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Chapter 3 Thermodynamic Properties of Fluids CHAPTER 3 THERMODYNAMIC PROPERTIES OF FLUIDS 3.1 Introduction There are two (2) main objectives of this chapter; the first is to develop the overall relation between thermodynamic properties and second, to apply the relations so numerical and meaningful values are acquired based on actual processes. 3.2 Thermodynamic Properties Function Relations Primary thermodynamic properties are P, V, T, U and S, while H, A and G are secondary properties given by the following relations; Enthalpy H U PV (3.1) Helmholtz A U TS (3.2) Gibbs G H TS (3.3) Chemical Engineering Thermodynamics 1 RAR Chapter 3 Thermodynamic Properties of Fluids From the first and second law of thermodynamics, for a closed system with changes occur only between equilibrium states, an equation is derived for reversible process d nU Td nS Pd nV (3.4) Based on the above equations, for 1 mole of a homogeneous fluid at constant composition, the fundamental property relations are as follows; dU TdS PdV (3.5) dH TdS VdP (3.6) dA PdV SdT (3.7) dG VdP SdT (3.8) 3.3 Table of Thermodynamic Properties Thermodynamic properties may be found also in form of tables. Unfortunately, only water has that of completed ones at various ranges of temperature and pressure. Other fluids have to be calculated. Let’s review the following examples. Example 1 Steam undergoes a change from initial state of 4500C and 3,000 kPa to a final state of 1400C and 235 kPa. Determine its enthalpy, H and entropy, S . Answer Initial condition P1 = 3,000kPa and T1 = 4500C Final condition P2 =235kPa and T2 = 1400C Chemical Engineering Thermodynamics 2 RAR Chapter 3 Thermodynamic Properties of Fluids From Table F2 (page 740), at P1 = 3,000kPa and T1 = 4500C, you will get H 1 3344.6 J and S1 7.0857 J . g gK From Table F2, at P2 =235kPa and T2 = 1400C (413.15K), interpolate between 225kPa & 250 kPa and 400K & 450K, you will get H 2 2744.5 J and S 2 7.2003 J g gK So, H H 2 H 2 2744.5 J 3344.6 J 600.1 J and g g g S S 2 S1 7.2003 J 7.0857 J 0.0115 J gK gK gK Exercise 1 Steam at 300psia and 5000F is throttled to 20psia. a) What is the temperature of steam in its final state? b) What is the entropy change? c) What would be the final temperature and entropy change for an ideal gas? 3.4 Residual Thermodynamic Properties Residual thermodynamic properties, M R also known as departure properties are the difference between real thermodynamic properties and ideal thermodynamic properties. For instance, at a certain pressure and temperature, a thermodynamic property is supposed to have an ideal value, but in reality (through measurements or calculation that has included noise factors) it does not. So, the difference is known as residual. In another words, residual property is how much the real property deviate from the ideal property. M R M M ig (3.9) Chemical Engineering Thermodynamics 3 RAR Chapter 3 Thermodynamic Properties of Fluids Where ; M R residual property M real property M ig ideal-gas property In this course, the two (2) main thermodynamics that we will focus on is enthalpy, H and entropy, S. However, please note that in thermodynamics, actual H and S values is relatively unimportant. What we want to find is actually their difference from initial to final state for application in industrial (real) processes. In this case H and S . Let say we have a fluid coming in a compressor and exiting at a certain pressure and temperature as follows; P1, T1 P2, T2 Initial State Final State Figure 1 Compressor How are you going to calculate the changes in its enthalpy, H and entropy, S ? Chemical Engineering Thermodynamics 4 RAR Chapter 3 Thermodynamic Properties of Fluids P1, T1 H , S P2, T2 H 1R , S1R H 2R , S 2R H ig , S ig Figure 2 Calculation Path Now that we know WHAT we are suppose to calculate, lets find the correlation (equations) to solve for enthalpy, H and entropy, S . The following table simplify on the derivation equation that will be used to calculate both H and S. Table 1 Derivation of Enthalpy Correlation M R M M ig (3.9) H H ig H R (3.10) Where T H H C ig dT ig ig 0 p (3.11) T0 H 0 is ideal-gas enthalpy at reference conditions ( T0 298.15 K , P0 1 bar ) ig Substitute (3.11) in equation (3.10) gives T H H 0 C ig dT H R ig p (3.12) T0 Note that Chemical Engineering Thermodynamics 5 RAR Chapter 3 Thermodynamic Properties of Fluids T C ig p dT R * ICPH R * MCPH * T T0 (3.13) T0 So, D T C ig dT R * T T0 * A T0 1 T02 2 1 2 B C T0 p 2 3 T0 (3.14) Equation (3.12) may be inserted in (3.13) to solve for H. The final equation is as follows; D ig B C H H 0 R * T T0 * A T0 1 T02 2 1 2 H R T0 (3.15) 2 3 How H can become H ? Remember that H is the difference between initial state and final state. So, H H 2 H1 (3.16) From equation (3.15), if D ig B C H 1 H 0 R * T T0 * A T0 1 T02 2 1 2 H 1R T0 (3.17) 2 3 And D ig B C H 2 H 0 R * T T0 * A T0 1 T02 2 1 2 H 2R T0 (3.18) 2 3 Equation (3.16) becomes; D H R * T T0 * A T0 1 T02 2 1 2 H 2R H 1R B C T0 (3.19) 2 3 Chemical Engineering Thermodynamics 6 RAR Chapter 3 Thermodynamic Properties of Fluids ig H 0 is cancelled out because it is ideal-gas enthalpy at reference conditions ( T0 298.15 K , P0 1 bar ). Table 2 Derivation of Entropy Correlation M R M M ig (3.9) S S ig S R (3.20) Where T dT P S ig S 0 C ig ig p R ln (3.21) T0 T P0 S ig 0 is ideal-gas entropy at reference conditions ( T0 298.15 K , P0 1 bar ) Substitute (3.21) in equation (3.20) gives T dT P S S 0 C ig ig p R ln S R (3.22) T0 T P0 Note that T dT C R * ICPS R * MCPS * ln ig p (3.23) T0 T So, T dT D 1 1 Cp ig R * ln * A BT0 CT02 2 (3.24) T0 T T0 2 ln Equation (3.24) may be inserted in (3.22) to solve for S. The final equation is as follows; D 1 1 S S 0 R * ln * A BT0 CT02 2 ig 2 ln S R (3.25) T0 Now that we know the overall equation that we need to solve, a new question arises. How to solve for the residual thermodynamic properties? Chemical Engineering Thermodynamics 7 RAR Chapter 3 Thermodynamic Properties of Fluids 3.4 Solving Residual Thermodynamic Properties Residual thermodynamic properties may be determined through the following methods; a) Virial EOS b) Cubic EOS c) Lee-Kesler Generalized Correlation d) Generalized Second Virial Coefficient Correlation 3.4.1 Virial EOS A common form of the virial EOS is when it is truncated to two (2) or three (3) terms. BP Z 1 B C 2 (3.26) RT Based on the above three-term virial equation, residual Gibbs energy, residual enthalpy and residual entropy may be found from the following equations; GR 3 i) 2 B C 2 ln Z (3.27) RT 2 HR B dB C 1 dC 2 ii) T (3.28) RT T dT T 2 dT SR B dB 1 C dC 2 ln Z T 2 T dT iii) (3.29) RT T dT 3.4.2 Cubic EOS The second applicable method to calculate residual thermodynamic properties is via the cubic EOS. However, there are two (2) cases to consider; Chemical Engineering Thermodynamics 8 RAR Chapter 3 Thermodynamic Properties of Fluids Case 1: 1 1 b 1 Z I 1 b ln Z ln (3.30) Or Case 2: b I (3.31) 1 b Z Thus, for residual Gibbs energy, residual enthalpy and residual entropy may be found from the following equations; GR Z 1 ln Z qI (3.32) RT HR d ln Tr Z 1 1 qI (3.33) RT d ln Tr SR d ln Tr ln z qI (3.34) RT d ln Tr But before you solve the equations, you need to find Z first (from Chapter 1) for either vapor phase or liquid phase. 3.4.3 Lee-Kesler Generalized Correlation Using Lee-Kesler generalized correlation, the following equations must be solve to get residual enthalpy and residual entropy’ Chemical Engineering Thermodynamics 9 RAR Chapter 3 Thermodynamic Properties of Fluids HR HR HR 0 1 (3.35) RTc RTc RTc SR SR 0 SR 1 (3.36) RTc RTc RTc 3.4.4 Generalized Second Virial Coefficient Correlation The fourth and final method is a personal favorite of mine. HR 0 dB 0 1 dB1 Pr B Tr B Tr (3.37) RTc dTr dTr SR dB 0 dB1 Pr dT (3.38) RTc r dTr Where; 0.422 B 0 0.083 (3.39) Tr1.6 0.172 B1 0.139 (3.40) Tr4.2 dB 0 0.675 2.6 (3.41) dTr Tr dB1 0.722 5.2 (3.42) dTr Tr Chemical Engineering Thermodynamics 10 RAR Chapter 3 Thermodynamic Properties of Fluids 3.5 Two-Phase Systems Preceding topics has concentrated only single (vapor) phase thermodynamic properties. That is when you have inlet and outlet of the same phase (vapor in, vapor out). So, how to solve for conditions where you have a liquid inlet and vapor outlet? 3.5.1 Liquid Vapor Pressure An important property to be determined before hand is the vapor pressure of liquids. To do this, there are four (4) methods available; a) Rough estimation, a nearly linear relation between vapor pressure and temperature. B ln P sat A (3.43) T b) Antoine equation B ln P sat A (3.44) T C Antoine equation is the most commonly method used to get vapor pressures. A, B and C are constant values for fluids. c) Wagner equation A B 1.5 C 3 D 6 ln P sat (3.45) 1 r Where 1 Tr Chemical Engineering Thermodynamics 11 RAR Chapter 3 Thermodynamic Properties of Fluids Wagner equation is less common, however it present a more accurate vapor pressure values. Its constant A, B, C and D for fluids also not available in your text book! d) Lee-Kesler correlation ln Prsat Tr ln Pr0 Tr ln Pr1 Tr (3.46) Where; ln Pr0 Tr 5.92714 6.09648 1.28862 ln Tr 0.169347Tr6 (3.47) Tr ln Pr1 Tr 15.2518 15.6875 13.4721ln Tr 0.43577Tr6 (3.48) Tr ln Prnsat ln Pr0 Trn and ln Pr1 Trn (3.49) Trn = reduced normal boiling point Prnsat = reduced vapor pressure 3.5.2 Phase Transition In case of two phase system, where you have an initial liquid inlet and vapor outlet (which is common in industrial processes) at constant pressure and constant pressure yields latent heat of phase transition (vaporization); H lv TS lv (Changes from liquid, l to vapor, v phase) (3.50) Chemical Engineering Thermodynamics 12 RAR Chapter 3 Thermodynamic Properties of Fluids An estimation of latent heats of vaporization for pure fluids (either liquid or vapor) at their respective normal boiling point is given by equation by Riedel; H n 1.092ln Pc 1.013 lv (3.51) RTn 0.930 Trn To estimate the latent heat of vaporization of a pure fluid at any temperature from a known value (normal boiling point) is given by equation by Watson; 0.38 H lv 1 Tr (3.52) H n 1 Trn lv Both of the latter equation may be use concurrently. 3.5.3 Vapor and Liquid In Equilibrium For vapor and liquid in equilibrium, i.e. saturated vapor and saturated liquid in equilibrium, the total value of any extensive property is equal to total property of the phase. M V ,U , H , S etc M 1 xv M l xv M v M M x v M lv l Where x v is vapor quality. 3.5.4 Calculational Path For Two Phase System Chemical Engineering Thermodynamics 13 RAR Chapter 3 Thermodynamic Properties of Fluids P1, T1 H , S H lv , S lv P2, T2 H 2R , S 2R H 1R , S1R H ig , S ig Figure 3 Calculation Path for Two Phase System Example 2 A steam power plant employs an adiabatic turbine. Steam enters the turbine at 6500C and 7,000 kPa and discharges at 20 kPa and 5000C. If you are asked to calculate the enthalpy and entropy change of the process sketch the relevant calculational path. Answer Chemical Engineering Thermodynamics 14 RAR Chapter 3 Thermodynamic Properties of Fluids P1=7,000 kPa, T1=6500C H , S P2=20 kPa, T2=5000C H 1R , S1R H 2R , S 2R H ig , S ig Exercise 2 Propane gas at 6500C and 5 bar is discharged to the atmosphere at 500C. If you are asked to calculate the enthalpy and entropy change of the process sketch the relevant calculational path. Example 3 Calculate compressibility factor, Z, residual enthalpy and residual entropy for ethanol at 450K and 5 bar using Peng-Robinson EOS. Answer a) Find Z To find Z, use equation from Chapter 3; Z Z 1 q Z Z bP P aT Tr Where r and q which is solved iteratively, starting with RT Tr brT Tr Z=1. Chemical Engineering Thermodynamics 15 RAR Chapter 3 Thermodynamic Properties of Fluids From the table of pure properties for ethanol; 0.645 Tc 513.9 K Pc 61.48 bar P 5 T 450 Pr 0.0813 Tr 0.8757 Pc 61.48 Tc 513.9 Pr 0.0813 So, 0.0778 0.0072 Tr 0.8757 Tr 0.37464 1.54226 0.26992 2 1 Tr0.5 1.1680 2 1 Tr 0.457241.1680 q 7.8388 Tr 0.07780.8757 Z 0.0072 Z 1 0.0072 7.83880.0072 Z 1 2 0.0072Z 1 2 0.0072 Z 0.0772 Z 1.0072 0.0564 Z 0.003Z 0.0174 Refer to Table 3.1. Chemical Engineering Thermodynamics 16 RAR Chapter 3 Thermodynamic Properties of Fluids For PREOS, it is Case 2, where . b I (3.31) 1 b Z Chemical Engineering Thermodynamics 17 RAR

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