Grupo metilo

Enthalpy-concentration Diagram

• McCabe Thiele method assumes constant molar flow rate because it considers equal latent heat of vaporization.

• Here we consider varying molar flow rate by solving simultaneous material and energy balances.

• In this case, the operating lines for the enriching and stripping section will be determined from simultaneous solution of mass and energy balance equations.

• To facilitate the solution of the heat balance equation, an enthalpy diagram can be constructed and used.

Enthalpy diagram

The liquid enthalpy

h = x CpA        A (T −To ) + (1− xA )CpB (T −To ) + ΔH sol         1

The vapor enthalpy

H = yA[λA +CvA(T −To )]+(1− yA)[λB +CvB (T −To )]         2

Correction for latent heat

λ A = CpA (TbA −To ) + λbA −CvA (TbA −To )         3

λ B = CpB (TbB −To ) + λbB −CvB (TbB −To )         4

Where  

T          is the boiling point for the mixture

TbA         The boiling point of pure A TbB         The boiling point of pure B To         reference temperature

λbA         latent heat of pure A at TbA 

Example:

Create the enthalpy diagram for Benzene-Toluene mixture

Let the reference temperature be 80.1 oC, thus the latent heat of A (Benzene)  does not need to be corrected.

For component B:

λ =B        167.5(110.6 −80.1) +33330−138.2(110.6−80.1) = 34224   kJ/kgmole

Create the liquid enthalpy line:

xA = 0,          T = 110.6

h = 0 + (1 − 0) (176.5)(110.6 − 80.1) =  5109 kJ/kg mole

xA = 0.3,         equilibrium diagram  ↱ T = 98 oC

h = 0.3(138.2)(98 ─ 80.1) + (1 − 0.3) (176.5)(98 − 80.1) =  2920 kJ/kg mole

xA = 0.5,         equilibrium diagram  ↱ T = 92 oC

h = 0.5(138.2)(92  − 80.1) + (1 − 0.5) (176.5)(92 − 80.1) =  1820 kJ/kg mole

xA = 0.8,         equilibrium diagram  ↱ T = 84 oC

h = 0.8(138.2)(84  − 80.1) + (1 − 0.8) (176.5)(84 − 80.1) =  562 kJ/kg mole

xA = 1.0,         equilibrium diagram  ↱ T = 80.1 oC

h = 1.0(138.2)(80.1  − 80.1) + 0 =  0 kJ/kg mole

Create the vapor enthalpy line

yA = 0,                   T = 110.6

H = 0 + (1 − 0) [33330 + 138.2(110.6 − 80.1)]  =  38439 kJ/kg mole

yA = 0.3,                  T =  

H = 0.3[30820 + 96.3( T – 80.1)] + (1 − 0.3) [34224 + 138.2(T − 80.1)] =  36268 kJ/kg mole

yA = 0.5,                  T = 98.8

H = 0.5[30820 + 96.3( 98.8 – 80.1)] + (1 − 0.5) [34224 + 138.2(98.8 − 80.1)] =  34716 kJ/kg mole

yA = 0.8,                  T =  

H = 0.8[30820 + 96.3( T – 80.1)] + (1 − 0.8) [34224 + 138.2(T − 80.1)] =  32380 kJ/kg mole yA = 1.0,                  T = 80.1

H = 1.0[30820 + 96.3( 80.1 – 80.1)] + 0  =  30820 kJ/kg mole

Column Design [pic 2][pic 3][pic 4]

Enriching Section  

The Mass balance:

        Vn+1 = Ln + D         5

The component balance

6

The enthalpy balance:         f

        Vn+1H n+1 = L hn        n + DhD + Qc         7

To eliminate the condenser duty is by heat balance around the condenser:

        +1        +1        +1        1 1

Stripping section

The Mass balance: [pic 5]

...