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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
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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
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Apakah yang disebut dengan Kerja? Kerja? `
Work (Kerja) adalah gaya F yang bekerja sepanjang pemindahan jarak /displacement x, searah dengan gaya. gaya
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Karena F = PA, kerja dapat diekspesikan dengan tekanan dan luas area:
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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.
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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?
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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.
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Konvensi disini: WA < 0: kerja dilakukan pada sistem
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S t System A
Sekeliling (Surroundings)
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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.
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Surroundings System B
B > 0: kerja dilakukan pada sekeliling
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Kerja,, Daya dan satuanKerja satuan-satuannya `
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Diagram ini menggambarkan difinisi kerja yang melibatkan kenaikan berat. Kerja mempunyai satuan Joules (J) dalam satuan SI. Æ1 J = 1 N*m
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Ide baru: apakah kecepatan yang mana kerja sedang lakukan?
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W menunjukkan kecepatan . Dalam satuan SI kecepatan adalah Joules per second atau Watts (W): 1 W =1J/s Watt adalah satuan tenaga (Power).
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American Engineering System: Power [=] Btu/s or ft-lbf /s or hP (horsepower)
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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
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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 `
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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.
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Boundary Work and its Process `
Kami sedang mengamati hubungan diantara kerja dan alur proses kami. kami
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Ingat bahwa kerja adalah produk dari daya gaya yang mendorong ekspansi dan memindahkan arah gaya.
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Jika piston tidak pernah bergerak cepat. Initial State: Final State: Dimana:
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P1 P2 P1>P2
V1 V2 V1
Alur proses ini dapat divisualisasikan pada diagram P-V.
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Positive and Negative Boundary `
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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.
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Expansion and Compression `
Kerja yang dilakukan oleh gas pada sekelilingnya selama ekspansi direpresentasikan oleh area dibawah P versus V.
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Exact vs. Inexact Differentials ` `
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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
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Mengintegrasikan data secara numerik untuk mencapai suatu estimasi W12. Bedanya berapa lama proses ini terjadi untuk melengkapi?
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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 `
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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. 18
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Quasi--Equilibrium Boundary Quasi `
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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. 20
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Isothermal Process for an Ideal Gas `
Ideal Gas EOS/equation of state: (Persamaan keadaan gas ideal)
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Batasan kerja untuk proses isotermal pada gas ideal:
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Boundary Work for a Polytropic `
Proses polytropic adalah proses dimana:
Dimana:
C & δ are constants
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Boundary work untuk proses polytropic:
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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.
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Bentuk lain dari kerja mekanik yang penting yaitu termasuk: 1. 2. 3. 4.
Shaft Work Gravitational Work Acceleration Work Spring Work
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What is Shaft Work? ` `
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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
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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
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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
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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: 26
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What is Gravitational Work ` ` `
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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:
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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:
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What is Spring Work? ` `
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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.
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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
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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.
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Data : T1=100oC Given: T1=100oC Find: ` `
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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.)
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Part a.)
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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 `
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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 36
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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.
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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 ` `
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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
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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
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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.
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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
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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.
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Heat Flow Representation SI System: Joules (J) American Engineering System: British thermal unit (Btu)
Q Cup of hot tea
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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
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Q
Udara sekeliling: Tsurr = 25oC
Q
Cup of i water ice t
Dalam perhitungan panas, yang mana sistem (cup) akan mempunyai Q = - 2 kJ ?
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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
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Dalam sistem adiabatic dengan tidak ada masa yang melintasi batasan sistem, kerja hanya bentuk energi yang dapat melintasi batasan sistem.
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Udara sekeliling: Tsurr T1 > Tsurr
Isolated vessel T1
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Adiabatic Processes & Thermal Equilibrium Q=0 Adiabatic saatt tidak tid k ada d panas ditransfer. dit f `
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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:
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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 50
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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:
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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.
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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
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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.
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Our system can consist O i off many devices.
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We are interested in the total energy within the system and energy that crosses the system boundary line: heat and work.
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We need to define a sign convention for heat and for work. work
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The First Law of Thermodynamics is the relationship between heat, work and the total energy of the system 53
First Law of Thermodynamics `
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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 ` ` `
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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?
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Work is not ALWAYS a Path Variable! State 2
Process Path
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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
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1st Law for a Non-Adiabatic, Closed System
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Components of the Total Energy of a System
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1st Law for Closed Systems: Recap
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Application of the 1st Law to a Stone Falling Into Water
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Differential or Rate Form of the 1st Law
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Summary Chapter 4, Lesson C - 1st Law of Thermodynamics `
CHAPTER 4, LESSON C - 1ST LAW OF THERMODYNAMICS
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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
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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
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The 1st Law for an adiabatic closed system: ΔE = ΔEK + ΔEP + ΔU = - Wad
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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.
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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.
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Imagine that there are two ways to reach the top of a staircase.
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There is the Easy Way and then the Hard Way.
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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.
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Each step is relatively simple and well-defined.
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The procedure will reduce the chance that you will overlook an important aspect of a problem.
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Procedure Overview ` `
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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.
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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.
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Once you understand the problem, you are ready to get something onto paper. 79
2- Draw a Diagram `
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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 `
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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 `
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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. 82
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5- Write Equations & Lookup Data `
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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 `
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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 `
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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 `
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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
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Read Carefully
Make sure you understand what the problem entails.
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Draw a Diagram
Draw a diagram with an appropriate boundary and sign conventions for heat and work.
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List Given Information
List all information found in the problem statement and assign variables. Also assign variables to unknown values.
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List All Assumptions
List all significant assumptions and justifications (if you can) upon which your solution is based.
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Write Equations and Lookup Data
Perform a Degree of Freedom (DOF) analysis. analysis
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Solve Equations
Use your math to solve for unknowns.
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Verify Assumptions Verify any assumptions that were not verified in step 4.
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Answers
Check the answer for correct units and interpret results.
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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 88
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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 `
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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:
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Recall from Lesson A, boundary work is defined as For an isobaric p process:
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Substituting for Wb:
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Form of the 1st Law for an Isobaric Process
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First Law for an Isochoric Process `
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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 `
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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:
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What is a Thermodynamic Cycle? `
A system completes a thermodynamic cycle
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when it undergoes two or more processes and the system returns to its initial state.
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This lesson is only an introduction to thermodynamic cycles. You will be learning more about them in each of the following chapters.
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In this lesson we will consider: ` Power Cycles ` Refrigeration Cycles ` Heat Pump Cycles
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Two ways to categorize thermodynamic cycles:
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Closed Cycles or Open Cycles
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Gas Cycle or Vapor Cycles
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Form of the 1st Law for an Isobaric Process
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A Closed, Gas Power Cycle
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Heat Engines
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First Law for a Heat Engine
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Thermal Efficiency of a Power Cycle
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What is a Refrigeration Cycle?
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First Law for a Refrigeration
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Coefficient of Performance for a Refrigeration Cycle
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What is a Heat Pump Cycle
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Coefficient of Performance
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Example #1
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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.
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