Tables of Thermodynamic Properties
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Tables of Thermodynamic Properties by Dr. M. A. Habib Professor, Mechanical Engineering Department, KFUPM. Procedure. The procedure of calculating the thermodynamic property is as follows: Define the state by two independent properties. Choose SI units tables.

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Procedure

Tables of Thermodynamic PropertiesbyDr. M. A. HabibProfessor,Mechanical Engineering Department, KFUPM


Procedure

Procedure

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.


Defining the state

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.


3 calculation of the thermodynamic properties 3 1 saturation region

3. Calculation of the thermodynamic properties3.1Saturation region

  • 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


3 1 saturation region

3.1 Saturation region


3 1 saturation region given p x

3.1 Saturation region_ Given P,x

  • 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 kParead T= 99.62 oC, vf =0.001043 m3/kgand and vfg= 1.67185 m3/kg . Then


3 1 saturation region given p x1

3.1 Saturation region_ Given P,x

  • v= vf + x (vfg)

  • = 1.01682 m3/kg


3 1 saturation region given t v

3.1 Saturation region_ Given T,v

  • Given T,vExample

  • 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


3 1 saturation region given t v1

3.1 Saturation region_ Given T,v

= 0.3947


3 1 saturation region given p v

3.1 Saturation region_ Given P,v

  • Given P,v

    Example.

  • Saturated water at P=100 kPa and v=0.1m3/kg. Determine T and x.

  • 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:


3 1 saturation region given p v1

3.1 Saturation region_ Given P,v


Given p t and given t v and t t c

Given P, T and Given T,v and T > Tc

  • 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


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, kPavf, m3/kgvfg, m3/kg

    100101.30.0010441.67185

    105120.80.0010471.41831


Interpolation1

Interpolation

  • 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:


Interpolation2

Interpolation

Thus, P102 =109.1 kPa


Superheated region

Superheated Region

Superheated region

  • Given P, T

    Example

  • Water vapor at

  • P=1 MPa,

  • T=300 oC.

  • Calculate v.


Superheated region1

Superheated Region

  • Solution

v=0.25794 m3/kg


Superheated region2

Superheated Region

  • GivenT, v

  • Example

  • Water vapor at T=300 oC, v=0.3 m3/kg. Determine P.

  • Solution


Superheated region3

Superheated Region

Interpolation gives:


Superheated region4

Superheated Region

  • GivenP, v

  • Example:

  • R134a vapor at P=0.8 MPa, v=0.035 m3/kg. Determine T.


Superheated region5

Superheated Region

  • Solution

Thus T= 98.55oC


Double interpolation

Double interpolation

  • 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.


Double interpolation1

Double interpolation


Procedure

  • Compressed-liquid region

  • GivenP,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


  • Procedure

    • 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.


  • Procedure

    • 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.


    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


    Compressed liquid table is not available1

    Compressed liquid table is not available

    • 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.


    Close to saturated vapor

    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:


    Procedure

    • Thus:

    Thus v= 0.247 m3/kg


    Defining the region

    Defining the region

    • Defining the region

    • 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.


    Procedure

    • The properties can then be determined according to the procedures given in Section 3 based on the region being compressed liquid, saturation or superheated.


    Given t x or p x

    Given(T, x) OR (P, x)

    • Given(T, x) OR (P, x)

  • The state has to be inside saturation region because x (having a value between 0 and1) indicates a mixture of liquid and vapor.


  • Given t v

    GivenT, v

    • GivenT, v

      From saturation tables, determine vf and vg at T. The region is determined based on the following conditional table:


    Given t v1

    Given T,v


    Given t v2

    Given T,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

    Given P, v

    Given P, v

    • From saturation tables determine vf and vg at Psat = P then:


    Given p v1

    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.


    Given p t

    Given P, T

    • Given 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:


  • Given p t1

    Given P, T

    Example:

    Water given at 100kPa and 40 oC


    Given p t2

    Given P, T

    • 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).


    Given p t3

    Given P, T


    Given p t4

    Given P, T

    • 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.


    Procedure

    • Property diagrams for case of P < Psat (The state is in the superheated region).


    Method 2 starting with temperature

    Method 2: Starting with temperature

    • b. Method 2: Starting with temperature:

    • From saturation tables determine Tsat at P, then, according to the following table, the region can be determined:


    Method 2 starting with temperature1

    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.


    Method 2 starting with temperature2

    Method 2: Starting with temperature

    • Property diagrams for case of T > Tsat (The state is in the superheated region).


    Method 2 starting with temperature3

    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.


    Method 2 starting with temperature4

    Method 2: Starting with temperature

    • Property diagrams for case of T < Tsat (The state is in the compressed-liquid region.


    Thermodynamic processes for a system

    Thermodynamic processes for a system

    • The following are examples of constant property processes of a system.

      • Constant-pressure process


    Constant volume process

    Constant-volume process

    • Work = = zero


    Process following the relation pv constant

    Process following the relation Pv = constant


    Isothermal process t constant

    Isothermal process (T=constant)


    Spring controlled process p cv

    Spring-controlled Process p=cv


    Isentropic process

    Isentropic Process


    Defining the region for u h or s

    Defining the region for u, h, or s

    • GivenT, h

  • From saturation tables, determine vf and vg at Tsat=T. Then use the following table to determine the region.


  • Given p h

    Given P, h

    • GivenP, h

  • From saturation tables determine hf and hg at Psat = P. Then use the following table to determine the region.


  • Calculation of properties for given u h s

    Calculation of properties for given u, h, s

    • 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).


  • Saturation region

    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

    = 0.3337


    Given p h1

    Given P, h

    • 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


    Given p h2

    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:


    Given p h3

    Given P, h

    Thus, from this table, T= Tsat = -22.58 oC and


    Superheated region6

    Superheated region

    • GivenT, h

    • R134a at T=100 oC, h= 467 kJ/kg


    Given t h

    Given T, h

    Substituting for h=467 kJ/kg gives P=2259.1 kPa.


    Given p h4

    GivenP, h

    • GivenP, h

    • Water P=1 MPa, h=3000 kJ/kg. Calculate T


    Given p h5

    GivenP, h

    Substituting for h=3000 kJ/kg gives T=276.4 oC


    Compressed liquid region

    Compressed liquid region

    • GivenP, 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.


    Pressure is very low

    pressure is very low

    • 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)


    Pressure is very low1

    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.


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