Download CHE3161 - Semester1 - 2011 - Solutions PDF

TitleCHE3161 - Semester1 - 2011 - Solutions
TagsBranches Of Thermodynamics Mathematical Physics Mechanical Engineering Statistical Mechanics
File Size717.0 KB
Total Pages12
Document Text Contents
Page 2

CHE 3161 (JUN 11)

Page 2 of 12

(2) For the two-steps process:

!!" ! ! !

!

!

!"

! !!!!

! !!!!!! ! !!!
! !!!!! ! !!!!

Since P 1 = P 2, hence

112212
V  P V  P W  +!=

Applying Ideal Gas Law:

)( 21

1212

T T  R

 RT  RT W 

!=

+!=

0

3

223

=

 ! "= PdV W 

Therefore,

)( 21

231213

T T  R

W W W Total 

!=

+=

For the isothermal expansion process:

1

4

114
ln
 P 

 P 
 RT W  =

If the two works have to be the same:

bar 



T T 
 P  P 



T T 

 P 

 P 

T T  R
 P 

 P 
 RT 

406.0

200

)600200(
exp3

)(
exp

)(
ln

)1()(ln

1

21
14

1

21

1

4

21
1

4
1

=

!
"

#
$
%

& '
=

!
"

#
$
%

& '
=

'
=

'=

(3) P increases since the left hand term of Eqn (1) will be multiplied by a factor of less that 1 (1
=100% efficiency).

Page 6

CHE 3161 (JUN 11)

Page 6 of 12

At the azeotrope, yi= xi, then,


i
=

P

P
i

sat 

Therefore,


1


2

=

P
2

sat 

P
1

sat 

Given the conditions,

ln! 
1
= Ax

2

2
; ln! 

2
= Ax

1

2

Then,

ln
 ! 

1


2

= A( x
2

2
! x

1

2
)

Therefore,

 A =

ln

1


2

 x
2

2
! x

1

2
=

ln
 P

2

sat 

P
1

sat 

 x
2

2
! x

1

2

Putting in the known numbers for satuation pressures and compositions at the azeotrope:

 A = -2.0998

 Next, at x1 = 0.6, x2 = 1-x1 = 0.4,


1 = exp( A x2

2
)=0.7146


2
= exp( A x

1

2
)= 0.4696

P  =  x
1

1
P
1

sat 
+  x

2

2
P
2

sat 
= 38.1898 kPa

The vapour composition y1 is:

 y
1
=

 x
1

1
P
1

sat 

P
=0.8443

Page 11

CHE 3161 (JUN 11)

Page 11 of 12

Thus, % = 0.664

 y1= 1 - % = 0.336

 y2= % = 0.664

(b) For an ideal solution:

!
"

#
$

%
&
'

(
=

i

v

o

v

ii


 P 

 P 
 y i)( ) 

For species 1 $n-C4H10: #1 = 0.200; T c,1= 425.1 K; P c,1= 37.96 bar

395.0
96.37

15
1,

1,1,

==

=!



cr 

 P 

 P  P  P 

1
1.425

425
1,

1,1,

==

=!



cr 



T T T 

Using Equation (3.65) to determine B
o

339.0
1

422.0
083.0

6.11
!=!=

o
 B

Using Equation (3.66) to determine B
1

033.0
1

172.0
139.0

2.4

1

1
!=!= B

Using Equation (11.68) to determine " 1.

{ } 872.0)033.0(2.0339.0
1

395.0
exp

1
=!"

#
$%

&
'+'=( 

For species 2 $iso-C4H10: #2 = 0.181; T c,2 = 408.1 K; P c,2= 36.48 bar

411.0
48.36

15
2,

2,2,

==

=!



cr 

 P 

 P  P  P 

041.1
1.408

425
2,

2,2,

==

=!



cr 



T T T 

Using Equation (3.65) to determine B
o

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