Procedure
Tables of
Thermodynamic
Properties
Dr. M. A. Habib
Professor,
Mechanical Engineering
Department, KFUPM
١
١
The procedure of calculating the thermodynamic
property is as follows:
Define the state by two independent properties.
Choose SI units tables.
Choose the table of the concerned substance.
Define the region being compressed liquid,
saturation or superheated.
Determine the required property from the table.
٢
3. Calculation of the thermodynamic properties
3.1
Saturation region
Defining the state
The state is defined by two independent
properties. It should be noted that, in the
saturation region, P and T are two
dependent properties.
٣
٢
Given T and x
Example:
Saturation water at T=100 oC and x=0.4.
Find P and v.
Solution:
From saturation table at T=100oC, read P=
101.3 kPa, vf = 0.001044 m3/kg and vfg=
1.67185 m3/kg, then:
v= vf + x (vfg) = 0.6698 m3/kg
٤
3.1 Saturation region
3.1 Saturation region_ Given P,x
٥
٣
Given P,x
Example
Saturated water at P= 100 kPa and x=0.6.
Find T and v.
Solution
From saturation table at P=100 kPa read
T= 99.62 oC, vf =0.001043 m3/kg and and
vfg= 1.67185 m3/kg . Then
٦
3.1 Saturation region_ Given P,x
3.1 Saturation region_ Given T,v
v= vf + x (vfg)
= 1.01682 m3/kg
٧
٤
Given T,v
Example
Saturated R134a at T=10 oC and v=0.02
m3/kg. Find x and P
Solution
From saturation table at T=10oC read P= Psat =
415.8 kPa, vf = 0.000794 m3/kg and vfg=
0.04866 m3/kg, then
٨
3.1 Saturation region_ Given T,v
3.1 Saturation region_ Given P,v
x=
Example.
Saturated water at P=100 kPa and v=0.1m3/kg.
Determine T and x.
v − vf
v fg
= ٠0٣٩٤٧
٩
٥
Given P,v
Solution
From saturation table at P=100 kPa read vf =
0.001043 m3/kg and vfg= 1.69296 m3/kg and T=
Tsat =99.62 oC. Then:
١٠
3.1 Saturation region_ Given P,v
Given P, T and Given T,v and T > Tc
x=
٦
v − vf
= 0.05845.
v fg
١١
Given P, T
Since the pressure and temperature are two
dependent properties inside the saturation
region, therefore, this state is not inside the
saturation region
Given T,v and T > Tc
The state is not inside the saturation region
١٢
Interpolation
Interpolation
,Interpolation is required when the given property is not found
but instead it lies between two given values in the tables
Example
Given saturated water at T= 102 oC, x=0.4.
Determine P and v.
T, oCP, kPa
vf, m3/kg vfg, m3/kg
100 101.3 0.001044 1.67185
105 120.8 0.001047 1.41831
١٣
٧
Solution
The available table provides only T=100 oC
and T= 105oC as shown. Therefore, the
student has to create another table having T
= 102 oC as shown in the table below. The
values P, vf and vfg at 102 oC can be obtained
by interpolation. As an example, P102 can be
obtained from the following equation:
١٤
Interpolation
Superheated Region
Superheated region
P102 − P100 102 - 100
P - 101.3 102 - 100
=
, thus 102
=
P105 - P100 105 - 100
120.8 - 101.3 105 - 100
Given P, T
Example
Water vapor at
P=1 MPa,
T=300 oC.
Calculate v.
Thus, P102 =109.1 kPa
T, oC
100
102
105
P, kPa
101.3
109.1
120.8
vf, m3/kg
0.001044
0.0010452
0.001047
vfg, m3/kg
1.67185
1.57043
1.41831
T, oC
300
P, kPa
v, m3/kg
0.25794
v= vf + x (vfg) = 0.6292 m3/kg
١٥
٨
١٦
Superheated Region
Superheated Region
Solution
v=0.25794
m3/kg
Given T, v
Example
Water vapor at T=300
oC, v=0.3 m3/kg.
Determine P.
Solution
T, oC
300
١٧
٩
P1=800 kPa
v1 , m3/kg
0.32411
P= ? kPa
v, m3/kg
0.3
P2=1000kPa
v2 , m3/kg
0.25794
١٨
Superheated Region
Superheated Region
Interpolation gives:
v − v1
P - P1
0.3 - 0.32411
P - 800
=
, thus
=
v 2 - v 1 P2 - P1
0.25794 - 0.32411 1000 - 800
Thus, P = 872.87 kPa
١٩
١٠
Given P, v
Example:
R134a vapor at P=0.8
MPa, v=0.035 m3/kg.
Determine T.
P = 0.8 MPa
T1= 90 oC
v1 = 0.03394 m3/kg
T= ? oC
v= 0.035 m3/kg
T2= 100 oC
v2= 0.03518 m3/kg
٢٠
Superheated Region
Double interpolation
Solution
v − v1
T - T1
0.035 - 0.03394
T - 90
=
, thus
=
0.03518 - 0.03394 100 - 90
v 2 - v 1 T2 - T1
Thus T= 98.55oC
٢١
١١
Double interpolation
Example
Given water at P= 1.1 MPa, T= 220 oC.
Solution
First, interpolate between v1a and v1b to get v1
Then, interpolate between v2a and v2b to get v2
Then, interpolate between v1 and v2 to get v.
٢٢
Double interpolation
Compressed-liquid region
T
T, oC
T1=200
T=220
T2=250
Pa =1.0 MPa
v1a =0.20596
v2a =0.23268
P=1.1 Mpa
v, m3/kg
v1 =0.18763
v=0.197584
v2 =0.21252
Pb==1.2 Mpa
v1b=0.1693
v2b =0.19235
٢٣
١٢
Given P,T
Example:
Compressed-liquid water at P=10 MPa and T=100
oC. Find v.
Solution:
From compressed-liquid table, at P = 10 MPa and
T= 100 oC read:
v= 0.001039 m3/kg
٢٤
Other Problems (Special
cases)
Case of compressed liquid water at low
pressure
Example:
Compressed liquid water at P=100 kPa,
T=20 oC. Calculate the specific volume.
٢٥
١٣
Solution:
The first available pressure in the
compressed liquid table is 5000 kPa. Thus
P<< 5000 kPa, therefore, the following
approximation will be performed:
Using the saturation table, read:
v=vf at T= 20 oC, thus v= 0.001002 m3/kg.
٢٦
Compressed liquid table is not available.
Compressed liquid table is not available
Case of state in the compressed liquid
region and the compressed liquid table is
not available.
Example:
Compressed liquid R134a at P=1.0 MPa
and T=10 oC. Determine v.
Solution
٢٧
١٤
The compressed-liquid table is not available,
therefore, the following approximation will be
performed:
Using the saturation table, read:
v= vf at T=10 oC, thus, v= 0.000794 m3/kg.
٢٨
Close to saturated vapor
A state in the superheated region and close
to saturated vapor.
Example
Water vapor at P= 800 kPa and T=180 oC.
Determine v.
Solution:
Interpolation between Saturation
temperature and T= 200 oC, thus:
٢٩
١٥
P, 800 kPa (170.43oC )
T, oC
v, m3/kg
Sat (T1=170.43oC ) v1= .24043
T= 180oC
v = ??
o
T2= 200 C
v2= 0.2608
v180 − v sat 180 − Tsat
=
v 200 − v sat 200 − Tsat
٣٠
Defining the region
Thus:
v - 0.24043
180 - 170.43
=
0.2608 - 0.24043 200 - 170.43
The region in which the state lies is
determined by its two independent
properties. The following are
examples. This section is only
concerned with defining the region.
Thus v= 0.247 m3/kg
٣١
١٦
٣٢
(T, x) OR (P, x)
The properties can then be determined
according to the procedures given in
Section 3 based on the region being
compressed liquid, saturation or
superheated.
٣٣
١٧
Given
The state has to be inside saturation
region because x (having a value
between 0 and1) indicates a mixture of
liquid and vapor.
٣٤
T, v
Given
Given T,v
From saturation tables, determine vf and vg
at T. The region is determined based on
the following conditional table:
Condition
v <vf
v >vg
vf ≤v ≤ vg
Region
Compressed liquid region
Superheated region
Saturation region
٣٥
١٨
٣٦
Given T,v
Given P, v
Example: R134a at T= 0 oC, v= 0.3 m3/kg. Define the region.
Solution: At T= 0 oC, vf = 0.000773 m3/kg and vg=0.06919 m3/kg, thus v > vg, then, the state
is in superheated region.
Given P, v
From saturation tables determine vf
and vg at Psat = P then:
Condition
v <vf
v >vg
≤
vf v≤vg
٣٧
١٩
Region
Compressed liquid region
Superheated region
Saturation region
٣٨
P, T
Given
Given P, v
Example:
Water at P= 150 kPa, v= 1.0 m3/kg. Define
the region.
Solution:
At P= 150 kPa, read vf=0.001053 m3/kg and
vg= 1.15933 m3/kg , Thus vf < v < vg , then
the state is in the saturation region.
٢٠
P, T
One of two alternative methods can be
used as follows:
a. Method 1: Starting with temperature:
From saturation tables determine Psat at T,
then, according to the following table the
region can be determined:
٣٩
Given
٤٠
P, T
Given
Condition
P > Psat
P < Psat
P, T
Region
Compressed liquid region
Superheated region
Example:
Given
Solution:
Psat (at T= 40 oC) = 7.384 kPa. Thus, P >
Psat, therefore, the state is in the compressed
liquid region.
Property diagrams for case of P > Psat (The
state is in the compressed-liquid region).
Water given at 100kPa and 40 oC
٤١
٢١
٤٢
P, T
Given
P, T
٤٣
٢٢
Given
Example
R134a at 1.0 MPa and T=50 oC. Define the
region.
Solution:
From saturation tables, determine Psat (at T=
50 oC). Thus Psat = 1318.1 kPa. Therefore,
P< Psat. Then, the state is in the superheated
region.
٤٤
Property diagrams for case of P < Psat (The
state is in the superheated region).
Method 2: Starting with temperature
٤٥
٢٣
b. Method 2: Starting with temperature:
From saturation tables determine Tsat at P, then,
according to the following table, the region can be
determined:
Condition
Region
T > Tsat
Superheated region
T< Tsat
Compressed liquid region
٤٦
Method 2: Starting with temperature
Method 2: Starting with temperature
Example:
Water at 300 oC and P=1 MPa. Define the
region.
Solution:
From saturation table at P=1 MPa, read Tsat
= 179.91oC, thus, T > Tsat . The state is in
the superheated region.
٤٧
٢٤
Property diagrams for case of T > TThe ( sat
.)state is in the superheated region
٤٨
Method 2: Starting with temperature
Method 2: Starting with temperature
Example
Water at 200 oC and P=3 MPa. Define the
region.
Solution
From saturation table at P= 3MPa get Tsat =
233.9. Thus, T< Tsat. Therefore, the state is
in the compressed-liquid region.
٤٩
٢٥
Property diagrams for case of T < Tsat (The state is
in the compressed-liquid region.
٥٠
Constant-volume process
Thermodynamic processes for a system
The following are examples of constant property
processes of a system.
Work = = zero
Constant-pressure process
Work =
٥١
٢٦
∫ pdv = zero
٥٢
Isothermal process (T=constant)
Process following the relation Pv = constant
Work)real substance =
Work =
٢٧
∫ pdv
= p1v1 ln
v2
v1
∫ pdv = ?
Work)ideal gas =
∫ pdv
= p1v1 ln
[P-v relation is not available].
٥٣
٥٤
v2
v1
Spring-controlled Process p=cv
1.1 Polytropic process Pvn = c
Work =
Work = area under process =
∫ pdv
=
p 2 v 2 − p1v1
1− n
P1 + P2
(v2-v1)
2
٥٥
٢٨
∫ pdv =
٥٦
Defining the region for u, h, or s
Isentropic Process
Real substance
Work =
∫ pdv =?? [p-v relation
Relation is not available].
٢٩
Ideal gas (cp = constant), Pvk=c
P v − P1 v 1
Work = ∫ pdv = 2 2
1− k
٥٧
Given T, h
From saturation tables, determine vf and vg at
Tsat=T. Then use the following table to determine
the region.
Condition
h <hf
h >hg
hf ≤h≤ hg
Region
Compressed liquid region
Superheated region
Saturation region
٥٨
Given P, h
Calculation of properties for given u, h, s
Given P, h
From saturation tables determine hf and hg at Psat
= P. Then use the following table to determine the
region.
Condition
h <hf
h >hg
hf ≤ h ≤ hg
Region
Compressed liquid region
Superheated region
Saturation region
٥٩
٣٠
Saturation region
Given T, h
Example.
Water T=200 oC, h=1500 kJ/kg. Calculate P
and x (if the state is in the saturation
region).
٦٠
Given P, h
Saturation region
Solution
From saturation table it can be found that hf < h < hg, thus
the state is in the saturation region. At T=100oC read P=
Psat, hf = 852.43 kJ/kg and hfg= 1940.75 kJ/kg. Thus,
P= Psat =1.5538 MPa
x=
h − hf
h fg
= 0.3337
٦١
٣١
Given P, h
Example.
R134a at P=120 kPa , h=200 kJ/kg. Calculate T
and x (if the state is in the saturation region).
Solution
From saturation table it can be found that hf < h <
hg, thus the state is in the saturation region. From
saturation table at P=120 kPa try to read T= Tsat, hf
and hfg
٦٢
Given P, h
Given P, h
In this case, because P=120 kPa is not given
in the table, interpolation has to be performed
between P=107.2 kPa and P=133.7 kPa to
create a row of data at P=120 kPa as shown
in the following table:
T
-25
-22.58
-20
٣٢
hf
167.38
170.45
173.74
hfg
215.57
214.0
212.34
Thus, from this table, T= Tsat = -22.58 oC and
x=
٦٣
P
107.2
120
133.7
h − hf
200 − 170.45
=
= 0.1381.
h fg
214.0
٦٤
Superheated region
Given T, h
h − h1
P − P1
=
.
h 2 − h1 P2 − P1
Given T, h
R134a at T=100 oC,
h= 467 kJ/kg
Thus
P1= 2000 kPa
h1= 471 kJ/kg
P?
h = 467 kJ/kg
P2= 2500 kPa
h2 = 463.28 kJ/kg
h − 471
P − 2000
=
463.28 − 471 2500 − 2000
Substituting for h=467 kJ/kg gives P=2259.1 kPa.
٦٥
٣٣
٦٦
P, h
Given
P, h
T − T1
h − h1
=
T2 − T1 h 2 − h1
Given P, h
Water P=1 MPa, h=3000 kJ/kg. Calculate T
T, oC
T1= 250
T?
T2= 300
Given
P = 1Mpa
h, kJ/kg
h1= 2942.59
h= 3000 kJ/kg
h2= 3051.15
thus
T − 250
h − 2942.59
=
300 − 250 3051.15 − 2942.59
Substituting for h=3000 kJ/kg gives T=276.4 oC
٦٧
٣٤
٦٨
Compressed liquid region
pressure is very low
Given P, T
Example
Compressed-liquid water at P=10 MPa,
T=100 oC. Calculate h
Solution
From the compressed-liquid table, get h=
426.48 kJ/kg.
٦٩
٣٥
Given P, T and the pressure is very low or
the compressed liquid tables are not
available
The following approximation can be done:
u= uf
h= hf + vf (P2-Psat at T)
٧٠
pressure is very low
Example
R 134a at P= 1MPa, T=10oC. Calculate
h.
Solution
h= hf + vf (P2-Psat at T), thus,
h= 213.58+ 0.000794(1000-415.8)=
214.04 kJ/kg.
٧١
٣٦