Wcycle =???kJ
Assumptions:
`
`
`
- The gas is a closed system
- Boundary work is the only form of work interaction- Changes in
kinetic and potential energies are negligible.
- Air
Ai behaves
b h
as an ideal
id l gas.
This must be verified at all three states.Part a.)
33
`
Part a.)
34
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`
`
`
`
`
Part b.)
Since Wcycle = W12 + W23 + W31,
we will work our way around the cycle and calculate each work term
along the way.
Step 1-2 is isobaric, therefore, the definition of boundary work
becomes:
` Eqn 1
We can simplify Eqn 1 using the fact that P2 = P1 and the Ideal Gas
EOS :
` Eqn 2
`
`
Eqn 3
We can determine the number of moles of air in the system from the
given mass of air and its molecular weight.
35
Eqn 4
y
y
y
y
y
y
y
y
MWair=29g/mole
N=68.97mole
R=8.314J/mole-K
W12=286.69kJ
Plug values into Eqn 3 :
Because the volume is constant
in step 2-3:W23=0kJ
Step 3-1 is isothermal, therefore, the definition of boundary work
becomes:
Eqn 5
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The problem is that we don't know either P3 or V3. Either one
would be useful in evaluating W31 because we know P1 and we can
determine V1 from the Ideal Gas EOS, Eqn 2.We can evaluate V3
using the fact that V3 = V2. Apply the the Ideal Gas EOS to state 2.
`
Eqn 6
Next, we can apply Eqn 6 to state 1 :
V1=1.070m3/mole
Now, we can plug values into Eqn 4 to evaluate W13 :
W31=-181.89kJ
Sum the work terms for the three steps to get Wcycle:
Wcycle =104.8kJ
37
Summary Chapter 4, Lesson A - Work
`
`
`
CHAPTER 4, LESSON A - WORK
Mempelajari difinisi kerja dalam termodinamik, yaitu kerja yang
dilakukan oleh sistem ke sekeliling jika efek tunggal terhadap sekeliling
dapat menjadi kenaikan berat.
Batasan untuk proses isothermal dan polytropic.
ISOTERMAL
POLITROPIK
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Summary Chapter 4, Lesson A - Work
Works
Descriptions
Shaft Work:
Kerja terkait dengan transmisi
energi melalui rotating shaft.
Gravitational
Work
Kerja yang dilakukan oleh atau
melawan medan gravitasi
(gravitational field).
Accelerational
Work
Kerja yang dilakukan untuk
mempercepat (accelerate).
(accelerate)
Spring Work
Kerja yang dilakukan untuk
menarik (stretch) atau menekan
(compress) pegas.
Formula
39
THE THERMODYNAMIC DEFINITION OF HEAT
Apa yang terjadi saat anda
meninggalkan
i
lk 1 cangkir
ki es di ruang
pada 25oC?
Apa yang terjadi saat anda
meninggalkan 1 cangkir air panas di
ruang pada 25oC?
`
`
`
`
Es akan meleleh dan kemudian air akan menjadi hangat.
Teh panas akan menjadi dingin.
Ini akan berlanjut hingga kedua cairan mencapai temperatur ruang.
Heat: Energi di transisi dari satu objek atau sistem ke yang lain yang
digerakkan oleh perbedaan temperatur diantara objek atau sistem.
40
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Internal Energy Revisited
Jangan bingung Heat dengan
energii internal
i t
l!
2 kJ
internal
energy
`
Internal Energy
`
Energi total dari molekul-molekul didalam suatu sistem. Ini termasuk
energi vibrational, rotational dan translational. Ada bentuk energi
internal yang kadang-kadang disebut "sensible" karena mereka berubah
saat perubahan temperatur. Tetapi energi internal juga termasuk energi
dari interaksi molekul, ikatan kimia dan energi nuklir.
41
Heat vs. Internal Energy
`
Internal Energy
Energi total dari molekul-molekul didalam
suatu sistem. Ini termasuk energi
vibrational rotational dan translational.
vibrational,
translational
Ada bentuk energi internal yang kadangkadang disebut "sensible" karena mereka
berubah saat perubahan temperatur.
Tetapi energi internal juga termasuk energi
dari interaksi molekul, ikatan kimia dan
energi nuklir.
`
Heat
Pada tingkat
g
molekuler, satu mekanisme
untuk perpindahan panas yaitu tumbukan
molekul-molekul yang mempunyai energi
vibrational, rotational dan translational
yang berbeda.
Perpindahan panas adalah aliran energi
karena perbedaan energi internal molekulmolekul yang sensibel.
Surrounding Air
2 kJ
internal
energy
Cup of
hot tea
2 kJ
heat
`
`
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Q is the Symbol for Heat
`
Surrounding Air
2 kJ
internal
energy
Cup of
hot tea
2 kJ
Q
`
`
Perpindahan panas
direpresentasikan oleh simbol,
simbol
Q.
Perpindahan panas per satuan
masa direpresentasikan oleh
simbol, Q .
Panas hanya ditransfer jika ada
perbedaan temperatur dan
sekali keseimbangan termal
tercapai (2 sistem pada
t
temperatur
t yang sama))
perpindahan panas berhenti.
43
Heat Flow Representation
SI System: Joules (J)
American Engineering System: British thermal unit (Btu)
Q
Cup of
hot tea
`
`
Surrounding Air:
Tsurr = 25oC
Q
Cup of
ice water
Heat flow akan direpresentasikan oleh garis panah merah
bergelombang di LearnThermo.com
Tanda panah akan selalu menunjukkan dari panas ke dingin untuk
menunjukkan arah aliran panas.
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Choosing a Sign Convention
`
`
`
Akankan aliran panas kedalam sistem positif atau keluar sistem yang
positif?
Kami sarankan menggambar panah pada diagram anda yang
merepresentasikan sign convention yang anda pilih.
Tanda panah menunjukkan arah aliran panas positif.
Cup of
hot tea
Q
Udara sekeliling:
Tsurr = 25oC
Q
Cup of
i water
ice
t
Dalam diagram yang ditunjukkan disini, kami melilih signconvention yang
merepresentasikan aliran panas kedalam sistem sebagai positif, dan aliran panas keluar
sistem sebagai negatif,
45
Default Sign Convention for ThermoThermo-CD
`
`
`
`
Kami akan menggunakan sign convention berikut:
Q > 0 Panas ditransfer ke sistem yaitu positif.
Q = 0 Proses disebut adiabatic saat tidak ada panas yang ditansfer.
Q < 0 Panas ditransfer dari sistem yaitu negatif.
Cup of
hot tea
46
Q
Udara sekeliling:
Tsurr = 25oC
Q
Cup of
i water
ice
t
Dalam perhitungan panas, yang mana sistem (cup)
akan mempunyai Q = - 2 kJ ?
23
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Introduction to Adiabatic
Q=0
Adiabatic
saat tidak ada panas ditransfer.
Insolasi pada vessel mengurangi
banyak jumlah aliran panas
melintasi batasan sistem.
Jika anda dapat mengisolasi vesel
secara sempurna, kemudian akan
menjadi aliran panas ZERO dan
proses akan menjadi adiabatic
`
`
Dalam sistem adiabatic dengan
tidak ada masa yang melintasi
batasan sistem, kerja hanya bentuk
energi yang dapat melintasi
batasan sistem.
`
Udara sekeliling: Tsurr
T1 > Tsurr
Isolated
vessel
T1
47
Adiabatic Processes & Thermal Equilibrium
Q=0
Adiabatic
saatt tidak
tid k ada
d panas ditransfer.
dit
f
`
`
`
`
`
Sistem yang mempunyai periode waktu yang
sangat lama, yang mencapai suatu titik,
dimana:
T2 = Tsurr
Saat sistem dan sekeliling dipertahankan
mempunyai temperatur sama, aliran panas
berhenti.
Q=0
Sistem dan sekeliling telah mencapai keadaan
dari keseimbangan termal (thermal
equilibrium)
Udara sekeliling: Tsurr
T2 = Tsurr
Cup of
water
T2
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The Differential of Heat
`
`
Hot Cup
of Tea
Heat
Aliran panas yaitu path
dependent. Ini bukan sifat
dari sistem. Turunan panas
yaitu tidak tepat.
JJumlah
l h energi
gi yang
g ditransfer
dit
f oleh
l h heat
h t
dalam suatu proses yang berpindah dari
State 1 ke State 2:
49
Heat Transfer Rate
`
Hot Cup
of Tea
adalah simbol kecepatan
perpindahan panasis
Kecepatan perpindahan panas
mempunyai satuan nits of energi
per satuan waktu atau (power).
J/s = W
BTU/h
Jika Kecepatan perpindahan panas
dikenal sebagai fungsi waktu untuk suatu proses, jumlah total dari
panas yang ditransfer
dit
f dapat
d
t dit
ditentukan
t k menggunakan:
k
Jika Kecepatan perpindahan panas
konstan untuk proses, kemudian jumlah total dari panas yang
ditransfer dapat ditentukan dengan menggunakan:
Dimana Δt, waktu yang dilewati untuk proses diberikan oleh
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Heat Flux
`
`
`
`
Q
Heat Flux: Heat transfer rate per unit area (A) Æ q =
A
Typical units for Heat Flux
Are: W/m2
BTU/(h • ft2)
When the heat flux is not a constant, but a known function of
Q
q = transfer, you can determine the
position across the area for for heat
A
heat transfer rate by integration:
51
Summary Chapter 4, Lesson B - Heat
CHAPTER 4, LESSON B - HEAT
`
Conduction is heat that is transferred from more energetic molecules to
adjacent less energetic molecules through molecular interactions.
Fourier'ss Law of Conduction
Fourier
k is the thermal conductivity and
it is the ability of a material to conduct heat.
`
Convection is a mode of energy transfer between a solid surface at one
temperature and an adjacent moving liquid or gas at a different temperature.
Energy is transferred as a result of the combined effects of conduction within
the fluid and the motion of the fluid.
N
Newton's
' Law
L
off Cooling
C li
h is the convection coefficient is determined experimentally
`
Radiation is a mode of energy transfer that results from changes in the
electronic configurations of atoms or molecules.
Stefan-Boltzmann Law
52
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INTRODUCTION TO THE 1ST LAW OF
THERMODYNAMICS
`
We begin by drawing a boundary
line to define the system we will
analyze.
`
Our system can consist
O
i off many
devices.
`
We are interested in the total
energy within the system and
energy that crosses the system
boundary line: heat and work.
`
We need to define a sign
convention for heat and for work.
work
`
The First Law of Thermodynamics
is the relationship between heat,
work and the total energy of the
system
53
First Law of Thermodynamics
`
`
`
`
`
The First Law of Thermodynamics
Energy cannot be created or
destroyed; it can only change form.
All energy
gy must be accounted for.
For example, accumulation of
energy, ΔE, occurs when energy
entering (shown as heat) is greater
than energy leaving (shown as
work).
We need to define a sign
convention for heat and for work.
The First Law of Thermodynamics is
the key to analyzing all of the
systems that we will consider in this
chapter and the next chapter.
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1st Law for an Adiabatic, Closed System
`
`
`
`
`
`
`
Application of the First Law of
Thermodynamics
Consider an adiabatic, closed system
Closed System
y
No mass crosses the boundary of the
system.
Adiabatic
No heat transfer occurs across the
boundary of the system (Q = 0)
Therefore, the only form of energy
which crosses the system boundary is
work.
The total energy of the system is a state
variable
variable.
But in this system, the only way to
change the total energy is through
work.em boundary is work.
So, can work still be a path variable?
55
Work is not ALWAYS a Path Variable!
State 2
Process Path
State 1
The total amount of work done
on or by
b a closed
l d system,
t
in
i an
adiabatic process, does not
depend on the process path! It
depends only on the initial and
final states
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Total Energy: A State Variable
57
1st Law for a Non-Adiabatic, Closed System
58
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Components of the Total Energy of a System
59
1st Law for Closed Systems: Recap
60
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Application of the 1st Law to a Stone Falling Into
Water
61
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63
64
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66
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67
68
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Differential or Rate Form of the 1st Law
69
70
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71
72
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73
74
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Summary Chapter 4, Lesson C - 1st Law of
Thermodynamics
`
CHAPTER 4, LESSON C - 1ST LAW OF THERMODYNAMICS
`
In this lesson we studied the First Law of Thermodynamics which tells us that
energy cannot be created or destroyed; it can only change forms.
We then applied
pp
the 1st Law to an adiabatic p
process taking
gp
place in a closed
system.
We learned that the total work done in such a process depends only on the
initial and final states of the system and not on the details of the process.
In this case Wtot is not a path variable.
Next, we introduced the three common forms of energy.
Potential energy is energy that the system possesses due to its position
withinin a potential field, such as gravity.
Kinetic energy is energy that the system possesses due to its net velocity, either
translational or rotational.
Internal energy is energy that the system possesses due to its molecular and
atomic structure and the random vibrational, rotational and translational
energies of its molecules.
The change in the total energy of the system is the sum of the changes in the
potential, kinetic, and internal energies
`
`
`
`
`
`
75
`
The change in the total energy of the system is the sum of the changes
in the potential, kinetic, and internal energies.
ΔE = ΔEK + ΔEP + ΔU
`
The 1st Law for an adiabatic closed system:
ΔE = ΔEK + ΔEP + ΔU = - Wad
`
The 1st Law for a non-adiabatic closed system:
ΔE = ΔEK + ΔEP + ΔU = Q - W
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Why Do We Need a Problem Solving Procedure ?
`
You have seen a few thermodynamics problems and
the tools used to solve them.
But, the problems in the chapters to come will be
larger and more complex.
`
So, now is a good time to develop a flexible, reliable
procedure that we can use to solve involved
problems without getting confused or overlooking
something important.
`
Imagine that there are two ways to reach the top of a
staircase.
`
There is the Easy Way and then the Hard Way.
`
Many engineering problems can be tackled with this
stair-step
t i t approach.
h The
Th advantages
d
t
are :
`
The procedure is flexible enough to be used for a very wide
variety of problems.
`
Each step is relatively simple and well-defined.
`
The procedure will reduce the chance that you will overlook
an important aspect of a problem.
77
Procedure Overview
`
`
`
`
`
There are 8 steps in our general problem
solving procedure.
The names of some of the steps make it
quite clear what they entail. While some
of the other steps will require quite a bit
more explanation. That is the purpose of
the remaining pages of this lesson.
It is important to understand each step in
the procedure. Then, we will present
example problems that apply the
procedure to some problems.
Keep in mind that the focus of this lesson
is the procedure and not the substance of
the thermodynamics problems
that
h we solve.
l
We will use this solution procedure, in
one form or another, for every example
problem in the rest of this course.
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1- Read Carefully
`
You must slowly and carefully read every
word in the problem statement and make
sure that you understand it very well
before proceeding. Most
people need to read a problem statement
more than once before
they are ready to begin Step 2.
`
This is one of those self-explanatory steps
that seems
impossible to mess up. Yet it is the main
reason for
many mistakes. If you don
don'tt understand
the
problem it is unlikely that you will solve it
correctly.
`
Once you understand the problem, you
are ready to get something onto paper.
79
2- Draw a Diagram
`
`
`
`
You will better understand almost every engineering problem when you draw
a clear, complete diagram. Don't try to save paper by cramping your
diagram. In addition, a diagram is
probably the best way to communicate the information about a process to
another
engineer. Basically, no problem solution is complete without a good diagram.
In most cases, you need to define the physical limits of your system by
drawing its boundaries. This is typically done by drawing a dashed or colored
line around the system in your diagram. As you probably already realize,
choosing the best boundary for your system can make
a problem much easier to solve. So, choose your
system boundary carefully.
You should also set your sign convention for heat and
work in your diagram. Draw an arrow indicating the direction for heat
transfer that will be considered
positive.
Draw another arrow indicating the positive
direction for work.
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3- List Given Information
`
`
`
`
You will usually do this step while you are
making your Diagram in Step 2. Re-read the
problem statement word by word and list
every bit of information. Place this list near
your diagram.
You should be able to put the problem
statement away once you have the Diagram
constructed and the Given Information listed.
Every numeric value in the problem statement
should be equated to a variable. Be sure to
include the appropriate units for each value.
Many non-numeric pieces of information can
be translated into variable assignments. For
example, for an isobaric process you might
write
P1 = P2.
You must also make a list of all of the variables
for which you must obtain values in order to
answer all of the
questions posed in the problem statement.
81
4- List All Assumptions
`
`
`
Making valid assumptions is the key to
solving most engineering problems. A good
assumption is one that simplifies the problem
without significantly reducing the accuracy of
the solution.
It is very important that you make a list of
all of the most significant assumptions upon
which your solution is based. Assumptions
regarding things like the curvature of
the Earth and the acceleration of gravity can
frequently be omitted from
this list unless they play a key role in the
solution of the problem.
The justifcation for each assumption should
be given as well.
well In many cases,
cases the
justification for an assumption will
consist of a short explanation. However,
some
assumptions cannot be justified until the
problem has been solved. See
Step 7 for more details.
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5- Write Equations & Lookup Data
`
`
`
`
`
`
n this step, you must write all of the independent
equations that describe the process you are analyzing.
In this course, most of the equations will be
independent, but that is NOT always the case.
An equation is NOT independent if the other
equations can be rearranged and
combined algebraically to obtain the same equation.
2 x + y = 11
x-y=1
x + 2 y = 10
Next, make a list of all of the unknown variables in
these equations. Decide which of the unknowns can
be determined from reference
data sources, such as the NIST WebBook, and look up
their
values. This leaves you with a shorter list of the true
unknowns for the problem.
You may not need to solve ALL of the independent
equations for ALL of the unknowns in order
to answer the questions that were posed
in the problem statement.
83
6- Solve Equations
`
`
`
In this step, you must apply your math skills
to solve a set of equations for the key
unkowns that will permit you to answer the
questions posed in the problem statement.
Techniques from algebra and calculus are
the main tools you will use to solve the
equations. Remember, a unique solution
may be obtained if
the number of correct, independent
equations is equal to the
number of unknown variables in those
equations.
If you are fortunate, the equations may be
solved one at a time for one unknown at a
time Often
time.
Often, however
however, more
than one equation must be solved
simultaneously
for an equivalent number of unknowns. We
will
learn more about this later in the course.
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7- Verify Assumptions
`
`
It is common to make an assumption that
depends on the value of one of the unknown
variables
in a problem. In this step, you must use the
values of the unknown variables that you
obtained in the previous step of this
procedure to verify that all of the
verifiable assumptions that you made were
indeed valid.
For example, you might assume that a gas
behaves as an ideal gas
throughout a problem, but you might not
know the molar volume
of the gas at the final state. While solving the
problem you
may have determined that the final molar
volume was
15 L/mole. The gas cannot be accurately
modeled
as an ideal gas at the final state. Therefore,
you need to solve the problem again
using a more sophisticated EOS.
85
8- Answers
`
`
`
In this step, you will use the values of
the unknowns, that you determined in
Step 6, to answer the questions that
were posed in the problem statement.
This often consists of simply
writing out the appropriate variable
with its value and appropriate units.
However,
some simple calculations or unit
conversions are often involved.
This overview of our Problem Solving
Procedure is brief. The only way to
really understand how to apply this
procedure is to do it yourself.
In order to prepare you to apply this
procedure yourself, we next present a
series of example problems. We
will solve the problems using the
Problem Solving Procedure
described in this lesson.
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Summary Chapter 4, Lesson D - Problem Solving
Procedure
`
CHAPTER 4, LESSON D - PROBLEM SOLVING PROCEDURE
8 Langkah
PROSEDUR
1
Read Carefully
Make sure you understand what the problem entails.
2
Draw a Diagram
Draw a diagram with an appropriate boundary and sign
conventions for heat and work.
3
List Given
Information
List all information found in the problem statement and
assign variables. Also assign variables to unknown values.
4
List All
Assumptions
List all significant assumptions and justifications (if you can)
upon which your solution is based.
5
Write Equations
and Lookup Data
Perform a Degree of Freedom (DOF) analysis.
analysis
6
Solve Equations
Use your math to solve for unknowns.
7
Verify Assumptions Verify any assumptions that were not verified in step 4.
8
Answers
Check the answer for correct units and interpret results.
87
Processes in Which One Intensive Variable is
Constant
`
`
Here is our generic process path depicting a series of states the system
passes through during a process.
When one intensive property remains constant during a process, it is
g
with an iso- p
prefix.
often designated
This lesson will explain the special
simplifications of the First Law of
Thermodynamics that are
applicable to Isobaric and
Isochoric processes
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Isobaric and Isochoric Processes
T-V Diagram : Isobaric
Process,, P = Constant
T-V Diagram :
Isochoric Process ,
V = Constant
89
First Law for an Isobaric Process
`
`
First Law of Thermodynamics for
an Isobaric, quasi-equilibrium ,
process in a closed system :
Assume changes
g in potential
p
and
kinetic energies are negligible:
`
Recall from Lesson A,
boundary work is defined as
For an isobaric p
process:
`
Substituting for Wb:
`
90
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Form of the 1st Law for an Isobaric Process
91
First Law for an Isochoric Process
`
`
`
`
First Law of Thermodynamics for
an Isobaric,quasi-equilibrium,
Isobaric quasi-equilibrium
process in a closed system:
Assume changes in potential and
kinetic energies are negligible:
Constant volume produces no bo
undary work:
If there is no
other
th work
k involved
i
l d
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Summary Chapter 4, Lesson E - Isobaric and
Isochoric Processes
CHAPTER 4, LESSON E - ISOBARIC AND ISOCHORIC PROCESSES
`
`
`
In this lesson, we studied isobaric and isochoric processes. We
simplified the first law of thermodynamics for these types of
processes. Thi
This yielded
i ld d ttwo much
h simpler
i l equations
ti
th
thatt we can use
to analyze isobaric and isochoric processes.
An isobaric process is a constant pressure process. If an isobaric, quasiequilibrium, process in a closed system involves only boundary work,
then:
An isochoric process is a constant volume process. If an isochoric,
quasi-equilibrium, process in a closed system involves only boundary
work , then:
93
What is a Thermodynamic Cycle?
`
A system completes a thermodynamic cycle
`
when it undergoes two or more processes and
the system returns to its initial state.
`
This lesson is only an introduction to
thermodynamic cycles. You will be learning
more about them in each of the following
chapters.
`
In this lesson we will consider:
` Power Cycles
` Refrigeration Cycles
` Heat Pump Cycles
`
Two ways to categorize thermodynamic cycles:
`
Closed Cycles or Open Cycles
`
Gas Cycle or Vapor Cycles
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Form of the 1st Law for an Isobaric Process
95
A Closed, Gas Power Cycle
96
48
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Heat Engines
97
First Law for a Heat Engine
98
49
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Thermal Efficiency of a Power Cycle
99
What is a Refrigeration Cycle?
100
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First Law for a Refrigeration
101
Coefficient of Performance for a Refrigeration
Cycle
102
51
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What is a Heat Pump Cycle
103
Coefficient of Performance
104
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Example #1
105
106
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107
108
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109
110
55
10/3/2011
Summary Chapter 4, Lesson F - Thermodynamic
Cycles
`
`
CHAPTER 4, LESSON F - THERMODYNAMIC CYCLES
In this lesson we introduced thermodynamic cycles. A system
undergoes a thermodynamic cycle when after the system moves
through a number of states and processes it returns to its initial state.
111
56