Presentasi berjudul: "ITK-233 Termodinamika Teknik Kimia I"— Transcript presentasi:
1 ITK-233 Termodinamika Teknik Kimia I 3 sks4 - The Second Law: EntropyDicky Dermawan
2 Process DirectionNo apparatus can operate in such a way that its only effect (in system & surroundings) is to convert heat absorbed by a system completely into work done by the systemIt is impossible by a cyclic process to convert the heat absorbed by a system completely into work done by the system.No process is possible which consist solely in the transfer of heat from one temperature level to a higher one.Partial conversion of heat to into work is the basis for nearly all commercial production of power.
3 Heat Engine Heat engines produce work from heat in a cyclic process Essential to all heat-engine cycles are:Absorption of heat into the system at a high temperatureRejection of heat to surroundings at a lower temperatureProduction of work
4 Conversion of heat to work always accompanied by heat release to surroundings
5 Heat Engine Thermal efficiency: For to be unity, QC must be zero. No engine has ever been built for which this is true; some heat is always rejected.
6 Carnot’s Theorem & Carnot Engine Carnot engine operates between two heat reservoirs in such a way that all heat absorbed is absorbed at the constant temperature of the hot reservoir and all heat rejected at the constant temperature of the cold reservoirFor two given heat reservoir, no engine can have a thermal efficiency higher than that of a Carnot engine.
7 Carnot CycleStep1: A system at TC undergoes a reversible adiabatic process that causes its temperature rise to TH Step 2: The system maintains contact with the hot reservoir TH, and undergoes a reversible isothermal process during which heat QH is absorbed from the hot reservoir Step 3: The system undergoes a reversible adiabatic process in the opposite direction of step 1 that brings its temperature back to that of the cold reservoir at TC.Step 4: The system maintains contact with the reservoir at TC, and undergoes reversible isothermal process in the opposite direction of the step 2 that returns it to its initial state with heat rejection of QC to the cold reservoir
8 Carnot Engine - Reversibility Any reversible engine operating between 2 heat reservoirs is a Carnot engine.An engine operating on a different cycle must necessarily transfer heat across finite temperature difference and therefore cannot be reversible.
9 Carnot Cycle with an Ideal Gas as Working Fluid Step1: a - b Reversible adiabatic compression until the temperature rise from TC to TH Step 2: b – c Reversible isothermal expansion to arbitrary point c with heat absorption of QH Step 3: c – d Reversible adiabatic expansion until the temperature decrease TC.Step 4: d – aReversible isothermal compression to the initial state with heat rejection of QC
10 EntropyOur analysis showed that or The equation suggests the existence of a property whose changes are given by the quantities Q/T When the isotermal steps are infinitesimal, the heat quantities become dQ: or Thus the quantities dQrev/T sum to zero for the arbitrary cycle, exhibiting the characteristic of a property. We call this property as entropy, S.There exists a property called entropy S, which is an intrinsic property of a system, functionally related to the measurable coordinates which characterize the system
11 Characteristic of Entropy Entropy owes its existence to the second law, from which it arises in much the same way as internal energy does from the first law.The change in entropy of any system undergoing a finite reversible process isWhen a system undergoes an irreversible process between two states, the entropy change of the system is evaluated to an arbitrary chosen reversible process that accomplishes the same change of state as the actual process. The integration is not carried out for the irreversible path.
12 Characteristic of Entropy Since entropy is a state function, the entropy changes of the irreversible and reversible processes are identical.The entropy change of a system caused by the transfer of heat can always be calculated by dQ/T whether the heat transfer is accomplished reversibly or irreversibly.When a process is irreversible on account of finite differences in other driving forces, such as pressure, the entropy changes is not caused solely by heat transfer, and for its calculations one must devise a reversible means of accomplishing the same change of state.
13 Calculation of Entropy For one mole of fluid undergoing a mechanically reversible process in a closed system, using the first law, the defining equation for enthalpy one find: For an ideal gas: Although derived for a mechanically reversible process, this equation is a general equation for the calculation of entropy changes, since it relates properties only and independent of process causing the change of state.
14 ExampleCalculate ΔS for each step of the cycle shown below. Assume ideal gas with constant Cp.
15 Problem 5.6A quantity of an ideal gas, Cp = 7/2 R at 20oC & 1 bar and having a volume of 70 m3 is heated at constant pressure to 25oC by the transfer of heat from a heat reservoir at 40oC. Calculate the heat transfer to the gas, the entropy change of the heat reservoir, the entropy change of the gas, and DStotal.
16 Problem 5.7A rigid vessel of 0.05 m3 volume contains an ideal gas, Cv = 5/2 R, at 500 K and 1 bar.If heat in the amount of J is transferred to the gas, determine its entropy changeIf the vessel is fitted with a stirrer that is rotated by a shaft so that work on the amount of J is done on the gas, what is the entropy change of the gas if the process is adiabatic? What is DStotal?
17 Example 5.3Methane gas at 550 K & 5 bar undergoes a reversible adiabatic expansion to 1 bar. Assuming methane to be an ideal gas at these conditions, determine its final temperature.
19 Heat Engine Thermal efficiency: For to be unity, QC must be zero. No engine has ever been built for which this is true; some heat is always rejected.
20 Mathematical Statement of The Second Law This mathematical statement of the second law affirms that every process proceeds in such a direction that the total entropy change associated with it is positive.The limiting value of zero being attained only by a reversible process.No process is possible for which the total entropy decreases.
21 Thermodynamic Efficiency In a process producing work, there is an absolute maximum work attainable which is accomplished by completely reversible process. For irreversible process, Wactual produced < Wideal In a process requiring work, there is an absolute minimum amount of work required which is accomplished by completely reversible process. For irreversible process, Wractual required > Wideal
22 Problem 5.12An ideal gas, Cp = 7/2 R, undergoes a cycle consisting of the following mechanically reversible steps:An adiabatic compression from P1, V1, T1 to P2, V2, T2An isobaric expansion from P2, V2, T2 to P3 = P2, V3, T3An adiabatic expansion from P3, V3, T3 to P4, V4, T4A constant-volume process from P4, V4, T4 to P1, V1 = V4, T1Sketch this cycle on a PV diagram and determine its thermal efficiency if T1 = 500 K, T2 = 800 K, T3 = 2000 K, and T4 = 1000 K
23 Problem 5.13A reversible cycle executed by 1 mol of an ideal gas for which Cp = 5/2 R consist of the following processes:Starting at 600 K & 2 bar, the gas is cooled at constant pressure to 300 KFrom 300 K & 2 bar, the gas is compressed isothermally to 4 barThe gas returns to its initial state along a path for which the product PT is constant.What is the thermal efficiency of the cycle?
24 Example 5.4A 40-kg steel casting (Cp = 0.5 kJ/(kg.K) at a temperature of 450oC is quenched in 150 kg of oil (Cp = 2.5 kJ/(kg.K) at 25oC. If there are no heat losses, what is the change of entropy of:The castingThe oilBoth considered together
25 Problem 5.11A piston/cylinder device contains 5 mol of an ideal gas, Cp = 5/2 R, at 20oC & 1 bar. The gas is compressed reversibly and adiabatically to 10 bar, where the piston is locked in position. The cylinder is then brought into thermal contact with a heat reservoir at 20oC, and heat transfer continues until the gas also reaches this temperature. Determine the entropy changes of the gas, the reservoir, and DStotal
26 IrreversibilityOne mole of an ideal gas, Cp = 7/2 R is compressed adiabatically in a piston/cylinder device from 2 bar & 25oC to 7 bar. The process is irreversible and requires 35% more work than a reversible adiabatic compression from the same initial state to the same final pressure. What is the entropy change of the gas?
27 Classical Lost Work & Process Analysis The ideal work is the maximum amount of work which can be done by the process by operating reversibly within the system and by transferring heat between the system and the surroundings reversibly. The lost work: For processes containing several units, lost-work calculations can be made for each unit and summed to determine the overall value.
28 Example 4.7Steam enters a turbine at 1.5 MPa & 500oC and exhausts at 0.1 MPa. The turbine delivers 85% of the shaft work of a reversible-adiabatic turbine although it is neither reversible nor adiabatic. Heat losses to the surroundings at 20oC are 9 kJ/kg steam. Determine temperature, entropy change of the steam leaving the turbine, & the lost work
29 Example 4.8Assume that 5000 kg/h of oil with a heat capacity of 3.2 kJ/kg.K is to be cooled from 220 to 40oC, using a large quantity of water which can be assumed to be at a constant temperature of 30oC. Determine the lost work in the process and the thermodynamic efficiency of the process.
30 Example 4.9Assume that 100 kg of methane gas/h is adiabatically compressed from 0.5 MPa & 300 K to 3.0 MPa & 500 K after which it is cooled isobarically to 300 K by a large amount of water available at 290 K. Determine the efficiency of the compressor. If the surroundings are assumed to be at 290 K, do a thermodynamic analysis of the process. Assume ideal gas. Cp = J/mol.K Cv = J/mol.K
31 8Steam dengan tekanan 12 bar dan temperatur 200 oC diekpansi satu tahap dalam sebuah turbin sehingga tekanannya menjadi 1,5 bar pada kondisi jenuhnya. Turbin bekerja secara adiabatik dengan efisiensi 85%. Perkirakan (dalam sistem satuan SI): Kualitas uap keluar turbin Kebutuhan uap air agar dihasilkan daya sebesar 50 MWatt Kerja ideal, kerja musnah dan perubahan entropi total, jika temperatur lingkungan 25 oC
35 Problem 5.8An ideal gas, Cp = 7/2 R, is heated in a steady-flow heat exchanger from 20oC to 100oC by another stream of the same ideal gas which enters at 180oC. The flow rates of the two streams are the same, and heat losses from the exchanger are negligible.Calculate the molar entropy changes of the two gas streams for both parallel and counter-current flow in the exchangerWhat is DStotal in each case?
36 Carnot CycleSuatu mesin Carnot menyerap panas sebesar 400 kJ dari sumber panas pada 525 oC dan melepas panas ke penerima panas pada 50 oC. Perkirakan kerja yang dihasilkan, perubahan entropy sumber panas, penerima panas dan total!
37 2nd lawSatu mol gas ideal dikompresi secara isotermal-ireversibel pada 130 oC dari 3 bar menjadi 6.5 bar dalam sebuah alat silinder berpiston. Kerja yang dibutuhkan untuk kompresi ini 30% lebih besar dari kerja untuk proses reversibelnya. Selama kompresi, sejumlah panas dibuang ke lingkungan yang memiliki temperatur konstan sebesar 25 oC. Hitung perubahan entropi dari gas, penerima panas dan total!
38 2nd lawUap jenuh dengan tekanan 5 bar dan temperatur 151,87 oC berada dalam silinder yang volumnya 0,750 m3. Silinder tersebut dilengkapi dengan piston yang dapat bergerak bebas. Uap air tersebut kemudian dikompresi secara reversibel sampai diperoleh uap air yang memiliki tekanan 12 bar dan temperatur 187,99 oC. Selama proses kompresi berlangsung, terjadi pembuangan panas ke lingkungan sejumlah 2000 kJ. Temperatur lingkungan diketahui sebesar 27 oC. Perkirakanlah: a. Prosentase uap air yang mengembun b. Perubahan entalpi dan energi dalam (dalam kJ/kg) Kerja yang dibutuhkan untuk kompresi tersebut (dalam kJ) Perubahan entropy uap air, lingkungan dan total
39 2nd lawGas nitrogen dengan laju 2 ton/jam mula-mula memiliki temperatur -168 oC dan tekanan 1 Mpa. Gas nitrogen tersebut kemudian dikompresi secara adiabatik dalam sebuah kompresor sehingga temperatur dan tekanannya masing-masing menjadi –63 oC dan 6 MPa. Dari deskripsi proses tersebut: Dengan mengasumsikan gas nitrogen sebagai gas ideal, perkirakan daya yang dibutuhkan (dalam kWatt), efisiensi mekanik (dalam %) dan perubahan entropi gas nitrogen pada proses kompresi tersebut! Ambil = 1,4. Dengan menggunakan diagram P-H gas nitrogen terlampir, jawablah kembali pertanyaan soal a!
40 7Karbon tetrakhlorida cair sebanyak 5 kg mengalami perubahan secara reversibel pada tekanan konstan sebesar 1 bar dalam sistem tertutup. Akibat perlakuan tersebut, temperatur fluida mengalami perubahan dari 0 oC menjadi 56 oC. Sifat-sifat dari CCl4 cair pada 1 bar dan 0 oC diasumsikan tidak dipengaruhi temperatur: = 1,2 x 10-3 K-1, CP = 0,84 kJ kg-1 K-1, dan = 1,590 kg/m3. Tentukanlah: a. Perubahan entropi dari karbon tetrakhlorida cair b. Temperatur lingkungan pada saat proses perubahan tersebut berlangsung (dalam oC) Jika proses berlangsung secara ireversibel, perkirakan efisiensi mekanik alat apabila kerja yang mampu dihasilkan sebesar 11,67 kJ! Berkaitan dengan soal c, hitunglah perubahan entropi total! Apa kesimpulan yang dapat diambil! Ambil nilai temperatur lingkungan sama dengan hasil perhitungan soal b!