Grupo metilo

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Lm =Vm+1 + B

The component balance

The enthalpy balance:

Vm+1H m+1 = (Vm+1 + B h) m − BhB + Qr 14

Overall heat balance:

Qr =DhD +Bh Q FhB + −c f 15

Substituting for Qr into the heat balance equation 14:

V Hm+1 m+1 = (Vm+1 +B h) m +DhD +Q Fhc − f 16

The feed condition

The q line remain the same

Solution procedure

- starting from xD draw the enriching operating line:

- select a value for x between xD and xf

- assume a constant molar flow rate; Ln= L, Vn+1 = V

- solve the component balance equation for yn+1

- solve the heat balance for Vn+1

- solve the mass balance for Ln

- resolve the component balance for yn+1

- if new value of yn+1 equals the old value go step h else go back to step c

- repeat steps a-g for another value for x.

- do the same for the stripping section starting from xB

Example:

Consider the separation of Benzene-Tolune mixture of example 11.4.1:

F = 100 kgmole/h xf = 0.45 xD = 0.95; xB = 0.1 R = 1.755; D = 41.2 kgmole/h; B = 58.8 kgmole/h

Tf = 54.5 oC ↱ q = 1.195

The stripping section:

L/D = 1.755 ❧ L = 1.755D = 72.3 kgmole/h

V1 = L + D = 72.3 + 41.2 = 113.5 kgmolr/h

Knowing xD and y1= xD using the boiling diagram gives T = 82.3 oC using the enthalpy diagram gives:

H1 = 31206 kJ/kg mole

hD = 139 kJ/kg mole

Let x = 0.55

from component balance ↱ yn+1 =[pic 6] 0.55 +[pic 7] 0.95 = 0.695

using the enthalpy chart:❧ at xn = 0.55 ↱ hn = 1590

at yn+1 = 0..695 ↱ Hn+1 = 33240

from the heat balance equation:

Vn+133240 = (Vn+1 − 41.2)1590 +113.5(31206) − 72.3(139)

Vn+1 = 109.5

From the mass balance equation: 109.5 = Ln + 41.2 ↱ Ln = 68.3

Resolving the component balance equation: yn+1 =[pic 8]0.55 +[pic 9]0.95 = 0.7

Since 0.7 is very close to 0.695 we stop the iteration and go for another value of x.

before proceeding to stripping section:

Qc =113.5(31206) −113.5(139) = 3526kJ h/

At the feed temperature and composition:

Either using the enthalpy chart for xf = 0.45 or enthalpy equation for xf = 0.45 & Tf = 54.4 oC we get:

hf = -3929 kJ/kgmole

using the enthalpy chart for xB = 0.1 we get hB = 4350 kJ/kgmolr

using the heat balance around the re-boiler:

Qr = 41.2(139) +58.8(4350) +3526+3929 = 4,180,500 kJ/h

Knowing the feed condition we calculate the molar flow rates in the stripping section:

Lm = Ln + qF = 72.3 + 1.195(100) = 191.8 kgmole/h

Vm+1 = Vn+1 - (1−q)F = 113.5 − (1 − 1.195)(100) = 133.0 kgmle/h

Srtipping section:

Let ym+1 = 0.207

Using the component balance: 0.207 =[pic 10]xm +[pic 11]0.1 ❧ xm = 0.174

Using the enthalpy chart:

at

xm = 0.174 ❧ hm= 3800

at

ym+1 = 0.207 ❧ Hm+1 = 37000

using the enthalpy balance equation:

Vm+137000 = (Vm+1 −58.8)380 + 58.8(139) + 3526 −100( 3929− )

❧ Vm+1 = 125

Using the mass balance: Lm = Vm+1 + B ❧ Lm = 183.8

Using the component balance: 0.207 =[pic 12]xm +[pic 13]0.1 ❧ xm = 0.173

Since the new value of x is close enough to old value

[pic 14]

Tray Efficiency

In McCabe Thiele method 1000% phase equiklibruum is assumed which is not totally correct: [pic 15]

Overall efficiency

Murphee Efficiency

EM = yynn* −−yynn++11 [pic 16]

Point Efficiency

y yn'− n+1' EMP = yn* −yn+1'[pic 17]

Correlations for the overall Efficiency [pic 18]

Drickamar and Bradford correlation

E= 0.17−0.616log(∑xfiμ μLi / L )

xf feed composition μL viscosity at average tower temperature μw viscosity of water at 293 K

CHU correlation

logE=1.67+0.3log( . )LV −0.25log(μLα)+0.3hL

L, V liquid and vapor molar flow rates μL viscosity of the liquid