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Basic Thermodynamics Syllabus:- Zeroth, first and second law of thermodynamics, thermodynamic system and processes, car not cycle, inerversibility and availability, behavior of ideal gas and real gases, properties of pure substance, calculation of work and heat in ideal processes, analysis of thermodynamic, cycles related to energy conversion. Gate → 2003 1. A 2 kw, 40 liters water heater is switched on for 20 minutes. The heat capacity CP for water is 4.2 kj/kg k. Assuming all the electrical energy has gone in to hearting the water, increase of the water temperature in degree centigrade is (a) 2.7 (c) 14.3 (b) 4.0 (d) 25.25 (Mark = 1) Solution Heat supplied by heater in 20 minutes 2 1000 20 60 2400000 J 2400 KJ As all the electric energy has gone into heating the water So Heat taken by water = 2400 KJ T1 But heat taken by water mcP dT T2 mcP T1 T2 Mass of water 40 10 3 1000 40 10 1000kg / m 1Litre 10 m 3 3 40kg cP Specific heat in kJ/kg K = 4.2. 3 3 40 4.2 T1 T2 240 2400 40 4.2 = 14.28 degree centigrade T1 T2 2. Considering the relationship Tds = Pdv between the entropy (s), internal energy (u), pressure (P), temperature (T) and Volume (v) which of the following statements is correct ? (a) It is applicable only for a reversible process. (b) For an irreversible process Tds > du + Pdv. (c) It is valid only for an ideal gas. 1 (d) It is equivalent to 1st Law, for a reversible process. ( Mark – 2) Solution. (c) Tds = du + Pdv Is valid only for an ideal gas both for reversible and irreversible process undergone by a closed system, since it is a relation among properties which are independent of the path. Common data question Nitrogen gas ( molecular weight 28 ) is enclosed in a cylinder by a piston, at the initial condition of 2 bar, 298 k and 1 m3. In a particular process, the gas slowly expands under isothermal condition, until the volume becomes 2 m3. Heat exchange occurs with the atmosphere at 298 k during this process. 3. The work interaction for the nitrogen gas is (a) 200 KJ (b) 138-6 KJ (c) 2 KJ (d) -200KJ (Mark – 2) 4. The entropy change for the universe during the process in KJ/K is (a) 0.4652 (b) 0.0067 (c) 0 (d) -0.6711 (Mark -2 ) Solution v2 Work done Pdv v1 In isothermal process T = Constant For an ideal gas Pv = RT Where R is constant So Pv = Constant Pv 1 1 P2 v2 P P2 v2 v v2 Work done v1 P2 v2 dv v v2 dv v v1 P2 v2 Pv 1 1 Loge v2 v1 2 v2 2m3v1 1m3 v2 Pv 1 1 Loge v1 Pv 1 1 Loge 2 2 105 1log e2 2 105 0.6931 138.629kJ Entropy change of the system d T { for reversible process} For isothermal process = du = 0 So from first law of thermodynamics d dw du d dw d Pdv S12 d Pdv T T As T is constant workdone S12 system T 138.629 298 0.4651 Entropy change of surrounding HeatExchangewithSurrounding SurroundingTemperature WorkdoneonSurrounding SurroundingTemperature = - 0.4641 d dw So entropy change of universe is zero. Gate 2004 5. A gas contained in a cylinder is compressed, the work required for compression being 5000 KJ. During the process, the heat interaction of 2000 KJ causes the surrounding to be heated. The change in internal energy of the gas during the process is (a) – 7000 KJ (c) + 3000 KJ (b) – 3000 KJ (d) + 7000 KJ (Mark – 1 ) Solution. (c) Using first law of thermodynamics. 3 w u w 5000KJ {work done is negative when it is done on the System } {Heat transferred is negative when it is taken from the system or it is rejected} 2000KJ 2000 5000 u 3000 KJ u 6. A Steel billet of 2000 kg mass is to be cooled from 1250 k to 450 k . The heat released during process is to be used as a source of energy. The ambient temperature is 303 k and specific heat of steel is 0.5 kJ/kg K. The available energy of this billet is. (a) 490.44 MJ (c) 10.35 MJ A 450 k (b) 30.95MJ (d) 0.1 MJ 1250 k B Available energy T D C To unavailable energy S Area under curve AB mcP 1250 450 2000 0.5 1250 450 800, 000 KJ 800MJ 1250 Entropy change from B to A 450 mc p dT T 1250 450 1250 2000 0.5 In 450 S B A 1021.65 KJ / Kg .K S B A mcP In Area under C-D = S B A To S B A 1021.65 303 309.56MJ So Area ABCD = 800 – 309.56 4 = 490MJ. Gate 2005 7. The following four figures have been drawn to represent a fictitious thermodynamic cycle, on the P –v and T – S planes. P T V S T P S V According to first Law of thermodymics, equal areas are enclosed by (a) Figure 1 and 2 (c) Figure 1 and 4 (b) Figure 1 and 3 (d) Figure 2 and 3 8. A Reversible thermodynamic cycle containing only three processes and producing work is to be constructed. The constraints are (i) There must be one isothermal process (ii) There must be one isentropic process (iii) The maximum and minimum cycle pressures and clearance volume are fined (iv) Polytrophic processes are not allowed. The number of possible cycle are (a) 1 (b) 2 (c) 3 (d) 4 (Mark – 2) Possible processes Constant pressure P Isothermal Compression 5 Isothermal Compression Isothermal Expansion P Isothermal Compression V 1 Isothermal Expansion P Constant Pressure Process 3 2 Isothermal Compression V Isothermal Compression P Constant Pressure V Isothermal Compression 6 9. Nitrogen at an initial state of 10 bar, 1 m3 and 300 K is expanded isothermally to a final volume of 2 m3 , the P – v – T relation is P P 2 v RT Where a > 0. The final pressure v (a) Will be slightly less than 5 bar. (b) Will be slightly more less than 5 bar. (c) Will be exactly 5 bar. (d) Can not be ascertained in the absence of the value of a. (Mark – 2) a Solution. (b) P 2 V RT V As process is isothermal – RT = constant a a P1 2 V1 P2 2 V2 V1 V2 P1 10bar V2 2m3 V1 1m3 T1 300 K 2 a a 10 1 P2 1 4 a 10 a 2 P2 2 a 2 P2 10 2 P2 5 a So P2 is slightly more than 5 bar as a is positive. The following table of properties was printed out for saturated liquid and saturated vapor of ammonia. The title for only the first two columns are available. All that we know that the other columns (columns 3 to 8 ) contain data on specific properties, namely, internal energy (KJ/kg ),enthalpy (KJ/kg ) and entropy (KJ/kg) t0 P(KPa) -20 190.2 88.76 0.3657 89.05 5.6155 1299.5 0 429.6 179.69 0.7114 180.36 5.3309 1318 1418 1442.2 7 20 587.5 272.89 1.0408 274.3 5.0860 1332.2 40 1554.9 368.74 1.3574 371.43 4.8662 1341.0 1460.2 1470.2 20. The specific enthalpy data are in columns (a) 3 and 7 (b) 3 and 8 (c) 5 and 7 (d) 5 and 8 10. when saturated liquid at 400C is throttled to – 200C the quality atenit will be (a) 0.189 (c) 0.231 (b) 0.212 (d) 0.788 10. (d) We know that h = u + PV h = enthalpy u = internal energy. So all the value in column 5 are greater than corresponding value in column 3 so column 5 represent specific enthalpy of saturate liquid and similarly all the values of column 8 are greater than corresponding values of column 7.50 the column 8 represent specific enthalpy of saturated steam. 11. Enthalpy of saturated liquid at 400C = 371.43 KJ/kg. Enthalpy of vapor at – 200C = ( let x be dry ness fraction) = 89.05+x (1418 – 89 -05). During throttling enthalpy remains constant 371.43 = 89.05 + x (1418 – 89.05) 282.38 = x (1418 – 89.05) x = 0.212. Gate → 2006 12. Given below is an extract from steam tables. Specific volume Temperature In oC 45 342.24 PSat (bar) Saturated Liquid 0.09593 150 Enthalpy (KJ/kg) Saturated Vapour Saturated Liquid Saturated Vapour 0.001010 15.26 188.45 2394.8 0.001658 0.010337 1610.5 2610.5 8 Specific enthalpy of water in KJ/kg at 150 bar and 450C is (a) 203.6 (b) 200.53 (c) 196.38 (d) 188.45 Solution. (b) Specific enthalpy at 150 and 450C 188.45 150 0.09593 105 0.001010 103 203.59 KJ / kg. 13. Match items from groups I, II, III, IV and V. Group II Group I E – Heat F work When added to the system is G Positive H Positive Group III Group IV Group V Differential Function Phenomenon I Enact K Path M Transient J in enact L Point N Boundary (d) Heat when added to the system is positive and is in enact differential because it is path dependent and is boundary phenomenon work when added to the system is negative and is in enact differential because it is path dependent and is transient phenomenon. Statement for linked answer question 14 and 15. A football was inflated to a gauge pressure of 1 bar when the ambient temperature was 150C. When the game started next day, the air temperature at the stadium was 50C. Assume that the volume of foot ball remains constant 2500 cm3. 14. The amount of heat lost by the air in the football and the gauge pressure of air in the football at the stadium respectively equal. (a) 30.6J, 1.94 bar (b) 21.8J, 0.93 bar (c) 61.1J, 1.94 bar d) 43.7J, 0.93 bar 15. Gauge pressure of air to which the ball must have been originally inflated so that it would be equal 1 bar gauge at the stadium is 9 (a) 2.23 bar (c) 1.07 bar (b) 1.94 bar (d) 1 bar Solution. 14 (d) Gauge pressure with in foot ball Pt Patm Pguage = 1.01 + 1 = 2.01 bat Temperature when total pressure is 2.01 bar = 150C or 2880k. Volume of air with in foot ball = 2500 cm3. Since the volume remains same PV PV 1 1 2 2 T1 T2 V1 V2 Next day temperature T2 5o C or 2780k. P1 P2 T1 T2 2.01 P2 288 278 P2 1.94 bar P2 Ptotal Patm P2 gauge 1.94 P2 gauge 1.94 1.01 = 0.93 bar. Heat lost = mCV dT PV Rm 1 1 T1 2.01 m 2500 100 3 105 288 287 0.000017447 105 / 287 1.7447kg 287 0.00607kg mCV dT 0.00607 0.72 15 5 Heat lost 0.04376 KJ 43.76 J 15 (c) Pgauge 1bar { final pressure} Pgauge 1 1.01 2.01 { final pressure} 10 P2 P1 T2 T1 2.01 P1 278 288 P1 2.08bar Pgauge 1.07bar Gate 2007 16. (d) Which of the following relationships is valid only for reversible processes under gone by a closed system of simple compressible substance ( neglect changes in kinetic and potential energy ) ? (a) du dw (b) Tds du Pdv (d) du Pdv (c) Tds du dw (d) du Pdv is true for reversible process and closed system. 37. Water has a critical specific volume of 0.003155 m3/kg. A closed and rigid steel tank of volume .025m3 contains a mixture of water and steam at 0.1 MPa. The mass of the mixture is 10 kg. The tank is now slowly heated. The liquid level inside the tank. (a) (b) (c) (d) Will rise. Will fall. Will remain constant. May rise or fall depending on the amount of heat transferred. Solution. Critical volume = 0.003155m3 / kg. 0.025 0.0025m3 / kg. Specific volume of mixture 10 Which is less than critical volume As it is a constant volume process Critical Volume v Saturated Liquid Saturated Volume v v P V 11 40. Which combination of the following statements is correct ? P : A gas cools upon expansion only. 12