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Pure Com ponent2 Pure Com ponent1 Mixture12 in W & out Q & Thermodynamics of separation t is the minimum work to separate a mixture into s pure components? Ex. Mining, Desalination, Materi ification, Recycling.

Thermodynamics of separation

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Thermodynamics of separation. What is the minimum work to separate a mixture into it’s pure components? Ex. Mining, Desalination, Material Purification, Recycling. Balance Eq’ns for Mass, Energy & Entropy. S irr. Minimum Work of Separation. Gibbs Free Energy of Mixing*. - PowerPoint PPT Presentation

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Page 1: Thermodynamics of separation

Pure Component 2

Pure Component 1

Mixture 12

inW&outQ&

Thermodynamics of separation

What is the minimum work to separate a mixture intoit’s pure components? Ex. Mining, Desalination, MaterialPurification, Recycling.

Page 2: Thermodynamics of separation

Pure Component 2

Pure Component 1

Mixture 12

inW&outQ&

dNi,sys

dt= &Ni,in − &Ni,out

dE

dt=−&Qout + &Win + &H12 − &H1 − &H2

&Win =(( &H1 + &H2 )− &H12 )−To(( &S1 + &S2 )−&S12 ) +To

&Sirr

&Win =−&N12 (Δhmix −T0Δsmix) +T0

&Sirr

Balance Eq’ns for Mass, Energy & Entropy

Sirr

Page 3: Thermodynamics of separation

&Win =−&N12Δgo

mix +T0&Sirr

wmin =

&Wmin

&N12

=−Δg∗mix

&Win =−&N12 (Δhmix −T0Δsmix) +T0

&Sirr

Minimum Work of Separation

Page 4: Thermodynamics of separation

Gibbs Free Energy of Mixing*

Δgomix = Δho

mix –T0 Δsomix.

Δgomix–T0 Δsmix =

–T0 (s12 –x1s1 – x2s2)

For non-interacting molecules entropy can dominateoften resulting in a negative Gibbs Free Energy and hence spontaneous mixing. I.e. Δgo

mix < 0

* at standard conditions

Page 5: Thermodynamics of separation

S = k ln

Boltzmann’s entropy equation

=n!

r!(n − r)!

How many ways can “r”atoms be positioned ina lattice with “n” locations?

Page 6: Thermodynamics of separation

wmin = T0Δsmix = k T0 (ln 12)

Ex. 4 atoms in 8 locations

Page 7: Thermodynamics of separation

wmin =−T0R(xlnx+ (1−x)ln(1−x))

Using Stirling’s Approximation

Where x is mol fraction r/n, and R = k Navo

ln N! = N ln N - N

Page 8: Thermodynamics of separation

Multi-component System

=n!

n1!n2 !.....n j !

wmin =−T0R xii=1

j

∑ lnxi

Page 9: Thermodynamics of separation

“Separation”

wmin =−T0R xii=1

n

∑ lnxi

Page 10: Thermodynamics of separation

))xln(NxlnN(RTW )N(min

i −+−= 1210

))1ln(ln)1(( 210)1(

min1 xNxNRTW N −+−−=−

wmin, 1 =T0R(ln1x1

)

“Extraction”

Page 11: Thermodynamics of separation

Separation Examples

• From the atmosphere

• From the Ocean

• Solutions– Polymer– Water based– Liquid metals (activity coef)

Page 12: Thermodynamics of separation

The minimum work to separate O2 from the atmosphere

ex,O2

o =T0R(ln1

xO2

) ≅−298(K )×8.314(J / molK )ln(0.212) =3.84(kJ / mol)

In wet air you get 3.97 kJ/mol : compare with Szargut

Table from the EngineeringToolBox.com

Page 13: Thermodynamics of separation

Energy

kg(target)=

kg(processed)kg(target)

gEnergy

kg(processed)

~1gg

Energykg(processed)

Page 14: Thermodynamics of separation

energy requirements for mining and milling, possible future trends

Chapman and Roberts p 113 & 116

underground ~ 1000/g (MJ/t metal)

open pit ~ 400/g (MJ/t metal)

Page 15: Thermodynamics of separation

Sherwood plot showing the relationship between the concentration of a target material in a feed stream and the market value of (or cost to remove) the target material [Grübler 1998].

Page 16: Thermodynamics of separation

Exergy of a Mixture

Page 17: Thermodynamics of separation

CRUST at To, po

Ore value at mine

Pure ore (e.g. Fe2O3)

Pure metal Metal alloy

Mixing in product

Mixing in waste stream

Further mixing and corrosion

Exergy

Purification Stages

Recycle to pure metal

Theoretical Exergy Values for a metal extracted from the earth’s crust shown at various stages of a product life cycle (not to scale)