10/3/2011

1

Chemical Engineering ThermodynamicsChemical Engineering ThermodynamicsPrepared byPrepared by:: DrDr NNINIEKINIEK FajarFajar PuspitaPuspita M Eng AugustM Eng August 20201111Prepared byPrepared by:: Dr. Dr. NNINIEKINIEK FajarFajar PuspitaPuspita, M.Eng August, , M.Eng August, 20201111

2011Gs_IV_2011Gs_IV_The First Law of ThermodynamicsThe First Law of Thermodynamics

1

Lesson 4Lesson 4

Lesson Topics Descriptions

Lesson 4ALesson 4A Work Mempresentasikan diskusi tentang difinisi-difinisi kerja.M ti b gk s c i gk s b t k b t k k jMempertimbangkan secara ringkas bentuk-bentuk kerjaberikut: work, boundary, shaft, gravitational, and spring

Lesson 4BLesson 4B Heat Mempresentasikan diskusi tentang difinisi-difinisi heat.Mempertimbangkan secara ringkas 3 mekanisme untukheat transfer: conduction, convection, and radiation.

Lesson 4CLesson 4C FLT_1st Law of

Mempelajari Hk I termodinamika dan dikenal sebagai prinsip-prinsip konservasi energi.

2

of Thermodynamics

p p p p gMempelajari penerapan Hk I pada sistem adiabatis tertutup dan kemudian sistem non-adiabatis tertutup

Lesson 4DLesson 4D Problem Solving Procedure

Mempelajari penggunaan prosedur yang akan membantumenyelesaikan persoalan secara sistematis. Ini untukmengingat mata kuliah thermo.

10/3/2011

2

Lesson 4Lesson 4

Lesson Topics Descriptions

Lesson 4ELesson 4E Isobaric and Mempelajari bagaimana menganalisa kasus-kasus khusus t j di d b d dIsochoric

Processesyang terjadi pada proses-proses yang berada padakondisi isobaric dan isochoric.

Lesson 4FLesson 4F Thermodynamic Cycles

Memperkenalkan siklus-siklus power, refrigeration, heat pump.

3

Apakah yang disebut dengan KerjaApakah yang disebut dengan Kerja??

Work (Kerja) adalah gaya F yang bekerja sepanjang pemindahan jarak /displacement x, searah dengan gaya searah dengan gaya.

Karena F = PA, kerja dapat diekspesikan dengantekanan dan luas area:

4

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 ataueskpansi Gas (atau

Cairan)

10/3/2011

3

DifinisiDifinisi TermodTermodiinaminamikk

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 d i i

5

didalam ruang dari garis putus-putus merupakan sistem yang tertutup dan yang lain dipertimbangkan sebagai lingkungan.

Arah mana kerja dipindahkan Arah mana kerja dipindahkan ??

Kami menggambarkan batasan dari 2 batasan dari 2 sistem, A and B.Kerja melintasi batas setiap sistem. Dapatkah anda katakan dari arah yang mana kerja bergerak? Kedalam atau

6

Kedalam atau keluar dari setiap sistem?

systems, A

systems, B

10/3/2011

4

Apakah yang disebut Konvensi tanda Apakah yang disebut Konvensi tanda ((Sign ConventionSign Convention))

??

Pertimbangkan System A: kincir angin mentrasfer kerja melintasi batas kedalam gas.gIni 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. S t A

7

penting menjadi konsisten.

Konvensi disini:WA < 0: kerja dilakukan pada sistem

System A

Sekeliling (Surroundings)

Apakah yang disebut Konvensi tandaApakah yang disebut Konvensi tanda ??

Pertimbangkan System B: generator mentransfer kerja melintasi batasan 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

8

memilih konvensi tanda untuk persoalan, hal ini penting untuk ditaati.

B > 0: kerja dilakukan pada sekeliling

System B

Surroundings

10/3/2011

5

KerjaKerja, Daya , Daya dan satuandan satuan--satuannyasatuannya

Diagram ini menggambarkan difinisi kerja yang melibatkan kenaikan berat. Kerja mempunyai satuan Joules (J) Kerja mempunyai satuan Joules (J) dalam satuan SI.

1 J = 1 N*m

Ide baru: apakah kecepatan yang mana kerja sedang lakukan?

9

W menunjukkan kecepatan .Dalam satuan SI kecepatan adalah Joules per second atau Watts (W): 1 W = 1 J / s Watt adalah satuan tenaga (Power).

American Engineering System: Power [=] Btu/s or ft-lbf /s or hP(horsepower)

Differentials of Properties are ExactDifferentials of Properties are Exact

The differential of every property is exact.Thi i i diff i l This is an inexact differential . So far we have only dealt with exact differentials

Exact Differentials :Volume, temperature pressure internal

State 2

10

temperature, pressure, internal energy, dan enthalpymerupakan fungsi-fungsi keadaan (mereka hanya tergantung pada keadaan dan tidak detail dari proses.The differential of every property is exact.

State 1

10/3/2011

6

Work is not a Property: It is InexactWork is not a Property: It is Inexact

Kerja adalah bukan sifat sistem atau sekelilingnya Kerja sekelilingnya. Kerja adalah variabel path dan variabel path mempuntai perbedaan yang tidak tepat (inexact differentials)

Untuk memahami ini lebih baik

State 2

11

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.

Boundary Work and its Process Boundary Work and its Process

Kami sedang mengamati hubungan diantara kerja dan alur proses kamialur proses kami.

Ingat bahwa kerja adalah produk dari daya gaya yang mendorong ekspansi dan memindahkan arah gaya.

Jika piston tidak pernah bergerak cepat.

12

bergerak cepat.

Initial State: P1 V1Final State: P2 V2Dimana: P1>P2 V1<v2

Alur proses ini dapat divisualisasikan pada diagram P-V.

10/3/2011

7

Positive and Negative Boundary Positive and Negative Boundary

Disini adalah diagram P-Vyang menunjukkan keadaan initial dan finalkeadaan initial dan finaldari sistem. Disini kami melihat proses kami dari State 1 (tekanan tinggi, volume kecil) keState 2 (tekanan rendah, volume besar).

13

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

Expansion and Compression Expansion and Compression

Kerja yang dilakukan oleh gas pada sekelilingnya selama ekspansi direpresentasikan oleh area dibawah P versus V.

14

10/3/2011

8

Exact vs. Inexact DifferentialsExact vs. Inexact Differentials

ExactIngat dari bab sebelumnya, bahwa entalpi d l h if t t i b l k d d adalah sifat atau variabel keadaan dan

turunannya (its differential) adalah tepat (exact): dH. Alur proses yang menghubungkan states 1dan 2 tidak ada jalan mempengaruhi perubahan entalpi dari state 1 to state 2.

InexactPada bab ini, kami telah mempelajari bahwa ke ja adalah va iabel ath dan

15

bahwa kerja adalah variabel path 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.

Menentukan batas aktual kerja secara eksperimentalMenentukan 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

16

Mengintegrasikan data secara numerik untuk mencapai suatu estimasi W12.Bedanya berapa lama proses ini terjadi untuk melengkapi?

The Actualexpansion of gas

(or liquid)

10/3/2011

9

NonNon--QuasiQuasi--Equilibrium Boundary Equilibrium Boundary

Mengapa ini beda, berapa cepat piston berberak?Key Reasons:Key Reasons:Kerja tergantung pada resistive forcedan bukan gaya yang diterapkan.Saat piston bergerak cepat, gas tidak tetap dalam keadaan kesetimbangan. Sebagai hasil, resistive force lebih besar dari padaP A.

17

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

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

QuasiQuasi--Equilibrium Boundary Equilibrium Boundary

Persamaan ini hanya benar untuk proses yang tidak pernah menyimpang dari keseimbangan.P i i d l h ki Persamaan ini adalah suatu perkiraan yang baik untuk quasi-equilibrium processes

Satu contoh dari quasi-equilibrium process adalah alat piston-silinder yang mana isi silinder dikmpresi secara pelan.

18

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.

10/3/2011

10

QuasiQuasi--Equilibrium Boundary Equilibrium Boundary

Contoh yang lebih baik dari quasi-equilibrium process ditunjukkan disini. Gas mula-mula pada keseimbangan. Kemudian, suatu waktu berat yang Kemudian, suatu waktu berat yang sedikit diambil dari belakang piston.Sebagaimana setiap berat diambil, gas terekspansi secara xpands sedikit dansistem 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

19

sand, akan lebi jauh mengurangi penyimpangan dari kesetimbangan.

• Dalam batas yang mana berat sangat kecil, sistem melewati deretan keadaan kesetimbangan.

• Dalam batas ini, gas secara aktual mengalami proses ekspansi quasi-equilibrium.

What is a Process Path?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 quasi-equilibrium.•Dua keadaan apapun dapat dihubungkan dengan jumlah alur proses

20

Dua keadaa apapu dapat d ubu gka de ga ju a a u p osesyang berbeda secara tidak terbatas (infinite).•Suatu tipe alur proses adalah biasa. Yang lain digunakan sebagai kasus-kasus benchmark untuk perbandingan dengan proses-proses nyata. Kami akan pertimbangkan proses-proses isotermal dan proces-proces polytropicpada halaman berikut.

10/3/2011

11

Isothermal Process for Isothermal Process for anan Ideal GasIdeal Gas

Ideal Gas EOS/equation of state: (Persamaan keadaan gas ideal)gas ideal)

Batasan kerja untuk proses isotermal pada gas ideal:

21

karena

Boundary Work for a Boundary Work for a PolytropicPolytropic

Proses polytropic adalah proses dimana:

C & δ are constantsDimana:

Boundary work untuk proses polytropic:

22

Untuk gas ideal (PV = nRT), persamaanboundary work disederhanakan menjadi:

10/3/2011

12

What Other Forms of Work 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. Shaft Work2. Gravitational Work3. Acceleration Work4. Spring Work

23

What is Shaft Work?What is Shaft Work?

Shaft Work: Kerja yang terkait dengan transmisi energi melalui poros 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

dan shaft bekerja

10/3/2011

13

Shaft Work and TorqueShaft Work and Torque

Shaft Work : Anda mungkin mengingat kembali dari pelajaran fisika anda nahwa

Trdari pelajaran fisika anda nahwa

gaya (F) yang memutar shaftmelalui tangan momen (r) menghasilkan torque (τ):

nor

25

Determining the Amount of Shaft WorkDetermining the Amount of Shaft Work

Gaya ini bertindak melalui jarak (s), yang terkait pada jumlah revolusi-revolusi dari perimeter, 2πr, dengan persamaan berikut:

Tr

g p

n

orDaman N adalah jumlah revolusi yang mana the shaft berputar.Ingat kembali bahwa kerja adalah gaya yang bertindak melalui suatu jarak. K it

26

Karena itu

Hitung kecepatan yang mana shaft berputar, mengharap dalam revolutions per minute (RPM), kami mendapatkan power tranmsi melalui shaft yang diekspresikan seperti dibawah ini:

10/3/2011

14

What is Gravitational WorkWhat is Gravitational Work

Gravitational Work: Kerja yang dilakukan oleh atau melawan gaya gravitasi.Gaya gravitasi yang bekerja pada ‘body’ F adalah: Gaya gravitasi yang bekerja pada body , F, adalah: Dimana:m = masa ‘body’g= percepatan gravitasigc = konstanta gravitasi(yang tergantung pada sistem satuan yang anda gunakan)

Tkerja yang diinginkan untuk

27

menaikkan berat dari z1 ke z2 adalah:

z1

z2

What is Acceleration What is Acceleration WorkWork

Acceleration Work: Suatu gaya harus diterapkan untuk mempercepat suatu ‘body’. Gaya yang diterapkan ini dikalikan dengan jarak yang ditempuh.

•Final state 50 km/h•Initial state 10 km/hAccelerational Work

28

10/3/2011

15

Determining the Amount of Acceleration WorkDetermining the Amount of Acceleration Work

Acceleration Work:Menggunakan Newton's Second Law of Motion, kami dapat menetapkan acceleration work yang diinginkan untuk acceleration work yang diinginkan untuk mempercepat dari satu keadaan ke keadaan yang lain.

•Final state 50 km/h•Initial state 10 km/hAccelerational Work

29

ds dihubungkan dengan kecepatan oleh :

Substitusi F dan ds kedalam persamaan kerja:

What is Spring Work?What is Spring Work?

Spring Work: Saat gaya diterapkan pada spring, spring merentang pada panjang yang baru Kerja spring ditentukan pada panjang yang 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 extension from rest (m)

30

p g ( )Konstanta spring dapat ditentukan jika perpanjangan spring diketahui untuk satu gaya yang diterapkan.

10/3/2011

16

Determining the Amount of Spring WorkDetermining the Amount of Spring Work

x diukur dari Spring Worktumpuan

x = 0 cm, F = 0 N

x1 = 1 cm Note: Spring work also occurs

31

x2 = 2 cm

F = 200 N

F = 400 N

Spring work also occurs when a spring is compressed to a length shorter than its resting length.

The displacement, x of a linear spring doubles when the force is doubled

Example #1Example #1

Work for a Cycle Carried Out in a Closed System4A-1 :

Work for a Cycle Carried Out in a Closed SystemAir undergoes a three-process cycle. Find the net work done for 2 kg of air if the processes are Process 1-2: constant pressure expansionProcess 2-3: constant volumeProcess 3-1: constant temperature compressionSketch the cycle path on a PV Diagram.

32

10/3/2011

17

Data : T1=100oC T2=600oC P1=200kPaGiven: T1=100oC T2=600oC P1=200kPa m2.0kgFind:Find:

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.

Ai b h id l

33

- Air behaves as an ideal gas.This must be verified at all three states.Part a.)

Part a.)

34

10/3/2011

18

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 :

35

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.

Eqn 4

MWair=29g/moleN=68.97moleR=8.314J/mole-KW12=286.69kJPlug values into Eqn 3 :

36

Because the volume is constant in step 2-3:W23=0kJStep 3-1 is isothermal, therefore, the definition of boundary work becomes:

Eqn 5

10/3/2011

19

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

37

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

Summary Chapter 4, Lesson A - WorkCHAPTER 4, LESSON A - WORKMempelajari difinisi kerja dalam termodinamik, yaitu kerja yang dilakukan oleh sistem ke sekeliling jika efek tunggal terhadap sekelilingdapat menjadi kenaikan berat. Batasan untuk proses isothermal dan polytropic.

ISOTERMAL POLITROPIK

38

10/3/2011

20

Summary Chapter 4, Lesson A - Work

WorksWorks DescriptionsDescriptions FormulaFormula

Shaft Work: Kerja terkait dengan transmisienergi melalui rotating shaft.

Gravitational Work

Kerja yang dilakukan oleh ataumelawan medan gravitasi(gravitational field).

Accelerational Work

Kerja yang dilakukan untukmempercepat (accelerate)

39

Work mempercepat (accelerate).

Spring Work Kerja yang dilakukan untukmenarik (stretch) atau menekan(compress) pegas.

THE THERMODYNAMIC DEFINITION OF HEATTHE THERMODYNAMIC DEFINITION OF HEAT

Apa yang terjadi saat anda i lk 1 ki di meninggalkan 1 cangkir es di ruang

pada 25oC?

Apa yang terjadi saat anda meninggalkan 1 cangkir air panas di ruang pada 25oC?

40

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.

10/3/2011

21

Internal Energy RevisitedInternal Energy Revisited

Jangan bingung Heat dengan i i t l !

Internal Energy

energi internal !

2 kJinternalenergy

41

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.

Heat vs. Internal Energy

Internal EnergyEnergi total dari molekul-molekul didalam suatu sistem. Ini termasuk energivibrational rotational dan translational 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.

HeatPada tingkat molekuler, satu mekanisme

2 kJinternalenergy

2 kJheat

Surrounding Air

Cup of

42

guntuk perpindahan panas yaitu tumbukan molekul-molekul yang mempunyai energivibrational, rotational dan translational yang berbeda.Perpindahan panas adalah aliran energi karena perbedaan energi internal molekul-molekul yang sensibel.

heatCup ofhot tea

10/3/2011

22

Q is the Symbol for Heat

Perpindahan panas direpresentasikan oleh simbol direpresentasikan oleh 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 t )

2 kJinternalenergy

2 kJQ

Surrounding Air

Cup of

43

temperatur yang sama) perpindahan panas berhenti.

QCup ofhot tea

Heat Flow RepresentationHeat Flow RepresentationSI System: Joules (J)

American Engineering System: British thermal unit (Btu)

Surrounding Air: Tsurr = 25oC

Q Q Cup ofhot tea

Cup ofice water

44

Heat flow akan direpresentasikan oleh garis panah merah bergelombang di LearnThermo.comTanda panah akan selalu menunjukkan dari panas ke dingin untuk menunjukkan arah aliran panas.

10/3/2011

23

Choosing a Sign ConventionChoosing a Sign Convention

Akankan aliran panas kedalam sistem positif atau keluar sistem yang positif?Kami sarankan menggambar panah pada diagram anda yang Kami sarankan menggambar panah pada diagram anda yang merepresentasikan sign convention yang anda pilih.Tanda panah menunjukkan arah aliran panas positif.

Udara sekeliling: Tsurr = 25oC

Q Q Cup ofhot tea

Cup ofi t

45

Dalam diagram yang ditunjukkan disini, kami melilih signconvention yang merepresentasikan aliran panas kedalam sistem sebagai positif, dan aliran panas keluar

sistem sebagai negatif,

hot tea ice water

Default Sign Convention for ThermoDefault Sign Convention for Thermo--CDCD

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.

Udara sekeliling: Tsurr = 25oC

Q Q Cup ofhot tea

Cup ofi t

46

Dalam perhitungan panas, yang mana sistem (cup) akan mempunyai Q = - 2 kJ ?

hot tea ice water

10/3/2011

24

Introduction to Adiabatic Introduction to Adiabatic Q = 0

Adiabaticsaat 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

Udara sekeliling: T Isolated

47

Dalam sistem adiabatic dengan tidak ada masa yang melintasi batasan sistem, kerja hanya bentuk energi yang dapat melintasi batasan sistem.

Udara sekeliling: Tsurr

T1 > T1 > TTsurrsurr

Isolated vessel

T1

Adiabatic Processes & Thermal EquilibriumAdiabatic Processes & Thermal Equilibrium

Q = 0 Adiabatic

t tid k d 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

saat tidak ada panas ditransfer.

Cup of

T2 = Tsurr

Udara sekeliling: Tsurr

48

mempunyai temperatur sama, aliran panas berhenti.Q = 0Sistem dan sekeliling telah mencapai keadaan dari keseimbangan termal (thermal equilibrium)

waterT2

10/3/2011

25

The Differential of HeatThe Differential of Heat

Heat Aliran panas yaitu path dependent. Ini bukan sifat dari sistem. Turunan panas yaitu tidak tepat.

Hot Cupof Tea

J l h gi g dit f l h h t

49

Jumlah energi yang ditransfer oleh heat dalam suatu proses yang berpindah dari State 1 ke State 2:

Heat Transfer RateHeat Transfer Rateadalah simbol kecepatan

perpindahan panasis

Hot CupHot Cupof Teaof Tea

Kecepatan perpindahan panas mempunyai satuan nits of energi per satuan waktu atau (power).

J/s = WBTU/h

Jika Kecepatan perpindahan panas dikenal sebagai fungsi waktu untuk suatu proses, jumlah total dari

dit f d t dit t k k

50

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

panas yang ditransfer dapat ditentukan menggunakan:

10/3/2011

26

Heat Flux

Heat Flux: Heat transfer rate per unit area (A) Typical units for Heat Flux

AQq =

Are: W/m2BTU/(h • ft2)When the heat flux is not a constant, but a known function of position across the area for for heat transfer, you can determine the heat transfer rate by integration:

AQq =

51

Summary Chapter 4, Lesson B - Heat

Conduction is heat that is transferred from more energetic molecules to adjacent less energetic molecules through molecular interactions.Fourier's Law of Conduction

CHAPTER 4, LESSON B - HEAT

Fourier s Law of Conductionk 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 ' L f C li

52

Radiation is a mode of energy transfer that results from changes in the electronic configurations of atoms or molecules.

Stefan-Boltzmann Law

Newton's Law of Cooling

h is the convection coefficient is determined experimentally

10/3/2011

27

INTRODUCTION TO THE 1ST LAW OF THERMODYNAMICS

We begin by drawing a boundary line to define the system we will analyze.

O i f Our system can consist of 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

53

convention for heat and for work.

The First Law of Thermodynamics is the relationship between heat, work and the total energy of the system

First Law of ThermodynamicsThe First Law of ThermodynamicsEnergy cannot be created or destroyed; it can only change form.All energy must be accounted for.gyFor 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

54

the key to analyzing all of the systems that we will consider in this chapter and the next chapter.

10/3/2011

28

1st Law for an Adiabatic, Closed SystemApplication of the First Law of ThermodynamicsConsider an adiabatic, closed systemClosed SystemyNo mass crosses the boundary of the system. AdiabaticNo 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

55

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?

Work is not ALWAYS a Path Variable!

State 2

Process Path

State 1 The total amount of work done b l d t i

56

on or by a closed system, in an adiabatic process, does not depend on the process path! It depends only on the initial and final states

10/3/2011

29

Total Energy: A State Variable

57

1st Law for a Non-Adiabatic, Closed System

58

10/3/2011

30

Components of the Total Energy of a System

59

1st Law for Closed Systems: Recap

60

10/3/2011

31

Application of the 1st Law to a Stone Falling Into Water

61

62

10/3/2011

32

63

64

10/3/2011

33

65

66

10/3/2011

34

67

68

10/3/2011

35

Differential or Rate Form of the 1st Law

69

70

10/3/2011

36

71

72

10/3/2011

37

73

74

10/3/2011

38

Summary Chapter 4, Lesson C - 1st Law of Thermodynamics

CHAPTER 4, LESSON C - 1ST LAW OF THERMODYNAMICSIn 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 the 1st Law to an adiabatic process taking place in a closed pp p g psystem.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.

75

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

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: ΔΔE = E = ΔΔEK + EK + ΔΔEP + EP + ΔΔU = U = -- WadWad

The 1st Law for a non-adiabatic closed system:ΔΔE = E = ΔΔEK + EK + ΔΔEP + EP + ΔΔU = Q U = Q -- WW

76

10/3/2011

39

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 t i t h Th d t

77

stair-step approach. The advantages 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.

Procedure OverviewThere 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 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 problemsh l

78

that we solve. We will use this solution procedure, in one form or another, for every example problem in the rest of this course.

10/3/2011

40

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. Mostbefore proceeding. Mostpeople need to read a problem statement more than once beforethey are ready to begin Step 2.

This is one of those self-explanatory steps that seemsimpossible to mess up. Yet it is the main reason formany mistakes. If you don't understand

79

many mistakes. If you don t understand theproblem it is unlikely that you will solve itcorrectly.

Once you understand the problem, you are ready to get something onto paper.

2- Draw a DiagramYou 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 isprobably the best way to communicate the information about a process to anotheranotherengineer. 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 makea problem much easier to solve. So, choose yoursystem boundary carefully.You should also set your sign convention for heat andwork in your diagram. Draw an arrow indicating the direction for heat transfer that will be considered

80

transfer that will be consideredpositive. Draw another arrow indicating the positivedirection for work.

10/3/2011

41

3- List Given InformationYou 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.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

81

writeP1 = P2.You must also make a list of all of the variables for which you must obtain values in order to answer all of thequestions posed in the problem statement.

4- List All AssumptionsMaking 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.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 ofthe Earth and the acceleration of gravity can frequently be omitted fromthis list unless they play a key role in the solution of the problem.The justifcation for each assumption should be given as well In many cases the

82

be given as well. In many cases, the justification for an assumption willconsist of a short explanation. However, someassumptions cannot be justified until theproblem has been solved. SeeStep 7 for more details.

10/3/2011

42

5- Write Equations & Lookup Datan 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 An equation is NOT independent if the other equations can be rearranged andcombined algebraically to obtain the same equation.2 x + y = 11 x - y = 1 x + 2 y = 10Next, make a list of all of the unknown variables in these equations. Decide which of the unknowns can be determined from referencedata sources, such as the NIST WebBook, and look up

83

data sources, such as the NIST WebBook, and look up theirvalues. This leaves you with a shorter list of the trueunknowns for the problem.You may not need to solve ALL of the independent equations for ALL of the unknowns in orderto answer the questions that were posedin the problem statement.

6- Solve EquationsIn 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 Techniques from algebra and calculus are the main tools you will use to solve the equations. Remember, a unique solution may be obtained ifthe number of correct, independent equations is equal to thenumber 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 however more

84

time. Often, however, morethan one equation must be solved simultaneouslyfor an equivalent number of unknowns. We willlearn more about this later in the course.

10/3/2011

43

7- Verify AssumptionsIt is common to make an assumption that depends on the value of one of the unknown variablesin a problem. In this step, you must use the values of the unknown variables that youvalues of the unknown variables that youobtained 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 gasthroughout a problem, but you might not know the molar volumeof the gas at the final state. While solving the problem you

85

problem youmay have determined that the final molar volume was15 L/mole. The gas cannot be accurately modeledas an ideal gas at the final state. Therefore,you need to solve the problem againusing a more sophisticated EOS.

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 simplyThis often consists of simplywriting 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 thisprocedure is to do it yourself.In order to prepare you to apply this

86

In order to prepare you to apply thisprocedure yourself, we next present aseries of example problems. Wewill solve the problems using theProblem Solving Proceduredescribed in this lesson.

10/3/2011

44

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.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 Perform a Degree of Freedom (DOF) analysis

87

5 Write Equations and Lookup Data

Perform a Degree of Freedom (DOF) 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.

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 often designated with an iso- prefix.g p

88

This lesson will explain the special simplifications of the First Law of Thermodynamics that are applicable to Isobaric and Isochoric processes

10/3/2011

45

Isobaric and Isochoric Processes

T-V Diagram : Isobaric Process, P = Constant,

89

T-V Diagram : Isochoric Process , V = Constant

First Law for an Isobaric ProcessFirst Law of Thermodynamics for an Isobaric, quasi-equilibrium , process in a closed system : Assume changes in potential and g pkinetic energies are negligible:

Recall from Lesson A,boundary work is defined asFor an isobaric process:

90

p

Substituting for Wb:

10/3/2011

46

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 an Isobaric,quasi-equilibrium, process in a closed system:Assume changes in potential andkinetic energies are negligible:Constant volume produces no boundary work:If there is no

th k i l d

92

other work involved

10/3/2011

47

Summary Chapter 4, Lesson E - Isobaric and Isochoric ProcessesCHAPTER 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

Thi i ld d t h i l ti th t processes. This yielded two much simpler equations that we can use to analyze isobaric and isochoric processes. An isobaric process is a constant pressure process. If an isobaric, quasi-equilibrium, process in a closed system involves only boundary work, then:

93

An isochoric process is a constant volume process. If an isochoric, quasi-equilibrium, process in a closed system involves only boundary work , then:

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 CyclesRefrigeration CyclesHeat Pump Cycles

94

Heat Pump Cycles

Two ways to categorize thermodynamic cycles:

Closed Cycles or Open Cycles

Gas Cycle or Vapor Cycles

10/3/2011

48

Form of the 1st Law for an Isobaric Process

95

A Closed, Gas Power Cycle

96

10/3/2011

49

Heat Engines

97

First Law for a Heat Engine

98

10/3/2011

50

Thermal Efficiency of a Power Cycle

99

What is a Refrigeration Cycle?

100

10/3/2011

51

First Law for a Refrigeration

101

Coefficient of Performance for a Refrigeration Cycle

102

10/3/2011

52

What is a Heat Pump Cycle

103

Coefficient of Performance

104

10/3/2011

53

Example #1

105

106

10/3/2011

54

107

108

10/3/2011

55

109

110

10/3/2011

56

Summary Chapter 4, Lesson F - Thermodynamic Cycles

CHAPTER 4, LESSON F - THERMODYNAMIC CYCLESIn 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