ENGINEERING THERMODYNAMICS STUDY MATERIAL

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 4. Write the statements of second law of thermodynamics. [MAY 2005] 4 Marks OR State and explain in brief Kelvin- Plank and Clausius statement of the second law of Thermodynamics [DEC 2006, DEC 2010] 4 Marks KELVIN-PLANCK STATEMENT “It is impossible to construct a heat engine to work in a cyclic process whose sole effect is to convert all the heat supplied to it into an equivalent amount of work”. CLAUSIUS STATEMENT “It is impossible to construct a device to work in a cyclic process whose sole effect is the transfer of heat from a body at a lower temperature to a body at a higher temperature”. Another form of the statement is “It is impossible for heat energy to flow from a body at a lower temperature to a body at a higher temperature without the aid of external work.” Both the statements are represented diagrammatically in Figure V O G C E 5. Establish the equivalence of Kelvin-Plank and Clausius statement of second law of thermodynamics. [MAY 2007] 4 marks OR State the Kelvin-Planck and Clausius statements of second law, also show their equivalence. [DEC 2007, MAY 2009, MAY 2011, MAY 2012] 6-10 marks Prof.S Venkatesh Rao 51

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EQUIVALENCE • The two statements of the second law lead to same conclusions even though they appear different. • It is impossible to have a system, which satisfies one of the statements and violates the other. • Consider a system as shown in Figure (a). = Fig (a) • It is obvious that, both the heat engine and heat pump obey the first law of thermodynamics but the pump obeys Clausius statement and engine violates Planck statement. • It we consider this system as a whole, we find that Q2 units of heat are continuously transferred from the heat sink at T2 to heat source at T1 (T1 > T2) without any external input to the system, which is against Clausius statement. • Thus a violation of Kelvin-Planck statement leads to the violation of Clausius statement. • Considering the system shown in Figure (b) the heat pump violates the Clausius statement while the engine satisfies the Kelvin-Planck statement. V O G C E Prof.S Venkatesh Rao 52

Vishwatmak Om Gurudev College Of Engineering Thermodynamics • But taken together the net effect is that an amount of heat (Q1-Q2) is taken from the source at T1 and is being completely and continuously converted into work (as no net exchange takes place with the sink) and this is against Kelvin Planck statement. • Thus a violation of Clausius statement leads to the violation of Kelvin-Planck statement. • Therefore it is obvious that the two statements are complementary to each other. • The truth of the first implied the truth of the second and vice-versa. 6. Explain Carnot engine and Carnot cycle [DEC 2008] 4 marks OR Explain Carnot engine and Carnot cycle. Prove that efficiency of Carnot engine is equal to V O G C E [MAY 2009] 10 Marks Give reasons why the Carnot cycle cannot be considered as the practical power plant cycle even-though it’s efficiency is maximum [MAY 2010] 4 Marks The Carnot cycle consists of four reversible processes, two frictionless isothermals and two frictionless adiabatic. A cyclic heat engine operating on the Carnot cycle is called a Carnot heat engine. A reversible cycle is an ideal hypothetical cycle in which all the processes constituting the cycle are reversible. Carnot cycle is a reversible cycle The working of the engine is represented on p-v & T-s diagram as shown in Figure and the sequence of operations is described below. Prof.S Venkatesh Rao 53

Vishwatmak Om Gurudev College Of Engineering Thermodynamics ProcessesV a-b Isothermal Heat AdditionO b-c Isentropic ExpansionG c-d Isothermal Heat RejectionC d-a Isentropic CompressionE EFFICIENCY OF CARNOT ENGINE Heat added during isothermal expansion process (a-b), 54 Heat rejected during isothermal expansion process (c-d) Negative sign indicates that heat is rejected from the system. But, Net work done during cycle, Thermal efficiency Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics The Carnot cycle cannot be performed in practice because of the following reasons: 1. It is impossible to perform a frictionless process. 2. It is impossible to transfer the heat without temperature potential. 3. Isothermal process can be achieved only if the piston moves very slowly to allow heat transfer so that the temperature remains constant. Adiabatic process can be achieved only if the piston moves as fast as possible so that the heat transfer is negligible due to very short time available. The isothermal and adiabatic processes take place during the same stroke therefore the piston has to move very slowly for part of the stroke and it has to move very fast during remaining stroke. This variation of motion of the piston during the same stroke is not possible. Reversed Carnot Cycle Carnot cycle can be used for a heat pump or a refrigerator when reversed, as shown in Figure. V O G C E If the purpose of reversed Carnot cycle is to extract the heat at temperature T2 (refrigerator) then the C.O.P. of the cycle is given by C .O.P . = Q2 = T2 where T1 > T2 W T1 − T2 If the purpose of the reversed Carnot cycle is to pump the atmospheric heat to a high temperature T1 (heat pump) then the C.O.P. of the cycle is given by C .O.P . = Q1 = T1 T1 > T2 W T1 − T2 Prof.S Venkatesh Rao 55

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 7. State and explain Carnot theorem [DEC 2008] 6 Marks OR State and prove Carnot Theorem [MAY 2011] 8 Marks 1. “No heat engine working in a cycle between two fixed temperatures can be more efficient than a reversible heat engine working between the same two fixed temperatures”. • The proof of the statement is given as follows – • Consider a reversible engine HEr working between two fixed temperatures T1 and T2 taking 100 kJ of heat from a source at temperature T1 and delivering 40 V O G C E kJ as work and the remaining 60 kJ being rejected to the sink as shown in Figure (a). • This gives 40% efficiency. • If the engine is reversed then 40 kJ work is required to pump 60 kJ from sink to the source. Totally 100 kJ will be pumped to the source. • Now assume that an irreversible engine HEi is more efficient than the reversible engine HEr. • Let the efficiency of the irreversible engine be 50% when working between the same temperatures limits T1 and T2 (higher than that of reversible engine). • Let this engine drive the reversible engine working between T1 and T2. • The heat taken by irreversible engine at temperature T1 to develop 40kJ of work = work/efficiency = 40/0.5 = 80 kJ and heat rejected to the sink is (80 – 40) = 40 kJ. • Therefore in this arrangement, the heat pumped by a reversible heat pump is 100 kJ and the heat taken by the irreversible heat engine to run the heat pump is 80 kJ. • Therefore Heat pumped to source by HEr = 100kJ • Heat taken from the source by HEi = 80 kJ • Heat pumped from the sink by HEr = 60 kJ • Heat rejected to the sink by HEr = 40 kJ Prof.S Venkatesh Rao 56

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Figure 6 Proof for Carnot Principle • Therefore the engine HEi and the reversed engine HPr taken together extract (60 – 40 ) = 20 kJ of heat from the low temperature sink at T2 and pump it (continuously) to the high temperature source at T1 without any other external inputs. • This is contradictory to the second law of thermodynamics. • Therefore, the assumption that HEi is more efficient than HEr should be incorrect. • So a reversible engine will be more efficient than an irreversible engine working between the same temperature limits. • HEr = Reversible Engine, HEi = Irreversible Engine. HPr = Reversible heat pump. V O G C E The second Carnot theorem is – 2. All reversible engines working between the same temperature limits have the same efficiency. • Replacing the irreversible engine by a reversible in the previous discussion, and applying the same reasoning. • It can be proved that one reversible engine cannot be more efficient than another reversible engine when working between the same temperature limits. • These two statements together with the proof lead to the following generation of the efficiency of reversible cycles. Prof.S Venkatesh Rao 57

Vishwatmak Om Gurudev College Of Engineering Thermodynamics • It was proved that for a Carnot cycle using perfect gas and working between T1 and T2 the efficiency is given by ������������ = .������������������������−������������������������ ������������������������ • It may also be noted that the Carnot cycle is a reversible cycle working between T1 and T2 should have the same efficiency as the Carnot cycle and the efficiency is not affected by the working fluid. The conclusion of the discussion is that the efficiency of reversible engines is dependent only on the working temperature and no other parameters. 8. Write short notes on: Thermodynamic temperature scale [MAY 2008] 5 Marks OR Write short notes on: absolute thermodynamic temperature scale [MAY 2010] 5 Marks V O G C E The temperature scales have been developed keeping the ice and steam points as references. The thermometers which are constructed using the physical properties of substances have got many drawbacks. The different thermometers may read different temperature except at the fixed point at which these are calibrated, because, the actual relationship between the thermometric properties of the substance and temperature (as volume – temperature or pressure temperature of resistance temperature relationship) may not agree at all points on the scale as shown in Figure. Prof.S Venkatesh Rao 58

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Different thermometers may read different temperatures in the intermediate region depending upon the properties of the substance used. Extrapolation also may be erroneous. Therefore, it is necessary to have some scale which will not be affected by the property of the thermometric substance. Such a scale can be developed by using the concept of reversible engines. “A temperature scale which is independent of the property of the thermometric substance is defined as thermodynamic temperature scale”. The efficiency of any heat engine cycle receiving heatQ1 and rejecting heat Q2 is givenV byO G ….. (1)C E By the second law, it is necessary to have a temperature difference (t1 - t2) to obtain work of any cycle. We know that the efficiency of all heat engines operating between the same temperature levels is the same, and it is independent of the working substance. Therefore, for a reversible cycle (Carnot cycle), the efficiency will depend solely upon the temperatures t1 and t2 at which heat is transferred, or ….. (2) Where f signifies some function of the temperatures. From Eqs (1) and (2) In terms of a new function F ….. (3) Prof.S Venkatesh Rao 59

Vishwatmak Om Gurudev College Of Engineering Thermodynamics If some functional relationship is assigned between t1, t2 and Q1/Q2, the equation becomes the definition of a temperature scale. Let us consider two reversible heat engines, E1 receiving heat from the source at t1 and rejecting heat at t2 to E2 which, in turn, rejects heat to the sink at t3 as shown in fig. Now E1 and E2 together constitute another heat engine E3 operating between t1 and t3 Now, V O G C E ….. (3) The temperatures t1, t2 and t3 are arbitrarily chosen. The ratio Q1 /Q2 depends only on t1 and t2, and is independent of t3. So t3 will drop out from the ratio on the right in equation (3). After it has been cancelled, the numerator can be written as φ (t1), and the denominator as φ (t2), where if> is another unknown function. Thus Since φ(t) is an arbitrary function, the simplest possible way to define the absolute thermodynamic temperature T is to let φ(t) = T, as proposed by Kelvin. Then, by definition ….. (4) The absolute thermodynamic temperature scale is also known as the Kelvin scale. Two temperatures on the Kelvin scale bear the same relationship to each other as do the heats absorbed and rejected respectively by a Carnot engine operating between two Prof.S Venkatesh Rao 60

Vishwatmak Om Gurudev College Of Engineering Thermodynamics reservoirs at these temperatures. The Kelvin temperature scale is, therefore, independent of the peculiar characteristics of any particular substance. The heat absorbed Q1 and the heat rejected Q2 during the two reversible isothermal processes bounded by two reversible adiabatics in a Carnot engine can be measured. In defining the Kelvin temperature scale also, the triple point of water is taken as the standard reference point. For a Carnot engine operating between reservoirs at temperatures T and Tt , Tt being the triple point of water (Fig.), arbitrarily assigned the value 273.16 K, V O G C E Prof.S Venkatesh Rao 61

Vishwatmak Om Gurudev College Of Engineering Thermodynamics FOR LAST MINUTE REVISION 1. Limitations of the first law of thermodynamics are (a) Processes can proceed in one direction only e.g. heat can flow from high tempera to low temperature body. However, heat cannot flow from low temperature bod3 high temperature body though it would not violate the first law. (b) All processes involving conversion of heat into work and vice-versa are not equivalent 2. A heat reservoir is source of infinite heat energy from which finite transfer of heat energy will not affect its temperature and other properties. 3. A heat engine is a device which can produce work continuously at the expense of heat en supplied to the engine. 4. A Carnot engine is a reversible heat engine or an ideal engine consisting of two isothermal and two reversible adiabatic processes. V O G C E 5. Carnot refrigerator and Carnot heat pump. 62 Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 6. (a) Clausius statement of second law of thermodynamics states that it is impossible to construct a device whose sole effect is transfer of heat from a cold body to hot body. (b) Kelvin-Planck statement of second law of thermodynamics states that it is impossible to construct a heat engine operating on a cycle whose sole effect is the transfer of heat from a single heat reservoir and its conversion into equal amount of work. 7. (a) Perpetual motion machine of first kind (PPM - 1) is a machine which will produce continuous work without receiving corresponding energy by the machine energy. It violates the first law of thermodynamics. (b) Perpetual motion machine of second kind (PPM - 2) is a machine which receives heat from a single heat reservoir and produces equal amount of work. It violates the second law of thermodynamics, hence, such a machine cannot be built in practice. V O G C E 8. Both the statements of second law of thermodynamics given above are equivalent. 9. Work is a high grade form of energy and heat is a low grade form of energy. 10. A reversible or ideal process is one which can be traced back to its initial state without leaving any change either in the system or surroundings 11. Causes of irreversibility are: (i) External irreversibility due to dissipative effects like friction and viscosity and due to finite temperature difference. (ii) Internal irreversibility. (iii) Chemical irreversibility. 12. Conditions for reversibility are (i) Process is carried out in thermodynamic equilibrium. (ii) The dissipative effects like friction and viscosity are not present in the system. 13. The Carnot Theorem states that when working between the given fixed temperature limits. (i) No engine operating on a cycle can be more efficient than reversible engine. (ii) All reversible engines are equally efficient. Prof.S Venkatesh Rao 63

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 14. Lord Kelvin in 1848 defined a temperature scale which is independent of thermodynamic properties of substance. It is called absolute thermodynamic temperature scale or Kelvin scale. V O G C E Prof.S Venkatesh Rao 64

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION QUESTIONS MAY 2005 Q 1 Fill in the blanks: (e) The absolute zero temperature cannot be attained without the violation of _______________of thermodynamics. 1 (g) Two Cannot engines producing different powers but working between same temperature limits will have ____________thermal efficiency. 1 Q 3(a) State both the statements of the second Law of thermodynamics. 4 DEC 2005 Q 1 (b) State TRUE or FALSE. If false, correct it. (ii) If the temperature. of source is increased, then the Carnot Cycle, efficiency decreases. 1 (iv) Two reversible engines working between same temperature limits can have different efficiencies. 1 (v) If it is possible to reject heat at absolute zero temperature, PPM-Il becomes theoretically possible. 1 Q 3(b) Show the COP of heat pump is equal to one plus COP of refrigerator when working between same temperature limits. 5 V O G C E MAY 2006 Q 3(a) Write the statements of second law of thermodynamics. 4 DEC 2006 Q 1 (c) State and explain in brief Plank-Kelvin and Clausius statement of the second law of Thermodynamics. 4 Q 2(b) Obtain an expression for efficiency of Carnot cycle. Why is the efficiency of an engine working on Carnot cycle maximum? 6 MAY 2007 Q1(b) Establish the equivalence of Kelvin-Plank and Clausius statement of second law of thermodynamics. 4 Q 2(b) Show that COP of heat pump is equal to one plus COP of refrigerator when working between same temperature limits. 4 Prof.S Venkatesh Rao 65

Vishwatmak Om Gurudev College Of Engineering Thermodynamics DEC 2007 Q1(b) Define the terms - Heat engine, Heat pump C.O.P. of refrigerator and C.O.P. of heat pump. 4 Q 2(b) State the Kelvin-Planck and Clausius statements of second law, also show their equivalence. 6 MAY 2008 Q 7 Write short notes on: (e) Thermodynamic temperature scale. 5 DEC 2008 (RC) Q 2 (b) State and explain Carnot theorem 6 Q 2 (c) Explain Carnot engine and Carnot cycle 4 MAY 2009 (RC) Q 3 (b) Explain Carnot engine and Carnot cycle. Prove that efficiency of Carnot engine is equal to V O G C E 10 DEC 2008 (RC) Q4(a) State the Kelvin plank and Clausius statements of the second law of thermodynamics, and establish the equivalence between them. 8 MAY 2010 (RC) Q1(b) Give reasons why the Carnot cycle cannot be considered as the practical power plant cycle even-though it’s efficiency is maximum. 4 Q3 (a) State and prove Carnot theorem. 7 Q7. Write short notes on: (a) Absolute thermodynamic temperature scale 5 Prof.S Venkatesh Rao 66

Vishwatmak Om Gurudev College Of Engineering Thermodynamics DEC 2010 (RC) Q1(d) Prove that (COP)HP = 1 + (COP)Ref 4 Q3(a) State and explain Kelvin plank and Clausius statements of second law of thermodynamics. 4 MAY 2011(RC) Q 1 (c) Explain with examples the limitations of First Law of Thermodynamics. 5 Q 2 (a) State and explain the equivalence of Kelvin Planck and Clausius statements of the second law of thermodynamics 6 Q 2 (b) Define C.O.P of Heat Pump and Refrigerator and derive the relationship between the two. 4 V O G C E DEC 2011 (RC) 1. (a) Fill in the blanks : (iii) Two carnot engines producing different powers but working between same temperature limits will have - -- -- - thermal efficiency. Q 3 (c) Write the limitations of first law of thermodynamics 4 Q 3 (a) State and prove Carnot Theorem 8 MAY 2012(RC) Q 1 (d) Define the heat engine, refrigerator, heat pump. 4 Q 2. (a) What are the limitation of first law of thermodynamics? 4 Q 3. (a) Write the statement of second law of thermodynamics and establish the equivalence between them. 10 Prof.S Venkatesh Rao 67

Vishwatmak Om Gurudev College Of Engineering Thermodynamics GRADED PROBLEMS TYPE I: DIRECT APPLICATION OF SECOND LAW 1. A cyclic heat engine operates between a source temperature of 1000 °C and sink temperature of 40 °C. Find the least rate of heat rejection per kW net output of the engine. 4[DEC 2008 (OC)] 2. COP of heat pump is 6. The power consumed by it is 6 kW. What is the refrigeration capacity of the heat pump? 3 [MAY 2008] 3. A reversible heat engine in a satellite operates between a hot reservoir at T1 and a radiating panel at T2. Radiation from the panel is proportional to its area and to T24. For a given work output and value of T1 show that the area of the panel will be minimum when V O G C E Determine the minimum area of the panel for an output of 1 kW If the constant of proportionality is 5.67 x 10-8 W/m2K4 and T1 is 1000 K. 8 [DEC 2004] TYPE II: INVENTORS CLAIM 4. A inventor claims that a new heat cycle will develop 400 W for a heat supply of 35.5 kJ/min. Is his claim possible? Given that source is at 2000 k and sink is at 850 k. 3 [DEC 2002] 5. An inventor claims to have developed a heat engine which has the following specifications:- Power developed =70kW Fuel burnt per hour = 7.2 Kg Calorific value of fuel = 44000 kJ/kg Temperature of the source = 727°C Temperature of the sink = 27°C Verify whether his claim is true. [MAY-92-8]: TYPE III: ENGINE OPERATING HEAT PUMP 6. A reversible power cycle is used to drive a reversible heat pump cycle. The power cycle takes in Q1 heat units at T1 and rejects Q2 at T2. The heat pump abstracts Q4 from the sink at T4 and discharges Q3 atT3. Develop an expression for the ratio Q4/Q3 terms of the four temperatures. 8 [DEC 2003] Prof.S Venkatesh Rao 68

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 7. A reversible heat engine operates between two reservoirs at temperatures of 600° C and 40° C. The engine drives a reversible refrigerator, which operates between reservoirs at temperature of 40 °C and —20° C. The heat transfer to the heat to the heat engine is 20000 kJ and the network output of the combined engine refrigerator plant is 360 kJ (i) Evaluate the heat transfer to the refrigerant and the net heat transfer to the reservoir at 400 C (ii) Reconsider (i) given that the efficiency of the heat engine and. the COP of The refrigerator are each 40% of their maximum possible values. 10 [MAY 2007, DEC 2002, MAY 2006, DEC 2008 (RC), DEC 2008 (OC)]] 8. An engine with 30% efficiency drives a refrigerator having COP of 5. What is the heat input into the engine if 10 MJ of heat is removed from the cold body by the refrigerator? Find total quantity of heat rejected to the surrounding. 6 [DEC 2006] TYPE IV: TWO OR MORE ENGINES IN SERIES V O G C E 9. Three Carnot engines R1, R2, R3 operate in series between two heat reservoirs which are at temperatures of 1000K and 300 K. Calculate intermediate temperatures if amount of work produced by these engines is in the proportion of 5 :4 :3. 10 [MAY 2009 (RC)] 10. Show that the thermodynamic temperature scale is independent of the working fluid. Two Carnot engine work in series between the source and sink temperature of 550 K and 350 K. If both engines develop equal power, determine the intermediate temperature. 10 [MAY 2004] Prof.S Venkatesh Rao 69

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION PROBLEMS (SOLVE AS HOME WORK) MAY 2005 NIL DEC 2005 NIL MAY 2006 Q 3(c) A reversible heat engine operates between two reservoirs at temperatures of 600°C and 40°C. The engine drives the refrigerator, which operates, between the reservoir at temperature of 40°C and -20°C. The heat transfer to. the engine is 2 MJ and the net work output of the combined engine and refrigerator plant is 360 kJ. Find heat transfer to the refrigerant and’ the net heat transfer to the reservoir at 40°C. 10 V O G C E DEC 2006 Q 2(b) An engine with 30% efficiency drives a refrigerator having COP of 5. What is the heat input into the engine if 10 MJ of heat is removed from the cold body by the refrigerator? Find total quantity of heat rejected to the surrounding. 6 MAY 2007 Q 4 (a) A reversible heat engine operates between two reservoirs at temperatures of 600° C and 40° C. The engine drives a reversible refrigerator, which operates between reservoirs at temperature of 40 °C and -20° C. The heat transfer to the heat to the heat engine is 20000 kJ and the network output of the combined engine refrigerator plant is 360 kJ (i) Evaluate the heat transfer to the refrigerant and the net heat transfer to the reservoir at 400 C (ii) Reconsider (i) given that the efficiency of the heat engine and. the COP of The refrigerator are each 40% of their maximum possible values. 10 DEC 2007 NIL MAY 2008 Q1 (d) COP of heat pump is 6. The power consumed by it is 6 kW. What is the refrigeration capacity of the heat pump? 3 Q 3(c) A Carnot engine operates between two reservoirs at temperature 800 K and 300 K.This engine drives a Carnot refrigerator, which operates, between reservoirs at temperature 255 K and 300 K. The heat transferred to the engine is 2000 kJ and the network output of, the combined engine and refrigerator system is 350 kJ. Find (i) Heat absorbed by the refrigerant. (ii) Net heat rejected to the reservoir at 300 K. 10 Prof.S Venkatesh Rao 70

Vishwatmak Om Gurudev College Of Engineering Thermodynamics DEC 2008 (RC) Q 3 (a) A heat engine operating between two reservoirs at 1200 K and 300 K is used to drive a heat pump which extracts heat from the reservoir at 300 K at a rate twice that at which the engine rejects heat to it. If the efficiency of the engine is 40% of the maximum possible and the C.O. P. of the heat pump is 50% of the maximum possible. What is the temperature of the reservoir to which the heat pump rejects heat? What is the rate of heat rejection from the heat pump if the rate of heat supply to the engine is 50 kW 12 MAY 2009 (RC) Q 4 (a) Three Carnot engines R1, R2, R3 operate in series between two heat reservoirs which are at temperatures of 1000K and 300 K. Calculate intermediate temperatures if amount of work produced by these engines is in the proportion of 5 :4 :3. 10 DEC 2009 (RC) Q4 (b) A reversible engine receives heat from two constant temperature sources at 870K and 580K. rejects 3000kJ/min to a sink at 290K. The engine develops 85 kW. Determine the heat supplied each source and the efficiency of the engine. 8 V O G C E MAY 2010 (RC) Q1(c) An inventor claims that a new heat cycle will develop 400 W for a heat supply of 355 kJ/min. Is his claim possible ? Given that source is at 200 K and sink is at 850 K. 4 Q4 (a) A reversible power cycle is used to drive a reversible heat pump cycle. The power cycle takes in Q1, heat units at temperature T1 and rejects Q2 heat units at temperature T2. The heat pump abstracts Q4 heat units from the sink at T4 and discharges Q3 heat units at temperature T3. Develop an expression for the ratio(Q4/Q3) in terms of four temperatures. 8 DEC 2010 (RC) Q4(b) A reversible heat engine absorbs 3000 kJ from a reservoir at 1000 K and rejects 2500 kJ at 500 K. Find the heat interchanged with the reservoir at 300 K and the net work output of the engine. 10 MAY 2011 (RC) Q 3 (a) A heat engine operating between two reservoirs at 1000 K and 300 K is used to drive a heat pump which extracts heat from the reservoir at 300 K at a rate twice that at which the engine rejects heat to it. If the efficiency of the engine is 40% of the maximum possible and the COP of the heat pump is 50% of the maximum possible what is the temperature of the reservoir to which the heat pump rejects heat? What is the rate of heat rejection from the heat pump if the rate of heat supply to the engine is 50 kW. 10 DEC 2011 (RC) NIL MAY 2012 (RC) NIL Prof.S Venkatesh Rao 71

Vishwatmak Om Gurudev College Of Engineering Thermodynamics DEC 2012 (RC) Q 3 (c) A reversible heat engine operates between two reservoirs at temperature of 600°C and 60°C the engine drives the refrigerator which operates between the reservoirs at temperature of 60°C and - 30°C. The heat transfer to the engine is 3 MJ and the net work output of the combined engine and refrigerator plant is 380 KJ. Find heat transfer to the refrigerant and the net heat transfer to the reservoir at 60°C. 10 V O G C E Prof.S Venkatesh Rao 72

Vishwatmak Om Gurudev College Of Engineering Thermodynamics CHAPTER 4 ENTROPY 1. What do you know by Clausius Inequality? [DEC 2009, MAY 2008, MAY 2010] 4-6 Marks OR State and prove Clausius inequality for an irreversible cycle. [DEC 2007, DEC 2009] 6 Marks CLAUSIUS THEOREM & INEQUALITY Consider a reversible cycle as shown in figure. It is divided into large number of strips by means of reversible adiabatics. Each strip may be closed at the top and bottom by means of reversible isotherms. The original closed cycle is thus replaced by a zigzag closed path, consisting of alternate adiabatic and isothermal process, such that the heat transferred ruing all the isothermal process is equal to the heat transferred in the original cycle. Now, for cycle 1-2-3-4. δQ1 is heat absorbed reversibly at T1 and δQ2 is heat rejected reversibly at T2 V O G C E Similarly for another elemental cycle 5-6-7-8 73 δQ1 + δQ 4 = 0 T3 T4 Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics If similar equations are written for all the elemental Carnot cycles, then for the whole (complete) original cycle  δQ1 + δQ 2  +  δQ 3 + δQ 4  + ............ =0 T1 T2 T3 T4 ∫or δQ = 0 T R δQ Thus the cyclic integral of T for a reversible cycle is equal to zero. This is known as Clausius Theorem FOR IRREVERSIBLE CYCLE According to Clausius Theorem, ∫ δQ = 0 T R In the II law, during Carnot’s theorem, we have proved that the efficiency of the Reversible engine is more than that of an Irreversible engine. SOURCE, AT T1>T2 Q1 H. W E Q2 SINK, AT T2 V O G C E Or ηR > ηI We also know that, η1 < ηR ….(1) Prof.S Venkatesh Rao 74

Vishwatmak Om Gurudev College Of Engineering Thermodynamics η of any engine = Heat supplied - Heat rejected Heat supplied = Q1 - Q2 Q1 or For small amount of heat δQ. η = δQ1 - δQ2 δQ1 = 1- δQ2 δQ1 Hence the Equation (1) will therefore be, But for reversible engine, Q1 = T1 Q1 T2  − δQ 2  = 1- T2 1 δQ1  T1 V  R O GBut for irreversible engine, C E 1−  δQ 2 1 < 1- T2 δQ1 T1  δQ 2 1 > T2 δQ1 T1 or opposite  δQ1  < T1 δQ 2 T2  δQ1 1 -  δQ 2 1 < 0 T1 T2 or We know that, heat added is +ve ∴ δQ1 is +ve And heat rejected is –ve ∴ δQ2 is –ve  δQ1 1 +  δQ 2 1 < 0 T1 T2 δQ , i.e. The algebraic sum of T for an irreversible cycle is always less than zero. ∫ δQ < 0 for an Irreversible cycle T 1 Prof.S Venkatesh Rao 75

Vishwatmak Om Gurudev College Of Engineering Thermodynamics ∫ δQ = 0 T And we know that from Clausius theorem for a reversible cycle. Combining results for reversible and irreversible cycle, we may write. ∫ δQ ≤ 0 T This expression is known as Clausius inequality. It implies whether any cyclic process is reversible or irreversible or impossible. 2. Define entropy and show that entropy is a property of system. [ MAY 2007, MAY 2009, MAY 2009] 4 Marks • The term 'entropy' which literally means transformation, was first introduced by Clausius. • It is an important thermodynamic property of a working substance which increases with the addition of heat, and decreases with its removal. • As a matter of fact, it is tedious to define the term entropy. • But it is comparatively easy to define change of entropy of a working substance. • In a reversible process, over a small range of temperature, the increase or decrease of entropy, when multiplied by the absolute ·temperature, gives the heat absorbed or rejected by the working substance. Mathematically, heat absorbed by the working substance, V O G C E ENTROPY – PROPERTY OF A SYSTEM To prove this, we have to prove that the change of entropy does not depend upon path but it depends upon end states. Then, we will able to say that, entropy is a property of the system. Prof.S Venkatesh Rao 76

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Consider a system, which changes its state from state point (1) to state point (2) by following the reversible path ‘a’ and returns from state point (2) to state point (1) by following the reversible path ‘b’. Then the two paths 1-a-2 and 2-b-1 together will form a cycle V ONow from Clausius theorem. G C ∫ δQ = 0 E (1− a − 2−b −1) T The above integral may be replaced as the sum of two integrals one for the path ‘a’ and other for the path ‘b’. R2 δQ 1 δQ = 0 +b ∫ ∫a TT R1 2 dQ (Note- may be read as integral T for the reversible path 1-a-2) ∫ ∫2δQ = 2 δQ a b TT IR IR δQ The magnitude of T (i.e. ‘ds’) is same for the paths ‘a’ and ‘b’ and it does not depend upon the path. So it depends upon the end states, hence it is point function and we know that properties are point functions, hence it is a property of the system. 3. State and explain the principle of increase of entropy [MAY 2006, MAY 2008] 4 Marks OR Write short notes on: Principle of increase of entropy [MAY 2010] 5 Marks Prof.S Venkatesh Rao 77

Vishwatmak Om Gurudev College Of Engineering Thermodynamics OR State the principle of increase of entropy of the universe and discuss any one application. [MAY 2011] 4 Marks Principle of increase in entropy states that the entropy of an isolated system (universe) either increases or remains constant. This is a corollary of the second law. We know that change in entropy in a reversible process is equal to ��������������������������������������������������. Let us now find the change in entropy in an irreversible process. V O G C E Consider a closed system undergoing a change from state 1 to state 2 by a reversible process 1-L-2 and returns from state 2 to the initial state 1 by an irreversible process 2- M-1 as shown in Fig. on the thermodynamic coordinates, pressure and volume. Since entropy is a thermodynamic property, we can write ….. (1) (Subscript I represents the irreversible process). Now for a reversible process, ….. (2) Substituting the value of ∫���������������������(��� ������������)(������������������������)������������ in eq (1), we get ….. (3) Prof.S Venkatesh Rao 78

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Again, since in eqn. (1) the processes 1-L-2 and 2-M-1 together form an irreversible cycle, applying Clausius equality to this expression, we get ….. (4) Now subtracting eqn. (4) from eqn. (3), we get, which for infinitesimal changes in states can be written as ….. (5) ������������������������ Eqn. (5) states that the change in entropy in an irreversible process is greater than ������������ Combining eqns. (2) and (5), we can write the equation in the general form as ….. (6) where equality sign stands for the reversible process and inequality sign stands for the irreversible process. It may be noted here that the effect of irreversibility is always to increase the entropy of the system. Let us now consider an isolated system. We know that in an isolated system, matter, work or heat cannot cross the boundary of the system. Hence according to first law of thermodynamics, the internal energy of the system will remain constant. Since for an isolated system, δQ = 0, from eqn. (6), we get V O G C E ….. (7) Eqn. (7) states that the entropy of an isolated system either increases or remains constant. This is a corollary of the second law. It explains the principle of increase in entropy. CHANGE IN ENTROPY OF THE UNIVERSE By including any system and its surrounding within a single boundary, as shown in Fig., an isolated system can be formed. The combination of the system and the surroundings within a single boundary is sometimes called the Universe. Prof.S Venkatesh Rao 79

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Hence, applying the principle of increase in entropy, we get ….. (8) Eqn. (8) states that the process involving the interaction of a system and the surroundings takes place only if the net entropy of the combined system increases or in the limit remains constant. Since all natural processes are irreversible, the entropy is increasing continually. V O G C E Prof.S Venkatesh Rao 80

Vishwatmak Om Gurudev College Of Engineering Thermodynamics FOR LAST MINUTE REVISION 1. Clausius inequality is given by, “When a system performs a reversible cycle, then but when the cycle is not reversible V O G C E 2. ‘Entropy’ is a function of a quantity of heat which shows the possibility of conversion of that heat into work. The increase in entropy is small when heat is added at a high temperature and is greater when heat addition is made at lower temperature. Thus for maximum entropy, there is a minimum availability for conversion into work and for minimum entropy there is maximum availability for conversion into work. 4. Entropy changes for a closed system (per kg) : Prof.S Venkatesh Rao 81

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION QUESTIONS MAY 2005 1 Q 1 Fill in the blanks: (c) Generated entropy is a __________ function and a ______________ differential. DEC 2005 NIL MAY 2006 6 Q 1 (e) State and explain the principle of increase of entropy. 3 Q 3(b) State and prove Clausius inequality for an irreversible cycle. DEC 2006V Q 4(a) What do you know by Clausius Inequality? 4O G C E MAY 2007 Q 1 (c) Show that entropy is property of system. 4 DEC 2007 Q 4 (a) State and prove Clausius inequality for an irreversible cycle. 6 MAY 2008 Q 1 (e) State and explain the principle of increase of entropy. 3 Q 7 Write short notes on: (b) Clausius in equality. 5 DEC 2008(RC) NIL MAY 2009(RC) Q 1(e) Show that entropy is a property of system. 5 DEC 2009(RC) 82 Q4(c) Define entropy and show that for an irreversible process Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics MAY 2010(RC) Q7. Write short notes on: (b) Principle of increase of entropy (d) Calusius Inequality 10 DEC 2010(RC) NIL MAY 2011(RC) Q 4 (a) State the principle of increase of entropy of the universe and discuss any one application. 4 DEC 2011 (RC) 1. (a) Fill in the blanks :( ii) Entropy is a ____________function an __________ differential 1 MAY 2011(RC) Q 1 (b) Show that entropy is property of system. 4 V O G C E Prof.S Venkatesh Rao 83

Vishwatmak Om Gurudev College Of Engineering Thermodynamics GRADED PROBLEMSV O 1. Air is throttled from 1.5 bar and 300C to 4 bar, adiabatically. Neglecting ∆KE, calculateG entropy change. 3 [DEC 2003]C E 2. A block of iron weighing 100 kg and having a temperature of 1000 C is immersed in 50 kg of water at a temp. of 200 C. What will be the change of entropy of the combined system of iron and water? Specific heat of iron = 0.45 kJ/kgK. 8 [DEC 2003] 3. Three identical finite bodies of constant heat capacity are at temperatures 300, 300 and 100 K. If n work or heat is supplied form outside, what is the highest temperature to which any one of the bodies can be raised by the operation of heat engines or refrigerators? 10 [DEC 2004] 4. If one kg of saturated water vapour at 100°C is condensed to saturated liquid at 100°C in a constant pressure process by heat transfer to the surrounding air at 25°C, what is the net increase in entropy of the water and the surroundings? 4[MAY 2005] 5. 600 kW of heat is supplied at a constant temperature of 291°C to a heat engine .a heat rejection takes place at 9°C. The following results were obtained: (i) 430 kW are rejected. (ii) 300 kW are rejected. (iii) 150 kW are rejected. . Using Clausius inequality classify which of the result report a reversible cycle irreversible cycle or impossible cycle. 9 [DEC 2005, DEC 2008 (OC)] 6. A small metallic object of 5 kg mass and Cp = 0.5 KJ/kg°K at a temperature of 227°C is thrown in a tank of water at 30°C temperature. Calculate change in entropy of the universe. 6 [DEC 2006] 7. One Kg of ice at —5° C is exposed to the atmosphere, which is a 20°C. The ice melts and comes into thermal equilibrium with the atmosphere. (a) Determine the entropy increase of the universe (b) What is the minimum amount of work necessary to convert the water back into ice at.—5° C? CP of ice is 2.093 kJ/kg K and the latent heat of fusion of ice is 333.3 kJ/kg. 10 [MAY 2007] 8. 4 kg of water at 270 C is mixed with 1 kg of ice at 0°C Assuming adiabatic mixing, determine the final temperature of mixture of ice and water Calculate the net change of entropy. Assume enthalpy of fusion of ice 335 kJ/kg. 10[DEC 2008(RC)] 9. A Carnot engine delivers 100 K.W of power by operating between temperature reservoirs at 100°C and 1000°C. Calculate the entropy change of each reservoir and net entropy change of the two reservoirs after 30 min. of operation. [DEC-96-7] Prof.S Venkatesh Rao 84

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION PROBLEMS(SOLVE AS HOME WORK) MAY 2005 Q 3(d) If one kg of saturated water vapour at 100°C is condensed to saturated liquid at 100°C in a constant pressure process by heat transfer to the surrounding air at 25°C, what is the net increase in entropy of the water and the surroundings ? 4 DEC 2005 Q 3(c) 600 kW of heat is supplied at a constant temperature of 291°C to a heat engine .a heat rejection takes place at 9°C. The following results were obtained: (i) 430 kW are rejected. (ii) 300 kW are rejected. (iii) 150 kW are rejected. . Using Clausius inequality classify which of the result report a reversible cycle irreversible cycle or impossible cycle. 9 V O MAY 2006 G NIL C E DEC 2006 Q 4 (b) A small metallic object of 5 kg mass and Cp = 0.5 KJ/kg°K at a temperature of 227°C is thrown in a tank of water at 30°C temperature. Calculate change in entropy of the universe. 6 MAY 2007 Q 4 (b) One Kg of ice at -5° C is exposed to the atmosphere, which is a 20°C. The ice melts and comes into thermal equilibrium with the atmosphere. (a) Determine the entropy increase of the universe (b) What is the minimum amount of work necessary to convert the water back into ice at.-5° C? CP of ice is 2.093 kJ/kg K and the latent heat of fusion of ice is 333.3 kJ/kg. 10 DEC 2007 NIL MAY 2008 NIL DEC 2008(RC) Q 4 (a) 4 kg of water at 27C is mixed with 1 kg of ice at 0°C Assuming adiabatic mixing, determine the final temperature of mixture of ice and water Calculate the net change of entropy. Assume enthalpy of fusion of ice 335 kJ/kg. 10 Prof.S Venkatesh Rao 85

Vishwatmak Om Gurudev College Of Engineering Thermodynamics MAY 2009(RC) NIL DEC 2009(RC) Q7(c) A lump of steel of mass 8 kg at 1000K is dropped in 80 kg of oil at 300K. Make calculations for the entropy change of steel, the oil and the universe. Take specific heats of steel and oil as 0.5 kJ/kg K and 3.5 kJ/kg K, respectively. 6 MAY 2010(RC) NIL DEC 2010(RC) Q3(b) A heat engine receives 1000 kW of heat at constant temperature of 285 °C. The heat rejected at 5°C. The possible heat rejected are: (i) 840 kW, (ii) 498 kW and (iii) 300 kW. Classify which of the result report a reversible cycle or irreversible cycle or impossible results. 8 Q3(c) Calculate the change in entropy of air, if it is throttled from 5 bar, 27°C to 2 bar adiabaticatly. 8 MAY 2011(RC) NIL DEC2011(RC) Q 3 (b) One kg of water of 0°C is brought into contact with a heat reservoir at 95°C. When the water has reached 95°C, find (i) Enthalpy change of water (ii) Enthalpy change of heat reservoir (iii) Enthalpy change of universe. 8 [ PRINT MISTAKE IN QP-ACTUALLY IT SHOULD BE ENTROPY CHANGE] V O G C E Prof.S Venkatesh Rao 86

Vishwatmak Om Gurudev College Of Engineering Thermodynamics CHAPTER 5 AVAILABILITY 1. Define i. Availability [MAY 2005, DEC 2005, MAY 2006, MAY 2007, MAY 2008 ….] ii. Available energy [ MAY 2006, DEC 2007, DEC 2010] iii. Unavailability. [MAY 2005, DEC 2005, MAY 2006, MAY 2007, MAY 2008 ….] iv. Unavailable energy [ MAY 2006, DEC 2007, DEC 2010] v. Dead state [MAY 2005, DEC 2005, MAY 2006, MAY 2007, MAY 2008 ….] vi. Irreversibility. [MAY 2005, DEC 2005, MAY 2006, MAY 2007, MAY 2008 ….] vii. Effectiveness. [ DEC 2007, DEC 2008, MAY 2010] V O G C E (i) Availability The theoretical maximum amount of work which can be obtained from a system at any state p1 and T1 when operating with a reservoir at the constant pressure and temperature p0 and T0 is called ‘availability’. (ii) Available energy ‘Available energy’ is the maximum portion of energy which could be converted into useful work by ideal processes which reduce the system to a dead state (a state in equilibrium with the earth and its atmosphere). Because there can be only one value for maximum work which the system alone could do while descending to its dead state, it follows immediately that ‘Available energy’ is a property. (iii) Unavailability or Unavailable energy The minimum energy that has to be rejected to the sink by the second law is called the unavailable energy (U.E.), or the unavailable part of the energy supplied. (iv) Dead state A system is said to be in the dead state when it is in thermodynamic equilibrium with the environment it is in. Prof.S Venkatesh Rao 87

Vishwatmak Om Gurudev College Of Engineering Thermodynamics At the dead state, a system is at the temperature and pressure of its environment (in thermal and mechanical equilibrium); it has no kinetic or potential energy relative to the environment (zero velocity and zero elevation above a reference level); and it does not react with the environment (chemically inert). Also, there are no unbalanced magnetic, electrical, and surface tension effects between the system and its surroundings, if these are relevant to the situation at hand. The properties of a system at the dead state are denoted by subscript zero, for example, P0, T0, h0, u0, and s0. Unless specified otherwise, the dead-state temperature and pressure are taken to be T0 = 25°C and P0 = 1 atm (101.325 kPa). A system has zero availability at the dead state V O G C E (v) Irreversibility The actual work which a system does is always less than the idealized reversible work, and the difference between the two is called the irreversibility of the process. Thus, Irreversibility, I = Wmax – W This is also sometimes referred to as ‘degradation’ or ‘dissipation’. (vi) Effectiveness Effectiveness is defined as the ratio of actual useful work to the maximum useful work. The useful output of a system is given by the increase of availability of the surroundings. Effectiveness, ������������ = ������������������������������������������������������������������������������������������������ ������������������������ ������������������������������������������������������������������������������������������������������������������������������������������������ ������������������������ ������������������������������������������������������������������������������������������������������������������������������������������������ ������������������������������������������������ ������������������������ ������������������������������������������������������������������������������������������������������������������������������������������������ ������������������������ ������������������������������������ ������������������������������������������������������������������������ For a compression or heating process the effectiveness is given by ������������ = ������������������������������������������������������������������������������������������������ ������������������������ ������������������������������������������������������������������������������������������������������������������������������������������������ ������������������������ ������������������������������������ ������������������������������������������������������������������������ ������������������������������������������������ ������������������������ ������������������������������������������������������������������������������������������������������������������������������������������������ ������������������������ ������������������������������������ ������������������������������������������������������������������������������������������������������������������������������������������������ ������������ = ������������������������������������������������������������������������������������ or ������������������������������������������������,������������������������������������������������������������������������ Prof.S Venkatesh Rao 88

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 2. Derive an expression for availability of a non-flow process [MAY 2006] 6 Marks Let us consider a system consisting of a fluid in a cylinder behind a piston, the fluid expanding reversibly from initial condition of p1 and T1 to final atmospheric conditions of p0 and T0. Imagine also that the system works in conjunction with a reversible heat engine which receives heat reversibly from the fluid in the cylinder such that the working substance of the heat engine follows the cycle O1LO as shown in Fig., where s1 = sL and T0 = TL (the only possible way in which this could occur would be if an infinite number of reversible heat engines were arranged in parallel, each operating on a Carnot cycle, each one receiving heat at a different constant temperature and each one rejecting heat at T0). V O G C E The work done by the engine is given by: Wengine = Heat supplied – Heat rejected ..……. ……(i) The heat supplied to the engine is equal to the heat rejected by the fluid in the cylinder. Therefore, for the fluid in the cylinder undergoing the process 1 to 0, we have ……(ii) Adding eqns. (i) and (ii), we get The work done by the fluid on the piston is less than the total work done by the fluid, since there is no work done on the atmosphere which is at constant pressure p0 Prof.S Venkatesh Rao 89

Vishwatmak Om Gurudev College Of Engineering Thermodynamics i.e., Work done on atmosphere = p0 (v0 – v1) The property, a = u + p0v – T0s (per unit mass) is called the non-flow availability function. 3. Define availability. Derive an expression of availability for steady flow system [MAY 2011] 10 Marks Consider a fluid flowing steadily with a velocity C1 from a reservoir in which the pressure and temperature remain constant at p1 and T1 through an apparatus to atmospheric pressure of p0. Let the reservoir be at a height Z1 from the datum, which can be taken at exit from the apparatus, i.e., Z0 = 0. For maximum work to be obtained from the apparatus the exit velocity, C0, must be zero. It can be shown as in previous question that a reversible heat engine working between the limits would reject T0 (s1 – s0) units of heat, where T0 is the atmospheric temperature. Thus, we have V O G C E In several thermodynamic systems the kinetic and potential energy terms are negligible The property, b = h – T0s (per unit mass) is called the steady-flow availability function. 4. Derive an expression for decrease in availability due to heat transfer with finite temperature difference. [MAY 2011] 5 Marks When heat is transferred through a finite temperature difference, there is always loss of available energy. Prof.S Venkatesh Rao 90

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Let us consider a heat reservoir at temperature T and the surroundings at temperature T0. Then a reversible heat engine cycle 1-2-3-4 could be devised drawing heat Q from the reservoir and rejecting heat Q0 to the surrounding as shown in Figure 3, such that. V O G C E Q = T ∆ S. Q = T ∆ S and A.E. corresponding to the heat Q is ( )0 0 A.E. = Wmax = Q - Q0 = T - T0 ∆ S Now, consider that this heat Q is transferred from the reservoir at T to another reservoir absorbing it at T’ where T’ < T. The resulting heat engine cycle now operating between T’ and T0 is 1’ – 2’ – 3’ – 4. Since the heat added Q is the same. Q = T ∆ S = T' ∆ S' where ∆S’ is the increase in entropy of the reservoir receiving heat at T’. Since T’ < T, ∆S’ > ∆S The heat rejected Q’0 for the new cycle is Q' = T ∆ S ′ > Q 0 00 and the available energy after, the heat transfer is ( )AE ′ - W' = Q - Q′ = T'-T ∆S < AE max 0 0 The loss or decrease of available energy is, AE - AE ′ = W - W ′ = Q′ - Q = T (∆S ′ - ∆S ) = T ∆S max max 0 0 0 0 net Prof.S Venkatesh Rao 91

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Which is equal to the product of the temperature of the surroundings and net increase in entropy. This decrease in available energy is equal to an increase in unavailable energy. 5. Show that the irreversibility of a process is given by — [DEC 2007] 6 Marks The actual work which a system does is always less than the idealized reversible work and the difference between the two is called as irreversibility of the process. Thus, irreversibility, I = Wmax - W This is also sometimes referred as degradation or dissipation. For non-flow process between the equilibrium states, when the system exchanges heat only with environment, then, V O G C E I = [(u - u )- T (S - S )]- [(u - u )+ ! ] ( )1 2 0 1 2 12 =T S −S -Q =T (∆S )system +T (∆S )surr 0 21 0 0 = T0  ∆S system + ∆S surr  Similarly, for steady flow process I =W -W max =   b V2 +   b V2 + gZ     V2 +   V2 +  + Q    +1 gZ  -  +2   -   h +1 gZ  -  h +2 gZ     1 1   2 2       2    2 2   1 2 1 2 2  ( )= T = (∆S )system +T (∆S )max 0 S -S -Q T 0 21 0 [ ]= T 0 (∆S )system + (∆S )surr = T ∆S 0 UNIVERSE [ ] (∆S )system + (∆S )surr = ∆S UNIVERSE Prof.S Venkatesh Rao 92

Vishwatmak Om Gurudev College Of Engineering Thermodynamics FOR LAST MINUTE REVISION 1. ‘Available energy’ is the maximum portion of the energy which could be converted into useful work by ideal processes which reduce the system to a dead state. 2. The theoretical maximum amount of work which can be obtained from a system at any state p1 and T1 when operating with a reservoir at the constant pressure and temperature p0 and T0 is called ‘availability’. 3. Energy is said to be degraded each time it flows through a finite temperature difference. That is, why the second law of thermodynamics is sometimes called the law of the degradation of energy, and energy is said to ‘run downhill’. 4. In non-flow systems: Maximum work available, V O G C E The property a = u + p0v – T0s is called the non-flow availability function. 5. In steady-flow systems: Maximum work available, The property, b = h – T0s is called the steady-flow availability function. 6. It may be noted that Gibb’s function g = (h – Ts) is a property of the system where availability function a = u + p0v – T0s is a composite property of the system and surroundings. When state 1 proceeds to dead state (zero state) 93 Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics a=b=g 7. The actual work which a system does is always less than the idealized reversible work, and the difference between the two is called the irreversibility of the process. This is also sometimes referred to as degradation or dissipation. Effectiveness is defined as the ratio of actual useful work to the maximum useful work. V O G C E Prof.S Venkatesh Rao 94

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION QUESTIONS MAY 2005 4 Q 4 (a) Define the following: — (i) Availability (ii) Unavailability. (iii) Dead state (iv) Irreversibility. DEC 2005 Q 1 (b) State TRUE or FALSE. If false, correct it. (iii) Power cannot be developed If the system is in dead state. 1 Q 5 (a) Define — (i) Availability, (ii) irreversibility 4 V O G C E MAY 2006 6 Q 4 (a) What do you mean by (i) Available energy (ii) Unavailable energy (iii) Availability (iv) Dead state Q 4 (b) Derive an expression for availability of a non-flow process. DEC 2006 NIL MAY 2007 Q 5 (b) Define (i) Availability (ii) irreversibility 4 DEC 2007 Q 4 (b) Explain: — (i) Available and unavailable energy (ii) Dead state (iii) Effectiveness. 6 Q 5 (b) Show that the irreversibility of a process is given by — 6 Prof.S Venkatesh Rao 95

Vishwatmak Om Gurudev College Of Engineering Thermodynamics MAY 2008 Q 5 (a) Define the following:— (i) Availability (ii) Unavailability (iii) Dead state (iv) Irreversibility. 4 DEC 2008(RC) Q 4 (b) Explain the following (i) Availability (ii) Irreversibility (iii) Effectiveness (iv) Dead State. 10 MAY 2009(RC) Q 6 (a) Define (i) availability (ii) unavailability (iii) Dead state (iv) Irreversibility. 8 V O G C E DEC 2009(RC) Q1(g) Define i) availability ii) unavailability iii) Dead state iv) Irreversibility 4 MAY 2010(RC) (b) Define: (i) Availability (ii) Dead state (iii) Irreversibility (iv) Effectiveness. 6 DEC 2010(RC) Q4(c) Define: (i) Available Energy 2 MAY 2011(RC) Q 1 (d) Derive an expression for decrease in availability due to heat transfer with finite temperature difference. 5 DEC 2011(RC) Q 4. (a) Define availability. Derive an expression of availability for steady flow system. 10 MAY 2012(RC) Q 2 (c) What do you mean by availability? 2 Prof.S Venkatesh Rao 96

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION PROBLEMS MAY 2005 Q 4 (b) An air compressor receives ambient air at 100 kPa, 25°C. It compresses the air to a pressure of 1 MPa, where It exits at temperature of 540 K. Since the air and compressor housing are hotter than the ambient, it loses 50 kJ/kg of air flowing through the compressor. Find (i) Reversible work (ii) Reversible heat transfer (iii) Irreversibility in the process. (For air at inlet hi= 298.82 kJ/kg; Se —Si = - 0.0573 kJ/kg.K and he=544.69 kJ/kg) 6 DEC 2005 Q 5 (c) A system at 500 K receives 7200 kJ/min heat from a source at 1000 K. The temperature of atmosphere is 300 K. Assuming that the temperatures of system and source remain constant during heat transfer, find out— (i) The entropy change during heat transfer (The net change in entropy of system and surroundings) (ii) The decrease in available energy after heat transfer. 6 V O G C E Q 4 (c) Air enters a compressor in a steady flow at 140 kPa, 17°C and 70 m/s and leaves at 350 KPa, and 110 m/s. The environment is at 100 KPa, 7°C calculate per kg of air — (i) the. actual amount of work, required. (ii). the minimum work required , (iii). the irreversibility of the process 8 DEC 2006 Q 4 (c) What do you know about Available energy and Availabi1ity? Certain gas having mass 3 kg and Cv = 0.81 KJ/kg°K, initially at 3 bar and 450 K receives 600 KJ of heat from an infinite source at 1500K. If surrounding temperature is 300K, find loss in available energy due to heat transfer. 10 MAY 2007 NIL DEC 2007 Q 4 (c) Air enters a compressor in a steady flow at 140 kPa, 17° C and 70 m/s and leaves at 350 kPa, 127° C and 110 m/s. The environment is at 100 kPa, 7°C. Calculate per kg of air (i) the actual amount of work required (ii) the minimum work required (iii) the irreversibility of the process. 8 MAY 2008 NIL Prof.S Venkatesh Rao 97

Vishwatmak Om Gurudev College Of Engineering Thermodynamics DEC 2008(RC) NIL MAY 2009(RC) NIL PROBLEMS FOR PRACTICE (1) [DEC-03 (RC)-6]: A centrifugal air compressor compresses air at the rate of 20 kg/min from 1 bar to 2 bar. The temperature increases from 20°C to 120°C during the compression. Determine actual and minimum power required to run the compressor. The surrounding air temp is 20°C. Neglect ΔKE and ΔPE. (2) [DEC-03 (RC)-6]: A metal piece of 1 kg mass with specific heat of 0.4 KJ/kgK is cooled from 200°C to 100°C by transferring heat to the surrounding air at 27°C. Determine availability and irreversibility for this process. V O G C E (3) [MAY-03(RC)-8]: Air enters a compressor in a steady flow at 140 kPa, 17°C and 70 m/s and leaves at 350 kPa, 127°C and 110 m/s. The environment is at 100 kPa, 7°C. Calculate per kg of air - i. the actual amount of work required, ii. the minimum work required, and iii. the irreversibility of the process. (4) [DEC-02(OC)-5]: 4 kg of gas initially at 2.5 bar and 400°C receives 700 kJ of heat from an infinite reservoir at 1200°C. The surrounding temperature is 17°C. Calculate the loss in available energy due to heat transfer. Cv = 0-81 kJ/kg/k. (5) [MAY-02-(REV)-6]: In a turbine the air expands from 7 bar, 600°C to 1 bar, 250°C. During expansion 9 kJ/kg of heat is lost to the surroundings which is at 1 bar, 15°C. Neglecting kinetic energy and potential energy changes, determine per kg of air:- i. The decrease in availability. ii. The maximum work. iii. The irreversibility. (6) [MAY-02-(REV)-5]: A pressure vessel has a volume of 1 m3 and contains air at 1.4 MPa, 175°C. The air is cooled to 25°C by heat transfer to the surroundings at 25°C. Calculate the availability in the initial and final states and the irreversibility of this process. (7) [MAY-99-8]: Superheated steam at 20 bar and 250°C enters a turbine and is expanded to 5 bar and dryness fraction 0.95. Compute the loss in availability for adiabatic expansion, if the surrounding temperature is 27°C. Prof.S Venkatesh Rao 98

Vishwatmak Om Gurudev College Of Engineering Thermodynamics (8) [MAY-99-6]: An amount of 90 kJ of heat is added to a tank of air that is originally at a temperature of26°C. The tank contains 2.8 kg of air. The temperature of the surrounding is 15°C.Determine the available energy that is added. Take Cv = 0.718 kJ/kg k. (9) [DEC-98-6]: 500 litres of furnace oil is heated from 30°C to 80°C per minute using exhaust steam coming out of steam engine at 1.25 bar and 0.8 dryness. Assuming only the enthalpy of evaporation of steam is used for heating the oil, determine the loss in available energy due to above heat transfer. Assume Specific gravity of oil = 0.8 Specific heat of oil = 3 kJ/kg k Saturation temperature of steam at 1.25 bar = 106°C Enthalpy of evaporation of steam at 1.25 bar = 2241 kJ/kg Atmospheric temperature =280 K. Neglect all other losses. V O G C E (10) [DEC-98-10]: Air exists in a tank at a pressure of 380 kPa and temperature 190°C. The mass of air present is 8.2 kg. The surroundings are at a pressure of 100 kPa and temperature 22°C. Determine the available energy of air present in the tank. (11) [DEC-97-6]: Air at the rate of 30 kg per minute is compressed in a centrifugal air compressor from 2 bar to 4 bar. The temperature increases from 27°C to 127°C during the compression. Determine actual and minimum power required. Surrounding temperature is 18°C. Neglect heat loss to surrounding and charges in potential and kinetic energy. (12) [MAY-97-7]: Air expands through a turbine from 4 bar, 500°C to 1 bar 350°C while an amount of 10 kJ/kg of heat is lost to atmosphere at 0.98 bar, 21 °C. The net change in kinetic energy is negligibly small. Find for this process - i. Decrease in system availability ii. maximum work iii. irreversibility. (13) [MAY-97-6]: 5 Kg of air at 600 K and 10 bar is enclosed in a closed system. Determine the availability of the system if the surrounding pressure and temperature are 1 bar and 300K. (14) [DEC-96-6]: 1 Kg of air expands in a non flow process from 10 bar and 167°C to 3 bar and 57°C.Calculate the maximum work that can be obtained from air. Assume To = 290 K and Po = 1 bar. (15) [DEC-95-8][DEC-93-5]: A heat source at 1200°K transfers heat at the rate of 150 KW to a system at 500°K temperature. The heat sink the is 300 K. Assuming that the temperatures remain constant, calculate :- i. Decrease in available energy after heat transfer Prof.S Venkatesh Rao 99

Vishwatmak Om Gurudev College Of Engineering Thermodynamics ii. Entropy gain or loss due to heat transfer. (16) [MAY-92-7]: Calculate the decrease in available energy when 25 Kg of water at 95°C mix with 35 Kg of water at 80°C Assume surrounding temperature as 27°C. (17) [DEC-91-10]: In a non flow system, air is at a pressure of 6 bar and temperature 290°K. If the surrounding conditions are P0=1 bar to T0 = 290°K. Calculate the availability of the system per kg of air. What should be the availability of the system if it were at 6 bar and 390°K? V O G C E Prof.S Venkatesh Rao 100