ocw chapter 10

8/18/2019 Ocw Chapter 10

1/64

Chemical Engineering Thermodynamics

!"#$%&'()*(+ -)*(.(/%(*01 234%$+*56$3 "3+ 7##.(5"6$3

Mohammad Fadil Abdul Wahab


2/64

8$% 9"#$% 0(:4*%;

ˆ f i

l = !

i x

i f

i

f̂ i

v = !̂

iy

i P

For liquid solution

VLE criteria (to be shown/derived in chapter 11),

ˆ f i

l =

ˆ f i

v

so !̂ i y

i P = "

i x

i f

i

where,

Consider a multicomponent system in a VLE

condition, the fugacity (to be defined in Chapter 11)

of species i for each phase is given by,

!̂ i fugacity coefficient species i in gas mixture

f i fugacity of pure species i

" i activity coefficient of species i in liquid solution


3/64

9"#$% 0(:4*%; (3

;)*(.(/%(*0 ?(4A (+;". .()*(+ >$.*6$3

and also for pure species in equlibrium and ideal gas vapor,

f i = f

i

l = f

i

v= P = P

i

sat

!̂

i = 1

! i = 1

we get, y

i P = x

i P

i

sat Raoult's Law (10.1)

equation becomes y

i P = x

i f

i


4/64

B$+(C;+ 0(:4*%; (3 ;)*(.(/%(*0 ?(4A

3$3D(+;". .()*(+ >$.*6$3

!̂ i y

i P = "

i x

i f

i

yi P = "

i x

i P

i

sat Modified Raoult's Law (10.5)

where " i is a function of T and x

i.

!̂ i = 1


5/64

ED9".*; F! " G

K i = y

i

xi

(10.10)

If Raoult’s Law is valid, yi P = x

i P

i

sat

K i = P i

sat

P (10.11)

If Modified Raoult’s Law is valid, y

i

P = xi

! i

P i

sat

K i =

! i P

i

sat

P (10.12)


6/64

ED9".*; H>(3@

I;J%(;>4;% KA"%4

For light hydrocarbon mixture (commonly found in industry),

K i is essentially function of T and P only.

K i are tabulated in a chart called the DePriester chart.


7/64


8/64

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RS(TQ

$%

!"!# %& '()*+)(,- ./01 (23 %4 506-2 .701 (23 8

89 *()*+)(,- ,:- % ;:-2 ,:- )(-(? (< ( ?-


9/64

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89 *()*+)(,- ,:- % ;:-2 ,:- E 9@ )0G+03H(>>-(? (< ( ?-


10/64

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N$ 5".5*."4; 4A; N ?A;3 4A; O>4 /*//.;

"##;"% "> " %;>*.4 $P (35%;">; (3 N "4

5$3>4"34 JQ 7.>$ 5".5*."4; 4A; 5$0#$>(6$3

RS(T $P 4A(> /*//.;Q

$%

N$ 5".5*."4; 4A; N ?A;3 4A; .">4 /*//.;

+(>"##;"% "> " %;>*.4 $P +;5%;">; (3 N "45$3>4"34 JQ 7.>$ 5".5*."4; 4A; 5$0#$>(6$3

RS(T $P 4A(> /*//.;Q


11/64

I-U N1 K".5*."4; R:(T "3+ NV @(9;3 RS(T "3+

J

N$ 5".5*."4; 4A; N ?A;3 4A; O>4 +;? F"

+%$# $P .()*(+G "##;"% "> " %;>*.4 $P

+;5%;">; (3 N "4 5$3>4"34 JQ 7.>$ 5".5*."4;

4A; 5$0#$>(6$3 R:(T $P 4A(> +;?Q

$%

N$ 5".5*."4; 4A; N ?A;3 4A; .">4 +;?

+(>"##;"% "> " %;>*.4 $P (35%;">; (3 N "45$3>4"34 JQ 7.>$ 5".5*."4; 4A; 5$0#$>(6$3

R:(T $P 4A(> +;?Q


12/64

Overall mole balance

T = L +V

I;%(9"6$3

V

L

Note: z i is overall composition.

Let T=1 mol, so V and L are mole fractions,

z i = Lxi +Vyi z

i = (1!V ) x

i +Vy

i (A)

Component mole balance,

Tz i = Lx

i +Vy

i


13/64

Substitute yi = K

i x

i into (A),

z i = (1!V ) x

i + K

i x

iV = x

i(1!V +VK

i) = x

i(1+V ( K

i !1))

xi =

z i

1+V ( K i !1)

Substitute xi =

yi

K i

into (A),

z i = (1!V )

yi

K i

+ yiV z

i K

i = (1!V ) y

i + y

iVK

i

yi =

z i K

i

1+V ( K i !1)

(10.16)


14/64

Also,

xi! " yi = 0!

z i

1+V ( K i !1)

! z

i K

i

1+V ( K i !1)

"" = 0

z i ! z

i K

i

1+V ( K i !1)

" = 0


15/64

L*//.;#$(34 K".5*."6$3

At bubble point (practically all liquid) L=1, V=0 and z i = xi

z i ! z

i K

i

1+V ( K i !1)

" = 0 becomes,

Bubblepoint criteria

( xi ! x

i K

i)" = 0

xi" = xi K i "

xi K

i " = 1 (10.13)


16/64


17/64

I;?#$(34 K".5*."6$3

At dewpoint (practically all vapor): L=0, V=1 and z i =

yi

zi ! z

i K

i

1+V ( K i !1)

" = 0 becomes,

Dewpoint criteria

yi ! yi K i

K i

" = 0

yi

K i

!" yi = 0"

yi

K i

=1 (10.14)"


18/64

If Raoult's Law valid,

yi

K i =!

yi

P i

sat

P

=1 P =1

yi

P i

sat !! (10.3),

see example 10.1

If Modified Raoult's Law valid,

yi

K i

=! y

i

" i P

i

sat

P

=1 P =1

yi

" i P

i

sat !

! (10.7),

see example 10.3


19/64

Relative Volatility

! ik =

yi

xi

yk

xk

=

K i

K k

at azeotrope ! ik =1

1= 1

!ik>1 Species i is relatively more volatile

!ik1

If Raoult's Law valid,

! 12

=

P 1

sat

P

P 2

sat

P

= P

1

sat

P 2

sat

If Modified Raoult's Law valid,

! 12

=

" 1 P

1

sat

P

" 2 P

2

sat

P

="

1 P

1

sat

" 2 P

2

sat


20/64

-:"0#.; OWQO


21/64

J.$4 J:OSO "4 NXYZ$K

ln P 1

sat / kPa = 14.2724!

2945.47

T / o

C + 224.00

ln P 2

sat / kPa = 14.2043!

2972.64

T / oC + 209.00

calculate at 75oC,

P 1

sat = 83.21kPa P

2

sat = 41.98kPa

Mixture: Acetonitrile(1)/Nitromethane(2)

Antoine Eqn,

Note: Acetonitrile(1) is more volatile.


22/64

Calculate P and y1, given a set of x1 and T=75oC.

This is BUBL P calculation.

xi K i ! = 1 (10.13) .

Let us assume Raoult's Law is valid , P = P b = x

i P

i

sat (10.2)!

P = x1 P

1

sat + x

2 P

2

sat = x

1 P

1

sat + (1" x

1) P

2

sat

P = ( P 1

sat " P

2

sat ) x1 + P

2

sat Eqn A note: a linear line (y=mx+c)

So,

Calculate P for a set of x1 (Eqn A) and then calculate y1 (Eqn B)

also, y1

=

x1 P

1

sat

P

Eqn B


23/64

T (oC) x1 P=Pb(kPa) y1

75 0 41.98 0

75 0.2 50.23 0.3313

75 0.4 58.47 0.5692

75 0.6 66.72 0.7483

75 0.8 74.96 0.8880 75 1 83.21 1 P 1

sat

P

2

sat

P = P b

= ( P 1

sat ! P

2

sat ) x1

+ P 2

sat y1 =

x1 P

1

sat

P

So now plot Px1 and Py1 on Pxy diagram!!

Point b

Given or SetCalculate


24/64

Ex: Calculate Pd and x1, given y1=0.6 and T=75oC

(i.e. what is the dew P for gas mixture at 75oC and 60% acetonitrile)

This is point c in previous Px1y1 diagram.Dew P calculation (Note: z1=y1).

Then calculate x1 using,

x1 =

y1 P

d

P 1

sat =

0.6(59.74)

83.21= 0.43

yi

K i

=1 (10.14),! If Raoult's Law valid, P d =1

yi

P i

sat !

(10.3)

P d =

1

0.6

83.21

+0.4

41.98

= 59.74kPaCompare with values

from Pxy diagram.


25/64

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#%;>>*%; 5".5*."6$3Q

• [;4 S(V 5".5*."4; J+ "3+ :(Q

• J.$4 J:S *>(3@ J+:(S(

• I2\]QQ


26/64

J.$4 N:OSO "4 JXYW^J"

T 1

sat / oC =

2945.47

14.2724 ! ln P / kPa! 224.00

T 2

sat / oC =

2972.64

14.2043! ln P / kPa! 209.00

so at 70kPa,

T 1

sat = 69.84

oC T

2

sat = 89.58

oC

Mixture: Acetonitrile(1)/Nitromethane(2)

Antoine Eqn,

As expected Acetonitrile(1) is more volatile


27/64


28/64

J.$4 N:OSO "4 JXYW ^J"

P (kPa) T=Tb(oC) x1 y1

70 69.84 1 (x2=0) 1 (y2=0)

70 74 0.7378 0.8484

70 78 0.5156 0.6759

70 82 0.3184 0.4742

70 86 0.1424 0.2401 70 89.58 0 (x2=1) 0 (y2=1) T 2

sat

T

1

sat

y1

=

x1 P

1

sat

P

So now plot Tx1and Ty1 on a Txy diagram!!

x1 =

P ! P 2

sat

P 1 sat

! P 2

sat

Given or Set


29/64

Ex: Calculate Tb and y1, given x1=0.6 and P=70kPa.

(i.e. calculate the bubble temperature at 70kPa and 60%

acetonitrile)

This is point b in previous Tx1y1 diagram. Note: z1=x1

Bubble temperature calculation!!

xi K

i ! = 1 (10.13),

The solution is not straightforward as T is unknown. Let’s

see how to solve mathematically,

For Raoult's Law, P b = x

i P

i

sat (10.2)!

P b = P

k

sat

P k

sat x

i P

i

sat = P

k

sat

! xi P i

sat

P k

sat ! = P k sat

xi"

ik !

where k is a component that arbitrarily chosen.


30/64

P b = P

k

sat x

i!

ik "

where ! ik =

P i

sat

P k

sat is relative volatility of i wrt k .

P k sat =

P b

xi!

ik " (A)

Also,

ln! ik = ln

P i

sat

P k

sat =ln P

i

sat -ln P k

sat = Ai #

Bi

T +C i

$

% &'

( ) - A

k +

Bk

T + C k

$

% &'

( )


31/64

Solution is through iteration,

1. Start with an initial guess of T as follows,

T = xi! T i

sat

2. Arbitrarily pick a component, e.g. Nitromethane so, k=2

3. Calculate !ik,(note: Number of !ik is equal to total number of component)

ln! ik = A

i " Bi

T +C i

# $ %

& ' ( - A

k + B

k

T +C k

# $ %

& ' (

T = 0.6(69.84) + 0.4(89.58) = 77.74oC

we get, !

12 = 1.9611

! 22

= 1


32/64

5. Calculate a new value of T using the Antoine eqn,

6. Stop if this T is equal or close to earlier value of T,

else use this value as a new guess. Repeat steps

3, 4 & 5 until converge.

T = B

k

Ak ! ln P

k

sat !C

k

T =2972.64

14.2043! ln44.3977! 209 = 76.53

oC

4. Calculate Pksat using eqn A, P

k

sat =

P

xi!

ik "

P 2

sat =

P

x1!

12 + x

2!

22

=70

0.6(1.9611)+0.4(1)=44.3977kPa


33/64

7. Finally, calculate yi using Raoult’s law (Use the Antoine Eqn

for Pisat)

y1=0.7472

T #12 P2sat

T

77.74 1.9611 44.39 76.53

76.53 1.9703 44.24 76.43

76.43 1.9717 44.22 76.42 Answer

(point b)


34/64

I-U N 5".5*."6$3

Calculate Td and x1, given y1 and P.

Example: Calculate Td and x1 for z1= y1 =0.6 and P=70kPa.

See page 356 for the solution (also by iteration) of DEW T

calculation.

Answer: Td

=79.58oCx1=0.4351


35/64

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K".5*."6$3 $P +;? #%;>>*%; "3+

/*//.; #%;>>*%; *>(3@ ED9".*; P%$0

I;J%(;>4;% 5A"%4Q

`$4;1 UAS 4A; >$.*6$3 (> /S 4%(". "3+ ;%%$%a


36/64

Flash Calculation


37/64

8.">A K".5*."6$3

73 (0#$%4"34 "##.(5"6$3 $P !'-bLiquid at pressure equal or higher than Pb

“flashes” or partially evaporates when the P is

reduced, thus producing a vapor and liquid.

{xi}, L {y

i},V

PTFlash calculation is to determine

V , L, {xi}, and {yi} at

T and P by assuming VLE.

Note: {xi} composition of liquid and {yi} composition of vapor


38/64

7> +;%(9;+ P$% !'- >S>4;0V

yi =

z i K

i

1+V ( K i !1)

(10.16)

yi" = 1, so:

z i K

i

1+V ( K i !1)

= 1 (10.17)"

Solution is by trial and error.

Guess V until the summation term equal to 1.

But………


39/64

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4A; >S>4;0 (> "54*"..S 4?$D#A">;Q

23 @;3;%". ]]

2P J+ c J c J/V 4?$ #A">;

$%

2P N/ c N c

N

+V 4?$ #A">;


40/64

-:"0#.;> $P

8.">A K".5*."6$3

-:"0#.; OWQZ1

8.">A 5".5*."6$3 P$% >S>4;0 ?A;%;


41/64

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`$3D5$3+;3>"/.; d">;>

NA(> (> " >#;5(". 5">; $P +;? N 5".5*."6$3Q

so P 2

sat = y

2 P

Solution is straightforward by the application of

Raoult’s Law to the condensable component H2O

(here identified as component 2).

The system contains a dew (liquid water) in VLE with

mixture of water vapor and non-condensable gases

(such as N2, CO2, O2 etc.).

y2 P = x

2 P

2

sat

The dew is 100% H2O, so x2=1


42/64

-:"0#.;

I;?#$(34 N $P K$0/*>6$3 J%$+*54> 8$% >4$(5A($0;4%(5 5$0/*>6$3 $P 0;4A"3;V 5".5*."4;

+;? #$(34 N $P 4A; 5$0/*>6$3 #%$+*54>Q

Ke_ f ghg f gFYi&gOG`g DDDDDDj Khg fgegh f gFYi&gOG`g

B$.; P%"56$3 $P egh (> g&FOfgfgFYi&gOGG X WQOi

P 2

sat = y2 P P

2

sat =0.19(101.325kPa)=19.25kPa

From steam table*, Td =T2sat= 59.5oC

*You could also use Antoine Eqn.


43/64

Henry’s Law


44/64

e;3%S=> '"? (> !'- %;."6$3 4A"4 (> 9".(+ P$%

(+;".D@"> 0(:4*%; (3 ;)*(.(/%(*0 ?(4A " +(.*4; >$.*6$3V

?A;%; ?; ?"34 4$ ^3$? 4A; 5$0#$>(6$3 $P +(>>$.9; @"> " (3

4A; +(.*4; >$.*6$3Q8$% ;:"0#.;V

!̂ i y

i P = "

i x

i f

i

yi P = x

i "

i f

i = x

i "

i P sat

i

let H i = "

i f

i = "

i P

i

sat

1no value for dissolved (supercritical)gas at VLE conditions

"G Khg "3+ egh >S>4;0Q

/G 7(% "3+ egh >S>4;0Q

! i is Henry’s constant (in bar) for dissolved gas (i ).

so y i P = x i ! i Henry’s Law


45/64

y i P = x i ! i

So at dilute solution,y i = ( ! i/P) x i

For constant system pressure P,

y i = (Constant)x i

If we plot y i vs x i , we get a straight linethrough the origin.

So Henry’s constant for dissolved gas (i ) canbe easily determined from experiment.


46/64

23 #%;9($*> ;:"0#.; $P +;?#$(34 P$% 5$0/*>6$3

#%$+*54V ?; ">>*0; 4A; .()*(+ (> ".. egh F:gXOGQ

What if we want to know the mole fraction of

dissolved CO2 (component 1) in the dew (liquid)?

U; 5$*.+ >$.9; 4A(> *>(3@ e;3%S=> '"? P$%

+(>>$.9;+ @">;> FKhgGQ


47/64

7##.S e;3%S=> ."? P$% 5$0#$3;34 OQ

H>; #O +"4" P%$0 N"/.; OWQO F3$4;1 4A(> "54*"..S 9".(+

"4 gZ$

KGQ

x1 =

y1 P

! 1

=(1 / (1+ 2 + 7.52))(1.013bar )

1670bar

=3.4622x10-5 " 0

As expected, only small amount of CO2 present

in liquid water.

See also example 10.2


48/64

2P 4A; P*@"5(4S $P " (3 .()*(+ #A">; (> @(9;3 /S e;3%S=> '"?V

ˆ f i = !

i x

i f

i = x

i"

i

ˆ

f iid =

! i yi P

! i y

i P = x

i"

i

2P 4A; @"> (> (+;". >$.*6$3 F';?(>& 9".(+GV

So we get the following version of Henry’s Law if gas

mixture is ideal solution ,


49/64

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J.$4 J:OSO "3+ :OSO +("@%"0> "4 NXkOlQOZE

8%$0 /*//.;#$(34 5".5*."6$3V

[;4 :O

K".5*."4; J/

K".5*."4; SO


50/64

x1 G1 G2 P y1

0.00 3.03 1.00 65.64 0.00

0.05 2.72 1.00 68.57 0.09

0.10 2.45 1.01 70.64 0.15

0.15 2.23 1.03 72.06 0.21

0.20 2.03 1.05 72.97 0.25

0.25 1.86 1.07 73.50 0.28

0.30 1.72 1.10 73.73 0.31

0.35 1.60 1.15 73.73 0.34

0.40 1.49 1.19 73.54 0.36

0.45 1.40 1.25 73.17 0.38

0.50 1.32 1.32 72.63 0.40

0.55 1.25 1.40 71.92 0.43

0.60 1.19 1.49 70.99 0.45

0.65 1.15 1.60 69.81 0.47

0.70 1.10 1.72 68.29 0.50

0.75 1.07 1.86 66.36 0.54

0.80 1.05 2.03 63.88 0.58

0.85 1.03 2.23 60.70 0.64

0.90 1.01 2.45 56.60 0.72

0.95 1.00 2.72 51.31 0.83

1.00 1.00 3.03 44.51 1.00

G1 is gamma1


51/64

'$5"4; "m;$4%$#(5 #%;>>*%; "4 NXkOlQOZE"3+ (4> 5$0#$>(6$3

T=318.15

40.00

45.00

50.00

55.00

60.00

65.00

70.00

75.00

80.00

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x1, Y1

K P a

Series1

Series2


52/64


53/64

T=318.15

40.00

45.00

50.00

55.00

60.00

65.00

70.00

75.00

80.00

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x1, Y1

K P a

Series1

Series2

Locate DEW P at T=318.15K, y1=0.6


54/64

!121

Locate azeotropic composition at 318.15K


55/64


56/64

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8%$0 ;:"0#.; OWQk/GV I;? J (> ngQli^J"

`$? 5".5*."4; /*//.; JV

P b = x

1!

1 P

1

sat + x

2!

2 P

2

sat = 71kPa

Pd


57/64

T=318.15

40.00

45.00

50.00

55.00

60.00

65.00

70.00

75.00

80.00

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x1, Y1

K P

a

Series1

Series2


58/64


59/64

K 1 =

P 1

sat !

1

P =

P 1

sat exp( A( x2)2 )

P

K 2 =

P 2

sat exp( A( x1)2 )

P

so (a) becomes,

66 = x1 P

1

sat exp( A(1" x1)2 ) + (1- x

1) P

2

sat exp( A( x1)2 )


60/64

66 = x

1(44.51)exp(1.107)(1! x

1)

2

)+

(1- x

1)(65.64)exp(1.107( x

1)

2

)

Guess x1,

x1 = 0.7 P=68.29

x1 = 0.8 P=63.88

x1 = 0.75 P=66.36

x1 = 0.76 P=65.91......good enough


61/64

Now we can calculate K 1 and K 2 for flash calculation,

K 1 =

P 1

sat !

1

P =

44.51exp(1.107(1" 0.76)2 )

66

= 0.719

K 2 =

65.64exp(1.107(0.76)2 )

66= 1.885


62/64

Substitute into eqn (10.17),

z i( K

i)

1+V ( K i !1)

= 1"

0.6(0.719)1+V (0.719 !1)

+ (1! 0.6)(1.885)1+V (1.885!1)

= 1 (10.17)

0.431

1! 0.281V +

0.754

1+ 0.885V = 1


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0.431

1! 0.281V +

0.754

1+ 0.885V = 1

Guess, " = 1

V=0.50 1.024

V=0.55 1.017V=0.70 1.002

V=0.75 0.999

V=0.73 1.000

So V=0.73 L=1-V=0.27


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yi =

z i K

i

1+V ( K i !

1) (10.16)

y1 = ..........

y2 = ..........

xi =

yi

K i

x1 = .........

x2 = .........

ocw chapter 10

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