Thermodynamic ProperDes for Fluids

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1 SKF2213: CHEMICAL ENGINEERING THERMODYNAMICS Thermodynamic roperdes for Fluids Mohammad Fadil Abdul Wahab

2 ObjecDves To develop fundamental property relations (FR) of U,H,S,G,A,V,,T from 1st and 2nd Law of Thermodynamic. Using FR to derive equations for thermodynamic properties such as H and S. To use the concept of residual property in the calculation of thermodynamic property. To develop and utilize the generalized correlation for the calculation of residual property. To be able to use diagram and table of thermodynamic property.

3 roperty RelaDons for Homogeneous hases Consider the following system, Closed system Homogeneous i.e. single phase and no change in composition! e.g. Air Initial conditions 1 and T 1 Final conditions 2 and T 2 No phase change!! 1st Law (Energy Balance) for closed system, d(nu) = Q + W

4 From definition of entropy (S),..see chapter 5 d(ns) = Q rev T For reversible process the EB becomes, d(nu) = Q rev + W rev d(nu) = Td(nS) d(nv ) ( 6.1)

5 We know that H=U+V, now let us also define, Differentiate these properties (H, A and G), d ( nh ) = d ( nu ) + d ( nv ) + nvd substitute eqn 6.1 for dnu, d ( nh ) = Td(nS) d(nv ) + d ( nv ) + nvd d ( nh ) = Td ( ns) + nvd

6 Continue, d( na) = d( nu ) Td ( ns) ( ns)dt substitute for dnu, ( ) = Td(nS) d(nv ) Td ( ns) ns ( ) = d ( nv ) ( ns)dt d na d na ( )dt d ( ng) = d ( nh ) Td ( ns) ( ns)dt substitute for dnh, ( ) = Td ( ns) + (nv )d Td ( ns) ns ( ) = ( nv )d ( ns)dt d ng d ng ( )dt

7 For n=1, ( ) ( ) ( ) ( ) du = TdS dv 6.7 dh = TdS + Vd 6.8 da = dv SdT 6.9 dg = Vd SdT 6.10 These are called fundamental property relations (FR). Note: The unit for TS and V is energy unit. Notice that the FR for Gibbs Energy is a function of commonly measured variable and T, G = G( T,)

8 All FRs are in the form of, df = MdX + NdY (6.11) where ( ) F = F X,Y Since F is a state function, we can differentiate F(X,Y) as follows, df = F X So let, Y dx + F Y X dy M = F X df = MdX + NdY Y N = F Y X

9 Differentiate one more time, M Y So, X = 2 F Y X N X Y = 2 F X Y M Y X = N X Y (6.12) Eqn 6.12 is the criterion of exactness for an exact differential expression of Eqn 6.11.

10 Apply criterion of exactness to FRs, du = TdS dv dh = TdS + Vd da = dv SdT dg = Vd SdT T V T S S T V T V = S = V S = S V T V = S T ( 6.13) ( 6.14) ( 6.15) ( 6.16) These (Eqn ) are known as Maxwell s equations

11 Application of FRs and Maxwell s Eqn: Use in the derivation of the general equation for H H = H(T, ) dh = H T dt + H Energy balance, T d (a) Q + W = du + de K + de Q dv = du Q = du + dv Note: for constant pressure process, dh = du + dv = du + dv + V d Q = (dh ) C p = Q dt = H dt So eqn (a) becomes, dh = C dt + H T d (b)

12 From FR, dh = TdS + Vd differentiate wrt at constant T, H H T T = T S = T S T T + V + V T From Maxwell eqn, V T = S T H Substitute into (b), T = T V T + V (6.19) dh = C dt + V T V T d ( 6.20) This general eqn for enthalpy is in term of measured variable,v,t

13 Application of FRs and Maxwell s Eqn - to derive general equation for S From FR, S = S(T, ) ds = S T dt + S T d (b) dh = TdS + Vd H T S T = T S T = C p T + V T = T S T = C p From Maxwell s equation, Substitute into (b), V T ds = C dt T = S V T T so, S T = V T d ( 6.21) This general eqn for entropy is in term of measured variable,v,t

14 ApplicaDon of General EquaDon of H (Eqn 6.20) and S (Eqn 6.21) for Homogenous and Constant ComposiDon Fluid. -For system with ideal gas, V = RT, V = RT, so V T = R Substitute into (6.20) and (6.21), dh ig = C ig dt + V T R d = C ig dt + V V d dh ig = C ig dt ( 6.23) ds ig ig dt = C T R d ( 6.24) These are similar to eqn 4.2 and eqn 5.14

15 -Alternative forms for liquids From Maxwell s eqn and the definition of volume expansivity, Also from 6.19, H T = V T V T = ( 1 βt )V (6.26) Substitute into 6.20, dh = C dt + V T V T d ( 6.20) dh = C dt + ( 1 βt )Vd (6.28) This is alternative form for liquid.

16 For entropy, S T = V T = βv (6.25) Substitute into eqn 6.21, ds = C dt T V T d ( 6.21) ds = C dt T βvd (6.29) This is alternative form for liquid.

17 See example 6.2 For incompressible liquid (β = κ = 0), C = C v

18 Let s look again at general equation for enthalpy for homogenous and constant composition system dh = C dt + V T V T d ( 6.20) To use eqn 6.20, we needs Initial and final T & VT relation or VT data C For gas phase however, most C are tabulated for ideal gas only i.e. C p ig (Table C.1) Unless we have actual C for gases, we can t use eqn 6.20 to calculate the enthalpy for real (actual) gas!!

19 How can we calculate the enthalpy of a real gas using an ideal gas heat capacity? Well, a. First we calculate H ig using C ig. b. Then we use VT relation or VT data to determine the residual enthalpy (H R ) which is the DIFFERENCE between real enthalpy (H) and ideal gas enthalpy (H ig ). Residual Enthalpy H R = H - H ig c. So, the real enthalpy is found by adding H R to the H ig. H = H ig + H R

20 Residual roperties Residual roperty = Real roperty - Ideal Gas roperty H R = H H ig S R = S S ig G R = G G ig etc. For example, to calculate H or H 2 -H 1 we shall use the hypothetical path.

21 ΔH = ΔH 1 + ΔH 2 + ΔH 3 ΔH = ΔH 1 + ΔH 2 ig + ΔH 3 ΔH = ( H ig 1 H ) ig 1 + C ( T T H 2 1) + ( ig H 2 H ) 2 ΔH = H 1 R + C ig T T R ( H 2 1) + H 2 ΔH = C ig ( T T H 2 1) + H R R 2 H 1 (6.93) Note, Now we need to figure out how to determine H R? C ig H is mean heat capacity and is given by equation (4.8)

22 To determine H R, let us start with the Gibbs Free Energy. G = G ( T,) divide by RT and differentiate, d G RT = 1 RT dg + G R d 1 T = 1 RT dg G RT 2 dt Substitute FR for dg, d G RT = 1 RT ( ) ( Vd SdT ) H TS RT 2 dt = V RT d S RT dt H RT 2 dt + S RT dt d G RT = V RT d H RT dt ( )

23 Let us do for residual Gibbs energy, G R = G G ig divide by RT and differentiate, d G R RT = d G RT d G ig RT Apply eqn 6.37, we get d G R RT ( ) = V V ig RT ( d H H ig ) dt RT 2 d G R RT = V R RT d H R dt RT ( )

24 Apply criterion of exactness. V R RT = H R RT = T ( ) R G RT R G RT T T ( ) ( 6.43) ( 6.44) Rearrange eqn 6.43 and integrate from ideal gas state (=0) to arbitrary (actual ), act. ( ) d G R = V R RT RT d ig 0

25 G R RT 0 = V R d = V V ig RT d RT 0 G R RT = ZRT RT 1 RT d ( ) ( ) G R RT = Z 1 d 6.49 Differentiate wrt. T at constant, ( ) dt d G R RT = ( ) ( ) Z 1 d = 1 Z T T T d = Z T d SubsDtute into eqn 6.44 for H R

26 We will get, H R RT =-T Z d T ( ) Similarly from, S R R = H R RT G R RT We will get, S R R = T 0 Z T ( ) d Z 1 0 d ( 6.48) Hence, as in eqn 6.46 and 6.48, we need either VT data or correlations for Z to solve for H R and S R.

27 Enthalpy for real gas H = H ig + H R H = H o ig + T T o C ig H = H o ig + C ig dt + H R H ( T T o ) + H R ( 6.52) note: ΔH ig = H ig H o ig = H ig = H o ig + T T o T T o C ig C ig dt dt Note: Here the reference state is ideal gas at T o, o. H = H 2 - H 1 is calculated as follows, H 1 = H o ig + C ig H 2 = H o ig + C ig H H T T R ( 1 o ) + H 1 T T R ( 2 o ) + H 2 ΔH = H 2 H 1 = C ig ( T T ) + H R R H H 1 (6.93)

28 Similarly, entropy for real gas S = S ig + S R S = S o ig + T T o C ig S = S o ig + C ig dt T Rln o + S R ln T Rln +S R ( 6.53) S T o o ΔS = C ig where C ig ln T 2 Rln 2 + S R R S T S 1 1 = R A + BT + CT 2 + D S o o τ 2 2 T o (6.94) τ + 1 τ 1 2 lnτ (5.17)

29 H R and S R by EOS Using two- term Virial EquaDon Combined Eqn 3.38, 6.49 and 6.44 will give, H R RT = R B T db dt (6.55) S R R = R db dt (6.56) Using three-term Virial Equation Combined Eqn 3.40, 6.49 in term of ρ H R RT = T B T db dt ρ + C T 1 2 dc dt ρ2 (6.62) Later we will show a much easier to use generalized virial-coefficient eqn..

30 Using Generic Cubic EOS H R RT = Z 1+ d lnα(t ) r 1 qi (6.67) d lnt r Where and S R R = ln(z β) + d lnα(t r ) d lnt r qi (6.68) β=ω r T r (3.53) q= Ψα(T ) r ΩT r (3.54) For ε = σ, I= β Z+εβ For ε σ, 1 Z + σβ I = ln σ ε 1+ εβ Note: We first have to solve for Z, using eqn 3.52 for vapor and vapor-like root.

31 H R and S R from Generalized CorrelaDons from = c r T = T c T r d = c d r dt = T c dt r Substitute into eqn 6.46 and 6.48, R = T 2 2 Z T c r T c T o r H R r r c d r c r r 2 Z = T r T o r r d r r (6.83) S R R = T c T r r o Z T c T r r r c d r c Z 1 r o ( ) r c d r Z = T c r r T o r r r d r Z 1 r o ( ) d r r (6.84)

32 The Lee/Kesler Correlation From Z = Z 0 + ωz 1 Z T r = Z o T r r + ω Z T r r Substitute into eqn 6.83 and 6.84, r H R RT c = T r r 2 Z o o T r r d r r r 2 Z - ωt r T r o r d r r r R = T r o S R r R = T r o S R Z o T r Z o T r + ω Z d r T r - ( Z o + ω Z 1) r r r r o d r + Z o 1 r Z - ω T r r r T + Z r o r d r r d r r ( ) 0 H R = H R RT c RT c ( ) 1 + ω H R (6.85) RT c S R R = S R ( ) 0 R + ω S R ( ) 1 R c (6.86)

33 Tables E5- E12 developed by Lee/Kesler give the values for the followings, ( H R ) 0 RT c, ( H R ) 1 R c T c, ( S R ) 0 R, S ( R ) 1 R c Substitute these values into equation 6.85, we get H R = ( H R ) 0 RT c ( ) 1 + ω H R RT c RT c

34 Generalized second- virial- coefficient Differentiate, Z =1 + B o r T r +ωb ' r T r ( ) Z o db 1 = T r + B d 1 o T r r dt r r T r dt r ( ) db ' 1 d 1 +ω r + B ' T r dt r T r dt r o Z db 1 = r Bo db ' 1 T r 2 dt r r T r T +ω r B' 2 r dt r T r T r Substitute into eqn 6.83 and 6.84, H R r B o = T RT r Bo c +ω B' T r T o r T r S R R = r o B o T r ω B' T r d r B' T r d r

35 As B 0 and B 1 are function of T only, the terms in parenthesis are constants, so integrate at constant T gives: H R B 0 B 1 = RT r B 0 T r + ω B 1 T c T r r T r (6.87) S R R = r B 0 T r ω B1 T r (6.88) Where B 0 = T r 1.6 (3.65) db 0 = (6.89) 2.6 dt r T r B 1 = T r 4.2 (3.66) db 1 dt r = T r 5.2 (6.90)

36 ( ) 0 H R = H R RT c RT c Where, Z, H R and S R for Mixtures Z = Z 0 + ωz 1 (3.57) + ω H ( R ) 1 (6.85) S R RT c R = S ω = R ( ) 0 y i ω i T pc = y i T ci pc = y i ci i i R + ω S R ( ) 1 (6.86) R c i T pr = T T pc pr = pc seudo-reduced parameters Refer to Table E5-E12 for values of, ( H R ) 0 RT c, ( H R ) 1, R c T c ( S R ) 0 ( ) 1 R, S R Refer to Table E1-E4 for values Z 0 and Z 1 R c

37 hase Equilibrium for System with ure Substance V L T,

38 Two- hase Systems Closed system Two phase in equilibrium (vapor-liquid equilibrium, VLE) at T,, n l, n v For this system, G(T,,n l,n V ) n v n l ( ) = ng d ng T dt +.n l,n v ng d + T,n l,n v ng n l T,,n v dn l + ng n v T,,n l dn v For constant T, and overall composition, 0 = G l dn l + G v dn v

39 Where we define partial molar property as, G l = ng n l T,,n v G v = ng n v T,,n l Since this is pure substance, the partial molar property of liquid is actually the property of liquid!! i.e. G l = G l and G v = G v

40 Note: DefiniDon of ardal Molar roperdes in Chapter 11 M i = ( nm ) n i (11.7),T,n j Where M V,U, H,S,G etc. M i is a property of pure species i M i is property of species i inside the mixture or solution

41 Continue, Also from mole balance, n v n l n l + n v = n t Note: Closed system dn l + dn v = dn t dn l = dn v

42 Substitute, 0 = G l dn l G v dn l ( G l G v )dn l = 0 G l = G v This is the phase equilibrium criteria (or vapor liquid equilibrium criteria) for pure substance. Let us check our steam table at T sat = 100 o C, G v = H V TS V = (7.3554) = 68.6 kj / kg G l = H l TS l = (1.3069) = 68.6 kj / kg

43 Consider system A at 2 different equilibrium conditions, in equilibrium at saturation conditions 1: T 1, 1, G l1, G v1 and in equilibrium at saturation conditions 2: T 2, 2, G l2, G v2 From equilibrium criteria, G l1 = G V1 Eqn 1 G l 2 = G V 2 Eqn 2

44 Let Eqn 2 Eqn 1, From FR, G l 2 G l1 = G V 2 G V1 dg l = dg v V l d sat S l dt = V v d sat S v dt d sat dt lv ΔS = ΔV lv dh = TdS + Vd But for vaporization of pure substance at constant, ΔH lv = TΔS lv + 0 Substitute, d sat dt lv ΔH = ersamaan Clapeyron (6.72) or (4.11) lv T ΔV

45 The equation could also be written as follows, d ln sat d( 1 T ) = ΔH lv Clapeyron Equation (6.74) lv RΔZ At low pressure (ideal gas), Clapeyron eqn 6.72 becomes, or, d sat sat dt T 2 = ΔH vap R d sat dt vap ΔH = RT 2 and sat V v = RT sat, and also V l << V v Clausius/Clapeyron equation d ln ΔH vap sat = R d 1 T To solve this equation, we only need either vapor pressure data or equation relating vapor pressure to boiling point.,

46 Vapor ressure ( sat ) vs Boiling oint Temp (T sat ) 1. Antoine equation, See Table B2 for the constants ln sat = A B T sat + C (6.76) 2. Wagner equation, ln r sat = Aτ + Bτ Cτ 3 + Dτ 6 1 τ (6.77) τ =1-T r 3. AIChE (1984) equation, ln sat = A + B T sat + ClnT sat + D(T sat ) E

47 4. Also, we could use Lee and Kesler Correlations Where, ln r sat ( T ) r ) = ln 0 ( r T r ) )+ωln 1 ( r T ) r ) (6.78) ln 0 ( r T r ) )= T r ln 1 ( r T r ) )= T r ω = ln sat rn lnT r T r lnT r T r ( ) ( T ) r ) ln 0 r T rn ln 1 r ( T rn ) Note: This correlation was developed for non-polar liquids.

48 Thermodynamic roperties for Two-hase Liquid/Vapor System M = M l + x v ΔM lv M V,U, H,S etc. x v is also known as quality, that is the mass fraction of vapor in the system, x v =m v /m T =n v /n T

49 Thermodynamic Diagrams Show the relationship of thermodynamic variables (T,,V, H, S) on a graph for a particular substance. For example: TS Diagram, H Diagram, V Diagram, HT Diagram HS Diagram (known as Mollier Diagram) Useful for analysis of thermodynamic processes as the paths of processes are easily traced and visualized.

50 Thermodynamic Tables Thermodynamic table enable us to read values of thermodynamic properties accurately. Usually requires interpolation (as well as extrapolation). For example: Ammonia Table, Freon Table, Methane Table, ropane Table Saturated Table (T vs, V sat liq, V lv, V sat vap, U sat liq, U lv, U sat vap, H sat liq, H lv, H sat vap, S sat liq, S lv, S sat vap ) Superheated Table ( and T vs V, U, H, S) Compressed Liquid Table After completion of chapter 3 and 6, you should be well equipped to develop thermodynamic table of a pure substance.

51 IAWS For further reading, The Interna2onal Associa2on for the roper2es of Water and Steam hcp:// rovide internadonally accepted formuladons for the properdes of steam, water and selected aqueous soludons for sciendfic and industrial applicadons. You can get sodware wricen based on IAWS formuladons

52 Reference Smith J.M., Van Ness H.C., and Abboc M. M., IntroducDon to Chemical Engineering Thermodynamics, 7 th EdiDon, McGraw Hill, New York, 2001.

For an incompressible β and k = 0, Equations (6.28) and (6.29) become:

For an incompressible β and k = 0, Equations (6.28) and (6.29) become: Internal Energy and Entropy as Functions of T and V These are general equations relating the internal energy and entropy of homogeneous fluids of constant composition to temperature and volume. Equation