THERMODYNAMICS PROPERTIES OF FLUIDS

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THERMODYNAMICS PROPERTIES
OF FLUIDS

CHAPTER VI

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Numerical values for thermodynamic
properties are essential to the calculation of
HEAT

and
WORK
.



Example :

work requirement for a compressor to
operate
ADIABATICALLY
, to rise the
pressure of gas from P
1

to P
2
:



Ws

=
Δ
H = H
2



H
1

By neglecting small kinetic and potential
energy changes. Thus the
SHAFT

work,
Ws
, is simply
Δ
H.

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PROPERTY RELATIONS for HOMOGENOUS PHASES


First law of thermodynamics, for a CLOSED system,
containing n moles :


d(
nU
) =
dQ

+
dW


For
REVERSIBLE

process :


d(
nU
) =
dQrev

+
dWrev

Where :

dWrev

=
p.d
(
nV
)



dQrev

=
T.d
(
nS
)





=

.




.

(

)


…(a)


Where :

U = molar internal energy; energy/mole



S = molar entropy; entropy/mole



V = molar volume; volume/mole

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Eq. (a) combines 1
st

law and 2
nd

law


contains
only
PROPERTIES of THE SYSTEM

(in the form of STATE function).


Eq. (a) derived for REVERSIBLE process, but the
application is
not restricted
just for reversible process;
since the eq. (a) already expressed in STATE VARIABLES.


Recall :


H


U + PV

Define :

1. Helmholtz energy


A


U


TS

2. Gibbs energy


G


H


TS


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From enthalpy :


nH

=
nU

+ P(
nV
)


d(
nH
) = d(
nU
) +
P.d
(
nV
)+(
nV
).
dP

…(b)

Equation (a) :



=

.




.

(

)

(a)


(b), results in :




=

.


+

.
𝑃


Similarly :


d(
nA
) = d(
nU
)


T.d
(
nS
)


nS.dT

…(c)

(a)


(c), results in :



𝐴
=


.






Analogue with that :

d(
nG
) = (
nV
)
dP



(
nS
).
dT


But remember :
all equations are written for the
ENTIRE
MASS
and
CLOSED system

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Application to a unit mass (or one mole) of homogenous
fluid with
CONSTANT COMPOSITION
:



dU

=
T.dS



p.dV



dH

=
T.dS

+
V.dP



dA

=
-
p.dV



S.dT



dG

=
V.dP



S.dT


From those equations, one can derive that :



𝜕
𝜕

=

𝜕𝑃
𝜕





𝜕
𝜕
𝑃

=
𝜕

𝜕
𝑃




𝜕
𝑃
𝜕


=

𝜕

𝜕






𝜕

𝜕

𝑃
=

𝜕

𝜕
𝑃


MAXWELL

Correlations

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Note :

Those math expression can be derived from :


if F = F(
x,y
), then the total differential,



dF

=
𝜕

𝜕



+
𝜕

𝜕





or,

dF

=
M.dx

+
N.dy

Where,

M =
𝜕
𝜕



and

N =
𝜕
𝜕


Further differentiation :

𝜕

𝜕

=
𝜕
𝜕
𝜕
𝜕

=
𝜕
2

𝜕𝜕


𝜕

𝜕


=
𝜕
𝜕

𝜕
𝜕


=
𝜕
2

𝜕

𝜕









=





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Example :


H = H(S,P)




dH

=
𝜕

𝜕



+
𝜕

𝜕
𝑃

𝑃


Using those above rule :

𝜕

𝜕
𝑃

=
𝜕
𝜕
𝑃
𝜕

𝜕

𝑃
=
𝜕
2

𝜕
𝑃
𝜕



𝜕

𝜕

𝑃
=
𝜕
𝜕

𝜕

𝜕
𝑃

=
𝜕
2

𝜕

𝜕
𝑃



Example :


G = G(P,T)





dG

=
𝜕

𝜕
𝑃


𝑃
+
𝜕

𝜕

𝑃



Further derivation results in :





𝑃
=




𝑃





𝑃

=




𝑃



etc.

T

V

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Consider :

H = H(P,T)




dH

=
𝜕
𝜕




+
𝜕
𝜕
𝑃

𝑃

From other equation :
dH

=
TdS

+
vdP



𝜕
𝜕𝑃

=

𝜕

𝜕𝑃

+




𝜕
𝜕

𝑃
=

𝜕
𝜕

𝑃

From Maxwell equation :


𝑃

=



𝑃

Hence,


𝜕
𝜕𝑃

=



𝜕

𝜕

𝑃

Putting back to the initial equation :


=
𝐶
.

+





𝑃
𝑃

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From the relationship :



𝜕
𝜕
𝑃
=

𝜕
𝜕
𝑃



𝜕
𝜕
𝑃
=




Entropy can be expressed as : S = S(T,P)



=
𝜕
𝜕
𝑃

+
𝜕
𝜕
𝑃

𝑃



total
entropy change



=
𝐶






𝑃
𝑃

Here, the correlation between V & T should
be
KNOWN
!

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Example :

For an ideal gas, the PVT behavior is expressed by
:


PV
ig

= RT

So that,
𝜕

𝑖𝑔
𝜕

=

𝑃


Putting back to general
dH

&
dS

equations:

dH
ig

=
Cp
ig
.dT

+

𝑖



𝑃
𝑃



dH
ig

=
Cp
ig
.dT

dS
ig

=
𝐶
𝑖




𝑃
𝑃



dSig

=
𝐶
𝑖




𝑃
𝑃


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Alternative form
:

1

𝜕
𝜕
𝑃
=
𝛽


(volume expansion
coeff
.)

From Maxwell equation, we get :


𝜕
𝜕
𝑃
=

𝜕
𝜕𝑃




𝜕
𝜕𝑃

=

𝛽

and,


𝜕
𝜕𝑃

=



𝜕
𝜕
𝑃



𝜕
𝜕𝑃

=
1

𝛽


The dependence of H and S on pressure, can then be
expressed as :



dH

=
Cp.dT

+ V(1
-
β
T)
dP

And,


dS

=
𝐶



𝛽𝑃


Since
β

and V are WEAK function of P; they can be
assumed constant at appropriate average values.

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The value of
β

usually applied only to liquids.



For
LIQUID
, far from the critical point : P
has little effect on S, H, and U.


For an incompressible fluid, usually
assumed that
β


0. In this case :

𝜕
𝜕
𝑃

0



𝜕
𝜕𝑃


0



𝜕

𝜕𝑃


0

However,
𝜕

𝜕𝑃


0
, since :



𝜕
𝜕𝑃

=



𝜕

𝜕

𝑃




𝟎

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Example 6
-
1

Determine the enthalpy and entropy changes
for liquid water for a change of state from 1
bar and 25
o
C to 1000 bar and 50
o
C. The
following data for water are available :







t/
o
C

P/bar

Cp
/J.mole
-
1
.K
-
1

V/cm
3
.mole
-
1

β
/K
-
1

25

1

73,305

18,075

256 x 10
-
6

25

1000

-

17,358

366 x10
-
6

50

1

75,314

18,240

458 x10
-
6

50

1000

-

17,535

568 x10
-
6

Copyright © Wondershare Software

Solution :

Δ
H

=
Cp.dT

+ V(1


β
T)
dP




Cp
avg
(T
2



T
1
) +
V
avg

(1


β
avg
.T
2
)(P
2



P
1
)

And

ΔS

=
Cp




-

β
.
V.dP




Cp
avg

ln


2

1

-

β
avg
.V
avg
.(
P
2



P
1
)

The path of integration :


H
1
;S
1

at 1 bar 25
o
C

1 bar

50
o
C

H
2
;S
2

At 1000
bar, 50
o
C

2


𝐶





𝐶




At 1 bar.



1

𝛽
𝑃


𝛽

.
𝑃



At 25
o
C

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Note :


ΔH & ΔS is state functions;


the path of integration can be
arbritrary
.


Cp

~

weak function of T

V &
β

~

weak function of P


Hence,

At P = 1 bar ;
Cp
avg

= ½ (73,305 + 75,314) = 75,310 J/mole/K

At t = 50
o
C ;
V
avg

= ½ (18,240 + 17,535) = 17,888 cm
3
/mole



β
avg

= ½ (458 + 568) x 10
-
6

= 513 x 10
-
6

K
-
1

Substitution gives :

ΔH = 75,310(323,15

298,15) +
17
,
888
1

513
10

6
323
,
15
10
(
1000

1
)



= 3,374 J/mole

ΔS = 75,310
ln

323
,
15
298
,
15

513
10

6
17
,
888
(
1000

1
)
10


= 5,14 J/mole/K

Can use
arithmetic
avg
.

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RESIDUAL PROPERTIES

Consider Gibbs free energy :


dG

=
VdP



SdT

[by definition : G = H


TS]

Can be expressed as : G = G(P,T)

The alternative form is :



=
 


ℎ 

 


𝑎

  𝑢 






1






2


(pure math derivation)




1

𝑃






2







𝑃



+



2




𝑃


+



2


Results in :





=


𝑃




2



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From the above equation, it can be derived :




=
𝛿
𝐺
𝑅𝑇
𝛿𝑃

;


and




=


𝜕
𝐺
𝑅𝑇
𝜕
𝑃


Other relationships for


;



etc. can be derived with the
same methods.

If the relation :


=

(

,
𝑃
)

is known, all the other
thermodynamics properties can be evaluated! (by simple
mathematical operations).



=


,
𝑃



GENERATING FUNCTION

for other
thermodynamics properties.

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However :


N
o convenient experimental method for determining the
values of G (or G/RT).


The equation of Gibbs are of
LITTLE PRACTICAL
use.


Define : the RESIDUAL GIBSS energy,


G
R



G


G
ig

In an analogues way, can be defined residual volume :


V
R

= V


V
ig




= V
-


𝑃

; and since : V =
 
𝑃

Then,


V
R

=

𝑃

(z


1)

Back to the Gibbs equation (the modified one) :





=


𝑃




2





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For ideal gas :




𝑅

=

𝑅

𝑃


𝑅


2


This is the
Fundamental Property Relation
for residual
properties; applicable to
constant composition fluids
.


From those equation we can derive :



𝑅

=
𝜕
𝐺
𝑅
𝑅𝑇
𝜕𝑃


And



𝑅

=


𝜕
𝐺
𝑅
𝑅𝑇
𝜕
𝑃



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At constant T :





𝑅

=

𝑅

𝑃




𝑅

𝑃


𝑅

𝑃

0
=


𝑅

𝑃
0
𝑃





=




𝑃

𝑃
0



































Recall that :

𝑅

=
1




𝑖
=
1
𝑃


1





𝑅

=

(


1
)
𝑃
𝑃

𝑃
0


at constant T!


Recall :

𝑅

=


𝜕
𝐺
𝑅
𝑅𝑇
𝜕
𝑃

Combination with the above eq. gives :



𝑅

=


𝛿
𝛿



1
𝑃
𝑃

𝑃
0






𝑅

=



𝛿
𝛿


0
𝑃
𝑃

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Similarly, as : G
R

= H
R



TS
R

, we obtain :


𝑅

=



𝛿
𝛿


0
𝑃
𝑃





1
𝑃
𝑃

𝑃
0


The calculation of Enthalpy & Entropy, is then :

H =
H
ig

+ H
R

and

S = S
ig

+ S
R


Where,

H
ig

= H
ig
o

+

𝐶
𝑖


0



; and



S
ig

=
S
ig
o

+

𝐶
𝑖


0




ln
𝑃
𝑃
0


The enthalpy and entropy eq. can be simply expressed as :



H =

0
𝑖
+
𝐶
ℎ
𝑖


T0
+





S =

0
𝑖
+
𝐶


𝑖
ln

0


ln
𝑃
𝑃0
+



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Where :


𝐶
ℎ
𝑖
=


𝑖𝑔

𝑇
𝑇
0


0




for :
Cp
ig

= A + BT + CT
2

+


2


= R [ A +
B.T
am

+

3
(4.T
am
2



T
0
.T) +

0
.

]


where Tam


½ (T + T
0
)



𝐶

𝑖


𝐶𝑝
𝑖𝑔
𝑑𝑇
𝑇
𝑇
𝑇
0
ln
𝑇
𝑇0

= R [
A+B.T
lm
+T
am
.T
lm

(C+

(
0
.

)
2
)]


where
Tlm

=


0
ln
𝑇
𝑇0



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Note :

The calculation of H & S for
REAL gas
is generally more convenient
using
RESIDUAL PROPERTIES
, i.e. :


H =
H
ig

+ H
R


and

S = S
ig

+ S
R

Where :



𝑅

=



𝛿
𝛿
𝑃
𝑃
0
𝑃
𝑃

;




𝑅

=



𝛿
𝛿

𝑃
0
𝑃
𝑃




1
𝑃
𝑃

𝑃
0


However for
LIQUID
, the
original equations
mostly more appropriate,
i.e. :


dH

=
Cp

dT

+ [v


T
𝛿
𝛿

] dP




=
Cp

dT

+ v (1


β
T)
dP

And,


dS

=
Cp




𝛿
𝛿

𝑃



=
Cp




-

β
v
dP


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Example 6
-
2

Calculate the enthalpy and entropy of
saturated
isobutane

vapor at 360 K. use the
following information :


1.
The vapor pressure of
isobutane

at 360 K
is 15,41 bar.

2.
Set
H
o
ig

= 18.115 J/mole &
S
o
ig

= 295,976
J/mole for ideal gas reference state at 300
K &1 bar.

3.

𝑖𝑔

=
1
,
7765
+
33
,
037
10

3



(

=

𝐾
)

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Solution :

There are 2 steps for calculating H & S of real gas, i.e. :


H =
H
ig

+ H
R


and

S = S
ig

+ S
R

1.
Calculation of RESIDUAL PROPERTIES :




𝑅

=



𝛿
𝛿
𝑃
𝑃
0
𝑃
𝑃


and






𝑅

=



𝛿
𝛿

𝑃
0
𝑃
𝑃




1
𝑃
𝑃

𝑃
0




The value of
𝛿
𝛿

should be calculated FIRST! But how?



𝛿
𝛿

given by the SLOPE of a plot of z
vs

T, at constant
pressure.

Numerical / graphical

Numerical / graphical

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(refer to table 6.1 for z data!)


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Z

T

0,1 bar

0,5 bar

2

bar

4

bar

𝛿
𝛿
=
 
!

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The result of :
1
𝑃

𝛿
𝛿

vs

P at 360 K, see table 6.2











From table 6.2


plot

1
𝑃

𝛿
𝛿

vs

P







1
𝑃

vs

P

Calculate the INTEGRAND
(numerical? Graphical? Up
to you)

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2. Calculate the ideal properties :
H
ig

& S
ig


H
ig

= H
ig
o

+

𝐶
𝑖


0





= 18.115 +


[
1
,
7765
+
33
,
037
10

3
360
3
00

]




= 24.439,8 J/mole

[R=8,314 J/mole/K]


S
ig

=
S
ig
o

+

𝐶
𝑖


0




ln
𝑃
𝑃
0



= 295,967 + 8,314

1
,
7765
+
33
,
067
10

3



360
3
00

8
,
314
ln
15
,
41
1



= 280,942 J/mole/K


Integration results :




𝛿
𝛿

𝑃
0
𝑃
𝑃

= 26,37x10
-
4

K
-
1






1
𝑃
𝑃

𝑃
0

=
-
0,2596

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Hence :




𝑅

=

360
26
,
37
10

4
=

0
,
9493




H
R

= (
-
0,9493) x 8,314 x 360 =
-
2.841,3 J/mole






𝑅

=

0
,
9493


0
,
2596
=

0
,
6897




S
R

= (
-
0,6897)(8,314) =
-
5,734 J/mole/K





H =
H
ig

+ H
R

= 24.439,8


2.841,3 = 21598,5 J/mole



S = S
ig

+S
R

= 292,411


5,374 = 286,676 J/mole/K


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TWO PHASE SYSTEM












Postulate

: during the phase transition (fusion/melting;
vaporization/condensation) :



dG

= 0

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Hence,

For two phases
α

and
β

of a pure species co
-
existing at
equilibrium :



G
α

= G
β

G
α

= molar Gibbs energy of phase
α

G
β

= molar
Gibbs energy of phase
β


And hence
:
dG
α

=
dG
β

Recall that :
dG

=
VdP



SdT

So that; V

α
dP
sat



S

α
dT

= V

β
dP
sat



S

β
dT

Results in :


𝑃


=










=







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= entropy change





= volume change

“which occurs when a unit amount of pure chemical is
transferred

from phase
α

to
β
, at eq. T,P”


Recall that :
dH

=
TdS

+
VdP

At constant T,P (
equilib
.) :
Δ
H
α
β

= T.
Δ
S
α
β



𝑃


=




.






Recall the
CLAYPERON

equation :



𝑃


=




.





From phase transition : liquid


vapor


𝑃


=


𝑣

.


𝑙𝑣


;


𝑣

= latent heat of vaporization

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For vaporization at low pressure :




𝑣




𝑣


𝑃


So that :


𝑃


=


𝑣
𝑅𝑇
𝑃


Or :


𝑃

𝑃




2

=


𝑣
















𝑣
=


 

𝑃


1
𝑇





Clausius

Clayperon

equation


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Clausius


Clayperon

:



𝑣

α

slope of plot of
ln

P
sat

vs

(1/T)


Experimentally :


𝑣

≠ f(T) at low pressure (P
↓)



Clausius



Clayperon

eqn. applies.


However at high pressures (P ↑) ;


𝑣


f(T).


𝑣
↓ if T↑ (from
triple point to critical point).


Semi empirical equations for
P
sat

vs

T :

(derived based on
Clausius



Clayperon
)


Ln
Psat

= A
-





[
Clausius



Clayperon
]


Ln
Psat

= A
-



+


[Antoine Equation]


(more accurate ; most popular one)


Ln
Psat

= A
-




+ D
ln

T + F.T6

[Riedel eqn.]


where A,B,D,F = constants


(wide temperature range)

Copyright © Wondershare Software

THERMO. PROPERTIES OF GASES : GENERALIZED
CORRELATION

Recall reduced properties :




𝑅

=



𝛿
𝛿
𝑃
𝑃
0
𝑃
𝑃


and



𝑅

=



𝛿
𝛿

𝑃
0
𝑃
𝑃




1
𝑃
𝑃

𝑃
0


Define : reduced pressure,
Pr

= P/Pc


dP

=
Pc.dPr

Reduced temp.,
Tr

= T/
Tc



dT

=
Tc.dTr


Putting back to the above eqns. :



𝑅


=


2

𝛿
𝛿
𝑃

𝑃

0
𝑃

𝑃





and,




𝑅

=



𝛿
𝛿
𝑃
𝑃

0
𝑃

𝑃





1
𝑃

𝑃


𝑃

0

Copyright © Wondershare Software

Refer to chapter 3 :


Z = Z
0

+
ω
Z
1


𝛿
𝛿
𝑃
=
𝛿

0
𝛿
𝑃
+

ω
𝛿

1
𝛿
𝑃

Result in :




=


2

𝛿

0
𝛿
𝑃
𝑃
0
𝑃
𝑃
+



2

𝛿

1
𝛿
𝑃
𝑃
0
𝑃
𝑃





=


0

+



1





𝑅

=



𝛿

0
𝛿
𝑃
𝑃
0
𝑃
𝑃



0

1
𝑃
𝑃

𝑃
0
+




𝛿

1
𝛿
𝑃
𝑃
0
𝑃
𝑃



1

1
𝑃
𝑃

𝑃
0





=


0

+



1


Copyright © Wondershare Software

More direct correlation between z and (
Pr,Tr
) :

(refer to chapter 3)


Z = 1 + B
0
𝑃


+
ω

B
1
𝑃




This results in :





Z
0

Z
1

Copyright © Wondershare Software

Where :







For a change of state of a real gas, from state
-
1 to state
-
2 :


H
2

= H
0
ig

+

𝐶
𝑖
2

0

+

2



H
1

= H
0
ig

+


𝐶
𝑖

1

0

+

1





Δ
H =


𝑖𝑔


2

1
+

2



1



Copyright © Wondershare Software

For the entropy, we obtain :



Δ
S
=


𝑖𝑔


2

1


ln
𝑃2
𝑃1
+

2



1



The schematic path to calculate
Δ
H &

Δ
S
for
real gas

is :


Copyright © Wondershare Software

Example :

Estimate
Δ
H
,
Δ
S, and
Δ
U to vaporize 1
-
butene from
its original saturated


liquid condition at 0
o
C, 1,273
bar to 200
o
C 70 bar (vapor).

The data available :

Tc

= 419,6 K

; Pc = 40,2 bar

;
ω

= 0,187



𝑖𝑔

=
1
,
967
+
31
,
630
10

3


9
,
837
10

6

2


(T in K)

Δ
Hv

at 0
o
C 1,273 bar = 21753 J/mole


Copyright © Wondershare Software

Solution :

The paths to estimate
Δ
H &
Δ
S can be describes as follow :



Copyright © Wondershare Software

Path :

(a) : vaporization at T
1
, P
1


(b) : transition REAL GAS


IDEAL GAS at T
1
, P
1

(c) : change from (T
1
, P
1
) to (T
2
, P
2
) in the IDEAL
GAS state

(d) : transition IDEAL GAS


REAL GAS at T
2
, P
2


Step (a) :


Δ
H
a =
Δ
Hv

= 21753 J/mole


Δ
S
a =
Δ
S
lv

=

𝑣
1

=
21753
273
,
15

= 79,64 J/mole/K

Step (b) :


Tr

=
273
,
15
419
,
6

= 0,651

;
Pr

=
1
,
273
40
,
2

= 0,0317


B
0

= 0,083
-

0
,
422
0
,
651
1
,
6

=
-

0,756



0

=
0
,
675
0
,
651
2
,
6
=
2
,
06

Copyright © Wondershare Software


B
1

= 0,139
-

0
,
172
0
,
651
4
,
2

=
-
0,904



1

=
0
,
722
0
,
651
5
,
2
=
6
,
73






=
0
,
0317

0
,
756

0
,
651 2
,
06
+
0
,
187

0
,
904

0
,
651 6
,
73


=
-
0,0978


H
b
R

= (
-
0,0978)(8,314)(419,6) =
-
341 J/mole






𝑅

=

0
,
0317
2
,
06
+
0
,
187 6
,
73
=

0
,
105



S
b
R

= (8,314)(
-
0,105) =
-
0,87 J/mole/K


Step (c) :



𝑖
=

(
1
,
987

+
31
,
630
10

3
473
,
15
2
73
,
15

9
,
837
10

6

)

 8
,
314



= 20564 J/mole

Copyright © Wondershare Software



𝑖
=
8
,
314


1
,
987

+
31
,
630
10

3
473
,
15
2
73
,
15

9
,
837
10

6

)


(
8
,
314
)

ln


70
1
,
273


= 22,16 J/mole/K


Step (d) :

Identical to step (b), with
Tr

=
473
,
15
419
,
6

=1,13 and

Pr

=
70
40
,
2

= 1,74

Results :


Δ
H
b
R

=
-
8582 J/mole




Δ
S
b
R

=
-
14,38 J/mole/K

Copyright © Wondershare Software

Hence :

(ΔH)real

=
ΔHa

+
ΔH
ig

+ (
H
d
R



H
b
R
)



= 21753 + 20564 + [
-
8582


(
-
341)]



= 34076 J/mole

(
ΔS)real

=
ΔSa

+
ΔS
ig

+
(
S
d
R



S
b
R
)



= 79,64 + 22,16 + (
-
14,38


(
-
0,87))



= 88,29 J/mole/K

(
ΔU)real

= (
ΔH)real


ΔPV =
(
ΔH)real


(P
2
V
2



P
1
V
1
)


(note : V
2

= Z
2
RT
2
/P
2

; Z
2

= Z
0

+
ω

Z
1

etc.)




Neglected!

Copyright © Wondershare Software

THERMODYNAMICS DIAGRAM







1


2 : Heating liquid





2


3 : vaporization process





3


4 : heating vapor







1= sub cooled liquid

4 = superheated vapor

Isentropic line

Critical

point

Isothermic


line

Copyright © Wondershare Software

Critical
point

Line of constant
pressure (isobaric
line)

Isenthal
pic line

b


a’

a



b’ : isentropic (adiabatic


reversible) compression

(S = const. ;
Δ
S = 0)

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Pressure enthalpy diagram for ammonia


Copyright © Wondershare Software

Copyright © Wondershare Software

THE
USE OF THERMO. DIAGRAM / DATA

Example
:



nozzle


isolated






adiabatic


reversible




(isentropic)

Determine the state of the steam existing the nozzle,
assuming that equilibrium is attained!

P
1

= 1000
kPa


T
1

= 260
o
C


If P
2

= 200
kPa

; T
2
? Condition/state?

Super
heat

steam

P
1

T
1

P
2

T
2

Copyright © Wondershare Software

Solution :

Refer to the steam table :

At P
1

= 1000
kPa



T
1

= 260
o
C

Since the process is adiabatic reversible :

ΔS = 0


S
1

= S
2


Hence :

S
2

= 6,9860 kJ/kg/K



P
2

= 200
kPa

Because at P
2

= 200
kPa


-

liquid water sat. :
S
l

= 1,5301


-

sat. steam :
S
v

= 7,1268



the state of fluid existing the nozzle
must be
a mixture of
liquid and vapor

H
1

= 2965,2 kJ/kg

S
1

= 6,9860 kJ/kg/K

At this condition, the
phase is
neither
liquid
nor

vapor

Refer to steam table
(T
2

= 120,23
o
C)

Copyright © Wondershare Software

If X
v

= mass fraction of vapor in existing stream, hence

S
2

= (1
-

X
v
).
S
l

+
X
v
.S
v

6,9860
= (1
-

X
v
) 1,530 + X
v

(7,1268)


X
v

= 0,9716


The exit stream will consist of 97,16%
wt

steam (vapor) and
2,84%
wt

liquid.


From steam table :

H
l

= 504,7 kJ/kg

H
v

= 2706,7 kJ/kg

H
2

= (0,9716)(2706,7) + (0,0284)(504,7)



= 2644,2 kJ/kg

Copyright © Wondershare Software

The process can be schematically described
as follow :


Isobaric line

260
kPa

T
1

= 260
o
C

T
2

1000
kPa

T

S

S
l

= 1,5301

S
v

= 7,1268

S = 6,9860

Copyright © Wondershare Software

260
o
C

Isothermal line

Isentropic
line

S = 6,9860

T
2

= 120,23
o
C

H
l

= 504,7

H
v

= 2706,7

H ?

P

H

P
2

= 200
kPa

P
1

= 1000
kPa