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DICKY DERMAWAN ITK-233 Termodinamika Teknik Kimia I 3 SKS 2 - PVT Behavior of Fluid, Equation of State.

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Presentation on theme: "DICKY DERMAWAN ITK-233 Termodinamika Teknik Kimia I 3 SKS 2 - PVT Behavior of Fluid, Equation of State."— Presentation transcript:

1 DICKY DERMAWAN ITK-233 Termodinamika Teknik Kimia I 3 SKS 2 - PVT Behavior of Fluid, Equation of State

2 P – T Diagram P – v Diagram

3 T-v Diagram

4 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: Equation of State

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

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

7 Problem 3.2 & 3.3 Generally, volume expansivity β 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 60 o C, how much work is required? At 60 o C, b=2700 bar and c = cm 3 g -1

9 Problem 3.5 Calculate the reversible work done in compressing 1 ft 3 of mercury at a constant temperature of 32 o F from 1 atm to 3000 atm. The isothermal compressibility of mercury at 32 o F is: κ/atm -1 = 3.9 x – 0.1 x 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 0 o C to 20 o C. Determine ΔV t, W, Q, and ΔU t The properties for liquid carbon tetrachloride at 1 bar & o o C may be assumed independent of temperature: β = 1.2 x K -1 C p = 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. a. Starting with the definition of κ, show that the path of the process is described by b. Determine an exact expression which gives the isothermal work done on 1 mol of this constant-κ substance

12

13 P – v Diagram

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

15 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.

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

17 Isothermal Process Closed System, Ideal Gas, Mechanically Reversible PV = constant

18 Isobaric Process Closed System, Ideal Gas, Mechanically Reversible TV -1 = constant

19 Isochoric (Constant V) Process Closed System, Ideal Gas, Mechanically Reversible TP -1 = constant

20 Adiabatic Process Closed System, Ideal Gas, Mechanically Reversible

21 Polytropic Process Closed 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 =

22 Example 3.2 Air is compressed from an initial condition of 1 bar & 25 o C to a final state of 5 bar & 25 o C by three different mechanically reversible processes in a closed system: a. Heating at constant volume followed by cooling at constant pressure b. Isothermal compression c. Adiabatic compression followed by cooling at constant volume. Assume air to be an ideal gas with the constant heat capacities: C V = 5/2 R, C P = 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

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

24 Problem 3.8 One mole of an ideal gas with C V = 5/2 R, C P = 7/2 R expands from P1 = 8 bar & T1 = 600 K to P2 = 1 bar by each of the following path: (a) Constant volume (b) Contant temperature (c) Adiabatically Assuming mechanical reversibility, calculate W, Q, U, and H for each of the three processes. Sketch each path in a single PV diagram

25 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. a. Sketch the cycle on a PV diagram b. Determine (where unknown) both T & P for states 1, 2, 3, and 4 c. Calculate W, Q, U, and H for each step of the cycle Data: C V = 5/2 R, C P = 7/2 R

26 Problem 3.10 An ideal gas, C V = 3/2 R & C P = 5/2 R is changed from P 1 = 1 bar & V t 1 = 12 m 3 to P 2 = 12 bar & V t 2 = 1 m 3 by the following mechanically reversible processes: a. Isothermal compression b. Adiabatic compression followed by cooling at constant pressure c. Adiabatic compression followed by cooling at constant volume d. Heating at constant volume followed by cooling at constant pressure e. Cooling at constant pressure followed by heating at constant volume Calculate W, Q, U t, and H t for each of these processes, and sketch the paths of all processes on a single PV diagram

27 Example 3.4 A 400 gram mass of nitrogen at 27 o C 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 & 27 o C. 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: a. The apparatus is immersed in an ice/water bath and is allowed to come to equilibrium b. A variable force is slowly applied to the piston so that the nitrogen is compressed reversibly at the constant temperature of 0 o C 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.

28 Example 3.4 (cont) c. The apparatus is removed from ice/water bath and comes to thermal equilibrium with the surrounding atmosphere at 27 o C d. 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, U & H for the nitrogen for each step of the cycle. Nitrogen may be considered an ideal gas for which C V = 5/2 R and C P = 7/2 R

29

30 Cubic Equation of State: van der Waals Equation (1873)

31 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)

32 Critical Constants & Accentric Factors: Paraffins

33 Critical Constants & Accentric Factors: Olefin & Miscellaneous Organics

34 Critical Constants & Accentric Factors: Miscellaneous Organic Compounds

35 Critical Constants & Accentric Factors: Elementary Gases

36 Critical Constants & Accentric Factors: Miscellaneous Inorganic Compounds

37 Cubic Equation of State: Redlich/Kwong Equation (1949)

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

39 Cubic Equation of State: Peng - Robinson Equation (1976)

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

41

42 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 virial eqn and are used in the petroleum & natural gas industries for light hydrocarbons.

43 Example 3.7 Reported values for the virial coefficients of isopropanol vapor at 200 o C are: B = -388 cm 3 mol -1 C = cm 6 mol -2 Calculate V and Z for isopropanol at 200 o C & 10 bar by: a. The ideal gas equation b. Two term truncated pressure expansion virial equation c. Three term truncated volume expansion virial equation

44 Pitzer Correlation for the Second Virial Coefficient

45 Z0Z0 PrPr TrTr Generalized Correlations 4 Gases: Pitzer type: Lee -Kessler

46 Z0Z0 PrPr TrTr

47 Z1Z1 PrPr TrTr

48 Z1Z1 PrPr TrTr

49 Real Gas, EOS Calculate Z and V for steam at 250 o C and 1,800 kPa by the following: a. Ideal gas equation b. Truncated virial equation with the following experimental values of virial coefficient: B = cm 3 mol -1 C = -5,800 cm 6 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

50

51 Correlation for Liquids: Rackett Equation

52 Correlation for Liquids: Lydersen, Greenkorn & Hougen

53 Problem 3.45 A 30 m 3 tank contains 14 m 3 of liquid n-butane in equilibrium with its vapor at 25 o C. Estimate the total mass of n-butane in the tank. The vapor pressure if n-butane at the given temperature is 2.43 bar.

54

55 Ideal Gas A cylinder pressure vessel having an inside diameter of 50 cm and height of 1.25 cm. Amount of the ammonia gas is confined in that a pressure vessel. Measured pressure and temperature of gas are 30 bars and 200 o C. What are the specific volume and amount of the ammonia gas in pressure vessel (in SI unit), assuming ammonia is an ideal gas?

56 Real Gas, EOS Calculate Z and for steam at 250oC and 1,800 kPa by the following: The ideal gas equation The truncated virial equation with the following experimental values of virial coefficient: B = cm3 mol-1 C = -5,800 cm6mol-2 The van der Waals equation The Redlich/Kwong equation The steam table Data: critical point of steam, TC = K; PC = bar; = 55.9 cm3mol-1

57 1 st law after diagram 1 kmole of ideal gas (Cv = 3R/2) undergoes a three-step mechanically reversible cycle in a closed system. Gas in initial state at 500 kPa and 27 oC is heated at constant pressure Followed by adiabatic expansion until the its pressure become 120 kPa Finally step, gas is isothermally compressed to initial condition From that process description: Illustrate the cycle process at P-V diagram in T parameter! b.Calculate H, U, Q and W for each step of process and total!


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