Chater 3 3.4- The Comressibility Fator Equatio of State The dimesioless omressibility fator, Z, for a gaseous seies is defied as the ratio Z = (3.4-1) If the gas behaes ideally Z = 1. The extet to whih Z differs from 1 is a measure of the extet to whih the gas is behaig oideally. The omressibility a be determied from exerimetal data where Z is lotted ersus a dimesioless redued ressure R ad redued temerature T R, defied as R = / ad T R = T/T I these exressios, ad T deote the ritial ressure ad temerature, resetiely. A geeralized omressibility hart of the form Z = f( R, T R ) is show i Figure 3.4-1 for 10 differet gases. The solid lies rereset the best ures fitted to the data. Figure 3.4-1 Geeralized omressibility hart for arious gases 10. It a be see from Figure 3.4-1 that the alue of Z teds to uity for all temeratures as ressure aroah zero ad Z also aroahes uity for all ressure at ery high temerature. If the,, ad T data are aailable i table format or omuter software the you should ot use the geeralized omressibility hart to ealuate,, ad T sie usig Z is just aother aroximatio to the real data. 10 Mora, M. J. ad Shairo H. N., Fudametals of Egieerig Thermodyamis, Wiley, 008, g. 11 3-19
Examle 3.4- ---------------------------------------------------------------------------------- A losed, rigid tak filled with water aor, iitially at 0 MPa, 50 o C, is ooled util its temerature reahes 400 o C. Usig the omressibility hart, determie (a) the seifi olume of the water aor i m 3 /kg at the iitial state. (b) the ressure i MPa at the fial state. Comare the results of arts (a) ad (b) with alues obtaied from the thermodyami table or software 11. Solutio ------------------------------------------------------------------------------------------ (a) The seifi olume of the water aor i m 3 /kg at the iitial state. Look u the ritial temerature T ad ritial ressure of water: Substae Chemial Formula M (kg/kmol) T (K) (bar) Water H O 18.0 647.3 0.9 0.33 Ealuate the redue ressured R ad redue temeratured T R R1 = 0/.09 = 0.91, T R1 = (50 + 73.15)/647.3 = 1.3 Figure E3.4- Geeralized Comressibility Chart With these alues for the redued ressure ad redued temerature, the alue of Z from Figure E3.4- is aroximately 0.83 Z = = M 1 1 = Z 1 M 1 `` 8314 N m/kmol K 793.15 K 1 = 0.83 18.0 kg/kmol 0 10 N/m 6 = 0.015 m3 /kg 11 Mora, M. J. ad Shairo H. N., Fudametals of Egieerig Thermodyamis, Wiley, 008, g. 113 3-0
This alue is i good agreemet with the seifi olume of 0.01551 m 3 /kg from CATT rogram (Table E3.4-). Table E3.4- Water roerties from CATT rogram Seifi Iteral Tem Pressure Volume Eergy State C MPa m 3 /kg kj/kg 1 50 0 0.01551 99 400 15.1 0.01551 739 (b) The ressure i MPa at the fial state. Sie both mass ad olume remai ostat, the water aor ools at ostat seifi olume ad thus at redued seifi olume ' R = ( 3 )( 6 ) = 0.015 m /kg.09 10 N/m 8314 N m 647.3 K 18.0 kg K ( ) = 1.1 The redued temerature at the fial state is T R = (400 + 73.15)/647.3 = 1.04 Figure E3.4- Geeralized Comressibility Chart Loatig the oit o the omressibility hart where ' R = 1.1 ad T R = 1.04, the orresodig alue for R is about 0.69. The fial ressure is the = ( R ) = (.09 MPa)(0.69) = 15.4 MPa This alue is i good agreemet with the ressure of 15.1 MPa from CATT rogram (Table E3.4-). 3-1
3.4-3 Virial Equatio of State A irial equatio of state exresses the quatity olume. as a ower series i the ierse of molar = 1 + B( T ) C( T ) + D( T ) + 3 + (3.4-8) I this equatio, B, C, ad D are alled irial oeffiiet ad are futios of temerature. We will show the use of a truated irial equatio with two terms. = 1 + B( T ) (3.4-9) I this equatio, B(T) a be estimated from the followig equatios: B(T) = (B 0 + ωb 1 ) (3.4-10) 0.4 B 0 = 0.083 1.6 T R 0.17, B 1 = 0.139 4. T R I equatio (3.4-10), ω is the Pitzer aetri fator, a arameter that reflets the geometry ad olarity of a moleule. The aetri fator for oer 1000 omouds a be obtaied from om4.exe rogram writte by T.K. Nguye. This rogram is aailable i the Distributio Folder for CHE30 ourse. Figure 3.4- Aetri fator for itroge from om4.exe rogram. 3-
Examle 3.4-3 ---------------------------------------------------------------------------------- A three-liter tak otais two gram-moles of itroge at 150.8 o C. Estimate the tak ressure usig the ideal gas equatio of state ad the usig the irial equatio of state truated after the seod term. Takig the seod estimate to be orret, alulate the eretage error that results from the use of the ideal gas equatio at the system oditios. Data for itroge: T = 16. K, = 33.5 atm, ad ω = 0.040 1. Solutio ------------------------------------------------------------------------------------------ From the ideal gas law, = 3.0 L/ mol = 1.5 L/mol, T = 150.8 + 73. = 1.4 K ideal = = ( ) 0.0806 L atm/mol K (1.4 K) 1.50 L/mol = 6.696 atm From the truated irial equatio, = 1 + B( T ) T R = 1.4/16. = 0.970 0.4 B 0 = 0.083 1.6 T R 0.4 = 0.083 1.6 0.97 = 0.360 0.17 B 1 = 0.139 4. T R 0.17 = B 1 = 0.139 4. 0.97 = 0.0566 B(T) = (B 0 + ωb 1 ) B(T) = ( ) 0.0806 L atm/mol K (16. K) [ 0.36 + (0.04)( 0.0566)] = 0.11 L/mol 33.5 atm = 0.11 1 1.5 = ( ) 1.50 L/mol 0.0806 L atm/mol K (1.4 K) (0.953) = 6.196 atm Error i usig ideal gas law ε = ideal 100 = 8.07 % 1 Felder R. M., Rousseau R. W., Elemetary Priiles of Chemial Proesses, Wiley, 005, g. 0 3-3
3.4-4 Soae-Redlik-Kwog (SRK) Equatio The Soae-Redlik-Kwog (SRK) equatio belogs to a lass of ubi equatios of state beause, whe exaded, they yield third-degree equatios for the seifi olume. The SRK equatio of state is = b αa ( + b) (3.4-11) I this equatio, the arameter a, b, ad α are emirial futios of the ritial temerature ad ressure, the Pitzer aetri fator, ad the system temerature. The followig orrelatios a be used to estimate these arameters: ( a = 0.4747 ) b = 0.08664 m = 0.48508 + 1.55171ω 0.1561ω α = + m( T ) 1 1 R Examle 3.4-4 ---------------------------------------------------------------------------------- A gas ylider with a olume of.50 m 3 otais 1.00 kmol of arbo dioxide at T = 300 K. Use the SRK equatio of state to estimate the gas ressure i atm. Data for arbo dioxide: T = 304. K, = 7.9 atm, ad ω = 0.5 13. Solutio ------------------------------------------------------------------------------------------ T R = 300/304. = 0.986 = (0.0806 L atm/mol K)(304. K) = 4.96 L atm/mol ( a = 0.4747 ) = 0.4747 ( 4.96 L atm/mol ) 7.9 atm = 3.6539 L atm/mol b = 0.08664 = 0.08664 4.96 L atm/mol 7.9 atm = 0.097 L/mol m = 0.48508 + 1.55171ω 0.1561ω = 0.863 13 Felder R. M., Rousseau R. W., Elemetary Priiles of Chemial Proesses, Wiley, 005, g. 03 3-4
α = 1+ m( 1 T ) R = 1+ 0.863( 1 0.986 ) = 1.0115 = b αa ( + b) 0.0806 L atm/mol K (300 K) = ( ) = 9.38 atm (.50 0.097) L/mol ( ) ( ) 1.0115 (3.654 L atm/mol ).50 L/mol (.50 + 0.097) L/mol Examle 3.4-5 ---------------------------------------------------------------------------------- A stream of roae at temerature T = 43 K ad ressure (atm) flows at a rate of 100.0 kmol/hr. Use the SRK equatio of state to estimate the olumetri flow rate V of the stream for = 0.7 atm, 7 atm, ad 70 atm. I eah ase, alulate the eretage differees betwee the reditios of the SRK equatio ad the ideal gas equatio of state. Data for roae: T = 369.9 K, = 4.0 atm, ad ω = 0.15 14. Solutio ------------------------------------------------------------------------------------------ We first alulate a, b, ad α from the followig exressios: flow ( a = 0.4747 ), b = 0.08664 m = 0.48508 + 1.55171ω 0.1561ω, α = m( T ) The SRK equatio is writte i the form 1+ 1 R f( ) = b + αa ( + b) = 0 is the alulated usig Newto s method: = f ( ) f '( ) = d, where f ( ) = ( ) b α a( + b) ( + b) The iitial alue for is obtaied from ideal gas law: ideal =. The iteratio roess stos whe /d is less tha 0.0001. The eretage differee betwee SRK ad ideal is 14 Felder R. M., Rousseau R. W., Elemetary Priiles of Chemial Proesses, Wiley, 005, g. 04 3-5
ideal 100% Oe is kow for a gie, the olumetri flow rate orresodig to a molar flow rate of 100.0 kmol/hr is obtaied as V flow (m 3 3 10 mol /hr) = (L/mol) kmol 1 m 3 3 10 L (100 kmol/hr) = 100 (L/mol) The alulatios are erformed usig the followig Matlab rogram: % Examle 3.4-5 T=369.9; % K =4.0; % atm w=0.15; % aetri fator Rg=0.0806; % L*atm/(mol*K) T=43; % K =iut('(atm) = '); Tr=T/T; a=0.4747*(rg*t)^/; b=0.08664*(rg*t)/; m = 0.48508 + 1.55171*w - 0.1561*w^; alfa=(1+m*(1-tr^0.5))^; ideal=rg*t/;=ideal; for i=1:0; f=-rg*t/(-b)+alfa*a/(*(+b)); df=rg*t/(-b)^-alfa*a*(*+b)/(*(+b))^; d=f/df; =-d; if abs(d/)<1e-4, break, ed ed Di=(ideal-)/*100; Flowrate=100*; fritf('ideal = %6.f, (L/mol) = %6.f, Peretage Differee = %6.3f\',ideal,,Di) fritf('flow rate (m3/hr) = %6.1f\', Flowrate) >> ex3d4d5 (atm) =.7 ideal = 49.588, (L/mol) = 49.406, Peretage Differee = 0.37 Flow rate (m3/hr) = 4940.6 >> ex3d4d5 (atm) = 7 ideal = 4.959, (L/mol) = 4.775, Peretage Differee = 3.86 Flow rate (m3/hr) = 477.5 3-6
>> ex3d4d5 (atm) = 70 ideal = 0.496, (L/mol) = 0.89, Peretage Differee = 71.57 Flow rate (m3/hr) = 8.9 The SRK equatio of state (ad eery other equatio of state) is itself a aroximatio. At 43 K ad 70 atm, the atual alue for is 0.579 L/mol. The eretage error i the SRK estimate ( = 0.89 L/mol) is 1%, ad that i the ideal gas estimate ( = 0.50 L/mol) is 9%. Reiew: The Newto-Rahso Method The Newto-Rahso method ad its modifiatio is robably the most widely used of all root-fidig methods. Startig with a iitial guess x 1 at the root, the ext guess x is the itersetio of the taget from the oit [x 1, f(x 1 )] to the x-axis. The ext guess x 3 is the itersetio of the taget from the oit [x, f(x )] to the x-axis as show i Figure 5.3-3. The roess a be reeated util the desired tolerae is attaied. f(x) f(x ) 1 B x 3 x x 1 Figure R-1 Grahial deitio of the Newto-Rahso method. The Newto-Rahso method a be deried from the defiitio of a sloe f (x 1 ) = f ( x x x 1 1) 0 x = x 1 f ( x1) f ' ( x ) 1 I geeral, from the oit [x, f(x )], the ext guess is alulated as x +1 = x f ( x ) f ' ( x ) 3-7
The deriatie or sloe f(x ) a be aroximated umerially as Examle f (x ) = f ( x + x) f ( x ) x Sole f(x) = x 3 + 4x 10 usig the the Newto-Rahso method for a root i the iteral [1, ]. Solutio From the formula x +1 = x f ( x ) f ' ( x ) f(x ) = 3 x + 4 x 10 f (x ) = 3 x + 8x x +1 = x 3 x + 4x 10 3x + 8x Usig the iitial guess, x = 1.5, x +1 is estimated as x +1 = 1.5 3 1.5 + 4 1.5 10 3 1.5 + 8 1.5 = 1.3733 The ext estimate is x +1 = 1.3733 The ext estimate is x +1 = 1.3653 3 1.3733 + 4 1.3733 10 3 1.3733 + 8 1.3733 3 1.3653 + 4 1.3653 10 3 1.3653 + 8 1.3653 = 1.3653 = 1.365 This alue is lose to the guessed alue of 1.3653, therefore we a aet it as the solutio x = 1.365 3-8