chapter 2 water steam

CHAPTER 2PROPERTY OF PURE SUBSTANCE(SIFAT-SIFAT BAHAN TULEN)

OBJECTIVES

General Objective

To define the properties of wet steam, dry saturated steam and superheated steam using information from the steam tables.

wet steam

(wap basah)

dry saturated steam(Wap tepu

kering)

superheated steam

(Wap panas lampau)

SPECIFIC OBJECTIVES

At the end of the CHAPTER 2, you will be

able to:

define the word phase and distinguish the solid, liquid and steam phases

understand and use the fact that the vaporization process is carried out at constant pressure

define and explain the following terms: saturation temperature, saturated liquid, wet steam, saturated steam, dry saturated steam, , dryness fraction and superheated steam

determine the properties of steam using the P-v diagram

understand and use the nomenclature as in the Steam Tables

apply single and double interpolation using the steam tables

locate the correct steam tables for interpolation, including interpolation between saturation tables and superheated tables where necessary

INTRODUCTION1. In thermodynamic systems, the working fluid can be in

the liquid, steam or gaseous phase.2. A substance that has a fixed composition throughout is

called a pure substance.3. Pure chemicals (H2O, N2, O2, Ar, Ne, Xe) are always

pure substances.4. A phase of substance can be defined as that part of a

pure substance that consists of a single, homogenous aggregate (totally) of matter.

5. Example: The three common phases for H2O that are usually used are solid, liquid and steam.

6. Studying phases or phase changes in thermodynamics: It is the concerned with the molecular structure and behavior of the different phases.

Water Steam is efficient and economic to generate.

Water is plentiful (banyak terdapat di mana-mana) and inexpensive. It is non-hazardous to health and environmentally sound. In its gaseous form, it is a safe and efficient energy carrier. Steam can hold five or six times as much potential energy as an equivalent mass of water.

When water is heated in a boiler, it begins to absorb energy. Depending on the pressure in the boiler, the water will evaporate at a certain temperature to form steam. The steam contains a large quantity of stored energy which will eventually be transferred to the process or the space to be heated.

It can be generated at high pressures to give high steam temperatures. The higher the pressure, the higher the temperature. More heat energy is contained within high temperature steam so its potential to do work is greater

Because of the direct relationship between the pressure and temperature of saturated steam, the amount of energy input to the process is easy to control, simply by controlling the saturated steam pressure. Modern steam controls are designed to respond very rapidly to process changes.

CHP-combined heat and power

THE MOLECULAR PHENOMENA INVOLVED IN EACH PHASE

Molecular bonds are strongest in solids and weakest in steams. One reason is that molecules in solids are closely packed together, whereas in steams they are separated by great distances.

The three phases of pure substances

Solid phase Liquid Phase

Steam Phase

SOLID PHASE

1. closely bound, therefore relatively dense.2. Arranged in a rigid three-dimensional pattern so

that they do not easily deform. An example of a pure solid state is ice.

Molecules arrangement in solid

In the solid phase, the molecules are:

LIQUID PHASE

1. closely bound, therefore also relatively dense and unable to expand to fill a space; but….

2. They are no longer rigidly structured so much so that they are free to move within a fixed volume. An example is a pure liquid state.

Molecules arrangement in liquid

In the liquid phase, the molecules are;

STEAM PHASE

In the steam phase, the molecules; 1. virtually do not attract each other. The

distance between the molecules are not as close as those in the solid and liquid phases;

2. are not arranged in a fixed pattern. There is neither a fixed volume nor a fixed shape for steam.

Molecules arrangement in steam

PHASE-CHANGE PROCESS

The distinction between steam and liquid.

1. Even though both will take up the shape of their containers, but…

2. Liquid will present a free surface if it does not completely fill its container.

3. Steam on the other hand will always fill its container.

With these information, let us consider the following system, let continue……

STATE 11. A container is filled with water, and a moveable, frictionless piston is

placed on the container. 2. As heat is added to the system, the temperature of the system will

increase. Note that the pressure on the system is being kept constant by the weight of the piston.

3. The continued addition of heat will cause the temperature of the system to increase until the pressure of the steam generated exactly balances the pressure of the atmosphere plus the pressure due to the weight of the piston.

4. At this point, the steam and liquid are said to be saturated.Moveable and frictionless piston

Liquid(Water)

PWeight

Isometric view of Cylinder and frictionless Piston assembly:

State 1

Experiment Background (conditions).

1. A mass of water at 1 atm pressure and room temperature.

2. At this temperature and pressure we may measure the specific volume (1/ = 1/1000 kg/m3).

3. Note that: v = 1/ = 1/1000 kg/m3= 1 liter/kg

4. We plot the above condition as a state of the point 1 on the diagram.

STATE 21. As more heat is added, the liquid that was at saturation

will start to vaporize State 2.2. The two-phase mixture of steam and liquid at State 2 has

only one degree of freedom, and as long as liquid is present, vaporization will continue at constant temperature.

3. As long as liquid is present, the mixture is said to be wet steam, and both the liquid and steam are saturated.

STATE 31. After all the liquid is vaporized, only steam is

present at State 3, and the further addition of heat will cause the temperature of steam to increase at constant system pressure.

STATE 41. The further addition of heat will cause the

temperature of steam to increase at constant system pressure. This state is called the superheated state, and the steam is said to be superheated steam as shown in State 4.

SUMMARY

W

W

W

W

Liquid Steam Superheated

Steam

STATE 1 STATE 2 STATE 3 STATE 4

PHASE= Intermolecular changeSTATE= Change when the system undergo process( heating, expanding, compression process etc)

Heating

Process

Heating

Process

Heating

Process

Phase Change

Phase Change

Saturated and Superheated Steam

Consider the following diagram in which the specific volume of H2O is presented as a function of temperature and pressure

20

100

300

1

2 3

4

T, oC

v, m3/kg

Compressed liquid

Saturated mixture

Superheated

steam

T-v diagram for the heating process of water at constant pressure

COMPRESSED OR SUB-COOLED LIQUID

20

100

300

1

2 3

4

T, oC

v, m3/kg

Compressed liquid

Saturated mixture

Superheated

steam

T-v diagram for the heating process of water at constant pressure

1. If we proceed to heat the water, the temperature will rise.In addition, water expands slightly as it is heated which makes the specific volume increase slightly.

2. We may plot the locus of such points along the line from State 1 to State 2. We speak of liquid in one of these conditions as being compressed or sub-cooled liquid.

3. State 2 is selected to correspond to the boiling point (100o C). We speak of State 2 as being the saturated liquid state, which means that all of the water is in still liquid form, but ready to boil.

4. As we continue to heat past the boiling point 2, a fundamental change occurs in the process. The temperature of the water no longer continues to rise. Instead, as we continue to add energy, liquid progressively changes to steam phase at a constant temperature but with an increasing specific volume. In this part of the process, we speak of the water as being a saturated mixture (liquid + steam). This is also known as the quality region.

5. 6. At State 3, all liquid will have been vaporized. This

is the saturated steam state.

7. As we continue to heat the steam beyond State 3, the temperature of the steam again rises as we add energy. States to the right of State 3 are said to be superheated steam.

In this experiment, we start with a mass of water at 1 atm pressure and room temperature. At this temperature and pressure we may measure the specific volume (1/ρ = 1/1000 kg/m3 =(0.001) . 1/(kg /m3) = v = 0.001 m3/ kg. We plot this state at point 1 on the diagram.

THE SAME EXPERIMENT CAN BE CONDUCTED AT SEVERAL DIFFERENT PRESSURES

T-v diagram of constant pressure phase change processes of a pure substance at various pressures for water

P = 1.01325 bar

P = 5 bar

P = 10 bar

P = 80 bar

P = 150 bar

P = 221.2 bar

Critical point

374.15

T, oC

v, m3/kg

Saturated

liquid

Saturated

steam

0.00317

CONNECTION THE LOCUS OF POINTS CORRESPONDING TO THE SATURATION CONDITION

T

v

Critical point

Saturated liquid line

Dry saturated steam line

P2 = const.

P1 = const. COMPRESS

LIQUID

REGION

WET STEAM

REGION

SUPERHEATED

STEAM

REGION

P2 > P1

T-v diagram of a pure substance

P-V DIAGRAM

P-v diagram of a pure substance

P

v

Critical point



T2 = const.

T1 = const.

COMPRESS

LIQUID

REGIONWET STEAM

REGION

SUPERHEATED

STEAM

REGION

T2 > T1

ACTIVITY 2A

Statement Phase

The molecules are closely bound, they are also relatively dense and unable to expand to fill a space. However they are no longer rigidly structured so that they are free to move within a fixed volume.

i._____________

The molecules are closely bound, they are relatively dense and arranged in a rigid three-dimensional patterns so that they do not easily deform.

ii.____________

The molecules virtually do not attract each other. The distance between the molecules are not as close as those in the solid and liquid phases. They are not arranged in a fixed pattern. There is neither a fixed volume nor a fixed shape for steam.

iii.____________

2.1 Each line in the table below gives information about phases of pure substances. Fill in the phase column in the table with the correct answer.

P

v

( vi )

( ii )

( iv )

T2 = const.

T1 = const.

( i )

( iii)

( v )

T2 > T1

2.2 Write the suitable names of the phases for the H2O in the P-v diagram below.

ACTIVITY 2A

Statement Phase

The molecules are closely bound, they are also relatively dense and unable to expand to fill a space. However they are no longer rigidly structured so that they are free to move within a fixed volume.

i.Liquid Phase

The molecules are closely bound, they are relatively dense and arranged in a rigid three-dimensional patterns so that they do not easily deform.

ii. Solid Phase

The molecules virtually do not attract each other. The distance between the molecules are not as close as those in the solid and liquid phases. They are not arranged in a fixed pattern. There is neither a fixed volume nor a fixed shape for steam.

iii. Steam Phase

2.1 Each line in the table below gives information about phases of pure substances. Fill in the phase column in the table with the correct answer.

P

v

( vi )

( ii )

( iv )

T2 = const.

T1 = const.

( i )

( iii)

( v )

T2 > T1

2.2 Write the suitable names of the phases for the H2O in the P-v diagram below.

2A.2 i.Compress liquid regionii.Saturated liquid lineiii.Wet steam regioniv.Dry saturated steam linev.Superheated steam regionvi.Critical point

PROPERTIES OF A WET MIXTURE

1. Between the saturated liquid and the saturated steam, there exist a mixture of steam plus liquid (wet steam region).

2. In liquid-steam mixture, the term describing the relative quantities of liquid and steam is dryness fraction (symbol x).

3. Thus, in 1 kg of wet mixture, there must be x kg of saturated steam plus (1 – x) kg of saturated liquid.

(1 - x ) kg of liquid

x kg of steamtotal mass = 1 kg

Liquid-steam mixtureP-v diagram

DRYNESS FRACTION

Dryness fraction= X= Quality= Pecahan kekeringan

mass totalsteam saturateddry of massfraction drynessX

total

steam

mmx

where mtotal = mliquid + msteam

mass totalsteam saturateddry of massfraction drynessX

total

steam

mmx

EXAMPLE Mass of 5kg of water in sealed vessel

was heated until 4.5 kg of steam presented. What is the quality value?

HOW TO FIND THE X VALUE?????

THE X VALUE

FOR A WET STEAM, THE TOTAL VOLUME OF THE MIXTURE IS GIVEN BY THE VOLUME OF LIQUID PRESENT PLUS THE VOLUME OF DRY STEAM PRESENT.

Specific volume (v)

steam wet of mass totalsteamdry of volumeliquid a of volume

v

Now for 1 kg of wet steam, there are (1 – x) kg of liquid and x kg of dry steam, where x is the dryness fraction as defined earlier, hence

v = vf(1 – x) + vgx

v = xvg

The volume of the liquid is usually negligibly small as compared to the volume of dry saturated steam

Where,vf = specific volume of saturated liquid (m3/kg)vg = specific volume of saturated steam (m3/kg)x = dryness fraction

1

steam wet of mass totalsteamdry of volumeliquid a of volume

v

EXAMPLE 1: Steam at 10 bar with dryness fraction of 0.9.Determine the specific volume.

EXAMPLE 2: Steam at 10 bar has a specific volume of 0.18 m3/kg. What is the quality of steam in this condition.

EXAMPLE 3:

A wet steam at 10 bar has a specific volume of 0.11664 m3/kg and the quality of 0.6 contained in the rigid vessel. What is the pressure of the vessel ?

Specific enthalpy(H per unit mass) h = u + Pv(kJ/kg)

Specific Enthalpy (h) Enthalpy (h) is the combination of properties H=U + PV (kJ)

The enthalpy of wet steam is given by the sum of the enthalpy of the liquid plus the enthalpy of the dry steam,

i.e. h = hf(1 – x) + xhg

h = hf + x(hg – hf ) h = hf + xhfg

Where,hf = specific enthalpy of saturated liquid (kJ/kg)hg = specific enthalpy of saturated steam (kJ/kg)hfg = difference between hg and hf (that is, hfg = hg - hf )2

SPECIFIC INTERNAL ENERGY (U)

Similarly, the specific internal energy of a wet steam is given by the internal energy of the liquid plus the internal energy of the dry steam,

i.e. u = uf(1 – x) + xug

u = uf + x(ug – uf ) Note : ufg is not tabulated

Where,uf = specific enthalpy of saturated liquid (kJ/kg)ug = specific enthalpy of saturated steam (kJ/kg)ug – uf = difference between ug and uf

3

SPECIFIC ENTROPY (S)Entropy is a property associated with the Second Law of Thermodynamics

The entropy of wet steam is given by the sum of the entropy of the liquid plus the entropy of the dry steami.e. s = sf(1 – x) + xsg

s = sf + x(sg – sf ) s = sf + xsfg

Where,sf = specific enthalpy of saturated liquid (kJ/kg K)sg = specific enthalpy of saturated steam (kJ/kg K)sfg = difference between sg and sf (that is, sfg = sg - sf )

4

REMEMBER!These equations are used very often

and are, therefore, important to remember!v = xvg

h = hf + xhfg

u = uf + x(ug – uf )

s = sf + xsfg

1

2

3

4

THE USE OF STEAM TABLES

Symbols Units Description

p bar Absolute pressure of the fluid

tsoC Saturation temperature corresponding to the

pressure p barvf m3/kg Specific volume of saturated liquid

vg m3/kg Specific volume of saturated steam

uf kJ/kg Specific internal energy of saturated liquid

ug kJ/kg Specific internal energy of saturated steam

hf kJ/kg Specific enthalpy of saturated liquid

hg kJ/kg Specific enthalpy of saturated steam

hfg kJ/kg Change of specific enthalpy during evaporation

sf kJ/kg K Specific entropy of saturated liquid

sg kJ/kg K Specific entropy of saturated steam

sfg kJ/kg K Change of specific entropy during evaporation

The property of steam tables

Below is a list of the properties normally tabulated, with the symbols used being those recommended by British Standard Specifications.

These steam tables are divided into two types:

Type 1: Saturated Water and Steam (Page 2 to 5 of steam tables)

Type 2: Superheated Steam (Page 6 to 8 of steam tables)

P

v

Critical point



T2 = const.

T1 = const.

COMPRESS

LIQUID

REGIONWET STEAM

REGION

SUPERHEATED

STEAM

REGION

T2 > T1

1. SATURATED WATER AND STEAM TABLES

The table of the saturation condition is divided into two parts.

Part 1Part 1 (Page 2) refers to the values of temperature from 0.01oC to 100oC, followed by values that are suitable for the temperatures stated in the table. The following partial view of the table is an example showing an extract from the temperature of 10oC.

t ps vg hf hfg hg sf sfg sg

0C bar m3/kg kJ/kg kJ/kg K

10 0.01227 106.4 42.0 2477.2 2519.2 0.151 8.749 8.900

Part 2 Part 2 (Page 3 to 5 of steam tables) is values of pressure from 0.006112 bar to 221.2 bar followed by values that are suitable for the pressures stated in the table. The following partial view of the table showing an extract from the pressure of 1.0 bar.

p ts vg uf ug hf hfg hg sf sfg sg

bar oC m3/kg kJ/kg kJ/kg kJ/kg K

1.0 99.6 1.694 417 2506 417 2258 2675 1.303 6.056 7.359

Page 2 to page 5

EXAMPLE 8.1Complete the following table for Saturated Water and Steam:

t Ps vg hf hfg hg sf sfg sg

oC bar m3/kg kJ/kg kJ/kg K

1 192.6

0.02337 8.666

100 1.01325

EXAMPLE 8.1Complete the following table for Saturated Water and Steam:

t Ps vg hf hfg hg sf sfg sg

oC bar m3/kg kJ/kg kJ/kg K

1 0.006566 192.6 4.2 2498.3 2502.5 0.015 9.113 9.128

20 0.02337 57.84 83.9 2453.7 2537.6 0.296 8.370 8.666

100 1.01325 1.673 419.1 2256.7 2675.8 1.307 6.048 7.355

EXAMPLE 8.2



0.045 31.0 2558

10 0.1944

311.0 5.615

Complete the missing properties in the following table for Saturated Water and Steam:

EXAMPLE 8.2



0.045 31.0 2558

10 0.1944

311.0 5.615


SOLUTION: From page 3 to page 5 of the steam tables, we can directly read:

EXAMPLE 8.2



0.045 31.0 31.14 130 2418 130 2428 2558 0.451 7.980 8.431

10 179.9 0.1944 762 2584 763 2015 2778 2.138 4.448 6.586

100 311.0 0.01802 1393 2545 1408 1317 2725 3.360 2.255 5.615


SOLUTION: From page 3 to page 5 of the steam tables, we can directly read:

EXAMPLE 8.3For a steam at 20 bar with a dryness fraction of 0.9, calculate the a) specific volumeb) specific enthalpyc) specific internal energy

EXAMPLE 8.3For a steam at 20 bar with a dryness fraction of 0.9, calculate the a) specific volume—(v)b) specific enthalpy— (h)c) specific internal energy—(u)


20 212.4 0.09957 907 2600 909 1890 2799 2.447 3.893 6.340

Solution to Example 8.3

An extract from the steam tables

Page 4



20 212.4 0.09957 907 2600 909 1890 2799 2.447 3.893 6.340



a) Specific volume (v),

v = xvg = 0.9(0.09957) = 0.0896 m3/kg

Page 4



20 212.4 0.09957 907 2600 909 1890 2799 2.447 3.893 6.340



b) Specific enthalpy (h),

h = hf + xhfg

= 909 + 0.9(1890) = 2610 kJ/kg Page

4



20 212.4 0.09957 907 2600 909 1890 2799 2.447 3.893 6.340



c) Specific internal energy (u), u = uf + x( ug -uf ) = 907 + 0.9(2600 - 907) = 2430.7 kJ/kg

Page 4


P

bar

v m3/kg

ts = 212.4 oC

vg

ug

hg

sg

20

uf

hf

sf

As the answer can be illustrated in the P-v diagram as follow:

The properties which is stated in the table The properties which

is stated in the table


P

bar

v m3/kg

ts = 212.4 oC

v

u

h

s

vg

ug

hg

sg

x = 0.9

20

uf

hf

sf

As the answer can be illustrated in the P-v diagram as follow:

The properties which is stated in the table The properties which

is stated in the table

The properties which is required by the question

EXAMPLE 8.4

Find the dryness fraction, specific volume and specific enthalpy of steam at 8 bar and specific internal energy 2450 kJ/kg.

EXAMPLE 8.4



8 170.4 0.2403 720 2577 721 2048 2769 2.046 4.617 6.663

Solution to Example 8.4An extract from the steam tables,

At 8 bar, ug = 2577 kJ/kg, since the actual specific internal energy is given as 2450 kJ/kg, the steam must be in the wet steam state ( u < ug).

Page 4

EXAMPLE 8.4



8 170.4 0.2403 720 2577 721 2048 2769 2.046 4.617 6.663



Page 4

From equation: u = uf + x(ug -uf) 2450 = 720 + x(2577 - 720)

x = 0.932

EXAMPLE 8.4



8 170.4 0.2403 720 2577 721 2048 2769 2.046 4.617 6.663



Page 4


x = 0.932

From equation v = xvg

= 0.932 (0.2403) = 0.2240 m3/kg

EXAMPLE 8.4



8 170.4 0.2403 720 2577 721 2048 2769 2.046 4.617 6.663



Page 4


x = 0.932


= 0.932 (0.2403) = 0.2240 m3/kg

From equation: h = hf + xhfg

= 721 + 0.932 (2048) = 2629.7 kJ/kg

EXAMPLE 8.4


x = 0.932


= 0.932 (0.2403) = 0.2240 m3/kg

From equation: h = hf + xhfg

= 721 + 0.932 (2048) = 2629.7 kJ/kg

P

bar

v m3/kg

ts = 170.4 oC

v

h

vg

x = 0.932

8

Extra Example

Q: 1-3 only

PROPERTY OF PURE SUBSTANCEFill in the blank with the appropriates values

ACTIVITY 2B

2B.1 The internal energy of wet steam is 2000 kJ/kg. If the pressure is 42 bar, what is the value of dryness fraction? [x = 0.6]

2B.2 Determine the specific volume, specific enthalpy and

specific internal energy of wet steam at 32 bar if the dryness fraction is 0.92. [v= 0.05746 m3/kg,h=2661kJ/kg, u= 2476 kJ/kg]

2B.5 Find the dryness fraction, specific volume and

specific internal energy of steam at 105 bar and specific enthalpy 2100 kJ/kg.[x = 0.52, v= 0.00882 m3/kg, u= 1998 kJ/kg]

SUPERHEATED STEAM TABLES It is a second part of the table. The values of the specific properties of a superheated steam are

normally listed in separate tables for the selected values of pressure and temperature.

A steam is called superheated when its temperature is greater than the saturation temperature corresponding to the pressure.

When the pressure and temperature are given for the superheated steam then the state is defined and all the other properties can be found.

For example, steam at 10 bar and 200 oC is superheated since the saturation temperature at 10 bar is 179.9 oC .

The equation: Degree of superheat = tsuperheat – tsaturation

The steam at this state has a degree of superheat of 200 oC – 179.9 oC = 20.1 oC .

SUPERHEATED STEAM TABLES The tables of properties of superheated steam

range in pressure from 0.006112 bar to the critical pressure of 221.2 bar.

At each pressure, there is a range of temperature up to high degrees of superheat, and the values of specific volume, internal energy, enthalpy and entropy are tabulated.

For the pressure above 70 bar, the specific internal energy is not tabulated. The specific internal energy is calculated using the equation:Specific internal energy, u = h – pv

SUPERHEATED STEAM TABLESFor example, from the superheated table at 10 bar and 200 oC, the specific volume is 0.2061 m3/kg and the specific enthalpy is 2829 kJ/kg.

A specimen row of values is shown in Table

EXAMPLE 8.5Complete the missing properties in the following table for Superheated Steam:

p(ts)

t 300 350 400 450

40(250.3)

vg 0.0498 v 0.0800

ug 2602 u 2921

hg 2801 h 3094

sg 6.070 s 6.364

EXAMPLE 8.5Complete the missing properties in the following table for Superheated Steam:

p(ts)

t 300 350 400 450

40(250.3)

vg 0.0498 v 0.0588 0.0664 0.0733 0.0800

ug 2602 u 2728 2828 2921 3010

hg 2801 h 2963 3094 3214 3330

sg 6.070 s 6.364 6.584 6.769 6.935

Solution to Example 8.5From page 7 of the steam tables, we can directly read

EXAMPLE 8.6Steam at 100 bar has a specific volume of 0.02812 m3/kg. Find the temperature, degree of superheat, specific enthalpy and specific internal energy.

EXAMPLE 8.6Steam at 100 bar has a specific volume of 0.02812 m3/kg. Find the temperature, degree of superheat, specific enthalpy and specific internal energy. Solution to Example 8.6First, it is necessary to decide whether the steam is wet, dry saturated or superheated.

EXAMPLE 8.6Steam at 100 bar has a specific volume of 0.02812 m3/kg. Find the temperature, degree of superheat, specific enthalpy and specific internal energy. Solution to Example 8.6First, it is necessary to decide whether the steam is wet, dry saturated or superheated.At 100 bar, vg = 0.01802 m3/kg. This is less than the actual specific volume of 0.02812 m3/kg. Hence, the steam is superheated. The state of the steam is at point A in the diagram below.

P

bar

v m3/kg

ts = 311.0 oC

100 425

oC

vg= 0.01802 v = 0.02812

A


P

bar

v m3/kg

ts = 311.0 oC

100 425

oC

vg= 0.01802 v = 0.02812

A

An extract from the superheated table, page 8.

p(ts)

t 425

100(311.0)

vg 0.01802 v x 10-2 2.812

hg 2725 h 3172

sg 5.615 s 6.321


P

bar

v m3/kg

ts = 311.0 oC

100 425

oC

vg= 0.01802 v = 0.02812

A


p(ts)

t 425

100(311.0)

vg 0.01802 v x 10-2 2.812

hg 2725 h 3172

sg 5.615 s 6.321

From the superheated table at 100 bar, the specific volume is 0.02812 m3/kg at a temperature of 425 oC. Hence, this is the isothermal line, which passes through point A as shown in the P-v diagram left.


P

bar

v m3/kg

ts = 311.0 oC

100 425

oC

vg= 0.01802 v = 0.02812

A


p(ts)

t 425

100(311.0)

vg 0.01802 v x 10-2 2.812

hg 2725 h 3172

sg 5.615 s 6.321


Degree of superheat = 425 oC – 311 oC= 114 oC


P

bar

v m3/kg

ts = 311.0 oC

100 425

oC

vg= 0.01802 v = 0.02812

A


p(ts)

t 425

100(311.0)

vg 0.01802 v x 10-2 2.812

hg 2725 h 3172

sg 5.615 s 6.321



So, at 100 bar and 425 oC, we havev = 2.812 x 10-2 m3/kgh = 3172 kJ/kg


P

bar

v m3/kg

ts = 311.0 oC

100 425

oC

vg= 0.01802 v = 0.02812

A


p(ts)

t 425

100(311.0)

vg 0.01802 v x 10-2 2.812

hg 2725 h 3172

sg 5.615 s 6.321




From equation u = h – Pv = PL EASE CHECK FOR THE UNIT

CONVERSION

= ??????????


P

bar

v m3/kg

ts = 311.0 oC

100 425

oC

vg= 0.01802 v = 0.02812

A


p(ts)

t 425

100(311.0)

vg 0.01802 v x 10-2 2.812

hg 2725 h 3172

sg 5.615 s 6.321





CONVERSION

= The Final Unit kJ/kg


P

bar

v m3/kg

ts = 311.0 oC

100 425

oC

vg= 0.01802 v = 0.02812

A


p(ts)

t 425

100(311.0)

vg 0.01802 v x 10-2 2.812

hg 2725 h 3172

sg 5.615 s 6.321





CONVERSION

= 2890.8 kJ/kg


P

bar

v m3/kg

ts = 311.0 oC

100 425

oC

vg= 0.01802 v = 0.02812

A


p(ts)

t 425

100(311.0)

vg 0.01802 v x 10-2 2.812

hg 2725 h 3172

sg 5.615 s 6.321




From equation u = h – Pv = 3172 kJ/kg – (100 x 102 kN/m2)(2.812 x

10-2 m3/kg) = 2890.8 kJ/kg

INTERPOLATIONFor properties which are not tabulated exactly in the tables, it is necessary to interpolate between the values tabulated as shown in Fig. below.

In this process it is customary to use a straight line that passes through two adjacent table points, denoted by and . If we use the straight line then it is called “interpolation”.

f(x)

x

Interpolation

Interpolation

The values in the tables are given in regular increments of temperature and pressure.

Often we wish to know the value of thermodynamic properties at intermediate values. It is common to use linear interpolation as shown in Fig below

y

x

y2

y

y1

x1 x x2

(x2 , y2)

(x , y)

(x1 , y1)

From the Fig. above, the value of x can be determined by:

1

12

121

12

12

1

1

xyy

xxyyx

yyxx

yyxx

There are two methods of interpolation:•single interpolation•double interpolation

Given data/value from the Question

The interpolate

d data/answ

er

1

12

121

12

12

1

1

xyy

xxyyx

yyxx

yyxx

SINGLE INTERPOLATION

Determine the saturation temperature at 77 bar.

Examples 8.7 .



Solution to Example 8.7The values of saturation temperature at a pressure of 77 bars are not tabulated in the Steam Tables. So, we need to interpolate between the two nearest values that are tabulated in the Steam Tables. ( Page 5)

Examples 8.7 .



Solution to Example 8.7The values of saturation temperature at a pressure of 77 bars are not tabulated in the Steam Tables. So, we need to interpolate between the two nearest values that are tabulated in the Steam Tables.

Examples 8.7 .

P

ts

80

77

75

290.5 ts 295

Please refer to table

Page 5 for the stated values for each axis




Examples 8.7 .

P

ts

80

77

75

290.5 ts 295

75805.290295

75775.290

st7580

5.2902957577

5.290

st




Examples 8.7 .

P

ts

80

77

75

290.5

ts 295

75805.290295

75775.290

st

55.290295

25.290

st

5.2905

5.42st

ts = 292.3 oC

75805.290295

75775.290

st

55.290295

25.290

st

5.2905

5.42st

Example 8.8

P

hg

105

103

100

2725 hg 2715

Determine the specific enthalpy of dry saturated steam at 103 bar.

Solution to Example 8.8Please refer to table


Example 8.8

hg

2725103 100

2715 2725105 100P

hg

105

103

100

2725 hg 2715



Often we wish to know the value of thermodynamic properties at intermediate values. Hence, It is common to use linear interpolation either in saturation or superheated phase for their intermediate properties values .



hg

2725103 100

2715 2725105 100

Example 8.8

hg

2725103 100

2715 2725105 100

hg

3 105

2725

2719gh

P

hg

105

103

100

2725 hg 2715

kJ/kg



Often we wish to know the value of thermodynamic properties at intermediate values. Hence, It is common to use linear interpolation either in saturation or superheated phase for their intermediate properties values .



hg

2725103 100

2715 2725105 100

hg

3 105

2725

2719gh

EXAMPLE 8.9Determine the specific volume of steam at 8 bar and 220oC. Solution to Example 8.9From the Steam Tables at 8 bar, the saturated temperature (ts) is 170.4 oC.The steam is at superheated condition as the temperature of the steam is 220oC > ts.


An extract from the Steam Tables,(Page 7)

p / (bar) (ts / oC)

t 200 220 250 (oC)

8 (170.4)

v 0.2610 v 0.2933



p / (bar) (ts / oC)

t 200 220 250 (oC)

8 (170.4)

v 0.2610 v 0.2933

P

v

250

220

200

0.2610 v 0.2933



p / (bar) (ts / oC)

t 200 220 250 (oC)

8 (170.4)

v 0.2610 v 0.2933

P

v

250

220

200

0.2610 v 0.2933

v

0 2610220 200

0 2933 0 2610250 200

. . .

v 0 27392. m3/kg

v

0 2610220 200

0 2933 0 2610250 200

. . .

v 0 27392.

DOUBLE INTERPOLATION

1. It is usually used in the Superheated Steam Table.

2. Double interpolation must be used when two of the properties (eg. temperature and pressure) are not tabulated in the Steam Tables.

3. Vertical or Horizontal Method

EXAMPLE 8.10Determine the specific enthalpy of superheated steam at 25 bar and 320oC

EXAMPLE 8.10Determine the specific enthalpy of superheated steam at 25 bar and 320oC1. To find the enthalpy of superheated steam at 25 bar and

320oC, an interpolation between 20 bar and 30 bar is necessary.

2. An interpolation between 300oC and 350oC is also necessary.




3. An extract from the Superheated Steam Tables (Page 7).





4. Construct a table as shown below. Arrange unknown Vertically

t(oC)p(bar)

300 320 350

20 3025 h1 3138 25

h

30 2995 h2 3117

The 3 unknown values is arranged in vertical





4. Construct a table as shown below.

5. Firstly, find the specific enthalpy (h1) at 20 bar and 320 oC;

t(oC)p(bar)

300 320 350

20 3025 h1 3138 25

h

30 2995 h2 3117

Interpolate within the Values from this virtual box of table







At 20 bar,

t(oC)p(bar)

300 320 350

20 3025 h1 3138 25

h

30 2995 h2 3117

T

h

350

320

300

3025 h1 3138

30035030253138

30032030251

h

2.30701 h kJ/kg

Interpolate within the Values from this virtual box , for T

and h values

The interpolated value will be an appropriate value of h1

in the constructed table.

30035030253138

30032030251

h

2.30701 h







At 20 bar,

t(oC)p(bar)

300 320 350

20 3025 h1 3138 25

h

30 2995 h2 3117

T

h

350

320

300

3025 h1 3138

30035030253138

30032030251

h

2.30701 h kJ/kg

6. Secondly, find the specific enthalpy (h2) at 30 bar and 320 oC; as the value taken from the highlighted blue arrow

At 30 bar,

T

h

350

320

300

2995 h2 3117

30035029953117

30032029952

h

8.30432 h kJ/kg

30035030253138

30032030251

h

2.30701 h

30035029953117

30032029952

h

8.30432 h







At 20 bar,

t(oC)p(bar)

300 320 350

20 3025 h1 3138 25

h

30 2995 h2 3117

T

h

350

320

300

3025 h1 3138

30035030253138

30032030251

h

2.30701 h kJ/kg

6. Secondly, find the specific enthalpy (h2) at 30 bar and 320 oC;

At 30 bar,

T

h

350

320

300

2995 h2 3117

30035029953117

30032029952

h

8.30432 h kJ/kg

7. Now interpolate between h1 at 20 bar, 320oC, and h2 at 30 bar, 320oC in order to find h at 25 bar and 320oC. As

the values highlighted by blue arrow

P

h

30

25

20

h1h h2

At 320oC,

20302025121

hhhh

h

3070 225 20

30438 3070 230 20

. . .

kJ/kg.h 3057

30035030253138

30032030251

h

2.30701 h

30035029953117

30032029952

h

8.30432 h

20302025121

hhhh

h

3070 225 20

30438 3070 230 20

. . .

h 3057

EXAMPLE 8.110.9 m3 of dry saturated steam at 225 kN/m2 is contained in a rigid cylinder. If it is cooled at constant volume process until the pressure drops to180 kN/m2, determine the following:a) mass of steam in the cylinderb) dryness fraction at the final stateSketch the process in the form of a P-v diagram.

THANK YOU

That is all for today

chapter 2 water steam

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