jj207 thermodynamics i chapter 2

JJ 207 THERMODYNAMICS I

1

Chapter 2

INTRODUCTION

In thermodynamic systems, the working fluid can be in the liquid, steam or gaseous

phase. In this unit, the properties of liquid and steam are investigated in some details

as the state of a system can be described in terms of its properties. A substance that

has a fixed composition throughout is called a pure substance. Pure chemicals (H2O,

N2, O2, Ar, Ne, Xe) are always pure substances. We all know from experience that

substances exist in different phases. A phase of substance can be defined as that part

of a pure substance that consists of a single, homogenous aggregate of matter. The

three common phases for H2O that are usually used are solid, liquid and steam.

When studying phases or phase changes in thermodynamics, one does not need to be

concerned with the molecular structure and behavior of the different phases.

However, it is very helpful to have some understanding of 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.

PHASES

The three phases of pure substances are: -

Solid Phase

In the solid phase, the molecules are;

(a) Closely bound, therefore relatively dense; and

(b) Arranged in a rigid three-dimensional pattern so that they do not easily

deform. An example of a pure solid state is ice.

Liquid Phase

In the liquid phase, the molecules are;

(a) Closely bound, therefore also relatively dense and unable to expand to fill a

space; but

(b) 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.

Steam Phase

In the steam phase, the molecules;

(a) Virtually do not attract each other. The distance between the molecules are

not as close as those in the solid and liquid phases;


2

(b) Are not arranged in a fixed pattern. There is neither a fixed volume nor a

fixed shape for steam.

The three phases described above are illustrated in Figure below. The following are

discovered:

(a) The positions of the molecules are relatively fixed in a solid phase;

(b) Chunks of molecules float about each other in the liquid phase; and

(c) The molecules move about at random in the steam phase.

The arrangement of atoms in different phases

PHASE-CHANGE PROCESS

The distinction between steam and liquid is usually made (in an elementary manner)

by stating that both will take up the shape of their containers. However liquid will

present a free surface if it does not completely fill its container. Steam on the other

hand will always fill its container.

A container is filled with water, and a moveable, frictionless piston is placed on the

container at State 1, as shown in Figure below. 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. 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.

At this point, the steam and liquid are said to be saturated. As more heat is added, the

liquid that was at saturation will start to vaporize until State 2. The two-phase

(a) (b) (c)

W W

W

W

Liqui

d

Steam Superheated

Steam

STATE 1 STATE 2 STATE 3 STATE 4

Heating water

and steam at

constant pressure


3

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. As long as liquid

is present, the mixture is said to be wet steam, and both the liquid and steam are

saturated. 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. This state is called the superheated state, and the steam is said to be

superheated steam as shown in State 4.

Saturated and Superheated Steam

While tables provide a convenient way of presenting precise numerical presentations

of data, figures provide us with a clearer understanding of trends and patterns.

Consider the following diagram in which the specific volume of H2O is presented as a

function of temperature and pressure:

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

Imagine that we are to run an experiment. 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/ = 1/1000 kg/m3). We plot this state at point 1

on the diagram.

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. 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 subcooled liquid.

State 2 is selected to correspond to the boiling point (100 oC). 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. As we continue to heat past the boiling point 2, a fundamental

20

100

300

1

2 3

4

T, oC

v, m3/kg

Compressed

liquid

Saturated

mixture

Superheated

steam


4

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.

At State 3, all liquid will have been vaporised. This is the saturated steam state.

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.

Summary of nomenclature:

Compressed or subcooled liquid (Between States 1 & 2)

A liquid state in which the fluid remains entirely within the liquid state, and below the

saturation state.

Saturated liquid (State 2)

All fluid is in the liquid state. However, even the slightest addition of energy would

result in the formation of some vapour.

Saturated Liquid-Steam or Wet Steam Region (Between States 2 & 3) Liquid and

steam exist together in a mixture.

Saturated steam (State 3)

All fluid is in the steam state, but even the slightest loss of energy from the system

would result in the formation of some liquid.

Superheated steam (The right of State 3)

All fluid is in the steam state and above the saturation state. The superheated steam

temperature is greater than the saturation temperature corresponding to the pressure.

The same experiment can be conducted at several different pressures. We see that as

pressure increases, the temperature at which boiling occurs also increases.

It can be seen that as pressure increases, the specific volume increase in the liquid to

steam transition will decrease.

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

0.0031

7

T-v diagram of

constant pressure

phase change

processes of a pure

substance at various

pressures for water.


5

At a pressure of 221.2 bar, the specific volume change which is associated to a phase

increase will disappear. Both liquid and steam will have the same specific volume,

0.00317 m3/kg. This occurs at a temperature of 374.15

oC. This state represents an

important transition in fluids and is termed the critical point.

If we connect the locus of points corresponding to the saturation condition, we will

obtain a diagram which allows easy identification of the distinct regions:

The general shape of the P-v diagram of a pure substance is very much like the T-v

diagram, but the T = constant lines on this diagram have a downward trend, as shown

in Fig. 8.2-4.

P-v diagram of a pure substance

P

v

Critical

point

Saturated liquid line

Dry saturated steam line

T2 = const.

T1 = const.

COMPRESS

LIQUID

REGION

WET STEAM

REGION

SUPERHEATED STEAM

REGION

T2 > T1

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


6

THE USE OF STEAM TABLES

The steam tables are available for a wide variety of substances which normally exist

in the vapour phase (e.g. steam, ammonia, freon, etc.). The steam tables which will

be used in this unit are those arranged by Mayhew and Rogers, which are suitable for

student use. The steam tables of Mayhew and Rogers are mainly concerned with

steam, but some properties of ammonia and freon-12 are also given.

Below is a list of the properties normally tabulated, with the symbols used being those

recommended by British Standard Specifications.

Symbols Units Description

p bar Absolute pressure of the fluid

ts oC Saturation temperature corresponding to the pressure p

bar

vf 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

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 ( to 8 of steam tables)


7

Complete the following table for Saturated Water and Steam:

t Ps vg hf hfg hg sf sfg sg

oC bar m

3/kg kJ/kg

kJ/kg K

0.01 206.1

0.02337 8.666

100 1.01325

Saturated Water and Steam Tables

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

Part 1

Part 1 refers to the values of temperature from 0.01oC to 100

oC, followed by values

that are suitable for the temperatures stated in the table. Table 8.4.1-1 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

Saturated water and steam at a temperature of 10 oC

Example 1

Solution

From page 2 of the steam tables, we can directly read:

t Ps vg hf hfg hg sf sfg sg oC bar m

3/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


8

Complete the missing properties in the following table for Saturated Water

and Steam:

p ts vg uf ug hf hfg hg sf sfg sg

bar oC m

3/kg kJ/kg kJ/kg kJ/kg K

0.045 31.0 2558

10 0.1944

311.0 5.615


bar oC m

3/kg kJ/kg kJ/kg kJ/kg K

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

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. Table

8.4.1-2 is an example showing an extract from the pressure of 1.0 bar.


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

Saturated water and steam at a pressure of 1.0 bar

f = property of the saturated liquid

g = property of the saturated steam

fg = change of the properties during evaporations

Example 2

Solution

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


9

PROPERTIES OF A WET MIXTURE

Between the saturated liquid and the saturated steam, there exist a mixture of steam

plus liquid (wet steam region). To denote the state of a liquid-steam mixture, it is

necessary to introduce a term describing the relative quantities of liquid and steam in

the mixture. This is called the dryness fraction (symbol x). Thus, in 1 kg of wet

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

Liquid-steam mixture

The dryness fraction is defined as follows;

where mtotal = mliquid + msteam

P-v diagram showing the location point of the dryness fraction

Specific volume

(1 - x ) kg of liquid

x kg of steam

total mass = 1 kg

At point A, x = 0

At point B, x = 1

Between point A and B, 0 x 1.0

Note that for a saturated liquid, x = 0;

and that for dry saturated steam, x = 1.

Sat. liquid

Sat. steam

Sat. liquid

P

v

ts

A B

x = 0.2 x = 0.8

vf vg

Sat. steam

mass total

steam saturateddry of massfraction dryness

total

steam

m

mx


10

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.

Therefore, the specific volume is given by,

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

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

saturated steam. Hence, for most practical problems,

v = xvg (8.2)

Where,

vf = specific volume of saturated liquid (m3/kg)

vg = specific volume of saturated steam (m3/kg)

x = dryness fraction

Specific enthalpy

In the analysis of certain types of processes, particularly in power generation and

refrigeration, we frequently encounter the combination of properties

U + PV. For the sake of simplicity and convenience, this combination is defined as a

new property, enthalpy, and given the symbol H.

H = U + PV (kJ)

or, per unit mass

h = u + Pv (kJ/kg)

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

enthalpy of the dry steam,

h = hf + xhfg

Where,

hf = specific enthalpy of saturated liquid (kJ/kg)

steam wet of mass total

steamdry of volumeliquid a of volume v


11

For a steam at 20 bar with a dryness fraction of 0.9, calculate the

a) specific volume

b) specific enthalpy

c) specific internal energy

hg = specific enthalpy of saturated steam (kJ/kg)

hfg = difference between hg and hf (that is, hfg = hg - hf )

8.3.3 Specific Internal Energy

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,

u = uf + x(ug – uf )

Specific Entropy

The entropy of wet steam is given by the sum of the entropy of the liquid plus the

entropy of the dry steam,

s = sf + xsfg

Summary:

v = xvg

h = hf + xhfg

u = uf + x(ug – uf )

s = sf + xsfg

Example 3

Solution

An extract from the steam tables


20 212.4 0.09957 907 2600 909 1890 2799 2.447 3.893 6.340

a) v = xvg

= 0.9(0.09957)

= 0.0896 m3/kg

b) h = hf + xhfg

= 909 + 0.9(1890)

= 2610 kJ/kg

c) u = uf + x( ug -uf )

= 907 + 0.9(2600 - 907)

= 2430.7 kJ/kg


12

Example 4

Find the dryness fraction, specific volume and specific enthalpy of steam at 8 bar and

specific internal energy 2450 kJ/kg.

Solution

An extract from the steam tables,


8 170.4 0.2403 720 2577 721 2048 2769 2.046 4.617 6.663

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).

u = uf + x(ug -uf)

2450 = 720 + x(2577 - 720)

x = 0.932

v = xvg

= 0.932 (0.2403)

= 0.2240 m3/kg

P

bar

v m3/kg

ts = 212.4 oC

v

u

h

s

vg

ug

hg

sg

x = 0.9

20

uf

hf

sf

P

bar

v m3/kg

ts = 170.4 oC

v vg

x = 0.932

8

h = hf + xhfg

= 721 + 0.932 (2048)

= 2629.7 kJ/kg


13

SUPERHEATED STEAM TABLES

The second part of the table is the superheated steam tables. 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 steam at this state has a degree of

superheat of 200 oC – 179.9

oC = 20.1

oC. The equation of degree of superheat is:

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:

For reference, the saturation temperature is inserted in brackets under each pressure in

the superheat tables and values of vg, ug, hg and sg are also given.

A specimen row of values is shown in Table 8.5.2. For example, from the

superheated table at 10 bar and 200 oC, the specific volume is 0.2061 m

3/kg and the

specific enthalpy is 2829 kJ/kg.

p

(ts)

t 200 250 300 350 400 450 500 600

10

(179.9)

vg

0.1944

v 0.206

1

0.232

8

0.258

0

0.282

5

0.306

5

0.330

3

0.354

0

0.401

0

ug 2584 u 2623 2711 2794 2875 2957 3040 3124 3297

hg 2778 h 2829 2944 3052 3158 3264 3370 3478 3698

sg 6.586 s 6.695 6.926 7.124 7.301 7.464 7.617 7.761 8.028

Superheated steam at a pressure of 10 bar

Degree of superheat = tsuperheat – tsaturation

u = h – pv


14

Complete 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 5

Solution

From the steam tables, we can directly read

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

Example 6

Steam 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

First, 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


15

An extract from the superheated table,

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 above.

Degree of superheat = 425 oC – 311

oC

= 114 oC

So, at 100 bar and 425 oC, we have

v = 2.812 x 10-2

m3/kg

h = 3172 kJ/kg

From equation 8.6,

u = h – Pv

= 3172 kJ/kg – (100 x 102 kN/m

2)(2.812 x 10

-2 m

3/kg)

= 2890.8 kJ/kg

Interpolation

The first interpolation problem that an engineer usually meets is that of “reading

between the lines” of a published table, like the Steam Tables. For properties which

are not tabulated exactly in the tables, it is necessary to interpolate between the values

tabulated as shown in Fig. 8.5-1 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”.

Interpolation

f(x)

x

Interpolation


16

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. 8.5-2.

Linear interpolation

From Figure above the value of x can be determined by:

1

12

121

12

12

1

1

xyy

xxyyx

yy

xx

yy

xx

There are two methods of interpolation:

i. Single interpolation

ii. Double interpolation

Single interpolation

Single interpolation is used to find the values in the table when one of the values is

not tabulated. For example, to find the saturation temperature, specific volume,

internal energy and enthalpy of dry saturated steam at 77 bar, it is necessary to

interpolate between the values given in the table.

Example 7

Determine the saturation temperature at 77 bar.

Solution

The 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.

y

x

y2

y

y1

x1 x x2

(x2 , y2)

(x , y)

(x1 , y1)


17

7580

5.290295

7577

5.290

st

5

5.290295

2

5.290

st

5.2905

5.42st

ts = 292.3 oC

Example 8

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

Solution

hg

2725

103 100

2715 2725

105 100

hg

3 10

52725

2719gh kJ/kg

Example 9

Determine the specific volume of steam at 8 bar and 220oC.

Solution

From 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,

p / (bar)

(ts / oC)

t 200 220 250

(oC)

8

(170.4)

v 0.2610 v 0.2933

v

0 2610

220 200

0 2933 0 2610

250 200

. . .

v 027392. m3/kg

P

ts

80

77

75

290.5 ts 295

P

hg

105

103

100

2725 hg 2715

P

v

250

220

200

0.2610 v 0.2933


18

Double Interpolation

In some cases a double interpolation is necessary, and it‟s usually used in the

Superheated Steam Table. Double interpolation must be used when two of the

properties (eg. temperature and pressure) are not tabulated in the Steam Tables. For

example, to find the enthalpy of superheated steam at 25 bar and 320oC, an

interpolation between 20 bar and 30 bar is necessary (as shown in example 8.9). An

interpolation between 300oC and 350

oC is also necessary.

Example 10

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

Solution

An extract from the Superheated Steam Tables:

t(oC)

p(bar)

300 320 350

20 3025 h1 3138

25 h

30 2995 h2 3117

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

At 20 bar,

300350

30253138

300320

30251

h

2.30701 h kJ/kg

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

300350

29953117

300320

29952

h

8.30432 h kJ/kg

T

h

350

320

300

3025 h1 3138

T

h

350

320

300

2995 h2 3117


19

Now interpolate between h1 at 20 bar, 320oC, and h2 at 30 bar, 320

oC in order to find

h at 25 bar and 320oC.

20302025

121

hhhh

h

3070 2

25 20

30438 3070 2

30 20

. . .

h 3057 kJ/kg.

Example 11

0.9 m3 of dry saturated steam at 225 kN/m

2 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 cylinder

b) dryness fraction at the final state

Sketch the process in the form of a P-v diagram.

Solution

Data: V1 = 0.9 m3

, P1 = 225 kN/m2 = 2.25 bar,

P2 = 180 kN/m

2 = 1.80 bar

a) Firstly, find the specific volume of dry saturated steam at 2.25 bar. Note that

the pressure 2.25 bar is not tabulated in the steam tables and it is necessary to use the

interpolation method.

From the Steam Tables,

vg at 2.2 bar = 0.8100 m3/kg

vg at 2.3 bar = 0.7770 m3/kg

vg1 at 2.25 bar,

20.230.2

8100.07770.0

20.225.2

8100.01

gv

vg1 0.7935 m3/kg

Mass of steam in cylinder,

1

1

gv

Vm (m

3 x kg/m

3)

= 1.134 kg

P

h

30

25

20

h1 h h2


20

b) At constant volume process,

Initial specific volume = final specific volume

v1 = v2

x1vg1 at 2.25 bar = x2vg2 at 1.8 bar

1(0.7935) = x2 (0.9774)

9774.0

)7935.0(12 x

= 0.81

P

bar

1.80

2.25

v m3/kg

1

2

0.7935 0.9774

v1 = v2


21

TUTORIAL

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.

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

3. Answer question below:

a. The internal energy of wet steam is 2000 kJ/kg. If the pressure is 42 bar, what is

the value of dryness fraction?

b. Determine the specific volume, specific enthalpy and specific internal energy of

wet steam at 32 bar if the dryness fraction is 0.92.

4. Find the dryness fraction, specific volume and specific internal energy of

steam at 105 bar and specific enthalpy 2100 kJ/kg.

P

v

( vi )

( ii )

( iv )

T2 = const.

T1 = const.

( i )

( iii)

( v )

T2 > T1


22

5. Steam at 120 bar is at 500 oC. Find the degree of superheat, specific volume,

specific enthalpy and specific internal energy.

6. Steam at 160 bar has a specific enthalpy of 3139 kJ/kg. Find the temperature,

degree of superheat, specific enthalpy and specific internal energy.

7 Determine the specific enthalpy of steam at 15 bar and 275oC.

8. Determine the degree of superheat and entropy of steam at 10 bar and 380oC.

9. A superheated steam at 12.5 MN/m2 is at 650

oC. Determine its specific

volume.

10. A superheated steam at 24 bar and 500oC expands at constant volume until the

pressure becomes 6 bar and the dryness fraction is 0.9. Calculate the changes in the

internal energy of steam. Sketch the process in the form of a P-v diagram.


23

DEFINITION OF PERFECT GASES

Did you know, one important type of fluid that has many applications in

thermodynamics is the type in which the working temperature of the fluid remains

well above the critical temperature of the fluid? In this case, the fluid cannot be

liquefied by an isothermal compression, i.e. if it is required to condense the fluid, then

cooling of the fluid must first be carried out. In the simple treatment of such fluids,

their behavior is likened to that a perfect gas. Although, strictly speaking, a perfect

gas is an ideal which can never be realized in practice. The behavior of many

„permanent‟ gases, e.g. hydrogen, oxygen, air etc is very similar to the behavior of a

perfect gas to a first approximation.

A perfect gas is a collection of particles that:

are in constant, random motion,

have no intermolecular attractions (which leads to elastic collisions in which

no energy is exchanged or lost),

are considered to be volume-less points.

You are more familiar with the term „ideal‟ gas. There is actually a distinction

between these two terms but for our purposes, you may consider them

interchangeable. The principle properties used to define the state of a gaseous system

are pressure (P), volume (V) and temperature (T). SI units (Systems International) for

these properties are Pascal (Pa) for pressure, m3 for volume (although liters and cm

3

are often substituted), and the absolute scale of temperature or Kelvin (K).

Two of the laws describing the behavior of a perfect gas are Boyle‟s Law and

Charles‟ Law.

BOYLE’S LAW

The Boyle‟s Law may be stated as follows:

Provided the temperature T of a perfect gas remains constant, then volume, V of a

given mass of gas is inversely proportional to the pressure P of the gas, i.e. P 1/V

(as shown in Fig. 3.1-1), or P x V = constant if temperature remains constant.

Graph P 1/V

P

1/V

P 1/V


24

If a gas changes from state 1 to state 2 during an isothermal process, then

P1 V1 = P2 V2 = constant

If the process is represented on a graph having axes of pressure P and volume V, the

results will be as shown in Fig. below. The curve is known as a rectangular

hyperbola, having the mathematical equation xy = constant.

P

P1 1

P2 2

3

P3

V1 V2 V3 V

P-V graph for constant temperature

Example 12

A quantity of a certain perfect gas is heated at a constant temperature from an initial

state of 0.22 m3

and 325 kN/m

2 to a final state of 170 kN/m

2. Calculate the final

pressure of the gas.

Solution

From equation P1V1 = P2V2

CHARLES’ LAW

The Charles‟s Law may be stated as follows:

Provided the pressure P of a given mass of gas remains constant, then the volume V of

the gas will be directly proportional to the absolute temperature T of the gas, i.e.

V T, or V = constant x T. Therefore V/T = constant, for constant pressure P.

If gas changes from state 1 to state 2 during a constant pressure process, then

PV = constant

3

2

23

2

112 m 0.421

kN/m 170

kN/m 325m 0.22 x

P

PVV


25

If the process is represented on a P – V diagram as before, the result will be as shown

in Fig. 3.2.

Example 13

A quantity of gas at 0.54 m3 and 345

oC undergoes a constant pressure process that

causes the volume of the gas to decreases to 0.32 m3. Calculate the temperature of the

gas at the end of the process.

Solution

From the question

V1 = 0.54 m3

T1 = 345 + 273 K = 618 K

V2 = 0.32 m3

constant2

2

1

1 T

V

T

V

1 2

P

V 0 V1 V2

P-V graph for constant pressure process

K 366

m 0.54

m 0.32 K 618

x

3

3

1

212

2

2

1

1

V

VTT

T

V

T

V


26

UNIVERSAL GASES LAW

Charles‟ Law gives us the change in volume of a gas with temperature when the

pressure remains constant. Boyle‟s Law gives us the change in volume of a gas with

pressure if the temperature remains constant.

The relation which gives the volume of a gas when both temperature and the pressure

are changed is stated as equation 3.3 below.

i.e.

No gases in practice obey this law rigidly, but many gases tend towards it. An

imaginary ideal that obeys the law is called a perfect gas, and the equation

is called the characteristic equation of state of a perfect gas.

The constant, R, is called the gas constant. The unit of R is Nm/kg K or J/kg K. Each

perfect gas has a different gas constant.

The characteristic equation is usually written

PV = RT

or for m kg, occupying V m3,

PV = mRT

Another form of the characteristic equation can be derived using the kilogram-mole as

a unit. The kilogram-mole is defined as a quantity of a gas equivalent to m kg of the

gas, where M is the molecular weight of the gas (e.g. since the molecular weight of

oxygen is 32, then 1 kg mole of oxygen is equivalent to 32 kg of oxygen).

From the definition of the kilogram-mole, for m kg of a gas we have,

m = nM

(where n is the number of moles).

Note: Since the standard of mass is the kg, kilogram-mole will be written simply as

mole.

Substituting for m from equations above

PV = nMRT or

RT

PV constant

RT

PV

nT

PVMR

2

22

1

11

T

VP

T

VP


27

Now Avogadro’s hypothesis states that the volume of 1 mole of any gas is the same as

the volume of 1 mole of any other gas, when the gases are at the same temperature

and pressure. Therefore V/n is the same for all gases at the same value of P and T.

That is the quantity PV/nT is constant for all gases. This constant is called the

universal gas constant, and is given the symbol Ro.

i.e.

or since MR = Ro then,

RR

M

o

Experiment has shown that the volume of 1 mole of any perfect gas at 1 bar and 1 oC

is approximately 22.71 m3. Therefore from equation 3.8

From equation 3.10 the gas constant for any gas can be found when the molecular

weight is known, e.g. for oxygen of molecular weight 32, the gas constant is

K J/kg 8.25932

4.8314

M

RR o

Example 14

0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m

2 and a

temperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m

2

and the temperature is 83oC. If the gas is assumed to be a perfect gas, determine:

a) the mass of gas (kg)

b) the final volume of gas (m3)

Given:

R = 0.29 kJ/kg K

Solution

From the question

V1 = 0.046 m3

P1 = 300 kN/m2

P2 = 1.27 MN/m2 = 1.27 x 10

3 kN/m

2

T1 = 45 + 273 K = 318 K T2 = 83 + 273 K = 356 K

R = 0.29 kJ/kg K

TnRPVnT

PVRMR oo or

K J/mole 8314.4273.15 x 1

22.71x 10 x 1 5

0 nT

PVR


28

From equation PV = mRT

the constant volume process i.e. V1 = V2

SPECIFIC HEAT CAPACITY AT CONSTANT VOLUME (CV)

The specific heat capacities of any substance is defined as the amount of heat energy

required to raise the unit mass through one degree temperature raise. In

thermodynamics, two specified conditions are used, those of constant volume and

constant pressure. The two specific heat capacities do not have the same value and it

is essential to distinguish them.

If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the

temperature of the gas by 1 degree whilst the volume of the gas remains constant, then

the amount of heat energy supplied is known as the specific heat capacity at constant

volume, and is denoted by Cv. The unit of Cv is J/kg K or kJ/kg K.

For a reversible non-flow process at constant volume, we have

dQ = mCvdT

For a perfect gas the values of Cv are constant for any one gas at all pressures and

temperatures. Equation above can then be expanded as follows :

Heat flow in a constant volume process, Q12 = mCv(T2 – T1)

Also, from the non-flow energy equation

Q – W = (U2 – U1)

mcv(T2 – T1) – 0 = (U2 – U1)

(U2 – U1) = mCv(T2 – T1)

i.e. dU = Q

Note:

In a reversible constant volume process, no work energy transfer can take place since

the piston will be unable to move i.e. W = 0.

kg 0.1496318 x 0.29

0.046 x 300

1

11 RT

VPm

K 1346300

10 x 1.27318

3

1

212

2

2

1

1

P

PTT

T

P

T

P


29

The reversible constant volume process is shown on a P-V diagram in Figure below:

P-V diagram for reversible constant volume process

Example 15

3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17

oC until

the temperature rose to 147 oC. If the gas is assumed to be a perfect gas, determine:

a) the heat flow during the process

b) the beginning pressure of gas

c) the final pressure of gas

Given

Cv = 0.72 kJ/kg K

R = 0.287 kJ/kg K

Solution

From the question

m = 3.4 kg

V1 = V2 = 0.92 m3

T1 = 17 + 273 K = 290 K

T2 = 147 + 273 K = 420 K

Cv = 0.72 kJ/kg K

R = 0.287 kJ/kg K

a) From equation, Q12 = mCv(T2 – T1)

= 3.4 x 0.72(420 – 290)

= 318.24 kJ

P2

P1 1

2

P

V V1 = V2


30

b) From equation, PV = mRT

Hence for state 1,

P1V1 = mRT1

2

3

1

11 kN/m 6.307

m 92.0

K 290kJ/kgK x 287.0 x kg 4.3

V

mRTP

c) For state 2,

P2V2 = mRT2

2

3

2

22 kN/m 5.445

m 92.0

K 042kJ/kgK x 287.0 x kg 4.3

V

mRTP

SPECIFIC HEAT CAPACITY AT CONSTANT PRESSURE (CP)

If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the

temperature of the gas by 1 degree whilst the pressure of the gas remains constant,

then the amount of heat energy supplied is known as the specific heat capacity at

constant pressure, and is denoted by Cp. The unit of Cp is J/kg K or kJ/kg K.

For a reversible non-flow process at constant pressure, we have

dQ = mCpdT

For a perfect gas the values of Cp are constant for any one gas at all pressures and

temperatures. Equation above can then be expanded as follows:

Heat flow in a reversible constant pressure process Q = mCp(T2 – T1)

RELATIONSHIP BETWEEN THE SPECIFIC HEATS

Let a perfect gas be heated at constant pressure from T1 to T2. With reference to the

non-flow equation Q = U2 – U1 + W, and the equation for a perfect gas

U2 – U1 = mCv(T2 – T1), hence,

Q = mCv(T2 – T1) + W

In a constant pressure process, the work done by the fluid is given by the pressure

times the change in volume, i.e. W = P(V2 – V1). Then using equation PV = mRT, we

have

W = mR(T2 – T1)

Therefore substituting,

Q = mCv(T2 – T1) + mR(T2 – T1) = m(Cv + R)(T2 – T1)


31

But for a constant pressure process from equation before,

Q = mCp(T2 – T1)

Hence, by equating the two expressions for the heat flow Q, we have

mCp(T2 – T1) = m(Cv + R)(T2 – T1)

Cp = Cv + R

Alternatively, it is usually written as

R = Cp - Cv

SPECIFIC HEAT RATIO ()

The ratio of the specific heat at constant pressure to the specific heat at constant

volume is given the symbol (gamma),

i.e. = v

p

C

C

Note that since Cp - Cv= R, from equation above, it is clear that Cp must be greater

than Cv for any perfect gas. It follows therefore that the ratio Cp/Cv = , is always

greater than unity. In general, is about 1.4 for diatomic gases such as carbon

monoxide (CO), hydrogen (H2), nitrogen (N2), and oxygen (O2). For monatomic gases

such as argon (A), and helium (He), is about 1.6, and for triatomic gases such as

carbon dioxide (CO2), and sulphur dioxide (SO2), is about 1.3. For some hydro-

carbons the value of is quite low (e.g. for ethane (C2H6), = 1.22, and for iso-

butane (C4H10), = 1.11.

Some useful relationships between Cp , Cv , R, and can be derived.

From equation above

Cp - Cv= R

Dividing through by Cv

vv

p

C

R

C

C1

Therefore = v

p

C

C, then,


32

vC

R1

)1(

RCv

Also, Cp = Cv hence substituting in equation above,

Cp = Cv = )1(

R

Cp = )1(

R

Example 16

A certain perfect gas has specific heat as follows

Cp = 0.846 kJ/kg K and Cv = 0.657 kJ/kg K

Find the gas constant and the molecular weight of the gas.

Solution

From equation R = Cp - Cv

i.e. R = 0.846 – 0.657 = 0.189 kJ/kg K

or R = 189 Nm/kg K

From equation M =R

R0

i.e. M = 44189

8314


33

NON-FLOW PROCESSES

A process occurs when a system‟s state (as measured by its properties) changes for

any reason. Processes may be reversible or actual (irreversible). In this context the

word „reversible‟ has a special meaning. A reversible process is one that is wholly

theoretical, but can be imagined as one which occurs without incurring friction,

turbulence, leakage or anything which causes unrecoverable energy losses. All of the

processes considered below are reversible and the actual processes will be dealt with

later.

Processes may be constrained to occur at constant temperature (isothermal), constant

pressure, constant volume, polytropic and adiabatic (with no heat transfer to the

surroundings).

Constant temperature (Isothermal) process (pV = C)

If the change in temperature during a process is very small then that process

may be approximated as an isothermal process. For example, the slow expansion or

compression of fluid in a cylinder, which is perfectly cooled by water may be

analysed, assuming that the temperature remains constant.

Constant temperature (Isothermal) process

The general relation properties between the initial and final states of a perfect

gas are applied as follows:

2

22

1

11

T

Vp

T

Vp

If the temperature remains constant during the process, T1 = T2 and the above relation

becomes

2211 VpVp

W

Q

P

v v1

v2

W

1

2


34

From the equation we can know that an increase in the volume results in a

decrease in the pressure. In other words, in an isothermal process, the pressure is

inversely proportional to the volume.

Work transfer:

Referring to the process represented on the p – V diagram in Figure above it is

noted that the volume increases during the process. In other words the fluid is

expanding. The expansion work is given by

2

1

pdVW

= 2

1

dVV

c (since pV = C, a constant)

= 2

1V

dVc

= 2

1

11V

dVVp

= 1

211 ln

V

VVp

lumesmaller vo

umelarger vol

= 1

21 ln

V

VmRT (since p1V1 = mRT1)

= 2

11 ln

p

pmRT (since

2

1

1

2

p

p

V

V )

Note that during expansion, the volume increases and the pressure decreases.

On the p – V diagram, the shaded area under the process line represents the amount of

work transfer.

Since this is an expansion process (i.e. increasing volume), the work is done

by the system. In other words the system produces work output and this is shown by

the direction of the arrow representing W.

Heat transfer:

Energy balance to this case is applied:

U1 + Q = U2 + W

For a perfect gas

U1 = mcvT1 and U2 = mcvT2

As the temperature is constant

U1 = U2


35

Substituting in the energy balance equation,

Q = W

Thus, for a perfect gas, all the heat added during a constant temperature

process is converted into work and the internal energy of the system remains constant.

Adiabatic process (Q = 0)

If a system is thermally well insulated then there will be negligible heat transfer into

or out of the system. Such a system is thermally isolated and a process within that

system may be idealised as an adiabatic process. For example, the outer casing of

steam engine, steam turbines and gas turbines are well insulated to minimise heat loss.

The fluid expansion process in such machines may be assumed to be adiabatic.

Adiabatic (zero heat transfer) process

For a perfect gas the equation for an adiabatic process is

pV = C

where = ratio of specific heat = v

p

C

C

The above equation is applied to states 1 and 2 as:

2211 VpVp

W

P

v v1

v2

W

1

2

Thermal insulation


36

2

1

1

2

V

V

p

p

Also, for a perfect gas, the general property relation between the two states is given

by the equation below

2

22

1

11

T

Vp

T

Vp

By manipulating 2 equations above the following relationship can be

determined:

1

2

1

1

1

2

1

2

V

V

p

p

T

T

By examining the equations the following conclusion for an adiabatic process on a

perfect gas can be drawn:

An increase in volume results in a decrease in pressure.

An increase in volume results in a decrease in temperature.

An increase in pressure results in an increase in temperature.

Work transfer:

Referring to the process represented on the p-V diagram it is noted that the

volume increases during the process.

In other words, the fluid expanding and the expansion work is given by the

formula:

2

1

pdVW

= 2

1

dVV

c

(since pV = C, a constant)

= 2

1

V

dVc

= 1

2211

VpVp [larger pV- small pV]

Note that after expansion, p2 is smaller than p1. In the p – V diagram, the

shaded area under the process represents the amount of work transfer.

As this is an expansion process (i.e. increase in volume) the work is done by

the system. In other words, the system produces work output and this is shown by the

direction of the arrow representing W.


37

Heat transfer:

In an adiabatic process, Q = 0.

Applying an energy balance to this case

U1 - W = U2

W = U1 – U2

Thus, in an adiabatic expansion the work output is equal to the decrease in

internal energy. In other words, because of the work output the internal energy of the

system decreases by a corresponding amount.

For a perfect gas, U1 = mcvT1 and U1 = mcvT1

On substitution

W = mcv(T1-T2) [larger T- smaller T]

We know

cp- cv = R

or

cv = 1

R

1

( )21

TTmRW

But, mRT2 = p2V2 and mRT1 = p1V1

Then the expression for the expansion becomes

1

2211

VpVpW

Example 17

In an industrial process, 0.4 kg of oxygen is compressed isothermally from 1.01 bar

and 22o C to 5.5 bar. Determine the work done and the heat transfer during the

process. Assume that oxygen is a perfect gas and take the molecular weight of oxygen

to be M = 32 kg/kmole.

Solution

Data: m = 0.4 kg; p1= 1.01 bar; t1= 22oC

p2 = 5.5 bar; W = ? Q = ?


38

From the equation

R =M

R0

= 32

8314

= 260 J/kgK

= 0.260 kJ/kgK

For an isothermal process

Work input,

W = mRTln1

2

p

p

= 01.1

5.5ln)27322( x 260.0 x 4.0

= 52 kJ

In an isothermal process all the work input is rejected as heat.

Therefore, heat rejected, Q = W = 52 kJ

Example 18

In a thermally insulated reciprocating air compressor, air at 0.98 bar and 20oC is

compressed into one sixth of its original volume. Determine the pressure and

temperature of the air after compression. If the compressor cylinder contains 0.05 kg

of air, calculate the required work input. For air, take = 1.4 and cv = 0.718

kJ/kgK.

Solution

Data : p1 = 0.98 bar; T1= 20 + 273 = 293 K

;6

1

1

2 V

V m = 0.05 kg; W = ?

As the cylinder is well insulated the heat transfer is negligible and the process

may be treated as adiabatic.

Considering air as a perfect gas

From equation,

2

1

1

2

V

V

p

p

p2 = 0.98 x 61.4

= 12 bar


39

From equation

1

2

1

1

2

V

V

T

T

T2 = 293 x 60.4

= 600 K

= 327oC

for an adiabatic compression process

W = mcv(T2-T1) [larger T- smaller T]

= 0.05 x 0.718 (600-293)

= 11 kJ

Polytropic process (pVn

= C)

This is the most general type of process, in which both heat energy and work

energy cross the boundary of the system. It is represented by an equation in the form

pVn

= constant

If a compression or expansion is performed slowly, and if the piston cylinder

assembly is cooled perfectly, then the process will be isothermal. In this case the

index n = 1.

If a compression or expansion is performed rapidly, and if the piston cylinder

assembly is perfectly insulated, then the process will be adiabatic. In this case the

index n = .

If a compression or expansion is performed at moderate speed, and if the

piston cylinder assembly is cooled to some degree, then the process is somewhere

between those discussed above. Generally, this is the situation in many engineering

applications. In this case the index n should take some value, which is between 1 and

depending on the degree of cooling.

Some practical examples include:

compression in a stationary air compressor (n = 1.3)

compression in an air compressor cooled by a fan (n = 1.2)

compression in a water cooled air compressor (n = 1.1)


40

Polytropic process

At states 1 and 2:

nn VpVp 2211

or

n

V

V

p

p

2

1

1

2

Also, for a perfect gas, the general property relation between the two states is given

by

2

22

1

11

T

Vp

T

Vp

By the manipulation of 2 equations above the following relationship can be

determined:

1

2

1

1

1

2

1

2

nn

n

V

V

p

p

T

T

By examining the equations the following conclusions for a polytropic process on a

perfect gas can be drawn as:

An increase in volume results in a decrease in pressure.

An increase in volume results in a decrease in temperature.

An increase in pressure results in an increase in temperature.

Work transfer:

Referring to the process represented on the p-V diagram it is noted that the

volume increases during the process.

In other words the fluid is expands and the expansion work is given by

W

Qloss

P

v v1 v2

W

1

2

P1

P2

pVn=C


41

2

1

pdVW

= 2

1

dVV

cn

(since pVn = C, a constant)

= 2

1

nV

dVc

= 1

2211

n


Note that after expansion p2 is smaller than p1. In the p – V diagram, the

shaded area under the process represents the amount of work transfer.

Since this is an expansion process (i.e. increase in volume), the work is done

by the system. In other words, the system produces work output and this is shown by

the direction of the arrow representing W.

Heat transfer:

Energy balance is applied to this case as:

U1 – Qloss - W = U2

Qloss = (U1 – U2) – W

or

W = (U1 – U2) - Qloss

Thus, in a polytropic expansion the work output is reduced because of the heat loses.

Example 19

The combustion gases in a petrol engine cylinder are at 30 bar and 800oC before

expansion. The gases expand through a volume ratio (1

2

V

V) of (

1

5.8) and occupy 510

cm3 after expansion. When the engine is air cooled the polytropic expansion index n =

1.15. What is the temperature and pressure of the gas after expansion, and what is the

work output?

Solution

State 1 State 2

P1= 30 bar

t1 = 800oC Qloss

W

V2 = 510 cm3

p2 = ?

t2 = ?


42

Data: p1 = 30 bar; T1 = 800 + 273 = 1073 K; n = 1.15

1

2

V

V= 8.5; V2 = 510 cm

3;

t2 = ? p2 = ? W = ?

Considering air as a perfect gas, for the polytropic process, the property relation is

given as:

1

2

112

n

V

VTT

=

115.1

5.8

11073

x

= 778.4 K

= 505.4oC

From equation

n

V

Vpp

2

112

=

15.1

5.8

1 x 30

= 2.56 bar

Now,

V2 = 510 cm3 = 510 x 10

-6 m

3

and,

1

2

V

V= 8.5

Then,

5.8

10510 6

1

x

V

= 60 x 10-6

m3

Work output during polytropic expansion is given as:

W = 1

2211

n


=115.1

)10510()1056.2()1060)(1030( 6565

xxxx

= 330 J

= 0.33 kJ


43

Constant volume process

If the change in volume during a process is very small then that process may be

approximated as a constant volume process. For example, heating or cooling a fluid in

a rigid walled vessel can be analysed by assuming that the volume remains constant.

a) Heating b) Cooling

Constant volume process (V2=V1)

The general property relation between the initial and final states of a perfect gas is

applied as:

2

22

1

11

T

Vp

T

Vp

If the volume remain constant during the process, V2 = V1 and then the above relation

becomes

2

2

1

1

T

p

T

p

or

1

2

1

2

p

p

T

T

From this equation it can be seen that an increase in pressure results from an increase

in temperature. In other words, in constant volume process, the temperature is

proportional to the pressure.

Work transfer:

Work transfer (pdV) must be zero because the change in volume, dV, during the

process is zero. However, work in the form of paddle-wheel work may be transferred.

Heat transfer:

Applying the non flow energy equation

Q – W = U2 – U1

gives Q – 0 = U2 – U1

i.e. Q = U2 – U1

p

v

2

1

Q

p

v

2

1

Q


44

This result, which is important and should be remembered, shows that the nett amount

of heat energy supplied to or taken from a fluid during a constant volume process is

equal to the change in the internal energy of the fluid.

5.3 Constant pressure process

If the change in pressure during a process is very small then that process may be

approximated as a constant pressure process. For example, heating or cooling a liquid

at atmospheric pressure may be analysed by assuming that the pressure remains

constant.

Constant pressure process

Consider the fluid in the piston cylinder as shown in Figure above. If the load on the

piston is kept constant the pressure will also remain constant.

The general property relation between the initial and final states of a perfect gas is

applied as:

2

22

1

11

T

Vp

T

Vp

If the pressure remain constant during the process, p2 = p1 and then the above relation

becomes

2

2

1

1

T

V

T

V

or

1

2

1

2

V

V

T

T

From this equation it can be seen that an increase in volume results from an increase

in temperature. In other words, in constant pressure process, the temperature is

proportional to the volume.

W

Q

P

v v1

v2

W

1 2

v2 – v1

p


45

Work transfer:

Referring to the process representation on the p-V diagram it is noted that the volume

increases during the process. In other words, the fluid expands. This expansion work

is given by

2

1

pdVW

2

1

dVp (since p is constant)

= p (V2 – V1) (larger volume – smaller volume)

Note that on a p-V diagram, the area under the process line represents the amount of

work transfer.

W = area of the shaded rectangle

= height x width

= p (V2 – V1) (larger volume – smaller volume)

Heat transfer:

Applying the non flow energy equation

Q – W = U2 – U1

or Q = (U2 – U1) + W

Thus part of the heat supplied is converted into work and the remainder is utilized in

increasing the internal energy of the system.

Substituting for W in equation

Q = (U2 – U1) + p(V2 – V1)

= U2 – U1 + p2 V2 – p1 V1 (since p2 = p1 )

= (U2 + p2 V2) – (U1 + p1 V1)

Now, we know that h = u + pv or H = U + pV

Hence

Q = H2 – H1 (larger H – smaller H)


46

Example 20

The specific internal energy of a fluid is increased from 120 kJ/kg to 180 kJ/kg during

a constant volume process. Determine the amount of heat energy required to bring

about this increase for 2 kg of fluid.

Solution

The non flow energy equation is

Q – W = U2 – U1

For a constant volume process

W = 0

and the equation becomes

Q = U2 – U1

Q = 180 – 120

= 60 kJ/kg

Therefore for 2 kg of fluid

Q = 60 x 2 = 120 kJ

i.e. 120 kJ of heat energy would be required.

Example 21

2.25 kg of fluid having a volume of 0.1 m3 is in a cylinder at a constant pressure of 7

bar. Heat energy is supplied to the fluid until the volume becomes 0.2 m3. If the initial

and final specific enthalpies of the fluid are 210 kJ/kg and 280 kJ/kg respectively,

determine

a) the quantity of heat energy supplied to the fluid

b) the change in internal energy of the fluid

Solution

Data: p = 7.0 bar; V1 = 0.1 m3 ; V2 = 0.2 m

3

a) Heat energy supplied = change in enthalpy of fluid

Q = H2 – H1

= m( h2 - h1 )

= 2.25( 280 – 210 )

= 157.5 kJ


47

b) For a constant pressure process

W = P(V2 – V1)

= 7 x 105 x ( 0.2 – 0.1)

= 7 x 104 J

= 70 kJ

Applying the non-flow energy equation

Q – W = U2 – U1

gives

U2 – U1 = 157.5 – 70

= 87.5 kJ


48

TUTORIAL

1. Study the statements in the table below. Mark the answers as TRUE or

FALSE.

STATEMENT TRUE or FALSE

i. Charles‟ Law gives us the change in volume

of a gas with temperature when the

temperature remains constant.

ii. Boyle‟s Law gives us the change in volume of

a gas with pressure if the pressure remains

constant.

iii. The characteristic equation of state of a

perfect gas is .

iv. Ro is the symbol for universal gas constant.

v. The constant R is called the gas constant.

vi. The unit of R is Nm/kg or J/kg.

2. 0.04 kg of a certain perfect gas occupies a volume of 0.0072 m3 at a

pressure 6.76 bar and a temperature of 127 oC. Calculate the molecular weight

of the gas (M). When the gas is allowed to expand until the pressure is 2.12

bar the final volume is 0.065 m3. Calculate the final temperature.

3. Two kilograms of a gas receive 200 kJ as heat at constant volume

process. If the temperature of the gas increases by 100 oC, determine the Cv of

the process.

4. A perfect gas is contained in a rigid vessel at 3 bar and 315 oC. The

gas is then cooled until the pressure falls to 1.5 bar. Calculate the heat

rejected per kg of gas.

Given: M = 26 kg/kmol and = 1.26.

5. A mass of 0.18 kg gas is at a temperature of 15 oC and pressure 130

kN/m2. If the gas has a value of Cv = 720 J/kg K, calculate the:

i.gas constant

ii.molecular weight

iii.specific heat at constant pressure

RT

PV


49

iv.specific heat ratio

6. 1 m3 of air at 8 bar and 120

oC is cooled at constant pressure process

until the temperature drops to 27 oC.

Given R = 0.287 kJ/kg K and Cp = 1.005 kJ/kg K, calculate the:

i. mass of air

ii. heat rejected in the process

iii.

iv. volume of the air after cooling.

7. A system undergoes a process in which 42 kJ of heat is rejected. If the

pressure is kept constant at 125 kN/m2 while the volume changes from 0.20 m

3

to 0.006 m3, determine the work done and the change in internal energy.

8. Heat is supplied to a gas in a rigid container.The mass of the container

is 1 kg and the volume of gas is 0.6 m3. 100 kJ is added as heat. If gas has

Cv = 0.7186 kJ/kg K during a process, determine the:

9. In the cylinder of a large engine, 1.0 kg of carbon dioxide at 527o C

and 20 bar expands isothermally to a pressure of 1.4 bar. What is the final

volume of the gas?

Take R = 189 Nm/kgK for carbon dioxide.

10. 1 kg of nitrogen (molecular weight 28) is compressed reversibly and

isothermally from 1.01 bar, 20oC to 4.2 bar. Calculate the work done and the

heat flow during the process. Assume nitrogen to be a perfect gas.

11. Air at 1.02 bar, 22oC, initially occupying a cylinder volume of 0.015

m3, is compressed reversibly and adiabatically by a piston to a pressure of 6.8

bar. Calculate the final temperature, the final volume, and the work done on

the mass of air in the cylinder.

12. 0.112 m3 of gas has a pressure of 138 kN/m

2. It is compressed to

690 kN/m2 according to the law pV

1.4 = C. Determine the new volume of the

gas.

13. 0.014 m3 of gas at a pressure of 2070 kN/m

2 expands to a pressure of

207 kN/m2 according to the law pV

1.35 = C. Determine the work done by the

gas during expansion.

14. A cylinder containing 0.07 kg of fluid has a pressure of 1 bar, a volume

of 0.06 m3 and a specific internal energy of 200 kJ/kg. After polytropic


50

compression, the pressure and volume of the fluid are 9 bar and 0.011 m3

respectively, and the specific internal energy is 370 kJ/kg.

Determine

a) the amount of work energy required for the compression

b) the quantity and direction of the heat energy that flows during the

compression.

15. The pressure of the gas inside an aerosol can is 1.2 bar at a temperature

of 25o C. Will the aerosol explode if it is thrown into a fire and heated to a

temperature of 600o C? Assume that the aerosol can is unable to withstand

pressure in excess of 3 bar.

a. 0.05 kg of air, initially at 130o C is heated at a constant pressure of 2

bar until the volume occupied is 0.0658 m3. Calculate the heat supplied and

the work done.

b. A spherical research balloon is filled with 420 m3 of atmospheric air at

a temperature of 10o C. If the air inside the balloon is heated to 80

oC at

constant pressure, what will be the final diameter of the balloon?

jj207 thermodynamics i chapter 2

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