Your SlideShare is downloading. ×
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
JJ207 Thermodynamics I Chapter 2
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

JJ207 Thermodynamics I Chapter 2

7,274

Published on

Published in: Technology, Business
0 Comments
3 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
7,274
On Slideshare
0
From Embeds
0
Number of Embeds
6
Actions
Shares
0
Downloads
287
Comments
0
Likes
3
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. JJ 207 THERMODYNAMICS I Chapter 2INTRODUCTIONIn thermodynamic systems, the working fluid can be in the liquid, steam or gaseousphase. In this unit, the properties of liquid and steam are investigated in some detailsas the state of a system can be described in terms of its properties. A substance thathas 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 thatsubstances exist in different phases. A phase of substance can be defined as that partof a pure substance that consists of a single, homogenous aggregate of matter. Thethree 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 beconcerned with the molecular structure and behavior of the different phases.However, it is very helpful to have some understanding of the molecular phenomenainvolved in each phase.Molecular bonds are strongest in solids and weakest in steams. One reason is thatmolecules in solids are closely packed together, whereas in steams they are separatedby great distances.PHASESThe three phases of pure substances are: -Solid PhaseIn 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 easilydeform. 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 aspace; but(b) They are no longer rigidly structured so much so that they are free to movewithin a fixed volume. An example is a pure liquid state.Steam PhaseIn the steam phase, the molecules;(a) Virtually do not attract each other. The distance between the molecules arenot as close as those in the solid and liquid phases; 1
  • 2. JJ 207 THERMODYNAMICS I(b) Are not arranged in a fixed pattern. There is neither a fixed volume nor afixed shape for steam.The three phases described above are illustrated in Figure below. The following arediscovered:(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. (a) (b) (c)The arrangement of atoms in different phasesPHASE-CHANGE PROCESSThe 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 willpresent a free surface if it does not completely fill its container. Steam on the otherhand will always fill its container.A container is filled with water, and a moveable, frictionless piston is placed on thecontainer at State 1, as shown in Figure below. As heat is added to the system, thetemperature of the system will increase. Note that the pressure on the system is beingkept constant by the weight of the piston. The continued addition of heat will causethe temperature of the system to increase until the pressure of the steam generatedexactly balances the pressure of the atmosphere plus the pressure due to the weight ofthe piston. STATE 1 STATE 2 STATE 3 STATE 4 W Heating water W W and steam at W constant pressure Steam Superheated Liqui Steam dAt this point, the steam and liquid are said to be saturated. As more heat is added, theliquid that was at saturation will start to vaporize until State 2. The two-phase 2
  • 3. JJ 207 THERMODYNAMICS Imixture of steam and liquid at State 2 has only one degree of freedom, and as long asliquid is present, vaporization will continue at constant temperature. As long as liquidis present, the mixture is said to be wet steam, and both the liquid and steam aresaturated. After all the liquid is vaporized, only steam is present at State 3, and thefurther addition of heat will cause the temperature of steam to increase at constantsystem pressure. This state is called the superheated state, and the steam is said to besuperheated steam as shown in State 4.Saturated and Superheated SteamWhile tables provide a convenient way of presenting precise numerical presentationsof 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 afunction of temperature and pressure: T, oC 300 4 Superheated steam Saturated 2 mixture 3 100 Compressed liquid 20 1 v, m3/kg T-v diagram for the heating process of water at constant pressureImagine that we are to run an experiment. In this experiment, we start with a mass ofwater at 1 atm pressure and room temperature. At this temperature and pressure wemay measure the specific volume (1/ = 1/1000 kg/m3). We plot this state at point 1on the diagram.If we proceed to heat the water, the temperature will rise. In addition, water expandsslightly as it is heated which makes the specific volume increase slightly. We mayplot the locus of such points along the line from State 1 to State 2. We speak of liquidin 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 asbeing the saturated liquid state, which means that all of the water is in still liquidform, but ready to boil. As we continue to heat past the boiling point 2, a fundamental 3
  • 4. JJ 207 THERMODYNAMICS Ichange occurs in the process. The temperature of the water no longer continues torise. Instead, as we continue to add energy, liquid progressively changes to steamphase at a constant temperature but with an increasing specific volume. In this part ofthe 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 againrises as we add energy. States to the right of State 3 are said to be superheatedsteam.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 thesaturation state.Saturated liquid (State 2)All fluid is in the liquid state. However, even the slightest addition of energy wouldresult in the formation of some vapour.Saturated Liquid-Steam or Wet Steam Region (Between States 2 & 3) Liquid andsteam 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 systemwould 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 steamtemperature is greater than the saturation temperature corresponding to the pressure.The same experiment can be conducted at several different pressures. We see that aspressure increases, the temperature at which boiling occurs also increases.T, oC P = 221.2 bar Critical point P = 150 bar P = 80 bar374.15 P = 10 bar T-v diagram of constant pressure P = 5 bar phase change P = 1.01325 bar processes of a pure substance at various pressures for water. Saturated Saturated liquid steamIt can be seen that as pressure increases, the specific volume increase in the liquid to v, 0.0031steam transition will7decrease. m3/kg 4
  • 5. JJ 207 THERMODYNAMICS IAt a pressure of 221.2 bar, the specific volume change which is associated to a phaseincrease 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 animportant transition in fluids and is termed the critical point.If we connect the locus of points corresponding to the saturation condition, we willobtain a diagram which allows easy identification of the distinct regions: Saturated liquid lineT Dry saturated steam line Critical point P2 = const. T-v diagram of a P2 > P1 pure substance COMPRESS P1 = const. LIQUID REGION SUPERHEATED STEAM REGION WET STEAM REGION vThe general shape of the P-v diagram of a pure substance is very much like the T-vdiagram, but the T = constant lines on this diagram have a downward trend, as shownin Fig. 8.2-4. Saturated liquid line P Dry saturated steam line Critical point SUPERHEATED STEAM REGION COMPRESS LIQUID REGION T2 = const. WET STEAM T2 > T1 REGION T1 = const. v P-v diagram of a pure substance 5
  • 6. JJ 207 THERMODYNAMICS ITHE USE OF STEAM TABLESThe steam tables are available for a wide variety of substances which normally existin the vapour phase (e.g. steam, ammonia, freon, etc.). The steam tables which willbe used in this unit are those arranged by Mayhew and Rogers, which are suitable forstudent use. The steam tables of Mayhew and Rogers are mainly concerned withsteam, 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 thoserecommended by British Standard Specifications. Symbols Units Description p bar Absolute pressure of the fluid o ts C 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 tablesThese 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) 6
  • 7. JJ 207 THERMODYNAMICS ISaturated Water and Steam TablesThe 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 valuesthat are suitable for the temperatures stated in the table. Table 8.4.1-1 is an exampleshowing an extract from the temperature of 10oC. t ps vg hf hfg hg sf sfg sg 0 C bar kJ/kg kJ/kg K m3/kg 10 0.01227 42.0 2477.2 0.151 8.749 106.4 2519.2 8.900 Saturated water and steam at a temperature of 10 oCExample 1 Complete the following table for Saturated Water and Steam: t Ps vg hf hfg hg sf sfg sg o 3 C bar m /kg kJ/kg kJ/kg K 0.01 206.1 0.02337 8.666 100 1.01325SolutionFrom page 2 of the steam tables, we can directly read: t Ps vg hf hfg hg sf sfg sg o 3 C bar m /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 7
  • 8. JJ 207 THERMODYNAMICS IPart 2Part 2 (Page 3 to 5 of steam tables) is values of pressure from 0.006112 bar to 221.2bar followed by values that are suitable for the pressures stated in the table. Table8.4.1-2 is an example showing an extract from the pressure of 1.0 bar. p ts vg uf ug hf hfg hg sf sfg sg bar oC kJ/kg kJ/kg kJ/kg K m3/kg 1.0 99.6 417 417 2258 2675 1.303 6.056 1.694 2506 7.359 Saturated water and steam at a pressure of 1.0 barf = property of the saturated liquidg = property of the saturated steamfg = change of the properties during evaporationsExample 2 Complete the missing properties in the following table for Saturated Water and Steam: p ts vg uf ug hf hfg hg sf sfg sg o bar C m3/kg kJ/kg kJ/kg kJ/kg K 0.045 31.0 2558 10 0.1944 311.0 5.615Solution From page 3 to page 5 of the steam tables, we can directly read: p ts vg uf ug hf hfg hg sf sfg sg o 3 bar C m /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 8
  • 9. JJ 207 THERMODYNAMICS IPROPERTIES OF A WET MIXTUREBetween the saturated liquid and the saturated steam, there exist a mixture of steamplus liquid (wet steam region). To denote the state of a liquid-steam mixture, it isnecessary to introduce a term describing the relative quantities of liquid and steam inthe mixture. This is called the dryness fraction (symbol x). Thus, in 1 kg of wetmixture, there must be x kg of saturated steam plus (1 – x) kg of saturated liquid. x kg of steam total mass = 1 kg (1 - x ) kg of liquid Liquid-steam mixtureThe dryness fraction is defined as follows; mass of dry saturated steam dryness fraction  total mass msteam x mtotalwhere mtotal = mliquid + msteam Sat. steam Sat. steam Sat. liquid At point A, x = 0 Sat. liquid At point B, x = 1 P x = 0.2 x = 0.8 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. A B ts vg v vf P-v diagram showing the location point of the dryness fractionSpecific volume 9
  • 10. JJ 207 THERMODYNAMICS IFor a wet steam, the total volume of the mixture is given by the volume of liquidpresent plus the volume of dry steam present. Therefore, the specific volume is given by, volume of a liquid  volume of dry steam v total mass of wet steamNow for 1 kg of wet steam, there are (1 – x) kg of liquid and x kg of dry steam, wherex is the dryness fraction as defined earlier. Hence, v = vf(1 – x) + vgxThe volume of the liquid is usually negligibly small as compared to the volume of drysaturated 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 fractionSpecific enthalpyIn the analysis of certain types of processes, particularly in power generation andrefrigeration, we frequently encounter the combination of propertiesU + PV. For the sake of simplicity and convenience, this combination is defined as anew property, enthalpy, and given the symbol H. H = U + PV (kJ)or, per unit massh = u + Pv (kJ/kg)The enthalpy of wet steam is given by the sum of the enthalpy of the liquid plus theenthalpy of the dry steam,h = hf + xhfgWhere,hf = specific enthalpy of saturated liquid (kJ/kg) 10
  • 11. JJ 207 THERMODYNAMICS Ihg = specific enthalpy of saturated steam (kJ/kg)hfg = difference between hg and hf (that is, hfg = hg - hf )8.3.3 Specific Internal EnergySimilarly, the specific internal energy of a wet steam is given by the internal energy ofthe liquid plus the internal energy of the dry steam,u = uf + x(ug – uf )Specific EntropyThe entropy of wet steam is given by the sum of the entropy of the liquid plus theentropy of the dry steam,s = sf + xsfgSummary:v = xvgh = hf + xhfgu = uf + x(ug – uf )s = sf + xsfgExample 3For a steam at 20 bar with a dryness fraction of 0.9, calculate thea) specific volumeb) specific enthalpyc) specific internal energySolutionAn extract from the steam tables p ts vg uf ug hf hfg hg sf sfg sg 20 212.4 0.09957 907 2600 909 1890 2799 2.447 3.893 6.340a) v = xvg c) u = uf + x( ug -uf ) = 0.9(0.09957) = 907 + 0.9(2600 - 907) = 0.0896 m3/kg = 2430.7 kJ/kgb) h = hf + xhfg = 909 + 0.9(1890) = 2610 kJ/kg 11
  • 12. JJ 207 THERMODYNAMICS I P bar x = 0.9 20 ts = 212.4 oC v m3/kg v vg uf u ug hf h hg sf s sgExample 4 Find the dryness fraction, specific volume and specific enthalpy of steam at 8 bar andspecific internal energy 2450 kJ/kg.SolutionAn extract from the steam tables, p ts vg uf ug hf hfg hg sf sfg sg 8 170.4 0.2403 720 2577 721 2048 2769 2.046 4.617 6.663At 8 bar, ug = 2577 kJ/kg, since the actual specific internal energy is given as 2450kJ/kg, the steam must be in the wet steam state ( u < ug). u = uf + x(ug -uf) h = hf + xhfg 2450 = 720 + x(2577 - 720) = 721 + 0.932 (2048)x = 0.932 = 2629.7 kJ/kgv = xvg= 0.932 (0.2403) P= 0.2240 m3/kg bar x = 0.932 8 ts = 170.4 oC v m3/kg v vg 12
  • 13. JJ 207 THERMODYNAMICS ISUPERHEATED STEAM TABLESThe second part of the table is the superheated steam tables. The values of thespecific properties of a superheated steam are normally listed in separate tables for theselected values of pressure and temperature.A steam is called superheated when its temperature is greater than the saturationtemperature corresponding to the pressure. When the pressure and temperature aregiven for the superheated steam then the state is defined and all the other propertiescan be found. For example, steam at 10 bar and 200 oC is superheated since thesaturation temperature at 10 bar is 179.9 oC. The steam at this state has a degree ofsuperheat of 200 oC – 179.9 oC = 20.1 oC. The equation of degree of superheat is: Degree of superheat = tsuperheat – tsaturationThe tables of properties of superheated steam range in pressure from 0.006112 bar tothe critical pressure of 221.2 bar. At each pressure, there is a range of temperature upto 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. Thespecific internal energy is calculated using the equation: u = h – pvFor reference, the saturation temperature is inserted in brackets under each pressure inthe 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 thesuperheated table at 10 bar and 200 oC, the specific volume is 0.2061 m3/kg and thespecific enthalpy is 2829 kJ/kg. p t 200 250 300 350 400 450 500 600 (ts) vg v 0.206 0.232 0.258 0.282 0.306 0.330 0.354 0.401 10 0.1944 1 8 0 5 5 3 0 0 (179.9) 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 13
  • 14. JJ 207 THERMODYNAMICS IExample 5Complete the missing properties in the following table for Superheated Steam: p t 300 350 400 450 (ts) vg 0.0498 v 0.0800 40 ug 2602 u 2921 (250.3) hg 2801 h 3094 sg 6.070 s 6.364SolutionFrom the steam tables, we can directly read p t 300 350 400 450 (ts) vg 0.0498 v 0.0588 0.0664 0.0733 0.0800 40 ug 2602 u 2728 2828 2921 3010 (250.3) hg 2801 h 2963 3094 3214 3330 sg 6.070 s 6.364 6.584 6.769 6.935Example 6Steam at 100 bar has a specific volume of 0.02812 m3/kg. Find the temperature,degree of superheat, specific enthalpy and specific internal energy.SolutionFirst, 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 of0.02812 m3/kg. Hence, the steam is superheated. The state of the steam is at point Ain the diagram below. P bar A 100 425 oC ts = 311.0 oC v m3/kg vg= 0.01802 v = 0.02812 14
  • 15. JJ 207 THERMODYNAMICS IAn extract from the superheated table, p t 425 (ts) vg 0.01802 v x 10-2 2.812 100 hg 2725 h 3172 (311.0) sg 5.615 s 6.321From the superheated table at 100 bar, the specific volume is 0.02812 m3/kg at atemperature of 425 oC. Hence, this is the isothermal line, which passes through pointA as shown in the P-v diagram above.Degree of superheat = 425 oC – 311 oC = 114 oCSo, 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/m2)(2.812 x 10-2 m3/kg)= 2890.8 kJ/kgInterpolationThe first interpolation problem that an engineer usually meets is that of “readingbetween the lines” of a published table, like the Steam Tables. For properties whichare not tabulated exactly in the tables, it is necessary to interpolate between the valuestabulated as shown in Fig. 8.5-1 below. In this process it is customary to use a straightline that passes through two adjacent table points, denoted by  and . If we use thestraight line then it is called “interpolation”. f(x) Interpolation x   Interpolation 15
  • 16. JJ 207 THERMODYNAMICS IThe 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. y y2 (x2 , y2) y (x , y) y1 (x1 , y1) x x1 x x2 Linear interpolationFrom Figure above the value of x can be determined by: x  x1 x 2  x1  y  y1 y 2  y1 x  y  y1 x2  x1   x  y 2  y1  1There are two methods of interpolation:i. Single interpolationii. Double interpolationSingle interpolationSingle interpolation is used to find the values in the table when one of the values isnot tabulated. For example, to find the saturation temperature, specific volume,internal energy and enthalpy of dry saturated steam at 77 bar, it is necessary tointerpolate between the values given in the table.Example 7Determine the saturation temperature at 77 bar.SolutionThe values of saturation temperature at a pressure of 77 bars are not tabulated in theSteam Tables. So, we need to interpolate between the two nearest values that aretabulated in the Steam Tables. 16
  • 17. JJ 207 THERMODYNAMICS I P t s  290.5 295  290.5 80  77  75 80  75 77 t s  290.5 295  290.5 75  2 5 ts 290.5 ts 295 24.5 ts   290.5 5 ts = 292.3 oCExample 8Determine the specific enthalpy of dry saturated steam at 103 bar.Solution P hg  2725 2715  2725  103  100 105  100 105 103 3 10 hg   2725 100 5 hg g2719 kJ/kg h 2725 hg 2715Example 9Determine the specific volume of steam at 8 bar and 220oC.SolutionFrom 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) t 200 220 250 (ts / oC) (oC) 8 v 0.2610 v 0.2933 (170.4) P 250v  0.2610 0.2933  0.2610 220 220  200 250  200 200 v  0.27392 m3/kg v 0.2610 v 0.2933 17
  • 18. JJ 207 THERMODYNAMICS IDouble InterpolationIn some cases a double interpolation is necessary, and it‟s usually used in theSuperheated Steam Table. Double interpolation must be used when two of theproperties (eg. temperature and pressure) are not tabulated in the Steam Tables. Forexample, to find the enthalpy of superheated steam at 25 bar and 320oC, aninterpolation between 20 bar and 30 bar is necessary (as shown in example 8.9). Aninterpolation between 300oC and 350oC is also necessary.Example 10Determine the specific enthalpy of superheated steam at 25 bar and 320oC.SolutionAn extract from the Superheated Steam Tables: t(oC) 300 320 350 p(bar) 20 3025 h1 3138 25 h 30 2995 h2 3117Firstly, find the specific enthalpy (h1) at 20 bar and 320 oC;At 20 bar, T h1  3025 3138  3025 350  320  300 350  300 320 300 h1  3070.2 kJ/kg h 3025 h1 3138Secondly, find the specific enthalpy (h2) at 30 bar and 320 oC; Th2  2995 3117  2995 350 320  300 350  300 320 300 h2  3043.8 kJ/kg h 2995 h2 3117 18
  • 19. JJ 207 THERMODYNAMICS INow interpolate between h1 at 20 bar, 320oC, and h2 at 30 bar, 320oC in order to findh at 25 bar and 320oC. P h  h1 h  h1  2 25  20 30  20 30h  3070.2 30438  3070.2 .  25 25  20 30  20 20 h  3057 kJ/kg. h h1 h h2Example 110.9 m3 of dry saturated steam at 225 kN/m2 is contained in a rigid cylinder. If it iscooled at constant volume process until the pressure drops to180 kN/m2, determinethe following:a) mass of steam in the cylinderb) dryness fraction at the final stateSketch the process in the form of a P-v diagram.SolutionData: V1 = 0.9 m3 , P1 = 225 kN/m2 = 2.25 bar, P2 = 180 kN/m2 = 1.80 bara) Firstly, find the specific volume of dry saturated steam at 2.25 bar. Note thatthe pressure 2.25 bar is not tabulated in the steam tables and it is necessary to use theinterpolation method.From the Steam Tables,vg at 2.2 bar = 0.8100 m3/kgvg at 2.3 bar = 0.7770 m3/kgvg1 at 2.25 bar,v g1  0.8100 0.7770  0.8100  2.25  2.20 2.30  2.20 vg1  0.7935 m3/kg V1 Mass of steam in cylinder, m  (m3 x kg/m3) vg1 = 1.134 kg 19
  • 20. JJ 207 THERMODYNAMICS Ib) 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) 1(0.7935) x2  0.9774 P = 0.81 bar 1 v1 = v2 2.25 1.80 2 v m3/kg 0.7935 0.9774 20
  • 21. JJ 207 THERMODYNAMICS I TUTORIAL1. Each line in the table below gives information about phases of puresubstances. 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 i._____________ no longer rigidly structured so that they are free to move within a fixed volume. The molecules are closely bound, they are relatively dense and arranged in a rigid three-dimensional patterns so that they ii.____________ do not easily deform. 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 iii.____________ pattern. There is neither a fixed volume nor a fixed shape for steam.2. Write the suitable names of the phases for the H2O in the P-v diagram below. ( ii ) P ( iv ) ( vi ) (v) (i) T2 = const. T2 > T1 ( iii) T1 = const. v3. Answer question below:a. The internal energy of wet steam is 2000 kJ/kg. If the pressure is 42 bar, what isthe value of dryness fraction?b. Determine the specific volume, specific enthalpy and specific internal energy ofwet steam at 32 bar if the dryness fraction is 0.92.4. Find the dryness fraction, specific volume and specific internal energy ofsteam at 105 bar and specific enthalpy 2100 kJ/kg. 21
  • 22. JJ 207 THERMODYNAMICS I5. 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 650oC. Determine its specificvolume.10. A superheated steam at 24 bar and 500oC expands at constant volume until thepressure becomes 6 bar and the dryness fraction is 0.9. Calculate the changes in theinternal energy of steam. Sketch the process in the form of a P-v diagram. 22
  • 23. JJ 207 THERMODYNAMICS IDEFINITION OF PERFECT GASESDid you know, one important type of fluid that has many applications inthermodynamics is the type in which the working temperature of the fluid remainswell above the critical temperature of the fluid? In this case, the fluid cannot beliquefied by an isothermal compression, i.e. if it is required to condense the fluid, thencooling 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 perfectgas 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 aperfect 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 whichno 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 distinctionbetween these two terms but for our purposes, you may consider theminterchangeable. The principle properties used to define the state of a gaseous systemare pressure (P), volume (V) and temperature (T). SI units (Systems International) forthese properties are Pascal (Pa) for pressure, m3 for volume (although liters and cm3are 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 andCharles‟ Law.BOYLE’S LAWThe Boyle‟s Law may be stated as follows:Provided the temperature T of a perfect gas remains constant, then volume, V of agiven 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. P P  1/V 1/V Graph P  1/V 23
  • 24. JJ 207 THERMODYNAMICS IIf a gas changes from state 1 to state 2 during an isothermal process, thenP1 V1 = P2 V2 = constantIf the process is represented on a graph having axes of pressure P and volume V, theresults will be as shown in Fig. below. The curve is known as a rectangularhyperbola, having the mathematical equation xy = constant.PP1 1 PV = constantP2 23P3 V1 V2 V3 V P-V graph for constant temperatureExample 12A quantity of a certain perfect gas is heated at a constant temperature from an initialstate of 0.22 m3 and 325 kN/m2 to a final state of 170 kN/m2. Calculate the finalpressure of the gas.SolutionFrom equation P1V1 = P2V2   325 kN/m 2 P1 V2  V1 x  0.22 m 3     0.421 m 3  2 P2  170 kN/m CHARLES’ LAWThe Charles‟s Law may be stated as follows:Provided the pressure P of a given mass of gas remains constant, then the volume V ofthe 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 24
  • 25. JJ 207 THERMODYNAMICS I V1 V2   constant T1 T2If the process is represented on a P – V diagram as before, the result will be as shownin Fig. 3.2. P 1 2 0 V V1 V2 P-V graph for constant pressure processExample 13A quantity of gas at 0.54 m3 and 345 oC undergoes a constant pressure process thatcauses the volume of the gas to decreases to 0.32 m3. Calculate the temperature of thegas at the end of the process.SolutionFrom the questionV1 = 0.54 m3T1 = 345 + 273 K = 618 KV2 = 0.32 m3 V1 V2  T1 T2 V2  T2  T1 x V1  0.32 m 3   618 K    0.54 m 3      366 K 25
  • 26. JJ 207 THERMODYNAMICS IUNIVERSAL GASES LAWCharles‟ Law gives us the change in volume of a gas with temperature when thepressure remains constant. Boyle‟s Law gives us the change in volume of a gas withpressure if the temperature remains constant.The relation which gives the volume of a gas when both temperature and the pressureare changed is stated as equation 3.3 below. PV  constant  R Ti.e. P1V1 P2V2  T1 T2No gases in practice obey this law rigidly, but many gases tend towards it. An PVimaginary ideal that obeys the law is called a perfect gas, and the equation R Tis 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. Eachperfect gas has a different gas constant.The characteristic equation is usually written PV = RTor for m kg, occupying V m3, PV = mRTAnother form of the characteristic equation can be derived using the kilogram-mole asa unit. The kilogram-mole is defined as a quantity of a gas equivalent to m kg of thegas, where M is the molecular weight of the gas (e.g. since the molecular weight ofoxygen 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 asmole.Substituting for m from equations above PV = nMRT or PV MR  nT 26
  • 27. JJ 207 THERMODYNAMICS INow Avogadro’s hypothesis states that the volume of 1 mole of any gas is the same asthe volume of 1 mole of any other gas, when the gases are at the same temperatureand 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 theuniversal gas constant, and is given the symbol Ro.i.e. PV MR  Ro  or PV  nRoT nTor since MR = Ro then, R R o MExperiment has shown that the volume of 1 mole of any perfect gas at 1 bar and 1 oCis approximately 22.71 m3. Therefore from equation 3.8 PV 1 x 10 5 x 22.71 R0    8314.4 J/mole K nT 1 x 273.15From equation 3.10 the gas constant for any gas can be found when the molecularweight is known, e.g. for oxygen of molecular weight 32, the gas constant is Ro 8314.4 R   259.8 J/kg K M 32Example 140.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m2 and atemperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m2and 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 KSolutionFrom the questionV1 = 0.046 m3P1 = 300 kN/m2 P2 = 1.27 MN/m2 = 1.27 x 103 kN/m2T1 = 45 + 273 K = 318 K T2 = 83 + 273 K = 356 KR = 0.29 kJ/kg K 27
  • 28. JJ 207 THERMODYNAMICS IFrom equation PV = mRT P1V1 300 x 0.046 m   0.1496 kg RT1 0.29 x 318the constant volume process i.e. V1 = V2 P1 P2  T1 T2  P   3  T2  T1  2   318 1.27 x 10      1346 K   P1   300 SPECIFIC HEAT CAPACITY AT CONSTANT VOLUME (CV)The specific heat capacities of any substance is defined as the amount of heat energyrequired to raise the unit mass through one degree temperature raise. Inthermodynamics, two specified conditions are used, those of constant volume andconstant pressure. The two specific heat capacities do not have the same value and itis essential to distinguish them.If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise thetemperature of the gas by 1 degree whilst the volume of the gas remains constant, thenthe amount of heat energy supplied is known as the specific heat capacity at constantvolume, 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 = mCvdTFor a perfect gas the values of Cv are constant for any one gas at all pressures andtemperatures. 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 equationQ – W = (U2 – U1)mcv(T2 – T1) – 0 = (U2 – U1) (U2 – U1) = mCv(T2 – T1)i.e. dU = QNote:In a reversible constant volume process, no work energy transfer can take place sincethe piston will be unable to move i.e. W = 0. 28
  • 29. JJ 207 THERMODYNAMICS IThe reversible constant volume process is shown on a P-V diagram in Figure below: P P2 2 P1 1 V V1 = V2 P-V diagram for reversible constant volume processExample 153.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC untilthe temperature rose to 147 oC. If the gas is assumed to be a perfect gas, determine:a) the heat flow during the processb) the beginning pressure of gasc) the final pressure of gasGivenCv = 0.72 kJ/kg KR = 0.287 kJ/kg KSolutionFrom the questionm = 3.4 kgV1 = V2 = 0.92 m3T1 = 17 + 273 K = 290 KT2 = 147 + 273 K = 420 KCv = 0.72 kJ/kg KR = 0.287 kJ/kg Ka) From equation, Q12 = mCv(T2 – T1) = 3.4 x 0.72(420 – 290) = 318.24 kJ 29
  • 30. JJ 207 THERMODYNAMICS Ib) From equation, PV = mRTHence for state 1,P1V1 = mRT1 mRT1 3.4 kg x 0.287 kJ/kgK x 290 KP1    307.6 kN/m 2 V1 0.92 m 3c) For state 2,P2V2 = mRT2 mRT2 3.4 kg x 0.287 kJ/kgK x 420 KP2    445.5 kN/m 2 V2 0.92 m 3SPECIFIC HEAT CAPACITY AT CONSTANT PRESSURE (CP)If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise thetemperature 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 atconstant 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 havedQ = mCpdTFor a perfect gas the values of Cp are constant for any one gas at all pressures andtemperatures. Equation above can then be expanded as follows:Heat flow in a reversible constant pressure process Q = mCp(T2 – T1)RELATIONSHIP BETWEEN THE SPECIFIC HEATSLet a perfect gas be heated at constant pressure from T1 to T2. With reference to thenon-flow equation Q = U2 – U1 + W, and the equation for a perfect gasU2 – U1 = mCv(T2 – T1), hence, Q = mCv(T2 – T1) + WIn a constant pressure process, the work done by the fluid is given by the pressuretimes the change in volume, i.e. W = P(V2 – V1). Then using equation PV = mRT, wehave W = mR(T2 – T1)Therefore substituting,Q = mCv(T2 – T1) + mR(T2 – T1) = m(Cv + R)(T2 – T1) 30
  • 31. JJ 207 THERMODYNAMICS IBut for a constant pressure process from equation before, 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 + RAlternatively, it is usually written as R = Cp - C vSPECIFIC HEAT RATIO ()The ratio of the specific heat at constant pressure to the specific heat at constantvolume is given the symbol  (gamma), Cpi.e. = CvNote that since Cp - Cv= R, from equation above, it is clear that Cp must be greaterthan Cv for any perfect gas. It follows therefore that the ratio Cp/Cv =  , is alwaysgreater than unity. In general,  is about 1.4 for diatomic gases such as carbonmonoxide (CO), hydrogen (H2), nitrogen (N2), and oxygen (O2). For monatomic gasessuch as argon (A), and helium (He),  is about 1.6, and for triatomic gases such ascarbon 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= RDividing through by Cv Cp R 1  Cv Cv CpTherefore  = , then, Cv 31
  • 32. JJ 207 THERMODYNAMICS I R  1  Cv R Cv  (  1)Also, Cp = Cv hence substituting in equation above, R Cp = Cv = (  1) R Cp = (  1)Example 16A certain perfect gas has specific heat as follows Cp = 0.846 kJ/kg K and Cv = 0.657 kJ/kg KFind the gas constant and the molecular weight of the gas.SolutionFrom equation R = Cp - Cvi.e. R = 0.846 – 0.657 = 0.189 kJ/kg Kor R = 189 Nm/kg K R0From equation M = R 8314i.e. M=  44 189 32
  • 33. JJ 207 THERMODYNAMICS INON-FLOW PROCESSESA process occurs when a system‟s state (as measured by its properties) changes forany reason. Processes may be reversible or actual (irreversible). In this context theword „reversible‟ has a special meaning. A reversible process is one that is whollytheoretical, but can be imagined as one which occurs without incurring friction,turbulence, leakage or anything which causes unrecoverable energy losses. All of theprocesses considered below are reversible and the actual processes will be dealt withlater.Processes may be constrained to occur at constant temperature (isothermal), constantpressure, constant volume, polytropic and adiabatic (with no heat transfer to thesurroundings).Constant temperature (Isothermal) process (pV = C) If the change in temperature during a process is very small then that processmay be approximated as an isothermal process. For example, the slow expansion orcompression of fluid in a cylinder, which is perfectly cooled by water may beanalysed, assuming that the temperature remains constant. P 1 W 2 W v v1 Q v2 Constant temperature (Isothermal) process The general relation properties between the initial and final states of a perfectgas are applied as follows: p1V1 p 2V2  T1 T2If the temperature remains constant during the process, T1 = T2 and the above relationbecomes p1V1  p2V2 33
  • 34. JJ 207 THERMODYNAMICS I From the equation we can know that an increase in the volume results in adecrease in the pressure. In other words, in an isothermal process, the pressure isinversely proportional to the volume. Work transfer: Referring to the process represented on the p – V diagram in Figure above it isnoted that the volume increases during the process. In other words the fluid isexpanding. The expansion work is given by 2 W   pdV 1 2 c = dV (since pV = C, a constant) 1 V 2 dV = c 1 V 2 dV = p1V1  1 V V  larger vol ume  = p1V1 ln 2  smaller volume  V1   V2 = mRT1 ln (since p1V1 = mRT1) V1 p V2 p = mRT1 ln 1 (since  1) p2 V1 p2 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 ofwork transfer. Since this is an expansion process (i.e. increasing volume), the work is doneby the system. In other words the system produces work output and this is shown bythe 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 34
  • 35. JJ 207 THERMODYNAMICS ISubstituting in the energy balance equation, Q=W Thus, for a perfect gas, all the heat added during a constant temperatureprocess 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 intoor out of the system. Such a system is thermally isolated and a process within thatsystem may be idealised as an adiabatic process. For example, the outer casing ofsteam 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. P 1 W 2 W v v1 Thermal insulation v2 Adiabatic (zero heat transfer) process For a perfect gas the equation for an adiabatic process is pV = C Cp where  = ratio of specific heat = Cv The above equation is applied to states 1 and 2 as: p1V1  p2V2 35
  • 36. JJ 207 THERMODYNAMICS I  p 2  V1    p1 V2 Also, for a perfect gas, the general property relation between the two states is givenby the equation below p1V1 p 2V2  T1 T2 By manipulating 2 equations above the following relationship can bedetermined:  1  1 T2  p 2   V      1 T1  p1   V2 By examining the equations the following conclusion for an adiabatic process on aperfect 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 thevolume increases during the process. In other words, the fluid expanding and the expansion work is given by theformula: 2 W   pdV 1 2 c =  dV (since pV = C, a constant) 1V 2 dV = c  1 V p V  p 2V2 = 1 1 [larger pV- small pV]  1 Note that after expansion, p2 is smaller than p1. In the p – V diagram, theshaded 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 bythe system. In other words, the system produces work output and this is shown by thedirection of the arrow representing W. 36
  • 37. JJ 207 THERMODYNAMICS I 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 ininternal energy. In other words, because of the work output the internal energy of thesystem 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 R cv =  1 mR(T1  T2) W  1 But, mRT2 = p2V2 and mRT1 = p1V1 Then the expression for the expansion becomes p1V1  p 2V2 W  1Example 17In an industrial process, 0.4 kg of oxygen is compressed isothermally from 1.01 barand 22o C to 5.5 bar. Determine the work done and the heat transfer during theprocess. Assume that oxygen is a perfect gas and take the molecular weight of oxygento be M = 32 kg/kmole.SolutionData: m = 0.4 kg; p1= 1.01 bar; t1= 22oC p2 = 5.5 bar; W=? Q=? 37
  • 38. JJ 207 THERMODYNAMICS IFrom the equation R R= 0 M 8314 = 32 = 260 J/kgK = 0.260 kJ/kgKFor an isothermal processWork input, p W = mRTln 2 p1 5.5 = 0.4 x 0.260 x (22  273) ln 1.01 = 52 kJIn an isothermal process all the work input is rejected as heat.Therefore, heat rejected, Q = W = 52 kJExample 18In a thermally insulated reciprocating air compressor, air at 0.98 bar and 20oC iscompressed into one sixth of its original volume. Determine the pressure andtemperature of the air after compression. If the compressor cylinder contains 0.05 kgof air, calculate the required work input. For air, take  = 1.4 and cv = 0.718kJ/kgK.Solution Data : p1 = 0.98 bar; T1= 20 + 273 = 293 K V2 1  ; m = 0.05 kg; W = ? V1 6 As the cylinder is well insulated the heat transfer is negligible and the processmay be treated as adiabatic. Considering air as a perfect gas  p 2  V1  From equation,   p1 V2  p2 = 0.98 x 61.4 = 12 bar 38
  • 39. JJ 207 THERMODYNAMICS I  1 T V  From equation 2   1  T1  V2  T2 = 293 x 60.4 = 600 K = 327oCfor an adiabatic compression processW = mcv(T2-T1) [larger T- smaller T] = 0.05 x 0.718 (600-293) = 11 kJPolytropic process (pVn = C) This is the most general type of process, in which both heat energy and workenergy 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 cylinderassembly is cooled perfectly, then the process will be isothermal. In this case theindex n = 1. If a compression or expansion is performed rapidly, and if the piston cylinderassembly is perfectly insulated, then the process will be adiabatic. In this case theindex n = . If a compression or expansion is performed at moderate speed, and if thepiston cylinder assembly is cooled to some degree, then the process is somewherebetween those discussed above. Generally, this is the situation in many engineeringapplications. 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) 39
  • 40. JJ 207 THERMODYNAMICS I P 1 pVn=C W P1 2 P2 W v1 v2 v Qloss Polytropic processAt states 1 and 2: p1V1n  p2V2n or n p 2  V1    p1 V2 Also, for a perfect gas, the general property relation between the two states is givenby p1V1 p 2V2  T1 T2 By the manipulation of 2 equations above the following relationship can bedetermined: n 1 n 1 T2  p 2  n V      1 T1  p1   V2 By examining the equations the following conclusions for a polytropic process on aperfect 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 thevolume increases during the process. In other words the fluid is expands and the expansion work is given by 40
  • 41. JJ 207 THERMODYNAMICS I 2 W   pdV 1 2 c = n dV (since pVn = C, a constant) 1V 2 dV = c n 1 V p V  p 2V2 = 1 1 [larger pV- small pV] n 1 Note that after expansion p2 is smaller than p1. In the p – V diagram, theshaded area under the process represents the amount of work transfer. Since this is an expansion process (i.e. increase in volume), the work is doneby the system. In other words, the system produces work output and this is shown bythe direction of the arrow representing W. Heat transfer: Energy balance is applied to this case as: U1 – Qloss - W = U2 Qloss = (U1 – U2) – Wor W = (U1 – U2) - QlossThus, in a polytropic expansion the work output is reduced because of the heat loses.Example 19The combustion gases in a petrol engine cylinder are at 30 bar and 800oC before V 8 .5expansion. The gases expand through a volume ratio ( 2 ) of ( ) and occupy 510 V1 1cm3 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 thework output?Solution V2 = 510 cm3 p2 = ? P1= 30 bar t2 = ? t1 = 800oC Qloss W State 1 State 2 41
  • 42. JJ 207 THERMODYNAMICS I Data: p1 = 30 bar; T1 = 800 + 273 = 1073 K; n = 1.15 V2 = 8.5; V2 = 510 cm3; V1 t2 = ? p2 = ? W=?Considering air as a perfect gas, for the polytropic process, the property relation isgiven as: n 1 V  T2  T1  1   V2  1.151  1  = 1073x    8.5  = 778.4 K = 505.4oC n V From equation p 2  p1  1  V 2  1.15  1  = 30 x    8.5  = 2.56 bar Now, V2 = 510 cm3 = 510 x 10-6 m3 and, V2 = 8.5 V1 Then, 510 x10 6 V1  8.5 = 60 x 10-6 m3 Work output during polytropic expansion is given as: p1V1  p 2V2 W = [larger pV- small pV] n 1 (30 x10 5 )(60 x10 6 )  (2.56 x10 5 )  (510 x10 6 ) = 1.15  1 = 330 J = 0.33 kJ 42
  • 43. JJ 207 THERMODYNAMICS IConstant volume processIf the change in volume during a process is very small then that process may beapproximated as a constant volume process. For example, heating or cooling a fluid ina rigid walled vessel can be analysed by assuming that the volume remains constant. Q p 2 p 1 1 2 v v Q a) Heating b) Cooling Constant volume process (V2=V1)The general property relation between the initial and final states of a perfect gas isapplied as: p1V1 p 2V2  T1 T2If the volume remain constant during the process, V2 = V1 and then the above relationbecomes p1 p 2  T1 T2or T2 p 2  T1 p1From this equation it can be seen that an increase in pressure results from an increasein temperature. In other words, in constant volume process, the temperature isproportional to the pressure. Work transfer:Work transfer (pdV) must be zero because the change in volume, dV, during theprocess 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 – U1gives Q – 0 = U2 – U1i.e. Q = U2 – U1 43
  • 44. JJ 207 THERMODYNAMICS IThis result, which is important and should be remembered, shows that the nett amountof heat energy supplied to or taken from a fluid during a constant volume process isequal to the change in the internal energy of the fluid.5.3 Constant pressure processIf the change in pressure during a process is very small then that process may beapproximated as a constant pressure process. For example, heating or cooling a liquidat atmospheric pressure may be analysed by assuming that the pressure remainsconstant. P W p 1 2 W v v1 v2 – v1 Q v2 Constant pressure processConsider the fluid in the piston cylinder as shown in Figure above. If the load on thepiston is kept constant the pressure will also remain constant.The general property relation between the initial and final states of a perfect gas isapplied as: p1V1 p 2V2  T1 T2If the pressure remain constant during the process, p2 = p1 and then the above relationbecomes V1 V2  T1 T2or T2 V2  T1 V1From this equation it can be seen that an increase in volume results from an increasein temperature. In other words, in constant pressure process, the temperature isproportional to the volume. 44
  • 45. JJ 207 THERMODYNAMICS I Work transfer:Referring to the process representation on the p-V diagram it is noted that the volumeincreases during the process. In other words, the fluid expands. This expansion workis given by 2 W   pdV 1 2  p  dV (since p is constant) 1 = p (V2 – V1) (larger volume – smaller volume)Note that on a p-V diagram, the area under the process line represents the amount ofwork 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 – U1or Q = (U2 – U1) + WThus part of the heat supplied is converted into work and the remainder is utilized inincreasing 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 + pVHence Q = H2 – H1 (larger H – smaller H) 45
  • 46. JJ 207 THERMODYNAMICS IExample 20The specific internal energy of a fluid is increased from 120 kJ/kg to 180 kJ/kg duringa constant volume process. Determine the amount of heat energy required to bringabout 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 212.25 kg of fluid having a volume of 0.1 m3 is in a cylinder at a constant pressure of 7bar. Heat energy is supplied to the fluid until the volume becomes 0.2 m3. If the initialand 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 fluidSolution Data: 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 46
  • 47. JJ 207 THERMODYNAMICS Ib) 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 equation Q – W = U2 – U1 gives U2 – U1 = 157.5 – 70 = 87.5 kJ 47
  • 48. JJ 207 THERMODYNAMICS ITUTORIAL 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 PV  R . T 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 48
  • 49. JJ 207 THERMODYNAMICS I iv.specific heat ratio6. 1 m3 of air at 8 bar and 120 oC is cooled at constant pressure processuntil 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 thepressure is kept constant at 125 kN/m2 while the volume changes from 0.20 m3to 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 containeris 1 kg and the volume of gas is 0.6 m3. 100 kJ is added as heat. If gas hasCv = 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 527 o Cand 20 bar expands isothermally to a pressure of 1.4 bar. What is the finalvolume of the gas? Take R = 189 Nm/kgK for carbon dioxide.10. 1 kg of nitrogen (molecular weight 28) is compressed reversibly andisothermally from 1.01 bar, 20oC to 4.2 bar. Calculate the work done and theheat flow during the process. Assume nitrogen to be a perfect gas.11. Air at 1.02 bar, 22oC, initially occupying a cylinder volume of 0.015m3, is compressed reversibly and adiabatically by a piston to a pressure of 6.8bar. Calculate the final temperature, the final volume, and the work done onthe mass of air in the cylinder.12. 0.112 m3 of gas has a pressure of 138 kN/m2. It is compressed to690 kN/m2 according to the law pV1.4 = C. Determine the new volume of thegas.13. 0.014 m3 of gas at a pressure of 2070 kN/m2 expands to a pressure of207 kN/m2 according to the law pV1.35 = C. Determine the work done by thegas during expansion.14. A cylinder containing 0.07 kg of fluid has a pressure of 1 bar, a volumeof 0.06 m3 and a specific internal energy of 200 kJ/kg. After polytropic 49
  • 50. JJ 207 THERMODYNAMICS Icompression, the pressure and volume of the fluid are 9 bar and 0.011 m3respectively, 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 thecompression.15. The pressure of the gas inside an aerosol can is 1.2 bar at a temperature oof 25 C. Will the aerosol explode if it is thrown into a fire and heated to atemperature of 600o C? Assume that the aerosol can is unable to withstandpressure in excess of 3 bar.a. 0.05 kg of air, initially at 130o C is heated at a constant pressure of 2bar until the volume occupied is 0.0658 m3. Calculate the heat supplied andthe work done.b. A spherical research balloon is filled with 420 m3 of atmospheric air ata temperature of 10o C. If the air inside the balloon is heated to 80oC atconstant pressure, what will be the final diameter of the balloon? 50

×