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ITK-233 Termodinamika Teknik Kimia I

ITK-233 Termodinamika Teknik Kimia I. Dicky Dermawan www.dickydermawan.net78.net dickydermawan@gmail.com. 3 sks. 2 - PVT Behavior of Fluid, Equation of State. P – T Diagram P – v Diagram. T-v Diagram. Equation of State.

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ITK-233 Termodinamika Teknik Kimia I

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  1. ITK-233Termodinamika Teknik Kimia I DickyDermawan www.dickydermawan.net78.net dickydermawan@gmail.com 3 sks 2 - PVT Behavior of Fluid, Equation of State

  2. P – T Diagram P – v Diagram

  3. T-v Diagram

  4. Equation of State For the regions of the diagram where a single phase exist: This means an equation of state exist relating P, V, T. An EOS may be solved for any one of the three quantities as a function of the other two, viz.: V = V(T,P) Volume expansivity: Isothermal compressibility:

  5. Example For liquid acetone at 20oC & 1 bar: Β = 1.487 x 10-3oC-1 κ = 62 x x 10-6 bar-1 V= 1.287 cm3 g-1 For aceton, find: a. The value of b. The pressure generated by heating at constant V from 20oC & 1 bar to 30oC c. The change in volume for a change from 20oC & 1 bar to 0oC & 10 bar.

  6. Problem 3.1 Express the volume expansivityβ and isothermal compressibility κ as functions of density ρ and its partial derivatives. The isothermal compressibility coefficient () of water at 50oC and 1 bar is 44.18 x 10-6 bar-1. To what pressure must water be compressed at 50oC to change its density by 1%? Assume that  is independent of P.

  7. Problem 3.2 & 3.3 Generally, volume expansivityβ and isothermal compressibility κ depend on T and P. Prove that The Tait equation for liquids is written for an isotherm as: where V is specific or molar volume, Vo is the hypothetical molar or specific volume at P = 0 and A & B are positive constant. Find an expression for the isothermal compressibility consistent with this equation.

  8. Problem 3.4 For liquid water the isothermal compressibility is given by: where c & b are functions of temperature only. If 1 kg of water is compressed isothermally & reversibly from 1 bar to 500 bar at 60oC, how much work is required? At 60oC, b=2700 bar and c = 0.125 cm3 g-1

  9. Problem 3.5 Calculate the reversible work done in compressing 1 ft3 of mercury at a constant temperature of 32oF from 1 atm to 3000 atm. The isothermal compressibility of mercury at 32oF is: κ/atm-1 = 3.9 x 10-6 – 0.1 x 10-9 P (atm)

  10. Problem 3.6 Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which the temperature change from 0oC to 20oC. Determine ΔVt, W, Q, and ΔUt The properties for liquid carbon tetrachloride at 1 bar & ooC may be assumed independent of temperature: β = 1.2 x 10-3 K-1 Cp = 0.84 kJ kg-1 K-1 ρ = 1590 kg m-3

  11. Problem 3.7 A substance for which κ is a constant undergo a mechanically reversible process from initial state (P1,V1) to final state (P2,V2), where V is molar volume. • Starting with the definition of κ, show that the path of the process is described by • Determine an exact expression which gives the isothermal work done on 1 mol of this constant-κ substance

  12. P – v Diagram

  13. Virial EOS & Ideal Gas Virial expansion: Alternative form: Compressibility factor For ideal gas: Z = 1, thus

  14. The Ideal Gas The internal energy of a real gas is a function of pressure and temperature. This pressure dependency is the result of forces between the molecules. In an ideal gas, such forces does not exist. No energy would be required to alter the average intermolecular distance, and therefore no energy would be required to bring about volume & pressure changes in an ideal gas at constant temperature. In other word, the internal energy of an ideal gas is a function of temperature only.

  15. Implied Property Relations for an Ideal Gas Thus H is also a funcion of temperature only

  16. Isothermal ProcessClosed System, Ideal Gas, Mechanically Reversible PV = constant

  17. Isobaric ProcessClosed System, Ideal Gas, Mechanically Reversible TV-1 = constant

  18. Isochoric (Constant V) ProcessClosed System, Ideal Gas, Mechanically Reversible TP-1 = constant

  19. Adiabatic ProcessClosed System, Ideal Gas, Mechanically Reversible

  20. Polytropic ProcessClosed System, Ideal Gas, Mechanically Reversible Polytropic process can be considered as general form of process. Isobaric process : δ = 0 Isothermal process : δ = 1 Adiabatic process : δ = γ Isochoric process : δ = ∞

  21. Example 3.2 Air is compressed from an initial condition of 1 bar & 25oC to a final state of 5 bar & 25oC by three different mechanically reversible processes in a closed system: • Heating at constant volume followed by cooling at constant pressure • Isothermal compression • Adiabatic compression followed by cooling at constant volume. Assume air to be an ideal gas with the constant heat capacities: CV = 5/2 R, CP = 7/2 R. Sketch the process in a PV diagram & calculate the work required, heat transferred, and the changes in internal energy & entalphy of the air for each processes

  22. Example 3.3 An ideal gas undergoes the following sequence of mechanically reversible processes in a closed system: • From an initial state of 70oC & 1 bar, it is compressed adiabatically to 150oC • It is then cooled from 150oC to 70oC at constant pressure • Finally, it is expanded isothermally to its original state Sketch the process in a PV diagram & calculate W, Q, DU, and DH for each of the three processes and for the entire cycle. Take CV = 3/2 R and CP = 5/2R

  23. Problem 3.8 One mole of an ideal gas with CV = 5/2 R, CP = 7/2 R expands from P1 = 8 bar & T1 = 600 K to P2 = 1 bar by each of the following path: • Constant volume • Contant temperature • Adiabatically Assuming mechanical reversibility, calculateW, Q, DU, and DH for each of the three processes. Sketch each path in a single PV diagram

  24. Problem 3.9 An ideal gas initially at 600 K & 10 bar undergoes a four-step mechanically reversible cycle in a closed system. In step 1-2 pressure decreases isothermally to 3 bar; in step 2-3 pressure decreases at constant volume to 2 bar; in step 3-4 volume decreases at constant pressure; and in step 4-1 the gas returns adiabatically to its initial step. • Sketch the cycle on a PV diagram • Determine (where unknown) both T & P for states 1, 2, 3, and 4 • Calculate W, Q, DU, and DH for each step of the cycle Data: CV = 5/2 R, CP = 7/2 R

  25. Problem 3.10 An ideal gas, CV = 3/2 R & CP = 5/2 R is changed from P1 = 1 bar & Vt1 = 12 m3 to P2 = 12 bar & Vt2 = 1 m3 by the following mechanically reversible processes: • Isothermal compression • Adiabatic compression followed by cooling at constant pressure • Adiabatic compression followed by cooling at constant volume • Heating at constant volume followed by cooling at constant pressure • Cooling at constant pressure followed by heating at constant volume Calculate W, Q, DUt, and DHtfor each of these processes, and sketch the paths of all processes on a single PV diagram

  26. Example 3.4 A 400 gram mass of nitrogen at 27oC is held in a vertical cylinder by a frictionless piston. The weight of the piston makes the pressure of the nitrogen 0.35 bar higher than that of the surroundings atmosphere, which is at 1 bar & 27oC. Thus the nitrogen is initially at a pressure of 1.35 bar, and is in mechanical & thermal equilibrium with its surroundings. Consider the following sequence of process: • The apparatus is immersed in an ice/water bath and is allowed to come to equilibrium • A variable force is slowly applied to the piston so that the nitrogen is compressed reversibly at the constant temperature of 0oC until the gas volume reaches one-half the value at the end of step a. At this point the piston is held in place by latches.

  27. Example 3.4 (cont’) • The apparatus is removed from ice/water bath and comes to thermal equilibrium with the surrounding atmosphere at 27oC • The latches are removed, and the apparatus is allowed to return to complete equilibrium with its surroundings Sketch the entire cycle on a PV diagram, and calculate Q, W, DU & DH for the nitrogen for each step of the cycle. Nitrogen may be considered an ideal gas for which CV = 5/2 R and CP = 7/2 R

  28. Cubic Equation of State:van der Waals Equation (1873)

  29. Theorem of Corresponding State;Accentric Factor All fluids, when compared at the same reduced temperature & reduced pressure, have approximately the same compressibility factor, and all deviate from ideal gas behavior to about the sam degree Accentric Factor (Pitzer)

  30. Critical Constants & Accentric Factors:Paraffins

  31. Critical Constants & Accentric Factors:Olefin & Miscellaneous Organics

  32. Critical Constants & Accentric Factors:Miscellaneous Organic Compounds

  33. Critical Constants & Accentric Factors:Elementary Gases

  34. Critical Constants & Accentric Factors:Miscellaneous Inorganic Compounds

  35. Cubic Equation of State:Redlich/Kwong Equation (1949)

  36. Cubic Equation of State:Soave/Redlich/Kwong Equation (1972)

  37. Cubic Equation of State:Peng - Robinson Equation (1976)

  38. Example 3.8 Given that the vapor pressure of n-butane at 350 K & 9.4573 bar, find the molar volumes of saturated vapor and saturated liquid n-butane at these conditions as given by: • Van der Waals • Redlich/Kwong • Soave/Redlich/Kwong • Peng-Robinson

  39. Application of The Virial Equations At low to moderate pressure, it is common to use truncated virial equation: At pressure above the range of applicability of the above eqn, the appropriate form is: Benedict/Webb/Rubin equation: and its modifications are inspired by volume expansion virialeqn and are used in the petroleum & natural gas industries for light hydrocarbons.

  40. Example 3.7 Reported values for the virial coefficients of isopropanol vapor at 200oC are:B = -388 cm3 mol-1 C = -26000 cm6 mol-2 Calculate V and Z for isopropanol at 200oC & 10 bar by: • The ideal gas equation • Two term truncated pressure expansion virial equation • Three term truncated volume expansion virial equation

  41. Pitzer Correlation for the Second Virial Coefficient

  42. Generalized Correlations 4 Gases: Pitzer type:Lee -Kessler Tr Z0 Pr

  43. Generalized Correlations 4 Gases: Pitzer type:Lee -Kessler Tr Z0 Pr

  44. Generalized Correlations 4 Gases: Pitzer type:Lee -Kessler Tr Z1 Pr

  45. Generalized Correlations 4 Gases: Pitzer type:Lee -Kessler Tr Z1 Pr

  46. Real Gas, EOS Calculate Z and V for steam at 250oC and 1,800 kPa by the following: a. Ideal gas equation b. Truncated virial equation with the following experimental values of virial coefficient: B = -152.5 cm3 mol-1 C = -5,800 cm6 mol-2 c. Pitzer correlation for virial equation d. Pitzer-type correlation of Lee – Kessler e. Steam table f. van der Waals equation g. Redlich/Kwong equation h. Soave/Redlich/Kwong equation i. Peng-Robinson Equation

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