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16 Years MADE ERSY Publications Solved Papers 2001-2016 Topicwise Previous Years Solved Papers .6L IAN r SERVICES EXAMINATIO Also useful for Civil Services Main Examination, Engineering Services Main Examination and various State Engineering Services Examinations /6(yeena4 \ 7 " PPrreevviioouuss YYeeaarrss SSoollvveedd PPaappeerrss ( L ' ~______. \ I' IInnddiiaann FFoorreesstt SSeerrvviicceess EExxaammiinnaattiioonn ((22000011--22001166)) MMeecchaniccaall EEnnggiinneeeerriinngg PPaappeerr--IIII r --------------. 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Basic Concepts, Heat and Work Q.1 Explain why PMMK I and PMMKII devices are not practicable. (Perpetual Motion Machine Kind) [IFS (Mains) 2002: 10 Marks] Solution: The first law states the principle of conservation of energy. Energy is neither created nor destroyed, but only gets transformed from one form to another. There can be no machine which would continuously supply mechanical work without some other form of energy disappearing simultaneously. Such fictitious machine is called a perpetual motion machine of the first kind or in brief PMM1. A PMM1 is thus impossible. The converse of the above statement is also true. There can be no machine which would continuously consume work without some other form of energy appearing simultaneously. The Kelvin Planck statement of second law states: It is impossible for a heat PMM1 engine to produce a net work in a complete cycle if it exchanges heat only with bodies at a single fixed temperature. T1 Thus, = - 0 1 and if 02 = 0 1 or 1-1 = 100%) the heat engine will produce n Q1 (i.e. Wnet = 0 net work in a complete cycle by exchanging heat with only one reservoir, thus violating the Kelvin-Planck statement. Such a heat engine is called a ~Q2= 0 perpetual motion machine of the second kind, PMM2. A PMM2 is thus PMM2 impossible. Q.2 State the Zeroth Law of Thermodynamics and highlight its significance. [IFS (Mains) 2013 : 5 Marks] Solution: When a body A is in thermal equilibrium with a body B, and also separately with a body C, than B and C will be in thermal equilibrium with each other. This is known as Zeroth law of Thermodynamics. It is the basis of temperature measurement. Significance: 1. It has wide applications in thermometry. 2. If a body C is a thermometer and it is used to measure temperatures of A and B. And if it shows both the readings as same, then (C) is infect showing its own temperature. Q.3 For an isothermal process, show that: 2 f pdv —12 Vdp .1 [IFS (Mains) 2013 : 5 Marks] 2 IFS (Mains) Previous Solved Papers Mechanical Engineering Solution: For an isothermal process, to prove: 2 2 fiPdV = -f VdP Isothermal process is defined as, PV = Constant = C Differentiating both sides, we get, PdV + VdP = 0 or PdV = -VdP When system undergoes change in state from state (1) to state (2). fi2PdV = -f2VdP Q.4 The readings of two thermometers A and B agree at ice point and steam point as 0°C and 100°C. The two temperature readings are related by the following expression: to = a+ btB + ctB where a, b and c are constants. In a constant temperature bath, the temperature are shown as 51°C on thermometer A and 50°C on thermometer B. Determine the reading on thermometer B when the thermometer A reads 65°C. Can you comment which of the two thermometers is correct? [IFS (Mains) 2016 : 20 Marks] Solution: Given: tA = a + btB + Ct82 As the reading of two thermometers A and B agree at ice point (0°C) and steam point (100°C). When tA = 0°C, tB is also 0°C = a + btB + CtB2 0 = a + b(0) + C(0)2 (a = 0) to = btB + CtB2 when, to = 100°C, tB is also 100°C 100 = b(100) + C(100)2 b + (100)C = 1 when, tB = 50°C, tA = 51°C tA = btB + CtB2 51 = (50)b + (50)2C From equation (i) and (ii) b = 1.04 C = -4 x 10-4 tA = 1.04 tB - 4 x 10-4 t82 when, tA reads 65°C 65 = 1.04tB - 4 x 10-4 tB2 or tB = 64.07°C None of the two thermometers are ideal. So we cannot comment on to which is more correct. 2. First Law of Thermodynamics Q.5 An insulated rigid chamber of 2.5 m3 capacity contains air at 25°C and 250 kPa. A paddle wheel inserted in the chamber does 900 kJ of work on the air. Assuming constant specific heats, calculate the entropy increase during the process. Take C = 0.717 kJ/kg-K and R = 0.287 kJ/kg-K. [IFS (Mains) 2001 : 10 Marks] Thermodynamics 3 IMIDE ERSY Solution: Given: Volume(V) = 2.5 m3, T= 25 + 273 = 298 k, P= 250 kPa, (Work done by paddle wheel), 900 kJ WPaddle = As the chamber is insulated one, the paddle wheel work will increase the internal energy of the liquid. do = dU + dW dQ = 0, and dW = -900 kJ (Work is done on the system) dU = 900 kJ mCv(T2- T1) = 900 kJ Also, pV = mRT 250x103 x2.5 m = - 7.3 kg 287 x 298 7.3 x 0.717 x (T2- 298) = 900 T2 = 469.76 K (As)system = Entropy increase of system rT2 MCvdT mCv Ini,R 1 = 7.3 x 0.717 x1446916) = 2.382 kJ/k T 298 Q.6 (i) An ideal gas is heated at constant volume until its temperature is 3 times the original temperature, then it is expanded isothermally till is reaches its original pressure, the gas is then restored to its original state. Determine the expression of net work done. (ii) 300 kJ/sec of heat is supplied at a constant fixed temperature of 290°C to a heat engine. The heat rejection takes place at 8.5°C. The following results were obtained: I. 215 kJ/s of heat is rejected II. 150 kJ/s of heat is rejected III. 75 kJ/s of heat is rejected Using Clausius inequality which of the results report a reversible cycle of irreversible cycle, or impossible result? [IFS (Mains) 2003 : 5 + 5 = 10 Marks] Solution: P 2 (7-= Constant) 41 3 1 V (i) Given: T2 = 3T1, T2 = T3 = 3T1 [constant volume process] W1-2 , W2_3 = P2 v 2 In P3 l 4 I IFS (Mains) Previous Solved Papers Mechanical Engineering = 3P1 and Pi = P3, = 3P3 P2 P2 = 3 P3 = P2V2 In(3) = 1.098 P2V2 = 1.098 mRT2 W2 - 3 = Pi( V3 - Vi) = mR( T3 - W3 - 1 = 2m RT2 = = mR(3 1 - Ti) = 2mRT1 0.66 mRT2 3 W3 - 1 = 0 + 1.098 mRT2 + 0.66 mRT2 = 1.764 mRT2 Wnet = W1-2 + W2-3 + = 5.294 mRT1 Wnet (ii) Given: Q1= 300 kJ/sec, Ti = 290 + 273 = 563 k, T2 = 8.5 + 273 = 281.5 k Clausius Inequality says, ,fid0 = 01 = 02 T T1 T2 Case I: Tt = 563 K, T2 = 281.5 K When W2 = 215 kJ/sec dQ 300 215 = 0.23 7 T 563 281.5 - r) As < 0 (Cycle is possible and irreversible) T Case II: When 02 = 150 kJ/sec aid oQ1 Q2 300 150 n (Cycle is reversible) T 1 T2 563 281.5 - Case III: When, 02 = 75 kJ/sec rficQ _ Q1 02 300 75 _ 0.266 T Ti T2 563 281.5 - d-O As ( > 0 (Cycle is impossible) Q.7 In a closed system, dry saturated steam at 100 bar expands isothermally and reversibly to a pressure of 10 bar. Calculate the heat supplied and work done per kg of steam during the process. [IFS (Mains) 2004: 20 Marks] Solution: Give: P1 = 100 bar (dry saturated steam), P2 = 10 bar Expansion is isothermal and reversible, from steam table, At P1 = 100 bar (saturated steam) = 311.06 °C u1 = 2544.4 kJ/kg h1 = 2724.7 kJ/kg v1 = vg = 0.018026 m3/kg MIME EPSY Thermodynamics I 5 As the expansion process is isothermal. PA T2= 311.06°C T1 and = 10 bar P2 100 bar 179.91 Tsat ••• (t2 > Tsat) -> (It is in superheated state) From steam table, 10 bar At, T300oc, U303,c = 2793.2 kJ/kg At, T350.c, u350°C = 2875.2 kJ/kg Using linear interpolation, method, 2875.2 - 2793.2 u- 2793.2 = (311.06 - 300) 350 - 300 = 2811.3384 kJ/kg 0.28247 - 0.25794 Similarly, (v- 0.25794) = (311.06 - 300) 350 - 300 v2 = 0.2633 m3/kg Work done/kg = Pv = (100 x 105) x (0.018026)14l—n 1 p2 415.063 kJ/kg W = and u2 - u1 = 2811.3384 - 2544.4 = 266.9384 kJ/kg Heat supplied/kg = Q = (u2- u1) + w Q = 266.9384 + 415.063 = 682.0014 kJ/kg Q.8 (i) An inventor claims to have developed an engine which draws 1000 kJ of heat energy per cycle each from two thermal reservoirs at temperatures 1500 K and 900 K and rejects 1600 kJ of heat energy per cycle to a thermal reservoir at 300 K while performing 400 kJ per cycle of work. Examine the validity of the claim using first law of thermodynamics and inequality of Clausius. (ii) Using appropriate T-ds relations, show that the slope of the constant volume line is higher than that of a constant pressure line passing through a given state represented on a temperature- entropy diagram for a perfect gas. Sketch the temperature-entropy diagram. [IFS (Mains) 2006 : 5 + 5 = 10 Marks] Solution: T2 (i) Given: Q1 = 1000 kJ, C)2 = 1600 kJ 900 k (Work) 400 kJ (Given) wgiven= Using 1st law of thermodynamics, work produced in a cycle 1 wnet wnet= 201 Q2 = 2 x 1000 - 1600 = 400 kJ = w (given) It satisfy the 1st law of thermodynamics. 300 k T3 According to clausius inequality. J 50 _ Q1+ 01 02 - 1000 1000 1600 -0.66+1.11-5.33 T T2 T3 1500 900 300 = -3.5633