Non-Ideal Gas Behavior P.V.T Relationships for Liquid and Solid:

Transcription

1 hemodynamis Non-Ideal Gas Behavio.. Relationships fo Liquid and Solid: An equation of state may be solved fo any one of the thee quantities, o as a funtion of the othe two. If is onsideed a funtion of and : = f (, ) It an be diffeentiated as: d ( ) d ( ) d...() olume expansivity: Isothemal Compessibility: Dividing eq. () by : d d ( ) d ( ( d k d () ) k d ) ( - β and k ae funtion of and they ineases as ineases. - β and k ae onstant when hange in and is elatively small. 3- When a fluid is inompessible β, k = 4- Integating eq. () : ln ( ) k ( )...(3) Example : Fo liquid aetone at o C and ba, β =.487 * -3 o C - k = 6* -6 ba - =.87 m 3 /gm ) 49

2 hemodynamis Find: a) he value of ( ) at o C and ba. b) he pessue geneated by heating at onstant fom o C and ba to 3 o C. ) he hange in volume fo a hange fom o C and ba to o C and ba. Solution : d d k d At onstant volume: d = d k d ( d ) d k ba / b) If we assume β and k ae onstants ( sine small inease in temp. ) then: d k d ( is onstant ) o C d k d = k = 4 () = 4 ba = + = + 4 = 4 ba ) ln ( ) k ( ) 3 6 ln (.487 ) ( ) (6 )( 9 )

3 hemodynamis =.97 (.87 ) =.49 m 3 /gm = = = -.38 m 3 /gm Behaviou of Real Gas : Ideal Gas Real Gas -No attation foes between moleules - hee is an attation foes Between the moleules - olume of moleules is negligible - he volume of moleules is Not negligible 3- essue is about 3 atm 3- he pessue is vey high < 3Ma 4- Equation of state is = nr 4- Equation of state is: =ZnR Z depends on ondition (, ) Equation of state fo Non- ideal gas (Real gas) : Non ideal gas behavio ous when the pessue is elatively high ( > 3 Ma fo many gases). - Compessibility Fato ( Z ): When a gases deviate fom ideal gas behavio signifiantly at states nea the satuation egion and the itial point. 5

4 hemodynamis peentage of eo = ( table ideal ) table whee steam an be teated as an ideal gas with less than peent eo. his deviation fom ideal gas behavio at a given tempeatue and pessue an auately be aounted fo by the intodution of a oetion fato alled the ompessibility fato Z defined as: Z n R Z = fo ideal gases. Fo eal gases Z an be geate than o less than unity. he ompessibility fato an be detemined fo any gas by using a genealized ompessibility hat. In the genealized hat the edued pessue and edued tempeatue must be used. hey ae alulated fom: R 5

5 hemodynamis * Note that is defined diffeently fom and. It is elated to and instead of. Lines of onstant ae also added to the ompessibility hats and this enables to detemine o without having to esot to time onsuming iteations. he following obsevations an be made fom the genealized ompessibility hat:. At vey low pessues ( >>), gases behave as an ideal gas egadless of tempeatue.. At high tempeatues ( <), ideal gas behavio an be assumed with good auay egadless of pessue (exept when <<). 3. he deviation of a gas fom ideal gas behavio is geatest in the neighbohood of the itial point. 53

6 hemodynamis - an de Waal s equation : R a b whee : a 7 R 64 3 Redlih Kwong equation : An impoved equation is : R a b ( b ) b R 8 whee : a.475 R.5 R b.867 A enti fato ( ω ): It is a paamete that uses in the elation of finding the ompessibility fato ( Z ) of gases whih ae unde elatively high pessue and it has a etain value fo eah gas. sat. ln ( ). 7 It was noted by expeiment that all omponents has the same fom of diagam and the diffeene is in thei itial point. ( All fluids having the same value of ω, when ompaed at the same and have about the same value of z, and all deviate fom ideal gas behavio to about the same degee ) he oelation fo Z developed by pitze & owokes takes the fom : Z Z Z he genealized ompessibility fato 54

7 hemodynamis whee : Note : Z & Z ae omplex funtions of both & Z an be found fom App. E table E. if > Z an be found fom App. E table E.3 if < Z an be found fom App. E table E. if > Z an be found fom App. E table E.4 if < ω,, an be found fom App. B. Geneal Coelation fo Liquids: - Rakett poposed a geneal eq. fo estimation the mola volume of sat. liquids: sat 3 m ( ) mol Z ( 7 ) Note: Only itial onditions ae equied. - Lydesen et al. developed a geneal method fo estimations of liquids volumes based on the piniples of oesponding states: 3- If a volume at etain onditions is known. hen we an find the new volume when the onditions diffe by: 55

8 hemodynamis whee : equied volume & : Known volume, ae edued densities ead fom fig. 3.6, p - as follow: Genealized iial Coeffiient Coelation: B B Z ( ) R R B B B R Z Z B he simplest fom of the viial equation B B B,.4.83,.6 B Z B

9 hemodynamis = Z n R n Z R Example : Detemine the mola volume of n-butane at 5 K and 5 ba by : a) he ideal gas b) he genealized ompessibility fato ) he genealized viial oeffiient oelation given that : = 45. K, = 38 ba, ω =.93 Solution: a) = R R 83.4( 5 ) 696 m 3 /mol 5 5 b ) Fom App. E. & E. Z =.86, Z =.38 Z Z Z = (.93)(.38) =.87 Z R (.87 )(83.4 )(5 ) 5 48 m 3 / mol ).4 B B

10 hemodynamis Z Z B B ( ) ( ) Z Z Z = (.34) =.878 Z R Example :.878( 83.4 )(5 ) 489 m 5 3 / mol What pessue is geneated when kmol of methane is stoed in a volume of.5 m 3 following: a) the ideal gas equation b) he Redlih Kwong equation at 5 o C? Base alulation on eah of the ) A genealized ompessibility fato oelation Solution: n R ( )(.834 )(33.5 ) a) 4.9 ba.5 b) he Redlih Kwong eq. : R a b ( b ) a.475 R R b (.834 ) ( 9.6 ) (.834 )(33.5 ) (.834 )(9.6 ) ( ) ba

11 hemodynamis ) In the absene of a known value fo, he iteation poedue must be followed as : sine: Z R Z 4.9 Z(.834 )( 33.5 ).5 Assume: Z = Z Z.4 Z , = and =.695 in table E.3 and E.4 by intepolated: Z =.8936, Z =.73 Z Z Z.8936 (. )(.73 ).8968 then alulate a new value of and new Z, he poedue ontinues until no signifiant hange ous fom one step to the next. he final value of Z is found.89 at = 4.4 and =.695 Z R (.89 )(.834 )(33.5 ).5 9.3ba Anothe solution: Compute value of using the following eq.: =.36 and =.695 R Fom Z-Chat, ead Z value =.89 Note: he expeiment value of pessue is 87.5 ba that is vey lose to the Redlih - Kwong eq. and genealized ompessibility eq., wheeas ideal gas eq. yield a esult is high by 4.6 % Example : a) Calulate the mola volume of satuated liquid ammonia at 3 K. 59

12 hemodynamis b) Calulate the mola volume of liquid ammonia at 3 K and ba. Given that : = 45.6 K, =.8 ba, = 7.5 m 3 /mol Z =.4, ω =.5 Solution : 3 a) = 7.5, Z =.4 sat Z.857 sat = 8.35 m 3 /mol ( ). 357 (7.5 )(.4 ) b) =.7643, Fom fig. 3.6 we have: ρ = m 3 /mol.857 6