SOLUTION TO HOMEWORK #7

#Problem 1

10.1-1

a. In order to solve this problem, we need to know what happens at the bubble point; at this

point, the first bubble is formed, so we can assume that all of the number of moles for the

components in the gas phase are close to zero (yet there is an equilibrium). Therefore, we

use the compositions given in the problem statement and we assume they are the

compositions of the liquid. (x=z, z being the composition given).

xET 0.05

xP 0.10

xNB 0.40

xMP 0.45

Now, we know, from Raoult’s law that:

yi xiPi

vap

P

and from the problem statement we know that the pressure is 5 bar. Therefore, we proceed to

calculate the bubble point temperature. The way the bubble point temperature is done is by doing

the following. We also know that:

y ii

1

and therefore, substituting

1

vap

i i

ii

i

x P

yP

Therefore:

1xETPET

vap xPPPvap xNBPNB

vap xMPPMPvap

P

Now, we do the following:

1

xET 10 A

T B

ET

xP 10 A

T B

P

xNB 10 A

T B

NB

xMP 10 A

T B

MP

P

From this equation, we can find the bubble point temperature. This equation can be solved for

the temperature, which comes out to be 294K. You can use bisection, guessing (better be clever

guessing), or Goal Seek in excel.

Now, we need to calculate the compositions. This is done by using Raoult’s law yielding:

yET 0.4167

yP 0.1730

yNB 0.1601

yMP 0.2502

One way of checking if the answer is correct is to use

y ii

1. If this equality is not correct,

then something was done wrong.

b. The dew point calculations are similar, but in this case the equations change slightly.

x i y iP

Pivap

also, since we are at the dew point (the point in which the first drop of liquid is formed), we can

say that almost no liquid is present.

1ii vap

i i i

yx P

P

which, when the correct terms are substituted in, we obtain:

1 PyET

10 A

T B

ET

yP

10 A

T B

P

yNB

10 A

T B

NB

yMP

10 A

T B

MP

where we use the composition given (z) as the composition of the vapor. Solving for the dew

point temperature yields that the temperature is 314K. Now, by using Raoult’s law, we get that

the compositions of the liquid phase are:

0.0039

0.0037

0.5215

0.4409

ET

P

NB

MP

x

x

x

x

c. In order to solve the flash problem, we need the help of other equations. First, we begin

by calculating the K-factors for each component. This is done by using the following

equation:

vap vap

i i i ii

i i

y P PK

x P P

Where we set the activity and fugacity coefficients to one because (Raoult law assumption).

Then, by calculating each K-factor, we obtain:

KET 10.185

KP 2.238

KNB 0.546

KMP 0.743

Now, in a flash, the feed stream is separated into the liquid and vapor streams. This is shown

below:

Therefore, we need to know what are the compositions of the vapor and liquid streams, so these

would be part of our equations used for solving this problem. We have:

Vapor Stream

y ii

1

Liquid Stream

x ii

1

Vapor

Liquid

Feed V

a

p

o

r

Now, by doing a mass balance around the flash, for each component, we can obtain four other

equations that could help us in solving this problem. These are (z’s represent the component’s

compositions of the feed):

zETF xETL yETV

Similarly for propane

zPF xPL yPV

for n-butane

zNBF xNBL yNBV

and for 2-methyl propane

zMPF xMPL yMPV

Now, we can use a basis of 1 mol for the feed (F=1), which will give us:

1 LV

Now, we can replace the V in all equations:

1

1

1

1

ET ET ET ET

P P P P

NB NB NB NB

MP MP MP MP

z F z x L y L

z F z x L y L

z F z x L y L

z F z x L y L

Now, we can solve this complex problem by doing the following: first, guess L (the liquid flow).

Once this value is guessed, we can proceed to solve the compositions by using the equations of

the K-factors and the mass balances. As stated before, from the K-factors we obtain:

K i y i

x i

so, for each component, we get:

yET 10.185xET

yP 2.238xP

yNB 0.546xNB

yMP 0.743xMP

Now, these relationships can be substituted into the mass balances as stated above as follows:

10.185 1

2.238 1

0.546 1

0.743 1

ET ET ET

P P P

NB NB NB

MP MP MP

z F x L x L

z F x L x L

z F x L x L

z F x L x L

which will allow us to calculate the compositions of each stream.

/[ 10.185 1 ]

/[ 2.238 1 ]

/[ 0.546 1 ]

/[ 0.743 1 ]

ET ET

P P

NB NB

MP MP

x z F L L

x z F L L

x z F L L

x z F L L

We are now left with only two equations, namely

y ii

1

and

x ii

1

Thus we equate them

i i

i i

y x

Which can be rewritten as:

0i i i

i i

K x x

or

( ) 0ET ET P P NB NB MP MP ET P NB MPK x K x K x K x x x x x

or

( 1) ( 1) ( 1) ( 1) 0ET ET P P NB NB MP MPK x K x K x K x

Thus substituting the x’s:

(10.185 1) /[ 10.185 1 ] (2.238 1) /[ 2.238 1 ]

(0.546 1) /[ 0.546 1 ] (0.743 1) /[ 0.743 1 ] 0

ET P

NB MP

z F L L z F L L

z F L L z F L L

By solving this equation for L (using F=1) , we obtain that:

Therefore, we have solved the flash problem.

#Problem 2

#Problem 3

#Problem 4

#Problem 5

Problem 6: a) Obtain an expression relating the minimum amount of work needed to separate a mixture

into its pure components (at constant T and P) as a function of the fugacities of the components

in the mixtures and the fugacities of the pure components.

b) Show that the expression can be written only in terms of temperature and the molar fractions,

when ideal mixture is assumed and the Lewis and Randall rule is used.

a) We start by writing (Chapter 4)

,rev revP TG W

Thus, if Gm is the gibbs free energy of the mixture of two components, G1 is the Gibbs free

energy of the stream of pure component 1 and G2 is the Gibbs free energy of the stream of pure

component 2, we write

1 2rev mW G G G

But

,1 1 ,2 2m m mG x G x G

However,

,

, ( , ) lnm io

m i i o

i

fG G P T RT

f

and 1 1 2 2( , ), ( , )o oG G P T G G P T

Substituting, we get:

1 2 ,1 ,1 ,2 ,2 1 1 2 2( , ) ( , )o o

rev m m m m mW G G G n G n G n G P T n G P T

After realizing that ,1 1mn n and

,2 2mn n , we get

,1 ,2

1 2

1 2

ln lnm m

rev o o

f fW RT n n

f f

Part b)

We can write ,

o

m i i if x f (Lewis and Randall) and therefore:

,1 1 ,2 2ln lnrev m mW RT n x n x

or

,1 1 ,2 2/ ln lnrev m mW n RT x x x x

Recognize this? It is somehow similar to the gibbs free energy of mixing of ideal mixtures,

right?

#Problem 7

#Problem 8

Problem #9

#Problem 10