Table Of Content

ÙetLe keâe@efcheefšMeve šeFcme SSC JE (Mains) petefveÙej FbpeerefveÙej cewkesâefvekeâue FbpeerefveÙeefjbie mee@uJ[ hesheme& mebheeove SJeb mebkeâueve cewkesâefvekeâue JE hejer#ee efJeMes<e%e meefceefle mebheeokeâerÙe keâeÙee&ueÙe ÙetLe keâe@efcheefšMeve šeFcme 12, ÛeÛe& uesve, Fueeneyeeo-211002 cees. : 9415650134 email : [email protected] Online test website : www.yctonline.com ØekeâeMeve Iees<eCee mecheeokeâ SJeb ØekeâeMekeâ Deevevo kegâceej cenepeve ves ef$eJesCeer Dee@Heâmesš, Fueeneyeeo mes cegefõle keâjJeekeâj, DeeOegefvekeâ ØekeâeMeve, 12, ÛeÛe& uesve, Fueeneyeeo-2 kesâ efueS ØekeâeefMele efkeâÙee~ Fme hegmlekeâ keâes ØekeâeefMele keâjves ceW mecheeokeâ SJeb ØekeâeMekeâ Éeje hetCe& meeJeOeeveer yejleer ieF& nw efHeâj Yeer efkeâmeer $egefš kesâ efueS Deehekeâe megPeeJe SJeb menÙeesie Dehesef#ele nw~ efkeâmeer Yeer efJeJeeo keâer efmLeefle ceW vÙeeefÙekeâ #es$e Fueeneyeeo nesiee~ cetuÙe : 195/- efJe<eÙe-metÛeer ,, SSC JE Gòej ØeosMe heer.Sme.meer. cewkesâefvekeâue cegKÙe hejer#ee (cid:1)(cid:1)(cid:1)(cid:1) Mechanical : hejer#ee hee"dÙe›eâce.............................................................................................. 3 (cid:1)(cid:1)(cid:1)(cid:1) SSC JE cegKÙe hejer#ee, 2016.................................................................................................. 5-20 Mechanical Engineering : JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) SSC JE cegKÙe hejer#ee, 2015............................................................................................... 21-38 Mechanical Engineering : JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) SSC JE cegKÙe hejer#ee, 2014............................................................................................... 39-51 Mechanical Engineering : JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) SSC JE cegKÙe hejer#ee, 2013............................................................................................... 52-63 Mechanical Engineering : JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) SSC JE cegKÙe hejer#ee, 2012................................................................................................ 64-76 Mechanical Engineering : JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) SSC JE cegKÙe hejer#ee, 2011................................................................................................ 77-89 Mechanical Engineering : JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) SSC JE cegKÙe hejer#ee, 2010................................................................................................ 90-99 Mechanical Engineering : JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) SSC JE cegKÙe hejer#ee, 2009............................................................................................ 100-107 Mechanical Engineering : JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) SSC JE cegKÙe hejer#ee, 2008............................................................................................ 108-117 Mechanical Engineering : JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) SSC JE cegKÙe hejer#ee, 2007............................................................................................ 118-125 Mechanical Engineering : JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) UPPSC IRRIGATION JE cegKÙe hejer#ee, 2008............................................................ 126-143 Mechanical Engineering Paper–I JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) UPPSC IRRIGATION JE cegKÙe hejer#ee, 2008............................................................ 144-158 Mechanical Engineering Paper–II JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) UPPSC IRRIGATION JE cegKÙe hejer#ee, 2005............................................................ 159-173 Mechanical Engineering Paper–I JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) UPPSC IRRIGATION JE cegKÙe hejer#ee, 2005............................................................ 174-188 Mechanical Engineering Paper–II JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) Uttarakhand PSC IRRIGATION JE cegKÙe hejer#ee, 2011...........................................189-197 Mechanical Engineering Paper–I JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) Uttarakhand PSC IRRIGATION JE cegKÙe hejer#ee, 2011.......................................... 198-210 Mechanical Engineering Paper–II JÙeeKÙee meefnle nue ØeMve-he$e (cid:1)(cid:1)(cid:1)(cid:1) MP PSC IRRIGATION JE cegKÙe hejer#ee, 2006.......................................................... 211-216 Mechanical Engineering : JÙeeKÙee meefnle nue ØeMve-he$e 2 YOUTH LIST OF TECHNICAL EXAM BOOKS (English-Hindi) (cid:1)(cid:1)(cid:1)(cid:1) SSC JE Electrical Solved Papers.............................................................................. cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) SSC JE Mechanical Solved Papers.......................................................................... cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) SSC JE Civil Solved Papers....................................................................................... cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) SSC JE (Pre) efmeefJeue FbpeerefveÙeefjbie Øewefkeäšme yegkeâ.......................................................... cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) SSC JE (Pre) Fuesefkeäš^keâue FbpeerefveÙeefjbie Øewefkeäšme yegkeâ..................................................... cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) SSC JE (Pre) cewkesâefvekeâue FbpeerefveÙeefjbie Øewefkeäšme yegkeâ....................................................... cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) SSC JE cesvme efmeefJeue/Fuesefkeäš^keâue/cewkesâefvekeâue mee@uJ[ hesheme&................................ ØelÙeskeâ cetuÙe : 195/- (cid:1)(cid:1)(cid:1)(cid:1) SSC JE hejer#ee efjøesâMej (Civil) ....................................................................................... cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) SSC JE Mechanical hejer#ee efjøesâMej............................................................................... cetuÙe : 120/- (cid:1)(cid:1)(cid:1)(cid:1) UPPSC mebÙegòeâ FbpeerefveÙeefjbie hejer#ee (efmebÛeeF&) efnvoer (Compulsory)............................. cetuÙe : 90/- (cid:1)(cid:1)(cid:1)(cid:1) UPPSC eEmeÛeeF& efJeYeeie (Irrigation) efmeefJeue FbpeerefveÙeefjbie mee@uJ[ hesheme&....................... cetuÙe : 150/- (cid:1)(cid:1)(cid:1)(cid:1) UPPSC eEmeÛeeF& efJeYeeie (Irrigation) cewkesâefvekeâue FbpeerefveÙeefjbie mee@uJ[ hesheme&.................. cetuÙe : 150/- (cid:1)(cid:1)(cid:1)(cid:1) Fuesefkeäš^keâue FbpeerefveÙeefjbie hejer#ee efjøesâMej............................................................................. cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) efHeâšj Fueskeäšê@efvekeäme hejer#ee efjøesâMej.................................................................................... cetuÙe : 90/- (cid:1)(cid:1)(cid:1)(cid:1) FuesefkeäšêfMeÙeve/kebâchÙetšj meeFbme hejer#ee efjøesâMej.................................................................... cetuÙe : 90/- (cid:1)(cid:1)(cid:1)(cid:1) Dee@šesceesyeeFue hejer#ee efjøesâMej.............................................................................................. cetuÙe : 90/- (cid:1)(cid:1)(cid:1)(cid:1) RRB JE-cewkesâefvekeâue/Fuesefkeäš^keâue/efmeefJeue/Fueskeäšê@efvekeäme/meeFbme efJeMes<eebkeâ.................. cetuÙe : 90/- (cid:1)(cid:1)(cid:1)(cid:1) ueeskeâes heeFueš-Dee@šesceesyeeFue/efHeâšj/Fuesefkeäš^keâue/Fueskeäšê@efvekeäme/ lekeâveerkeâer efJe%eeve efJeMes<eebkeâ..................................................................................................cetuÙe : 90/- (cid:1)(cid:1)(cid:1)(cid:1) UPPCL JE Fuesefkeäš^keâue FbpeerefveÙeefjbie hejer#ee efjøesâMej...................................................... cetuÙe : 70/- (cid:1)(cid:1)(cid:1)(cid:1) UPPCL JE Fuesefkeäš^keâue FbpeerefveÙeefjbie hejer#ee mee@uJ[ hesheme&..............................................cetuÙe : 120/- (cid:1)(cid:1)(cid:1)(cid:1) UPPCL FuessefkeäšêfMeÙeve š^s[ hejer#ee efjøesâMej....................................................................... cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) UPPCL FuessefkeäšêfMeÙeve š^s[ mee@uJ[ hesheme&.......................................................................... cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) DRDO STA `B' cewkesâefvekeâue FbpeerefveÙeefjbie Øewefkeäšme yegkeâ.................................................. cetuÙe : 90/- (cid:1)(cid:1)(cid:1)(cid:1) DRDO STA `B' cewkesâefvekeâue FbpeerefveÙeefjbie efjøesâMej........................................................... cetuÙe : 120/- (cid:1)(cid:1)(cid:1)(cid:1) DRDO STA `B' Fueskeäšêsefvekeäme Sb[ keâcÙegefvekesâMeve efjøesâMej............................................. cetuÙe : 120/- (cid:1)(cid:1)(cid:1)(cid:1) DRDO STA `B' keâchÙetšj meeFbme hejer#ee efjøesâMej.............................................................cetuÙe : 120/- (cid:1)(cid:1)(cid:1)(cid:1) DMRC/LMRC JE efmeefJeue hejer#ee mhesMeue.................................................................... cetuÙe : 150/- (cid:1)(cid:1)(cid:1)(cid:1) DMRC/LMRC JE cewkesâefvekeâue hejer#ee mhesMeue.............................................................. cetuÙe : 150/- (cid:1)(cid:1)(cid:1)(cid:1) DMRC/LMRC JE Fuesefkeäš^keâue hejer#ee mhesMeue............................................................ . cetuÙe : 120/- (cid:1)(cid:1)(cid:1)(cid:1) DMRC/LMRC JE ceWvšsvej hejer#ee mhesMeue.................................................................... cetuÙe : 150/- (cid:1)(cid:1)(cid:1)(cid:1) DMRC/LMRC SC/TO hejer#ee mhesMeue........................................................................ cetuÙe : 150/- (cid:1)(cid:1)(cid:1)(cid:1) DMRC/LMRC CRA hejer#ee mhesMeue............................................................................ cetuÙe : 150/- (cid:1)(cid:1)(cid:1)(cid:1) UPRVUNL JE efmeefJeue FbpeerefveÙeefjbie mee@uJ[ hesheme& Sb[ Øewefkeäšme yegkeâ............................. cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) UPRVUNL JE Fuesefkeäš^keâue FbpeerefveÙeefjbie mee@uJ[ hesheme& Sb[ Øewefkeäšme yegkeâ...................... cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) UPRVUNL JE cewkesâefvekeâue FbpeerefveÙeefjbie mee@uJ[ hesheme& Sb[ Øewefkeäšme yegkeâ....................... cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) UPRVUNL efHeâšj mee@uJ[ hesheme& Sb[ Øewefkeäšme yegkeâ........................................................... cetuÙe : 100/- (cid:1)(cid:1)(cid:1)(cid:1) ceOÙe ØeosMe Sub. Engineer Practice Book (Civil)..................................................... cetuÙe : 90/- (cid:1)(cid:1)(cid:1)(cid:1) BSNL TTA hejer#ee mhesMeue............................................................................................. cetuÙe : 495/- (cid:1)(cid:1)(cid:1)(cid:1) BSNL TTA Electronics & Communication Refresher...................................... cetuÙe : 90/- (cid:1)(cid:1)(cid:1)(cid:1) BSNL TTA Solved Papers ....................................................................................... cetuÙe : 150/- (cid:1)(cid:1)(cid:1)(cid:1) CIVIL/MECHANICAL/ELECTRICAL Formula Book........................cetuÙe : 100/- each (cid:1)(cid:1)(cid:1)(cid:1) CIVIL/MECHANICAL/ELECTRICAL 10,000 Pointer........................cetuÙe : 150/- each (cid:1)(cid:1)(cid:1)(cid:1) UPSSSC JE efmeefJeue/cewkesâefvekeâue/Fuesefkeäš^keâue/ke=âef<e mee@uJ[ hesheme& Sb[ Øewefkeäšme yegkeâ........ØelÙeskeâ 100/- (cid:1)(cid:1)(cid:1)(cid:1) SSC JE efmeefJeue/cewkesâefvekeâue/Fuesefkeäš^keâue DeOÙeeÙeJeej SJeb Keb[Jeej hejer#ee hueevej.................... 395/- 4 SSC JE Mains Mechanical Engineering Syllabus (cid:1)(cid:1)(cid:1)(cid:1) Theory of Machines and Machine Design : Concept IC Engine Performance, IC Engine Combustion, of simple machine, Four bar linkage and link motion, IC Engine Cooling & Lubrication. Flywheels and fluctuation of energy, Power (cid:1)(cid:1)(cid:1)(cid:1) Rankine cycle of steam : Simple Rankine cycle plot transmission by belts–V-belts and Flat belts, on P-V, T-S, h-s planes, Rankine cycle efficiency Clutches–Plate and Conical clutch, Gears–Type of with & without pump work. gears, gear profile and gear ratio calculation, (cid:1)(cid:1)(cid:1)(cid:1) Boilers; Classification; Specification; Fittings & Governors–Principles and classification, Riveted Accessories : Fire Tube & Water Tube Boilers. joint, Cams, Bearings, Friction in collars and pivots. Air Compressors & their cycles; Refrigeration (cid:1)(cid:1)(cid:1)(cid:1) Engineering Mechanics and Strength of Materials cycles; Principle of a Refrigeration Plant; : Equilibrium of Forces, Law of motion, Friction, Nozzles & Steam Turbines Concepts of stress and strain, Elastic limit and elastic Fluid Mechanics & Machinery constants, Bending moments and shear force diagram, (cid:1)(cid:1)(cid:1)(cid:1) Properties & Classification of Fluid : ideal & real Stress in composite bars, Torsion of circular shafts, fluids, Newton’s law of viscosity, Newtonian and Bucking of columns–Euler’s and Rankin’s theories, Non-Newtonian fluids, compressible and Thin walled pressure vessels. incompressible fluids. (cid:1)(cid:1)(cid:1)(cid:1) Thermal Engineering : Properties of Pure (cid:1)(cid:1)(cid:1)(cid:1) Fluid Statics : Pressure at a point. Substances : p-v & p-T diagrams of pure substance (cid:1)(cid:1)(cid:1)(cid:1) Measurement of Fluid Pressure : Manometers, U- like H2O, Introduction of steam table with respect to tube, Inclined tube. steam generation process; definition of saturation, wet (cid:1)(cid:1)(cid:1)(cid:1) Fluid Kinematics : Stream line, laminar & turbulent & superheated status. Definition of dryness fraction of flow, external & internal flow, continuity equation. steam, degree of superheat of steam. H-s chart of (cid:1)(cid:1)(cid:1)(cid:1) Dynamics of ideal fluids : Bernoulli’s equation, steam (Mollier’s Chart). (cid:1)(cid:1)(cid:1)(cid:1) 1st Law of Thermodynamics : Definition of stored Total head; Velocity head; Pressure head; Application of Bernoulli’s equitation. energy & internal energy, 1st Law of Thermodynamics (cid:1)(cid:1)(cid:1)(cid:1) Measurement of Flow rate Basic Principles : of cyclic process, Non Flow Energy Equation, Flow Venturimeter, Pilot tube, Orifice meter. Energy & Definition of Enthalpy, Conditions for (cid:1)(cid:1)(cid:1)(cid:1) Hydraulic Turbines : Classifications, Principles. Steady Flow, Steady State Steady Flow Energy (cid:1)(cid:1)(cid:1)(cid:1) Centrifugal Pumps: Classifications, Principles, Equation. Performance. (cid:1)(cid:1)(cid:1)(cid:1) 2nd Law of Thermodynamics : Definition of Sink, Production Engineering Source Reservoir of Heat, Heat Engine, Heat Pump & (cid:1)(cid:1)(cid:1)(cid:1) Classification of Steels : mild steal & alloy steel, Refrigerator, Thermal Efficiency of Heat Engines & Heat treatment of steel, Welding – Arc Welding, Gas coefficient of performance of Refrigerators, Kelvin– Planck & Clausius Statements of 2nd Law of Welding, Resistance Welding, Special Welding Techniques i.e. TIG, MIG, etc. (Brazing & Thermodynamics, Absolute or Thermodynamic Scale Soldering), Welding Defects & Testing; NDT, of temperature, Clausius Integral, Entropy, Entropy Foundry & Casting – methods, defects, different change calculation of ideal gas processes. Carnot casting processes, Forging, Extrusion, etc, Metal Cycle & Carnot Efficiency, PMM-2; definition & its cutting principles, cutting tools, Basic Principles of impossibility. (cid:1)(cid:1)(cid:1)(cid:1) Air standard Cycles for IC engines : Otto cycle; machining with (i) Lathe (ii) Milling (iii) Drilling (iv) Shaping (v) Griding, Machines, tools & plot on P-V, T-S Planes; Thermal Efficiency, Diesel manufacturing processes. Cycle; Plot on P-V, T-S planes; Thermal efficiency. 3 SSC JE cegKÙe hejer#ee, 2016 MECHANICAL JÙeeKÙee meefnle nue ØeMve-he$e 1.(a) Draw and explain the P–T (Pressure – heefjJeefle&le veneR nesleer nw~ Fme Øe›eâce ceW oer ieF& T<cee keâes efJeefMe° Temperautre) diagram for a pure substance. T<cee keânles nw~ Megæ heoeLe& keâe P–T (oeye –leeheceeve) DeejsKe yeveeFS peue kesâ efueS P–T [eÙe«eece– Deewj Gmekeâer JÙeeKÙee keâerefpeS–15 Gòej-1(a) : Megæ heoeLe& (Pure Substance)– Megæ heoeLe& Ssmes heoeLe& nesles nw pees Skeâ mes DeefOekeâ DeJemLee ceW jnles nw~ Megæ heoeLeex keâer mechetCe& õJÙeceeve keâer jemeeÙeefvekeâ mebIešve meceeve nesleer nw DeLee&le peue keâes Megæ heoeLe& (Pure substance) ceevee pee mekeâlee nw~ (b) With the assumptions, derive the Steady Flow Energy Equation (SFEE). ceevÙeleeDeeW kesâ DeeOeej hej, efvejblej (DeheefjJeleea) ØeJeen GoenjCe kesâ efueS yeHe&â keâe Skeâ šgkeâ[Ì e pees –150C hej nw Gmes 2500 C keâer Yeehe yeveeves keâe jsKeeefÛe$e Tpee& meceerkeâjCe (Sme.Sheâ.F&.F&.) JÙeglheVe keâerefpeS~ 15 Gòej-1(b) : DeheefjJeleea ØeJeen kesâ efueS ceevÙeleeS :– T P (i) ØeJeen keâer oj Fveuesš (Inlet) leLee DeeGšuesš (outlet) hej atm meceeve nesveer ÛeeefnS~ 2500C 6 (ii) T<cee mLeeveevlejCe keâer oj efveÙele nesveer ÛeeefnS~ (iii) keâeÙe& mLeeveevlejCe keâer oj efveÙele nesveer ÛeeefnS~ 4 1000C (iv) efveef§ele efyevog leLee efveef§ele meceÙe ceW ef›eâÙeekeâejer heoeLe& keâer efmLeefle 5 meceeve nesveer ÛeeefnS~ 00C 2 (v) efmemšce keâer jemeeÙeefvekeâ mebjÛevee heefjJeefle&le vener nesveer ÛeeefnS~ 3 1 ∆∆∆∆Q –150C Øe›eâce 1-2:→→→→ Øe›eâce 1 mes 2 ceW –150C kesâ yeHe&â keâes T<cee oer peeleer nw efpememes –150C keâe yeHe&â 00C kesâ yeHe&â ceW heefjJeefle&le neslee nw~ Fme Øe›eâce ceW yeHe&â keâes efJeefMe° T<cee oer peeleer nw~ Øe›eâce 2–3: →→→→ Fme Øe›eâce ceW 00C keâer yeHe&â keâes T<cee osves hej 00C keâe heeveer yevelee nw~ Fme Øe›eâce ceW DeJemLee heefjJeefle&le veneR nesleer nw~ Fme Øe›eâce ceW oer ieF& T<cee keâes ieghle T<cee keânles nw~ efÛe$e ceW Skeâ Keguee efvekeâeÙe ØeoefMe&le efkeâÙee ieÙee nw efpemeceW Øe›eâce 3–4: →→→→ Fme Øe›eâce ceW 00C kesâ heeveer keâes T<cee oskeâj keâeÙe&keâejer heoeLe& keâe efvejblej ØeJeen nes jne nw~ keâeÙe&keâejer heoeLe& 1000C keâe peue yeveeÙee peelee nw Fme Øe›eâce ceW peue keâe leeheceeve meskeämeve 1 keâer lejHeâ mes ØeJesMe keâj jne nw Deewj meskeämeve 2 keâer heefjJeefle&le neslee nw~ Fme Øe›eâce ceW oer ieF& T<cee keâes efJeefMe° T<cee lejHeâ mes efvekeâue jne nw keânles nw~ p = ØeJesMe kesâ meceÙe keâeÙe&keâejer heoeLe& keâe oeye (N/m2) Øe›eâce 4–5:→→→→ Fme Øe›eâce ceW 1000C peue T<cee oskeâj 1000C keâer 1 V = ØeJesMe kesâ meceÙe keâeÙe&keâejer heoeLe& keâe efJeefMe° DeeÙeleve (m3/kg) Yeehe yeveeÙeer peeleer nw Fme Øe›eâce ceW DeJemLee heefjJeefle&le nesleer nw~ s1 V = ØeJesMe kesâ meceÙe keâeÙe&keâejer heoeLe& keâe Jesie (m/sec) Fme Øe›eâce ceW oer ieF& T<cee keâes ieghle T<cee keânles nw~ 1 u = ØeJesMe kesâ meceÙe keâeÙe&keâejer heoeLe& keâe efJeefMe° Deevleefjkeâ Tpee& Øe›eâce 5–6:→→→→ Fme Øe›eâce ceW 1000C keâer Yeehe keâes T<cee oskeâj 1 (J/kg) 2500C keâer Yeehe yeveeÙeer peeleer nw~ Fme Øe›eâce ceW DeJemLee 5 z = Fveuesš keâer [sšce melen mes TÛeeF& (m) Gòej-(c) (i) 1 P , Vs , V , u , leLee z = efve<keâeefmele meskeämeve hej keâeÙe&keâejer heoeLe& 2 2 2 2 2 kesâ ceeve q....1 -2 = efmemšce Éeje efJeleefjle keâeÙe& m= õJÙeceeve ØeJeen keâer oj W = efmemšce Éeje efJeleefjle keâeÙe& 1-2 keâeÙe&keâejer heoeLe& kesâ 1kg õJÙeceeve hej efJeÛeej keâjves hej e = Deevleefjkeâ Tpee& + mLeeveevleefjle Tpee& + ieeflepe Tpee& + efmLeeflepe 1 Tpee& + Øeoeve keâer ieF& T<cee v2 P = 1.2 bar e =u +p V + 1 +gz +q (in J/kg) 1 1 1 s1 z 1 1−2 Q = 50 kJ V = 0.14 m3 Fmeer Øekeâej efve<keâeefmele meskeämeve hej Øe›eâce 1-2 kesâ efueS (DeeFmeesyesefjkeâ Øe›eâce) e =u +p V + v22 +gz +w (in J/kg) Q = 50 kJ 2 1 2 s2 z 2 1−2 V –V = ∆V = 0.14 m3 2 1 efveJesMe leLee efve<keâemeve kesâ oewjeve Ùeefo Tpee& keâe keâesF& #ejCe ve nes lees P = 1.2 × 105 N/m2 e =e Øe›eâce 1-2 Éeje efkeâÙee ieÙee keâeÙe& = JeeÙegceb[ueerÙe oeye kesâ efJe™æ 1 2 v2 v2 Øemeej ceW efkeâÙee ieÙee keâeÙe& + 90 kg õJÙeceeve Yeej keâes 5.5m G"eves u1+p1Vs1+ 21 +gz1+q1−2 =u2+p2vs2 + z2 +gz2+w1−2 ceW efkeâÙee ieÙee keâeÙe& W = P.∆V + mgh 1–2 nce peeveles nw W1–2 = 1.2×105×0.14 + 90×10×5.5 (g = 10m/sec2) W = 21.75 kJ u +p v =h leLee u +p v =h 1–2 1 1 s1 1 2 2 s2 2 Deevleefjkeâ Tpee& ceW heefjJele&ve (∆u) = Q – W v2 v2 ∆u = 50 – 21.7 h + 1 +gz +q =h + 2 +gz +w 1 2 1 1−2 2 2 2 1−2 ∆u=28.25kJ h+ke +pe +q =h +ke +pe +w Gòej-(c) (ii) 1 1 1−2 2 2 2 1−2 v2 v2 m(h + 1 +gz +q )=m(h + 2 +gz +w ) 1 1 1−2 2 2 1−2 z z (c) A system receives 50 kJ of heat while expanding with volume change of 0.14 m3 against an atmosphere of 1.2 ×××× 105 N/m2. A mass of 90 kg in the surroundings is also lifted Øe›eâce 2-1' (™æes<ce mebheer[ve) Éeje Deheveer ØeejefcYekeâ DeJemLee ceW Dee through a distance of 5.5 m. peelee nw~ ™æes<ce Øe›eâce ceW T<cee mLeeveevlejCe MetvÙe neslee nw~ Skeâ efmemšce 1.2 ×××× 105 N/m2 kesâ JeeÙegceb[ue kesâ Øeefle Deevleefjkeâ Tpee& ceW heefjJele&ve = efmemšce Éeje efkeâÙee ieÙee keâeÙe& 0.14 m3 kesâ DeeÙeleve heefjJele&ve kesâ meeLe Øemeeefjle nesles (∆u) = 110 kJ meceÙe 50 kJ T<cee «enCe keâjlee nw~ Deemeheeme (heefjJesMe) Gòej-(c) (iii) ceW 90 kg keâe õJÙeceeve Yeer 5.5 m keâer otjer mes GlLeeefhele mechetCe& Deevleefjkeâ Tpee& ceW heefjJele&ve (∆u) = (∆u ) +(∆u ) neslee nw~ 1 1−2 2 2−1' ∆u = Øe›eâce 1-2 ceW Deevleefjkeâ heefjJele&ve = 28.25 kJ 1 (i) Calculate the change in energy of the system. ∆u = Øe›eâce 2-1' ceW Deevleefjkeâ Tpee& ceW heefjJele&ve = 110 kJ efmemšce keâer Tpee& ceW heefjJele&ve keâe heefjkeâueve keâerefpeS~ 2 ∆u = 28.25 + 110 (ii) The system is returned to its initial volume by an adiabatic process which requires 110 kJ of ∆u=138.25kJ work. Find the change in energy of the system. (d) (i) Define the second law of thermodynamics efmemšce ™æes<ce Øe›eâce Éeje Deheves ØeejbefYekeâ DeeÙeleve ceW using Calusius and Kelvin-Planck Dee peelee nw efpemekesâ efueS 110 kJ keâeÙe& keâer statements. T<ceeieeflekeâer kesâ efÉleerÙe efveÙece keâes keäueeefmeÙeme Deewj DeeJeMÙekeâlee nesleer nw~ efmemšce keâer Tpee& ceW heefjJele&ve kesâequJeve-hueebkeâ kesâ keâLeveeW Éeje heefjYeeef<ele keâerefpeS~ %eele keâerefpeS~ Gòej-(d) kewâefuJeve hueebkeâ keâLeve (Kelvin-planck statement) (iii) For the combined processes of (i) and (ii), (cid:1) ‘‘Fme Øekeâej keâer efkeâmeer Yeer Ùegefòeâ keâes yeveevee DemecYeJe nw pees calculate the change in energy of the system. (i) Deewj (ii) kesâ mebÙegkeäle ØekeâceeW kesâ efueS efmemšce keâer efkeâmeer hetjs Ûe›eâ ceW efheC[ mes T<cee ueskeâj efvekeâeÙe ceW keâesF& Tpee& ceW heefjJele&ve keâe heefjkeâueve keâerefpeS~ 15 heefjJele&ve efkeâS efyevee, Gmes hetjer lejn mes keâeÙe& ceW yeoue os~’’ 6 (cid:1) T<cee Fvpeve ceW keâeÙe&keâejer heoeLe& iece& efheC[ (œeesle) mes T<cee ueslee nw, Gmekeâe kegâÚ Yeeie Ùeeefv$ekeâ keâeÙe& ceW heefjJeefle&le keâjlee nw, Deewj Mes<e Yeeie keâes "C[s efheC[ (efmebkeâ) keâes os oslee nw~ Ghejesòeâ DeejsKe kesâ Devegmeej (cid:1) T<cee keâes keâeÙe& ceW heefjJeefle&le keâjves kesâ efueS œeesle Je efmebkeâ kesâ 1–2 = efmLej leehe hej T<cee efveie&ce (Input) leeheeW ceW Devlej nesvee DeeJeMÙekeâ nw~ Deye lekeâ keâesF& Yeer Ssmee 2–3 = ™æes<ce Øemeej Fvpeve veneR yeveeÙee pee mekeâlee nw pees œeesle mes T<cee ueskeâj Gmes 3–4 = efmLej leehe hej T<cee keâe efvemleejCe (Rejection) hetjer lejn mes Ùeeefv$ekeâ keâeÙe& ceW yeoue os~ keâgÚ T<cee efmebkeâ keâes 4–1 = ™æes<ce mebheer[ve osvee DeeJeMÙekeâ neslee nw~ keâeveexš Ûe›eâ keâer o#elee ØeÛeeueve kesâ leeheceeve heefjefOe (jWpe) hej efveYe&j (cid:1) kewâefuJeve-hueebkeâ kesâ keâLeveevegmeej Fme Øekeâej keâer efkeâmeer Yeer Ùegefòeâ keâes keâjlee nw~ yeveevee DemecYeJe nw pees efkeâmeer hetjs Ûe›eâ ceW Skeâ efheC[ mes T<cee peneB hej Q = Heat added 1 ueskeâj efvekeâeÙe ceW keâesF& heefjJele&ve efkeâS efyevee, Gmes hetjer lejn mes keâeÙe& ceW Q2 = Heat rejected yeoue os~ T Dele: η =1− 2 keäueeefmeÙeme keâLeve (Classius statement):– carnot T 1 (cid:1) T<cee keâe ØeJeen Skeâoce GÛÛe leehe mes efvecve leehe keâer Deesj mJele: η =f(T and T ) carnot 1 2 ner neslee nw~ Ùee peneB T = GÛÛe leeheceeve 1 (cid:1) T<cee keâe ØeJeen, efvecve leehe mes GÛÛe leehe keâer Deesj leye lekeâ veneR T = efvecve leeheceeve 2 nes mekeâlee peye lekeâ efvekeâeÙe keâes keâesF& Jee¢e keâeÙe& ve efoÙee peeÙe~ keâeveexš Ûe›eâ cebs oes meceleeheer Øe›eâce leLee oes DeeFmesvšêefhekeâ Øe›eâce nesles nw~ meceleeheerÙe efJemleej– Q = p v L (v /v ) =mRTL r 1–2 1 1 n 2 1 1 n DeeFmesvšêefhekeâ efJemleej– mR(T −T ) W = 1 2 2−3 γ−1 meceleeheerÙe mecheer[ve– Q = p v L (v /v ) =mRT L r 3–4 3 3 n 3 4 2 n DeeFmesvšêefhekeâ mebheer[ve– (cid:1) Skeâ jsøeâerpejsšj ceW keâeÙe&keâejer heoeLe& "C[s (efmebkeâ) mes T<cee ueslee mR(T −T) W = 2 1 nw, efkeâmeer yee¢e œeesle mes keâeÙe&keâejer heoeLe& hej keâeÙe& efkeâÙee peelee 4−1 γ−1 nw, leye Ùen iece& efheC[ (œeesle) keâes DeefOekeâ cee$ee ceW T<cee oslee ∴ efkeâÙee ngDee keâeÙe& · oer ieÙeer T<cee – Úes[Ì er ieÙeer T<cee nw~ DeLee&led efyevee efkeâmeer yee¢e Spesvmeer keâer meneÙelee kesâ T<cee efvecve =mRTL r−mRT L r 1 n 2 n leehe mes GÛÛe leehe keâer Deesj veneR yen mekeâleer~ keäuee@efmeÙeme kesâ =mRL r(T −T ) keâLeve keâes efvecve Øekeâej mes JÙeòeâ efkeâÙee pee mekeâlee nw~ n 1 2 (cid:1) Skeâ Ssmeer Ùegefòeâ yeveevee DemecYeJe nw pees Skeâ Ûe›eâ ceW efyevee keâesF& ∴ η= keâ=le keâeÙe& = mRLnr(T1−T2) yee¢e keâeÙe& keâer meneÙelee kesâ Skeâ "C[s efheC[ mes T<cee ueskeâj iece& oer ieÙeer T<cee mRTL r 1 n efheC[ keâes os os~ Ùen mhe° nw efkeâ T<cee meleled ™he mes "C[s T η=1− 2 efheC[ mes iece& efheC[ keâer Deesj veneR yen mekeâleer~ T 1 (ii) Describes the working of the Carnot cycle. (iii) What do you mean by the term "Entropy"? keâeveexš Ûe›eâ keâer keâeÙe&ØeCeeueer keâe JeCe&ve keâerefpeS~ ‘‘Svšê@heer’’ mes Deehe keäÙee mecePeles nQ 15 Gòej-(d) (ii) keâeveexš Ûe›eâ ceW T<cee mLeveevlejCe efmLej leehe hej neslee Svšê@heer (Entropy)– nw~ keâeveexš Fbpeve keâer o#elee meYeer FbpeveeW mes DeefOekeâ nesleer nw Fmekeâer (cid:1) Fmes efkeâmeer efvekeâeÙe keâer DeCegDeeW keâer DeefveÙeefcelee keâer e[f «eer kesâ ™he o#elee meesme& SJeb efmebkeâ kesâ leeheceeve hej efveYe&j neslee nw~ FmeceW 4 ceW heefjYeeef<ele efkeâÙee peelee nQ Ùeefo DeCegDeeW keâe efJekeâej Øeef›eâÙee nesleer nw~ (Disorder) yeÌ{sieer leye Svšêheer ceW heefjJele&ve Yeer yeÌ{siee~ keâeveexš Ûe›eâ (cannot cycle) : keâeveexš Ûe›eâ Skeâ ØeefleJeleea Ûe›eâ (cid:1) Svšêheer keâe mebyebOe T<cee kesâ uesve-osve mes nw (vee ekfeâ leehe mes) Ùeefo (reversible cycle) nw~ Fme Ûe›eâ ceW oes meceleeheerÙe leLee oes ™æes<ce efkeâmeer efvekeâeÙe keâes T<cee oer peeÙe lees Gmekeâer Svšêheer yeÌ{sieer~ Ùeefo Øe›eâce nesles nw~ efmemšce mes T<cee efvekeâeueer peeS lees Gmekeâer Svšêheer Iešsieer~ 7 (cid:1) T<cee ieeflekeâer kesâ efÉleerÙe efveÙece kesâ keäueeefmeÙeme keâLeve Svšêheer keâes Deešes Ûe›eâ (otto cycle):– heefjYeeef<ele keâjlee nw~ DeLee&led keäueeefmeÙeme kesâ Devegmeej- (cid:1)dQ (cid:1) ∫ ≤0 (Fme meceerkeâjCe keâes keäueeefmeÙeme Inequality Yeer T keânles nQ) (cid:1)dQ (i) ∫ =0 [ØeefleJelÙe& Ûe›eâ kesâ efueS] T (cid:1)dQ (ii) ∫ <0 [DeØeefleJelÙe& Ûe›eâ kesâ efueS] 1–2:– ØeefleJelÙe& ®æes<ce mebheer[ve T 2–3:– efmLej DeeÙeleve hej T<cee Ùeeis e (cid:1)dQ (iii) ∫ >0 [DemecYeJe] 3–4:– ØeeflJele&Ùe ®æes<ce Øemeej T 4–1:– efmLej DeeÙeleve hej T<cee evfekeâeme (cid:1) Svšê@heer efJemleejerÙe iegCeOece& (Extensive property) nw, Svšê@heer Dee@šes Ûe›eâ keâer o#elee :– keâe cee$ekeâ KJ/K (cid:1) efkeâmeer efvekeâeÙe keâer Svšê@heer efyevog heâueve (Point function) nesleer nw~ (cid:1) efJeueie efvekeâeÙe keâer Svšêheer Ùee lees ye{Ì sieer Ùee efveÙele jnsieer hejvleg Iešsieer veneR~ T.ds = du+pdv T.ds = dh–vdp du = Deevleefjkeâ Tpee& ceW heefjJele&ve dh= SvLewuheer ceW heefjJele&ve pdv leLee vdp · efkeâÙee ieÙee keâeÙe& Ûe›eâerÙe Øe›eâce kesâ efueS– efJeefYeVe Øe›eâceeW ceW Svšê@heer heefjJele&ve (Entrop changes in (cid:1) (cid:1) various process) : ∫dQ=∫dw (T<ceeieeflekeâer keâs ØeLece evfeÙece mes) Øe›eâce Svšêheer heefjJele&ve Q −Q =w (Process) (Entropy change) A R Q −Q (i) Ø(meI›esceâeocDec eeh Ùoelereifvcek eâ (i) S2−S1 =2.3mCvlogPP12  ηotto = AQQA R η =1− R process) otto Q A (ii) meceoeyeer V  Øe›eâce 2–3 kesâ efueS– Øe›eâce (ii) S −S =2.3mC log 2  (Isobaric 2 1 v  V1  dQ=Td<cwee +Ùeedsieu e fveÙece DeeÙeleve hej process) ∆v=0, pdv=0 (iii) mØee›ceeâlceee he er (iii)S −S =2.3m(C −C )logV2  dQ=du (Isothermal 2 1 P v  V1  du=mcvdt Q =mc (T −T ) process) V  A v 3 2 =2.3mRlog 2  Fmeer Øekeâej Øe›eâce t–1 kesâ efueS  V1  Q =mC (T −T) (iv) ®æes<ce Øe›eâce (iv) 0 R v 4 1 (T −T ) (Adiabatic η =1− 4 1 otto process) (T3−T1) (v) hee@ueeršêefhekeâ γ−n T  Øe›eâce 1–2 ceW, Ø(eP›eoâcley tropic (v)S2−S1 =2.3mCv n−1logT12  T1 = P1 γγ−1 =V2 γ−1=ρ1 γ−1 process) T2 P2   V1  ρ2  2. (a) With the help of P–V and T–s diagrams mebheer[ve Devegheele derive the thermal efficiency expression  V  for air-standard Otto cycle. rc = 1  P–V Deewj T–s DeejsKeeW keâer meneÙelee mes JeeÙeg V2  ceevekeâ Dee@šes Ûe›eâ kesâ T<ceerÙe o#elee JÙebpekeâ keâes T1 = 1 JÙeglheVe keâerefpeS~ 15 T2 (rc)γ−1 8 T =T.(r )γ−1 Øe›eâce (1-2) ceW 2 1 c Fmeer Øekeâej Øe›eâce 3–4 kesâ efueS Ûetbefkeâ TVγ–1 = efveÙeleebkeâ ∴ TVγ–1 = TVγ–1 γ−1 1 1 2 2 T V  4 = 3  V γ−1 T3 V4  T2 = T1 1  V =V , V =V V2  1 4 2 3 T = 300 (14)1.4–1 T 1 2 4 = =r(1−γ) T =862.12K T3 rc(γ−1) c 2 T =T r (γ−1) 3 4 c (ii) (T −T ) T = 2500oC = 2773 K η =1− 4 1 3 otto Øe›eâce DeeFmeesyesefjkeâ Øe›eâce nw~ (T −T ) 3 2 T V η =1− (T4−T1) 2= 2 otto r γ−1(T −T ) T3 V3 c 4 1 V T 1 3 = 3 ] η =1− V T otto r γ−1 2 2 c V 2773 2 = =3.216 (cid:1) Dee@šes Ûe›eâ keâer o#elee mebheer[ve Devegheele leLee γ yeÌ{eves mes yeÌ{leer V 862.12 3 nw~ V (cid:1) Dee@šes Ûe›eâ kesâ efueS mebheer[ve Devegheele 6–12 lekeâ jKee peelee nw, V3 = ρ (cutofratio) Fmemes DeefOekeâ nesves hej Fbpeve ceW vee@efkebâie (Knocking) keâe iegCe 2 P=3.216 yeÌ{ peelee nw~ (cid:1) Dee@šes Ûe›eâ keâe jesue cee@[ue hesšêsue Fbpeve neslee nw~ (iii) (cid:1) Dee@šes Ûe›eâ keâes efmLej DeeÙeleve Ûe›eâ Yeer keânles nQ, keäÙeeWefkeâ T<cee  ργ −1  keâe Ùeesie leLee efvekeâeme oesveeW efmLej DeeÙeleve hej neslee nw~ T<ceerÙe o#elee (ηth) = 1−rc1−γγ(ρ−1) (b) An air-standard Diesel cycle has a compression  3.2161.4−1  ratio of 14. The air conditions before = 1−14(1−1.4)  compression are 1 bar and 27oC. The 1.4(3.216−1) maximum temperature of the cycle is 2500oC. η =53.64% Determine the th Skeâ JeeÙeg ceevekeâ [erpeue Ûe›eâ keâe mebheer[ve Devegheele 14 (c) Derive the network output and thermal nw~ mebheer[ve mes henues JeeÙeg keâer oMeeSB 1 bar Deewj 27oC efficiency expressions for a simple Rankine nw~ Ûe›eâ keâe DeefOekeâlece leeheceeve 2500oC nw~ cycle with schematic and T – s diagrams. Ùeespeveeyeæ SJeb T–s DeejsKe yeveeles ngS meeOeejCe jweqvkeâve (i) temperature and pressure at salient points of Ûe›eâ kesâ efueS vesšJeke&â efveie&le Deewj T<ceerÙe o#elee JÙebpekeâ the cycle. Ûe›eâ kesâ mecegVele efyebogDeeW hej leeheceeve Deewj oeye %eele keâes JÙeglheVe keâerefpeS~ 15 keâerefpeS~ jWefkeâve Ûe›eâ (Rankine cycle):– (ii) network output per unit mass of air. heeJej hueevšme kesâ efueS jWefkeâve Ûe›eâ Deeo&Me Ûe›eâ neslee nw~ efkeâmeer vesšJeke&â efveie&le Øeefle Ùetefveš JeeÙeg õJÙeceeve %eele keâerefpeS~ Yeehe hueevš keâer hejheâejcesvme keâer leguevee kesâ efueS jsefkeâve Ûe›eâ Deeo&Me (iii) thermal efficiency. Ûe›eâ neslee nw~ jsefkeâve Ûe›eâ keâeveexš Jeshej Ûe›eâ keâe mebmeesefOele ™he nw~ T<ceerÙe o#elee %eele keâerefpeS~ 15 jsefkeâve Ûe›eâ kesâ Ieškeâ:– (i) (i) yeeÙeuej, (ii) šjyeeFve, (iii) kebâ[ms ej, (iv) heche P = 1 bar, T = 27oC = 3000K 1 1 T = 2773oK, r = 14 3 c 9