JJ207 Thermodynamic Topic 3 Properties of Pure Substances

JJ207-THERMODYNAMICS 1 Topic 3- Properties of Pure Substances

TOPIC 3 PROPERTIES OF PURE SUBSTANCES

At the end of the topic you will be able to:

Review the properties of pure substances Analyze the state of steam using the properties of pure substances Calculate saturation temperature and pressure, specific enthalpy, specific

volume, dryness fraction. Determine steam/water properties by referring to steam table (any condition

except supercritical) Relate the ideal gas model to pure substances Explain the gas constant, universal gas constant and general gas equation. Describe the law of ideal gas. Explain the specific heat at constant pressure and constant volume. Describe the characteristic of gas equation (equation of state) constant pressure

process, constant volume process, isothermal process, polytrophic process and adiabatic process.

3.0 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.

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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 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 PhaseIn 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;(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 Fig. 3.0 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.

Source: Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles

Figure 3.0 The arrangement of atoms in different phases

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(a) (b) (c)


3.1 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.

With this information, let us consider the following system:

A container is filled with water, and a moveable, frictionless piston is placed on the container at State 1, as shown in Fig. 3.1. 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.

Figure 3.1 Heating water and steam at constant pressure

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 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.

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WW

W

W

Liquid Steam SuperheatedSteam

STATE 1 STATE 2 STATE 3 STATE 4


3.2 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 pressure1:

Figure 3.2-1 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 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

1 Figures from Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles

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20

100

300

1

2 3

4

T, oC

v, m3/kg

Compressed liquid

Saturatedmixture

Superheatedsteam


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.

Saturation Pressure The pressure at which the liquid and vapour phases are in equilibrium at a given temperature.

Saturation Temperature The temperature at which the liquid and vapour phases are in equilibrium at a given pressure.

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The same experiment can be conducted at several different pressures. We see that as pressure increases, the temperature at which boiling occurs also increases.2

Figure 3.2-2 T-v diagram of constant pressure phase change processes of a pure substance at various pressures for water.

It can be seen that as pressure increases, the specific volume increase in the liquid to steam transition will decrease.

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.


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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

Saturatedliquid

Saturatedsteam

0.00317


If we connect the locus of points corresponding to the saturation condition, we will obtain a diagram which allows easy identification of the distinct regions3:

Figure 3.2-3 T-v diagram of a pure substance

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. 3.2-4.

Figure 3.2-4 P-v diagram of a pure substance


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P

v

Critical point

Saturated liquid line

Dry saturated steam line

T2 = const.

T1 = const.

COMPRESSLIQUIDREGION

WET STEAMREGION

SUPERHEATEDSTEAMREGION

T2 > T1

T

v

Critical point

Saturated liquid line

Dry saturated steam line

P2 = const.

P1 = const. COMPRESSLIQUIDREGION

WET STEAMREGION

SUPERHEATEDSTEAMREGION

P2 > P1


3.3 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.

Figure 3.3-1 Liquid-steam mixture

The steam dryness fraction (x) is defined as follows;

where masstotal = massliquid + masssteam

The quality is zero for the saturated liquid and one for the saturated steam (0 ≤ x ≤ 1).Consider a mixture of saturated liquid and saturated steam. The liquid has a mass mf

and occupies a volume Vf. The steam has a mass mg and occupies a volume Vg.

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(3.1)

(1 - x ) kg of liquid

x kg of steam

total mass = 1 kg

Figure 3.3-2 P-v diagram showing the location point of the dryness fraction

At point A, x = 0At point B, x = 1Between 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


3.3.1 Specific volume

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 (3.2)

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

3.3.2 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 massh = u + Pv (kJ/kg) (3.3)

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 (3.4)

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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 )

3.3.3 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,i.e. u = uf(1 – x) + xug

u = uf + x(ug – uf ) (3.5)

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

Equation 3.5 can be expressed in a form similar to equation 3.4. However, equation 3.5 is more convenient since ug and uf are tabulated. The difference is that, ufg is not tabulated.

3.3.4 Specific Entropy

A person looking at the steam tables carefully will notice two new properties i.e. enthalpy h and entropy s. Entropy is a property associated with the Second Law of Thermodynamics, and actually, we will properly define it later. However, it is appropriate to introduce entropy at this point.

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

s = sf + x(sg – sf ) s = sf + xsfg (3.6)

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 )

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3.4 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.

Table 3.4 The property of steam tables

Symbols Units Description

p bar Absolute pressure of the fluid

tsoC 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

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Pbar

v m3/kg

ts

vuhs

vg

ug

hg

sg

v = xvg

h = hf + xhfg

u = uf + x(ug – uf )s = sf + xsfg

uf

hf

sf

Figure 3.3-3 P-v diagram showing the location point v, u, h, s at wet steam.

Complete the following table for Saturated Water and Steam:

t Ps vg hf hfg hg sf sfg sg

oC kPa m3/kg kJ/kg kJ/kg K

0.01 206.1

2.3392 8.6661

100 101.42


sg kJ/kg K Specific entropy of saturated steam

sfg kJ/kg K Change of specific entropy during evaporation

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)

3.4.1 Saturated Water and Steam Tables

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

Part 1Part 1 refers to the values of temperature from 0.01oC to 100oC, followed by values that are suitable for the temperatures stated in the table. Table 3.4.1-1 is an example showing an extract from the temperature of 10oC.

Table 3.4.1-1 Saturated water and steam at a temperature of 10 oC

t ps vg hf hfg hg sf sfg sg

0C kPa m3/kg kJ/kg kJ/kg K

10 1.2281 106.32 42.022 2477.2 2519.2 0.1511 8.7488 8.8999

Example 3.1

Solution to Example 3.1

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

p ts vg uf ug hf hfg hg sf sfg sg

kPa oC m3/kg kJ/kg kJ/kg kJ/kg K

5.0 32.87 2560

1000 0.1944

311.0 5.615


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

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 3.4.1-2 is an example showing an extract from the pressure of 1.0 bar.

Table 3.4.1-2 Saturated water and steam at a pressure of 100 kPa


kpa oC m3/kg kJ/kg kJ/kg kJ/kg K

100 99.61 1.694 417 2506 417 2258 2675 1.303 6.056 7.359

Note the following subscripts:f = property of the saturated liquidg = property of the saturated steamfg = change of the properties during evaporations

Example 3.2

Solution to Example 3.2From page 3 to page 5 of the steam tables, we can directly read:

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t Ps vg hf hfg hg sf sfg sg

oC kPa m3/kg kJ/kg kJ/kg K

0.01 0.6117 206.1 0.001 2500.9 2500.9 0.000 9.155 9.1556

20 2.3392 57.76 83.91 2453.5 2537.5 0.2965 8.369 8.666

100 101.42 1.672 419.1 2256.4 2675.6 1.3072 6.047 7.354

For a steam at 2000 kPa with a dryness fraction of 0.9, calculate the a) specific volumeb) specific enthalpyc) specific internal energy


kPa oC m3/kg kJ/kg kJ/kg kJ/kg K

5 31.0 31.14 130 2418 130 2428 2558 0.451 7.980 8.431

1000 179.9 0.1944 762 2584 763 2015 2778 2.138 4.448 6.586

10000 311.0 0.01802 1393 2545 1408 1317 2725 3.360 2.255 5.615


Example 3.3


An extract from the steam tables


20000 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

b) Specific enthalpy (h), h = hf + xhfg

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

c) Specific internal energy (u),u = uf + x( ug -uf )

= 907 + 0.9(2600 - 907)= 2430.7 kJ/kg

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Pbar

v m3/kg

ts = 212.4 oC

vuhs

vg

ug

hg

sg

x = 0.9

20

uf

hf

sf

Find the dryness fraction, specific volume and specific enthalpy of steam at 800 kPa and specific internal energy 2450 kJ/kg.


Example 3.4


An extract from the steam tables,


800 170.4 0.2403 720 2577 721 2048 2769 2.046 4.617 6.663

At 800 kPa, 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).

From equation 8.5, u = uf + x(ug -uf)

2450 = 720 + x(2577 - 720) x = 0.932

From equation 8.2, v = xvg

= 0.932 (0.2403)= 0.2240 m3/kg

From equation 8.4, h = hf + xhfg

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

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Pbar

v m3/kg

ts = 170.4 oC

v vg

x = 0.932

8


3.4.2 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:

Degree of superheat = tsuperheat – tsaturation (3.7)

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:

u = h – pv (3.8)

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 3.4.2. For 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.

Table 3.4.2 Superheated steam at a pressure of 10 barp

(ts)t 200 250 300 350 400 450 500 600

1000(179.9)

vg 0.1944 v 0.2061 0.2328 0.2580 0.2825 0.3065 0.3303 0.3540 0.4010

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

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

p(ts)

t 300 350 400 450

4000(250.3)

vg 0.0498 v 0.0800

ug 2602 u 2921

hg 2801 h 3094

sg 6.070 s 6.364

Steam at 10000 kPa has a specific volume of 0.02812 m3/kg. Find the temperature, degree of superheat, specific enthalpy and specific internal energy.


Example 3.5


From page 7 of the steam tables, we can directly read

p(ts)

t 300 350 400 450

4000(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 3.6


First, it is necessary to decide whether the steam is wet, dry saturated or superheated.

At 10000 kPa, 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.

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An extract from the superheated table,

p(ts)

t 425

10000

(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 10000 kPa and 425 oC, we havev = 2.812 x 10-2 m3/kgh = 3172 kJ/kg

From equation 3.8,u = h – Pv

= 3172 kJ/kg – (100 x 102 kN/m2)(2.812 x 10-2 m3/kg)= 2890.8 kJ/kg

Note that equation 3.8 must be used to find the specific internal energy for pressure above 70 bar as the specific internal energy is not tabulated in steam tables.

8.5 Interpolation

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Pbar

v m3/kg

ts = 311.0 oC

100 425

oC

vg= 0.01802

v = 0.02812

A


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. 3.5-1 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”.

Figure 3.5-1 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. 3.5-2.

From Fig. 3.5.2, the value of x can be determined by:

There are two methods of interpolation:i. single interpolationii. double interpolation

3.5.1 Single interpolation

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f(x)

x

Interpolation

y

x

y2

y

y1

x1 x x2

(x2 , y2)

(x , y)

(x1 , y1)

Figure 3.5-2 Linear interpolation

Determine the saturation temperature at 7700 kPa.

Determine the specific enthalpy of dry saturated steam at 10300 kPa.

Determine the specific volume of steam at 800 kPa and 220oC.


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 3.6


The values of saturation temperature at a pressure of 7700 kPa are not tabulated in the Steam Tables. So, we need to interpolate between the two nearest values that are tabulated in the Steam Tables.

ts = 292.3 oC

Example 3.7


hg

3 10

52725

kJ/kg

Example 3.8

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P

ts

80

77

75

290.5 ts 295

P

hg

105

103

100

2725 hg 2715



From the Steam Tables at 800 kPa, 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 / (kPa) (ts / oC)

t 200 220 250 (oC)

800 (170.4)

v 0.2610 v 0.2933

v

0 2610

220 200

0 2933 0 2610

250 200

. . .

v 0 27392. m3/kg

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P

v

250

220

200

0.2610 v 0.2933

Determine the specific enthalpy of superheated steam at 2500 kpa and 320oC.


3.5.2 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 2000 kPa and 3000 kPa is necessary (as shown in example 8.9). An interpolation between 300oC and 350oC is also necessary.

Example 3.8


An extract from the Superheated Steam Tables:

t(oC)p(kpa)

300 320 350

2000 3025 h1 3138

2500 h

3000 2995 h2 3117

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

At 20 bar,

kJ/kg

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T

h

350

320

300

3025 h1 3138

0.9 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 state

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


Secondly, find the specific enthalpy (h2) at 3000 kPa and 320 oC;

At 30 bar,

kJ/kg

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

At 320oC,

h

3070 2

25 20

30438 3070 2

30 20

. . .

h 3057 kJ/kg.

Example 3.9

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T

h

350

320

300

2995 h2 3117

P

h

30

25

20

h1 h h2



Data: V1 = 0.9 m3 , P1 = 225 kN/m2 = 2.25 bar, P2 = 180 kN/m2 = 180 kpaa) 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 22 0 kPa= 0.8100 m3/kgvg at 230 kPa = 0.7770 m3/kg

vg1 at 225 kPa,

vg1 0.7935 m3/kg

Mass of steam in cylinder, (m3 x kg/m3)

0 9

0 7935

.

. = 1.134 kg

b) At constant volume process,Initial specific volume = final specific volume

v1 = v2

x1vg1 at 225 kPa = x2vg2 at 180 kPa 1(0.7935) = x2 (0.9774)

= 0.81

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Pbar

1.80

2.25

v m3/kg

1

2

0.7935 0.9774

v1 = v2


3.1 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.____________

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

3.3 The internal energy of wet steam is 2000 kJ/kg. If the pressure is 42 bar, what is the value of dryness fraction?

3.4 Determine the specific volume, specific enthalpy and specific internal energy of wet steam at 32 bar if the dryness fraction is 0.92.

| KBD/JKM/PUO 72

Activity 3

P

v

( vi )

( ii )

( iv )

T2 = const.

T1 = const.

( i )

( iii)

( v )

T2 > T1


3.5 Find the dryness fraction, specific volume and specific internal energy of steam at 105 bar and specific enthalpy 2100 kJ/kg.

3.6 Steam at 120 bar is at 500 oC. Find the degree of superheat, specific volume, specific enthalpy and specific internal energy.

3.7 Steam at 160 bar has a specific enthalpy of 3139 kJ/kg. Find the temperature, degree of superheat, specific enthalpy and specific internal energy.

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

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

3.10 A superheated steam at 12.5 MN/m2 is at 650oC. Determine its specific volume.

3.11 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.

Feedback To Activity 3

3.1 i) Liquid Phaseii) Solid Phaseiii) Steam Phase

3.2 i) Compress liquid regionii) Saturated liquid lineiii) Wet steam regioniv) Dry saturated steam linev) Superheated steam regionvi) Critical point

3.3 Dryness fraction (x),u = uf + x(ug -uf)

2000 = 1097 + x(2601 - 1097)x = 0.6

| KBD/JKM/PUO 73


3.4 Specific volume (v),v = xvg = 0.92(0.06246)

= 0.05746 m3/kg

Specific enthalpy (h), h = hf + xhfg

= 1025 + 0.92(1778) = 2661 kJ/kg

Specific internal energy (u),u = uf + x( ug -uf )

= 1021 + 0.92(2603 - 1021)= 2476 kJ/kg

3.5 Dryness fraction (x),h = hf + x hfg

2100 = 1429 + x(1286)x = 0.52

Specific volume (v),v = xvg = 0.52(0.01696)

= 0.00882 m3/kg

Specific internal energy (u),u = uf + x( ug -uf )

= 1414 + 0.52(2537 – 1414)= 1998 kJ/kg

3.6 From the superheated table at 120 bar, the saturation temperature is 324.6 oC. Therefore, the steam is superheated.

Degree of superheat = 500 oC – 324.6 oC = 175.4 oC

So, at 120 bar and 500 oC, we havev = 2.677 x 10-2 m3/kgh = 3348 kJ/kg



3.7 At 160 bar, hg = 2582 kJ/kg. This is less than the actual specific enthalpy of

| KBD/JKM/PUO 74


3139 kJ/kg. Hence, the steam is superheated.

From the superheated table at 160 bar, the specific enthalpy of 3139 kJ/kg is located at a temperature of 450 oC.

The degree of superheat = 450 oC – 347.3 oC = 102.7 oC

At 160 bar and 450 oC, we have v = 1.702 x 10-2 m3/kg



3.8

kJ/kg

3.9 Degree of superheat = 380oC – 179.9oC = 200.1oC

kJ/kg K

3.10 An extract from the superheated steam table:

| KBD/JKM/PUO 75

T

h

300

275

250

2925 h 3039

T

s

400

380

350

7.301 s 7.464


t(oC)p(bar)

600 650 700

120 3.159 x 10-2 v1 3.605 x 10-2

125 v

130 2.901 x 10-2 v2 3.318 x 10-2

Firstly, find the specific volume (v1) at 120 bar and 650 oC;At 120 bar,

m3/kg

Secondly, find the specific volume (v2) at 130 bar and 650 oC;At 130 bar,

v2 = 3.1095 x 10-2 m3/kg

Now interpolate between v1 at 120 bar, 650oC, and v2 at 130 bar, 650oC in order to find v at 125 bar and 650oC.

At 650oC,

v = 3.246 x 10-2 m3/kg

3.11 Data: P1 = 24 bar, T1 = 500oC

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T

v

700

650

600

3.159 x 10-2 v1 3.605 x 10-2

T

v

700

650

600

2.901 x 10-2 v2 3.318 x 10-2

P

v

130

125

120

v v2v1


P2 = 6 bar, x2 = 0.9

Firstly, find the initial internal energy at 24 bar, 500oC. Note that the pressure 24 bar is not tabulated in the Superheated Steam Tables and it is necessary to use the interpolation method to find the changes in the internal energy of steam.

At 500oC,

kJ/kg

Secondly, find the final internal energy at 6 bar where x = 0.9,u2 = uf2 + x2( ug2 -uf2 )

= 669 + 0.9(2568 - 669) = 2378.1 kJ/kg

The changes in the internal energy of steam is, (u2 – u1) = 2378.1 – 3112.8

= - 734.7 kJ/kg

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P

u

30

24

20

3116 u1 3108

Pbar

6

24

v m3/kg

1

2

221.8oC

v1 = v2

500oC

158.8oC 500oC 221.8oC

158.8oC

v1 = v2


1. With reference to the Steam Tables,i. determine the specific volume, specific enthalpy and specific internal

energy of wet steam at 15 bar with a dryness fraction of 0.9.ii. determine the degree of superheat, specific volume and specific internal

energy of steam at 80 bar and enthalpy 2990 kJ/kg.iii. complete the missing properties and a phase description in the

following table for water;

Pbar

toC

x vm3/kg

ukJ/kg

hkJ/kg

skJ/kg K

Phasedescription

2.0 120.2 6.4

12.0 1 2784

175 354.6 0.9

200 425

2. With reference to the Steam Tables,i. find the dryness fraction and specific entropy of steam at 2.9 bar and

specific enthalpy 2020 kJ/kg.

ii. determine the degree of superheat and internal energy of superheated steam at 33 bar and 313oC.

iii. determine the enthalpy change for a process involving a dry saturated steam at 3.0 MN/m2 which is superheated to 600 oC and carried out at constant pressure.

3. Steam at 7 bar and dryness fraction 0.9 expands in a cylinder behind a piston isothermally and reversibly to a pressure of 1.5 bar. Calculate the change of internal energy and the change of enthalpy per kg of steam.

| KBD/JKM/PUO 78

SELF-ASSESSMENT


Have you tried the questions????? If “YES”, check your answers now.

1. i. v = 0.11853 m3/kgh = 2600 kJ/kgu = 2419.8 kJ/kg

ii. degree of superheat = 55 oCv = 2.994 x 10-2 m3/kgu = 2750.48 kJ/kg

iii.

P

bar

toC

x v

m3/kg

u

kJ/kg

h

kJ/kg

s

kJ/kg K

Phase

description

2.0 120.2 0.87 0.7705 2267 2421 6.4 Wet steam

12.0 188 1 0.1632 2588 2784 6.523 Dry sat.

steam

175 354.6 0.9 0.007146 2319.8 2448.1 5.0135 Wet steam

200 425 - 0.001147 2725.6 2955 5.753 Superheated

steam

2. i. x = 0.68s = 5.2939 kJ/kg K

ii. Degree of superheat = 73.85oCu = 2769 kJ/kg

iii. h2 – h1 = 879 kJ/kg

3. u2 –u1 = 218 kJ/kg h2 – h1 = 246 kJ/kg

| KBD/JKM/PUO 79

Feedback to Self-Assessment


3.6 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 behaviours is likened to that a perfect gas. Although, strictly speaking, a perfect gas is an ideal which can never be realized in practice. The behaviour of many ‘permanent’ gases, e.g. hydrogen, oxygen, air etc is very similar to the behaviour 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 litres and cm3

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

Two of the laws describing the behaviours of a perfect gas are Boyle’s Law and Charles’ Law.

3.7 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.

Figure 3.1-1 Graph P 1/V

| KBD/JKM/PUO 80

P

1/V

P 1/V

A quantity of a certain perfect gas is heated at a constant temperature from an initial state of 0.22 m3 and 325 kN/m2 to a final state of 170 kN/m2. Calculate the final pressure of the gas.


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

P1 V1 = P2 V2 = constant (3.1)

If the process is represented on a graph having axes of pressure P and volume V, the results will be as shown in Fig. 3.1-2. 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

Figure 3.1-2 P-V graph for constant temperature

Example 3.1


From equation P1V1 = P2V2

3.8 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.

| KBD/JKM/PUO 81

PV = constant

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.


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

If the process is represented on a P – V diagram as before, the result will be as shown in Fig. 3.2.

Example 3.2

Solution to Example 3.2From the questionV1 = 0.54 m3 T1 = 345 + 273 K = 618 K V2 = 0.32 m3

3.9 Universal Gases Law

| KBD/JKM/PUO 82

constant2

2

1

1 T

V

T

V(3.2)

1 2

P

V0V1 V2

Figure 3.2 P-V graph for constant pressure process


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. (3.4)

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 writtenPV = RT (3.5)

or for m kg, occupying V m3,PV = mRT (3.6)

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 (3.7)

(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 equation 3.7 in equation 3.6,

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RT

PVconstant (3.3)

0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m2

and a temperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m2 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


PV = nMRT or (3.8)

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. (3.9)

or since MR = Ro then,

(3.10)

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

Example 3.3

| KBD/JKM/PUO 84



From the questionV1 = 0.046 m3 P1 = 300 kN/m2

T1 = 45 + 273 K = 318 K P2 = 1.27 MN/m2 = 1.27 x 103 kN/m2

T2 = 83 + 273 K = 356 K R = 0.29 kJ/kg K

From equation 3.6PV = mRT

From equation 3.4, the constant volume process i.e. V1 = V2

3.10 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 (3.11)

For a perfect gas the values of Cv are constant for any one gas at all pressures and temperatures. Equations (3.11) can then be expanded as follows :

Heat flow in a constant volume process, Q12 = mCv(T2 – T1) (3.12)

Also, from the non-flow energy equation

| KBD/JKM/PUO 85

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:

c) the heat flow during the processd) the beginning pressure of gase) the final pressure of gas

GivenCv = 0.72 kJ/kg KR = 0.287 kJ/kg K


Q – W = (U2 – U1)mcv(T2 – T1) – 0 = (U2 – U1) (U2 – U1) = mCv(T2 – T1) (3.13)

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.

The reversible constant volume process is shown on a P-V diagram in Fig. 3.4.

Figure 3.4 P-V diagram for reversible constant volume process

Example 3.4


From the question | KBD/JKM/PUO 86

P2

P1 1

2

P

VV1 = V2


m = 3.4 kgV1 = V2 = 0.92 m3 T1 = 17 + 273 K = 290 K T2 = 147 + 273 K = 420 K Cv = 0.72 kJ/kg KR = 0.287 kJ/kg K

a) From equation 3.13,Q12 = mCv(T2 – T1)

= 3.4 x 0.72(420 – 290) = 318.24 kJ

b) From equation 3.6,PV = mRT

Hence for state 1,P1V1 = mRT1

c) For state 2,P2V2 = mRT2

3.11 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 (3.14)

For a perfect gas the values of Cp are constant for any one gas at all pressures and temperatures. Equation (3.14) can then be expanded as follows:

Heat flow in a reversible constant pressure process Q = mCp(T2 – T1) (3.15)

| KBD/JKM/PUO 87


3.12 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 gasU2 – 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)

But for a constant pressure process from equation 3.15,Q = mCp(T2 – T1)

Hence, by equating the two expressions for the heat flow Q, we havemCp(T2 – T1) = m(Cv + R)(T2 – T1)Cp = Cv + R

Alternatively, it is usually written as

R = Cp - Cv 3.16

3.13 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. = (3.17)

Note that since Cp - Cv= R, from equation 3.16, 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.

| KBD/JKM/PUO 88


From equation 3.17

Cp - Cv= R

Dividing through by Cv

Therefore using equation 3.17, = , then,

3.18

Also from equation 3.17, Cp = Cv hence substituting in equation 3.18,

Cp = Cv =

Cp = 3.19

Example 3.5


From equation 3.16

R = Cp - Cv

i.e. R = 0.846 – 0.657 = 0.189 kJ/kg Kor R = 189 Nm/kg KFrom equation 3.10

| KBD/JKM/PUO 89

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.


M =

i.e. M =

3.14 Characteristic of Gas Equation

Once a fluid has entered a system, it may be possible for it to undergo a series of processes in which the fluid does not flow. An example of this is the cylinder of an internal combustion engine. In the suction stroke, the working fluid flows into the cylinder in which it is then temporarily sealed. Whilst the cylinder is sealed, the fluid is compressed by the piston moving into the cylinder, after which heat energy is supplied so that the fluid possesses sufficient energy to force the piston back down the cylinder, causing the engine to do external work. The exhaust valve is then opened and the fluid is made to flow out of the cylinder into the surroundings. Processes which are undergone by a system when the working fluid cannot cross the boundary are called non-flow process.

The equation for non-flow process is given as follows:

U1 + Q = U2 + W

or, U2 – U1 = Q –WThis process occurs during the compression and the working stroke as mentioned in the above example (refer to Fig. 3.8).

3.15.1 Constant temperature (Isothermal) process (pV = C)

| KBD/JKM/PUO 90

SUCTIONSTROKE COMPRESSION

STROKE

WORKINGSTROKE EXHAUST

STROKE

Figure 3.8 The cycle of an internal combustion engine


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.

Figure 3.8.1 Constant temperature (Isothermal) process

The general relation properties between the initial and final states of a perfect gas are applied as follows:

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

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 Fig.3.8.1 it is noted that the volume increases during the process. In other words the fluid is expanding. The expansion work is given by

= (since pV = C, a constant)

=

=

| KBD/JKM/PUO 91

W

Q

P

vv1

v2

W

1

2

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.


=

= (since p1V1 = mRT1)

= (since ) (3.20)

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

Substituting in the energy balance equation,

Q = W (3.21)

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.

Example 3.6


Data: m = 0.4 kg; p1= 1.01 bar; t1= 22oCp2 = 5.5 bar; W = ? Q = ?

| KBD/JKM/PUO 92


From the equation

R =

=

= 260 J/kgK = 0.260 kJ/kgK

For an isothermal processWork input,

W = mRTln

=

= 52 kJIn an isothermal process all the work input is transferred as heat.Therefore, heat transferred, Q = W = 52 kJ

3.15.2 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.

Figure 3.8.2 Adiabatic (zero heat transfer) processFor a perfect gas the equation for an adiabatic process is

pV = C

| KBD/JKM/PUO 93

W

P

vv1

v2

W

1

2

Thermal insulation


where ratio of specific heat, =

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

(3.22)

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

(3.23)

By manipulating equations 3.22 and 3.23 the following relationship can be

determined:

(3.24)

By examining equations 3.22 and 3.24 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 (Fig.4.3) 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:

= (since pV = C, a constant)

=

= (3.25)

| KBD/JKM/PUO 94

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.


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 (as shown in Fig 3.8.2).

Heat transfer:In an adiabatic process, Q = 0.Applying an energy balance to this case (Fig.3.8.2):

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 substitutionW = mCv(T1-T2) (3.26)

We know

Cp- Cv = R or Cv =

Substituting in equation 3.26

(3.27)

But, mRT2 = p2V2 and mRT1 = p1V1

Then the expression for the expansion becomes

(3.28)

Referring to the process represented on the p-V diagram it is noted that during this process the volume increases and the pressure decreases. For a perfect gas, equation 3.24 tells that a decrease in pressure will result in a temperature drop.

Example 3.7

| KBD/JKM/PUO 95


Solution to Example 3.7Data : p1 = 0.98 bar; T1= 20 + 273 = 293 K

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 4.4,

p2 = 0.98 x 61.4

= 12 bar

From equation 4.6,

T2 = 293 x 60.4

= 600 K = 327oC

Re-writing equation 4.8 for an adiabatic compression process

W = mcv(T2-T1) = 0.05 x 0.718 (600-293) = 11 kJ

3.15.3 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 (3.29)

| KBD/JKM/PUO 96


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)

Figure 5.1 Polytropic process

Equation 5.1 is applied at states 1 and 2 as:

Or

(3.30)

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

(3.31)

| KBD/JKM/PUO 97

W

Qloss

P

vv1 v2

W

1

2

P1

P2

pVn=C


By the manipulation of equations 3.30 and 3.31 the following relationship can be determined:

(3.32)

By examining equations 5.2 and 5.4 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 (Fig.5.1) it is noted that the volume increases during the process.In other words the fluid is expands and the expansion work is given by

= (since pVn = C, a constant)

=

= (3.33)

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 as shown in Fig. 5.1.

Heat transfer:

Energy balance is applied to this case (Fig.5.1) as:

U1 – Qloss - W = U2 Qloss = (U1 – U2) – W

or | KBD/JKM/PUO 98


W = (U1 – U2) - Qloss

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

Referring to the process represented on the p–V diagram (Fig.5.1) it is noted that during this process the volume increases and the pressure decreases. For a perfect gas, equation 5.4 tells us that a decrease in pressure will result in a temperature drop.

| KBD/JKM/PUO 99

For adiabatic process: W=

For polytropic process:W=

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

before expansion. The gases expand through a volume ratio ( ) of ( )

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?


Example 3.8


State 1 State 2

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

= 8.5; V2 = 510 cm3;

t2 = ? p2 = ? W = ?

Considering air as a perfect gas, for the polytropic process, the property relation is given by equation 5.4 as:

=

= 778.4 K = 505.4oC

| KBD/JKM/PUO 100

P1= 30 bart1 = 800oC Qloss

W

V2 = 510 cm3

p2 = ?t2 = ?


From equation 5.2

=

= 2.56 bar

Now,V2 = 510 cm3 = 510 x 10-6 m3

and,

= 8.5

Then,

= 60 x 10-6 m3

Work output during polytropic expansion is given by equation 5.5 as:

W =

=

= 330 J = 0.33 kJ

| KBD/JKM/PUO 101


TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

3.1 0.112 m3 of gas has a pressure of 138 kN/m2. It is compressed to 690 kN/m2 according to the law pV1.4 = C. Determine the new volume of the gas.

3.2 0.014 m3 of gas at a pressure of 2070 kN/m2 expands to a pressure of 207 kN/m2 according to the law pV1.35 = C. Determine the work done by the gas during expansion.

3.3 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 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.Determinea) the amount of work energy required for the compressionb) the quantity and direction of the heat energy that flows during the

compression.

| KBD/JKM/PUO 102

Activity 3A


3.15.4 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

Figure 5.2 Constant volume process (V2=V1)

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

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

or

(5.6)

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.

| KBD/JKM/PUO 103

p

v

2

1

Q

p

v

2

1

Q

INPUTINPUT

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.


Heat transfer:Applying the non flow energy equation

Q – W = U2 – U1

gives Q – 0 = U2 – U1

i.e. Q = U2 – U1 (5.7)

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.

Example 5.2


The non flow energy equation isQ – W = U2 – U1

For a constant volume processW = 0

and the equation becomesQ = 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.

3.15.5 Constant pressure process

| KBD/JKM/PUO 104


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.

Figure 5.3 Constant pressure process

Consider the fluid in the piston cylinder as shown in Figure 5.2. 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:

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

or

(5.8)

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.

Work transfer:

| KBD/JKM/PUO 105

W

Q

P

vv1

v2

W

1 2

v2 – v1

p


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

(since p is constant)

= p (V2 – V1) (larger volume – smaller volume) (5.9)

Note that on a p-V diagram, the area under the process line represents the amount of work transfer. From Figure 5.3

W = area of the shaded rectangle = height x width = p (V2 – V1) (larger volume – smaller volume)

This expression is identical to equation 5.9

Heat transfer:Applying the non flow energy equation

Q – W = U2 – U1

or Q = (U2 – U1) + W (5.10)

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 5.10

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 + pVHence

Q = H2 – H1 (5.11)

Referring to the process representation on the p-v diagram shown in Figure 5.3, it is noted that heating increases the volume. In other words, the fluid expands. For a perfect gas, equation 5.8 tells us that an increase in volume will result in corresponding increase in temperature.

| KBD/JKM/PUO 106

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, determinea) the quantity of heat energy supplied to the fluidb) the change in internal energy of the fluid


Example 5.3

Solution to Example 5.3Data: p = 7.0 bar; V1 = 0.1 m3 ; V2 = 0.2 m3

a) Heat energy supplied = change in enthalpy of fluid Q = H2 – H1

= m( h2 - h1 ) = 2.25( 280 – 210 ) = 157.5 kJ

b) For a constant pressure processW = P(V2 – V1) = 7 x 105 x ( 0.2 – 0.1) = 7 x 104 J = 70 kJ

Applying the non-flow energy equationQ – W = U2 – U1

givesU2 – U1 = 157.5 – 70

= 87.5 kJ

SUMMARY | KBD/JKM/PUO 107


R = gas constant. (unit Nm/kg K or J/kg K)M = molecular weight (unit kg/kmol)Ro = universal gas constant (unit 8.3144 kJ/kmol K or 8314 J/kmol K))Cv = specific heat capacity at constant volume (unit J/kg K or kJ/kg K)Cp = specific heat capacity at constant pressure (unit J/kg K or kJ/kg K) = specific heat ratio

PV = mRT

R = Cp - Cv =

| KBD/JKM/PUO 108



5.3 The pressure of the gas inside an aerosol can is 1.2 bar at a temperature of 25 o

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.

5.4 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.

| KBD/JKM/PUO 109

Activity 3B


Feedback To Activity 3B

5.4 Data: p1 = 1.2 bar; T1= 25 + 273 = 298 KT2 = 600 + 273 = 873 K; p2 = ?

We can idealize this process at constant volume heating of a perfect gas. Applying the general property relation between states 1 and 2

in this case V2 = V1

Hence,

or

= 1.2 x

= 3.52 bar

Since the aerosol cannot withstand pressures above 3 bar, it will clearly explode in the fire.

5.5 Data: m = 0.5 kg; p = 2 bar; V2 = 0.0658 m3;

T1 = 130 + 273 =403 K

Using the characteristic gas equation at state 2

T2 =

=

= 917 K

For a perfect gas undergoing a constant pressure process, we have

Q = mcp(T2 – T1)

i.e. Heat supplied = 0.05 x 1.005(917 – 403) = 25.83 kJ

| KBD/JKM/PUO 110


W = p (V2 – V1)

From equation pV = RT

Work done = R (T2 – T1) = 0.287(917 – 403)

i.e. Work done by the mass of gas present = 0.05 x 0.287 x 514 = 7.38 kJ

| KBD/JKM/PUO 111


You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback to Self-Assessment on the next page. If you face any problem, discuss it with your lecturer. Good luck.

1. A receiver vessel in a steam plant contains 20 kg of steam at 60 bar and 500oC. When the plant is switched off, the steam in the vessel cools at constant volume until the pressure is 30 bar. Find the temperature of the steam after cooling and the heat transfer that has taken place.

2. 0.25 kg of combustion gas in a diesel engine cylinder is at temperature of 727oC. The gas expands at constant pressure until its volume is 1.8 times its original value. For the combustion gas, R = 0.302 kJ/kgK and cp = 1.09 kJ/kgK. Find the following:a) temperature of the gas after expansionb) heat transferredc) work transferred

3. A quantity of gas has an initial pressure and volume of 0.1 MN/m2 and 0.1 m3, respectively. It is compressed to a final pressure of 1.4 MN/m2 according to the law pV1.26 = constant. Determine the final volume of the gas.

4. A mass of 0.05 kg of air at a temperature of 40oC and a pressure of 1 bar is compressed polytropicly following the law pV1.25 = C. Determine the following:a) final temperatureb) final volumec) work transferd) heat transfere) change in internal energy

| KBD/JKM/PUO 112

SELF-ASSESSMENT



1. 233.8oC; 14380 kJ rejected

2. 1527oC; 218 kJ added; 60.4 kJ output

3. 0.01235 m3

4. 158.9oC; 12390 cm3; 6.82 kJ input; 2.56 kJ rejected; 4.26 kJ increase

| KBD/JKM/PUO 113




3.1 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.

3.2 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.

| KBD/JKM/PUO 114

Activity 3A


Feedback To Activity 3A

3.1 i. Falseii. Falseiii. Trueiv. Truev. Truevi. False

3.2 From the question,m = 0.04 kgV1 = 0.072 m3 V2 = 0.072 m3

P1 = 6.76 bar = 6.76 x 102 kN/m2 P2 = 2.12 bar = 2.12 x 102 kN/m2 T1 = 127 + 273 K = 400 K

From equation 3.6P1V1 = mRT1

Then from equation 3.10

RR

Mo

i.e. Molecular weight = 27

| KBD/JKM/PUO 115


From equation 3.6P2V2 = mRT2

i.e. Final temperature = 1132.5 – 273 = 859.5 oC.


3.3 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.

3.4 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.

3.5 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 constantii. molecular weight

iii. specific heat at constant pressureiv. specific heat ratio

| KBD/JKM/PUO 116

Activity 3B


Feedback To Activity 3B

3.3 From the question,m = 2 kgQ = 200 kJ(T2 – T1) = 100 oC = 373 K

Q = mCv(T2 – T1)

3.4 From the question,P1 = 3 barT1 = 315 oC = 588 KP2 = 1.5 bar M = 26 kg/kmol = 1.26

From equation 3.10,

From equation 3.18,

During the process, the volume remains constant (i.e. rigid vessel) for the mass of gas present, and from equation 3.4,

| KBD/JKM/PUO 117


Therefore since V1 = V2,

Then from equation 3.12,Heat rejected per kg gas, Q = Cv(T2 – T1)

= 1.230(588 – 294)= 361.6 kJ/kg

3.5 From the questionm = 0.18 kgT = 15 oC = 288 KV = 0.17 m3

Cv = 720 J/kg K = 0.720 kJ/kg K

i. From equation 3.6,PV = mRT

ii. From equation 3.10,

iii. From equation 3.16,R = Cp - Cv

Cp = R + Cv = 0.426 + 0.720 = 1.146 kJ/kg K

iv. From equation 3.17,

CONGRATULATIONS, IF YOUR ANSWERS ARE CORRECT YOU CAN PROCEED TO THE NEXT INPUT…..

| KBD/JKM/PUO 118

SELF-ASSESSMENT



1. 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 airii. heat rejected in the process

iii. volume of the air after cooling.

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

3. 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:

i. change in temperature ii. change in internal energy


| KBD/JKM/PUO 119

Feedback To Self-Assessment


1. i. m = 7.093 kgii. Q = 663 kJiii. V2 = 0.763 m3

2. W = -24.25 kJ(U2 – U1) = -17.75 kJ

3. i. (T2 – T1) = 139.2 Kii. (U2 – U1) = 100 kJ


4.1 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.

4.2 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.

4.3 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.

| KBD/JKM/PUO 120

Activity 4


Feedback to Activity 4

4.1 Data: m = 1.0 kg; T1= 527 + 273 = 800 Kp1 = 20 bar; p2= 1.4 bar; V2 = ?

Carbon dioxide is a perfect gas and we can apply the following characteristic gas equation at state 1.

p1V1 = mRT1

V1 =

=

= 0.0756 m3

Applying the general property relation between state 1 and 2

For an isothermal process T1 = T2

Hence,

V2 =

V2 = 1.08 m3

4.2 Data: m=1kg; M= 28 kg/kmole p1 = 1.01 bar;T1 = 20 + 272 = 293 K; p2 = 4.2 bar

From equation

| KBD/JKM/PUO 121


R =

=

= 0.297 kJ/kgK

The process is shown on a p-v diagram below. When a process takes place from right to left on a p-v diagram the work done by the fluid is negative. That is, work is done on the fluid.

From equation 4.2

W =

= 1 x 0.297x293x ln

= -124 kJ/kg

For an isothermal process for a perfect gas,Q = W = -124 kJ/kg

4.3 Data: p1=1.02 bar; T1=22 + 273 = 295 K;v1= 0.015 m3; p2= 6.8 bar

From equation 4.6

| KBD/JKM/PUO 122

p

v

4.2

1.01

pV=C


T2 = 295 x

= 507.5 K(where for air = 1.4)i.e. Final temperature = 507.5 – 273 = 234.5oC

From equation 4.4

or

i.e. Final volume v2 = 0.0038 m3

For an adiabatic process,W = u1 – u2

and for a perfect gas, W = cv(T1- T2) = 0.718(295-507.5) = - 152.8 kJ/kg

i.e. work input per kg = -152.8 kJ

The mass of air can be found using equation pV = mRT

i.e. Total work done = 0.018 kg x -152.8 kJ/kg = -2.76 kJ

| KBD/JKM/PUO 123



3.1 0.112 m3 of gas has a pressure of 138 kN/m2. It is compressed to 690 kN/m2 according to the law pV1.4 = C. Determine the new volume of the gas.

3.2 0.014 m3 of gas at a pressure of 2070 kN/m2 expands to a pressure of 207 kN/m2 according to the law pV1.35 = C. Determine the work done by the gas during expansion.

3.3 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 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.Determinec) the amount of work energy required for the compressiond) the quantity and direction of the heat energy that flows during the

compression.

| KBD/JKM/PUO 124

Activity 3A


Feedback To Activity 3A

3.1 Since the gas is compressed according to the law pV1.4 = C, then,

or

from which,

=

= 0.012 x

= 0.0348 m3

3.2 The work done during a polytropic expansion is given by the expression:

W =

In this problem V2 is, as yet, unknown and must therefore be calculated.Now

or V2 = 0.014 x

V2 = 0.077 m3

Work done =

= 37.3 x 103 Nm= 37.3 x 103 J

= 37.3 kJ

3.3 a) For a polytropic process, | KBD/JKM/PUO 125


In the given case1 x 0.06n = 9 x 0.011n

n = 1.302

W =

=

= -13.2 kJ

The negative sign indicates that work energy would flow into the system during the process.

b) The non-flow energy equation gives

Q – W = U2 – U1

Q – (- 13.2) = ( 370 x 0.07 ) – ( 200 x 0.07 ) Q = - 1.3 kJ

The negative sign indicates that heat energy will flow out of the fluid during the process.


1. 0.05 m3 of a perfect gas at 6.3 bar undergoes a reversible isothermal process to a pressure of 1.05 bar. Calculate the heat flow to or from the gas.

2. Nitrogen (molecular weight 28) expands reversibly in a perfectly thermally insulated cylinder from 2.5 bar, 200oC to a volume of 0.09 m3. If the initial volume occupied was 0.03 m3, calculate the work done during the expansion. Assume nitrogen to be a perfect gas and take cv = 0.741 kJ/kg K.

| KBD/JKM/PUO 126

SELF-ASSESSMENT


3. A mass of 0.05 kg of air at a temperature of 40oC and a pressure of 1 bar is compressed adiabatically to 5 bar. Determine the following:a) final temperatureb) final volumec) work transferd) heat transfere) change in internal energy

4. A quantity of gas occupies a volume of 0.3 m3 at a pressure of 100 kN/m2 and a temperature of 20oC. The gas is compressed isothermally to a pressure of 500 kN/m2 and then expanded adiabatically to its initial volume. For this quantity of gas determine the following:a) the heat received or rejected (state which) during the compression,b) the change of internal energy during the expansion,c) the mass of gas.


1. 56.4 kJ

2. 9.31 kJ

3. 222.7oC, 14230 cm3, 6.56 kJ input, 0 kJ, 6.56 kJ increase.

4. – 48.3 kJ (heat rejected), -35.5 kJ, 0.358 kg

| KBD/JKM/PUO 127
