8 assere se 8 e PART-A SHORT QUESTIONS WITH SOLUTIONS ~ pune tho term PUTO Substance with examples 1 ‘yutanet ‘ ie tance is either ed a The phe on oe Substance ot a mixture of substances, which have the same chemical composition pis MASS. ' be different, but there is no change in molecular structure. pa water or mixture of z ai Sept Sr tatencan OF Water and steam are pure substances. But, mixture of ai and steam, mixture of ait and Wat do you understand by triple point? yer! gipeoint : ‘he state in which liquid solid and vapour phases ofa substance co-exists in equilibrium is called the triple point ofthat i sane Figure shows the P-T diagram containing the triple point. tt Model Papert, 1(0) Figure: PT Diagram The triple point can also be defined as the point of intersection of melting curve, vapourization curve and the itiaton curve: 1 What Is the difference between critical point and triple point? tnover 1 ; bterence between the Critical Point and Triple Point Critical Point | L]) Wis defined as a limiting state where the latent | | heat of vapourization becomes zero on increasing, the pressure, 21 Only process possible below critical point is ‘Sublimation process. “ ‘Triple Point ] 1] tis defined as the point where solid, liquid and vapour phases co-exist in equill From triple point every process is possible i.e., melting sublimation and vapoutization,Q4. What is normal bolling point and oxplain its significance? Answer : ’ Normat Boiling Point Itis the temperature at which the vapour pressure iS equal to 760 mm of Hg at sea level. This ean be achieved by adding a non-volatile solute to. liquid which reduces liquid vapour pressure and by increasing the temperature greater thant normal boiling point, Atmospheric boiling point is also called as normal boiling point. At this temperature, liquid and vapour [pressure becomes adequate to avoid atmospheric pressure which lifts the liquid to form bubbles. ; 1. The temperature and pressures of a vapour depends on each other i.e, higher the temperature, higher the vapour pressure It is useful in evaluating the saturation temperature and saturation pressure of a gas. Q5. Isittrue that water boils at higher temperatures at higher pressures? Explain. Answe Nov/Dee.18, (R16), 10) ‘Yes,-water boils at higher temperature at higher pressures. From the phase diagram of water, the line between the liquid pahse and vapour phase indicates the boiling point corresponding to different pressure and temperature combination. As the line moves upwards, the pressure and temperature are also increasing. This shows that the water boils at higher temperatures at higher pressures. CZ Figure Sion irae dpetice Mc Q6. Write the Clausius Clapeyron equation and its significance. Now Dee.16, (R15), a1(0) OR Write the clapeyron equation and point out its utility. Answer = * NovJDec.-18, (R16), Q7(a) The clausius ~ clapeyron equati moDYNAMICS 8 THER! Where, inthalpy of Vapoutisation n= 5h tot e ¥,7 ‘av~ specific volume of saturate, ‘re Saturation temperature of vapoy, _ slope of saturation curve, ance Oe ant equation is used f0 calculate the ined from rate of h®h, peyronith the data obtained € of yration, Wi shane of ee ressure with tempera é 7, piscuss the Importance of Clauisius cj av. equation. layer, Answer Model Papert 0 DE¥6. stance of Clausius Clayperon Equation is of system containing two, rtisused in the analysis 0 ina state of equilibrium. Pag, from rate of change oF prey used to determine the eta t ‘Impor Li ‘The data obtained with temperature is evaporation. 4, Atany temperature, the vapour Pressure of liquid Fe determined using this equation. +4, Itis applicable for any phase change in a syitem, ‘The slope of sublimation curve in a phase transitoncy be obtained. Answer Mollier Chart Mollier diagram is a graphical representation of te steam tables, Its also named as enthalpy-entropy (is) diagna, in which the enthalpy (h) is plotted along the ordinate and te ‘entropy (s) along abscissa. Enthalpy and entropy of waterani dry saturated steam, for any particular pressure are obtained from the steam tables. These values of enthalpies and entrops are plotted and then liquid line and dry saturated line i obtained Both these lines meet at a point C (ie., critical point).Fr wae Substances and Proportos of Gases Go the Isobars on Mol & on one another? tpMothe diagram, a8 the pressure of the ud ince Ss 280, jon temperature corresponding tit als increases ay rte figure. Therefor, itis clear from the diagram thet Trews on the mole diagram diverge from one another, x baEethalpy) —— Git. Why do the Isobars on Mollier diagram diverge from one another? Why do isotherms on Mollior diagram become horizontal in superheated region at low pressures? answer t bars on Mollier Diagram Foranswer refer Unit-ll, Q9. otherms on Mollier Diagram NowsDec.-17, (R16), 140) Entep.s —> Figure: Enthalpy-Entropy Diagram of Water ‘As entropy increases beyond saturated vapour line, tathalpy remains constant thereby isotherms becomes ferizontal, because temperature isa function of enthalpy. Qt. Whatdo you understand by a ‘thermodynamic process’? Answer ¢ ‘Thermodyniamics is a branch of science that deals with tte conversion of energy from one form to other. The process ‘ofchange of energy is known as thermodynamic process. The ‘bermodynamic process is classified as follows, (i) " Non-flow processes Flow processes. 3.3, Q12. Whatare the different types of thermodynamic In gases? Model Papers, (f) The different types of thermodynamic processes are as follows, ()_Isochorte Process In | this process, the volume remains constant ‘and hence i also known as constant volume process. (ii) Isobarie Process In this process, the pressure remains constant and hence, itis also known as constant pressure process. (il) Isothermal Process In this process, the temperature remains constant and hence, itis also known as hyperbolic process. (iv) Adiabatic Process In this process, there is no heat energy transfer takes lace in or out of the system. This process is also wn as isentropic process. Polytropic Process In this process, both work and heat energy transfer takes place across the boundary of the system. o) Q13. Represent the following expansion process. ‘on P-V diagram, {a)Constant pressure process (b)Adiabatic process and (c)lsothermal process. Figure: P-V Diagram pr Se Oo ire ee ee Q14. Represent the following processes on the ‘same temperature-entropy diagram: (a) Constant volumé process (b) Constant pressure process (c) Adiabatic process SPECTROM ALL-IN-ONE JOURNAL, FOR ENGINEERING STUDENTS si |A GROUPwe process, constant |e uiabatic process on a TS dingrary, ee ss a : = pv ppreasuie proces Q15. Distinguish betweon adiabatic and isentropic proces! Answer t Noel Peree a oy The process in which there is no heat transfer to and | 1. | The process is which there fs no change in entropy ig . Se el : | from the fluid is defined as an adiabatic process. defined as an isentropic process; | [An adiabatic process can be either reversible or | 2, | Isentropic process isan reversible adiabatic process. || or irreversible process, ae Foran adiabatic process, there should be availability] 3. } This process require no irreversibilities and ‘of perfect thermal insulation forthe system, it can perform as an ideal process. | For an adiabatic process, energy is transferred as | 4, | Foran isentropic process, the energy is transferred as heat, Q=0. eae heat from the surroundings to the system is given by, | dQ=Tds ‘ | ‘Whar i Q-Heat transfer | T= Temperature | Entropy. , | Q16. What is steam calorimetry? List the methods used for measuring dryness fraction of steam. oR Discuss the various methods for determining the dryness fraction of steam with relative merits and | | Answer! ¥ Nov.45, R13, 018. Steam Calorimetry . | The principle of measuring the dryness fraction of steam is termed as steam calorimetry, : ‘Steam calorimeter is used to measure the dryness fraction of steam and it can be explained by any one ofthe fallowlt | methods. r (H:- Throttling calorimeter (ii) Bucket calorimeter (iii) Separate - throttling calorimeter. : : 5 i {,dok forthe tAGROUP Loco. Meontne TITLE COVER did the WATERMARK PAGE lisitie before you |iris defined as the gas wh, ich does evat forces of attraction. Generally Ot have any sero allthe gas laws at, fred ideal pre peat. BES of pressures eal gxs obeys the law, PY =RT ule’ In shows that internal : wore its femperature, “BY OF an ideal gas yeh u.states thatthe “change in internal ener i BY ofa perfe gsdiecly oporonal tothe change of temperate ee eat dU «dT «dU = mCdT dU=mCT,-T) ‘Where, m~ Mass of one kg of gas ” T, ~ Final temperature of gas T, ~ Initial temperature of gas C-Constant, {if Define equation of a state for a perfect gas. saswer : mise The functional relationship between the independent sopertes such as pressure (P), temperature (7) and specific ‘lune (v) of pure substance is known as equation of state. Mabematically itis expressed ag, APT.) =0 Inthe above expression, if two of the properties are ‘town then the third property can be determined using equation se, {i8, Explain the non flow process. Maswer : [Model Paper-tt, Q1(f) | Nov/Dec. 16, (R15), Q1(el] ___ The thermodynamic process in which the matter (or) {itrotatoned to ow in rout ofthe system icaled 2 How process, Generally all non-flow processes occur only ‘chsed systems, In this process flow energy is neglected. The EM of Control surface” plays a major role than concept trol yolume”, in non-flow process. Examples of non- Process are air heated in a closed room, piston-cylinder Seenent, a tea cup with a lid on it PERG aux. ON JOURNAL FOR ENGINEERD 20. List out the various non flow proc the first law of thermodynamics is applied to Closed system. Answer . A nnon-flow process is considered for a closed system, ‘here work and heat interactions takes place. A gas in a closed system can undergo the following on-flow processes, @ Constant pressure process ( P = Constant) Gi Constant volume process (V = Constant) (ii) Constant temperature process (T= Constant) (iv) Adiabatic process (PV"= Constant) (®)_Polytropffic process (PV* = Constant) Q21. Determine the pressure of saturated steam at 40°C if at 35°C the pressure is 5.626 kPa, the ‘enthalpy of vaporization is 2418.6 kJ/kg and the Specific volume is 25.22 m'/kg. The enthalpy of vaporization is essentially constant over this temperature range? Use Claussius Clayperon equation. Answer : Given that, Final temperature, T, = 40°C = 40+ 273 = 313k Initial temperature, 7, = 35°C = 35 + 273 = 308k . Initial pressure, P, = 5.628 kPa Specific enthalpy of vaporization, 1, = 2418.6 ki/kg, Specific volume, v, = 25.22 m'skg The Clausius-Clapeyron equation is given as, Pa) _ hye ( tod Fa). Mef tod fl a(t t A). 2i86( 1b 0.287 (308 . 313 5.628. Since R= 0.287 kiikg-k, ee.-14, (R13), Q7(0) In P, | . 5.186 « wf) 8427.17 x 5.186 « 10%) P, —2.| = “lees wer P, : Pee age cr ease P, = 5.628 * 1,548 = 8.712 kpa P= 8.712 kpa (or) 0.087 bar ‘Therefore, the pressure of saturated steam at 40°C is 8.712 kpa i : =NAMICS LJNTU-HYDERa, MODY! te On i nee eS PART-B UTIONS ESSAY QUESTIONS WITH SOU 3.4, PURE SUBSTANCES Moltier Charts 3.4.4 PAV-T Burtaces, T-s and hes Diagrams ram fOF 2 Pure subg _ Temperature) diagram f* ta 22. Whats a pure substance? Draw and explain P-T(Preasuie ~ Tome Motes Pee Answers Pore Suttance 4, which have the same chemical compo, A single substance or a mixture of substance : region ts baa freee be ateat et she molecular structure doesnot Vary Example * Ait, steam, water or mixture of water and steam are pure sul Jagd are not pare substances. ‘Mathematically, a pure substance can be written as, Parry) Fefen T=fP) Where P, Tand rare the intensive properties. P-T Dingram for a pure substance ‘ en ‘The phase changes taking place between three phases of water are shown by curves separating the solid, liquid and vapa, phases. The P-T diagram is also known as phase diagram, tarsi yo sutances, But, mixture of air and steam and mixture fyi oP Pressure Temperate TT, er Figure: P-T Diagram for a Pure Substance \Vapourization curve separates the liquid phase from the vapour phase, The points on this curve represents the equiv of liquid and vapour phase. This curve is also known as boiling point curve. At critical point vapourization terminates and there is no distinto® between liquid and vapour phases. Above erica pint, there willbe a continuous change of state without phase transom ‘The slope of vapourization curve is always positive, Thus the boiling point ofthe liquid water wil i ith incre? {nthe pressure. Melting curve separates solid phase from the liquid phase. This curve is ae isch Saas or freezing point curve. The point on this curve represents the equilibrium of solid and liquid. ba eee Ail these three curves meets ata point, which is known as triple point and itr sol “as : z =presents the state of equilibrium of 3 it 0.01 = 4 ee The triple point of water is at eparce 0.6113 kpa, pressure and temperature at critical state of water are 37 Look for the SUA ‘GROUP LOGO {7} on the TITLE COVER arid the WATERMARK PAGE iriside before you DYY3.7 and ET 0, and discuss the comparisons, ec 14, (R43).O7(a) ‘Surface of CO, Figure (1) Where, C-Critical point C~Critical point» TPL - Triple point line ‘TPL = Triple point line LV-~ Liquid and vapour = Constant ii SV Solid and vapour CPL ~Constant pressure line LV - Liquid and vapour SV ~ Solid and vapour 1, | Itrepresents expansion of water upon frezing.| 1. | t represents contraction of CO, upon frezing. 2. | The slope of fusion curve is negative. 2. | The slope of fusion curve is positive. 3. | Triple point temperature of Water is 273.16 k. | 3. | Triple point temperature of CO, is 216.55 k. 4, | Triple point pressure of water is 610 Pa. 4, | Triple point pressure of CO, is 517 * 10° Pa, 5, | The slope ofthe sublimation and vaporization | 5. | The slope ofthe sublimation and vaporization curve curve of water is positive. of CO, is also positive. 6, | The solid-liquid surface slope is opposite to | 6. | The solidtiquid surface slope is also opposite to that that of CO, of water {d, Draw the phase equilibrium diagram for a pure substance on T-s diagram with constant property li Asner:3.8 PV diagram t0 nbiamned by platting many number of ‘sorberma lines ae shown in figure (1) Consider the expansion oF Fqui fioen the point F to the point 4 theough the path 1234 The watet between the states 1-2 is an unsaturated fiquid. At Point 21s known as saturated quid, Hetween the point 2103 tte mnivtere of Trquid and vapour known as wet vapour, Point S represents a saturated vapour and fim the point 3 as it moves Yo point ite known as uncaturated oF supetheated vapen Ieimall the isothermal processes the states representing the saturated figud and vapour are joined, a curve ABC is obsained. The curve AB is the saturated liquid line and the Fevtion BC is the saturated vapour line, The meeting ofthe two tines AB and BC takes place at the ctitical point. Therefore, critical point gives the highest temperature and pressure where both the tiquti and vapour phases can coexist, ___ Figore (2h T-s Diagram for Water TS diagram is obtained by plotting many constant pressure process as thown in figure (2). if all the saturated Jicquid and vapour points of all the constant pressure processes ‘ae joined, we get the curve ABC which isa combination of two ‘curves AB and BC. Therefore, the change in entropy between. the two saturated points decreases as pressure increases and finally becomes zero at the critical point. 25. Draw and explain P-T diagram for H,O and label various phases. Answer : P-T Diagram for H,O Nov-10, Set2, a4(a) DYNAMICS WA HYDEn, of water om # emp oO follows. . wae rom —10°C (0 0°C (1-2) t (hy, eating jean 0°C solid 10 Water ay ‘ Melting ate ron OC 10 100°C (3%) atin ater at 100°C 10 SEEAM At 109%, fea 10 250°C (5-6) be reversed FOM SEG 9 yyy a qi) He cr ww (wy Heating of Process will cooling ough the points (2, 3) jing throug! ) is ey, Curve pass he passing through the points (4, s);, 4, vicapourization curve” os apr pressure of a Solid 1S MASUEd a gp If vapour Pitted, “sublimation curve” jg Obtain temperature and is plotted, ‘The point where fusion CUrVe, vapoutization sublimation eurve meets is termed as “triple point”, ‘Slopes of sublimation and eppatizato cues positive for al substances, Slope of fusion Curve for mat Xe substances is postive, but for wate it is negative, Significance of Vapourization Curye Itgives the saturation temperatures at different and it shows that the temperature at which liquid bois jg sensitive to pressure Significance of Fusion Curve Its small slope indicates that, the temperature a ny ‘a solid melts is not a strong function of pressure, Values at Triple Point for Water Pressure = 4,58 mm of Hg and ‘Temperature = 273.16 K. Q26. Explain Moliier diagram (h-s diagram). Model Papera a5 wea, Answer = Mollier Diagram Mollier diagram is a graphical represent: of te steam tables. Iis also named as enthalpy-entropy (h-s)digran in which the enthalpy (h) is plotted along the ordinate and entropy (8) along abscissa, Enthalpy and entropy of waters dry saturated steam, for any particular pressure are obuit from the steam tables. These values of enthalpies and entropit are plotted and then liquid line and dry saturated line is obainet Both these lines meet at a point C (j.e., critical point) sid H ; om] + // Teves | Subteation | - repon TIPCOT TOE T. Figure: P-T Diagram for H, ea ae a Bang) 2 : Look for the SIA GROUP Loco Figure: Mollier onthe TITLE COVER ahd the WATERMARK PAGE inside before you™‘expansion and compre’ er ewe portions, bya lines peer (ie Wet) TEBION, the emg eS “Saturat fp a given Pressure. In the faa team ee temperature Of steam increagen nH 5 one Moller diagram has the fo Constant pressure lines Constant temperature lines © Syyness fraction lines Joothermal lines tsentropic lines ‘Throttling lines. Show that on a Mollier a the slope of a constant pre, * with temperature in the s inereann ‘perheated region, According to first and second law dh =Tds + Pay Itean also be written as, (2) -7 Equation (1) forms base for Mol lowing Be @ @ of thermodynamics, =) a Mier diagram. It can be dexlyseenthat | Gp Ji slope of constant pressure line on elie diagtam. : : Change in enthalpy for superheated vapour is given by, aC, 0,,,-T.) ‘Change in entropy, Frup \” ds =C,1n| cae Then, substituting the above values in equation (1), (3) Cy Coup Tear) Teun ~ Tea’ * 4) (Tr) (Tr Brig sur — owe) (2) Since, 7, is greater than T., from the above equation its, ett, the slope of pressure line increases withthe inerease in ‘tetenperature of superheated vapour at constant pressure. PPPS Figure: hs Diagram ALL-IN-ONE JOURNAL FOR ENGINEERING STUD! ; 1.2 Pha: at cr Transformations ; Triple Point fitical State Properties During Change of Phase, Dryness Fraction 228. Describe the process of formation of steam and give its graphical representation. Answer : To study the phase transformation of water from solidice 10 steam, consider a certain mass of ice is at a temperature less than 0°C. This ice is kept in a cylinder and pressure is applied ‘n it, Now, heat is supplied slowly to the cylinder containing ice. Then, the phase transformation of ice takes place and it is shown in figure (1). {en oe ; [een : Ln ol A ea Tewee ¢ | rate ead igure (2) Figure (2), shows the graph drawn between temperature and heat supplied, Point-1, in the graph shows the initial temperature of the ice before the heat is supplied. Due to the supply of heat the temperature of ice increases to 0°C. This state is represented by point:2. The line *1-2" shows the supply of +heat. On further heating of ice, the temperature of ice remains constant, but the ice changes’to water at 0°C. The total heat supplied is utilized for changing the state of ice. Point-3, represents the water at 0°C. The line *2-3° represents heat of fusion. On further heating, the temperature of water,increases upto 100°C. Point-4, on the graph represents water at 100°C. The line ‘3-4” shows the heating process of water. Further heating the water temperature remains constant until the water gets vapourized. The heat required to vapourize water isknown as latent heat of vaporization. The line *4-5° represents the process of conversion of water to steam. Point-5, represents saturation condition of steam. The steam gets superheated on further heating. Point-6, represents superheated steam. With the increase in pressure the boiling point of water increases, but the latent heat of vapourization decrease: SIA GROUP g‘using 4 T-v diagram, Anawer Tos understand the concept of phase equilibrium of Pare ewbetance on temperature epecifie volume diagram, ssomender 1 hg of sce exostined in a cylinder ata temperate 40°C moaiotained at atmospheric pressure, When the hroat t sunt 16 the cylinder, tansformation of solid state ‘bs guid state occurs in following steps and is shown in figure. Daring the first state, ie is iniilly’at sti state (8) which 4s shown in figure (1). Ar this stage, the temperature of ice weat-are 0 @ Gil) w) tv) wi) ‘Figore (1): Water in Different Thermodynamic States. ‘When the temperature of water increases from ~40°C 100°C the ice starts melting and changes its phase from “8 to “Fas shown in figure 1(b). If the heat is supplied continuously, the phase ‘s" changes 101, at constant temperature (0°C) and the transformation ‘of solid to liquid occurs. At this stage, the specific volume ‘of mixture (s+ .) decreases, as shown in figure 1(c) During the process /, and /,, the temperature of water increases from 0°C to 100°C. Increase in the temperature shout 4°C causes the specific volume decreases initially and then increases. As shown in figure 1(d) and (0). A state of liquid water (L) as shown in figure {() is heated at 60°C. If the heat is supplied continuously, liquid (1) starts evaporating (v) at a constant temperature (100°C). During this stage, specific volume of mixture (/,* ¥) increases and transformation of liquid to vapour occurs, shown in figure 1(g), (On further heating, the liquid (|) is completely evaporated and changes into vapour as shown in figure 1(h. During the process (v to V), further heating of vapour, increases the temperature fromi 100 to 300°C and the cylinder is completely occupied with vapour. Te specific ‘volume also increases in large amount, But the vapour ‘V” ‘occupies more volume than vapour‘ figure (2) shows, different regions on temperature, specific volume diagram. 1, Senate Heating Iris the pre the tempera! the phases ‘ * 12), sensible heating of ice ( From the Hite) OC, ensble heating of wae cs °C to 100°C) and sensible (i,t) from °C 10 i me race one $00°C) are considered. Latent Heating Itis the process in whict temperature. From the Figure 2) considered, 51,9 0°C (Melting of ice into water) eon 100°C (Waporization of liquid watering vapour). Sensible cooling can be obtained if the process jy followed in opposite direction. Py arid Ts diagram for water sta 030. Paw fe iquid phase to superheated steam, below ~40°C, i cess which occs 40°C, in, vane af substance changes Without changing Prose, ie nthe phase changes at contin, Iatent heating forthe processes ‘Answer: [Model Papert, 6a) | Mayidune-09, Set, say 7 Vinita) —> Figure (1): P-V Diagram for Water P-v diagram is obtained by plotting many number of isothermal ines as shown in figure (1). Consider the expansion of liquid from the point | to the point 4 through the path 1234. The ‘water between the states 1-2is an unsaturated liquid. At point 28 ‘known as saturated liquid. Between the point 2to 3 itis amixtst of liquid and vapour known as wet vapour. Point 3 represents® saturated vapour and from the point 3 as it moves to point 48 4nown as unsaturated or superheated vapour. Fin all the isothermal processes, the states represent’ the saturated liquid and vapour are joined, a curve ABC objained, The curve AB is the saturated liquid line and portion BCs the saturated vapour line. The meeting of he"? lines AB and BC takes place at the critical point. Th critical point gives the highest temperature and pressute WH both the liguid and vapour phases can_—s Fig 2:1 Diagram fr Wty Fa diagram is obtained by yy eos peta shown in fg (yf {Eta vapour points ofl the constant we get the curve ABC which, 6 AB ad BC: Therefor, the change ro sstuated points decreases 1 1 many fa any constant py tecomes 210 a the ri What crtcal point? What procona Postlble the critical below foe Dect, 462,036) How to estimate the critical point condi a pure substance? at answer + (eal Point The rtical point can be defined asa limiting tate where {elateatheat of vapourization becomes 2er0 on increasing the ese The properties of steam ata critical point is given by, Dee.44, R53, a9) Cia pressure, P= 57 = 221.2 Where, a and b are constants expressed in terms of ‘esperature and pressure, Critical temperature or critical isotherm, 7, 373.15°C, «tzal volume, V,= 0.000317 m’/kg, Figure shows the critical ‘ittin the P-v diagram, The process possible below the critical ‘isthe sublimation process, It is a process where a solid fae changes directly to a vapour state without first going ‘emgh aliquid state, ate Figure: P-V Diagram 7 arcane samme Fa Branaame aoe sia GROUP aA eT O22, Winai do you undereland ty Wiple point? Oto the prossure and tornperture of water at He Wiple point, Ame OES Wea tina 48, RAS Che) Wr hia AD, 4, OM) Shipte Poin 1 sate in ich solid Squid and vapor phases of Wbatace coexists in equilib i alle the wip yr of ‘hat substance, Vig sbvrns the PT dingram containing the triple poins, Figure: P-T Diagram The triple point can also be defined as the point of intersection of melting curve, vapourization curve and the sublimation curve, ‘The melting curve separates the solid phase and the liquid phase, The vapourization curve separates the liquid phase and the vapour phase, Sublimation curve separates the solid hase and vapour phase. When a solid is heated at a pressure below the triple point, it directly gets converted into vapour ice, it undergoes sublimation. Different substances have different triple points. For water, triple point is at temperature 0.01°C and pressure 0.611 KPa SS eee 33. Explain the significance of triple point in caso of pure substance, « Answers Model Papers, Q&(b) Triple Point * ‘the point at which the three phases i., solid, liquid and gaseous phases ofa substance coexist in equilibrium is known astiple point. ‘Triple point for water occurs ata pressure of 4.58 mm of Hg and at a temperature of 273.16 K as shown in figure. mek “Temperate (1) > Figure: Triple Point of Water e3.12 ofthe ce 10Pe8 oF sublimation, vaporization curve Tor most Water mePstanees including the fusion curve'is positive. For he slope of fusion curve is negative. Thermodynamic inde Properties cannot be varied in i is pendent ‘manner at the critical pe they are said t ‘ariant. However, any single property can be varied independently along the curves 04, OR OC. The system thermo Rte phase region issaid to be bivaiant ie, two lynamie properties can be varied independently AS measurement of properties like intemal energy, entropy aid enthalpy’ ofa system cannot be done direetly i» change in these properties of a system is measured. For is purpose, triple point is made as the reference state where the entropy and internal energy of saturated liquid water are assigned a value equal to zero. temper tiple point for a substance occurs at a fixed temperature and pressure, Any variation of properties like Pressure, temperature lads to disappearance of atleast one ase. : Number of iriple points in case of a pure substance ‘may be more than one but the equilibrium of solid, liquid and vapour phases occurs at only one point. At other tripe points, equilibrium’ of two phases of solid and one phase of liquid oF ‘Wo phase of solid and one phase of vapour or equilibrium of three phases of solid may.take place. Q34, Explain the following terms relating to steam formation, () Sensible heat of water fi (ii) Latent heat of steam. Answer : (Sensible Heat of Water It is the process of heat transfer in which change in temperature is sen without the change in phase. Sensible heat of water is the heat absorbed hy water in being heated from freezing point tothe saturation temperature at constant pressure. Let, T,Saturation temperature at atmospheric pressure. Sensible heat = C,(7, - 0) Se » Sensible heat of water = 4,187 ki/kg, Specific heat of water, C, = 4.187 ki/kg Sensible heat of water is also known as “enthalpy cof water” and is denoted by *H,. . (i) Latent Heat of Steam ‘The amount of heat required to convert 1 kg of water from all water to steam is called latent heat of steam or enthalpy of evaporation (h,). If only a partis evaporated, the heat required is x.h, (x is called dryness fraction). Enthalpy of wet steam is given by, h=htxh, When 'x = 0, the condition is saturated Water and when ni Saturated steam. oe 2 4 NAMICS [JNTU-HYDER,, ay my ‘THERMODY' Dry steam 0 Dee acon fhm {uy Enthatpy of wet steam (iv) Superheated steam. Answer + ae heated at constant pressure cal Ha conch et ee sertpion of heat does not witness any rise init tempeanes ae converting the water particles into steam. This pa continues until all the water particles convert compleey sam. This steam is known as dry steam which haga dyn. fraction equal to one. (ii) Dryness Fraction of Steam Itis defined as the ratio of mass of actual dry steam yg total mass of steam containing i. It is usually denoted by p 1m, Mass of dry steam contained in steam im. — Weight of water particular in suspension y * ‘the steam Thus, m, +m, = Total mass of steam a (ii) Enthalpy of Wet Steam It is the sum of the enthalpy of water and enthalpy. ‘evaporation (latent heat) It may also be defined asthe amt of heat transfer required to convert 1 kg of water at freeing injo wet steam, hahy thy Where, it Enthalpy of wet steam fy ~ Enthalpy of water ‘hg — Enthalpy of evaporation. ‘Superheated Steam When the steam temperature is above the salurse temperature corresponding to the steam pressure, the stexns superheated. Heat transfer per kg of steam during superbesing given as=C,(T.,~T) Where, tS ee G,,~- Specific heat of superheated steam T, ~ Saturation temperature atthe pressure considered T.,,~ Temperature of superheated steam. Enthalpy of superheated steam is, Nyt hg + Co Toy Ty) Where (7 iy), )) —Degree of supesheit” 1A. GROUP L0G0 £9 or the TITLE COVER ant the WATERMARK PAGE inside Beiore you)ywer * wr vont 0 or answer reer Unit, ag, 1 Temperture and Critical Een hie Point where 7 ed dry steam Tine meets, At ye Mid line ang Fropaisation BECOMES Zero, Tempera rena gat beat Seespondineoitical ointarekrowmar ent PESSres eee eSUr Fepectvely Foran cre PEE 415°C and crtcal pressures 29) 9 eq pel emPeTaure he etal temperate, resus aaa Bay without vaporisation, peers t 40°C Is cont Water a ntinuousl BF i ng 300°C per hour. Ata section dawre the pressure is 3 bar, the qualiy is tobe oe Find the rate of water spray in rgine ‘ pore Given that, “evmeea 8,c4 Pressure, P, = S bar ‘Temperature, 7, ~ 300°C Pressure, P, =3 bar m, = 5000 kg, Quality of the steam = 95% = 0,95 * s7,=40°C Pi i = 5000 ke Biatel Faget PMP) LMovi0ee 16 0, ao Tessure saturate ie = (msm) Figure From steam tables, at P, ~ 5 bar, 7, = 3000C 1h, = 3064.2 ki/kg, AUP, =3bar, 7, = 133.5°C : And, h, = 561.4 +0.95 (2163.2) ic steady flow, 30642-26648 m, = $000 | "261644-1676 £914.23 kg/hr Rate of water spray, m, ECTRUM aicin-ONE JOURNAL FOR ENGINEERING x 938. Find the enthalpy and entropy of the prossuro is 2 MPa and the specific volume 130.09 miskg. Answer : Given that, Pressure of steam = 2 MPa Specific volume = 0.09 m'kg. From steam tables, the properties of steam at given Pressure of 2 MPa i.e,,2 * 10* Pa = 20 bar are, ¥, = 0.001177 mike ‘ ¥_ "0.0996 m'/kg = 908.8 kikg ig = 1890.7 ki/kg h, = 2799.5 kikg AAT KSikg-K 5p, 3.894 kiikg-K 5,7 6341 ike Given specific volume is less than v,. ++ Steam in wet condition and the dryness fraction is, Maylsune-t2, Sat, O4(b) ++, Dryness fraction, Enthalpy’of steam at given pressure and specific volume is, hahtrh, =908.8 + 0.903 x 1890.7 = 2616.102 Kirke And Entropy of steam at given pressure and specific volume is, 548, 2.447 + 0.903 «3.894 963 kifke K. rating and throttling calorimeter the im after throttling is 10 bar. ‘The pressure and temperature of steam after throttling Is 1.1 bar and 110°C respectively. At the separator 0.6 kgs of water is trapped and 3.4 kgs of condensed water is collected from the condenser. Determine the dryness fraction of steam in the main pipeline. Take C,, . for superheated steam 2.1 Kulkg k. ‘Answer + Now/Dee.-18, (R16), a7(b) Given that, For throttling Calorimeter, 2 Pressure before throttl Pressure after throttling, P,= 1.1 bar ‘Temperature after throttling, T, = 110°C % € e SIA-GROUP Qi. yiMacs oC trapped water, m, = 06 ke Macs of condensed water, m, = 34 ky. Consider, (1) Separating Calorimeter ‘The dryness fraction of wet steam is piven as, 42 08S (H) Throttling calorimeter {®) Before Throttling From steam tables, at P, =10 bar, Specific enthalpy of saturated liquid, hy, = 762.81 Wiikg ‘Specific enthalpy of evaporation, hg, P2778 Kg (@) After Throttling From steam tables, at P, = 1.1 bar, By interpolation method, From steam tables, in between the temperatures: 100°C and 110°C ‘Saturation temperature, T,,, = 102.05°C and Specific enthaply of saturated vapour, fh, 7 2679-257 kitkg, Enthalpy Before Throttling Ayn hig thy, = 762.81 +x, (2778.1) (I) Enthalpy After Throttling hy = hg + Cy (Ty Tr) = 2679.257 + 2,1(110~ 102.05) 1, = 2695,952 kilkg 2 (2) Equating, the equations (1) and (2) 762.81 + x, (2778.1) = 2695.952 From the above relation, 762.81 778.1 24, 0.6958 . “The final dryness fraction is given as, KEN eH = 0.85 «0.6985 =0.59 “The final drayness draction (x) is 0.5 wieatoam In each Ca feed water to be converted into stear, m= 10kg 5 MNimt ‘Mass of ‘ater is Gonverted to Wet Steam wit, w ee n of x= 0.95. Dryness Fractiot From steam tables, ALP = 1S bar hy = 844.7 KT, fy = 1945.2 KO a J 22.36145 KiIKg, 5, 7 4.1216 Kk Enhalpy of steam, f= 2,42 s 2 844.7 2095 « 19452 b= 2692.64 kikg Entropy, 5=5,+*5, 5 =2.3145 +0.95 x 4.1261 (i) Superheated Steam at 300°C ‘ From steam tables, ‘Atpressure P= 15 ba, 7,= = 6.234UigK 198.3°C 1.3K 1 y= 844.7 Kk | yg = 1945.2 Kk 5,7 2.3145 kiikg K 54,7 4.1261 ki/kg K Enthalpy of steam, | Ingg = ht hg CuTg = T) = 844.7 + 1945.2 + 2.2 x (573-4713) (Take C= 220K) | g™ 3013.64 ki/kg | Entropy of steam, i hy “6 sea) tase, bo ( ) Tue s20(23) 19452 4713 = 2.286 + 4.128 + 0.43 ag" 6844 EI K. | - | Look forthe SIA GROUP LOGO onthe TITLE COVER and the WATERMARK PAGE inside before you | Ded‘Transfer — Steam enna re 4, Dorive Clapeyron, off significance. “AUatio, aewer cer" Equation Figure Consider a single component si a imple system containi wo phases (say liquid and vapour) in a state of, equity as shown in figure. A system is said to be in equilibrium if it sass the conditions of thermal, mechanical and chemical equilibria ie. the temperature and pressure of the two phases tusto be identical and the chemical potential should be sane inboth the phases. In ordes to analyse this problem, ay phase rrasition occurs at constant temperature and pressure, Gibbs fire energy is used. eet as Change in Gibbs free energy is given by, dg =~ SdT + VaP Consider a system consisting of a liquid phase at state 1 and vapour phase at state 1’, in a state of equilibrium. If the temperature of the system is changed from 7, to T, along the vaporization curve, the system attains the state 2 and", where 2 ‘represents the saturation lic equilibrium with the vapour in state 2', For phase transition from 1 t0 1’, dg=~SdT + VaP =0 8, = 8,88 = By ‘Similarly, for phase transition from 2 to 2', B78, In reaching state 2 from state 1, the change in the Gibbs fee energy of the liquid. phase is given by, 8,78, = -SAT+ Vee Similarly, change in Gibbs free energy of vapour phase ‘reaching the state 2' from state 1” is given by, 8-8, =- SAT Vp saturation curve, ‘The above equation is known as the clapeyron equation. This equation is valid for any phase change. Significance Clapeyron equation is used to calculate the enthalpy of evaporation, with the data obtained from rate of change of pressure with temperature. Q42. Explain briefly the following processes as applied to steam. (i) Polytropic process (li) Isothermal process. Answer : ‘Model Paper-til, Q6(b) @ _ Polytropic Process Applied to Steam 1 wy y . PV Diagram TS Diagram Figure (1) Appolytropic process is represented by, PV "= Constant wae ? n—Polytropic index. ‘In this process, there is a change in entropy. Hence, itis ‘known as polytropic process. Heat interaction is given by, = W+ (u,~1,) ‘Work done in a polytropic process is, W= ABA and w= h=PY Q= Aaah + he PK) h, PY) SIA\GROUP (5.Figure (2) During evaporation or condensation, (i. it is same as constant pressure process. Q =W4(u,-u). Work done, W=PA,-P,V,andu=h-PY 2O=[PV,- PV} + Mi PVD) oh, Once, the steam enters superheated region, it behaves ike a perfect gas obeying gas laws, ie., isothermal expansion in this region is hyperbolic (i.c., PY = Constant). O=W+ uu) ie, Q=PV, of?) (nh). po Ne Q43. Write short notes on the following, (i) Throttling calorimeter (il) Bucket calorimeter. ” answer : Mode! Paper-Hi, 7(b) @ Throttling Calorimeter ‘Throttling calorimeter is used for determining the’ ‘dryness fraction of the steam. The principle of this calorimeter is throttling the wet steam, so that it becomes superheated. The throttling calorimeter is shown in figure (1). Figure (1k Throttling Calorimeter “The wet steam is allowed to enter the sampling tube from ‘the main pipe of steam. This i passed through throttle valve and js throttled. This throttled steam is passed in the inner chamber of the throttling calorimeter Look for the SUA GROUP LOGO. g on the TITLE COVER and the WATERMARK PAGE inside before you THERMODYNAMICS [INTU-HYDERARAD, Chamber shouldbe superheated prasad by th ne epee eau) Semone respect pe saturation €emPerat cs ma ore than the st" ates superheated condition of bya manometer Let, P, ~ Initial pressure of wet steam WP, p,- Pressure of throttled steam 1,, — Saturation temperature of throttled steam ts "Temperature ofa superheated throtle steam, 1, ~ Latent heat of Vapourization at P,- .~ Specific heat of superheated temperature of throttled steam. iy, —Enthalpy of water at P, ‘x — Dryness fraction of wet steam : sipy before throttling = Enthalpy after throttling ie, y= hae hy, XL, = Enthe Dryness fraction, x= Also, hag + ly = + Clog —bad =A i tea) — hw, : ye BEM hy zt. Dryness fraction, he +Cyltag~tasd—he, _ he, hb eer) By using above equations dryness fraction can be found out. (ii) Bucket Calorimeter Bucket calorimeter is used to determine the dryness fraction of steam. ‘Aknown mass of steam is passed through a known mast cof water and steam is condensed completely. Heat lost by steam is equal to heat gained by water. Working __ Steam is passed through sampling tube into bucket calorimeter containing a known mass of water weight of cle- rimeter with water before and after mixing steam are obtai by weighting. : "Temperature of water before and after mixing withistea are measured by a mercury thermometer. Pressure before and after mixing water with steam st measured by pressure gauge. 51 Pn cnoetme Le} i — rey coment vombiremmamibnin ae Figure (2k: Bucket Calorimeter P, ~ Pressure of gauge (bar) P, — Pressure of atm (bar) x—Dryness fraction 1h, ~ Latent heat of steam Cr, Specific heat of water Cy, ~Specific heat of calorimeter 1m, ~ Mass of calorimeter (kg) ‘mc, ~Mass of calorimeter and water 1m,=Mass in calorimeter (mm) (kg) {,~ Duration temperature of steam, At equilibrium, heat lost by steam = Heat gained by water and calorimeter (~My * Cr A= ta N= (me, =m) Cr tag =.) +, Ce * Uy hy) Thus, x can be determined from above equation Dryness fraction obtained is inacurated due to losses (convection and radiation losses) which are neglected. (Dryness fraction),,..,.,< (Dryness fraction)... (Q4, Why cannot a throttling calorimeter measure the quality if the steam Is very wot? . Answer : ‘Throttling calorimeter cannot be used to measure the quality of the steam when the steam is very wet because throttling ir) to one fhase substance (vapour) i.e., the ,n. IF the steam is very wet than throttling to seam afler undergoing the throttling process should be in a super heated con atmospheric pressure may not ensure the superheated condition of the steam at exit as shown in the figure, Figure: h-s Diagram Tthas been found that a minimum of 5°C:supetheat is required for the steam to be present in a single phase. ‘ i 6. Clearly steam at point 6 isnot ina superheated condi Consider, the throttling of steam (wet) from point S tothe point 6 s tina super ion ‘nd therefore, itis not omsbleto measure its quality by the use of throttling calorimeter, It can be done in separating and throttling 28 of calorimeter. SIA GROUP 2%Prinelpte of Operation lectrical calorimeter is used to ineasure the quality of Wet steam. Ifinsteady flow steam is allowed to flow through an clectric heater, The input energy should be iti adequate amount to transfer the steam into super-heated region where pressure (P)) and temperature (1) are measured, Consider J and V be the current and voltage flowing ‘across the heater respectively. Then, the electrical energy inpit at steady state is given as, QO» VIx 10° KW _ (fm is the mass flow rate of steam, then by applying SFE (Steady Flow Energy Equation) to elect heat TA + Qe tighy a+ 2aa, ” am +h, can be computed when h,, Q and #, are known. Also, hy = hy, *81hy,, Q46. A spherical shell of diameter 50 cm contains ‘steam at a pressure of 40 bar and 0.85 dryness fraction.Calculate the mass of water and steam. Answer : Given that, Diameter of shell, d= 50cm =0.5 m xd _ x05? Volume of shell ¥= = = 0.0654 m? Pressure of steam, P= 40 bar Dryness fraction, x ~ 0.85 From steam tables At pressure, P = 40 bar V, = 0.04975 (m'/kg) 4h, = 1087.40 (ki/kg) 4h, = h, ~h,= 2800.3 ~ 1087.40 = 17129 kg Density of steam is given by, 1 Density = 597 = G85KO04975. 23.648 kg/m? THERMODYNAMICS (UNTU- HYDERABy, Mass of 0,0654 im! of wt = 0.0658 * 23.648 sons of team = 1537 8 ‘Also dryness fractions 085" 757 1,306 + 0.85 m, = 1.537 0,85m,= 0.231 ke m2 0.272 ky 272k Mass of water ng calorimeter has stoam ontering ofr. Urey ‘and coming out of It at 0.05 WPaat 100°C. Determine dryness fraction of steam, ‘Answer t Given that, “Pressure of steam entering the calorimeter, P,=10MPa= 10x10" Pa Pressure of steam coming out of calorimeter, P,=0.05 MPa = 0.05x10* Pa. “Temperature of steam coming out of ealorimete, T, = 100°C. Calorimeter used is throttling calorimeter, 4. The process during which the steam underwent i throtting. . =) From steam tables, Enthalpy of steam ata pressure of 0.5 bar (i.e, 0.05MP) and 100°C temperature is h, = 2682.5 kJ/kg, Properties of steam at a pressure of 100 bar are, hy = h,= 1407.6 kitkg : Moy hy, = 137A kikg, Substituting the values of Ay, Fp, «and, in equation (1) ‘26825-14076 137A 0988pressure of stoam samnt™ of steam at ext = 4 ton exit = 150°C. Suerimeter = 0.8 nernarae from separ throttling calorimeter = 10 ikgracharge from iperatur the dryness fractioy in. Deter : mn ofthe sample ween Given that, Nov-t0, Set2, a5(0) Pressire of steam sample, p =P Pressure of steam at exit, P= | by jaber Temperatute of, mae of steam atthe exit, 1 ; . = 150°C Discharge from separating. calorimeter m™=0Skgmin Discharge from throttling calorimeter, m= l0kghtin From steam tables, AMP,=P,= 12bar, hy, = 798.4 kIkg, hg, = 1984.3 Kikg AtP, = 1 bar and 150°C And, hop, = 27764 kik Fora throttling process 2-3, Aah, bint fa” hn 798.4 -+ x{19843)= 27764 “27164-1984 ® 19843 i978 > jopaa 79997 Now, quality of steam supplied, 99710 Q49, Sample of ste¢ is put through a throttling which pressure calorimeter In ‘and température are found to be 0.1 MPa, 120°C. Find the quality of sample taken from the boller. Answer : Mode! Papers, 7) | NowsDee-7, (R46), O74 Given that, Pressure of steam leaving the boiler, MPa=30 bar steam in the calorimeter, Abar ‘Temperature in the calorimeter, aes elo Entry — > = Figure : Throttling Process As throttling process occurs at constant enthalpy, hh, gt ah, = hy, * ED) From steam tables, property values, =) hg = 1008.4 kihkg hg, = 1794 ekg Corresponding to P, = 1 bar, h,, =2675.4 1k q, = 99.75°C = 372.75 K (On substituting the above values in equation (1), 1008.4 +x, (1794) =2675.4 + 2.1 (393-372.6) (--For steam, C= 2.1 ki/kg) _ 2675.4 +2.1(204) ~ 1008.4 " 1794 x,20.953 :. The quality of steam leaving the boiler, x, = 0.953 ‘@50. Steam flows In a pipe line at 1.5 MPa. After expanding to 0.1 MPain throttling calorimeter, the temperature is found to be 120°C. Find the quality of steam In the pipe line. What is the maximum moisture at 1.5 MPa that can be determined with this set-up tat least 5°C of superheatis required aftor throttling for accurate readings?” NoviDec.-09, Set, 8(0) Answer Figure: hs Diagram of Steam © ope 120°0 = 120+273 = 393K — ————=SIAGROUP 445 DEcrman ALLIn-ONE JOURNAL FOR ENGINEERINGICS [JNTU-HYDE} oe THER OO Ne CF ey Toner temperature 7, = 2 From steam tables, 7,7 553K At p= LS MPa, h,= 844.89 kihkg ¥, 2 0.045 m! Initial volume, hy, 1947.3 klik p= P,= 7 bare 700 KN/m? nF ‘When. p = 0.1 MPa, t= 120°C (State 2) mn, = 2716.2 kik, Ay ti = hy 844,89 + x, > 1947.3 = 2716.2 X,* 1947.3 = 2716.2~ 844.89 _ 2716.2-844.89 x 19H73 y= BAB 1 19473 ! £0961 ‘Now, atp=0.1 MPa and t= 99,63 + 5 = 104.63°C (State-3) Since, h, = 2685.5 kI/kg And, h, = hy, 2685.5 = 844.89 +x, x 1947.3 = 20855-84489 eo 973 i = 1840.61 6 987.3 2 x= 0.945 ‘The amount of moisture absorbed is (1 ~0.946) = 0.055 oat Figure (2) ++ The maximum moisture that can be determined with this setup is only 5.5% Q51. A certain quantity of air initially at a pressure of 7 bar and 280°C has a volume of 0.045 m?. It undergoes the following processes in the following sequence in cycle. = 710° x0.045 Answer + {a)_ Expands at constant pressure to 0.1m?” red (b)_. Follows polytrophic process with n= 1.4 “ and 4 m= 0.198 kg (c). A constant temperature process (which P,V,=mRT, completes) evaluate the following, “eh (i) Heat received In cycle pete a (ii) Heat rejected in cycle ax10° x01 (iii) Efficiency of cycle. Now-10, et, 08, Given that, Pressure, P,= 7 bar 710° Nim? =700 Nim? Polytropic index, n= 2° 0.1985%287 + T,=1228.728K Also, PV} 4 = Ry}4 magne _ (7 Yi 38 (R eo Oe Look ortie SLA GROUP LOGO 29 on the TITLE COVER and ihe WATERMARK PAGE insido before youl!Pure Substances and Propet 8 Of Gas, pw (2.22204 “TF = 0.428 hap AS Py = RT, 0.428 *10°* T= 0.1985 « 2g , , 0.1985x287x553, 2 Oazei08 = 0.736 m’ Heat Received in the Cycte 7 x $53 a {Assume C,= 0.71 i3/kg Ky 95.234 + 38.500 133.734, Applying fist law to reversible olytropic process 23, Op) = AE, + Wy 0, met, —1,)+ BBN 0, = mC4T,- 7.) + ™D—T) Q,,= 0.1985 * 0.71 » (553 ~ 1228.728) + 0.1985 (1228728-553) 14-1 Qy=-95.234 + 96.240 & Qy™ 1.006 KF [Heat received] Total heat received in the cycle, 2=0,,+0, = 133.734 + 1.006 1 Q= 134.7410 Heat Rejected in the Cycle * 0.287 Applying first law of thermodynamics to reversible ‘ahemal process 3-1, 2), = AE, + W,, Dy, = 0+ My, Gil) “Efficiency of the Reversible Thermod; Cycle Meat received Heat rejected Me Heat received | EMI4=#8 001 Nowe 3474 ot Ny 7 or 34.7%. 952. Steam initially at 0.3 MPa, 250°C Is cooled at constant volume, () At what temperature will steam become superheated vapour? (i) What Is the quality of steam at 80°C? (li) Whatis tho heat transferred per kg of steam In cooling from 250°C to 80°C. Answer : (Mode! Paper, 7a) | Nov/Dee.-18; (R16), Q6(2 Given that, Pressure, P=0.3 MPa Temperature, T= 250°C From steam tables, at 0.3 MPa pressure, T,, = 133.55C (@ Saturation Temperature of Steam ‘As, T>T., (ie 133.55 > 250) the steam i in superheated state. <=. From the properties of supetheated steam in steam tables at 0.3 MPa and 250°C Specific Volume, ¥, = 0.7964 m'/kg [~~ j= 4 V,= V, = 0.7964 m/kg Enthalpy of steam, When, ¥, (sat. vapour) = 0.8919 m*/kg, 7.5 = 120°C V,= 0.17076, 7,,,= 125°C At, V,= 0.7964, By linear interpolation, 7,,= 123.9°C + The steam becomes saturated vapour at 123.9°C (T.,) (i) Steam Quality at 80°C At 80°C, Specific volume, ¥,= 0.001029 m’/kg . V,= 3.407 mkg Enthalpy, h,= 334.91 ki/kg, fh, = 2308.8 kivkg Saturated pressure, P= 4739 kPa anton, 0.7964 = Frac? *3 Varwe 001029 +x, V, sre Vg) .001029 +x,(3.407 - 0.001029) 0.7964 = 0.001029 + x, (3.406) x2 SRG = 0.901029, = 0.2333.22 ree ee 32491 4 0.238 © (2308.8) Ae RTSORING (uy Heat Transferred per he of Steam Prom the fst law of thermody armies, aed + Pay tay,» Thea, fa SOc mt, = PD 6 ~300% 0.796) = (§759=47.3 ¥0.79%5)~ 2967 = 838.2492 = 2728.8 Heat transferred Q, , = -1890.5508 KIKS Figure G53. Find the intemal energy of 1 kg if steam pressure of 10 bar, when the condition of steam Is, (Wet with a dryness fraction of 0.85 (i) Dry and saturated (ii) Superheated, the degree of superheat 50°C. ‘The specific heat of superheated steam at constant pressure is 2.01 ki/kgK. Answer: May/June-13, (R09), Q4(b) Given that, ‘Mass of steam, m= 1 kg. Pressure, P= 10 bar ~ 10 100kPa = 1000 kPa Dryness fraction, x= 0.85 Thy ~ T= 50°C C= 2.01 kaikg-K ‘When the Steam is Wet With a Dryness Fraction of 0.85 “The properties of steam at J0 bar from steam tables are, T,,=179.9°C, ¥, = 0.1943 mkg f= 162.6 kik yg = 2013.6 Kies @ b= 27762 kik, Look for the SIA GROUP LoGo @ ai) on the TITLE COVER and the WATERMARK PAGE| THER apy of wet stan nha lP +E oly gr 08s «20136 A716 kke tee ven ste internal eneTB) vo wet steam is Biven Dy, eke PF Where: _gpecifie volume at dryness fraction of 0.45 oat, = 0,85 * 0. 1943 = 0.165 mike yu ah-P vo U = 2474.16- 1000 * 0.165 © 22309.16 Kikg Internal energy of wet steam with a dryness fraction of 0.85 js 2309.16 kKhkg- Dry and Saturated Steam ; “The internal eneray of SFY ‘and saturated steam is given by, u=h,— PY, =2716.2- 1000 * 0.1943 = 2581.9 kikg ste interal enerey of ay saturated steam, = 2581.9 ky If the Steam is Superheated Tag Toa = 50°C Tug 179.9= 50°C” 2: Tyg = 29IC ‘The specific enthalpy of siperheated steam i given by, Digg hg + Cy ag ~ Fa) =2776.2+2.01 x50 = 2876.7 kilkg Specific volume of superheated steam is, 4, Tonlin) Ta) = 0.19435.22994273 179.94+273 ¥, = 0.216 mikg. The internal energy of superheated steam is given bY = hhgy~ Ph og = 2876.7 — 1000 x 0.216 = 2660.7 kik. The internal energy of superheated steam when # Ksiks.Volume, ¥) = 3m Mass of wot steam, m= § hg ,= 0.67758" Enthalpy, d= H+ x, hy = 504.7 + 0.67758 (2201.6) h, = 1996.5 kiikg Entropy, u,= A, —P,y, = 1996.5 -200 x 0.6 u, = 1876.5 kihkg Since the rigid tank heated until the steam becomes dry sorted, ¥,, = 0.6 mike From steam tables, AtP=300 kPa, v, = 0.606 m'/kg ALP=310 kPa, v, = 0.587 m/kg From steam tables, h,= 27.25 ki/kg aye heP = 2725 = (309.7) (0.587) au, = 2543.2 Kiikg ‘eat transfer to the tank, = m(u,~m)) = 5 (2543.2 - 1876.5) 3.2 prorenties OF GASE 3.2.1 Portect Gas Laws - Equation of Stato, Specific and Universal Gas Constants Q55. Exptain pertoct gas laws. Answer 1 Mode Papert, 7s) The following are the perfect gas laws, L Hoyle's Law For a given mass of perfect gas, the volume is inversely Proportional to the absolute pressure when temperature is Constant, this is called Boyle's law and it is given as, fue <5 PY = Constant 2. Charle’s Law (At constant 7, n) For a given mass of perfect gas, the volume is directly proportional to the absolute temperature‘at constant pressure, this is called Charle’s aw and itis given as, V oT (Atconstant P,n) ¢ = Constant 3. Gay-Lussac’s Law For a given mass of perfect gas, the pressure is directly proportional to the absolute temperature at constant volume, this is called Gay-lussac’s law and itis given as, Pe T (Atconstant V,n) , 2 cons {Q56. Derive the expression for entropy change of an ideal g ‘The equation for change in entropy of an ideal gas between two states | and 2 can be derived by using the relations. Td, = dU+ Pav ol) Tas = dH VAP (2) According to ideal gas equation, dU =CdT di = Cdr pv =RT By substituting these values in equation (1), TdS=C, dT + tw aT R as=¢, 40,8 vty Integrating above equations under limits 1 and 2,24 Uy substituting the values of d#7 and pin equation ( P R ar ER ip pe T Integrating the above equation under limits 1 and 2, SHC Q57. Derive the relationship between the two princip! specific heats and characteristic gas constant for a perfect gas. Answer : Derivation of the Relationship between the C,, C, of & Perfect Gas : Consider a perfect gas being heated from temperature T, to T, at a constant pressure. From the non-flow equation, *Q=U,-U,+W o (1) Where U,, U, are initial and final internal energies and “Wis the work done. For a perfect gas, U~U,=mC,(T,-7) ‘Substituting this value in equation (1), Q=mC,(T,-1) +0. ‘The work done by the fluid in a constant pressure process is given by, i W=pidry= WV,-¥) =mR(T,-T) According to ideal gas equation, PU, V) = R(T, 7) W.= mR(T,-T)) And, @ =mC,(Z,~1,) For a constant pressure process, ‘Substituting all the values from equations (2), (3) and. (4) in equation (1), mC,(T,=7,) + R(T, ~7,)= mCT,-T,) 1) =) 4) m(C, +R) (T,~1,)= mC{T,-T) C,+R =G, : CpG Rh On dividing both sides by C,, Sey and a G 7 ¥ 2 caw Similarly, C, = Where, Gy C,- Specific heat capacities R= Characteristic gas constant, pywamics (JNTU-HYDERARa,, Answer Expression for Heat ‘Trans! Volume Process sera gasin non-flow process is heated ta consay Corea a3 SbOWD in figure: Due to this, mesic yo rire wll increase as there is no change in it volun tempera inet ‘Therefore no wor fer in a Non-flow Constan, wrk is done by the gas. supplied (Q) Sa Figure (1) — Votuine (ms) > Diagram = Pressure (bar) = 7 — Temp.°c) ——» Figure (3: PT Diagram ___ All the heat supplied (Q) to the gas is stored in the for of internal energy. Consider m kg of a certain gas being heated at const Volume from initial state 1 to a final state 2, Look for the SIA. GROUP Loco @ ‘on the TITLE COVER and the WATERMARK PAGE inside before you DU!ud perature OF BA OF Fina state (2) ‘he process is shown on the pressure vol stite-volume (Psy scoe-temperature(P-7) diagram in gues (1) ang a) Pressure, volume Sine the ss isheated at constant volume, (ie, =) (ies Py) Fa Boe Five 2 7 7 ©) Z'Constan Workdone by the Gas AS, 8W = Pav ‘On integrating from state (1) to state (2), Change in Internal Energy of Gas Change in internal energy, dU=mC,.dT On integrating from state (I) to state (2), ieee te U,-U,=m6,(0,-7) () Heat Supplied or Heat Transfer For non-flow process, 8 = dU + 5W [First law of thermodynamics] On integrating from state (1) to state (2), Jeo = Jewsfow ee 25 W,-U)* Is Qa" (=U) +0 0,,=W,-U)+mC,(1,-7) ._. This indicates that all the heat supplied to the gas is \tlized in increasing the internal energy of the gas. Hence, the heat transfer in a non-flow constant volume 59. “Agas oxpands necording tothe law PY"= Constant \'P.V, arothe intial conditions as rogards prossure° and volume rospoctivoly and r Is the ratio of ‘xPansion, show that the work done by gas during the changes is oqual to mil and 2 | that thie heat received by the gas is equal to Whore ys the ratio of the specific heat of the gas. Answer ‘Workdone when the gas expands is given as, 1 puec cs" -cven =ntl Wo Bu, C= Py : Recess is equal to m C, (T,- T)). SPECTROM ALLAN-ONE JOURNAL FOR ENGINEERING26 Veal received, Q, Atm, =o ranye Mile s) Hence proved Q60 . Explain, : (i) Throttling process (ii) Free expansion process. Answer : Model Papers, 70) @ Throttling Process 5 ‘Throttling process is one of the processes in thermo- dynamic analysis of steam. ft is constant enthalpy or constant total heat process. Steam is said to be throttled when it passes ‘through a restricted opening such as a narrow aperture or a slightly opened valve. Example Refrigerant expansion in throttle valve of a vapour ‘compression refrigerator, leakage of air through a punctured hole and leakage of fluid through a crack in the container, @ oO ( iy : “ai = PEN 0 dai Figure (1): Theatling Process i j ; ‘The fluid is forced out by its pressure throu eh ananpy, arited opening o aperture as shown i Figwe The pen? ea ow that du to frictional resistance betwen hg * wa the sides ofthe aperture, the velocity ofthe oucomingat K.Eis converted into hea, most reduced to zer0. The’ i. fhtion 1F Steam before the throttle is wet, then af hoe it may become dry steam. Ifit is dry before throttle, they it superheated after throttle. During the throttling process, 2 No heat transfer (Q,, = 0) [No work done (W¥,, = 0) due to exparision “No change of internal energy, (Au = 0) i ii) ii) iv) Enthalpy or total heat of fluid remains const (Ah=0) ive, hy = hy Free Expansion Process Free expansion is one of the thermodynamic proces, in which the internal energy of two systems remains constn, ‘The free expansion (or unresisted expansion) process is ap inreversible non-flow process. It occurs when a fluid is allowed to expand suddenly into a vacuum chamber through an orice of large dimensions. wi Figure (2: Before Expansion Figure (3: After Expansion Figure 2) (Before Expansion) shows an insulated vesel with a partition made into two chambers (4) and (B). Let be ‘chamber (4) contain a perfect gas having pressure (P,), volume (V,)and temperature 7, and the chamber (B)is perfectly evice ated (ie., vacuum chamber). The insulated vessel has no het transfer across the boundary of the system. Ifthe patton removed, then the perfect gas will expand freely and occupy the whole space shown in figure (3) (After Expansion). By this, the volume of gas increases from V, to ¥,, pressure decreas fom Fw sad temperature may also decrease foal to 7,, But there is no change of boundary (ie.,no expansion bindayjoldeasleetqneam ee Look forthe SUA GROUP L060 {fon the TITLE COVER andthe WATERMARK PAGE inside bbfore youbU!r Substances, foe Sn Proportion of oomteann sina a? OLA tater Boer (otis) —p agra (hntern 4: BY Diagram (Internal Enargy Versus Volome Diagram) ‘The following points may be considered for free exy fect gas (superheated st eee se : steam may be considered of Expansion is free from any resistance and hen © ipdone by the system, ee ig Nobeat is transferred from/or into the system, Fint law of thermodynamics, internal tem remains constant. once a Reeexpansion cannot be called Isothermal or Adib process because in an actual process, work is done by the perfect gas during expansion. dU20,dT=0 ot T,~T,, ic., there is no change in temperature of the system and hence df= 0. (a) Theenthalpy of the system ina fre expansion process remains constant (H, = H,). Thus the free expansion process may also be called as constant enthalpy process * ee Ee @. Demonstrate, using the second law, that free ‘expansion Is Irreversible. Asswer Figure (1: Free Expansion Consider a perfect gas having volume (V,) pressure (Py ‘limpet (7) contained in chamber A which s separated ten chamber B by a partition as shown in figure ()- There ‘stoheat transfer from the surroundings as the chambers are Petty insulated, ‘When the partition Sceuying the entire space 8 volume of gas increases to Fy Press Aeth temperature may decrease t0-7,, A8 the system (6 tie ‘boundary expansion occurs so workdone i 76F0- 2 Qyyr 0, Wg oand dv =O athe fice expansion ofthe ga takes place suddeny from equilibrium sae {to equilibrium state 2, There Ht bbe any intermediate equilibrium state as shown inP V-diagram- Thus, the process is irreversible and expansion is Known as irreversible adiabatic expansion. ‘The free expansion is reversible by second law of ‘thermodynamics, which can be ‘demonstrated by installing a0 engine o insulated contain between chamber “A” and chamber *B’ and allowing the gas to expand from ‘A’ to ‘B" as ‘shown in figure 2). Insulated container Figure (2) ‘Anengine div generates work outputs at the expense of internal energy of gas. Heat is transferred to chamber °B” from ‘source for restoring the internal energy of gasin chamber ‘B'- “The internal energy of gas can be restorgd to its initial state. Inchamber A and vacuum ‘B? chamber by use of reversed free ‘expansion. This results a cyclic process, in which work output canbe restored by exchanging heat. This justifies kelvin-planck staternent, Therefore, free expansion is ireversible. Q62, Show that the work done during a flow process is -fvep. * Answer: ‘The steady flow equation for unit mass given by, 8q-Sw=dh + ake) + dpe) (1) Also, h=utpy ) «dh = du dtpv) dh= du pdv+ vdp From first law, we have, 8q = du + pdy aan gle3.28 ‘in equation (1 p= Bye = Bp + udp + ke + ape) = Bw ve + che) + dpe) By neglecting the changes in kinetic and potential ‘enemies ie. Ake) © 0 and dpe) 0, = b= vip By applying integration on both sides, Ju --}e the importance of this process, Answer : Dec, (R13), 68) Free Expansion Proceis For answer refer Unit-IIl, Q46, Topic: Free Expansion Process. The free expansion process is important as, 1, Its used to analyse the state of a substance that undergoes free expansion in an enclosed area. 2. Therelation between intemal energy (u) and volume (¥) ‘at any point in the process can be obtained. 3. By applying second law of thermodynamics, the free expansion process can be made reversible or cyclic. Hence, it is used in analysis of cyclic processes. 4, Its used to obtain the P-V diagram, which represents the initial and final equilibrium states of a substance. When a closed system executes a certain non- flow process the work and heat interactions per degree rise in temperature at each temperature attained are given by, aw dQ Fr 7 (4-0.08T) ks IK; $2 = 4.00 Kak Calculate for the Increase (or) decrease in the Internal energy of the system if itis to operate between the temperature limits of 200°C and 500°C, Answer : Nov.-10, Set-3, Q5(b) Given hat, ow ~ 4-008 aw=(4-008D)aT ea) d And, 22 =1 THERMODYNAMICS LINTUHYDERAR d Qo dT > 7, 200°C = 2001273 = 473 K fl 300°C $004273 = 773 K A : ‘According to the frst Jaw of thermodynamics, Aged ~) substitute equations (1) and (2) i equation (>) i dt =dU+ (4-008 d= dT ~(4-0.08T)AT inegaton on bot sides wit iis 7, amy [rave fiar-[a-o08rer jf, = far [vy 4-oonnar fa Integrating te above equation with respect to 7, ost? : ve {o-]) 009078 —4n3 San aaa = 300 ~(1200 ~ 14952) 65. A gas occupies 0.034 m’ at 600 kPa and 85°C. is expanded in the non-flow process according to the law PV'2= constant toa pressure of 60kPa after which itis heated at constant pressure back to Its original temperature. Sketch the process on the P-v and T-S diagrams and calculate for the whole process the work done, the heat transferred. Take C, = 1.047 and C, = 0.775 kdlkg k for the gas. Answer Figure: PV Diagram.” Figure: Given that, Initial pressure, P, = 600 kPa = 6 bar Initial temperature, 7, = 85°C = 398 K Initial volume, ¥, = 0,034 m* js Diagrampve Supstances and Proportion of Gago ete Ty NPA OS bap ~ ‘ 3 -pemnperatue, 7, cr HOT WR K G7 OTS KKK PASC ASK Y refs} 00m ¥, 70.2316 m? _ ened forthe oes 1-2 is given by, ma" ow Raritan a - [i av [RY = Ry! = py oan = (600 * 0.034") J ae pape s oss. Wyq 23252 KI Heat transfer for the process 1-2 is given by, os-(F)" -(42} 32.52 = 16.261) ( y=14,0= 12) Forthe process 1-2 Ba) *(3) a T= 358 x ey 7, $243.9 K = 244K For the process 2-3, Salk RoW i au wenrh Q66. A fluid Is confined In a cylinder by Answer 3.29 Me 7h xoane = 0.3398 m! Work done in the proves 2-3+ WY, = 0.6 x 10°(0.3398 ~0.2316) & Wy 62 KS Heat transferred in the process 23+ 15 "TT = 1,047(358 ~244) Oy 2119351 ‘Work done in the whole process: 3 WD= M+ = 3252+ 6.492 c WD = 39.0120 Heat transferred in the whole process, = 2,2*0s- = 16.26 + 119.35 35.61 spring loaded friction less piston so that the pressure in the fluid Is a linear function of the volume (p= a+bV). The Internal energy of the fluld Is given by the following equation U = 32 + 3.15 pV. Where U Is In ka, p In kPa and V in cuble meter, if the fluld changes from arvinitial state of 120 kPa, 0.025 m* to a final state of 300 kPa, 10,056 m?, with no work other than that done on the piston, find the direction and magnitude of the work and heat transfer. = Nov.10, Set2, 8 Given that, PeatbV U=32+3.15 PV Initial pressure, P,= 120 kPa Initial volume, ¥;~ 0.025 m* Initial pressure, P,= 300 kPa ).056 m-- ON Dec Ee ee DECTROM ALL-IN-ONE JOURNAL FOR ‘ENGINEERING STUIChange in the internal ‘energy of the fluid Press fe given as, Un 32+ 318py, Vr aeaispy, ASP, - Pr) UUs 3.190300 » 0.056 ~ 120 « 0.025) U-Usaa70 But, Pratby Pza+by, 120=a+ 0.0255 And, P=a+ bY, 300= a+ 0.0566 Q Solving equation (1) and equation (2), a= 25.16 Kin? b= 5806.45 KNim? Now, work transfer involved during the process. To hilaa! an Shee ms sayy of -25.16(0.056 ~ 0.025) (0.056? ~0.025" + 5806.45 W,,=-0.78+7.29 WH, =6510 t ‘As, the magnitude is being positive. Hence work is done by the system. ‘The heat transferred by the system is given as, 2a“ Y-U)+M, Q, =43.47+6.51 O,. = 49.98 KI THERMODYNAMICS JNTU-HYDERg, Teylinder machine containing i dit h a nd a sfritionioss and itis held down gy” we tuld duo to the atmospheric pregy, 401.325 kpa. The stirring device Is tumeg 0 revolutions with an average torque againg fluid of 1.5 N-m. Meanwhile the piston of diamotor moves out 0.8 m. Find the net transfer for the system. hp, Answer t NovtO 803.9 Given that, Diameter of piston, d= 0.5 m Stroke of piston, L = 0.8 m Number of revolutions of stirring device, N= 12000 rpm ‘Torque, T = 1.5 N-m ‘Atmospheric pressure, p = 101.325 kpa Let, W, Work done by the stirer device upon the system, This is negative work for the system. ,,—Work done by the system upon the surroundings, This is positive work forthe system. W,=2nNT =2%n% 12000 1.5 = 113097.336 (N-m) of (Joule). 3.097 kd W,=p.A.L = xf Ex00.5)% = ms( fas! }on = 15.916 k-N-m 2W,= 15.916 k3 ‘Net work transfer for the system, Wg=-W,+W, (EW, = Negative work for the system) 113.097 + 15,916 Heat is transferred into the system due to magnitude being positive.[rr Per [ aa a mepinis OWromoved,ajpe' Piston ea The cylinder contains sstho piston against the stor plat gcePSY of 2000 gn on 2 at 200 KPa and ambient Move and aftera wile themes Se a caea anpsretre lem cannot be ‘ris po! solved wis ‘ 'ed without (Model Paper, a7(b) | NovsDec.-17, (R16), Q6(e)] FS 5 sidering the gi ‘he dimensions Of the cylinder are conde gn” Sion ofthe yin Given that, Cylinder pressure, P.= 200 kPq ‘Temperature, T= 290 K Density of piston, p = 8000 kg/m! Atmospheric pressure, P,= 101 kPg ® Volume of CO,, ¥,= fd" x =F Oly V, = 7.853 «104m? x Volume of piston, ¥,= 3 dlp x “| (0.17 0.1) ¥, 853 x 10+m? Mass of piston, m, = V, x p= 7.853 10 x 8000 = 6.282 kg w Pressure due to pistion weight™= “7 _mxe x qe 6.282*9.81 Rosy? qo = 7,846 kPa i pe ARS ae LA CD "VECTROM ALLciR-OME JOURNAL FOR ENGINEERING STODENTS— ‘S1A,GROUP fsA aS eee ee as ressure| THERMODYNAMICS [JNTU. oy, ight Toso piston wel exerted by the Piston on the gas” = Atmospheri pressure + Presa = 101 + 7.846 = 108.846 kPa moves up, as pressure ins tinder is greater than pressure plton Coa etameved then the pson moves up Pres? inside the oy Sey the ‘Considering piston is moved up until it touches the ends. Then, 4 Volumne of cylinder, ¥, = $ 0.1? x O15 Y= L178 10m Pressure inthe eylinder, P,= ite 4 LaTexio? P, = 133 kPa 7 inst the stops. + As the pressure in the cylinder is more than the piston pressure, piston rests against the stops. ‘pok fot the SIA GROUP Loco on the TITLE COVER and the WATERMARK PAGE irisidé before you0 rag