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# 7.34: Untitled Page 179

## Chapter 7

( M

)  ( M

)

CO2 5

N2 5

(31)

CY   C 

5

( M

)

( 1

)

Y

 

C H

3

Y

5

 2  2 1 

2

4

1  ( M

) CY  ( 1  C )

Y

Y

C

 

2H4 5

Substitution of this result into Eq. 30 leads to an equation for the molar flow rate of ethylene ( C H ) in Stream 35. This result can be expressed 2

4

in the compact form

( M

)

C H

5

  

(32)

  

2

4

   0

1

 ( M

C CY

) 

C

2H4 5 

where the two parameters are given by

3  5 Y 1

3  5 Y 2 1

1

Y

2

2

   

 

 

,

 

(33)

CY

Y

Y

At this point we are ready to use a trial‐and‐error procedure to first solve for ( M

) and then solve for the parameter  .

C2H4 5

Picard’s method

We begin by defining the dimensionless molar flow rate as

( M

)

C H

2

4 5

x

(34)

so that the governing equation takes the form

x

H( x)    

(35)

 C

 0

1

CY  x

In order to use Picard’s method (see Appendix B4), we define a new function according to

Definition:

f ( x)  x H( x)

(36)

and for any specific value of the dependent variable, x , we can define a i

new value, x

, by

i1

Definition:

x

f x ,

i

, , ,... ,

(37)

1

( )

1 2 3

i

i

Material Balances for Complex Systems

335

To be explicit, we note that the new value of the dependent variable is given by

x

i

x

x

,

i

, , , ... ,

i

  

1

1 2 3

i

(38)



1C

CY  ix 

In terms of the parameters for this particular problem

  37 . 6190 ,

  15 . 1678 ,

C  0 . 7 ,

Y  0 . 5

(39)

we have the following iterative scheme

15 . 1678 x

x

x

37 6190

i

.

,

i

1 , 2 , 3 , ... ,

i

1

i

(40)

0 8571

.

i

x

By inspection, one can see that x  0 . 8571 thus we choose our first guess to be x  0 . 62 and this leads to the results shown in Table 7.8a.

o

Table 7.8a. Iterative Values for Dimensionless Flow Rate

(Picard’s Method)

i

xi

xi+1

0

0.6200

– 1.4237

1

– 1.4237

45.6633

2

45.6633

98.7402

3

98.7402

151.6598

4

151.6598

204.5328

5

204.5328

257.3835

6

257.3835

…..

7

…..

…..

Clearly Picard’s method does not converge for this case and we move on to Wegstein’s method (see Appendix B5).

Wegstein’s method

In this case we replace Eq. 37 with

Definition:

i

x

1

1 qf( x )  qx ,

i  1 , 2 , 3 , ... ,

(41)

i

i

336