Thermodynamics

704 | Thermodynamicsthe second-law efficiency is always less than 1 since the actual separationprocess requires a greater amount of work input because of irreversibilities.Therefore, the minimum work input and the second-law efficiency provide abasis for comparison of actual separation processes to the “idealized” onesand for assessing the thermodynamic performance of separation plants.A second-law efficiency for mixing processes can also be defined as theactual work produced during mixing divided by the maximum work potentialavailable. This definition does not have much practical value, however, sinceno effort is done to produce work during most mixing processes and thus thesecond-law efficiency is zero.Special Case: Separation of aTwo Component MixtureConsider a mixture of two components A and B whose mole fractions are y Aand y B . Noting that y B 1 y A , the minimum work input required to separate1 kmol of this mixture at temperature T 0 completely into pure A andpure B is, from Eq. 13–54,w min,in = –R u T 0 ln y A (kJ/kmol A)A + By A , y BA + B1 kmoly A , y BSeparationunitSeparationunitpure A(1 kmol)A + B(a) Separating 1 kmol of A froma large body of mixture wmin,in = –R u T 0 ( y A ln y A + y B ln y B )(kJ/kmol mixture)(b) Complete separation of1 kmol mixture into itscomponents A and Bpure Apure BFIGURE 13–23The minimum work required toseparate a two-component mixture forthe two limiting cases.orw min,in R u T 0 1y A ln y A y B ln y B 21kJ>kmol mixture2or, from Eq. 13–55,W min,in R u T 0 1N A ln y A N B ln y B 21kJ2W # min,in N # mR u T 0 1y A ln y A y B ln y B 2m # mR m T 0 1y A ln y A y B ln y B 21kW2(13–58a)(13–58b)(13–58c)Some separation processes involve the extraction of just one of the componentsfrom a large amount of mixture so that the composition of the remainingmixture remains practically the same. Consider a mixture of two componentsA and B whose mole fractions are y A and y B , respectively. The minimum workrequired to separate 1 kmol of pure component A from the mixture of N m N A N B kmol (with N A 1) is determined by subtracting the minimum workrequired to separate the remaining mixture R u T 0 [(N A 1)ln y A N B ln y B ]from the minimum work required to separate the initial mixture W min,in R u T 0 (N A ln y A N B ln y B ). It gives (Fig. 13–23)w min,in R u T 0 ln y A R u T 0 ln 11>y A 21kJ>kmol A2(13–59)The minimum work needed to separate a unit mass (1 kg) of component A isdetermined from the above relation by replacing R u by R A (or by dividing therelation above by the molar mass of component A) since R A R u /M A . Eq.13–59 also gives the maximum amount of work that can be done as one unitof pure component A mixes with a large amount of A B mixture.An Application: Desalination ProcessesThe potable water needs of the world is increasing steadily due to populationgrowth, rising living standards, industrialization, and irrigation in agriculture.There are over 10,000 desalination plants in the world, with a total desalted