South Valley University

Faculty of Engineering
Department of Mechanical Engineering

Thermodynamics MPEG 113

Dr. Mostafa Abouelsoud

 Presentation layout
 Goal of today PPT
 Pressure measurements
 Importance of Dimensions and Units
 Temperature and Zeroth law
 Energy
 Work
 Heat
Evaluating the W for Ideal Gas in Closed Systems
 Properties of Pure Substance
1 Definition of Pure Substance
2 Phases of a Pure Substance
3 Phase Change of Pure Substance
4 Subcooled or Compressed Liquid
5 Liquid Vapor region
6 Phase Change Terminology and Definitions
7 p-T (Pressure-Temperature)
8 Property Diagrams in Common Use
9 Formation of Steam
10 Important Terms Relating Steam Formation
11 Thermodynamic Properties of Steam and Steam Tables
12 Ideal, Perfect and Real Gases
13 The Equation of State for a Perfect Gas
14 Internal Energy and Enthalpy of a Perfect Gas
15 Is Water Vapor an Ideal Gas?
 Pure Substance

 is a substance that that has a fixed chemical composition throughout

 Example of pure substances; water, nitrogen, helium, and carbon dioxide.

 A mixture of various chemical elements or compounds also qualifies as a pure substance as

long as the mixture is homogeneous.

Air, for example, is a mixture of several gases, but it is often considered to be a pure substance
because it has a uniform chemical composition.

 A mixture of two or more phases of a pure substance is still a pure substance as long as the
chemical composition of all phases is the same.
 A mixture of ice and liquid water, for example, is a pure substance because both phases have
the same chemical composition. A mixture of liquid air and gaseous air,
 However, is not a pure substance since the composition of liquid air is different from the
composition of gaseous air, and thus the mixture is no longer chemically homogeneous.
 Phases of a Pure Substance

 We all know from experience that substances exist in different phases.

Phase‐Change Processes of Pure Substances

1. Solid
2. Mixed phase of liquid and solid
3. Sub‐cooled or compressed liquid
4. Wet vapor or saturated
liquid‐vapor mixture
5. Superheated vapor
 Phase Change of Pure Substance

 As the water is heated slowly its temperature rises until the temperature of the liquid water
becomes 100°C.

Phase change of water at constant pressure from liquid

to vapour phase
 Property Diagrams for Phase Change Processes

 If the pressure of a solid is measured at various temperatures until the triple point is reached and
then that of the pressure of liquid is measured until the critical point is reached, the result when
plotted on a p-T diagram appears. Triple point of water is 611.2 Pa and 0.01 oC.
Triple Point

 The triple point is merely the point of intersection of sublimation, fusion and vaporization curves.

 At the triple point all three phases of a pure substance coexist.

 Triple-point data for some interesting substances are;

p(kPa)

7.04

17.1

43.26

12.6

0.152

6.06

517

1.67

0.611
Liquid - Vapour Terms

 Saturation temperature. The phase change temperature corresponding to the saturation pressure.
Sometimes called the boiling temperature.
 Saturation pressure. The phase change pressure.
 Compressed liquid. Liquid whose temperature is lower than the saturation temperature.
Sometimes called a sub-cooled liquid.
 Dry (saturated) vapour. Vapour which has just completed evaporation. The pressure and
temperature of the vapour are the saturation values. Dry vapour is represented by a point g
on a diagram.
 Wet vapour. The mixture of saturated liquid and dry vapour during the phase change.
 Superheated vapour. Vapour whose temperature is greater than the saturation temperature
corresponding to the pressure of the vapour.
 Degree of superheat. The term used for the numerical amount by which the temperature of a
superheated vapour exceeds the saturation temperature.
 Saturated liquid. Liquid at the saturation temperature corresponding to the saturation pressure.
That is liquid about to commence evaporating,

 The amount of heat required to convert the liquid water completely into vapour under
this condition is called the heat of vaporization.
 Thermodynamic properties of Steam and Steam Tables

 In engineering problem, for any fluid which is used as working fluid, the six basic

thermodynamic properties required are:

1. p (pressure),
2. T (temperature),
3. v (volume),
4. u (internal energy),
5. h(enthalpy) and
6. s (entropy).

 The values of these properties are determined theoretically or experimentally and are tabulated in

the form of tables which are known as ‘Steam Tables’.

 It may be noted that steam has only one saturation temperature at each pressure.
 Following are the thermodynamic properties of steam which are tabulated in the form of table;
 p= Absolute pressure (bar or kPa) ;
 ts = Saturation temperature (°C) ;
 vf= Specific volume of saturated liquid (m3 /kg) ;
 vg= Specific volume of saturated vapour (steam) (m3 /kg).
 hf= Enthalpy of saturated liquid (kJ/kg) ;
 hfg= Enthalpy or latent heat of vaporization (kJ/kg) ;
 hg= Enthalpy of saturated vapour (steam) (kJ/kg) ;
 sf = Entropy of saturated liquid (kJ/kg K) ;
 sfg= Entropy of vapourization (kJ/kg K) ;
 sg = Entropy of saturated vapour (steam) (kJ/kg K) ;
 vfg = vg – vf Change of volume during evaporation.

 hfg = hg – hf Change of enthalpy during evaporation.

 sfg = sg – sf Change of entropy during evaporation.

 ufg = ug – uf Change of internal energy during evaporation.

Saturated Liquid‐Vapor Mixture
Dryness Fraction is the mass of dry saturated vapor in (1 kg) of a mixture of liquid and vapor
3 Dryness fraction (x)

 The term dryness fraction is related with wet steam.

 It is defined as the ratio of the mass of actual dry steam to the mass of steam containing it. It is
usually expressed by the symbol ‘x’.

Where;
 ms= Mass of dry steam contained in steam considered,
 mw= Mass of water particles in suspension in the steam considered,
T

V= Vf+ Vg
mtot v = mf vf+ mg vg
4 Total heat or enthalpy of wet steam (h)

 It is defined as the quantity of heat required to convert 1 kg of water at 0°C into wet steam at

constant pressure. It is the sum of total heat of water and the latent heat and this sum is also called

enthalpy.

h= hf +x hfg where hfg = hg- hf

 If steam is dry and saturated, then x = 1

 Then

h= hf + hfg = hf + (hg – hf )= hg
 If T1 and pressure P1 are known.
 We compare the pressure (P1) with saturation pressure at (P1).
 P1  Psat. at T1 Then the point 1 at compressed liquid.
 P1  Psat. at T1 Then the point 1 at superheated vapour.

 Or we compare the temperature (T1) with saturation temperature at (T1).

 T1  Tsat. at P1 Then the point 1 at compressed liquid.
 T1  Tsat. at P1 Then the point 1 at superheated vapour.
Subcooled or Compressed Liquid
 It may be noted that if the temperature of the liquid water becomes lower than the saturation
temperature for the given pressure, the liquid water is called a subcooled liquid (point1).
 Further, at point ‘1’ the temperature of liquid is 20°C and its saturation pressure (Psat) is 2.3339 kPa,
which is lower than the pressure on the liquid water, which is 1 atmosphere.

T
 If point 1 at
100 kPa and 20 0C
We must compare the pressure P1 with saturation pressure at T1.
100 kPa  Psat. at 20 0C 100 kPa  2.3339 kPa
Then the point 1 at compressed liquid

 If point 1 at
100 kPa and 20 0C
We must compare the temperature (200C) with saturation temperature at 100 kPa.
200C  Tsat at 100 kPa or 200C  1000C
Superheated Vapour

 When all the liquid has been evaporated completely and heat is further added, the temperature of

the vapour increases.

 At atmospheric pressure (1 bar) When the temperature increases above the saturation temperature

(100°C), the vapour is known as the superheated vapour and the temperature at this state is called

the superheated temperature.

 The difference between the superheated temperature and the saturation temperature at the given

pressure is called the degree of superheat.

Supercritical State.

 The curve 12-13 represents a constant pressure heating process,

when the pressure is greater than the critical pressure.

 At this state, the liquid water is directly converted into superheated

steam.
Pv diagram
 Property Diagrams in Common Use

 Besides p-V diagram which is useful because pressure and volume are easily visualised and the

T-s chart which is used in general thermodynamic processes, there are other charts which are

of practical use for particular applications.

 The specific enthalpy - specific entropy chart is used for steam plant work.

 The pressure - specific enthalpy chart is used in refrigeration work.

 Sketches of these charts are shown are drawn for H2O (water and steam).
The h-s Diagram or Mollier Diagram
 Important Terms Relating Steam Formation

1 Sensible heat of water (hf )

 Qs=mcpt=1*4.18 (100-0)=418 kJ

2 Latent heat or hidden heat (hfg)

 It is the amount of heat required to convert water at a given temperature and pressure into steam

at the same temperature and pressure.

 It is expressed by the symbol hfg and its value is available from steam tables.

 The value of latent heat is not constant and varies according to pressure variation.
5 Superheated steam.

 When steam is heated after it has become dry and saturated, it is called superheated steam and the

process of heating is called superheating.

 Superheating is always carried out at constant pressure.

 The additional amount of heat supplied to the steam during superheating is called as ‘Heat of

superheat’.

 Superheated steam behaves like a gas and therefore it follows the gas laws.

 The advantages obtained by using ‘superheated’ steam are as follows:

i. By superheating steam, its heat content and hence its capacity to do work is increased without

having to increase its pressure.

ii. Superheating is done in a superheater which obtains its heat from waste furnace gases

which would have otherwise passed uselessly up the chimney.

iii. High temperature of superheated steam results in an increase in thermal efficiency.

6 Volume of wet and dry steam

 If the steam has dryness fraction of x, then 1 kg of this steam will contain x kg of dry steam and (1
–x ) kg of water.
 If vf is the volume of 1 kg of water and vg is the volume of 1 kg of perfect dry steam (also known as
specific volume),
 volume of 1kg of wet steam=volume of dry steam + volume of water then = Vf + Vg
Then mv= (mv)f + (mv)g so v= (mf/m)vf + (mg/m)vg
so v=(1-x)vf + xvg
v=vf -xvf + xvg
v= vf + x (vg- vf)
v= vf +x vfg
Is Water Vapor an Ideal Gas?
 Compressibility Factor

 The ideal-gas equation is very simple and thus very convenient to use.

 Gases deviate from ideal-gas behavior significantly at states near the saturation region and the

critical point.

 This deviation from ideal-gas behavior at a given temperature and pressure can accurately be

accounted by the introduction of a correction factor called the compressibility factor Z defined

as;

 For ideal gases Z equals 1 where, for real gases Z can be greater than or less than unity.

 The farther away Z is from unity, the more the gas deviates from ideal-gas behavior.
 Example

 Four kilograms of water are placed in an enclosed volume of 1 m3. Heat is added until the

temperature is 150°C. Find the (a) pressure, (b) mass of the vapor, and (c) volume of the vapor.
 Solution
 Known: m= 4 kg, V = 1 m3., Tf = 150°C
 Required: p, mg, and vg.
 Analysis;
From Steam Table, the specific volume of water in an enclosed system is
v = V/m = ¼ = 0.25 m3/kg
And the saturated volume of vapor at 150°C is vg = 0.3928 m3/kg.

 the vapor mass;

. .
 Example

 1000 kg of steam at a pressure of 16 bar and 0.9 dry is generated by a boiler per hour. The

steam passes through a superheater via boiler stop valve where its temperature is raised to

380°C. If the temperature of feed water is 30°C, determine; (i) The total heat supplied to feed

water per hour to produce wet steam. (ii) The total heat absorbed per hour in the superheater.
 Solution
 Known: m= 1000 kg, p=16 bar, T1 = 30°C, T3 = 380°C, x2= 0.9.
 Required: QB, Qsup..
 Assumption: constant pressure process.
 Analysis;

1 2 3
Boiler Superheater
For water h1= 4.18*30 = 125.4 kJ/kg
Using saturated steam tables with interpolation at p= 16 bar
hf = 858.4 kJ/kg , hg=2793.8 kJ/kg
h2 = hf + x. hfg =858.4 + 0.9 *(2793.8- 858.4) = 2600 kJ/kg
Using superheated steam tables with interpolation at p=16 bar and 380°C
h3=3210 kJ/kg
The total heat supplied to feed water per hour to produce wet steam QB
QB =h2-h1 =2600-125.4 =2474.6 kJ/kg
(ii) The total heat absorbed per hour in the superheater. Qsup.
Qsup.= h3-h2= 3210- 2600 = 610 kJ/kg
Thanks for your attention