' Chapter Second Law of Thermodynamics 7 mark QUESTIONS ‘Sobution: oe Fer CE, 2 _o Tr Ty 2. Te= 300K Yow som a | ce} —> w. | yo For CE, . 7 Q. SOMI 500 Henoe, the correct answer is 50. centworempersmre = RA evened ¢ temperature of -5°C cycle refrigerator maintains a The ambient air temperature is ©. The heat gained by the refrigerator at a continuous rate is. 2.5 kJ's, The power (in watt) required to pump this. heat out continuously is {2014-84} Solution: 873.1 W 4. The following four figures have been drawn to represent 4 fictitious thermodynamic cycle, on the P-V and T-S planes {2005} [2018-82], (=! 7.22 | Thermodynamics Fig.3 Fig. 4 Acconding to the first law of thermodynamics, equal areas are enclosed by (a) figures 1 and 2 (b) figures 1 and 3 (©) figures 1 and 4 (a) figures 2 and 3 Solution: (a) All areas are equal in magnitude Clockwise loop in P-V diagram shows work is done by system, Work done by system in a cycle requires net heat addition to system. This is area of 7-5 diagram. The loop in T-S diagram has to clockwise as it would then mean net heat added to system in eycle which in turn produces net work from system. Hence, both P-V’ and T-S loops have to clockwise, Figures 1 and 2 is one case which covers equal area (option given). Figure 3 and 4 is other case which covers equal area =8373W asc | | @ lew To=25t 5c For refrigerator which is reversible, (5 +273) _ a (COP) 67 reversible In terms of Heat (cop) = 2 = 67 Ww W = Q,/6.7 = 2.5 * 10°/6.7 = 373.1 Hence, the correct option is (a). 5. An industrial heat pump operates between the temperatures of 27°C and 13°C. The rates of heat addition and heat rejection are 750 W and 1000 W, respectively. The COP for the heat pump is 12003) (a) 75 (b) 6.5 (c) 4.0 (d) 3.0 Solution: (c) It may not be a reversible hy leat data to get COP Pury P80, H (COP ny = HEM Pieteg 1000 1000 — 759 > 4 are | 4 000 Heat pump |*~w 4 750w [ 196 Hence, the correct option is (c) 6 Consider a refrigerator and a heat pn reversed Carnot eycle between the limits. Which of the following is corre. (a) COP of refrigerator () COP of refrigerator = COP of heatp. (c) COP of refrigerator = COP of heat Pury (4) COP of refrigerator = inverse pump ~1 Solution: (c) COP of heat pun i 1 of the (Op, (COP ,chigertor Preis (COP susp Qy=O.+ (COP) yyy W= (COP), + IF (COP) puyy)= (COP) Li Hence, the correct option is (c). stem uni 7. In the case of a refrigeration syste inreversible eyele, § 52 iso? Solution: <0 $e <0 forineversibie ev? Tpermodynamic ¥CIE Operating between typ vo Me limits #8 reversible if the product of ine ote wher PETAL AS 2 eat engine and the Cow ween 8 FIBA equal to |. Ty ggqy oN of 1 ty Ee ' } ne {Mow | REF : w, + Y Grn _qversible engine, W ‘output Heat input 7, Mae = _qeersible refrigerator Qresies W, (COP)pe = ne* (COP) aT, Ty tce, product for reversible thermodynamic cycle is not aqualto 1. 4. reversible heat transfer demands: [1993] i) thetemperature difference causing heat transfer tends tozero (b) the system receiving heat must be at a constant temperature. (c) the system transferring out heat must be at a constant temperature, (@) both interacting systems must be at constant temperatures Solution: (a) Reversible Heat transfer depends that heat transfer should secur across very small (infinitesimal) temperature difference. Hee, the correct option is (a. 'W Acondenser ofa reftigeration system rejects heat at a rate Of 120 kW, while its compressor consumes a power of 30 W.The coefficient of performance of the system would be [1992] fy 14 (b) 4 9 3 (d) 3 : (d) (COp)g = Qovaporior 7.23 Chapter 4. Second Law of Thermodynamics | Create ~ rang 130-30 w Hence, the correct option is (d) See eee Two-marks QUESTIONS |. The figure shows a heat engine (HE) working between wo reservoirs. The amount of heat (Q,) rejected by the heat engine is drawn by a heat pump (HP). The heat pump Feceives the entire work output (1¥) of the heat engine Mf temperatures, 7, > 7, > T,, then the relation between the efficiency (7) of the heat engine and the coefficient of Performance (COP) of the heat pump is (2019-82) Ty, th T r Mg Q, (ve}_*_.e ¥ bs a Th uy (b) COP= 7! 1 (© COP=1+7 (@) COP=7 Solution: (a) W=nxQne 7.24 | Thermodynamics 2.A heat pump absorbs 10 kW. of heat from outside environment at 280 K while absorhing 16 KW of work Te delivers the heat to a room that must be kept warm at 200K The Coefficient of Performance (COP) of the heat pumps [2017-81] Solution: The Coefficient of Performance can be calculated as COP = Head delivered to room work input 2s kw ESE IG Iskw Here Heat delivered = Heat taken + work input Hence. the correct answer 1s (1.67). 3. The COP of a Carnot heat pump operating between 6°C and 37°C 18 [2015-83] 37-4273) Solution: COP = — 10 73) -(6+273)] Hence. the correct answer is 9.8 to 10.2, 4. reversible heat engine receives 2 kJ of heat from a reservoir at 1000 K and a certain amount of heat from a reservoir at 800 K. It rejects 1 kJ of heat to a reservoir at 400 K. The net work output (in kJ) of the cycle is [2014-81] (a) 08 (b) 1.0 (c) 14 (d) 20 Solution: (c) . 1000 k ¥ 2k=0, Reversible (He) —»w | = 42 pine 00k | [400K | It's reversible Heat engine. Hence, entropy chan; universe (system + surroundings) is zero. (M8), = (AS) yp, 4 (AS)yys + (AS) yy + (AS) yp Heat engine is cyclic. Entropy change of Heat engine is zero, 0, oo 0-H 1, My N, I, 2 2 4,! 000 800 ' 400 Q,=04k) Using First law for eyclie device, 2+O,= Ot W 24042140 (ay 50 (b) 250 (c) 300 (4) 60 Solution: (¢) High - temperature’ 1 0,= 100% Irreversible| ba t LO 4 0, = 50% T [tow ree temp. sink : 6. An irreversible heat engine oxy, tract Weiaki Hence, the correct option ise) temperature soutce ata rate of jy) toa sink at a rate of SO kW The the heat engine is used to drive», operating between a set of indepey,) reservoirs at 17°C and 19°C 1), the heat pump delivers heat , , sink is Using First law for Heat engine W=0,-Q, = 100 - 50 = 50kW Goesres _ 2, (COP)puny = see = =! w input As pump is reversible, we find 1's Cop by temperature limits (COP)reverible Ty Qu Ww Q,, = 6 x $0= 300 kW Hence, the correct option is (c) A heat transformer is a device tha heat, supplied to it at an intermedia high temperature reservoir while te}e0!2 Part to a low temperature heat sith | —_ § transformer, L000 KJ of heat 1s supplisd MAXIMUM amount of heat un KI that 6a 400K, when the rest is rejected to a he 1250 te) 34.33 tb) L429 (a) 57.4 — 2, = 100 ay |tanstormer(A) rn heat that can be transfer ouitina situation When entire ger will 1S Fevers wf gs when entropy change is zero, “ersible. This Ted to 4oy, Seat = AN coy + (AS), + a) fist Law 0,-2,+2, . sop Equation (1) and (2), we eliminate g, Qo oo oe 2 , 100-9, 400-300 3009, + 40000 - 4009, 120000 0,=57.16 kJ ‘transferred to 400 K reservoir is 57.16 kJ. owe, the correct option is (d). engine having an efficiency of 70% is used to drive ‘ngerator having a coefficient of performance of 5 ‘Teenegy absorbed from low temperature reservoir by ‘fhigerator for each kJ of energy absorbed from high “perature source by the engine is [2004] 0.2857 = (b) 0.71 i wo (@) 71k ton; () =0.70 a) k Qy First lay, Som 0, = w Q) Heat taken Work input nal A solar collector receiving solar radiatio 7.25 ics | .rmoayna™ Chapter 4 Second Law of The! ) Qs W Using Equation (1) and (3). 2 _ SW _ sor) Qn Oy gine 3.5 kJ for each kJ absorbed by end! the correct option is (¢) oe Of 0.6 kW/m? transforms it to the internal aad Of @ uid at an overall efficiency of $0% The ult heated to 350 K is used to run a heat engine yihich Tejects heat at 313 K. If the heat engine is to i 2.5 kW power, then minimum area of the solar collec required would be : oe (a) 8.33 m? (b) 16.66 (©) 39.68 m? (d) 79.36 Solution: (d) Let A be area of solar collector Hence, fluid gets heat by amount Q =[(0.6 kWim?) x (Area)]n, = (0.64)n, kW a 2 | Steady fuid s onpetee +350 k | re | (e) > Work output (1%) 7a, 313k | Efficiency of Heat engine = Work output _ 9" Heatsupplied Efficiency of Heat engine is less than reversible heat engine operating between same temperature limits TS Ne, 350 0223.64 Using Equation (1), (0.6)(A)(0.5) = 23.64 Ge ee (0.5)(0.6)7.26 | Thermodynamics A> 78,82 me? Minimum area is close to 78.82 m?. Hence, the correct option is (d). 9A Carnot eycle s having an efficiency of 0.75. - e temperature of the high temperature reservoir is 727°C, What is the temperature of low temperature reservoir [2002] (a) 23°¢ (by 28°C ©) oc (a) 250% Solution: (b) Maryn = 0.75 Ty, = 727°C +273 = 1000 K i -1, ae = 075 007 1000 1000-7, = 750; 7, = 250K T, = 280-273 =-23°¢ Hence, the correct option is (b), 10. A cyclic heat engine does 50 kl of work per cycle. If efficiency of the heat engine is 75%, then heat rejected 4 Per cycle is [2001 . 2 1 (@) 16=Ky (b) 332K) 3 3 1 Z © 37L13 @) 66213 2 3 Solution: (a) Given data W=50k; 1=0.75 10 Qasdea 5 0.75 = %0 ade 50. = —— = 66.66 kI Quases = 575 rejected = Qassea~ W = 66.66 ~ 50 = 16.66 KI =162k3 3 Hence, the correct option is (a), It 12. For two cycles coupled in series, the torpi"s A solr eneray based eat engin wg heat at 100°C and releets 704 of at 30°C is to be designed. The thermay ce Aen engine is coe (a) 70% (b) 18.8% (c) 12.5% (d) indeterminate : Solution: (c) Let ws layout of Heat nn it | 100 + 273 L=s73k Q, 7804s | © | ow @ ¥ 70K 103K Work output Thermal efficiency,n a Heat supplied Ld 9, = 0.187 8.7% Hence, this is possible (r| vag co 10. CLL Chapner 5 Entropy | 7-29 (by §e0 0, gee >0 (4) 8 0 Also it satisfies from second law that become of irreversible engine §2 <0 F Hence, the correct option is (a) One kilogram of water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is 2010) (a) equal to entropy change of the reservoir (b) equal to entropy change of water (c) equal to zero (d) always positive Solution: (d) ‘As Heat transfer takes place across finite temperature difference, process is irreversible. Hence, entropy change of universe is positive. Also according to the principle of entropy increase (ds), 2 0. Hence, the correct option is (d). ). Ifa closed system is undergoing an irreversible process, the entropy of the system (a) must increase (b) always remains constant (©) must decrease (@) can increase, decrease or remain constant Solution: (d) Entropy change of a process for a closed system is 12009) $42 + cagynemal As = JF + (As)inevesies [prion Entropy of a system can increase, decrease or remain ao constant. It all depends on. a and As term do Term ia can be negative and large enough to cancel, (As)ireverble All possibilities exist for entropy change for a system when it undergoes a process. Hence, the correct option is (d). 2 moles of oxygen are mixed adiabatically with, another 2 moles of oxygen in mixing chamber so that the final total pressure and temperature of the mixture become same a those of the individual constituents at their initial states The universal gas constant is given as R. The change i entropy due to mixing, per mole of oxygen, is given by [2008 (a) -Rin2 (b) 0 (c) Rin2 (d) Rind7.30 | Thermodynamics Solution: (b) hen similar gases are mixed, entropy of mixing 1s Zero. fence, the correct option is (b) Which of the following relationships is valid only for Teversible process undergone by closed system of simple Compressible substance (neglect changes in kinetic and Potential energy)? (@) 8Q=au+sw [2007] Solution: (a) Similar to Question no. 10 Hence, the correct option is (d) For an ideal gas the expression as bs T)—| -7{ [a] )] is always equal to (a) zero OR Solution: (c) Z 11997] ) GC, (d) RT Tas equation will be used to get (3) lp . Tds = dh ~ vdp At constant Pressure, P=c; dp=0; Tds=dh Tds = G dt (for ideal gas, dh = G dT) [4] -& @ . lar), "7 Tds = du + pdv At constant volume, Tds = dv Tds = [4| =o @) Using Equation (1) and (2) in given expression, = C, -C,=R Hence, the correct option is (c). 13. A system undergoes a state change from I to 2. According to the Second law of thermodynamics, for the process to ible, the entropy change, S ~ S; of the system be feasible, a (a) is positive or zero (b) is negative or zero (c) is zero (d) can be positive, negative or zero Solution: (d) Entropy change for a process og zero, There are no restrictions Posie Hence, the correct option is 4) 14, The slopes of constant Volume the Fs diagram are ang ay, TT eg, Solution: — 2 Let us use Tds equations Constant volume Tas = du + pdv, y Tds = du For ideal gas, change in internal energy ds = HT T Slope in Ts diagram, Constany, Constant pressure Ts = dh ~ vdp; P= constant Tds = dh = Slope, < =— 15. A 1500 W electrical heater is used to tex water (C, = 4186 J/kg-K) in an insulated by 4 temperature of 30°C to 80°C. Ifthe het is only infinitesimally larger than the wate during the process, the change in entropy fi WK and for water ___ JK Solution: For water, change in entropy is As = fas ms dT =|" = msin( 2 =n 2 wo 2 = 20x 4186x na ¥ 2786.9 = 12787) K 1 For heater, thee is work input and Het ‘o water. Work input des not bring change of water is only due to heat tt dQ _ p-mdl. 7 rset T as=+f fervo wat [Heat transfert" inter heatet {-) sign means heat rejected from 1, 1) = -12787 JK. == msin|Howing relat ong the fol a Feocess UNUETEONE by a page ie ou ys (b) Tas aes, pa 11993) Orde + pas b acd) jo i du + Sw 1s First law, vatig a= du + pdv {Or any Procesy caly FOF TeverSIble process. ag an (5 = PAV) is derived assuming a res uasi-static Tds = du + pdv sonot properties and hence, vatig he correct option is (d), for any process st EXECUEES aN itTEVETSble cycle ) fasso (@) $2 >0 casio: (8) 2 <0 is equation for ireversible cycle _ the correct option is (a), — - TWO-MARKS QUESTIONS szount of 100 KW of heat is transferred through a seady state. One side of the wall is maintained at ~¢ and the other side at 27°C. The entropy generated =WK) due to the heat transfer through the wall is [2014-3] Sinton: (see figure) ‘ow using the relation §, = 5,42 +r specific heat §T¥o identical metal blocks Land MGSPEITE re = 04 Ks/kg-K), each having a mass of ts initially at so KA reversible refrigeraicl ja antl heat from block L and rejects heat to block © the temperature of block L reaches 293 K. 2014-S4] temperature (in K) of block Mis i Solution: 7, ~ 313K, T,~ 29312 Work done (d8)sorne 0 for minimum work done Now we have (ds), = (ds), (ds), " 3343 K, 4. An ideal gas of mass m and temperature T, undergoes a reversible isothermal process from an initial pressure P. to final pressure P,. The heat loss during the process is 0 The entropy change As of the gas is (2012) B @ man (b) mR in Solution: (b) (4S) seternay = MR In (#) ‘Common Data for Questions 5 and 6: In an experimental set-up, air flows betwees P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station Pare 150 kPa and 350 ‘The temperature at station Q is 300 K. The followin, the properties and relations pertaining to air. sere 1 two stations ee a7.32 | Thermodynamics Specific heat at constant pressure, C, = 1.005 ike K Specific heat at constant volume C, = 0.718 kdikg K, Characteristic gas constant, 287 hike K, If the air has to flow from station P to station Q, the maximum possible value of pressure in kPa at station J is close to (2011) (a) $0 (c) 128 Solution: (b) Condition at P P, = 150 kPa T, =350K Conditions at @ T, =300L Flow is from P to Q Entropy change of air is calculated (b) 87 (d) 150 by Tds equation + Tas = dh vdp _ dh _vdp ear CT (apy T P (using gdh = C, dt and ideal as state equation] ds C,qT 27 as =f _fp( 2) ie) qr, As =C, in - Rin T, Py If process is reversible, then entropy change of universe is zero (AD ennene = A= 53~5, L = Cyn Rin r Py C, in = Rin 7 PB [300 (0.718 + 0287) 1n| 22] = 0.287In 22. 150 py = 83.43 = 87 kPa Hence, the correct option is (b) 6. If the pressure at station Q is 50 kPa, the change in sic (Sp—5,) in ki’kg Kis . 55 (b) 0 B cial (d) 0355 c) 0. Solution: (¢) 2 = 50 kPa As =C, In Rin’: Sy 5, = (0.718 + 0.259 0.16 kike k Hence, the correct option is (c) 7. Consider the following two proce, 1A heat source at 1209 K joy. sink at 800 K ou I. Abheat source at 800 K loses 244), at S00 K : Which of the following statement (a) Process I is more irreversi (b) Process Il is more irreversible ae (c) Irreversibility associated in 1 Solution: (by Case! Case 1200 aon ¥ 25000 Y 206 Process which is more irreversib! entropy of universe by larger amour Process I (As), nese = (AS):359 ~ (AS -2,2 T, 7, Process I] (AS)anene = = 1SKIK (As) > (As), Process II is more irreversible than POSS Hence, the correct option is (b) A cyclic device operates between thre? shown in the figure. Heat is transfert! device. It is assumed that heatee and the eyelie device takes place across erat sure difference. Interactions between meg he respective thermal reservoirs that gviee ae are all in the form of heat transfer. Cae te 12008] 300 K [300% | me : 50K ” soon! ,/ cyelo)) 604d device <= sevice cam De » “Spe heat engine “ye heat pump oF a reversible refrigeration este eat engine mtnie heat pump or an jsatot o ask entropy change of system (cyclic device) coding (reservoirs). =(38)jc00 + (AS)so0 + (AS)a00 + (AS)eyetic device irreversible D2, %8 5 jG [cyclic device has zero entropy change] 100 50, 60. 000 500 300 ~0.1-0.1+0.2 =0 “cage =0°=9 Reversible device vansfer, = 100 +50 - 60 =90 kJ “teat is added to device in cycle. So, it is heat rect option is (a). Smption refrigeration system is a heat pump al reservoirs shown in the figure. A ct of 100 W is required at 250 K when sc aldilable is at 400 K. Heat rejection occurs ‘nihimum value of heat required (in W) is [2005] | 400k | (b) 100, (d) 20 a Using First law, 10, Considering the relationship Chapter 5 Entropy | Solution: (c) = Q,+ 100 =100 9, Q +100 250 400° 300 Q, 280K) 7.33 |_aox_| | 30% |@ a} (HP) + ’ o troow [aor Io Entropy change of universe is always positive or equal to zero As20 (AS), + (As), + (AS), + (AS)up 20 Bi Bio o9 a (Heat pump is cyclic device so, no entropy change) =100_ Q | a 2 250 400 * 300 Hence, its entire thing is reversible, then heat transfer from source is 80 kJ. So, minimum value is 80 kJ Hence, the correct option is (c). TaS = dU + PdV between the entropy (S), Internal energy (U), pressure (P), ‘Temperature (7) and volume (V) Which of the following statements is correct? (a) Itis applicable only for a reversible process (b) For an irreversible process TdS > dU + Pd (c) Itis valid only for an ideal gas (a) It is equivalent to first law, for a reversible proc Solution: (d) It is equivalent to first law, for a reversible process. Tas = dQ (for reversible process) (1 dw = pdv (for reversible proc Tas = dv + pdv ising Equation (1) and (2), dQ = dv + dw (First law) Hence, the correct option is («) Common Data for Questions 11 and 12: ) {2003} ) ) Nitrogen gas (molecular weight 28) is enclosed in a cylinder by a piston, at the initial condition of bar are 298 K and Im‘. Ina particular proc the gas lowly expands under isothermal condition, until the volume becomes 2 m?, Heat exchange occurs with the atmosphere at 298 K during this process, 12003}a 7.34 | Thermodynamics UL The work interaction for the Nitrogen gas is @) 200K) (b) 138.6) ©) 20 () 200K) Solution: (b) Po=2 bar; T= 208K Wye Process ts isothermal T= 298K; v,=2m' Work done by system for isothermal process (py = 0) is We pyytn = 2x10 Vin \ = 138629 38.0) Hence, the correct option is (b) 12. The entropy change for the s ystem during the process in WKis (2) 0.4652 one? wu (@) 06711 Solution: (a) ‘iropy change is found by Tas equation Tas = du + pds For isothermal case, du = 0 (du = C,dT for ideal gas) Tas = pdv ds = Pav T as = joe T ; Jpav (T = constant) as — Work done T 38.6 298 Hence, the correct option is (a), 13. One kilomole of an ideal gas is throtth Pressure of 0.5 MPa to 0.1 MPa. The is 300 K. The entropy change of the 4652 KK Universe is [1995] (a) 13.38KIK (b) 4014.3 KK (©) 0.0446 kK (@) 0.0446 IK Solution: (a) ges Initial conditions ; p, = 0.5 mPa; T,=300K led from an initial initial temperature Exit conditions Py 1 mPa: Ts =p “Wy n-th _ vp a7 Enthalpy is constant during throttling hy =hy ah=0 For ideal g dh=Cydt; Car=9 aT=0 T=c Hence, Equation (1) becomes, ==nR in 22 y, 1X8.314 x 10° In| 2 = 13.3810 YK 3.38 KIK, Hence, the correct ‘option is (a), 0 '4. Figure below shows a reversible heat engine £, heat interactions with three constant temperature Calculate the thermal efficiency of the hea Solution: Let us draw diagram of reser engineee ——=— 1735 Chapter 5 Enropy | ie 1s reversible, We Need to find entropy ¢ cat ynd it should be zero anche oe oan ney Nar Heat inp’ 0 “ae . o@ a @..8.0. 0 @ +o, (Heat engine is cyctic device Ce. Zero entropy change] or ~ 50 , 2 9 me $00” 300 150 0.6 = 60%eT ONE-MARK QUESTIONS Which one of the following statements is correct for a superheated vapour? 2018-81} (2) Its pressure is less than the saturation pressure at a given temperature, (b) Its temperature is less than the saturation temperature ata given pressure. (©) Its volume is less than the volume of the saturated vapour at a given temperature. (d)_Its enthalpy is less than the enthalpy of the saturated vapour at a given pressure. Solution: For superheated vapour, the pressure is less than the saturated pressure at a given temperature. P Py). but Quality Constant pressure lines in the superheated \ollier diagram will have 2) apasitive slope 5) anegative slope «) zet0 slope 4) both positive and negative slope Salution: (a) Nollier diagram is a plot in A-s coordinates. (enthalpy- entopy) Letususe one of Tas equation region of the [1995] dh ~vdp = dh ~ 0 (constant pressure) =fhil ey aT ce, the correct option is (a). 2erelationship (@T/@p),, = 0 holds good for 2) anideal gas at any state 5) areal gas at any state ©) any gas at its critical state “) any gas at its inversion point Solution: (a) aT 11993) that at_constant dp =0. This relation means i ». This is valid “athalpy; temperature does not depend 0” P ‘or ideal gas, ce, the correct option is (a) Dating the phase change of a pure subs 8) dG=9 (b) dP=0 (a) av tance (119931 pure 38 erat During and temp “nstant dP=0,dT=0 Phase change, pressure cances | Chapter 6 Property of Pure = v= function) AG = -sdr + VaP (variation of GBPS =0 Hence, the correct option is (a) and (b). : number 0! At the triple point of a pure substance, the 11993] degrees of freedom is (a) 0 () 1 © 2 (a) 3 Solution: (a) Gibbs phase rule for pure substance is F+p=C+2 Atriple point, p= 3 and C= 1 F=C+2-p F=1+2-3; F This means that it’s a unique point and occurs at a single Value of pressure and temperature. Hence, the correct option is (a). — Two-marks QUESTIONS 1, A tank of volume 0.05 M° contains a mixture of saturated Water and saturated steam at 200°C. The mass of the liquid present is 8 kg. The entropy (in kJ/kg K) of the mixture is (correct to two decimal places). [2018-81] Property data for saturated steam and water are: At 200°C. Py, = 1.5538 MPa V,= 0.001157 m’ ‘kg, V, Sjp=4.1014kIikg K, S, Solution: 0.12736 mkg 33.9 kikg K Volume of tank V’= 0.05 m> At 200°C. Pog, = 1.5538 MPa V,= 0.001137 m3 kg, 12736 mkg. 1O14k kg K mass of specific entropy of mixture +5 my +m, my XO tm, Xt 0.05 = 8x 0.001157 +m, x 9 mn, = 0.08074 0.12736 mm, = 0.3198 ke 0.3198 0319848 eetap, RSE a Thermodynamics ¥= 0.0384 3309 + 0.0384 x 4.1084 $= 2488 KI kg K. Hence, the correct answer is 2488, 2.48. For water at 25°C, dp/dT, = 0.189 kPalK (P. 18 = Saturation pressure in kPa and 7, is the saturation vanberature in K) and the specific vohume of dry saturated NaPOr ts 43.38 m'/kg. Assume that the specific volume of liquid is negligible in ‘comparison with that of vapor. Using the Cla Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25° (in klk) is Tas {2016-S1] ap, a, 0.189 x 19? = __“s__ 2984338 neg = 2443.24 kikg, Hence, the correct answer is 2400 — 2500. 43. The Vander Waals equation of state is [2015-83] |p+ ] (v-b)=RT. J |erS]o-o where p is the pressure, v is the specific volume, T'is the temperature, and R is the characteristic gas constant, The SI unit ofa is (a) Sikg-K (b) miikg (c) mi/kg-s? (d) Pakg a g The unit of 5) should be equal to ki/kg ike) A rigid container of volume 9,5 KN-m_ im? R = Nos aan = Nem m)_mé 5 or @ hex igh Fas my Hence, the correct option is (c), : water at 120°C (vj = 0. 0106 ming Oni, The state of water is sy, Oh (a) compressed liquid (b) saturated liquid (c) a mixture of saturated fj (d) superheated vapor i, Ly Ui and te Solution: V, = 0.5 mm =1 kg. 8 , = 0.8908 mikg hos ~ Diag Ue 1 7 r+36, > 015 = 0.00106 + (0.89055, = x= 0.56 Since dryness fraction, x < 1, the state ture of saturated liquid and saturated y Hence, the correct option is (C) Common Data for Questions 5 and 6: In the figure shown, the system is a pure sun in a piston-cylinder arrangement. The system a two-phase mixture containing 1 kg ofl kg of vapour ata pressure of 100 kPa, Initial, rests on a set of stops, as shown in the fue 4p 0f 200 kPa is required to exactly balance te w piston and the outside atmospheric pressue He" takes place into the system until its volume inc 50%. Heat transfer to the system occurs in such" that the piston, when allowed to move, dos 01 slow(quasi-static/quasi-equilibrium) process. Th reservoir from which heat is transfered toi as a temperature of 400°C. Average temperstst System boundary can be taken as 75°. He 8 of at por - wr (| Specific volume of liquid (77) 294 VPN well as values of saturation tempera! table below:‘Saturation |», | temperature, | (m3/kg) Typ CO) || 0.001 O1 200 | 0.0015 200 0.002 aye ond of the Process, which one of the yg will be true? came vapor will be left in the system | 2) Svapour will be left in the system Myiquid + Vapour mixture will be left in the system * ghe mixture Will exist at a dry saturate vapour state ction: gal conditions p, = 100 kPa T, = 100°C (saturated mixture) ¥,= [ny v+ my Viele 100 kPa (1) 0.001) + (0.03) (0.1) x 103 m3 sal volume is 1.5 times initial volume Vgnat = 1-Sv,= 1.5 x 4 x 103 = 6 x 1073 m3 specific volume in final condition is Vinal m following y= _ Moral = Olay ~ 1.03 = 5.825 x 103 mi/kg Vp > (%Q)200 Kea 5825 x 103 > 2x 103 = Vapour is present in super heated condition. Hience, the correct option is (a). cork done by the system during the process is, 2) OL KI (b) 0.2 kJ 6) 0.3 kJ (d) O.4kI Solution: (d) La us draw entire process in TV diagram system oes from point 1 to point a with volume being Constant until pressure at point a becomes 200 kPa uflcient to raise piston. After that (point a—point 2). ‘is constant pressure process. Nk done by system in process 1-2 sakes plans et ‘hen piston starts rising in a quasi-static way (23°): “onstant pressure process from point a to Pow Chapter 6 Property of Pure Substances | W = J pdv =P (¥2—¥Q) =P (¥2—%1) = 200 = 103 (6 x 103-1 x 104) 400 = 0.4 kJ Hence, the correct option is (d). - The net entropy generation (considering the) system and the thermal reservoir together) during the process is closest to [2008] (a) 7.5.I/K (c) 8.5K Solution: (¢) Net entropy generation is change in entropy of universe (AS) gen = (AS) universe = (A8)reservoir + (AS), =¢. (b) 7.7K (@) 10K eyster + (A)eystem =— 4s 104K (400 + 273) =1000 = +10 = 8.5 JK Hence, the correct option is (c). 8. Given below is an extract from steam tables: [2006] ‘Temp.| Psa | Specific Volume | Enthalpy €C) | (bar) (mike) | (kshkg) ‘Saturated [Saturated | Saturated| Saturated liquid | vapour | liquid | vapour 45 [0.9593 |o.001010| 15.26 | 18845 | 23978 | 342.24 150 | 0.001658 | 0.010337 | 1610.5 | 261.5 Specific enthalpy of water in ki/kg at ISO bar and 45°C is (a) 203.60 (b) 200.53 (c) 196.38 (a) 188.45 Solution: (a) ‘Common Data for Questions,5 and 6: The following table of properties was printed out for saturated liquid and saturated vapour of ammonia. The titles for only the first two columns are available. All that we know is that the other columns (columns 3 to 8) contain data on specific properties, namely, internal energy (kJ/kg), enthalpy (ki/kg) and entropy (kJ/kg: K). [2005] TCO |P (kPa) | | | =20 | 1902 | 88.76 [0.3657] 89.05 12995] 14180 | 0 | 4296 14 180.36 | 13180) 14412 | 20 | 8575 1.0408] 274.30 1332.2| 146022 | 49 [272.89[ 13574] 371.43 | 48662 [1381.0] 14702 | 9. The specific enthalpy data are in columns (a) 3and7 (b) 3and8 (©) Sand7 (d) Sand87.40 10. Fe Tee ea eee errs | Thermodynamics Solution: (a) We use following facts to find out columns: (Specific internal energy), pou > (Specifi (Specific enthalpy), ~( (Specific enthalpy pecific internal ENTLY syne wnt * (Specific internal energy) jaya Also for entropy, ra Using all these facts, we can see that columns 5 and 8 are specific enthalpy columns Hence, the correct option is (4). When saturated liquid at 40°C is throttled to -20°C, the quality at exit will be (a) 0.189 (b) 0.212 (©) 0.231 (4) 0.788 Solution: (b) During throttling, enthalpy of substance remains constant f= hy Aiguiasorc = hy + XIyy ) sy. 371.43 = 89.05 + x (1418 — 89.05) 2125 150 bar (0.9593 bar We need to find specific enthalpy of compressed liquid at point 4. From A to B, temperature is same as 45°C, there is a different pressure. Specific internal energy of compressed liquid depends only on temperature. (a), = Wy © (tty iy 9593 bar (Ay = PY )o.aso3 bur 0.9593 x 10° x 0.001010 = 188.45 5 — 10 = 188.35 kik (Ay =a Pa 7188.35 + 50105044) 1g (Specific volume does not ch, = 203.6 ki/kg Hence, the correct option is (by, of 11. A vessel of volume 1.0 m? contains a water and steam in equilibrium ay (Me 90% of the volume is occupied by i, dryness fraction of the mixture, a." Sm, v,= 0.001 m¥/kg and v, = 1.7m) kg, sume, ay Solution: Total volume of container he v=lm Volume occupied by gas is 0.90V = 0.9 m3 Volume occupied by liquid is O.1V=0.1 m3 Veas = (Meas) X (specific volume) 0.9 = mga, X17 : Mogg = 0.529 kg tiquia = 1 X (Specific volume), 0.1 =m, x 0.001 m, = 100 kg Dryness fraction _ Mass of vapour content total mass om, +m, 0.520) © 100 + 0.529 = 5.266 = 10° 12. In the vicinity of the triple point, the vapour p= liquid and solid ammonia are respectively £82" In P = 15.16 — 3063/T and In P= 18.70-3754T where P is in atmospheres and Tis in Keb" What is the temperature at the triple point? Solution: Triple point is unique to pure sus Pressure and temperature value. 3063 In P = 15.16 = In P =18.70 - =3063 3754 _ 5.79 -15.16 T T= 195.2K Triple point temperature is 195.2 K-