ENGINEERING THERMODYNAMICS STUDY MATERIAL

Vishwatmak Om Gurudev College Of Engineering Thermodynamics CHAPTER 6 PROPERTIES RELATIONS INTRODUCTION In the previous chapters, we have studied a number of properties like pressure, specific volume, temperature, mass, internal energy, enthalpy, entropy and specific heats. Some of these are measured experimentally and the remaining are calculated from experimental data. In order to calculate the properties that cannot be measured directly, certain thermodynamic relations are required. These relations are developed and discussed in this chapter. HELMHOLTZ AND GIBBS FUNCTION: GIBBSION EQUATIONS The Helmholtz and Gibbs functions are new functions used in problems related to chemical equilibrium and combustion. The Helmholtz and Gibbs functions are extensive properties. These are expressed as Helmholtz Function F = U -TS ….. (1) Gibbs Function G = H – TS …… (2) The corresponding specific values f = u -Ts …….. (3) g = h – Ts …….(4) In order to derive Gibbsian equation using first and second law of thermodynamics, V O ……(5) G 101 C E Using Eq. (5) Further differentiating Eq. (1) and Eq. (2) Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics …… (7) …… (8) Rewriting the above derived equations, we get the following set of Gibbsian relations: V O G C E MAXWELL RELATIONS The equations that relate the partial derivative of properties of p, v, T and s of a compressible fluid are called Maxwell relations. The four Gibbsian relations for a unit mass are Since u, h, f and g are the properties thus point functions and the above relations can be expressed in the form Applying the cyclic relation as Thus replacing M, N, y and x by T, p, v, s of each of Gibbsian equations in cyclic order, we get the following: The first relation leads to Prof.S Venkatesh Rao 102

Vishwatmak Om Gurudev College Of Engineering Thermodynamics The second relation leads to The third relation leads to And the fourth relation leads to These derived relations are called the first, second, third and fourth Maxwell relations, respectively. These are extremely important in thermodynamics and provide means of determining change in entropy. THERMODYNAMICS SQUARE Figure shows a thermodynamic square, a mnemonic diagram which is very useful to obtain the Maxwell and Gibbsian relations. The corners are levelled with properties T, p, s and v, respectively. The sequence of properties T, p, s, v can be kept in mind by remembering a single word (Tpsv), which starts from left top corner of the square and runs clockwise. The sides of the square are levelled by potential u,f, g and h, starting from the bottom side and run clockwise. The square is also stamped with two diagonal arrows heading towards the left corners. V O G C E Prof.S Venkatesh Rao 103

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Gibbsian Equations The Gibbsian relations can be obtained by using the following rule: (i) Each potential is flanked with its natural variable as u = f(v, s), h = φ(p, s), etc. The differential of these variables are multiplied by its properties lying on their diagonally opposite corners. For example, du= Tds—pdv (ii) If the arrow is pointing away from the natural variable then a positive sign is used before the quantity and if the arrow is pointing towards the quantity then a negative sign should be used before the quantity. For example, if the arrow is pointing away from s, a positive sign is considered for T ds, and the arrow is pointing towards v, a negative sign for p dv. V O G C E Maxwell Relations The Maxwell relations can also be written by using thermodynamic square or mnemonic diagram as follows: (i) The partial derivative of the properties lying on a corner of a side with respect to the other corner property is equated with partial derivative of the property on a corner opposite with respect to the other corner property of the opposite side. For example, (ii) If the partial derivative of properties lying on the horizontal line are equated then a positive sign is used in the relation. If the partial derivatives of properties lying on vertical lines are equated then a negative sign appears on either side of equations. For example, partial derivative lying on vertical lines Prof.S Venkatesh Rao 104

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Table : Gibbsian and Maxwell relations derived by using a thermodynamic square VOLUMETRIC EXPANSIVITY, ISOTHERMAL AND ISENTROPIC COMPRESSIBILITY The equation of state for any thermodynamic system is expressed as f (p, v, T) =0 Any two properties of the system are independent and the value of the third property is fixed by the values of the other two properties. The cyclic relation for their partial derivative can be written as V O G C E Coefficient of Volumetric Expansion The coefficient of volumetric expansion β is defined as the ratio of fractional change in volume to change in temperature when pressure is kept constant. It is expressed as In other words the coefficient of volumetric expansion is an indication of change in volume with respect to change in temperature, while pressure remains constant. It is also called coefficient of thermal expansion. The quantity ��������������������������������������������������������������is the slope of the curve on v-T diagram. Coefficient of Isothermal Compressibility The coefficient of isothermal compressibility, ������������ is defined as the ratio of fractional change in volume to change in pressure when temperature remains constant or in other words, the Prof.S Venkatesh Rao 105

Vishwatmak Om Gurudev College Of Engineering Thermodynamics isothermal compressibility is an indication of the change in volume with change in pressure, when the temperature is kept constant. It is expressed as where the quantity �������������������������������������������������������������� is the slope of the curve on a v-p diagram. Since volume decreases with an increase in pressure, therefore, negative sign is used in the above relation, which makes w positive for all substances in all phases. The coefficient of volumetric expansion and isothermal compressibility are thermodynamic properties like specific volume, Isentropic Compressibility The isentropic compressibility ������������ is the ratio of fractional change in volume to change in pressure, when entropy remains constant. It is an indication of change in volume with change in pressure for an isentropic process. It is given as V O G C E The isentropic compressibility is measured in kPa-1, a reciprocal of pressure unit. JOULE-THOMPSON COEFFICIENT-THE POROUS PLUG EXPERIMENT The porous plug experiment was designed to measure the resulting temperature change, when a fluid flows steadily through a porous plug, which is inserted in a thermally insulated horizontal pipe as shown in Fig. The steady flow through such a restricted passage is called throttling process or isenthalpic process. It is an irreversible process. A gas at initial pressure p1 temperature T1 and enthalpy h1 flows steadily through a porous plug located in an insulated, horizontal tube and emerges into a space, which is maintained at pressure P2 Consider the region enclosed by dotted lines as control volume. Since the pipe is insulated and it does not experience any work transfer; Q =0, and W = 0. If the kinetic and potential energy changes are negligible, then the steady-flow energy equation reduces to h1=h2 Prof.S Venkatesh Rao 106

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Joule—Thompson conducted experiments on porous plug for the same gas with given initial pressure p1 and temperature T1 but with different downstream pressures etc. The different exit temperatures were obtained after expansion as respectively. The V O G C E downstream pressure was altered by controlling the opening of the throttle valve. The plot of exit pressures and temperatures for a gas yields an isenthalpic curve as shown in Fig. (b). During the throttling process, the temperature may decrease, increase or remains constant. The Joule—Thompson coefficient is the slope of isenthalpic curve. It is designated as µ and defined as Joule and Thomson conducted a series of experiments for real gases on a porous plug with constant inlet pressure and temperature. The gas is forced to flow through different sizes of porous plugs, each giving a different set of T2 and P2 When these experimental data are plotted on a T—p diagram, it gives a family of curves as shown in Fig. It is observed that each curve passes through a maximum point referred as inversion point, a curve passing through the inversion point is known as an inversion curve. Prof.S Venkatesh Rao 107

Vishwatmak Om Gurudev College Of Engineering Thermodynamics The temperature on the point is known as inversion temperature. The slope of isenthalpic curve ������������ = ��������������������������������������������������������������is positive. If the state after throttling falls within the inversion curve then the throttling produces a drop in temperature with decrease in pressure. Therefore, the region within the inversion curve is known as the cooling zone. At room temperature and at low moderate pressures, the state of most gases after throttling falls inversion curve. Similarly, if the state after throttling falls out of the inversion curve where the slope isenthalpic curve ������������ = �������������������������������������������������������������� is negative, the temperature increases with decrease in pressure and there is a heating effect after throttling. Therefore, the region outside the inversion curve is known as heating zone. The Joule—Thomson coefficient µ decides the heating or cooling effect with decrease in pressure. V O G C E The state of any ideal gas after throttling always falls. at the point on the curve, where slope ������������ = �������������������������������������������������������������� = ������������ Therefore, the temperature remains-constant after throttling for ideal gases. (∵ h = const., ∴ T= const.) CLAUSIUS-CLAPEYRON EQUATION The Clausius- Clapeyron equation relates saturation pressure, saturation temperature, enthalpy of vaporisation and specific volume of two phases of saturated fluid during phase change. Considering entropy as a function of temperature and specific volume s =f(T v) and During the phase change of a pure substance, the saturated temperature remains constant, thus dT = 0 and the above equation reduces to Using the third Maxwell relation, we get 108 Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics ……(1) For two phases of a pure substance in equilibrium, the pressure and temperature are independent of specific volume. Thus the derivative in Eq, (1) may be written as ……(2) the pressure and temperature during phase change, remain constant, thus V ……(3) O GUsing Eq. (2), we get C E This equation is known as Clapeyron equation. The derivative dp/dT represents the slope on the pressure-temperature curve. Equation (3) can also be applied to solid-gas and solid-liquid phase changes. At temperatures relatively high but below critical point, vg>> vf and pure substance follows the ideal gas behaviour The above equation becomes This equation is referred as Clausius Clapeyron equation. Rearranging the above equation, On integration, 109 Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics This equation can be used more frequently. V O G C E Prof.S Venkatesh Rao 110

Vishwatmak Om Gurudev College Of Engineering Thermodynamics FOR LAST MINUTE REVISION 1. The thermodynamic relations are used for those properties which cannot be measured directly. 2. The Gibbsian relations are 3. The Maxwell relations are equations that relate the partial derivative of properties p, v, T and s of a simple compressible fluid to each other. These Maxwell relations derived from Gibbsian relations are listed below: V O G C E 4. The coefficient of volumetric expansion β is defined as 5. The coefficient of isothermal compressibility is defined as 111 6. The coefficient of isentropic compressibility is defined as Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 7. The general expressions for internal energy, enthalpy and entropy in terms of p, v, T and specific heat data 8. The general relation for Cv and Cp V O G C E 9. The Joule—Thomson coefficient The Joule—Thomson coefficient µ decides the heating or cooling effect after throttling 10. For an ideal gas, during throttling h = C, thus T = Constant 112 11. The coefficient of constant temperature is defined as Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 12. The Clausius—Clapeyron equation yields V O G C E Prof.S Venkatesh Rao 113

Vishwatmak Om Gurudev College Of Engineering Thermodynamics QUESTIONS 1. What are Maxwell relations and why they are important in thermodynamics? 2. Sketch the thermodynamic mnemonic diagram and explain its use to obtain Gibbsian equations and Maxwell relations. 3. Define Helmholtz and Gibbs functions. 4. Show that the work done by the system during an adiabatic process is equal to decrease in internal energy of the system. 5. What is Clapeyron equation? 6. What are assumptions made in obtaining Clausius equation from Clapeyron equation? 7. State Helmholtz and Gibbs function and then derive Gibbsian relations. 8. State Gibbsian relations and then derive Maxwell relations. 9. Define (a) Coefficient of volumetric expansion, (b) Coefficient of isothermal compressibility. (c) Isentropic compressibility. 10. Establish the relationship between 8 and w for an ideal gas. 11. Considering specific entropy as function of pressure, sp. volume and temperature, derive three Tds relations. Using Tds relations, prove that V O G C E 12. What is Joule Thompson coefficient? Discuss the zone of heating and cooling with help of inversion curve. 13. Considering u =f(T, v), prove that Prof.S Venkatesh Rao 114

Vishwatmak Om Gurudev College Of Engineering Thermodynamics PROBLEMS 1. For a perfect gas, show that where β is the co-efficient of cubical/volume expansion. 2. Find the value of co-efficient of volume expansion β and isothermal compressibility K for a Van der Waals’ gas obeying V O G C E 3. Prove that the internal energy of an ideal gas is a function of temperature alone. 4. Prove that specific heat at constant volume (Cv) of a Van der Waals’ gas is a function of temperature alone. 5. Determine the following when a gas obeys Van der Waals’ equation, (i) Change in internal energy; (ii) Change in enthalpy ; (iii) Change in entropy. 6. The equation of state in the given range of pressure and temperature is given by where C is constant. Derive an expression for change of enthalpy and entropy for this substance during an isothermal process. 7. For a perfect gas obeying pv = RT, show that Cv and Cp are independent of pressure. 8. Using the first Maxwell equation, derive the remaining three. Prof.S Venkatesh Rao 115

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 9. Derive the following relations: where a = Helmholtz function (per unit mass), and g = Gibbs function (per unit mass).V 10. Find the expression for ds in terms of dT and dp.O 11. Derive the following relations:G where β = Co-efficient of cubical expansion, and K = Isothermal compressibility.C 12. Derive the third Tds equationE and also show that this may be written as : 13. Using Maxwell relation derive the following Tds equation 14. Derive the following relations Prof.S Venkatesh Rao 116

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 15. Prove that for any fluid Show that for a fluid obeying van der Waal’s equation where R, a and b are constants where f(T) is arbitrary. 16. Derive the following relations: VWith the aid of eqn, (ii) show that O G C E The quantity ������������������������ ��������������������������������������������������ℎis known as Joule-Thomson cooling effect. Show that this cooling effect for a gas obeying the equation of state 17. The pressure on the block of copper of 1 kg is increased from 20 bar to 800 bar in a reversible process maintaining the temperature constant at 15°C. Determine the following : (i) Work done on the copper during the process, (ii) Change in entropy, (iii) The heat transfer, (iv) Change in internal energy, and (v) (Cp – Cv) for this change of state. Given : β (Volume expansitivity = 5 × 10–5/K, K (thermal compressibility) = 8.6 × 10–12 m2/N and v (specific volume) = 0.114 × 10–3 m3/kg. 18. Using Clausius-Claperyon’s equation, estimate the enthalpy of vapourisation. The following data is given : 19. An ice skate is able to glide over the ice because the skate blade exerts sufficient pressure on the ice that a thin layer of ice is melted. The skate blade then glides over this thin melted water layer. Determine the pressure an ice skate blade must exert to allow smooth ice skate at – 10°C. The following data is given for the range of temperatures and pressures involved: Prof.S Venkatesh Rao 117

Vishwatmak Om Gurudev College Of Engineering Thermodynamics hfg(ice) = 334 kJ/kg ; vliq. = 1 × 10 m3/kg ; vice = 1.01 × 103 m3/kg. 20. For mercury, the following relation exists between saturation pressure (bar) and saturation temperature (K) : Calculate the specific volume vg of saturation mercury vapour at 0.1 bar. Given that the latent heat of vapourisation at 0.1 bar is 294.54 kJ/kg. Neglect the specific volume of saturated mercury liquid. V O G C E Prof.S Venkatesh Rao 118

Vishwatmak Om Gurudev College Of Engineering Thermodynamics CHAPTER 7 PROPERTIES OF STEAM 1. Define & Explain 1. Pure substance 2. Triple point 3. Saturation temperature 4. Saturation pressure 5. Critical point 6. Compressed or sub cooled liquid 7. Saturated liquid (water) 8. Wet steam & Dryness fraction, Quality of steam 9. Dry and saturated steam 10. Superheated steam &Degree of superheat, 11. Enthalpy of evaporation V O G C E Pure substance A pure substance is a system which is (i) homogeneous in composition, (ii) homogeneous in chemical aggregation, and (iii) invariable in chemical aggregation. — “Homogeneous in composition” means that the composition of each part of the system is the same as the composition of every other part. “Composition means the relative proportions of the chemical elements into which the sample can be analyzed. It does not matter how these elements are combined. Triple point The triple point is merely the point of intersection of sublimation and vapourisation curves. It must be understood that only on p-T diagram is the triple point represented by a point. On p-V diagram it is a line, and on a U-V diagram it is a triangle. Prof.S Venkatesh Rao 119

Vishwatmak Om Gurudev College Of Engineering Thermodynamics — The pressure and temperature at which all three phases of a pure substance coexist may be measured with the apparatus that is used to measure vapour pressure. V O G C E Saturation temperature The temperature at which vapourisation takes place at a given pressure is called the saturation temperature (or Boiling Point) and denoted by tsat and the given pressure is called the saturation pressure. The saturation temperature increases with pressure. At 1 atm it is 1000 C for water and it is 179.90 at 10 bar 81.350 C at 0.5 bar Critical point As the pressure increases, latent heat of evaporation decreases. The pressure corresponding to 0 kJ/kg latent heat is known as Critical point. OR. It is the point where saturated liquid line and saturated vapour line meet. Prof.S Venkatesh Rao 120

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Compressed or sub cooled liquid The liquid having temperature below saturation temperature is known as sub cooled liquid or compressed liquid For e.g. water at 1 atm pressure and 80°C The difference between saturation temperature and the temperature of the subcooled liquid is known as ‘Degree of sub cooling’. In above ex. the ‘Degree of sub cooling’ is (100-80) =200 C Saturated liquid (water) Vapour at saturation temperature is known as Saturated vapour. For e.g. steam at 1 atm and 100°C, containing no suspended water particles. The steam at this state is also known as dry steam (vapour) or saturated steam. Refer Figure state B V O G C E Wet steam & Dryness fraction, Quality of steam The steam at any state between saturated liquid represent saturated vapour state is known as Wet steam. In this state, there are water particles in suspension. Thus the wet steam is a mixture of liquid and vapour. As the heat is added, these suspended particles also gets evaporated and finally giving saturated and dry steam. Dry and saturated steam Vapour at saturation temperature is known as Saturated vapour. For e.g. steam at 1 atm and 100°C, containing no suspended water particles. The steam at this state is also known as dry steam (vapour) or Dry and saturated steam. Superheated steam The steam having temperature above saturated temperature is known as superheated steam. Prof.S Venkatesh Rao 121

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Degree of superheat For superheated steam, difference between its temperature and corresponding saturation temperature is known as Degree of superheat. So degree of superheat = tsup – tsat. Dryness fraction, Quality of steam This is the number which represents the quality of steam. It gives percentage of vapour present in the steam. If mv = mass of vapour in the given mass of steam, and ml = mass of liquid in the given mass of steam. Then the dryness fraction ‘x’ can be obtained by, m x= v m +m v1 Thus if in 1 kg of wet steam 0.9 kg is the dry steam and 0.1 kg water particles then x = 0.9. V O G C E Enthalpy of evaporation Amount of heat required to convert 1 kg of liquid from saturated liquid state to saturated vapour state is known as Latent heat, or Enthalpy of evaporation. Latent heat also depends on pressure, i.e. it decreases with pressure for e.g. latent heat of water at 1 atm 2258 kJ/kg pressure is and at 10 atm pressure is 2013 kJ/kg It is denoted by hfg is 2. Write short notes on (c) Explain p—T diagram for water. [MAY 2005, MAY 2005] 5 Marks The state changes of a pure substance, upon slow heating at different constant pressures. If these state changes are plotted on p-T coordinates, the diagram, as shown in Fig., will be obtained. If the heating of ice at-10°C to steam at 250°C at the constant pressure of 1 atm is considered. 1-2 is the solid (ice) heating, 2-3 is the melting of ice at 0°C, 3-4 is the liquid heating, 4-5 is the vaporization of water at 100°C, and 5-6 is the heating in the vapour phase. The process will be reversed from state 6 to state 1 upon cooling. The curve passing through the 2, 3 points is called the fusion curve, and the curve passing Prof.S Venkatesh Rao 122

Vishwatmak Om Gurudev College Of Engineering Thermodynamics through the 4, 5 points (which indicate the vaporization or condensation at different temperatures and pressures) is called the vaporization curve. If the vapour pressure of a solid is measured at different temperatures, and these are plotted, the sublimation curve will be obtained. The fusion curve, the vaporization curve, and the sublimation curve meet at triple point. V O G C E The slopes of the sublimation and vaporization curves for all substances are positive. The slope of the fusion curve for most substances is positive, but for water, it is negative. The temperature at which a liquid boils is very sensitive to pressure, as indicated by the vaporization curve which gives the saturation temperatures at different pressures, but the temperature at which a solid melts is not such a strong function of pressure, as indicated by the small slope of the fusion curve. 3. Draw the following chart for steam: (i)T-s and (ii)h-s [DEC 2010] 4 Marks Figure:T-s Diagram Prof.S Venkatesh Rao 123

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Figure: H-s Diagram V O G C E Prof.S Venkatesh Rao 124

Vishwatmak Om Gurudev College Of Engineering Thermodynamics FOR LAST MINUTE REVISION 1. A pure substance is a system which is (i) homogeneous in composition, (ii) homogeneous in chemical aggregation, (iii) invariable in chemical aggregation. 2. The triple point is merely the point of intersection of sublimation and vapourisation curves. It must be understood that only on p-T diagram is the triple point represented by a point. On p- V diagram it is a line, and on a U-V diagram it is a triangle. 3. Steam as a vapour does not obey laws of perfect gases unless and until it is highly in super dry condition. 4. Dryness fraction is the ratio of the mass of actual dry steam to the mass of steam containing it. V O G C E where, ms = Mass of dry steam contained in steam considered mw = Mass of water particles in suspension in the steam considered. 5. Superheated steam behaves like a gas and therefore, it follows gas laws. The law for adiabatic expansion is pv1.3 = C. 6. External work of evaporation = p(vg – vf) Prof.S Venkatesh Rao 125

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION QUESTIONS MAY 2005 Q 1 Fill in the blanks : (a) At critical pressure, the latent heat of any pure substance is ______________kJ/kg. 1 Q 7 (iv) Explain P-T diagram for water. 5 DEC 2005 Q 1 (b) State TRUE or FALSE. If false, correct it. (i) In the vicinity of critical point, the specific volumes of liquid and vapour are almost equal. 1 MAY 2006 Q 5 (c) Define: (i) Wet steam (ii) Superheated steam (iii) Dryness fraction (iv) Saturation temperature 4 DEC 2006 Q 5 (b) Define the term dryness fraction and explain working of throttling calorimeter. 6 V MAY 2007 O NIL G C E DEC 2007 Q 5 (a) Define (i) Saturation temperature (ii) Wet steam (iii) Superheated steam (iv) Dryness Fraction. 4 MAY 2008 Q 7 Write short notes on (c) Explain p—T diagram for water. 5 DEC 2008(RC) NIL MAY 2009(RC) NIL Prof.S Venkatesh Rao 126

Vishwatmak Om Gurudev College Of Engineering Thermodynamics DEC 2009(RC) NIL MAY 2010(RC) Q4(c) Define : (i) Wet steam (iii) Subcooled liquid (ii) Quality of steam (iv) Triple point of water. 4 DEC 2010(RC) Q1(f) What do you understand by the degree of superheat and the degree of subcooling? 4 Q4(a) Draw the following chart for steam: (i)T-s and (ii)h-s 4 MAY 2011(RC) Q 6 (a) Define Dryness fraction, Critical Point, Triple Point and Degree of Superheat 4 V DEC 2011(RC) O6 Q 5 (b) Define- G (i) Saturation temperature C (ii) Dryers fraction E (iii) Latent heat of evaporation MAY 2011(RC) Q 5. (a) Define : (a) wet steam (b) Superheated steam (c) Dryness fraction (d) Saturation temperature. 8 Prof.S Venkatesh Rao 127

Vishwatmak Om Gurudev College Of Engineering Thermodynamics GRADED PROBLEMS TYPE I: PROPERTIES OF STEAM 1. Calculate the enthalpy, entropy and specific volume of steam at 3 MPa and 350°C using steam table. 4[MAY 2009(OC)] 2. Calculate the enthalpy, entropy and specific volume of steam at 2.5 MPa and 320°C using steam table. 3 [DEC 2008(OC), MAY 2006, MAY 2003] 3. 2 kg of steam is at 8 bar and 0.8 dry Determine its enthalpy and volume 3[DEC 2008(RC)] 4. Find the enthalpy and entropy of steam when the pressure is 2MPa and the specific volume is 0.09 m3/Kg. 4 [MAY 2007] V O G C E 5. Find the enthalpy, entropy, and volume of one kg of steam at 1.4MPa, 380°C. 4[DEC 2006] 6. 5 kg of wet steam contains 3.5 kg of dry steam at 10 bar pressure. Find dryness fraction of the steam. Also, find enthalpy, entropy and specific volume of — (i) Liquid present in wet steam (ii) Vapour present in the wet steam (iii) Wet steam. 8 [DEC 2005] 7. 10 kg of water with a quality of 40% is contained in a tank at 1 MPa. What is the volume of the tank? [DEC 2003] 8. Using steam table, find enthalpy and entropy of the steam when P = 3 MPa and v = 0.055 m3/kg. 3 [DEC 2002] TYPE II: THERMODYNAMIC PROCESSES (A): V = C 9. A rigid tank of 0.06 m3 volume contains the mixture of water and water vapour at 80 KPa. The mass of mixture in the tank is 16kg. Calculate the heat added and the quality of the mixture when the pressure inside the tank is raised to 7MPa. 10 [MAY 2004] Prof.S Venkatesh Rao 128

Vishwatmak Om Gurudev College Of Engineering Thermodynamics (B): P = C 10. One kg of steam at a pressure of 17 bar and dryness 0.95 is heated at a constant pressure, until it is completely dry. Determine (i) Increase in volume (ii) Quantity of heat added. 4 [MAY 2009(RC)] 11. Determine amount of heat that should be supplied to 2kg of water at 25°C to convert it into steam at 5 bar and 0.9 dry. 4 [DEC 2007] (C): Isentropic 12. Steam at a temperature 2000 C expands isentropically to a pressure of 1 bar and dryness fraction 0.96. Determine inlet (initial) pressure. 4 [DEC 2003] V O G C E 13. Steam at 10 bar and 0.9 dryness fraction is available. Find the final dryness fraction of steam in each of the following two cases – i. 170 kJ of beat is removed per kg of steam at constant pressure ii. Steam expands isentropically to a pressure 0.5 bar in a turbine in a flow process. The turbine develops 300 kJ of work per kg of steam. [MAY-96-7] (D): Throttling 14. 5 kg steam is throttled from 12 bar initial pressure to 0.5 bar final pressure. The temperature at the final state is 1070 C. Find following for the initial state. (i) Dryness fraction (iv) Mass of liquid (ii) Specific entropy (v) Mass of vapour. (iii) Specific volume 10[MAY 2005, DEC 2008(OC)] 15. 5 kg of steam at a pressure of 10 bar and dryness fraction of 0.98 is throttled to 1 bar. Find the final condition of the steam and change of entropy. Also find out pressure at which throttled steam will be just dry and saturated. [MAY-97-4] Prof.S Venkatesh Rao 129

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION PROBLEMS (SOLVE AS HOME WORK) MAY 2005 Q 5 (a) 5 kg steam is throttled from 12 bar initial pressure to 0.5 bar final pressure. The temperature at the final state is 1070 C. Find following for the initial state. (i) Dryness fraction (iv) Mass of liquid (ii) Specific entropy (v) Mass of vapour. (iii) Specific volume 10 DEC 2005 Q 4 (a) 5 kg of wet steam contains 3.5 kg of dry steam at 10 bar pressure. Find dryness fraction of the steam. Also, find enthalpy, entropy and specific volume of — (i) Liquid present in wet steam (ii) Vapour present in the wet steam (iii) Wet steam. 8 V O G C E MAY 2006 Q 1 (g) Calculate the enthalpy, entropy and sp-volume of steam at 2.5MPa and 320°C using steam table 3 DEC 2006 Q 1 (d) Find the enthalpy, entropy, and volume of one kg of steam at 1.4MPa, 380°C. 4 MAY 2007 Q 1 (c) Find the enthalpy and entropy of steam when the pressure is 2MPa and the specific volume is 0.09 m3/Kg. 4 DEC 2007 Q 1 (c) Determine amount of heat that should be supplied to 2kg of water at 25°C to convert it into steam at 5 bar and 0.9 dry. 4 MAY 2008 Q 1 (g) Calculate the enthalpy, entropy and sp-volume of steam at 3 MPa and 315°C using steam table. 4 Prof.S Venkatesh Rao 130

Vishwatmak Om Gurudev College Of Engineering Thermodynamics DEC 2008(RC) Q 1 (e) 2 kg of steam is at 8 bar and 08 dry Determine its enthalpy and volume 3 MAY 2009(RC) Q 1 (b) One kg of steam at a pressure of 17 bar and dryness 0.95 is heated at a constant pressure, until it is completely dry. Determine (i) Increase in volume (ii) Quantity of heat added. 4 DEC 2009(RC) Q1(b) Steam at 15 bar and 300°C is throttled to 10 bar before supplying to steam turbine. It then undergoes isentropic expansion to 1 bar in the turbine. Determine isentropic heat drop and the condition of steam at exit from the turbine. Use enthalpy- entropy chart. 4 V O G C E MAY 2010(RC) Q1 (d) 2 kg of steam is at 8 bar and 0.8 dry. Determine its enthalpy and volume. 4 Q5 (b) A rigid cylinder of volume 0.028 m3 contains steam at 60 bar and 350 °C. The cylinder is cooled until the pressure is 50 bar. Calculate (i) The state of steam after cooling. (ii) The amount of heat rejected by the steam. 8 DEC 2010(RC) NIL MAY 2011(RC) NIL MAY 2011(RC) Q 1 (c) 2 kg of steam is at 10 bar and 0·8 dry. Determine enthalpy, entropy and volume (use steam table) 5 Prof.S Venkatesh Rao 131

Vishwatmak Om Gurudev College Of Engineering Thermodynamics CHAPTER 8 VAPOUR POWER CYCLE 1. Describe the different operations of Rankine cycle and derive an expression for its efficiency. Vapour power cycles use vapour as a working substance and are used to produce power from conversion of heat into work. The steam power is one of the successful systems for conversion of heat into mechanical work. V O G C E The steam power plant consists of a boiler, steam turbine (or steam engine), steam condenser and feed pump. Figure 1 shows the steam power plant. The heat energy released due to the combustion of fuel will be utilized in the boilers for converting water into steam and this steam is then expanded into the steam engines/steam turbines to obtain useful work. The steam after producing useful work, is generally condensed in the condensers. If the condensate is pure, then it is pumped as feed water to the boiler by means of a feed pump. Prof.S Venkatesh Rao 132

Vishwatmak Om Gurudev College Of Engineering Thermodynamics The Rankine cycle consists of the following processes. Process 1-2 • During 5-1, feed water is heated to saturated temperature (Tsat) in economizer/boiler itself. • During 1-2, heat is added (qi) at constant temperature and pressure P1 converts water into steam in the boiler. In brief, process 5-1-2 is the constant pressure heat addition in the boiler. Process 2-3 • During this process, isentropic expansion of the steam takes place in the turbine from boiler pressure P1 to condenser pressure (back pressure) Pb, which results into turbine work ‘WT’. V O G C E Process 3-4 • The steam is condensed at constant pressure Pb in the condenser. • The steam rejects heat qT to the cooling water in the condenser. Process 4-5 • It is pumping process, where water at condenser pressure Pb is raised to boiler pressure P1. This pumping process is isentropic. • At point 5 temperature is lower than Tsat. i.e. point 5 is in subcooled state. • It should be noted that, the steam generated in the boiler or steam entering the turbine may be wet, dry-saturated or superheated. • Correspondingly the expansion processes in the turbine will be 2’-3’, 2-3’ 2’’ – 3’’ respectively. ANALYSIS OF RANKINE CYCLE Applying SFEE to boiler, turbine, condenser and pump: (Assuming 1 kg of steam). From Equation (2), q-w =∆h Prof.S Venkatesh Rao 133

Vishwatmak Om Gurudev College Of Engineering Thermodynamics For boiler In the boiler, heat qi converts water into steam. ( )q = h - h kJ/kg (W = 0) ∴ i 21 For turbine In the turbine when the steam expands, work will be produced ( )W = h - h kJ/kg ( q = 0) ∴ T 23 For Condenser In the condenser condensation will be completed and qr amount of heat will be rejected to cooling water. q = h -h ∴ r 43 V ( )q = - h - h kJ/kg O∴r 34 G CNegative sign indicates that heat is rejected E For feed pump 0 − W = h - h (q = 0) ∴ P 54 ( )W = - h - h kJ/kg. ∴P 54 Negative sign indicates that work to be supplied pump. Also, pump is a steady flow equipment, we know that work done in a steady flow system. Wp = P1 VdP ∫ Pb The specific volume of water remains constant, since the effect of pressure is negligible. [ ]∴ P1 W = -V ∫ dp = - V P - P p Pb 1b [ ]W = V P - P kJ/kg Magnitude of p 1b Note Pressure P1 and Pb should be in kPa and V = Vf at condenser pressure in m3/kg. Now, Rankine cycle efficiency, Prof.S Venkatesh Rao 134

Vishwatmak Om Gurudev College Of Engineering Thermodynamics η = Shaft work W = net Rankine Heat supplied q i = WT - WP q ( ) ( )i h -h - h -h η =2 3 5 4 Rankine h2 - h5 Since pump work, WP = h5 – h4 is very small, hence can be neglected h -h ∴η =2 3 Rankine h -h 24 V O G C E 2. Sketch the Rankine cycle on p-v and T-s plots. [MAY 2011] 4 Marks Figure: Schematic Diagram of Rankine Cycle Prof.S Venkatesh Rao 135

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Figure: P-V Diagram V O G C E Figure: T-s Diagram Figure: h-s Diagam Prof.S Venkatesh Rao 136

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 3. Give reasons why Carnot cycle cannot be considered as the theoretical cycle for steam power plant even through its efficiency is maximum. [DEC 2010] 4Marks V O G C E • Through Carnot cycle is thermodynamically simple and has the highest thermal efficiency for given limits of temperature T1 and T2, is extremely difficult to operate in practice due to the following reasons. • It is difficult to compress a wet vapour isentropically to the saturated state as required by the process 4-1. • It is difficult to control the quality of condensate coming out of the condenser so that the state 4 is exactly obtained. • Due to incomplete condensation in condenser, feed pump has to handle wet steam with large volume. • Hence it consumes large power thus reducing the work ratio considerably. • Due to incomplete condensation in condenser, feed pump has to handle wet steam with large volume. • Hence it consumes large power thus reducing the work ratio considerably. • Due to poor work ratio, power output of the plant decreases with high S.S.C. • The cycle is more difficult to operate in practice with superheated steam due to the necessity of supplying the superheat at constant temperature instead of constant pressure. • Due to the above limitations, Carnot cycle is replaced by practical cycle called Rankine Cycle and is the usually accepted ideal cycle for steam plant Prof.S Venkatesh Rao 137

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 4. Discuss the differences between Carnot and Rankine Cycle [MAY 2011] 5 Marks i. In Rankine cycle condensation of steam is complete, whereas in case of Carnot cycle it is incomplete. ii. Rankine cycle is practical cycle whereas Carnot cycle is theoretical one. iii. As the condensation is complete in Rankine cycle, feed pump handles condensate (water) having very small volume. Therefore it consumes negligible power. iv. Between the same temperature limits, Rankine cycle provides a higher specific work output than Carnot cycle. v. Due to use of superheated steam in Rankine cycle, steam remains fairly dry even towards the end of expansion in turbine. Hence erosion of turbine blades is minimized, whereas Carnot cycle cannot be operated on superheated steam and hence after expansion steam is wet and erosion of turbine blades is more. V O G C E 5. Draw a simple schematic diagram of a thermal power plant with one reheater. Also represent this on a T—s diagram. [MAY 2009, DEC 2010] 4 Marks Figure 2: Reheat cycle. 138 Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Figure 3: Ideal reheating process on T-s and h-s chart 6. Draw schematic diagram of regenerative Rankine cycle. Also represent this on T-S diagram. [DEC 2009] 4 Marks V O G C E 7. Explain as how reheating and regeneration in Rankine Cycle is beneficial. [MAY 2006, DEC 2006] 4 Marks Benefit of Reheating 1. There is increased output of the turbine 2. Erosion and corrosion problems in the steam turbine are eliminated. 3. Final dryness fraction of the steam is improved 4. Overall efficiency of plant increases Prof.S Venkatesh Rao 139

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Benefit of Regeneration 1. The heating process in the boiler tends to become reversible. 2. The thermal stresses set up in the boiler are minimised. This is due to the fact that temperature ranges in the boiler are reduced. 3. The thermal efficiency is improved because the average temperature of heat addition to the cycle is increased. 4. Heat rate is reduced. 5. The blade height is less due to the reduced amount of steam passed through the low pressure stages. 6. Due to many extractions there is an improvement in the turbine drainage and it reduces erosion due to moisture. 7. A small size condenser is required. V O G C E 8. Explain regenerative cycle with schematic, P-V and T-S diagrams. [MAY 2008] 8 Marks • It is observed that in Rankine cycle the condensate which is fairly at low temperature has irreversible mixing with hot boiler water and this results in decrease of cycle efficiency. • Therefore, some methods are adopted to heat the feed water from the hot well of condenser irreversibly by interchange of heat within the system and thus improving the cycle efficiency. • This method of heating is called regenerative feed water heating and the cycle is called ‘Regenerative cycle’. • The principle of regeneration can be practically utilized by extracting steam from the turbine at several locations and supplying it to the regenerative heaters. • The resulting cycle is known as regenerative or bleeding cycle. This bleed steam is used to preheat the feed water. • Figure (a) shows a layout of a condensing steam power plant of a regenerative cycle with open type of feed water heater. Figure (b) shows T-S diagram for open type feed water heating. Prof.S Venkatesh Rao 140

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Advantages of Regenerative Cycle 1. Heating process in the boiler tends to become reversible. 2. Thermal efficiency is improved 3. Thermal stresses set up in the boiler are minimized 4. Heat rate reduced Disadvantages of Regenerative Cycle 1. For a given power, a large capacity boiler is required. 2. Plant is more complicated. 3. Because of addition of heaters greater maintenance is required. 4. The cost of plant is more. V O G C E 9. What is the effect of regeneration on the — (i) Specific output (ii) Mean temperature of heat addition (iii) Cycle efficiency (iv) Steam rate (v) Heat rate of steam power plant. [DEC 2007] 6 Marks (i) Specific output: Decreases, because the part of steam is removed for feed water heating. (ii)Mean temperature of heat addition: Increases, because the heat addition decreases. (iii) Cycle efficiency: Increases as heat addition decreases, (iv) Steam rate: Increases as specific work output decreases. (v) Heat rate of steam: increase as efficiency increases. Prof.S Venkatesh Rao 141

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 10. What is the effect of reheat on (i) specific output (ii) cycle efficiency (iii) steam rate (iv) heat rate of steam power plant? [MAY 2009] 8 Marks (i) Specific output: Increases (ii) Cycle efficiency: May increase, decrease or remain same (iii) Steam rate: decreases as work out put increases (iv) Heat rate of steam: Increases as extra heat is supplied for reheating 11. What do you understand by the mean temperature of the heat addition? What is metallurgical limit? [DEC 2010] 4 Marks In the Rankine cycle, heat is added reversibly at a constant pressure, but at infinite temperatures. If Tm1 is the mean temperature of heat addition, as shown in Fig., so that the area under 4s and 1 is equal to the area under 5 and 6, then heat added V O G C E Prof.S Venkatesh Rao 142

Vishwatmak Om Gurudev College Of Engineering Thermodynamics The maximum temperature of steam that can be used is fixed from metallurgical considerations (i.e., the materials used for the manufacture of the components which are subjected to high-pressure, high-temperature steam like the super heaters, valves, pipelines, inlet stages of turbine, etc.). When the - maximum temperature is fixed, as the operating steam pressure at which heat is added in the boiler increases from p 1 to p2, the mean temperature of heat addition increases, since T ml between 7 sand 5 is higher than that between 4s and 1. V O G C E Prof.S Venkatesh Rao 143

Vishwatmak Om Gurudev College Of Engineering Thermodynamics FOR LAST MINUTE REVISION 1. Carnot cycle efficiency 2. Rankine cycle is the theoretical cycle on which steam prime movers work. V O G C E 3. The thermal efficiency of Rankine cycle is increased by (i) Increasing the average temperature at which heat is added to the cycle. (ii) Decreasing the average temperature at which heat is rejected to the cycle. 4. Thermal efficiency of regenerative cycle Prof.S Venkatesh Rao 144

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION QUESTIONS MAY 2005 Q 1 Fill in the blanks: (f) As the number of feed water heaters increase, the gain in thermal efficiency in successive regenerative heating _____________________ 1 DEC 2005 Q 1 Fill in the blanks: (iii) _____________ heaters are used to increase the efficiency of a Rankine Cycle. 1 MAY 2006 Q 5 (a) Explain as how reheating and regeneration in Rankine Cycle is beneficial. V 6 O G C E DEC 2006 Q5(a) How do reheating and regeneration benefit in Rankine cycle. How do reheating and regeneration benefit in Rankine cycle. 4 MAY 2007 NIL DEC 2007 Q 6 (b) What is the effect of regeneration on the — (i) Specific output (ii) Mean temperature of heat addition (iii) Cycle efficiency (iv) Steam rate (v) Heat rate of steam power plant. 6 MAY 2008 Q 4 (a) Explain regenerative cycle with schematic, P-V and T-S diagrams. 8 DEC 2008(RC) NIL Prof.S Venkatesh Rao 145

Vishwatmak Om Gurudev College Of Engineering Thermodynamics MAY 2009(RC) Q 1 (c) Draw a simple schematic diagram of a thermal power plant with one reheater. Also represent this on a T—s diagram. 4 Q 5 (b) What is the effect of reheat on (i) specific output (ii) cycle efficiency (iii) steam rate (iv) heat rate of steam power plant. 8 DEC 2009(RC) Q7(a) Draw schematic diagram of regenerative Rankine cycle. Also represent this on T-S diagram. 4 MAY 2010(RC) NIL DEC 2010(RC) Q1(b) Draw a simple schematic of thermal plant with one reheater. Also represent it on T-s diagram. 4 Q1 (e) What do you understand by the mean temperature of the heat addition? What is metallurgical limit? 4 Q6 (a) Give reasons why Carnot cycle cannot be considered as the theoretical cycle for steam power plant even through its efficiency is maximum. 4 V O G C E MAY 2011(RC) Q 1 (c) Discuss the differences between Carnot and Rankine Cycle 5 DEC 2011(RC) Q 5 (b) Define- (iv) Work ratio (v) Specific steam consumption. 2 MAY 2012(RC) Q 1 (f) Sketch the Rankine cycle on p-v and T-s plots. 4 Prof.S Venkatesh Rao 146

Vishwatmak Om Gurudev College Of Engineering Thermodynamics GRADED PROBLEMS TYPE I: SIMPLE CYCLE 1. In a steam power plant boiler pressure is 60 bars and condenser pressure is 0.08 bars. The steam temperature at boiler outlet is 500°C. Determine:— (i) Turbine work per kg (ii) Heat transfer in condenser per kg (iii) Cycle efficiency (iv) Mass flow rate of steam to provide 5 MW. 12[MAY 2008, DEC 2007] 2. Steam at 20 bar, 360°C is expanded in a steam turbine to 0.08bar. It then enters a condenser, where it is condensed to saturated liquid water. The pump feed back the water in to the boiler. (a) Assuming the ideal processes, find per kg of steam the net work and cycle efficiency.(b) If the turbine and the pump have each 80% efficiency, find the percentage reduction in the net work and cycle efficiency. Sketch p-v and T-s diagrams. 10 [DEC 2006] V O G C E 3. In a Rankine Cycle, the steam at inlet to the turbine is at 10 MPa and 450°C. The exhaust pressure is 0.6bar. Determine - (i) the Pump work (ii) the Turbine work(iii) the Rankine efficiency (iv) the Condensate heat flow (v) the Dryness fraction at the end of expansion (solve the problem with the help of steam table only) 10[MAY 2006] TYPE II: REHEAT CYCLE 4. A steam Power station uses the fallowing cycle. Steam at boiler outlet — 150 bar and 550°C, reheat at 40 bar to 550°C, condenser at 0.1 bar. Find (i) Quality at turbine exhaust (ii) Cycle efficiency (iii) Steam rate. 12[MAY 2009(RC)] 5. A steam power plant works on Rankine cycle between pressure ratio 20 bar and 0.05 bar. Steam supplied to the turbine is dry and saturated. Find the thermal efficiency, work ratio and specific steam consumption. What would be efficiency and work ratio in case of Carnot cycle operating in the pressure limits. 10 [DEC 2008(OC)] Prof.S Venkatesh Rao 147

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 6. In a reheat cycle steam at 500°C expands in H.P turbine till it is saturated vapour. It is reheated at constant pressure to 400°C and then expands in L.P. turbine to 40°C. If the maximum moisture content is limited to 15% at the turbine exhaust find (i) reheat pressure (ii) the pressure of steam at inlet to H.P turbine (iii) net specific work output (iv) cycle efficiency (v) steam rate. Assume all ideal process and neglect feed pump work. 12[DEC 2008(RC)] TYPE III: REGENERATIVE CYCLE 7. In a steam power plant the condition of steam at inlet to the steam generator is 20 bar and 300° C and the condenser pressure is 0.1 bar. Two feedwater heaters operate at optimum temperature. Determine (a).the quality of steam at exhaust, (b) network per kg of steam (c) cycle efficiency and (d) the steam rate. Neglect pump work 10 [MAY 2007] V O G C E 8. A boiler feeds an engine at 56 bar and 600 0 C. Before being passed on to the condenser at 30°C, the steam is bled for regenerative feed heating at 65 bar. For an ideal regenerative cycle and 1 kg/s of throttle steam, determine — (i) The amount of bled steam. (ii) Net work done (iii) The thermal efficiency of the cycle (iv) Steam rate per kW-hr. (v) Mean temperature of heat addition. 10 [MAY 2005] 9. A thermal power plant operates on Rankine cycle with regenerator. Superheated steam at 3.5 MPa and 500°C enters the turbine. A suitable fraction of the steam is allowed to expand to the condenser pr. of 10 kPa. The extracted steam is mixed with the condensate after it has been raised to 0.6 MPa and the mixture is fed to the feed pump, calculate % increase, in the thermal efficiency of the plant, and the % decrease in the net work obtained from the plant. 9 [JUNE- 95] Prof.S Venkatesh Rao 148

Vishwatmak Om Gurudev College Of Engineering Thermodynamics TYPE IV : REHEAT+ REGENERATIVE CYCLE 10. A steam power plant equipped with regenerative as well as reheat arrangement is supplied with steam to the H.P turbine at 80 bar and 470°C. For Feed heating, a part of steam is extracted at 7 bar and the remainder of steam is reheated to 350°C in a reheater and then expanded in L.P. turbine down to 0.035 bar. Determine the following:--(i) Amount of steam bled off for feed heating (ii). Amount of steam in L.P. turbine (iii) Heat supplied in the boiler and reheater (iv) Output of the turbine (v) Cycle efficiency 12 [DEC 2004, DEC 2005 V O G C E Prof.S Venkatesh Rao 149

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION PROBLEMS (SOLVE AS HOME WORK) MAY 2005 Q 5 (b) A boiler feeds an engine at 56 bar and 600 0 C. Before being passed on to the condenser at 30°C, the steam is bled for regenerative feed heating at 65 bar. For an ideal regenerative cycle and 1 kg/s of throttle steam, determine — (i) The amount of bled steam. (ii) Net work done (iii) The thermal efficiency of the cycle (iv) Steam rate per kW-hr. (v) Mean temperature of heat addition. 10 V O G C E DEC 2005 Q 4 (b) In a reheat-regenerative Rankine Cycle the steam enters the HP turbine at 40 bar, 3600 C and expands up to 6 bar. At this stage, some part of the steam is used for feed water heating while the rest is reheated up to 360°C. This steam is then expanded in the LP turbine up to 0.2 bar pressure. For this cycle— (i) Represent the cycle on T—S diagram (ii) Find Enthalpies at all salient points (iii) Find the total pump work (iv) Find total turbine work (v) Find thermal efficiency of Rankine Cycle (vi) Steam flow rate/MW power generated. 12 MAY 2006 Q 5 (b) In a Rankine Cycle, the steam at inlet to the turbine is at 10 MPa and 450°C. The exhaust pressure is 0.6bar. Determine - (i) the Pump work (ii) the Turbine work(iii) the Rankine efficiency (iv) the Condensate heat flow (v) the Dryness fraction at the end of expansion (solve the problem with the help of steam table only) . 10 DEC 2006 Q 5 (b) Steam at 20 bar, 360°C is expanded in a steam turbine to 0.08bar. It then enters a condenser, where it is condensed to saturated liquid water. The pump feed back the water in to Prof.S Venkatesh Rao 150