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Basic Concepts S K Mondal’s 1. Chapter 1 Basic Concepts Theory at a Glance (For GATE, IES & PSUs) Intensive and Extensive Properties Intensive property: Whose value is independent of the size or extent i.e. mass of the system. These are, e.g., pressure p and temperature T. Extensive property: Whose value depends on the size or extent i.e. mass of the system (upper case letters as the symbols). e.g., Volume, Mass (V, M). If mass is increased, the value of extensive property also increases. e.g., volume V, internal energy U, enthalpy H, entropy S, etc. Specific property: It is a special case of an intensive property. It is the value of an extensive property per unit mass of system. (Lower case letters as symbols) eg: specific volume, density (v, ρ). Thermodynamic System and Control Volume • In our study of thermodynamics, we will choose a small part of the universe to which we will apply the laws of thermodynamics. We call this subset a SYSTEM. • The thermodynamic system is analogous to the free body diagram to which we apply the laws of mechanics, (i.e. Newton’s Laws of Motion). • The system is a macroscopically identifiable collection of matter on which we focus our attention (e.g., the water kettle or the aircraft engine). System Definition • System: A quantity of matter in space which is analyzed during a problem. • Surroundings: Everything external to the system. • System Boundary: A separation present between system and surrounding. Classification of the system boundary:• Real solid boundary • Imaginary boundary 1 B Basic C Conce epts S K Mondal’s Cha apter 1 Th he system bou undary may be b further cla assified as:: • Contrrol Mass Systtem. • Contrrol Volume Sy ystem. Th he choice of booundary depends on the problem p beiing analyzed d. Ty ypes off Syste em Closed System S (Contro ( ol Mass System m) 1. It’s a system of fixed masss with fixed identity. 2. Thiss type of system s is ussually referred d to as “closed system”.. 3. Therre is no masss transfer across a the systtem boundarry. 4. Energy transferr may take place into or out of the system. Fig. A Co ontrol Masss System or Closed C Systtem 2 B Basic C Conce epts S K Mondal’s Cha apter 1 O Open Sy ystem (Contr rol Vollume System S ) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Its a system of fixed f volum me. Thiis type of system s is ussually refeerred to as "open system m” or a "c control volu ume" Mass transfer can take place acrooss a control volume. Eneergy transfer may also occur intoo or out of the system. A control volum me can be seen n as a fixeed region accross which mass and d energy tran nsfers are stu udied. Con ntrol Surfa ace – Its the bou undary of a control voolume Fig. A Co ontrol Volum me System acrooss which th he transfer off both or r Open System mass and energ gy takes placee. Thee mass of a co ontrol volum me (open systeem) may or may m not be fix xed. Wh hen the net in nflux of mass across the control surfa ace equals zeero then the mass of the systtem is fixed and a vice-verssa. Thee identity of mass in a control c volum me always ch hanges unlik ke the case for f a control mass system (cllosed system)). Most of the engiineering deviices, in generral, represent an open sysstem or contrrol volume. Ex xample: • Heat exchanger - Fluid enters an nd leaves thee system con ntinuously with w the transfer of heat across the sy ystem bound dary. • Pump - A continuous c fllow of fluid takes place th hrough the system s with a transfer off mechanical energy from m the surroun ndings to the system. 3 B Basic C Conce epts S K Mondal’s Cha apter 1 Issolated d Syste em 1. 2. 3. It is a sysstem of fixed d mass with h same identity and fixed energy. e No intera action of masss or energy takes place beetween the system and d the surround dings. In more informal words w an isoolated system iss like a closed shop amiidst a busy marrket. Fig. An n Isolated System S Q Quasi-Sttatic Prrocess Th he processees un nrestrained can be e restrained d or Wee need restra ained processses in practice. Aq quasi – staticc process is one o in which • The dev viation from m thermodyn namic equilibriu um is infinite esimal. • All statess of the syste em passes th hrough are equillibrium state es. • If we rem move the weig ghts slowly one o by Fig. A quasi q – static process one the pressure of the gass will displace the piston gradually. It is quasista atic. • On the other hand if we remove all a the weights at once th he piston will be kicked up p by the gas pressure. (This is un nrestrained expansion) but b we don’tt consider th hat the work k is done – because it i is not in a sustained manner. • In both cases c the systtems have un ndergone a ch hange of statte. • Another e.g., if a person climbs down d a ladder from roof to t ground, it is a quasista atic process. On the otther hand if he jumps theen it is not a quasistatic process. p 4 B Basic C Conce epts S K Mondal’s Cha apter 1 La aws of Thermodynam mics h Law deals with therrmal equilib brium and provides a means for measuring • The Zeroth temperaturees. L deals with w the consservation of energy and introduces the concept of internal • The First Law energy. modynamics provides witth the guideliines on the coonversion heeat energy of • The Second Law of therm matter into work. It also o introduces the t concept of o entropy. L of therm modynamics defines the absolute zerro of entropy y. The entrop py of a pure • The Third Law crystalline substance s at absolute zeroo temperaturre is zero. Summattion of 3 Laws s • Firstly, there isn’t a meaningful m teemperature of the sourcce from whiich we can get g the full conversion of o heat to wo ork. Only at infinite tem mperature onee can dream of getting th he full 1 kW work outputt. • Secondly, more m interestiingly, there isn’t i enough work availab ble to producce 0 K. In oth her words, 0 K is unattaiinable. This is i precisely th he Third law w. • Because, wee don’t know what 0 K loooks like, we haven’t got a starting pooint for the temperature t scale!! That is why all te emperature scales are at best b empirica al. Yo ou can’t get something for nothing g: To get work k output you must m give som me thermal energy. e Yo ou can’t get something for very litttle: To get some e work outputt there is a minimum m amount of therm mal energy th hat needs to be given. 5 B Basic C Conce epts S K Mondal’s Cha apter 1 Yo ou can’t get every thing g: However mu uch work you u are willing to give 0 K can’t c be reach hed. Vio olation of all a 3 laws: Try to get ev verything forr nothing. eroth Law L of Thermo T odynam mics Ze • If two systeems (say A an nd B) are in thermal equ uilibrium witth a third sy ystem (say C)) separately (that is A an nd C are in th hermal equillibrium; B an nd C are in th hermal equillibrium) then n they are in thermal equ uilibrium themselves (tha at is A and B will be in thermal equilib brium). • All tem mperature measuremen m nts are base ed on Zeroth h law of the ermodynam mics In nternatiional Te empera ature Sc cale To provide a standard s forr temperaturre measurem ment taking g into accoun nt both theooretical and pra actical consid derations, th he Internation nal Tempera ature Scale (IITS) was adoopted in 1927 7. This scale has been refin ned and exte ended in several revisioons, most reecently in 1990. The In nternational Temperature Scale S of 1990 (ITS-90) is defined in su uch a way th hat the temp perature mea asured on it con nforms with the thermod dynamic temp perature, thee unit of which is the kellvin, to withiin the limits of accuracy of measuremen nt obtainablle in 1990. The T ITS–90 is based on the assigneed values of tem mperature off a number of o reproducib ble fixed poin nts (Table). In nterpolation between thee fixed-point tem mperatures is i accomplisshed by form mulas that give g the rela ation between readings of o standard insstruments an nd values of the ITS. In the t range froom 0.65 to 5.0 K, ITS-90 is defined by b equations giv ving the temp perature as functions f of the t vapor prressures of pa articular helium isotopess. The range froom 3.0 to 24.5561 K is ba ased on meassurements ussing a helium m constant-voolume gas th hermometer. In the range frrom 13.8033 to 1234.93 K, K ITS-90 is defined by means m of certtain platinum m resistance theermometers. Above 1234 4.9 K the tem mperature iss defined usiing Planck’s equation for blackbody rad diation and measuremen m nts of the inteensity of visib ble-spectrum m radiation th he absolute temperature t 6 B Basic C Conce epts S K Mondal’s Cha apter 1 ale. The absoolute temperrature scale is also know wn as Kelvin n temperaturre scale. In defining d the sca Keelvin tempera ature scale also, the triple point of wa ater is taken as the stand dard referencce point. For aC Carnot engin ne operating between reseervoirs at tem mperature θ and θtp, θtp being the triiple point of wa ater arbitrariily assigned the t value of 273.16 K. Tim me Constan nts: The time constant iss the amountt of time req quired for a thermocouple t e to indicate 63.2% of step change c in te emperature of o a surround ding media. Some of thee factors influ uencing the meeasured timee constant are sheath walll thickness, degree of inssulation com mpaction, and d distance of jun nction from the t welded ca an on an ung grounded theermocouple. In I addition, the t velocity of o a gas past thee thermocoup ple probe gre eatly influencces the time constant c mea asurement. In general, tim me constants for f measurem ment of gas can c be estima ated to be ten times as loong as those forr measuremeent of liquid. The time con nstant also varies v inverseely proportional to the sq quare root of thee velocity of the t media. 7 B Basic C Conce epts S K Mondal’s Cha apter 1 W Work a path p fun nction Woork is one of the basic mo odes of energy y transfer. Th he work donee by a system m is a path fu unction, and nott a point fun nction. There efore, work is not a pr roperty of th he system, and it cann not be said tha at the work k is posses ssed by the system. It is an intera action acrosss the bounda ary. What is stoored in the system is ene ergy, but nott work. A decrease in energy of the system s appea ars as work don ne. Therefore, work is energy in transit and d it can be identified i o only when the t system un ndergoes a process. p Free Exp pansion n with Zero Z W Work Tra ansfer Fr ree Expansio on Let us con nsider an insu ulated containeer (Figure) wh hich is divided d into two com mpartments A and d B by a thin n diaphragm. Compartmen nt A contains a mass of ga as, while com mpartment B is i completely eva acuated. If thee diaphragm is punctured, the gas in A will expand into B until the t pressures in A and B beccome equal. This T is known n as free or unrestrained expansion. T The process of free ex xpansion is irr reversible.. Also work done is zeroo during freee expansion n. Free Expa ansion pd dV-worrk or Displacement Work W Let the gas in the t cylinder (Figure show wn in below) be a system having initia ally the pressure p1 and vollume V1. Th he system is in thermody ynamic equilibrium, the state of wh hich is described by the cooordinates p1 , V1. The pistton is the on nly boundary which moves due to gas pressure. Leet the piston moove out to a new final po osition 2, wh hich is also a thermodyn namic equilib brium state specified by preessure p2 an nd volume V2. At any inteermediate poiint in the tra avel of the piiston, let the pressure be pa and the volu ume V. This must m also bee an equilibrium state, siince macrosccopic propertties p and V 8 B Basic C Conce epts S K Mondal’s Cha apter 1 sig gnificant only. m states. Whe en the piston n moves an in nfinitesimal distance dl, and if ‘a' bee the area of forr equilibrium thee piston, the force F actin ng on the pistton F = p.a. and a the infinitesimal amoount of work done by the gass on the pistoon. dW W= F ⋅ dl d = pad dl = pdV V wh here dV = ad dl = infinitesimal displa acement volu ume. The diffferential sig gn in dW with the line dra awn at the toop of it will be explained later. l Wh hen the pistoon moves out from positioon 1 to positioon 2 with thee volume cha anging from V1 to V2, the am mount of work k W done by the t system will w be W1−2 = ∫ V2 V1 9 pd dV B Basic C Conce epts S K Mondal’s Cha apter 1 Th he magnitudee of the worrk done is given g by thee area underr the path 1--2, as shown n in Fig. Sin nce p is att all times a thermod dynamic cooordinate, all the states pa assed through h by the sysstem as the volume cha anges from V1 to V2 mu ust be equiliibrium statess, and the path p 1-2 mu ust be qua asi-static. The T piston moves inffinitely slow wly so that every state passed thrrough is an equilibrium e state. Th he integration n ∫ pdV can n be perform med only on a quasi-sta atic path. Fig. Quasi-Static pdV Work k Heat Tra ansfer-A A Path Functio on Heeat transfer is i a path fu unction, thatt is, the amoount of heat transferred when a systtem changes froom state 1 to state 2 depe ends on the intermediate i e states throu ugh which th he system pa asses, i.e. its patth. Thereforee dQ is an ine exact differen ntial, and wee write ∫ 2 1 dQ Q = Q1−2 or 1 Q2 ≠ Q2 − Q1 Th he displacemeent work is given g by W1−2 = ∫ 2 1 dW = ∫ 2 1 PR ROBLEM MS & SO OLUTION NS pdV ≠ W2 − W1 Ex xample 1 In a closed systtem, volume changes from m 1.5m3 to 4.5 m3 and hea at addition iss 2000 kJ. Ca alculate the cha ange in interrnal energy given g the presssure volumee relation as 10 ⎞ ⎛ W Where p is in n kPa and V is in m3. p = ⎜V 2 + V ⎟⎠ ⎝ So olution: 10 Basic Concepts S K Mondal’s Work done = V2 ∫ p.dV = V1 Chapter 1 V2 ⎛ ∫ ⎜⎝ V V1 2 + 10 ⎞ dV V ⎟⎠ ⎡1 V ⎤ = ⎢ V23 − V13 + 10ln 2 ⎥ V1 ⎦ ⎣3 4.5 ⎤ ⎡1 = ⎢ 4.53 − 1.53 + 10ln 1.5 ⎥⎦ ⎣3 ( ( ) ) ⎡1 ⎤ = ⎢ ( 91.125 − 3.375 ) + 10ln3⎥ 3 ⎣ ⎦ = [29.250 + 10.986] = 40.236 kJ First Law of Thermodynamics:Q = W + ΔU 2000 = 40.236 + ΔU ∴ ΔU = 2000 – 40.236 = 1959.764 kJ Example 2. A fluid is contained in a cylinder piston arrangement that has a paddle that imparts work to the fluid. The atmospheric pressure is 760 mm of Hg. The paddle makes 10,000 revolutions during which the piston moves out 0.8m. The fluid exerts a torque of 1.275 N-m one the paddle. What is net work transfer, if the diameter of the piston is 0.6m? Solution: Work done by the stirring device upon the system W1 = 2πTN = 2π × 1.275 × 10000 N-m = 80kJ This is negative work for the system. (Fig.) Work done by the system upon the surroundings. W2 = p.dV = p.(A × L) π (0.6)2 × 0.80 = 22.9kJ 4 This is positive work for the system. Hence the net work transfer for the system. W = W1 + W2 = - 80 + 22.9 = - 57.l kJ. = 101.325 × 11 Basic Concepts S K Mondal’s Chapter 1 ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20-Years GATE Questions GATE-1. List-I A. Heat to work B. Heat to lift weight C. Heat to strain energy D. Heat to electromagnetic energy 5. Thermal radiation 6. Bimetallic strips Codes: A B C D (a) 3 4 6 5 (c) 3 6 4 2 1. 2. 3. 4. (b) (d) List II [GATE-1998] Nozzle Endothermic chemical reaction Heat engine Hot air balloon/evaporation A 3 1 B 4 2 C 5 3 D 6 4 Open and Closed systems GATE-2. An isolated thermodynamic system executes a process, choose the correct statement(s) form the following [GATE-1999] (a) No heat is transferred (b) No work is done (c) No mass flows across the boundary of the system (d) No chemical reaction takes place within the system GATE-2a. Heat and work are (a) intensive properties (c) point functions [GATE-2011] (b) extensive properties (d) path functions Quasi-Static Process GATE-3. A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and 0.015 m3. It expands quasi-statically at constant temperature to a final volume of 0.030 m3. The work output (in kJ/kg) during this process will be: [GATE-2009] (a) 8.32 (b) 12.00 (c) 554.67 (d) 8320.00 Free Expansion with Zero Work Transfer GATE-4. A balloon containing an ideal gas is initially kept in an evacuated and insulated room. The balloon ruptures and the gas fills up the entire room. Which one of the following statements is TRUE at the end of above process? (a) The internal energy of the gas decreases from its initial value, but the enthalpy remains constant [GATE-2008] (b) The internal energy of the gas increases from its initial value, but the enthalpy remains constant (c) Both internal energy and enthalpy of the gas remain constant (d) Both internal energy and enthalpy of the gas increase 12 Basic Concepts S K Mondal’s GATE-5. Chapter 1 Air is compressed adiabatically in a steady flow process with negligible change in potential and kinetic energy. The Work done in the process is given by: [GATE-1996, IAS-2000] (a) –∫Pdv (b) +∫Pdv (c) –∫vdp (d) +∫vdp pdV-work or Displacement Work GATE-6. In a steady state steady flow process taking place in a device with a single inlet and a single outlet, the work done per unit mass flow rate is given by ω=− outlet ∫ vdp , where v is the specific volume and p is the pressure. The inlet expression for w given above: (a) Is valid only if the process is both reversible and adiabatic (b) Is valid only if the process is both reversible and isothermal (c) Is valid for any reversible process (d) Is incorrect; it must be ω = − [GATE-2008] outlet ∫ vdp inlet GATE-7. A gas expands in a frictionless piston-cylinder arrangement. The expansion process is very slow, and is resisted by an ambient pressure of 100 kPa. During the expansion process, the pressure of the system (gas) remains constant at 300 kPa. The change in volume of the gas is 0.01 m3. The maximum amount of work that could be utilized from the above process is: [GATE-2008] (a) 0kJ (b) 1kJ (c) 2kJ (d) 3kJ GATE-8. For reversible adiabatic compression in a steady flow process, the work transfer per unit mass is: [GATE-1996] (a) ∫ pdv (b) ∫ vdp (c) ∫ Tds (d ) ∫ sdT Previous 20-Years IES Questions IES-1. Which of the following are intensive properties? [IES-2005] 1. Kinetic Energy 2. Specific Enthalpy 3. Pressure 4. Entropy Select the correct answer using the code given below: (a) 1 and 3 (b) 2 and 3 (c) 1, 3 and 4 (d) 2 and 4 IES-2. Consider the following properties: [IES-2009] 1. Temperature 2. Viscosity 3. Specific entropy 4. Thermal conductivity Which of the above properties of a system is/are intensive? (a) 1 only (b) 2 and 3 only (c) 2, 3 and 4 only (d) 1, 2, 3 and 4 IES-2a. Consider the following: 1. Kinetic energy 2. Entropy 13 [IES-2007, 2010] Basic Concepts S K Mondal’s Chapter 1 3. Thermal conductivity 4. Pressure Which of these are intensive properties? (a) 1, 2 and 3 only (b) 2 and 4 only (c) 3 and 4 only (d) 1, 2, 3 and 4 IES-3. Which one of the following is the extensive property of a thermodynamic system? [IES-1999] (a) Volume (b) Pressure (c) Temperature (d) Density IES-4. Consider the following properties: [IES-2009] 1. Entropy 2. Viscosity 3. Temperature 4. Specific heat at constant volume Which of the above properties of a system is/are extensive? (a) 1 only (b) 1 and 2 only (c) 2, 3 and 4 (d) 1, 2 and 4 IES-4a Consider the following: 1. Temperature 2. Viscosity 3. Internal energy 4. Entropy Which of these are extensive properties? (a) 1, 2, 3 and 4 (b) 2 and 4 only (c) 2 and 3 only (d) 3 and 4 only. [IES-2010] Thermodynamic System and Control Volume IES-5. Assertion (A): A thermodynamic system may be considered as a quantity of working substance with which interactions of heat and work are studied. Reason (R): Energy in the form of work and heat are mutually convertible. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false [IES-2000] (d) A is false but R is true IES-5a A control volume is [IES-2010] (a) An isolated system (b) A closed system but heat and work can cross the boundary (c) A specific amount of mass in space (d) A fixed region in space where mass, heat and work can cross the boundary of that region Open and Closed systems IES-6. A closed thermodynamic system is one in which [IES-1999, 2010, 2011] (a) There is no energy or mass transfer across the boundary (b) There is no mass transfer, but energy transfer exists (c) There is no energy transfer, but mass transfer exists (d) Both energy and mass transfer take place across the boundary, but the mass transfer is controlled by valves IES-7 Isothermal compression of air in a Stirling engine is an example of 14 Basic Concepts S K Mondal’s (a) (b) (c) (d) Chapter 1 Open system Steady flow diabatic system Closed system with a movable boundary Closed system with fixed boundary [IES-2010] IES-8. Which of the following is/are reversible process(es)? [IES-2005] 1. Isentropic expansion 2. Slow heating of water from a hot source 3. Constant pressure heating of an ideal gas from a constant temperature source 4. Evaporation of a liquid at constant temperature Select the correct answer using the code given below: (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 4 IES-9. Assertion (A): In thermodynamic analysis, the concept of reversibility is that, a reversible process is the most efficient process. [IES-2001] Reason (R): The energy transfer as heat and work during the forward process as always identically equal to the energy transfer is heat and work during the reversal or the process. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-9a Which one of the following represents open thermodynamic system? (a) Manual ice cream freezer (b) Centrifugal pump (c) Pressure cooker (d) Bomb calorimeter [IES-2011] IES-10. Ice kept in a well insulated thermo flask is an example of which system? (a) Closed system (b) Isolated systems [IES-2009] (c) Open system (d) Non-flow adiabatic system IES-10a Hot coffee stored in a well insulated thermos flask is an example of (a) Isolated system (b) Closed system (c) Open system (d) Non-flow diabatic system [IES-2010] IES10b A thermodynamic system is considered to be an isolated one if (a) Mass transfer and entropy change are zero (b) Entropy change and energy transfer are zero (c) Energy transfer and mass transfer are zero (d) Mass transfer and volume change are zero [IES-2011] IES-10c. Match List I with List II and select the correct answer using the code given below the lists: [IES-2011] List I List II A. Interchange of matter is not possible in a 1. Open system B. Any processes in which the system returns to 2. System its original condition or state is called C. Interchange of matter is possible in a 15 Basic Concepts S K Mondal’s Chapter 1 D. The quantity of matter under consideration in thermodynamics is called Code: (a) (c) A 2 2 B 1 4 C 4 1 D 3 3 (b) (d) A 3 3 B 1 4 3. Closed system 4. Cycle C 4 1 D 2 2 Zeroth Law of Thermodynamics IES-11. Measurement of temperature is based on which law of thermodynamics? [IES-2009] (a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Third law of thermodynamics IES-12. Consider the following statements: [IES-2003] 1. Zeroth law of thermodynamics is related to temperature 2. Entropy is related to first law of thermodynamics 3. Internal energy of an ideal gas is a function of temperature and pressure 4. Van der Waals' equation is related to an ideal gas Which of the above statements is/are correct? (a) 1 only (b) 2, 3 and 4 (c) 1 and 3 (d) 2 and 4 IES-13. Zeroth Law of thermodynamics states that [IES-1996, 2010] (a) Two thermodynamic systems are always in thermal equilibrium with each other. (b) If two systems are in thermal equilibrium, then the third system will also be in thermal equilibrium with each other. (c) Two systems not in thermal equilibrium with a third system are also not in thermal equilibrium with each other. (d) When two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other. International Temperature Scale IES-14. Which one of the following correctly defines 1 K, as per the internationally accepted definition of temperature scale? [IES-2004] (a) 1/100th of the difference between normal boiling point and normal freezing point of water (b) 1/273.15th of the normal freezing point of water (c) 100 times the difference between the triple point of water and the normal freezing point of water (d) 1/273.15th of the triple point of water IES-15. In a new temperature scale say °ρ, the boiling and freezing points of water at one atmosphere are 100°ρ and 300°ρ respectively. Correlate this scale with the Centigrade scale. The reading of 0°ρ on the Centigrade scale is: [IES-2001] (a) 0°C (b) 50°C (c) 100°C (d) 150°C 16 Basic Concepts S K Mondal’s Chapter 1 IES-16. Assertion (a): If an alcohol and a mercury thermometer read exactly 0°C at the ice point and 100°C at the steam point and the distance between the two points is divided into 100 equal parts in both thermometers, the two thermometers will give exactly the same reading at 50°C. [IES-1995] Reason (R): Temperature scales are arbitrary. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-17. Match List-I (Type of Thermometer) with List-II (Thermometric select the correct answer using the code given below the List-I List-II 1. Pressure A. Mercury-in-glass B. Thermocouple 2. Electrical resistant C. Thermistor 3. Volume D. Constant volume gas 4. Induced electric voltage Codes: A B C D A B (a) 1 4 2 3 (b) 3 2 (c) 1 2 4 3 (d) 3 4 Property) and [IES 2007] C 4 2 D 1 1 IES-18. Pressure reaches a value of absolute zero (a) At a temperature of – 273 K (b) Under vacuum condition (c) At the earth's centre (d) When molecular momentum of system becomes zero [IES-2002] IES-19. The time constant of a thermocouple is the time taken to attain: (a) The final value to he measured [IES-1997, 2010] (b) 50% of the value of the initial temperature difference (c) 63.2% of the value of the initial temperature difference (d) 98.8% of the value of the initial temperature difference Work a Path Function IES-20. Assertion (A): Thermodynamic work is path-dependent except for an adiabatic process. [IES-2005] Reason(R): It is always possible to take a system from a given initial state to any final state by performing adiabatic work only. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-20a Work transfer between the system and the surroundings (a) Is a point function (b) Is always given by ∫ pdv (c) Is a function of pressure only [IES-2011] (d) Depends on the path followed by the system 17 Basic Concepts S K Mondal’s Chapter 1 Free Expansion with Zero Work Transfer IES-21. Match items in List-I (Process) with those in List-II (Characteristic) and select the correct answer using the codes given below the lists: List-I List-II [IES-2001] A. Throttling process 1. No work done B. Isentropic process 2. No change in entropy C. Free expansion 3. Constant internal energy D. Isothermal process 4. Constant enthalpy Codes: A B C D A B C D (a) 4 2 1 3 (b) 1 2 4 3 (c) 4 3 1 2 (d) 1 3 4 2 IES-22. The heat transfer, Q, the work done W and the change in internal energy U are all zero in the case of [IES-1996] (a) A rigid vessel containing steam at 150°C left in the atmosphere which is at 25°C. (b) 1 kg of gas contained in an insulated cylinder expanding as the piston moves slowly outwards. (c) A rigid vessel containing ammonia gas connected through a valve to an evacuated rigid vessel, the vessel, the valve and the connecting pipes being well insulated and the valve being opened and after a time, conditions through the two vessels becoming uniform. (d) 1 kg of air flowing adiabatically from the atmosphere into a previously evacuated bottle. pdV-work or Displacement Work IES-23. One kg of ice at 0°C is completely melted into water at 0°C at 1 bar pressure. The latent heat of fusion of water is 333 kJ/kg and the densities of water and ice at 0°C are 999.0 kg/m3 and 916.0 kg/m3, respectively. What are the approximate values of the work done and energy transferred as heat for the process, respectively? [IES-2007] (a) –9.4 J and 333.0 kJ (b) 9.4 J and 333.0 kJ (c) –333.0 kJ and –9.4 J (d) None of the above IES-24. Which one of the following is the correct sequence of the three processes A, B and C in the increasing order of the amount of work done by a gas following idealgas expansions by these processes? 18 Basic Concepts S K Mondal’s IES-25. Chapter 1 (a) A – B – C (b) B – A – C An ideal gas undergoes an isothermal expansion from state R to state S in a turbine as shown in the diagram given below: [IES-2006] (d) C – A – B (c) A – C – B The area of shaded region is 1000 Nm. What is the amount is turbine work done during the process? (a) 14,000 Nm (b) 12,000 Nm (c) 11,000 Nm (d) 10,000 Nm IES-26. [IES-2004] Assertion (A): The area 'under' curve on pv plane, ∫ pdv represents the work of reversible non-flow process. [IES-1992] Reason (R): The area 'under' the curve T–s plane ∫ Tds represents heat of any reversible process. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-27. If ∫ pdv ∫ and − vdp for a thermodynamic system of an Ideal gas on valuation give same quantity (positive/negative) during a process, then the process undergone by the system is: [IES-2003] (a) Isomeric IES-28. (b) Isentropic (c) Isobaric (d) Isothermal Which one of the following expresses the reversible work done by the system (steady flow) between states 1 and 2? [IES-2008] 2 (a) ∫ pdv 1 2 2 (b) − ∫ vdp (c) − ∫ pdv 1 1 2 (d) ∫ vdp 1 Heat Transfer-A Path Function IES-29. Assertion (A): The change in heat and work cannot be expressed as difference between the end states. [IES-1999] Reason (R): Heat and work are both exact differentials. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true 19 Basic Concepts S K Mondal’s Chapter 1 Previous 20-Years IAS Questions Thermodynamic System and Control Volume IAS-1. The following are examples of some intensive and extensive properties: 1. Pressure 2. Temperature [IAS-1995] 3. Volume 4. Velocity 5. Electric charge 6. Magnetisation 7. Viscosity 8. Potential energy Which one of the following sets gives the correct combination of intensive and extensive properties? Intensive Extensive (a) 1, 2, 3, 4 5, 6, 7, 8 (b) 1, 3, 5, 7 2, 4, 6, 8 (c) 1, 2, 4, 7 3, 5, 6, 8 (d) 2, 3, 6, 8 1, 4, 5, 7 Zeroth Law of Thermodynamics IAS-2. Match List-I with List-II and select the correct answer using the codes given below the lists: [IAS-2004] List-I List-II A. Reversible cycle 1. Measurement of temperature B. Mechanical work 2. Clapeyron equation C. Zeroth Law 3. Clausius Theorem D. Heat 4. High grade energy 5. 3rd law of thermodynamics 6. Inexact differential Codes: A B C D A B C D (a) 3 4 1 6 (b) 2 6 1 3 (c) 3 1 5 6 (d) 1 4 5 2 IAS-3. Match List-I with List-II and select the correct answer: [IAS-2000] List-I List-II A. The entropy of a pure crystalline 1. First law of thermodynamics substance is zero at absolute zero temperature B. Spontaneous processes occur 2. Second law of thermodynamics in a certain direction C. If two bodies are in thermal 3. Third law of thermodynamics equilibrium with a third body, then they are also in thermal equilibrium with each other D. The law of conservation of energy 4. Zeroth law of thermodynamics. Codes: A B C D A B C D (a) 2 3 4 1 (b) 3 2 1 4 20 Basic Concepts S K Mondal’s (c) Chapter 1 3 2 4 1 (d) 2 3 1 4 International Temperature Scale IAS-4. A new temperature scale in degrees N is to be defined. The boiling and freezing on this scale are 400°N and 100°N respectively. What will be the reading on new scale corresponding to 60°C? [IAS-1995] (c) 220°N (d) 280°N (a) 120°N (b) 180°N Free Expansion with Zero Work Transfer IAS-5. In free expansion of a gas between two equilibrium states, the work transfer involved [IAS-2001] (a) Can be calculated by joining the two states on p-v coordinates by any path and estimating the area below (b) Can be calculated by joining the two states by a quasi-static path and then finding the area below (c) Is zero (d) Is equal to heat generated by friction during expansion. IAS-6. Work done in a free expansion process is: (a) Positive (b) Negative (c) Zero IAS-7. In the temperature-entropy diagram of a vapour shown in the given figure, the thermodynamic process shown by the dotted line AB represents (a) Hyperbolic expansion (b) Free expansion (c) Constant volume expansion (d) Polytropic expansion [IAS-2002] (d) Maximum [IAS-1995] IAS-8. If ∫ pdv ∫ and − vdp for a thermodynamic system of an Ideal gas on valuation give same quantity (positive/negative) during a process, then the process undergone by the system is: [IAS-1997, IES-2003] (a) Isomeric (b) Isentropic (c) Isobaric (d) Isothermal IAS-9. For the expression ∫ pdv to represent the work, which of the following conditions should apply? [IAS-2002] (a) The system is closed one and process takes place in non-flow system (b) The process is non-quasi static (c) The boundary of the system should not move in order that work may be transferred (d) If the system is open one, it should be non-reversible 21 Basic Concepts S K Mondal’s IAS-10. IAS-11. Chapter 1 Air is compressed adiabatically in a steady flow process with negligible change in potential and kinetic energy. The Work done in the process is given by: [IAS-2000, GATE-1996] (a) –∫pdv (b) +∫pdv (c) –∫vdp (d) +∫vdp Match List-I with List-II and select the correct answer using the codes given below the lists: [IAS-2004] List-I List-II A. Bottle filling of gas 1. Absolute Zero Temperature B. Nernst simon Statement 2. Variable flow C. Joule Thomson Effect 3. Quasi-Static Path D. ∫pdv 4. Isentropic Process 5. Dissipative Effect 6. Low grade energy 7. Process and temperature during phase change. Codes: A B C D A B C D (a) 6 5 4 3 (b) 2 1 4 3 (c) 2 5 7 4 (d) 6 1 7 4 pdV-work or Displacement Work IAS-13. Thermodynamic work is the product of (a) Two intensive properties (b) Two extensive properties (c) An intensive property and change in an extensive property (d) An extensive property and change in an intensive property [IAS-1998] Heat Transfer-A Path Function IAS-14. Match List-I (Parameter) with List-II using the codes given below the lists: List-I A. Volume B. Density C. Pressure D. Work Codes: A B C D (a) 3 2 4 1 (c) 2 3 4 1 22 (Property) and select the correct answer List-II 1. Path function 2. Intensive property 3. Extensive property 4. Point function A B C (b) 3 2 1 (d) 2 3 1 [IAS-1999] D 4 4 Basic Concepts S K Mondal’s Chapter 1 Answers with Explanation (Objective) Previous 20-Years GATE Answers GATE-1. Ans. (a) GATE-2. Ans. (a, b, c) For an isolated system no mass and energy transfer through the system. dQ = 0, dW = 0, ∴ dE = 0 or E = Constant GATE-2a. Ans. (d) ⎛ V2 ⎞ ⎟ ⎝ V1 ⎠ ⎛V ⎞ = P1V1 ln ⎜ 2 ⎟ ⎝ V1 ⎠ GATE-3. Ans. (a) Iso-thermal work done (W) = RT1 ln ⎜ ⎛ 0.030 ⎞ = 800 × 0.015 × ln ⎜ ⎟ ⎝ 0.015 ⎠ = 8.32 kJ/kg GATE-4. Ans. (c) It is free expansion. Since vacuum does not offer any resistance, there is no work transfer involved in free expansion. 2 Here, ∫ δω = 0 and Q1-2=0 therefore Q1-2 = ΔU + W1-2 so, ΔU = 0 1 GATE-5. Ans. (c) For closed system W = + ∫ pdv , for steady flow W = − ∫ vdp GATE-6. (c) GATE-7. Ans. (b) W = Resistance pressure. Δ V = 1 × Δ V = 100 × 0.1 kJ = 1kJ GATE-8. Ans. (b) W = − ∫ vdp Previous 20-Years IES Answers IES-1. Ans. (b) IES-2. Ans. (d) Intensive property: Whose value is independent of the size or extent i.e. mass of the system. Specific property: It is a special case of an intensive property. It is the value of an extensive property per unit mass of system (Lower case letters as symbols) e.g., specific volume, density (v, ρ). IES-2a. Ans. (c) Kinetic energy 1 mv 2 depends on mass, Entropy kJ/k depends on mass so 2 Entropy is extensive property but specific entropy kJ/kg K is an intensive property. 23 Basic Concepts S K Mondal’s Chapter 1 IES-3. Ans. (a) Extensive property is dependent on mass of system. Thus volume is extensive property. IES-4. Ans. (a) Extensive property: Whose value depends on the size or extent i.e. mass of the system (upper case letters as the symbols) e.g., Volume, Mass (V, M). If mass is increased, the value of extensive property also increases. IES-4a Ans. (d) The properties like temperature, viscosity which are Independent of the MASS of the system are called Intensive property IES-5. Ans. (d) • But remember 100% heat can’t be convertible to work but 100% work can be converted to heat. It depends on second law of thermodynamics. • A thermodynamic system is defined as a definite quantity of matter or a region in space upon which attention is focused in the analysis of a problem. • The system is a macroscopically identifiable collection of matter on which we focus our attention IES-5a Ans. (d) IES-6. Ans. (b) In closed thermodynamic system, there is no mass transfer but energy transfer exists. IES-7. Ans. (c) IES-8. Ans. (d) Isentropic means reversible adiabatic. Heat transfer in any finite temp difference is irreversible. IES-9. Ans. (a) The energy transfer as heat and work during the forward process as always identically equal to the energy transfer is heat and work during the reversal or the process is the correct reason for maximum efficiency because it is conservative system. IES-9a. Ans. (b) IES-10. Ans. (b) Isolated System - in which there is no interaction between system and the surroundings. It is of fixed mass and energy, and hence there is no mass and energy transfer across the system boundary. IES-10a Ans. (a) IES-10b. Ans. (c) IES-10c. Ans. (d) IES-11. Ans. (a) All temperature measurements are based on Zeroth law of thermodynamics. IES-12. Ans. (a) Entropy - related to second law of thermodynamics. Internal Energy (u) = f (T) only (for an ideal gas) Van der Wall's equation related to => real gas. IES-13. Ans. (d) IES-14. Ans. (d) IES-15.Ans. (d) C −0 0 − 300 = ⇒ C = 150°C 100 − 300 100 − 0 24 Basic Concepts S K Mondal’s Chapter 1 IES-16. Ans. (b) Both A and R are correct but R is not correct explanation for A. Temperature is independent of thermometric property of fluid. IES-17. Ans. (d) IES-18. Ans. (d) But it will occur at absolute zero temperature. IES-19. Ans. (c) Time Constants: The time constant is the amount of time required for a thermocouple to indicated 63.2% of step change in temperature of a surrounding media. Some of the factors influencing the measured time constant are sheath wall thickness, degree of insulation compaction, and distance of junction from the welded cap on an ungrounded thermocouple. In addition, the velocity of a gas past the thermocouple probe greatly influences the time constant measurement. In general, time constants for measurement of gas can be estimated to be ten times as long as those for measurement of liquid. The time constant also varies inversely proportional to the square root of the velocity of the media. IES-20. Ans. (c) IES-20a Ans. (d) IES-21. Ans. (a) IES-22. Ans. (c) In example of (c), it is a case of free expansion heat transfer, work done, and changes in internal energy are all zero. ⎛m ⎝ ρ2 IES-23. Ans. (a) Work done (W) = P Δ V = 100 × (V2 – V1) = 100 × ⎜ − m⎞ ⎟ ρ1 ⎠ 1 ⎞ ⎛ 1 = 100 kPa × ⎜ − ⎟ = –9.1 J 999 916 ⎝ ⎠ IES-24. Ans. (d) WA = ∫ pdV = 4 × (2 − 1) = 4 kJ 1 WB = ∫ pdV = × 3 × (7 − 4) = 4.5 kJ 2 WC = ∫ pdV = 1× (12 − 9) = 3kJ IES-25. Ans. (c) Turbine work = area under curve R–S = ∫ pdv = 1 bar × ( 0.2 − 0.1) m3 + 1000 Nm = 105 × ( 0.2 − 0.1) Nm + 1000Nm = 11000 Nm IES-26. Ans. (b) IES-27. Ans. (d) Isothermal work is minimum of any process. pv = mRT pdv + vdp = 0[ ∵ T is constant] ∫ pdv = − ∫ vdp 2 ∫ IES-28. Ans. (b) For steady flow process, reversible work given by − vdp . 1 IES-29. Ans. (c) A is true because change in heat and work are path functions and thus can't be expressed simply as difference between the end states. R is false because both work and heat are inexact differentials. 25 Basic Concepts S K Mondal’s Chapter 1 Previous 20-Years IAS Answers IAS-1. Ans. (c) Intensive properties, i.e. independent of mass are pressure, temperature, velocity and viscosity. Extensive properties, i.e. dependent on mass of system are volume, electric charge, magnetisation, and potential energy. Thus correct choice is (c). IAS-2. Ans. (a) IAS-3. Ans. (c) IAS-4. Ans. (d) The boiling and freezing points on new scale are 400° N and 100°N i.e. range is 300°N corresponding to 100°C. Thus conversion equation is °N = 100 + 3 × °C = 100+ 3 × 60 = 100 + 180 = 280 °N IAS-5. Ans. (c) IAS-6. Ans. (c) Since vacuum does not offer any resistance, there is no work transfer involved in free expansion. IAS-7. Ans. (b) IAS-8. Ans. (d) Isothermal work is minimum of any process. IAS-9. Ans. (a) IAS-10. Ans. (c) For closed system W = + ∫ pdv , for steady flow W = − ∫ vdp IAS-12. Ans. (b) Start with D. ∫PdV only valid for quasi-static path so choice (c) & (d) out. Automatically C-4 then eye on A and B. Bottle filling of gas is variable flow so A-2. IAS-13. Ans. (c) W = ∫ pdv where pressure (p) is an intensive property and volume (v) is an extensive property IAS-14. Ans. (a) Pressure is intensive property but such option is not there. 26 Fir rst Law w of Therm T odyna amics S K Mondal’s 2. Cha apter 2 Firstt Law w of T herm modyn namic cs Theo ory at a Glance (Fo or GAT TE, IES S & PS SUs) First Law w of Th hermody ynamic cs Sttatement: • When a closed syste em executes a complete cy ycle the sum of heat interacttions is equa al to the sum of work interacttions. Mathem matically • (Σ Q)cyclle = (Σ Σ W)cycle The sum mmations be eing over the entire cycle. Allternate sttatement:: Wh hen a closed system undergoes a cyclle the cyclic integral of heat h is equal to the cyclicc integral of woork. Ma athematicallly ∫ δQ = ∫ δW In other word ds for a two process cycle Q A1 A − 2 + QB2 −1 = WA1− 2 + WB 2 −1 In nternal Energy y – A Prroperty y of Sys stem W Which can be e written ass 28 Fir rst Law w of Therm T odyna amics S K Mondal’s • • • Cha apter 2 Since A and B are arbitrarily a ch hosen, the con nclusion is, as a far as a prrocess is conccerned (A or B) the difference δQ – δW rema ains a consta ant as long as a the initial and the fina al states are the sam me. The diffe erence depends only on th he end points of the proccess. Note that Q and W themseelves depend on the path followed. f Butt their differeence does nott. This im mplies that th he difference between thee heat and woork interactioons during a process is a propertty of the system. This prroperty is called the energ gy of the systtem. It is dessignated as E and is equa al to some of all the energies e at a given state. Wee enunciate the t FIRST LA AW for a proccess as δQ δ – δW W = dE • • • • • An isollated system m which do oes not inter ract with th he surround dings Q = 0 and W = 0. Thereffore, E rema ains constan nt for such a system. Let us reconsider r th he cycle 1-2 along a path A and 2-1 alon ng path B as shown s in fig. Work done d during the path A = Area A under 1-A-2-3-4 1 Work done d during the path B = Area A under 1-B-2-3-4 1 Since these two are eas are not equal, e the neet work interraction is tha at shown by the shaded area. 29 Fir rst Law w of Therm T odyna amics S K Mondal’s • • • • • • • • • • • Cha apter 2 The nett area is 1A2B1. Therefoore some worrk is derived by the cycle. First la aw compels that this iss possible only o when there t is also o heat interaction betweeen the systtem and the surrounding gs. In oth her words, if i you have e to get wo ork out, yo ou must give e heat in. Thus, the t first law can be consttrued to be a statement of o conservatioon of energy - in a broad sense. In the example e show wn the area under u curve A < that und der B The cyccle shown ha as negative work outputt or it will receive work from the su urroundings. Obviously, the net heat h interacttion is also neegative. Thiss implies that this cycle will w heat the nment. (as pe er the sign coonvention). environ For a process we can n have Q = 0 or W = 0 We can n extract worrk without su upplying heat (during a process) bu ut sacrificing g the energy of the system. s We can n add heat to the system without doin ng work (in process) p wh hich will go too increasing the eneergy of the sy ystem. Energy y of a system m is an exte ensive prop perty he internal en nergy depend ds only upon the initial an nd final statees of the systtem. Internall energy of a Th sub bstance does not include any energy that t it may possess p as a result r of its macroscopic m c position or moovement. Tha at so why in SFEE S (Stead dy flow energ gy equation) C2/2 and gz iss there. Reecognize that h and similarly = u + pv p from whicch u2 + p2 v2 = h2 u1 + p1v1 = h1 Q Q–W = [(h2 + C22/2 + gZ2) - (h1+C C12/2 + gZ1)] = [(h2 - h1) + (C22/2 - C12/2) + g(Z2 - Z1)] or [W Where C = Vellocity (m/s), h = Specific enthalpy e (J/k kg), z = elevattion (m) Bu ut Reme ember: Miicroscopic vieew of a gas is a collection of particles in random mootion. Energy y of a particle consists of tra anslational energy, rottational ene ergy, vibrattional energ gy and speciific electron nic energy. Alll these energ gies summed d over all thee particles off the gas, form m the specifiic internal en nergy, e , of thee gas. 30 Fir rst Law w of Therm T odyna amics S K Mondal’s Cha apter 2 Perpetua al Motio on Mac chine off the First Kind-PMM M1 Th he first law sttates the gen neral principlle of the consservation of energy. e Energ gy is neither created nor desstroyed, but only gets tra ansformed froom one form to t another. here can be no n machine which would d continuoussly supply mechanical m woork without some other Th forrm of energy disappearing g simultaneoously (Fig. sh hown in below w). Such a fictitious mach hine is called ap perpetual mootion machine e of the first kind, or in brrief, PMM1. A PMM1 is thus t impossible. he converse of the above statementt is also tru ue, i.e. theree can be no machine which w would Th con ntinuously coonsume work k without som me other form m of energy appearing a sim multaneously y (Fig.). A PMM1 P The e Converse of o PMM1 Enthalpy y Th he enthalpy of a substance e H is defined d as H = U + PV P It iis an extenssive propertty of a system m and its uniit is kJ. h = u + pv p Specific Enthallpy m and its unitt is kJ/kg. It iis an intensiive propertty of a system Intternal energy y change is equal to the heat transfeerred in a coonstant volum me process involving i no woork other tha an pdv work. It is possiblle to derive an a expression n for the hea at transfer in n a constant preessure process involving no work oth her than pdv v work. In su uch a processs in a closed d stationary sysstem of unit mass m of a pure substancee. d Q = du + pdv At constan nt pressure pdv = d(pv) Therefoore ( dQ ) = du + d ( pv ) P or ( dQ ) or P = d(u+pv) ( dQ ) P = dh W Where H = U + PV is the enthalpy, a property of system. s Specific enth halpy h = H/m m, kJ/kg and also h = u + pdv Where h = sp pecific enthalpy, kJ/kg 31 First Law of Thermodynamics S K Mondal’s Chapter 2 u = specific internal energy, kJ/kg dv = change in specific volume, m3/kg. Specific heat at constant volume The specific heat of a substance at constant volume Cv is defined as the rate of change of specific internal energy with respect to temperature when the volume is held constant, i.e., ⎛ ∂u ⎞ Cv = ⎜ ⎟ ⎝ ∂T ⎠v For a constant volume process ( Δu )v = T2 ∫ C .dT v T1 The first law may be written for a closed stationary system composed of a unit mass of a pure substance. Q = Δu + W or d Q = du + d W For a process in the absence of work other than pdv work d W = pdv Therefore d Q = du + pdv Therefore, when the volume is held constant (Q )v = ( Δu )v T2 (Q )v = ∫ Cv .dT T1 ⋅ Since u, T and v are properties, Cv is a property of the system. The product m Cv is called the heat capacity at constant volume (J/K). Specific heat at constant pressure The specific heat at constant pressure Cp is defined as the rate of change of specific enthalpy with respect to temperature when the pressure is held constant. ⎛ ∂h ⎞ CP = ⎜ ⎟ ⎝ ∂T ⎠ P For a constant pressure process ( Δh )P = T2 ∫C P .dT T1 The first law for a closed stationary system of unit mass dQ = du + pdv Again, h = u + pv Therefore dh = du + pdv + vdp = d Q + vdp Therefore dQ = dh – vdp Therefore ( dQ )P = dh 32 First Law of Thermodynamics S K Mondal’s or (Q ) p Chapter 2 = ( Δh) p ∴ Form abow equations ( Q )P Cp = T2 ∫C P .dT T1 is a property of the system, just like Cv. The heat capacity at constant pressure is equal to m C p (J/K). Application of First Law to Steady Flow Process S.F.E.E S.F.E.E. per unit mass basis (i) C 12 C 22 dQ dW + gz1 + = h2 + + gz2 + h1 + 2 dm 2 dm [h, W, Q should be in J/Kg and C in m/s and g in m/s2] (ii) C12 dQ C22 dW gZ1 gZ 2 h1 + + + = h2 + + + 2000 1000 dm 2000 1000 dm [h, W, Q should be in KJ/Kg and C in m/s and g in m/s2] S.F.E.E. per unit time basis ⎛ ⎞ dQ C12 w1 ⎜ h1 + + z1 g ⎟ + 2 ⎝ ⎠ dτ ⎛ ⎞ dWx C22 = w2 ⎜ h2 + + z2 g ⎟ + dτ 2 ⎝ ⎠ Where, w = mass flow rate (kg/s) Steady Flow Process Involving Two Fluid Streams at the Inlet and Exit of the Control Volume 33 Fir rst Law w of Therm T odyna amics S K Mondal’s Cha apter 2 Ma ass balance w A 1C v1 1 + 1 A 2C v2 + w 2 2 = = w A 3C v3 3 3 + w + 4 A 4C v4 4 here v = speccific volume (m ( 3/kg) Wh En nergy balan nce ⎛ ⎞ ⎛ ⎞ dQ C12 C22 w1 ⎜ h1 + + Z1 g ⎟ + w2 ⎜ h2 + + Z2 g ⎟ + 2 2 ⎝ ⎠ ⎝ ⎠ dτ ⎛ ⎞ ⎛ ⎞ dWx C2 C2 = w3 ⎜ h3 + 3 + Z3 g ⎟ + w4 ⎜ h4 + 4 + Z4 g ⎟ + 2 2 dτ ⎝ ⎠ ⎠ ⎝ ome examplle of steady y flow proce esses:So Th he following examples e illu ustrate the applications of o the steady flow energy equation in some of the eng gineering sysstems. No ozzle and Diiffuser: An nozzle is a device d which increases th he velocity or o K.E. of a fluid at thee expense of its pressure droop, whereas a diffuser in ncreases the pressure of a fluid at thee expense of its K.E. Figu ure show in bellow a nozzle which is insu ulated. The steady s flow en nergy equation of the con ntrol surface gives dWx C2 C2 dQ = h2 + 2 + Z 2 g + h1 + 1 + Z1 g + dm dm 2 2 F Fig. 34 First Law of Thermodynamics S K Mondal’s Chapter 2 dWx dQ = 0; = 0, and the change in potential energy is zero. The equation reduces to dm dm C2 C2 h1 + 1 = h2 + 2 (a) 2 2 The continuity equation gives AC AC (b) w= 1 1 = 2 2 v1 v2 When the inlet velocity or the ‘velocity of approach’ V1 is small compared to the exit velocity V2, Equation (a) becomes Here C22 h1 = h2 + 2 or C2 = 2(h1 − h2 )m / s where (h1 – h2) is in J/kg. Equations (a) and (b) hold good for a diffuser as well. Throttling Device: When a fluid flows through a constricted passage, like a partially opened value, an orifice, or a porous plug, there is an appreciable drop in pressure, and the flow is said to be throttled. Figure shown in below, the process of throttling by a prettily opened value on a fluid flowing in an insulated pipe. In the steady-flow energy equationdWx dQ = 0, =0 dm dm And the changes in P. E. are very small and ignored. Thus, the S.F.E.E. reduces to C2 C2 h1 + 1 = h2 + 2 2 2 (Fig.- Flow Through a Valve) Often the pipe velocities in throttling are so low that the K. E. terms are also negligible. So h1 = h2 or the enthalpy of the fluid before throttling is equal to the enthalpy of the fluid after throttling. Turbine and Compressor: Turbines and engines give positive power output, whereas compressors and pumps require power input. For a turbine (Fig. below) which is well insulated, the flow velocities are often small, and the K.E. terms can be neglected. The S.F.E.E. then becomes 35 First Law of Thermodynamics S K Mondal’s Chapter 2 (Fig.-. Flow through a Turbine) h1 = h2 + dWx dm Wx = h1 − h2 m The enthalpy of the fluid increase by the amount of work input. or Heat Exchanger: A heat exchanger is a device in which heat is transferred from one fluid to another, Figure shown in below a steam condenser where steam condense outside the tubes and cooling water flows through the tubes. The S.F.E.E for the C.S. gives wc h1 + ws h2 = w c h3 + ws h4 or , ws ( h2 − h4 ) = w c (h3 − h1 ) Here the K.E. and P.E. terms are considered small, there is no external work done, and energy exchange in the form of heat is confined only between the two fluids, i.e. there is no external heat interaction or heat loss. Fig. Figure (shows in below) a steam desuperheater where the temperature of the superheated steam is reduced by spraying water. If w1, w2, and w3 are the mass flow rates of the injected water, of the steam entering, and of the steam leaving, respectively, and h1, h2, and h3 are the corresponding enthalpies, and if K.E. and P.E. terms are neglected as before, the S.F.E.E. becomes w1h1 + w2 h2 = w3 h3 and the mass balance gives w1 + w2 = w3 36 Fir rst Law w of Therm T odyna amics S K Mondal’s Cha apter 2 Th he above law w is also callled as stead dy flow ene ergy equatiion. This can n be applied d to various pra actical situattions as work k developing system and work absorp ption system. Let the ma ass flow rate un nity. (1)) Work deve eloping systtems (a) Water turbines In this casse Q = 0 and ΔU = 0 and equation e becomes 2 C p1 v1 +z1g + 1 = z2g + p2 v2 + W 2 (b) Steam or gas turbin nes In this casse generally ΔZ can be assumed to be zero and thee equation becomes ⎛ C 2 − C22 ⎞ W = ( h1 – h2 ) + ⎜ 1 ⎟ + ΔQ 2 ⎝ ⎠ orbing syste ems (2)) Work abso (a) Centrifuggal water pum mp The system m is shown in the Figure below b Fig. he energy equ uation now becomes, b In this sysstem Q = 0 and ΔU = 0; th C22 p1 v1 +z1g + W = z2g + p2 v2 + 2 al compress sor – In this system s Δz = 0 and the equ uation becom mes, (b) Centrifuga 2 2 C1 C + h1 + W – Q = 2 + h2 2 2 (c) Blowers – In this case we have Δ z = 0, p1 v1 = p2 v 2 and Q = 0; now thee energy simp plifies to 37 Fir rst Law w of Therm T odyna amics S K Mondal’s Cha apter 2 C2 u1 +W = u2 + 2 ass C2 C1 2 f the tem mperature rise is very small and heat loss is negleected (i.e.) Δh h = 0, q = 0 (d)) Fans – In fans and hence th he energy equ uation for fan ns becomes, C2 W= 2 2 essor – In a reciprocating g compressorr ΔKE and ΔP PE are neglig gibly energy (e) Reciprocatting compre equation app plied to a reciprocating coompressor is h1 – Q = h 2 – W or W = Q + ( h2 – h1 ) g and absorb bing system ms (3)) Non-work developing (a) Steam boiiler – In thiss system we neglect ΔZ, ΔKE Δ and W (i.e.) ΔZ = ΔK KE = W = 0;; the energy equation for a boiler beco omes Q = (h2 – h1) (b) Steam con ndenser – In n this system m the work doone is zero and a we can also a assume ΔZ Δ and ΔKE are very sma all. Under ste eady conditioons the chang ge in enthalp py is equal too heat lost by y steam. Q = (h1 – h2) and d this heat is also equal to the changee in enthalpy y of cooling water w circulatted (i.e.) the heat lost by steam will be e equal to heat gained by the cooling water. w (c) Steam nozz zle: In this systeem we can assume ΔZ an nd W to be zero z and hea at transfer which w is nothing but any possible heatt loss also zero. The eneergy equation n for this casse becomes. C2 C2 h1 + 1 = h2 + 2 2 2 or C2 = C12 + 2(h1 − h2 ) (viii) Unsteady y Flow Anallysis:Ma any flow proccesses, such as a filling up and a evacuatiing gas cylind ders, are not steady, Such h processes can n be analyzed d by the conttrol volume teechnique. Coonsider a dev vice through which w a fluid d is flowing un nder non-stea ady state cond ditions (Figu ure-shown in below). The rate r at which h the mass off fluid witthin the conttrol volume iss accumulateed is equal too the net ratee of mass flow w across the control c surrface, as giveen below: 38 Fir rst Law w of Therm T odyna amics S K Mondal’s Cha apter 2 F Fig. dmv dm1 dm d 2 = w1 − w2 = − dτ dτ dτ here mv is th he mass of flu uid within th he control volume at any instant. i Wh Over an ny finite periiod of time Δmv = Δm1 – Δm2 Th he rate of acccumulation off energy with hin the contrrol volume iss equal to thee net rate of energy flow acrross the control surface. If I Ev is the en nergy of fluid d within the control c volum me at any instant, Ra ate of energy increase = Rate R of energy y inflow – Ra ate of energy outflow. ⎞ dQ ⎛ ⎛ ⎞ dW dEv C2 C2 − w2 ⎜ h 2 + 2 +Z2g ⎟ − = w1 ⎜ h1 + 1 +Z Z1g ⎟ + equatiion... A dτ 2 2 ⎠ dτ ⎝ ⎝ ⎠ dτ ⎞ ⎛ mC2 Ev = ⎜ U + + mgZ m ⎟ 2 ⎠v ⎝ Wh here m is thee mass of fluiid in the conttrol volume at a any instantt 2 ⎞ dE d ⎛ mC C ∴ v = + mgZ ⎟ = ⎜U + dτ dτ ⎝ 2 ⎠v ⎞ dm1 dQ ⎛ ⎛ ⎞ dm 2 dW C12 C2 + − ⎜ h 2 + 2 +Z2g ⎟ − +Z1g ⎟ ⎜ h1 + 2 2 dτ dτ dτ ⎝ ⎠ dτ ⎠ ⎝ ( equaation........B ) Following Figu ure shows all these energy y flux quantitties. For any y time interva al, equation (B) ( becomes 2 2 ⎛ ⎞ ⎛ ⎞ C C ΔEv = Q − W + ∫ ⎜ h1 + 1 +Z1g ⎟dm m1 − ∫ ⎜ h2 + 2 +Z2g ⎟dm2 2 2 ⎝ ⎠ ⎝ ⎠ Fig. Equatioon (A) is gene eral energy equation. e Forr steady flow, dEv =0 dτ and the equatioon reduces For F a closed system s w1 = 0, 0 w2 = 0, theen from equattion (A), dEv dQ dW = − dτ dτ dτ Orr dEv = dQ d – dW or dQ = dE + dW d 39 Fir rst Law w of Therm T odyna amics S K Mondal’s Cha apter 2 Floow Processess Ex xample of a variable flo ow problem m: Va ariable flow processes p may y be analyzed d either by th he system tecchnique or th he control vollume tecchnique, as illlustrated below. Consideer a process in i which a ga as bottle is filled from a pipeline p (Figu ure shown in below). In thee beginning the t bottle con ntains gas of mass m1 at state s p1, T1, v1, h1 and u1. The valve iss opened and gas flows in nto the bottle e till the masss of gas in th he bottle is m2 at state p2, T2, v2, h2 an nd u2. The sup pply to the piipeline is verry large so th hat the state of gas in the pipeline is constant at p p ,TP , v P , h P , u P and v P . Sy ystem Techn nique: Assum me an enveloope (which is extensible) of o gas in the pipeline p and the tube wh hich would ev ventually entter the bottlee, as shown in n Figure abov ve. Energy y of the gas be efore filling. ⎛ C2 ⎞ E1 = m1 u1 + ( m2 – m1 ) ⎜ P + uP ⎟ 2 ⎝ ⎠ Wh here ( m2 – m1 ) is the ma ass of gas in the t pipeline and a tube whiich would enter the bottlee. E2 = m2 u2 ⎛ C2 ⎞ ΔE = E2 – E1 = m2 u2 – m1 u1 ( m1 – u1 ) ⎜ P + uP ⎟ ⎠ ⎝ 2 The P.E. terms t are neglected. The gas in the boottle is not in n motion, and d so the K.E. terms have beeen omitted. Now, therre is a chang ge in the volume of gas because of the t collapse of the envellope to zero vollume. Then the t work done W = p (V2 – V1 ) p = p p ⎡⎣0 – ( m2 – m1 ) v P ⎤⎦ = – ( m2 – m1 ) p p v P Ussing the firstt for the process Q = ΔE + W ⎡C2 ⎤ = m2 u2 – m1 u1 – ( m2 – m1 ) ⎢ P + uP ⎥ − ( m2 – m1 ) pP vP ⎣ 2 ⎦ 40 Fir rst Law w of Therm T odyna amics S K Mondal’s Cha apter 2 ⎛ C2 ⎞ = m2 u2 – m1 u1 – ( m2 – m1 ) ⎜ P + h P ⎟ ⎝ 2 ⎠ hich gives thee energy bala ance for the process. p wh Co ontrol Volum me Techniq que: Assume a control vollume bounded by a controol surface as shown in Fig gure above, Applying A the energy equa ation in this case, the folllowing energy y balance ma ay be wrritten on a tim me rate basiss dEv dQ ⎛ C 2 ⎞ dm m = + ⎜ hP + P ⎟ dτ dτ ⎝ 2 ⎠ dττ Sin nce hP and CP are constan nt, the equatiion is integra ated to give for f the Total process 2 ⎛ C ⎞ ΔEv = Q + ⎜ h P + P ⎟ ( m2 − m1 ) 2 ⎠ ⎝ Noow ΔEv = U 2 – U1 = m2 u2 – m1 u1 ⎛ C2 ⎞ Q = m2 u2 – m1 u1 – ⎜ h P + P ⎟ ( m2 − m1 ) 2 ⎠ ⎝ D Discharg ging an nd Charrging a Tank Let us considerr a tank disccharging a flu uid into a su upply line (Fiigure). Sincee dWx = 0 an nd dmin = 0, app plying first la aw to the con ntrol volume,, ⎛ ⎞ C2 + gz ⎟ dmout dUV = dQ Q + ⎜h + 2 ⎝ ⎠ out Assuming K.E. and P.E. of the t fluid to be b small and dQ = 0 d(mu) = hdm mdu+ udm = udm+ pv dm d Ag gain or dm du = m pv onst. V = vm = co vdm + mdv = 0 dm dv =− m v du dv =− pv v d ( u + pv ) = 0 or dQ = 0 wh hich shows that t the process is adiab batic and quasi-static. For cha arging the tan nk 41 Charging and Disc charging a Tank First Law of Thermodynamics S K Mondal’s Chapter 2 ∫ ( hdm ) in = ΔUV = m2u2 − m1u1 mphp = m2u2 − m1u1 where the subscript p refers to the constant state of the fluid in the pipeline. If the tank is initially empty, m1 = 0. m p h p = m2 u2 Since mp = m2 hp = u2 If the fluid is an ideal gas, the temperature of the gas in the tank after it is charged is given by c pTp = cvT2 T2 = γTp or PROBLEMS & SOLUTIONS Example 1 The work and heat transfer per degree of temperature change for a closed system is given by dW 1 dQ 1 = kJ / ° C; = kJ / ° C dT 30 dT 10 Calculate the change in internal energy as its temperature increases from 125ºC to 245ºC. Solution: dT dW = 30 W = T2 dT 1 1 = (T2 − T1 ) = ( 245 − 125 ) 30 30 T1 30 ∫ dQ = Q= T2 dT 10 dT ∫ 10 T1 = 1 ( 245 − 125 ) = 12 kJ 10 Applying First Law of Thermodynamics Q = W + ΔU ΔU = Q – W = 12 – 4 = 8kJ. Example 2 Air expands from 3 bar to 1 bar in a nozzle. The initial velocity is 90 m/s. the initial temperature is 150ºC. Calculate the velocity of air at the exit of the nozzle. Solution: The system in question is an open one. First Law of Thermodynamics for an open system gives ⎡ ⎤ ⎡ ⎤ C2 C2 w1 ⎢h1 + 1 + Z1 g ⎥ + Q = w2 ⎢h2 + 2 + Z2 g ⎥ + W 2 2 ⎣ ⎦ ⎣ ⎦ Since the flow is assumed to be steady. w1 = w2 Flow in a nozzle is adiabatic flow. 42 First Law of Thermodynamics S K Mondal’s Chapter 2 Hence Q = 0 Also W = 0 The datum can be selected to pass through axis; then Z1 = Z2. Hence C2 C2 h1 + 1 = h2 + 2 2 2 C22 C2 ( or ) = ( h1 − h2 ) + 1 2 2 and γ −1 ⎛p ⎞γ T2 = T1 ⎜ 2 ⎟ ⎝ p1 ⎠ γ for air = 1.4 T1 = 150 + 273 = 423 0.4/1.4 ⎛1 ⎞ ∴ T2 = 423 ⎜ ⎟ = 309 K ⎝3⎠ For air Cp = 1.005 kJ/kgºC Cv = 0.718 kJ/kgºC. R = 287 J/kg K = 0.287 kJ/kg K S.F.E.E. : - We have (h1 – h2) = Cp(T1 - T2) C22 C2 902 = ( h1 − h2 ) + 1 = 1.005 × 103 × (423 − 309) + 2 2 2 or, C2 = 487 m / s. Example 3 An evacuated cylinder fitted with a valve through which air from atmosphere at 760 mm Hg and 25°C is allow to fill it slowly. If no heat interaction is involved, what will be the temperature of air in the bottle when the pressure reaches 760 mm Hg? Use the following: (1) Internal energy of air u = u0 + 0.718T kJ/kg where T is temperature in °C. (2) R = 0.287 kJ/kg K. Solution: Applying first law, ignoring potential and kinetic energy terms, to the vessel as control volume. 43 First Law of Thermodynamics S K Mondal’s Chapter 2 Q + m i h i = m e h e + m2 u2 – m1 u1 + W Here Q = 0, W = 0, m e = 0 ( no mass leaving from control vol.) m1 = 0 ( evacuated ) ∴ m2 = mi ∴ hi = u2 hi = ui + pv = u0 + 0.718Ti + 0.287 (Ti + 273 ) = u0 + 0.718 × 25 + 0.287 × 298 = u0 + 103.48kJ / kg = u2 or u2 − u0 = 103.48kJ / kg u2 = u0 + 0.718T2 T2 = u2 − u0 103.48 = = 144.2°C 0.718 0.718 Example 4 A system whose mass is 4.5 kg undergoes a process and the temperature changes from 50° C to 100°C. Assume that the specific heat of the system is a function of temperature only. Calculate the heat transfer during the process for the following relation ship. 80 cn = 1.25 + kJ / kg °C [t is in oC] t + 160 Solution: 100 100 80 ⎞ ⎛ = = Q mc dt 4.5 1 2 ∫50 n ∫50 ⎜⎝1.25 + t + 160 ⎟⎠ dt 100 ⎧⎪100 ⎫⎪ dt = 4.5 ⎨ ∫ 1.25dt + ∫ ⎬ 0.0125t + 2.0 ⎭⎪ 50 ⎩⎪ 50 100 100 ⎪⎧ ⎡ 1 ⎤ ⎪⎫ = 4.5 ⎨[1.25t ]50 + ⎢ ln ( 0.0125t + 2.0 )⎥ ⎬ ⎣ 0.0125 ⎦50 ⎭⎪ ⎩⎪ ⎧ ⎫ ⎡ 1 = 4.5 ⎨[1.25 × 50] + ⎢ (ln (1.25 + 2.0 ) − ln (0.625 + 2 )) ⎤⎥ ⎬ 0.0125 ⎣ ⎦⎭ ⎩ 1 3.25 ⎧ ⎫ = 4.5 ⎨62.5 + ln ⎬ = 358 kJ 0.0125 2.625 ⎭ ⎩ 44 Fir rst Law w of Therm T odyna amics S K Mondal’s Cha apter 2 ASKED D OBJEC CTIVE QUESTIONS (G GATE, IES, IAS) P Previou us 20-Y Years GATE E Ques stions A Applicatiion of First F Law w to Ste eady Flo ow Proc cess S.F F.E.E Co ommon Data for Qu uestions Q1 Q and Q2 2: [GA ATE-2009] Th he inlet and d the outle et conditions of ste eam for an n adiabatic c steam tur rbine are as indic cated in the t figure. The otations are as usually followed. no ATE-1. GA If mass m flow rate r of steam through the turbine is 20 kg/ss the power r output of the e turbine (in n MW) is: (a) 12.157 GA ATE-2. [GA ATE-2009] (b) 12.941 (cc) 168.001 (d) 168.785 Asssume the ab bove turbin ne to be par rt of a simp ple Rankine e cycle. The density of water at the inlet to the e pump is 1000 kg/m3. Ignoring kinetic k and d potential ene ergy effects s, the specifi fic work (in kJ/kg) supp plied to the pump is: [G GATE-2009] (a) 0.293 (b) 0.35 1 (c)) 2.930 (d) 3.510 3 GA ATE-3. The e following g four figures have been dra awn to re epresent a fictitious the ermodynam mic cycle, on n the p-v and d T-s planess. [G GATE-2005] Acc cording to the t first law w of thermodynamics, equal e areas are enclose ed by (a) Figures 1and d 2 (b) Fig gures 1and 3 (c) Figuress 1and 4 (d) Figures 2 and 3 45 First Law of Thermodynamics S K Mondal’s Chapter 2 Internal Energy – A Property of System GATE-4. A gas contained in a cylinder is compressed, the work required for compression being 5000 kJ. During the process, heat interaction of 2000 kJ causes the surroundings to the heated. The change in internal energy of the gas during the process is: [GATE-2004] (a) – 7000 kJ (b) – 3000 kJ (c) + 3000 kJ (d) + 7000 kJ GATE-4a. The contents of a well-insulated tank are heated by a resistor of 23 Ω in which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive. The rates of heat (Q), work (W) and change in internal energy ( ΔU) during the process in kW are [GATE-2011] (a) Q = 0, W = –2.3, Δ U = +2.3 (b) Q = +2.3, W = 0, Δ U = +2.3 (c) Q = –2.3, W = 0, Δ U = –2.3 (d) Q = 0, W = +2.3, Δ U = –2.3 Discharging and Charging a Tank GATE-5. A rigid, insulated tank is initially evacuated. The tank is connected with a supply line through which air (assumed to be ideal gas with constant specific heats) passes at I MPa, 350°C. A valve connected with the supply line is opened and the tank is charged with air until the final pressure inside the tank reaches I MPa. The final temperature inside the tank (A) Is greater than 350°C (B) Is less than 350°C (C) Is equal to 350°C (D) May be greater than, less than, or equal to 350°C, depending on the volume of the tank Previous 20-Years IES Questions First Law of Thermodynamics IES-1. Which one of the following sets of thermodynamic laws/relations is directly involved in determining the final properties during an adiabatic mixing process? [IES-2000] (a) The first and second laws of thermodynamics (b) The second law of thermodynamics and steady flow relations (c) Perfect gas relationship and steady flow relations (d) The first law of thermodynamics and perfect gas relationship 46 First Law of Thermodynamics S K Mondal’s Chapter 2 IES-2. Two blocks which are at different states are brought into contact with each other and allowed to reach a final state of thermal equilibrium. The final temperature attained is specified by the [IES-1998] (a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Third law of thermodynamics IES-3. For a closed system, the difference between the heat added to the system and the work done by the system is equal to the change in [IES-1992] (a) Enthalpy (b) Entropy (c) Temperature (d) Internal energy IES-4. An ideal cycle is shown in the figure. Its thermal efficiency is given by IES-5. ⎛ v3 ⎞ ⎜ − 1⎟ v (a)1 − ⎝ 1 ⎠ ⎛ p2 ⎞ ⎜ − 1⎟ ⎝ p1 ⎠ ⎛ v3 ⎞ ⎜ − 1⎟ 1 ⎝ v1 ⎠ (b) 1 − γ ⎛ p2 ⎞ ⎜ − 1⎟ ⎝ p1 ⎠ (c)1 − γ (b) 1 − ( v3 − v1 ) p1 ( p2 − p1 ) v1 1 ( v3 − v1 ) p1 γ ( p2 − p1 ) v1 [IES-1998] Which one of the following is correct? The cyclic integral of (δQ − δW ) for a process is: (a) Positive (b) Negative (c) Zero [IES-2007] (d) Unpredictable IES-6. A closed system undergoes a process 1-2 for which the values of Q1-2 and W1-2 are +20 kJ and +50 kJ, respectively. If the system is returned to state, 1, and Q2-1 is [IES-2005] 10 kJ, what is the value of the work W2-1? (a) + 20 kJ (b) –40 kJ (c) –80 kJ (d) +40 kJ IES-7. A gas is compressed in a cylinder by a movable piston to a volume one-half of its original volume. During the process, 300 kJ heat left the gas and the internal energy remained same. What is the work done on the gas? [IES-2005] (a) 100kNm (b) 150 kNm (c) 200 kNm (d) 300 kNm IES-8. In a steady-flow adiabatic turbine, the changes in the internal energy, enthalpy, kinetic energy and potential energy of the working fluid, from inlet to exit, are -100 kJ/kg, -140 kJ/kg, -10 kJ/kg and 0 kJ/kg respectively. Which one of the following gives the amount of work developed by the turbine? [IES-2004] (a) 100 kJ/kg (b) 110 kJ/kg (c) 140 kJ/kg (d) 150 kJ/kg 47 First Law of Thermodynamics S K Mondal’s Chapter 2 IES-9. Gas contained in a closed system consisting of piston cylinder arrangement is expanded. Work done by the gas during expansion is 50 kJ. Decrease in internal energy of the gas during expansion is 30 kJ. Heat transfer during the process is equal to: [IES-2003] (a) –20 kJ (b) +20 kJ (c) –80 kJ (d) +80 kJ IES-10. A system while undergoing a cycle [IES-2001] A – B – C – D – A has the values of heat and work transfers as given in the Table: Process Q kJ/min W kJ/min A–B B–C C–D D–A +687 -269 -199 +75 +474 0 -180 0 The power developed in kW is, nearly, (a) 4.9 (b) 24.5 (c) 49 IES-11. (d) 98 The values of heat transfer and work transfer for four processes of a thermodynamic cycle are given below: [IES-1994] Process 1 2 3 4 Heat Transfer (kJ) Work Transfer (kJ) 300 Zero -100 Zero 300 250 -100 -250 The thermal efficiency and work ratio for the cycle will be respectively. (a) 33% and 0.66 (b) 66% and 0.36. (c) 36% and 0.66 (d) 33% and 0.36. IES-12. A tank containing air is stirred by a paddle wheel. The work input to the paddle wheel is 9000 kJ and the heat transferred to the surroundings from the tank is 3000 kJ. The external work done by the system is: [IES-1999] (a) Zero (b) 3000 kJ (c) 6000 kJ (d) 9000 kJ Internal Energy – A Property of System IES-13. For a simple closed system of constant composition, the difference between the net heat and work interactions is identifiable as the change in [IES-2003] (a) Enthalpy (b) Entropy (c) Flow energy (d) Internal energy IES-14. Assertion (A): The internal energy depends on the internal state of a body, as determined by its temperature, pressure and composition. [IES-2006] Reason (R): Internal energy of a substance does not include any energy that it may possess as a result of its macroscopic position or movement. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true 48 First Law of Thermodynamics S K Mondal’s Chapter 2 IES-15. Change in internal energy in a reversible process occurring in a closed system is equal to the heat transferred if the process occurs at constant: [IES-2005] (a) Pressure (b) Volume (c) Temperature (d) Enthalpy IES-16. 170 kJ of heat is supplied to a system at constant volume. Then the system rejects 180 kJ of heat at constant pressure and 40 kJ of work is done on it. The system is finally brought to its original state by adiabatic process. If the initial value of internal energy is 100 kJ, then which one of the following statements is correct? [IES-2004] (a) The highest value of internal energy occurs at the end of the constant volume process (b) The highest value of internal energy occurs at the end of constant pressure process. (c) The highest value of internal energy occurs after adiabatic expansion (d) Internal energy is equal at all points IES-17. 85 kJ of heat is supplied to a closed system at constant volume. During the next process, the system rejects 90 kJ of heat at constant pressure while 20 kJ of work is done on it. The system is brought to the original state by an adiabatic process. The initial internal energy is 100 kJ. Then what is the quantity of work transfer during the process? [IES-2009] (a) 30 kJ (b) 25 kJ (c) 20 kJ (d) 15 kJ IES-17a A closed system receives 60 kJ heat but its internal energy decreases by 30 kJ. Then the [IES-2010] work done by the system is (a) 90 kJ (b) 30 kJ (c) –30 kJ (d) – 90 kJ IES-18. A system undergoes a process during which the heat transfer to the system per degree increase in temperature is given by the equation: [IES-2004] dQ/dT = 2 kJ/°C The work done by the system per degree increase in temperature is given by the equation dW/dT = 2 – 0.1 T, where T is in °C. If during the process, the temperature of water varies from 100°C to 150°C, what will be the change in internal energy? (a) 125 kJ (b) –250 kJ (c) 625 kJ (d) –1250 kJ IES-19. When a system is taken from state A to state B along the path A-C-B, 180 kJ of heat flows into the system and it does 130 kJ of work (see figure given): How much heat will flow into the system along the path A-D-B if the work done by it along the path is 40 kJ? (a) 40 kJ (b) 60 kJ (c) 90 kJ (d) 135 kJ IES-20. [IES-1997] The internal energy of a certain system is a function of temperature alone and is given by the formula E = 25 + 0.25t kJ. If this system executes a process for which the work done by it per degree temperature increase is 0.75 kJ/K, then the heat interaction per degree temperature increase, in kJ, is: [IES-1995] (a) –1.00 (b) –0.50 (c) 0.50 (d ) 1.00 49 First Law of Thermodynamics S K Mondal’s IES-21. Chapter 2 When a gas is heated at constant pressure, the percentage of the energy supplied, which goes as the internal energy of the gas is: [IES-1992] (a) More for a diatomic gas than for triatomic gas (b) Same for monatomic, diatomic and triatomic gases but less than 100% (c) 100% for all gases (d) Less for triatomic gas than for a diatomic gas Perpetual Motion Machine of the First Kind-PMM1 IES-22. Consider the following statements: [IES-2000] 1. The first law of thermodynamics is a law of conservation of energy. 2. Perpetual motion machine of the first kind converts energy into equivalent work. 3. A closed system does not exchange work or energy with its surroundings. 4. The second law of thermodynamics stipulates the law of conservation of energy and entropy. Which of the statements are correct? (a) 1 and 2 (b) 2 and 4 (c) 2, 3 and 4 (d) 1, 2 and 3 Enthalpy IES-23. Assertion (A): If the enthalpy of a closed system decreases by 25 kJ while the system receives 30 kJ of energy by heat transfer, the work done by the system is 55 kJ. [IES-2001] Reason (R): The first law energy balance for a closed system is (notations have their usual meaning) ΔE = Q − W (a) (b) (c) (d) Both A and R are individually true and R is the correct explanation of A Both A and R are individually true but R is NOT the correct explanation of A A is true but R is false A is false but R is true Application of First Law to Steady Flow Process S.F.E.E IES-24. IES-25. Which one of the following is the steady flow energy equation for a boiler? (a) h1 + v12 v2 = h2 + 2 2 gJ 2 gJ (b) Q = ( h2 − h1 ) (c) h1 + v12 v2 + Q = h2 + 2 2 gJ 2 gJ (d) Ws = ( h2 − h1 ) + Q [IES-2005] A 4 kW, 20 litre water heater is switched on for 10 minutes. The heat capacity Cp for water is 4 kJ/kg K. Assuming all the electrical energy has gone into heating the water, what is the increase of the water temperature? [IES-2008] (a) 15°C (b) 20°C (c) 26°C (d) 30°C Discharging and Charging a Tank 50 First Law of Thermodynamics S K Mondal’s IES-26. Chapter 2 An insulated tank initially contains 0.25 kg of a gas with an internal energy of 200 kJ/kg .Additional gas with an internal energy of 300 kJ/kg and an enthalpy of 400 kJ/kg enters the tank until the total mass of gas contained is 1 kg. What is the final internal energy(in kJ/kg) of the gas in the tank? [IES-2007] (a) 250 (b) 275 (c) 350 (d) None of the above Previous 20-Years IAS Questions IAS-1. A system executes a cycle during which there are four heat transfers: Q12 = 220 kJ, Q23 = -25kJ, Q34 = -180 kJ, Q41 = 50 kJ. The work during three of the processes is W12 = 15kJ, W23 = -10 kJ, W34 = 60kJ. The work during the process 4 1 is: [IAS-2003] (a) - 230 kJ (b) 0 kJ (c) 230 kJ (d) 130 kJ IAS-2. Two ideal heat engine cycles are represented in the given figure. Assume VQ = QR, PQ = QS and UP =PR =RT. If the work interaction for the rectangular cycle (WVUR) is 48 Nm, then the work interaction for the other cycle PST is: (a) 12Nm (b) 18 Nm (c) 24 Nm (d) 36 Nm IAS-2001] IAS-3. A reversible heat engine operating between hot and cold reservoirs delivers a work output of 54 kJ while it rejects a heat of 66 kJ. The efficiency of this engine is: [IAS-1998] (a) 0.45 (b) 0.66 (c) 0.75 (d) 0.82 IAS-4. If a heat engine gives an output of 3 kW when the input is 10,000 J/s, then the thermal efficiency of the engine will be: [IAS-1995] (a) 20% (b) 30% (c) 70% (d) 76.7% IAS-5. In an adiabatic process, 5000J of work is performed on a system. The system returns to its original state while 1000J of heat is added. The work done during the non-adiabatic process is: [IAS-1997] (a) + 4000J (b) - 4000J (c) + 6000J (d) - 6000J IAS-6. In a thermodynamic cycle consisting of four processes, the heat and work are as follows: [IAS-1996] Q: + 30, - 10, -20, + 5 W: + 3, 10, - 8, 0 The thermal efficiency of the cycle will be: (a) Zero (b) 7.15% (c) 14.33% (d) 28.6% 51 First Law of Thermodynamics S K Mondal’s Chapter 2 IAS-7. Match List-I (Devices) with List-II (Thermodynamic equations) and select the correct answer using the codes below the lists: [IAS-1996] List-I List-II A. Turbine 1. W = h2 – h1 B. Nozzle 2. h1 = h2 C. Valve 3. h1 = h2 + V2/2 D. Compressor 4. W = h1 – h2 Codes: A B C D A B C D (a) 4 3 2 1 (b) 2 3 1 4 (c) 4 3 1 2 (d) 3 2 4 1 IAS-8. Given that the path 1-2-3, a system absorbs 100kJ as heat and does 60kJ work while along the path 14-3 it does 20kJ work (see figure given). The heat absorbed during the cycle 1-4-3 is: (a) - 140 Kj (b) - 80 kJ (c) - 40kJ (d) + 60 kJ IAS-9. The given figure shows the variation of force in an elementary system which undergoes a process during which the plunger position changes from 0 to 3 m. lf the internal energy of the system at the end of the process is 2.5 J higher, then the heat absorbed during the process is: (a) 15 J (b) 20 J IAS-10. [IAS 1994] (c) 25 J [IAS-1994] (d) 30 J The efficiency of a reversible cyclic process undergone by a substance as shown in the given diagram is: (a) 0.40 (b) 0.55 (c) 0.60 (d) 0.80 [IAS 1994] Internal Energy – A Property of System IAS-11. Which one of the following is the correct expression for change in the internal energy for a small temperature change Δ T for an ideal gas? [IAS-2007] (a) ΔU = Cv × ΔT (b) ΔU = C p × ΔT 52 First Law of Thermodynamics S K Mondal’s (c) ΔU = Cp Cv Chapter 2 ( × ΔT ) (d) ΔU = C p − Cv × ΔT IAS-12. The heat transferred in a thermodynamic cycle of a system consisting of four processes is successively 0, 8, 6 and -4 units. The net change in the internal energy of the system will be: [IAS-1999] (a) – 8 (b) Zero (c) 10 (d) –10 IAS-13. During a process with heat and work interactions, the internal energy of a system increases by 30 kJ. The amounts of heat and work interactions are respectively [IAS-1999] (a) - 50 kJ and - 80 kJ (b) -50 kJ and 80 kJ (c) 50 kJ and 80 kJ (d) 50 kJ and - 80 kJ IAS-14. A mixture of gases expands from 0.03 m3 to 0.06 m3 at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is: [IAS 1994] (a) 30 kJ (b) 54 kJ (c) 84 kJ (d) 114 kJ IAS-15. In an adiabatic process 6000 J of work is performed on a system. In the nonadiabatic process by which the system returns to its original state 1000J of heat is added to the system. What is the work done during non-adiabatic process? [IAS-2004] (a) + 7000 J (b) - 7000 J (c) + 5000 J (d) - 5000 J Enthalpy IAS-16. The fundamental unit of enthalpy is: (a) MLT-2 (b) ML-2T-1 (c) ML2T-2 [IAS 1994] (d) ML3T-2 Application of First Law to Steady Flow Process S.F.E.E IAS-17. In a test of a water-jacketed compressor, the shaft work required is 90 kN-m/kg of air compressed. During compression, increase in enthalpy of air is 30 kJ/kg of air and increase in enthalpy of circulating cooling water is 40 kJ/ kg of air. The change is velocity is negligible. The amount of heat lost to the atmosphere from the compressor per kg of air is: [IAS-2000] (a) 20kJ (b) 60kJ (c) 80 kJ (d) 120kJ IAS-18. When air is compressed, the enthalpy is increased from 100 to 200 kJ/kg. Heat lost during this compression is 50 kJ/kg. Neglecting kinetic and potential energies, the power required for a mass flow of 2 kg/s of air through the compressor will be: [IAS-1997] (a) 300 kW (b) 200 kW (c) 100 kW (d) 50 kW 53 First Law of Thermodynamics S K Mondal’s Chapter 2 Variable Flow Processes IAS-19. Match List-I with List-II and select the correct answer using the codes given below Lists: [IAS-2004] List-I List-II A. Bottle filling of gas 1. Absolute zero temperature B. Nernst Simon statement 2. Variable flow C. Joule Thomson effect 3. Quasistatic path D. ∫ pdv Codes: (a) (c) IAS-20. 4. Isenthalpic process A 6 2 B 5 5 C 4 7 5. Dissipative effect 6. Low grade energy 7. Process and temperature change A B C (b) 2 1 4 (d) 6 1 7 D 3 4 during phase D 3 4 A gas chamber is divided into two parts by means of a partition wall. On one side, nitrogen gas at 2 bar pressure and 20°C is present. On the other side, nitrogen gas at 3.5 bar pressure and 35°C is present. The chamber is rigid and thermally insulated from the surroundings. Now, if the partition is removed, (a) High pressure nitrogen will get throttled [IAS-1997] (b) Mechanical work, will be done at the expense of internal energy (c) Work will be done on low pressure nitrogen (d) Internal energy of nitrogen will be conserved 54 First Law of Thermodynamics S K Mondal’s Chapter 2 Answers with Explanation (Objective) Previous 20-Years GATE Answers C12 C2 gZ1 gZ2 dQ dW + + = h2 + 1 + + 2000 1000 dm 2000 1000 dm 2 160 9.81× 10 100 2 9.81× 6 dW 3200 + + = 2600 + + + 2000 1000 2000 1000 dm dW 600 + 7.8 + 0.04 = + dm GATE-1. Ans. (a) h1 + GATE-2. Ans. (c) W = ν (P2 − P1 ) = 1 ( 3000 − 70 ) × kJ/kg = 2.93 1000 GATE-3. Ans. (a) Fig-1 & 2 both are power cycle, so equal areas but fig-3 & 4 are reverse power cycle, so area is not meant something. GATE-4. Ans. (c) dQ = du + dw Q = u2 − u1 + W or − 2000 = u2 − u1 − 5000 or u2 − u1 = 3000kJ GATE-4a. Ans. (a) Q = 0, W = –2.3, ΔU = I +2.3 Tank is well insulated so Q = 0 Work is given to the system in the form of electric current. R I So, W = − I2 R = − 102 × 23 = –2300 W = –2.3 kW By 1st Law of Thermodynamics Q1 − 2 = U2 − U1 + W1 − 2 0 = U 2 − U1 − 2.3 ΔU = 2.3 kW GATE-5. Ans (a) The final Temp. (T2)= γT1 Previous 20-Years IES Answers IES-1. Ans. (a) If we adiabatically mix two liquid then perfect gas law is not necessary. But entropy change in the universe must be calculated by Second law of thermodynamics. Final entropy of then system is also a property. That so why we need second law. IES-2. Ans. (b) Using conservation of energy law we may find final temperature. IES-3. Ans. (d) From First law of thermodynamics, for a closed system the net energy transferred as heat Q and as work W is equal to the change in internal energy, U, i.e. Q – W = U 55 Fir rst Law w of Therm T odyna amics S K Mondal’s Cha apter 2 IES-4. Ans. (c)) Total heat addition a is coonstant volum me heat addiition, Q12 = cv (T2 − T1 ) Tottal heat rejection is consta ant pressure heat rejectioon, Q31 = c p (T3 − T1) Now w from equattion of state P P1 P2 = (∵ v = const.) or T2 = 2 × T1 P1 T1 T2 and d v v1 v 3 = (∵ p = const.)) or T3 = 3 × T1 v1 T1 T3 c (T − T ) Q31 (T − T ) = 1− p 3 1 = 1− γ 3 1 cv (T2 − T1) Q12 (T2 − T1) Effiiciency, η = 1 − ⎛ v3 ⎞ ⎜ × T1 − T1 ⎟ v ⎠ = 1 − γ (v 3 − v1) p1 or η = 1 − γ ⎝ 1 ( p2 − p1) v1 ⎛ P2 ⎞ ⎜ × T1 − T1 ⎟ ⎝ P1 ⎠ IES-5. Ans. (c c) It is du = đQ đ – đW, as u is a thermoodynamic prooperty and itts cyclic integ gral must be zeroo. W or Q1− 2 + Q2 −1 = W1− 2 + W2 −1 IES-6. Ans. (b) ΣdQ = ΣdW or 20 + ( −10 ) = 50 + W2−1 or W2−1 = −40kJ IES-7. Ans. (d) dQ = du + dw as u = const. Theerefore du = 0 or dQ = dw d = 300kNm IES-8. Ans. (d) ⎞ ⎛ V2 Q − Wx = Δ ⎜ h + + gz ⎟ 2 ⎝ ⎠ O − Wx = −140 − 10 + 0 or Wx = 150 kJ / kg Cha ange of internal energy = -100 kJ/kg is superfluous data. IES-9. Ans. (b) Q = Δ E+ Δ W Δ E = –30 kJ (ddecrease in in nternal energ gy) ( done by the system m) Δ W = + 50 kJ (work Q = –30 + 50 = + 20 kJ IES-10. Ans. (a a) Net work = 74 – 180 kJ/m min = 294 kJ J/min = 294/6 60 kJ/s = 4.9 kW dW =47 ∑ IES-11. Ans. (b b) ηth = Work done 3000 − 100 = 0..66 = 300 heaat added Woork ratio = ∑ ( + w) − ∑ ( − w) = 550 − 350 = 0.366 5 550 ∑ ( + w) IES-12. Ans. (a) This iss a case of constant voolume process or an n is isochooric process.. By perrforming worrk on the sysstem tempera ature can n be raised. In an irrev versible constant volu ume process, the system m doesn't perrform worrk on the surrrounding at the expense of its inteernal energy. 56 Fir rst Law w of Therm T odyna amics S K Mondal’s Cha apter 2 IES-13. Ans. (d d) IES-14. Ans. (a) ( The interrnal energy depends d only y upon the in nitial and fin nal states of the system. Inteernal energy y of a substa ance does noot include an ny energy th hat it may possess p as a resu ult of its ma acroscopic position or movement. m T That so why in SFEE v2/2 / and gz is there. If in nternal energ gy include poosition or movement then n why this v2/2 / and gz term ms is there. Bur Remembe er: Miccroscopic view of a gas is a collectioon of particlles in randoom motion. Energy E of a parrticle consists of transla ational energy, rotatio onal energy y, vibration nal energy and d specific ellectronic en nergy. All th hese energies summed oveer all the parrticles of the gass, form the sp pecific interna al energy, e , of the gas. IES-15. Ans. (b b) dQ = dU + pdV if V is con s tan t ( dQ )v = ( dU)v IES-16. Ans. (a a) Q2 = 180 0kJ = Δu + ΔW = Δu + ( −40)) ∴U1 = 100kJ, U2 = 100 0 + 170 = 270 0 kJ, U3 = 270 − 180 + 40 = 130 kJ IES-17. Ans. (d d) IES-17a. For the process p 1-2 dQ = +85 , dW = 0 For the process p 2-3 dQ = -90 kJ, dW = -20kJ For the process p 3-1 dQ = 0, dW = ? For a cycclic process ∑ dQ = ∑ dW ⇒ 85-90+ +0 = 0-20+ dW W ⇒ -5 = -20+ dW ⇒ 5 = +15kJ dW = -20+5 An ns. (a) dQ = dU + dW d or 60 = – 30 + dW or dW = 90 KJ 57 First Law of Thermodynamics S K Mondal’s Chapter 2 IES-18. Ans. (c) dQ = du + dw 2.dt = du + ( 2 − 0.1T ) dT 150 0.1 0.1 ⎡1502 − 1002 ⎤⎦ = 625kJ × ⎡⎣T 2 ⎤⎦ = 100 2 2 ⎣ IES-19. Ans. (c) Change of internal energy from A to B along path ACB = 180 - 130 = 50 kJ. It will be same even along path ADB. ∴ Heat flow along ADB = 40 + 50 = 90 kJ. or ∫ du = ∫ 0.1TdT = IES-20. Ans. (d) dQ = dE + dW Given: E = 25 + 0.25t kJ and or dQ dE dW = + dt dt dt dW = 0.75 kJ / k dt dE = 0.25 kJ / K dt dQ dE dW Therefore = + = 0.25 + 0.75 kJ / K = 1.00 kJ / K dt dt dt then IES-21. Ans. (a) IES-22. Ans. (a) A closed system does exchange work or energy with its surroundings. option ‘3’ is wrong. 4. “The law of conservation of entropy” is imaginary so option ‘4’ is also wrong. IES-23. Ans. (a) v2 v2 dQ dw IES-24. Ans. (b) h1 + 1 + gz1 + = h2 + 2 + gz 2 + =0 2 dm 2 dm dw =0 For boiler v1, v2 is negligible and z1 = z2 and dm dQ or = ( h2 − h1 ) dm IES-25. Ans. (d) mCP ΔT = ( 4 × 10 × 60 ) ⇒ 20 × 4 × ΔT = 2400 ⇒ ΔT = 30C° IES-26. Ans. (c) Enthalpy of additional gas will be converted to internal energy. Uf= miui+(mf-mi)hp = 0.25x200+(1-0.25)x400 = 350 kJ As total mass = 1kg, uf=350 kJ/kg Note: You cannot simply use adiabatic mixing law here because it is not closed system. This is a problem of variable flow process. If you calculate in following way it will be wrong. Final internal energy of gas(mixture) is m u + m2 u2 u= 1 1 m1 + m2 kJ ⎞ kJ ⎞ ⎛ ⎛ (0.25kg) ⎜ 200 ⎟ + (0.75kg) ⎜ 300 kg ⎟ kg ⎝ ⎠ ⎝ ⎠ u= (0.25 + 0.75)kG u = 275 kJ kg 58 First Law of Thermodynamics S K Mondal’s Chapter 2 It is valid for closed system only. Previous 20-Years IAS Answers IAS-1. Ans. (b) ∑ dQ = ∑ dW or 220 -25 -180 +50 = 15 -10 +60 +W4-1 IAS-2. Ans. (c) Area under p-v diagram is represent work. Areas Δ PTS= IAS-3. Ans. (a) η = 1 1 Area (WVUR) ∴ Work PTS= × 48 =24 Nm 2 2 work output work out put 54 = = = 0.45 Heat input work output + heat rejection 54 + 66 IAS-4. Ans. (b) Thermal efficiency = W 3 × 103 watts = = 0.3 = 30% Q 10.000 J/s IAS-5. Ans. (c) Q1− 2 = (U2 − U1 ) + W1−2 or 0 = (U2 − U1 ) + ( −5000 ) or (U2 − U1 ) = 5000 J Q2−1 = (U1 − U2 ) + W2−1 or W2−1 = Q2−1 − (U1 − U2 ) = Q2−1 + (U2 − U1 ) = 1000 + 5000 = 6000 J IAS-6. Ans. (c) Net work output = 3 + 10 – 8 = 5 unit Therefore efficiency, η = and Heat added = 30 + 5 = 35 unit 5 × 100% = 14.33% 35 IAS-7. Ans. (a) IAS-8. Ans. (d) Q123 = U13 + W123 or, 100 = U13 + 60 or, U13 = 40 kJ And Q143 = U13 + W143 = 40+20 = 60 kJ IAS-9.Ans. (b) Total work = 5 × 3 + IAS-10. Ans. (c) Efficiency = = 1 × 5 × 1 = 17.5 J or δW = du + δW = 2.5 + 17.5 = 20 J 2 Area under 500 and 1500 Area under 0 and 1500 1 × {(5 − 1) + (4 − 2)} × (1500 − 500) 2 1 × {(5 − 1) + (4 − 2)} × (1500 − 500) + (5 − 1) × 500 2 = 3000 = 0.6 5000 IAS-11. Ans. (a) IAS-12. Ans. (b) Internal energy is a property of a system so ∫ du = 0 IAS-13. Ans. (a) dQ = du + dW if du = +30kJ then dQ = −50kJ and dW = −80kJ IAS-14. Ans. (b) δW = du + δW = du + pdV or 84 × 103J = du + 1 × 106 × (0.06 – 0.03) = du +30 kJ or du = 83 – 30 = 54 kJ 59 First Law of Thermodynamics S K Mondal’s Chapter 2 IAS-15. Ans. (a) Q1-2 = U2 –U1 +W1-2 Or 0 = U2 –U1 - 6000 or U2 –U1 = +6000 Q2-1 = U1-U2+W2-1 or W2-1 = Q2-1 - (U1-U2) =1000+6000=7000J IAS-16. Ans. (c) IAS-17. Ans. (a) Energy balance gives as dW dQ = ( Δh )air + ( Δh ) water + dm dm dQ or = 90 − 30 − 40 dm = 20kJ / kg of air compressed. IAS-18. Ans. (a) dQ dw = m ( h2 ) + dt dt dw dQ or = m ( h1 − h2 ) + = 2 × (100 − 200 ) − 50 × 2 = −300kW dt dt i.e. 300kW work have to given to the system. m ( h1 ) + IAS-19. Ans. (b) IAS-20. Ans. (a) 60 Second Law of Thermodynamics S K Mondal’s 3. Chapter 3 Second Law of Thermodynamics Theory at a Glance (For GATE, IES & PSUs) The first law of thermodynamics states that a certain energy balance will hold when a system undergoes a change of state or a thermodynamic process. But it does not give any information on whether that change of state or the process is at all feasible or not. The first law cannot indicate whether a metallic bar of uniform temperature can spontaneously become warmer at one end and cooler at the other. All that the law can state is that if this process did occur, the energy gained by one end would be exactly equal to that lost by the other. It is the second law of thermodynamics which provides the criterion as to the probability of various processes. Regarding Heat Transfer and Work Transfer (a) Heat transfer and work transfer are the energy interactions. A closed system and its surroundings can interact in two ways: by heat transfer and by work transfer. Thermodynamics studies how these interactions bring about property changes in a system. (b) The same effect in a closed system can be brought about either by heat transfer or by work transfer. Whether heat transfer or work transfer has taken place depends on what constitutes the system. (c) Both heat transfer and work transfer are boundary phenomena. Both are observed at the boundaries of the system, and both represent energy crossing the boundaries of the system. (d) It is wrong to say 'total heat' or 'heat content' of a closed system, because heat or work is not a property of the system. Heat, like work, cannot be stored by the system. Both heat and work are the energy is transit. (e) Heat transfer is the energy interaction due to temperature difference only. All other energy interactions may be termed as work transfer. (f) Both heat and work are path functions and inexact differentials. The magnitude of heat transfer or work transfer depends upon the path the system follows during the change of state. (g) Heat transfer takes place according to second law of thermodynamics as it tells about the direction and amount of heat flow that is possible between two reservoirs. 62 Seco ond La aw of Therm modyn namicss S K Mondal’s Cha apter 3 Q Qualitatiive Diffference e between Hea at and Work W • • Thermoodynamic deffinition of woork: Positivee work is done by a system when thee sole effec ct external to t the system m could be reduced d to the rise of o a weight. Thermoodynamic deffinition of heeat: It is th he energy in n transition n between th he system an nd the surrou undings by virtue v of the differen nce in temperrature. Siign Conv ventions s • • • • Work done d BY the system s is +ve e Obviously work don ne ON the sysstem is –ve Heat giiven TO the system s is +v ve Obviously Heat reje ected by the system s is –v ve Heeat and work are not comp pletely intercchangeable forms f of energy. When woork is convertted into hea at, we alway ys have W =Q but when heat is converted into work in n a complete closed c cycle process p Q >W Th he arrow indicates the direction of energy transform mation. Wo ork is said too be a high grade g energ gy and heat is i low grade e energy. Th he complete conversion c of low grade en nergy into high grade eneergy in a cycle is impossib ble. HEAT T and WORK K are NOT pr roperties beecause they depend d on thee 63 Second Law of Thermodynamics S K Mondal’s Chapter 3 path and end states. HEAT and WORK are not properties because their net change in a cycle is not zero. Heat and work are inexact differentials. Their change cannot be written as differences between their end states. Thus Similarly ∫ ∫ 2 1 2 1 δQ ≠ Q2 − Q1 and is shown as 1 Q2 or Q1−2 δW ≠ W2 − W1 and is shown as 1W2 or W1−2 Note. The operator δ is used to denote inexact differentials and operator d is used to denote exact differentials. Kelvin-Planck Statement of Second Law There are two statements of the second law of thermodynamics, the Kelvin-Planck statement, and the Clausius statement. The Kelvin-Planck statement pertains to heat engines. The Clausius statement pertains to refrigerators/heat pumps . Kelvin-Planck statement of second law It is impossible to construct a device (engine) operating in a cycle that will produce no effect other than extraction of heat from a single reservoir and convert all of it into work. Mathematically, Kelvin-Planck statement can be written as: Wcycle ≤ 0 (for a single reservoir) Clausius’ statement of second law It is impossible to transfer heat in a cyclic process from low temperature to high temperature without work from external source. Reversible and Irreversible Processes A process is reversible with respect to the system and surroundings if the system and the surroundings can be restored to their respective initial states by reversing the direction of the process, that is, by reversing the heat transfer and work transfer. The process is irreversible if it cannot fulfill this criterion. 64 Seco ond La aw of Therm modyn namicss S K Mondal’s Cha apter 3 C Clausius s' Theo orem Let a system be b taken from m an equilib brium state i to another equilibrium state f by foollowing the rev versible path h i-f(Figure). Let a reverssible adiabatiic i-a be draw wn through i and anotheer reversible adiiabatic b-f bee drawn thro ough f. Then a reversiblee isothermal a-b is drawn n in such a way w that the areea under i-a-b-f is equal to t the area under u i-f. App plying the firsst law for Processs i − f Qi − f = U f − U i + Wif Processs i − a − b − f Qiabf = U f − U i + Wiabf i Since Wif = Wiabf ∴ Qi f = Qiabf = Qia + Qab + Qbf Since Qia = 0 and d Qbf = 0 Qif = Qab Fig. Rev versible Patth Substitutted by Two o Reversible e Adiabatics and a Reversible e Isothermall Heeat transferreed in the process i-f is equ ual to the hea at transferreed in the isoth hermal proceess a-b. Th hus any reveersible path may m be subsstituted by a reversible zigzag path, between th he same end sta ates, consistiing of a rev versible adia abatic follow wed by a rev versible isotthermal and then by a rev versible adia abatic, such that t the hea at transferred during thee isothermal process is the t same as tha at transferred during the original proccess. Let a smooth closed c curve representing g a reversiblee cycle (Fig below.) b be considered. Leet the closed cyccle be divided d into a larg ge number off strips by means m of reversible adiaba atics. Each strip may be cloosed at the toop and bottom m by reversib ble isotherma als. The origiinal closed cy ycle is thus reeplaced by a zig gzag closed path p consistiing of altern nate adiabatiic and isotheermal processses, such th hat the heat tra ansferred durring all the isothermal prrocesses is eq qual to the heat h transferrred in the orriginal cycle. Th hus the origin nal cycle is replaced by a large number of Carnot cycles. If th he adiabatics are close to onee another an nd the numbe er of Carnot cycles c is larg ge, the saw-tooothed zig-za ag line will cooincide with thee original cyccle. 65 Seco ond La aw of Therm modyn namicss S K Mondal’s Cha apter 3 F Fig. A Reverssible Cycle Split S into a Large Num mber of Carn not Cycles a dQ1 heat is absorb bed reversiblly at T1, an nd dQ2 heat is rejected For the elemeental cycle abcd versibly at T2 rev dQ1 dQ2 = T1 T2 If h heat supplied d is taken as positive and d heat rejecteed as negative dQ1 dQ2 + =0 T2 T1 Sim milarly, for th he elementall cycle efgh dQ3 dQ4 + =0 T3 T4 If ssimilar equattions are wriitten for all th he elementall Carnot cycles, then for the t whole original cycle dQ3 dQ4 d Q1 dQ2 + + + + ... = 0 T1 T2 T3 T4 or ∫ R dQ d =0 T Th he cyclic inttegral of d Q/T for a re eversible cy ycle is equall to zero. Th his is known as a Clausius' theeorem. The leetter R emph hasizes the fa act that the equation is va alid only for a reversible cycle. c 66 Second Law of Thermodynamics S K Mondal’s Chapter 3 Thus the original cycle is replaced by a large number of Carnot cycles. If the adiabatics are close to one another and the number of Carnot cycles is large, the saw-toothed zig-zag line will coincide with the original cycle. Refrigerator and Heat Pump [with RAC] Equivalence of Kelvin-Planck and Clausius Statements II Law basically a negative statement (like most laws in society). The two statements look distinct. We shall prove that violation of one makes the other statement violation too. Let us suspect the Clausius statement-it may be possible to transfer heat from a body at colder to a body at hotter temperature without supply of work Let us have a heat engine operating between T1 as source and T2 as a sink. Let this heat engine reject exactly the same Q2 (as the pseudo-Clausius device) to the reservoir at T2. To do this an amount of Q1 needs to be drawn from the reservoir at T1. There will also be a W = Q1 – Q2. Combine the two. The reservoir at T2 has not undergone any change (Q2 was taken out and by pseudo-Clausius device and put back by the engine). Reservoir 1 has given out a net Q1-Q2. We got work output of W. Q1-Q2 is converted to W with no net heat rejection. This is violation of KelvinPlanck statement. 67 Second Law of Thermodynamics S K Mondal’s • Chapter 3 Let us assume that Clausius statement is true and suspect Kelvin-Planck statement Pseudo Kelvin Planck engine requires only Q1–Q2 as the heat interaction to give out W (because it does not reject any heat) which drives the Clausius heat pump Combining the two yields: • The reservoir at T1 receives Q1 but gives out Q1–Q2 implying a net delivery of Q2 to it. • Q2 has been transferred from T2 to T1 without the supply of any work!! • A violation of Clausius statement Moral: If an engine/refrigerator violates one version of II Law, it violates the other one too. All reversible engine operating between the same two fixed temperatures will have the same η and COP. If there exists a reversible engine/ or a refrigerator which can do better than that, it will violate the Clausius statement. Let us presume that the HP is super efficient!! For the same work given out by the engine E, it can pick up an extra Δ Q from the low temperature source and deliver over to reservoir at T1 . The net effect is this extra has Δ Q been transferred from T2 to T1 with no external work expenditure. Clearly, a violation of Clausius statement!! SUM UP • Heat supplied = Q1; Source temperature = T1 ; Sink temperature = T2 • Maximum possible efficiency = W/Q1= (T1 – T2)/T1 • Work done = W = Q1(T1 – T2)/T1 68 Seco ond La aw of Therm modyn namicss S K Mondal’s Cha apter 3 Carnot Engine E with w sam me effic ciency or o same e work output o T1 Q1 = T2 Q2 Sin nce, T1 − T2 Q1 − Q2 = T2 Q2 T1 − T2 = (Q1 − Q2 ) T2 Q2 T2 − T3 = (Q2 − Q3 ) T3 T = (Q2 − Q3 ) 2 Q2 Q3 • For sam me work ou utput W1 = W2 Q1 – Q 2 = Q2 – Q3 T1 – T2 = T2 – T3 • For sam me efficiency η1 = η2 orr 1 − T T2 = 1− 3 T1 T2 o T2 = T1 × T3 or • Whic ch is the t mo ore effe ective way w to increase the efficiency off a Carn not engiine: to increase e T1 keeping T2 stant; orr to decrrease T2 keeping g T1 con nstant? cons Th he efficiency of o a Carnot engine is giveen by η = 1− T2 T1 If T2 is constan nt ⎛ ∂η ⎞ T2 ⎜ ⎟ = 2 ⎝ ∂T1 ⎠T2 T1 ⎛ ∂η ⎞ AS S T1 increasess, η increasess, and the sloope ⎜ d (Fiigure). If T1 is i constant, ⎟ decreases ⎝ ∂T1 ⎠T2 69 Seco ond La aw of Therm modyn namicss S K Mondal’s Cha apter 3 ⎛ ∂η ⎞ 1 ⎜ ⎟ = − T1 ⎝ ∂T2 ⎠T1 ⎛ ∂η ⎞ As T2 decreasess, η increasess, but the slo ope ⎜ r consttant (Figure)). ⎟ remains ⎝ ∂T2 ⎠T1 Also Since ⎛ ∂η ⎞ ⎛ ∂η T2 T1 η⎞ ⎟ =− 2 ⎜ ⎟ = 2 and ⎜ T1 ⎝ ∂T1 ⎠T2 T1 ⎝ ∂T2 ⎠T1 ⎛ ∂η ⎞ ⎛ ∂∂η ⎞ T1 > T2 , ⎜ ⎟ >⎜ ⎟ ⎝ ∂T2 ⎠T1 ⎝ ∂T1 ⎠T2 So, the more effective e way y to increasse the efficieency is to deecrease T2. Alternatively A y, let T2 be deccreased by ΔT with T1 rem maining the same s η1 = 1 − T2 − ΔT T1 If T1 is increase ed by the sam me ΔT, T2 rem maining the same s η2 = 1 − T2 T1 + ΔT The en η1 − η 2 = T2 T − ΔT − 2 T1 + ΔT T1 (T = 1 Sin nce − T2 ) ΔT + ( ΔT ) 2 T1 (T1 + ΔT ) T1 > T2 , ( η1 − η 2 ) > 0 Th he moree effectiv ve way to increease thee cycle efficiency e y is d decrea ase T2. 70 to Second Law of Thermodynamics S K Mondal’s Chapter 3 PROBLEMS & SOLUTIONS Example 1. An inventor claims that his petrol engine operating between temperatures of 2000°C and 600°C will produce 1 kWhr consuming 150g of petrol having 45000 kJ/kg calorific value. Check the validity of the claim. Solution: By Carnot's theorem, the thermal efficiency of a reversible cycle engine which cannot be exceeded is given by T − T2 2273 − 873 ηmax = 1 = = 0.616 or 61.6% 2273 T1 Actual thermal efficiency is given by 1 ×103 × 3600 = 0.53 or 53% ηt = 0.15 × 45000 ×103 Since actual efficiency is less than the maximum obtainable, the inventor's claim is feasible. Example 2. Two reversible engines takes 2400 kJ per minute from a reservoir at 750 K and develops 400 kJ of work per minute when executing complete cycles.The engines reject heat two reservoirs at 650 K and 550 K. Find the heat rejected to each sink. Solution: QA + QB = 2400 QA2 + QB2 = 2000 QA − QA2 QA = 750 − 650 = 0.1333 750 QA = 1.1539QA 2 71 Second Law of Thermodynamics S K Mondal’s Chapter 3 Similarly QB = 1.3636QB 2 i.e. 1.1539Q A 2 + 1.3636QB 2 = 2400kJ Q A 2 = 2000 − QB 2 ( ) i.e. 1.1539 × 2000 − QB 2 + 1.3636QB 2 = 2400kJ . 0.2091QB 2 = 92 ∴ QB 2 = 440kJ and QA 2 = 1560kJ Example 3. A solar powerd heat pump receives heat from a solar collector at temperature Th, uses the entire energy for pumping heat from cold atmosphere at temperature ‘Tc’ to a room at temperature ‘Ta’. The three heat transfer rates are Qh, Qa and Qc respectively. Derive an expression for the minimum ratio Qh/Qc, in terms of the three temperatures. If Th = 400 K, Ta = 300 K, Tc = 200 K, Qc = 12 kW, what is the minimum Qh? If the collector captures 0.2 kW/m2, what is the minimum collector area required? Solution: Let Qh, Qa and Qc be the quantity of heat transferred from solar collector, room and atmosphere respectively. Qa = Qh + Qc Qa − Qc = Qh or, COPHP = Ta Qa Q + Qc = h and also Ta − Tc Qa − Qc Qh Qh + Qc Ta = Qh Ta − Tc ∴ Qc Ta T − Ta + Tc Tc = −1 = a = Qh Ta − Tc Ta − Tc Ta − Tc ∴ Qh = Qh (Ta − Tc ) = Qc Tc Qc × (Ta − Tc ) Area = Tc = 12 × ( 300 − 200 ) 200 = 6kW 6 = 30m2 0.2 72 Second Law of Thermodynamics S K Mondal’s Chapter 3 Example 4. A reversible engine works between 3 thermal reservoirs A, B and C. The engine absorbs an equal amount of heat from the reservoirs A and B, maintained at temperatures of T1 and T2 respectively and rejects heat to the thermal reservoir C maintained at T3. The efficiency of this engine is α times the efficiency of reversible engine operating between reservoirs A and C only. Show that T1 2T = ( 2α − 1) + 1 (1 − α ) T2 T3 Solution: η1 = W1 2Q1 η2 = T1 − T3 T1 ⎛ T1 − T3 ⎞ ⎟ ⎝ T1 ⎠ η1 = αη2 = α ⎜ The cycle is reversible so ΔS = 0 Q Q Q ∴ 1+ 1 = 2 T1 T2 T3 Also, 2Q1 = W1 + Q2 Combining above equations we have Q1 Q1 2Q1 − W1 + = T1 T2 T3 ⎛ T − T3 ⎞ W1 =α⎜ 1 ⎟ 2Q1 ⎝ T1 ⎠ 73 Second Law of Thermodynamics S K Mondal’s Chapter 3 ⎛ T − T3 ⎞ W1 = 2Q1α ⎜ 1 ⎟ ⎝ T1 ⎠ α ( T1 − T3 ) ⎞ Q1 Q1 2Q1 ⎛ + = ⎜⎜1 − ⎟⎟ T1 T2 T3 ⎝ T1 ⎠ 1 1 2 2α 2α + = − + T1 T2 T3 T3 T1 T1 2T1 = (1 − α ) + ( 2α − 1) T2 T3 Example 5. Ice is to be made from water supplied at 15°C by the process shown in figure. The final temperature of the ice is -10°C, and the final temperature of the water that is used as cooling water in the condenser is 30°C. Determine the minimum work required to produce 1000 kg of ice. Solution: Quantity of ice produced = 1000 kg. Specific heat of ice = 2070 J/kg K Specific heat of water = 4198 J/kgK Latent heat of ice = 335 kJ/kg 15 + ( −10 ) Mean temperature of ice bath = = 2.5º C 2 15 + 30 = 22.5° C Mean temperature of condenser bath = 2 Q2 = 1000 × 4198(15 - 2.5) + 1000 × 335 × 103 +1000 × 2070(2.5 + 10) = 413.35MJ For a reversible refrigerator system T Q T Q 1− L = 1− 2 ; L = 2 ; TH Q1 TH Q1 Q1 = Q2 TH ⎛ 22.5 + 273 ⎞ = 413.35 × 106 ⎜ ⎟ = 443.36 MJ TL ⎝ 2.5 + 273 ⎠ Minimum work required = Q1 – Q2 = 443.36 - 413.35 = 30.01 MJ. 74 Second Law of Thermodynamics S K Mondal’s Chapter 3 75 Second Law of Thermodynamics S K Mondal’s Chapter 3 ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20-Years IES Questions IES-1. Which one of the following is correct on basis of the second law of thermodynamics? [IES 2007] (a) For any spontaneous process, the entropy of the universe increases (b) ∆S =qrev/T at constant temperature (c) Efficiency of the Stirling cycle is more than that of a Carnot cycle (d) ∆E=q+w (The symbols have their usual meaning) IES-2. Assertion (A): Second law of thermodynamics is called the law of degradation of energy. [IES-1999] Reason (R): Energy does not degrade each time it flows through a finite temperature difference. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-3. Heat transfer takes place according to (a) Zeroth Law of Thermodynamics (c) Second Law of Thermodynamics IES-3a. Consider the following statements: [IES-2010] 1. Slow heating of water from an electric heater. 2. Isentropic expansion of air. 3. Evaporation of a liquid from a heat source at the evaporation temperature. 4. Constant pressure heating of a gas by a constant temperature source. Which of these processes is/are reversible? (a) 3 only (b) 2 and 3 only (c) 2 and 4 only (d) 1, 2, 3 and 4 IES-3b. Consider the following statements: [IES-2010] 1. Boiling of water from a heat source at the same boiling temperature. 2. Theoretical isothermal compression of a gas. 3. Theoretical polytropic compression process with heat rejection to atmosphere. 4. Diffusion of two ideal gases into each other at constant pressure and temperature. Which of these processes are irreversible? (a) 1, 2, 3 and 4 (b) 1 and 4 only (c) 2, 3 and 4 only (d) 3 and 4 only 76 [IES-1996] (b) First Law of Thermodynamics (d) Third Law of Thermodynamics. Second Law of Thermodynamics S K Mondal’s Chapter 3 Kelvin-Planck Statement of Second Law IES-4. Consider the following statements: The definition of 1. Temperature is due to Zeroth Law of Thermodynamics. 2. Entropy is due to First Law of Thermodynamics. 3. Internal energy is due to Second Law of Thermodynamics. 4. Reversibility is due to Kelvin-Planck's statement. Of these statements (a) 1, 2 and 3 are correct (b) 1, 3 and 4 are correct (c) 1 alone is correct (d) 2 alone is correct [IES-1993] Clausius' Statement of the Second Law IES-5. Assertion (A): Heat cannot spontaneously pass from a colder system to a hotter system without simultaneously producing other effects in the surroundings. Reason (R): External work must be put into heat pump so that heat can be transferred from a cold to a hot body. [IES-1999] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true Clausius' Theorem IES-6. A steam power plant is shown in figure, (a) The cycle violates first and second laws of thermodynamics. (b) The cycle does not satisfy the condition of Clausius inequality. (c) The cycle only violates the second laws of thermodynamics (d) The cycle satisfies the Clausius inequality [IES-1992] IES-7. An inventor says that his new concept of an engine, while working between temperature limits of 27°C and 327°C rejects 45% of heat absorbed from the source. His engine is then equivalent to which one of the following engines? (a) Carnot engine (b) Diesel engine [IES-2009] (c) An impossible engine (d) Ericsson engine 77 Second Law of Thermodynamics S K Mondal’s IES-7a Chapter 3 An inventor states that his new engine rejects to the sink 40% of heat absorbed from the source while the source and sink temperatures are 327º C and 27º C respectively. His engine is therefore equivalent to [IES-2010] (a) Joule engine (b) Stirling engine (c) Impossible engine (d) Carnot engine Equivalence of Kelvin-Planck and Clausius Statements IES-8. Assertion (A): Efficiency of a reversible engine operating between temperature [IES-2002] limits T1 and T2 is maximum. Reason (R): Efficiency of a reversible engine is greater than that of an irreversible engine. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true Carnot Engine with same efficiency or same work output IES-9. A reversible engine operates between temperatures T1, and T2, The energy rejected by this engine is received by a second reversible engine at temperature T2 and rejected to a reservoir at temperature T3. If the efficiencies of the engines are same then the relationship between T1, T2 and T3 is given by: [IES-2002] (a) T2 = (T1 + T3 ) 2 (b) T2 = (T 2 1 + T32 ) (c) T2 = T1T3 (d) T2 = (T1 + 2T3 ) 2 IES-10. A reversible engine operates between temperatures 900 K & T2 (T2 < 900 K), & another reversible engine between T2 & 400 K (T2 > 400 K) in series. What is the value of T2 if work outputs of both the engines are equal? [IES-2005] (a) 600 K (b) 625 K (c) 650 K (d) 675 K IES-10a. An engine operates between temperature limits of 900 K and T2 and another between T2 and 400 K. For both to be equally efficient, the value of T2 will be (a) 700 K (b) 600 K (c) 750 K (d) 650 [IES-2010] IES-11. Two reversible engine operate between thermal reservoirs at 1200 K, T2K and 300 K such that 1st engine receives heat from 1200 K reservoir and rejects heat to thermal reservoir at T2K, while the 2nd engine receives heat from thermal reservoir at T2K and rejects heat to the thermal reservoir at 300 K. The efficiency of both the engines is equal. [IES-2004] What is the value of temperature T2? (a) 400 K (b) 500 K IES-12. (c) 600 K (d) 700 K A series combination of two Carnot’s engines operate between the temperatures of 180°C and 20°C. If the engines produce equal amount of work, then what is the intermediate temperature? [IES-2009] 78 Second Law of Thermodynamics S K Mondal’s (a) 80°C Chapter 3 (b) 90°C (c) 100°C (d) 110°C IES-13. An engine working on Carnot cycle rejects 40% of absorbed heat from the source, while the sink temperature is maintained at 27°C, then what is the source temperature? [IES-2009] (a) 750°C (b) 477°C (c) 203°C (d) 67.5°C IES-13a A Carnot engine rejects 30% of absorbed that to a sink at 30º C. The temperature of the heat source is [IES-2010] (a) 100º C (b) 433º C (c) 737º C (d) 1010º C IES-14. A reversible heat engine rejects 50 percent of the heat supplied during a cycle of operation. If this engine is reversed and operates as a heat pump, then what is its coefficient of performance? [IES-2009] (a) 1.0 (b) 1.5 (c) 2.0 (d) 2.5 IES-15. A heat engine is supplied with 250 kJ/s of heat at a constant fixed temperature of 227°C; the heat is rejected at 27°C, the cycle is reversible, then what amount of heat is rejected? [IES-2009] (a) 250 kJ/s (b) 200 kJ/s (c) 180 kJ/s (d) 150 kJ/s IES-16. One reversible heat engine operates between 1600 K and T2 K, and another reversible heat engine operates between T2K and 400 K. If both the engines have the same heat input and output, then the temperature T2 must be equal to: [IES-1993] (a) 1000 (b) 1200 (c) 1400 (d) 800 Previous 20-Years IAS Questions Kelvin-Planck Statement of Second Law IAS-1. Assertion (A): No machine would continuously supply work without expenditure of some other form of energy. [IAS-2001] Reason (R): Energy can be neither created nor destroyed, but it can only be transformed from one form into another. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true Equivalence of Kelvin-Planck and Clausius Statements IAS-2. A heat engine is supplied with 250 KJ/s of heat at a constant fixed temperature of 227°C. The heat is rejected at 27°C. The cycle is reversible, if the amount of heat rejected is: [IAS-1995] (a) 273 KJ/s (b) 200 KJ/s (c) 180 KJ/s (d) 150 KJ/s. 79 Second Law of Thermodynamics S K Mondal’s IAS-3. Chapter 3 A reversible engine En as shown in the given figure draws 300 kcal from 200 K reservoir and does 50 kcal of work during a cycle. The sum of heat interactions with the other two reservoir is given by: (a) Q1 + Q2 = + 250 kcal (b) Q1 + Q2 = –250 kcal (c) Q1 + Q2 = + 350 kcal (d) Q1 + Q2 = –350 kcal [IAS-1996] Carnot Engine with same efficiency or same work output IAS-4. Consider the following statements: [IAS-2007] 1. Amount of work from cascaded Carnot engines corresponding to fixed temperature difference falls as one goes to lower absolute level of temperature. 2. On the enthalpy-entropy diagram, constant pressure lines diverge as the entropy increases. Which of the statements given above is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 IAS-5. In a cyclic heat engine operating between a source temperature of 600°C and a sink temperature of 20°C, the least rate of heat rejection per kW net output of the engine is: [IAS 1994] (a) 0.460 kW (b) 0.505 kW (c) 0.588 kW (d) 0.650 kW Answers with Explanation (Objective) Previous 20-Years IES Answers IES-1. Ans. (a) IES-2. Ans. (c) A is true but R is false. IES-3. Ans. (c) Heat transfer takes place according to second law of thermodynamics as it tells about the direction and amount of heat flow that is possible between two reservoirs. IES-3a. Ans. (b) All spontaneous processes are irreversible. Statement-1 and statement-4 heat is transferred with a finite temperature difference they are irreversible. IES-3b. Ans. (d) Any natural process carried out with a finite temperature gradient is an irreversible process. All spontaneous processes are irreversible. Statement -4 is a spontaneous process. IES-4. Ans. (c) Out of 4 definitions given, only first definition is correct and balance three are wrong. IES-5. Ans. (b) A and R are true. A is the Clausius statement of second law of thermodynamics. Spontaneously means without change in surroundings. If question comes like following then answer will be (a). 80 Second Law of Thermodynamics S K Mondal’s Chapter 3 Assertion (A): External work must be put into heat pump so that heat can be transferred from a cold to a hot body. Reason (R): Heat cannot spontaneously pass from a colder system to a hotter system without simultaneously producing other effects in the surroundings. IES-6. Ans. (d) 300 T = 0.5 IES-7. Ans. (c) Carnot efficiency of engine = η = 1 − 2 = 1 − 600 T1 IES-7a But according to the inventor’s Claim Efficiency of engine = 1-0.45 = 0.55 ∵ Efficiency of Actual Engine cannot be greater then Carnot efficiency. So this is an impossible engine. Ans. (c) We know Carnot efficiency ηcarnot = 1 − T2 T1 ⇒1− 300 1 = = 0.5 600 2 ηcarnot = 50% But inventor claims 60% efficiency (means 40% heat rejection). It is then impossible. IES-8. Ans. (a) IES-9. Ans. (c) IES-10. Ans. (c) Figure from another question W1 = W2 or Q1 − Q2 = Q2 − Q3 or T1 − T2 = T2 − T3 or T2 = IES-10a. Ans. (b) When equally efficiency 1− T T2 = 1− 3 T1 T2 or T2 = T1T3 = 900 × 400 = 600 K IES-11. Ans. (c) η1 = η2 or 1 − T2 300 = 1− 1200 T2 or T2 = 1200 × 300 = 600K 81 T1 + T3 900 + 400 = = 650K 2 2 Seco ond La aw of Therm modyn namicss S K Mondal’s Cha apter 3 IES-12. Ans. (c c) Sou urce Tempera ature = T1, Inte ermediate Temperature = T Sin nk Temperatu ure = T2 W1 = W2 ∵ ⎛ T⎞ ⎛ T ⎞ Q1 ⎜ 1 − ⎟ = Q2 ⎜ 1 − 2 ⎟ T⎠ ⎝ ⎝ T1 ⎠ T1 ⎛ T1 T2 T ⎞ ⎛ T2 ⎞ ⎜1− ⎟ = ⎜1− ⎟ ⇒ − 1 = 1− T ⎝ T1 ⎠ ⎝ T⎠ T T T1 + T2 T + T2 180 + 20 0 =2 = = 100°C ⇒ ⇒T = 1 2 2 T IES-13. Ans. (b b) Sink temp perature = 27 7°C = 27 + 27 73 = 300K It is given that engine rejeccts 40% of ab bsorbed heat from the sou urce Q Q Forr a carnot cyccle engine 1 = 2 T2 T1 ⇒ Q 0.4 4Q 30 00 = ⇒T= = 750K = 477°C T 30 00 0..4 IES-13a Ans. (c) Q1 Q2 = T1 T2 or T1 = T2 × Q1 Q = 30 03 × 1 = 1010 K = 737 o C Q2 0.3Q1 Q1 Q2 = T1 T2 Q 0.5 Q1 ⇒ 1 = T1 T2 IES-14. Ans. (c c) ⇒ T2 = 0.5 T1 If the engine is reversed an nd operated as a the Heat Pum mp. Theen COP coeffficient of perfformance = T1 = T1 − T2 1 1 = =2 T2 5 1 − 0.5 1− T1 IES-15. Ans. (d d) Heat supp plied by the Heat H Engine = Q1 = 250 kJ/sec k Sou urce tempera ature = 227°C = 500 K Sin nk temperatu ure = 27°C = 300K 3 82 Second Law of Thermodynamics S K Mondal’s Chapter 3 250 Q2 = ⇒ Q2 = 250 × 0.6 = 150kJ / sec 500 300 IES-16. Ans. (d) Two reversible heat engines operate between limits of T2 and 400K 1600K and T2; Both have the same heat input and output, T1 − T2 1600 − T2 T2 − 400 is same for both or = or T2 = 800 K T1 1600 T2 i.e. Previous 20-Years IAS Answers IAS-1. Ans. (a) IAS-2. Ans. (d) IAS-3. Ans. (b) Q1 Q2 = T1 T2 ∑Q = ∑W 300 + Q1 + Q2 = 50 IAS-4. Ans. (b) For reversible cycle T1 T2 T3 = = Q1 Q2 T3 or T1 − T2 Q1 − Q2 = T2 Q2 or T1 − T2 = (Q1 − Q2 ) × Similarly T2 Q2 T2 − T3 = ( Q2 − Q3 ) × T3 Q3 If T1 − T2 = T2 − T3 then Q1 − Q2 = Q2 − Q3 or W1 = W2 IAS-5. Ans. (b) Reversible engine has maximum efficiency where Therefore least heat rejection per kW net output, W 1 Q2 = × T2 = × 293 = 0.505 kW T1 − T2 873 − 293 83 Q1 Q2 Q1 − Q2 W = = = T1 T2 T1 − T2 T1 − T2 Enttropy S K Mondal’s 4. Cha apter 4 Entro opy Theo ory at a Glance (Fo or GAT TE, IES S & PS SUs) Tw wo Reve ersible Adiabati A c Paths cannot Intersec ct Each Other Let it be assum med that two o reversible adiabatics a AC and BC in ntersect each h other at point C (Fig.). A be drawn in such a wa ay that it inteersects the reeversible adiabatics at A Let a reversiblee isotherm AB e processes AB, A BC, and CA C together constitute c a reversible cy ycle, and the and B. The three reversible r th he net work output in a cycle. But such s a cycle is impossiblle, since net areea included represents woork is being produced p in a cycle by a heat h engine by b exchangin ng heat with a single reseervoir in the proocess AB, which w violattes the Kelv vin-Planck statement s o the secon of nd law. Theerefore, the asssumption of the intersection of the reversible r ad diabatics is Fig. F wrong. Through T one point, there can n pass only one o reversiblee adiabatic. nce two consstant propertty lines can never intersect each otther, it is infferred that a reversible Sin adiiabatic path must represe ent some prooperty, which h is yet to be identified. i 83 Entropy S K Mondal’s Chapter 4 The Property of Entropy ⎛ dQ ⎞ ds = ⎜ ⎟ ⎝ T ⎠ • • Reversible dS is an exact differential because S is a point function and a property. The subscript R in dQ indicates that heat dQ is transferred reversibly. S s = J / kg K m • It is an extensive property, and has the unit J/K. The specific entropy is an intensive property and has unit J/kgK • The change of entropy may be regarded as a measure of the rate of availability of heat for transformation into work. If the system is taken from an initial equilibrium state i to a final equilibrium state f by an irreversible path, since entropy is a point or state function, and the entropy change is independent of the path followed, the non-reversible path is to be replaced by a reversible path to integrate for the evaluation of entropy change in the irreversible process Sf − Si = ∫ i f dQrev = ( ΔS )irrev path T Integration can be performed only on a reversible path. 84 Enttropy S K Mondal’s Cha apter 4 Te empera ature-Entropy Plot The Ineq quality of Clau usius Th hen for any cy ycle ∫ dQ ≤ T ∫ ds Sin nce entropy is i a property and the cycliic integral off any property is zero ∫ dQ T ≤ 0 Th his equation is i known as the t inequalitty of Clausiu us. It providess the criterion of the reverrsibility of a cyccle. If ∫ ∫ dQ = 0 , the cyclle is reversibble, T d dQ < 0, the cycle isi irreversibble and possible T 85 Entropy S K Mondal’s ∫ Chapter 4 dQ > 0 , the cycle is impossible, since it violates the second law. T Entropy Change in an Irreversible Process Flow of current through a resistance – when a battery discharges through a resistance heat is dissipated. You can’t recharge the battery by supplying heat back to the resistance element!! Pickpocket !!!Marriage!!!!.............................................are irreversible Process. Applications of Entropy Principle (S1-S2)irreversible > (S1-S2)reversible An irreversible process generates more entropy than a reversible process. An irreversible engine can’t produce more work than a reversible one. • An irreversible heat pump will always need more work than a reversible heat pump. • An irreversible expansion will produce less work than a reversible expansion • An irreversible compression will need more work than a reversible compression. 86 Enttropy S K Mondal’s Cha apter 4 Maximum Work W Obtainable fr rom two Finite F Bod dies at tem mperature es T1 and T2 Let us consideer two inden ntical finite bodies b of con nstant heat capacity at temperaturee T1 and T2 resspectively, T1 being high her than T2. If the two bodies b are meerely brough ht together in nto thermal con ntact, deliverring no work,, the final tem mperature Tf reached wou uld be the maximum T + T2 Tf = 1 2 ne is operated d between th he two bodies acting as thermal t enerrgy reservoirrs (shown in If a heat engin Fig g. below), parrt of the hea at withdrawn n from body 1 is converted to work W by the heat engine, and thee remainder is rejected to t body 2. Th he lowest atttainable fina al temperaturre Tf corresp ponds to the dellivery of the largest possiible amount of o work, and is associated d with a reverrsible process. A As work is deelivered by th he heat engin ne, the temp perature of body 1 will bee decreasing and that of bod dy 2 will be increasing. i When W both th he bodies atta ain the final temperaturee Tf, the heatt engine will stoop operating. Let the bodiies remain att constant pressure and undergo u no ch hange of phase. Fig. Total heat with hdrawn from body 1 Q1 = Cp ( T1 − Tf ) Wh here Cp is thee heat capaciity of the twoo bodies at coonstant presssure. Total heat rejeccted to body 2 Q2 = Cp (Tf – T2) ∴A Amount of tootal work dellivered by thee heat enginee W = Q1 – Q2 = Cp (T1 + T2 - 2T Tf) For given valuees of Cp, T1 and a T2, the magnitude m off work W dep pends on Tf. Work obtain nable will be ma aximum when n Tf is minim mum. Noow, for body 1, 1 entropy ch hange ΔS1 is given g by T Tf dT f ΔS1 = ∫ C p = C pln T1 T1 T For body 2, enttropy change ΔS2 would be Tf Tf dT Δ S2 = ∫ C p = C pln T2 T T2 nce the work king fluid ope erating in thee heat engine cycle does not undergo any entropy y change, ΔS Sin of tthe working fluid in heat engine = ∫ dS d = 0 . Applying the entrropy principlle 87 Enttropy S K Mondal’s Cha apter 4 ΔSuniv ≥ 0 ∴ C pln Tf T1 C pln + C pln Tf2 T1T2 Tf T2 ≥0 ≥0 nimum From abovee Eq. for Tf to be a min C pln n Tf2 T1 .T2 =0 Tf2 = 0 = ln 1 or ln ∴ Tf = T1 .T2 T1T2 For W to be e a maximum m, Tf will be T1 .T2 . From m abow Equattion Wmax = C p (T1 + T2 − 2 T1T2 ) = C p ( T1 − T2 )2 Th he final temp peratures of the t two bodiees, initially at a T1 and T2, can range from f (T1 + T2)/2 with no dellivery of work k to T1 .T2 with w maximu um delivery of o work. • Maximu um Work Obtainable from a Finite Bo ody and a TER :- Let one of the bodies consiidered in thee previous seection be a thermal t enerrgy reservoirr. The finite bod dy has a therrmal capacity y Cp and is at a temperaturre T and the TER is at teemperature T0, such that T > T0. Let (Q - W). Then F Fig. – Ma aximum Worrk Obtainable e When One of the Bodiess is a TER. 88 Enttropy S K Mondal’s Cha apter 4 T0 ΔSBody = ∫ C p T T dT = C pln n 0 T T ΔSHE = ∫ dss = 0 ΔSTER = Q −W To T0 Q − W + T To By the entropy principle,. ∴ ΔSuniv = C pln l ΔSuniv ≥ 0 C pln T0 Q − W ≥0 + T To or C pln T0 W − Q ≥ To T or T W −Q ≤ C pln l 0 T To or W ≤ Q + T0C pln ∴ Wmax = Q + T0C pln or ⎡ T⎤ Wmax = C p ⎢(T − T0 ) − T0ln n ⎥ To ⎦ ⎣ To T To T • Processses Exhibiiting Exte ernal Mec chanical Ir rreversibility :on of Work :: Let us con nsider the isoothermal disssipation of work w through [i] Isothermall Dissipatio a ssystem into the internall energy of a reservoir, as a in the flow of an elecctric current through a ressistor in conttact with a re eservoir (Fig.in below.) At A steady statte, the intern nal energy of the resistor and hence its teemperature is i constant. So, S by first la aw: W=Q he flow of currrent represents work trransfer. At steady s state the work is dissipated isothermally Th intto heat transsfer to the surroundings. Since the surrroundings absorb a Q unitt of heat at temperature t T, 89 Enttropy S K Mondal’s Cha apter 4 (Fig.. ) Q W = T T At stea ady state, ΔSsys = 0 ΔSsurr = W T he irreversible process is thus t accompa anied by an entropy e increease of the un niverse. Th ∴ ΔSuni u v = ΔSsys + ΔSsurr = [ii]] Adiabatic c Dissipation of Work :: Let W be the t stirring work w supplied to a viscou us thermally inssulated liquid, which is dissipated adiabatically a into interna al energy inccrease of thee liquid, the tem mperature off which incre eases from Ti to Tf ( show wn in fig beloow). Since th here is no flow w of heat to or from the surrroundings. ( (Fig. ) To calculate th he entropy ch hange of the system, thee original irreversible patth (dotted lin ne) must be rep placed by a reversible r on ne between the t same end d states, i an nd f. Let us replace the irreversible perrformance off work by a reversible r isoobaric flow off heat from a series of resservoirs rang ging from Ti to Tf to cause th he same chan nge in the sta ate of the sysstem. The entropy changee of the systeem will be C p dT Tf dQ ΔSsys = ∫ f =∫f = C p ln T Ti Ri T Ri here Cp is thee heat capaccity of the liq quid. wh ΔSuniv = ΔSsys + ΔSsurr = C pln n Tf Ti Entropy Genera ation • Irreversible Processses increase the t entropy of o the universse. • Reversiible Processe es do not effecct the entrop py of the univ verse. • Impossible Processe es decrease th he entropy off the universse. ΔSuniverse = 0 90 Entropy S K Mondal’s Chapter 4 Entropy Generation in the universe is a measure of lost of work. ΔS Universe = ΔS System + ΔS Surroundings The losses will keep increasing. The sin keeps accumulating. Damage to environment keeps increasing. When the entropy of the universe goes so high, then some one has to come and set it right. HE SAYS HE WILL COME. Every religion confirms this. Let us all wait. Cheer up, things are not that bad yet!! Entropy and Direction: The Second Law a Directional law of Nature The second law indicates the direction in which a process takes place. A process always occurs in such a direction as to cause an increase in the entropy of the universe. 91 Entropy S K Mondal’s Chapter 4 Summary 1. Clausius theorem: 2. S f − Si = f ∫ i ⎛ dQ ⎞ =0 T ⎟⎠ rev. ∫ ⎜⎝ dQ rev = ( Δ s ) irrev . path . T Integration can be performed only on a reversible path. 4. dQ ≤0 T At the equilibrium state. The system is at the peak of the entropy hill. (isolated) 5. Tds = du + Pdv 6. Tds = dh – Vdp 7. Famous relation S = K lnW 3. 8. Clausius Inequality: ∫ where K =Boltzmann constant W = thermodynamic probability General case of change of entropy of a Gas. ⎧ P V ⎫ S2 − S1 = m ⎨Cv ln 2 + CP ln 2 ⎬ P1 V1 ⎭ ⎩ Initial condition of gas P1, V1, T1, S1 and Final condition of gas P2, V2, T2, S2 9. Process and property change Table: 92 Entropy S K Mondal’s Chapter 4 PROBLEMS & SOLUTIONS Example 1. 5 kg of air is compressed in a reversible polytrophic process from 1 bar and 40°C to 10 bar with an index of compression 1.25. Calculate the entropy change during the process. Solution:T2 ⎛ p2 ⎞ =⎜ ⎟ T1 ⎝ p1 ⎠ n −1 n n −1 0.25 ⎛p ⎞n ⎛ 10 ⎞ 1.25 = ( 273 + 40 ) ⎜ ⎟ T2 = T1 ⎜ 2 ⎟ p ⎝1 ⎠ ⎝ 1⎠ Therefore T2 = 496 K 2 2 2 2 dT vdp dT dp −∫ = Cp ∫ − R∫ T T T p 1 1 1 1 φ2 − φ1 = C p ∫ ⎛T ⎞ ⎛p ⎞ =1.005ln ⎜ 2 ⎟ − R ln ⎜ 2 ⎟ T ⎝ 1⎠ ⎝ p1 ⎠ 496 = 1.005ln − 0.287 ln(10) 313 = 0.4627 − 0.6608 = − 0.1981 kJ / kg K . Total change in entropy = 5 × (-0.1981) = - 0.9905 kJ/K (Reduction). (Hence heat is rejected during the process). Example 2. Two compartments of an insulated vessel each of 3 m3 contain air at 0.7 MPa, 95°C and 0.35 MPa, 205°C. If the removed, find the change in entropy, if the two portions mix completely and adiabatically. Solution: 93 Entropy S K Mondal’s Chapter 4 m1 = p1V1 0.7 ×106 × 3 = = 19.883 kg RT1 287 × ( 273 + 95 ) m2 = p2V2 0.35 ×106 × 3 = = 7.654 kg RT2 287 × ( 273 + 205 ) Assuming specific heat to be constant Tf = Cv (m1T1 + m2T2 ) Cv (m1 + m2 ) ΔS1 = m1Cv ln Tf T1 = 398.6 K = 125.6°C = 19.883 × 1005 × ln 398.6 = 1595J / K 368 398.6 = − 1396.6J / K 478 ΔS = 1595 − 1397 = 198J / K ΔS2 = 7.654 ×1005 × ln 94 Entropy S K Mondal’s Chapter 4 ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20-Years GATE Questions Applications of Entropy Principle GATE-1. A 1500 W electrical heater is used to heat 20 kg of water (Cp = 4186 J/kg K) in an insulated bucket, from a temperature of 30°C to 80°C. If the heater temperature is only infinitesimally larger than the water temperature during the process, the change in entropy for heater is….. J/k and for water ............. J/K. [GATE-1994] Entropy Generation in a Closed System GATE-1A An ideal gas of mass m and temperature T1 undergoes a reversible isothermal process from an initial pressure P1 to final pressure P2. The heat loss during the process is Q. The entropy change ΔS of the gas is ⎛P ⎞ ⎛P⎞ ⎛P ⎞ Q (a) mR ln ⎜ 2 ⎟ (b) mR ln ⎜ 1 ⎟ (c) mR ln ⎜ 2 ⎟ − (d ) zero [GATE-2012] P P ⎝ 1⎠ ⎝ 2⎠ ⎝ P1 ⎠ T1 Data for Q2 and Q3 are given below. Solve the problems and choose correct answers. 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 2m3. Heat exchange occurs with the atmosphere at 298 K during this process. GATE-2. The work interaction for the Nitrogen gas is: (a) 200 kJ (b) 138.6 kJ (c) 2 kJ [GATE-2003] (d) –200 kJ GATE-3. The entropy change for the Universe during the process in kJ/K is: [GATE-2003] (a) 0.4652 (b) 0.0067 (c) 0 (d) –0.6711 GATE-4. If a closed system is undergoing an irreversible process, the entropy of the system [GATE-2009] (a) Must increase (b) Always remains constant (c) Must decrease (d) Can increase, decrease or remain constant Entropy and Direction: The Second Law a Directional law of Nature GATE-5. One kilogram of water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is: (a) Equal to entropy change of the reservoir [GATE-2010] 95 Entropy S K Mondal’s Chapter 4 (b) Equal to entropy change of water (c) Equal to zero (d) Always positive Common Data for Questions GATE-6 and GATE-7: In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K. The following are the properties and relations pertaining to air: Specific heat at constant pressure, cp = 1.005 kJ/kg K; Specific heat at constant volume, cv = 0.718 kJ/kg K; Characteristic gas constant, R = 0.287 kJ/kg K. Enthalpy, h = cpT Internal energy, u = cvT GATE-6. If the pressure at station Q is 50 kPa, the change in entropy (sQ – sP ) in kJ/kg K is (a) – 0.155 GATE-7. (b) 0 (c) 0.160 [GATE-2011] (d) 0.355 If the air has to flow from station P to station Q, the maximum possible value of pressure in kPa at station Q is close to [GATE-2011] (a) 50 (b) 87 (c) 128 (d) 150 Previous 20-Years IES Questions Two Reversible Adiabatic Paths cannot Intersect Each Other IES-1. The relation ds = dQ , where s represents entropy, Q represents heat and T T represents temperature (absolute), holds good in which one of the following processes? [IES-2009] (a) Reversible processes only (b) Irreversible processes only (c) Both reversible and irreversible processes (d) All real processes IES-2. IES-2a. Which of the following statement is correct? [IES-2008] (a) The increase in entropy is obtained from a given quantity of heat transfer at a low temperature. (b) The change in entropy may be regarded as a measure of the rate of the availability of heat for transformation into work. (c) The entropy represents the maximum amount of work obtainable per degree drop in temperature (d) All of the above A heat engine receives 1000 kW of heat at a constant temperature of 285°C and rejects 492 kW of heat at 5°C. Consider the following thermodynamic cycles in this regard: [IES-2000] 1. Carnot cycle 2. Reversible cycle 3. Irreversible cycle 96 Entropy S K Mondal’s Chapter 4 Which of these cycles could possible be executed by the engine? (a) 1 alone (b) 3 alone (c) 1 and 2 (d) None of 1, 2 and 3 The Property of Entropy IES-3. Assigning the basic dimensions to mass, length, time and temperature respectively as M, L, T and θ (Temperature), what are the dimensions of entropy? [IES-2007] (a) M LT-2 θ (b) M L2 T-1 θ-1 (c) M L2 T-2θ-1 (d) M L3T-2 θ -1 IES-4. A Carnot engine operates between 327°C and 27°C. If the engine produces 300 kJ of work, what is the entropy change during heat addition? [IES-2008] (a) 0.5 kJ/K (b) 1.0 kJ/K (c) 1.5 kJ/K (d) 2.0 kJ/K Temperature-Entropy Plot IES-4a Isentropic flow is [IES-2011] (a) Irreversible adiabatic flow (c) Ideal fluid flow IES-5. (b) Reversible adiabatic flow (d) Frictionless reversible flow A system comprising of a pure substance executes reversibly a cycle 1 -2 -3 -4 -1 consisting of two isentropic and two isochoric processes as shown in the Fig. 1. Which one of the following is the correct representation of this cycle on the temperature – entropy coordinates? [IES-2002] 97 Enttropy S K Mondal’s IES-6. Cha apter 4 A cycle of pressure p – volume dia agram is sh hown in th he given Fig g. I, Same S cyc cle on tem mperature-e entropy diagram willl be represe ented by: [IES-1995] IES-7. An ideal air standard cycle c is given sho own in the tem mperature-e entropy diagram. [IES-1997] 98 Enttropy S K Mondal’s Cha apter 4 The e same cyc cle, when re epresented on the pre essure-volum me coordin nates takes the e form IES-8. Ma atch figures s of Colum mn-I with th hose given in Column n-II and se elect given bellow the colu umns: [IES-1994] Collumn-I (p-v diagram) Colum mn-II (T-s diagram) Cod des: (a) (c) IES-9. A 1 3 B 2 1 C 3 2 (b) (d) A 2 3 B 3 2 C 1 1 A cyclic c proce ess ABCD shown s in the e V-T diagra am perform med with a constant c ma ass of an id deal gas. The e process of o p-V diagr ram will be as shown in n [IES-1992] 99 Enttropy S K Mondal’s IES-10. Thr ree process ses are repr resented on n the p-v an nd T-s diagr rams in the e following figu ures. Match h processess in the tw wo diagramss and selec ct the corre ect answer usiing the code es given bellow the diag grams: [IES-1994] Cod des: (a) (c) IES-11. Cha apter 4 A 1 3 B 2 2 C 3 1 (b) (d) A 2 1 B 3 3 C 1 2 Tw wo polytropiic processes undergon ne by a perffect gas are e shown be elow in the pre essure-volum me co-ordin nates. [IES-2008] Wh hich represe entation sho ows correcttly the abov ve processess on the tem mperature– enttropy co-ord dinates? 100 0 Entropy S K Mondal’s IES-12. Chapter 4 Assertion (A): If a graph is plotted for absolute temperature as a function of entropy, the area under the curve would give the amount of heat supplied. Reason (R): Entropy represents the maximum fraction of work obtainable from heat per degree drop in temperature. [IES-1998] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true The Inequality of Clausius IES-13. For real thermodynamic cycle: (a) IES-14. ∫ ∫ (c) ∫ dQ =0 T δQ T =0 (b) ∫ δQ T <0 (c) ∫ δQ T >0 For an irreversible cycle: (a) IES-16. (b) [IES-2005] dQ <0 T (d) For a thermodynamic cycle to be irreversible, it is necessary that (a) ∫ IES-15. dQ > 0 but < ∞ T ∫ dQ ≤0 T (b) (d) ∫ dQ =∞ T [IES-1998] ∫ δQ T ≥0 [IES-1994, 2011] ∫ dQ >0 T (c) ∫ dQ <0 T (d) ∫ dQ ≥0 T If a system undergoes an irreversible adiabatic process, then (symbols have usual meanings) [IES-1997] dQ = 0 and ΔS > 0 T dQ (c) ∫ > 0 and ΔS = 0 T (a) dQ = 0 and ΔS = 0 T dQ (d) ∫ < 0 and ΔS < 0 T ∫ (b) 101 ∫ Entropy S K Mondal’s Chapter 4 Entropy Change in an Irreversible Process IES-17. Consider the following statements: [IES-1998] In an irreversible process 1. Entropy always increases. 2. The sum of the entropy of all the bodies taking part in a process always increases. 3. Once created, entropy cannot be destroyed. Of these statements (a) 1 and 2 are correct (b) 1 and 3 are correct (c) 2 and 3 are correct (d) 1, 2 and 3 are correct IES-18. Consider the following statements: [IES-1997] When a perfect gas enclosed in a cylinder piston device executes a reversible adiabatic expansion process, 1. Its entropy will increase 2. Its entropy change will be zero 3. The entropy change of the surroundings will be zero Of these statements (a) 1 and 3 are correct (b) 2 alone is correct (c) 2 and 3 are correct (d) 1 alone is correct IES-19. A system of 100 kg mass undergoes a process in which its specific entropy increases from 0.3 kJ/kg-K to 0.4 kJ/kg-K. At the same time, the entropy of the surroundings decreases from 80 kJ/K to 75 kJ/K. The process is: [IES-1997] (a) Reversible and isothermal (b) Irreversible (c) Reversible (d) Impossible IES-20. Which one of the following temperature entropy diagrams of steam shows the reversible and irreversible processes correctly? [IES-1996] Applications of Entropy Principle IES-21. A Carnot engine operates between 27°C and 327°C. If the engine produces 300 kJ of Work, What is the entropy change during heat addition? [IES-2005] 102 Entropy S K Mondal’s Chapter 4 (a) 0.5 kJ/K (b) 1.0 kJ/K (c) 1.5 kJ/K (d) 2.0 kJ/K IES-22. The entropy of a mixture of ideal gases is the sum of the entropies of constituents evaluated at: [IES-2005] (a) Temperature and pressure of the mixture (b) Temperature of the mixture and the partial pressure of the constituents (c) Temperature and volume of the mixture (d) Pressure and volume of the mixture IES-23. The heat added to a closed system during a reversible process is given by Q = αT + β T 2 , where α and β are constants. The entropy change of the system as its temperature changes from T1 to T2 is equal to: [IES-2000] β ⎡ ⎤ (b) ⎢α (T2 − T1 ) + T22 − T12 ⎥ / T1 2 ⎣ ⎦ ( (a) α + β (T2 − T1 ) ⎛ T2 ⎞ ⎟ + 2β (T2 − T1 ) ⎝ T1 ⎠ β ⎡α ⎤ (c) ⎢ T22 − T12 + T23 − T13 ⎥ / T12 2 ⎣2 ⎦ ( ) ( ) ) ( d ) α ln ⎜ IES-24. One kg of air is subjected to the following processes: [IES-2004] 1. Air expands isothermally from 6 bar to 3 bar. 2. Air is compressed to half the volume at constant pressure 3. Heat is supplied to air at constant volume till the pressure becomes three fold In which of the above processes, the change in entropy will be positive? (a) 1 and 2 (b) 2 and 3 (c) 1 and 3 (d) 1, 2 and 3 IES-25. A reversible heat engine receives 6 kJ of heat from thermal reservoir at temperature 800 K, and 8 kJ of heat from another thermal reservoir at temperature 600 K. If it rejects heat to a third thermal reservoir at temperature 100 K, then the thermal efficiency of the engine is approximately equal to: [IES-2002] (a) 65% (b) 75% (c) 80% (d) 85% IES-26. A reversible engine exceeding 630 cycles per minute drawn heat from two constant temperature reservoirs at 1200 K and 800 K rejects heat to constant temperature at 400 K. The engine develops work 100kW and rejects 3200 KJ heat per minute. The ratio of heat drawn from two reservoirs (a) 1 IES-27. (b) 1.5 (c) 3 Q1200 is nearly. Q800 [IES-1992] (d) 10.5 In which one of the following situations the entropy change will be negative (a) Air expands isothermally from 6 bars to 3 bars [IES-2000] (b) Air is compressed to half the volume at constant pressure (c) Heat is supplied to air at constant volume till the pressure becomes three folds (d) Air expands isentropically from 6 bars to 3 bars 103 Enttropy S K Mondal’s Cha apter 4 Entropy and Direction n: The Second Law a Direc ctional aw of Nature la IES-28. IES-29 A mass m M of a fluid at tem mperature T1 is mixed d with an eq qual mass of o the same fluiid at tempe erature T2. The T resultan nt change in n entropy of o the universe is: [IES-1992] (a) Zero (b) Negliigible (c) Alwayss negative (d) Always positive p Increase in n entropy off a system represents r (a) Increase in availabiliity of energy (b) Increasse in tempera ature e in pressure (d) Degrad dation of energy (c) Decrease [IES-2011] Previo ous 20 0-Years s IAS Quest Q ions Tw wo Reve ersible Adiabati A c Paths cannot Intersec ct Each Other IA AS-1. Wh hich one of the t followin ng is the cor rrect statem ment? Tw wo adiabatic c will: (a) Intersect at absolute zerro temperature (b) Never interssect (c) Become orth hogonal at ab bsolute zero temperature t (d) Become parallel at absollute zero tem mperature [IAS-2007] The Prop perty of o Entro opy IA AS-2. Heat flows be etween two o reservoirss having te emperature es 1000 K and a 500 K, resspectively. If I the entrop py change of o the cold reservoir r iss 10 kJ/K, th hen what is the e entropy ch hange for th he hot reser rvoir? [IAS-2004] (a) –10 kJ/K (b) –5 kJ/K (c)) 5 kJ/K (d) 10 kJ/K Te empera ature-Entropy Plot IA AS-3. An ideal cyclle is shown n in the giv ven pressure e-volume diiagram: [IAS-1997] 104 4 Enttropy S K Mondal’s Cha apter 4 The e same cyclle on temperature-entr ropy diagram will be re epresented as: IA AS-4. The e therma al efficien ncy of t the hyp pothetical heat h engine e cycle show wn in the t given figure is: (a) 0.5 (b) 0.45 0 (c) 0.35 (d) 0.25 [IAS-2000] AS-5. IA Wh hich one of the followiing pairs be est expresses a relatio onship similar to that exp pressed in the pair r “pressure e-volume” for a the ermodynam mic system und dergoing a process? [IAS-1995] (a) Enthalpy-en ntropy (b) Pressu ure-enthalpy P mperature (d) Tempeerature-entroopy (c) Pressure-tem IA AS-6. An ideal gas contained c in n a rigid ta ank is cooled c such h that T2 < and P2 <P1 In the e given ure entro opy temperatu dia agram, th his processs path is rep presented by the line la abelled. (a) A (b) B (c) C (d) D [IAS-1999] 105 5 Enttropy S K Mondal’s IA AS-7. Cha apter 4 In the T-S diiagram shown in the figu ure, which one of the e following is represented r d by the area a under the e curve? (a) Total work done d during the process (b) Total heat absorbed during the process d the (c) Total heatt rejected during process (d) Degree of irreversibility [IAS-2004] The Ineq quality of Clau usius IA AS-8. Cla ausius inequ uality is sta ated as (a) IA AS-9. ∫ δQ < 0 (b) [IAS-2001] ∫ δQ = 0 (c)) Q ∫δ T >0 (d) For r real therm modynamic cycle: (a) ∫ dQ > 0 butt < ∞ T (b) ∫ Q ∫δ T ≤0 [IAS-2003] dQ <0 T (c)) ∫ dQ =0 T (d) ∫ dQ =∞ T IA AS-10(i). If a system undergoes an a irreversiible adiabatic processs, then (sym mbols have usu ual meaning gs) [IAS-1999] dQ = 0 an nd ΔS > 0 T dQ (c) ∫ > 0 an nd ΔS = 0 T (a) ∫ dQ = 0 and ΔS = 0 T dQ (d) ∫ < 0 and ΔS < 0 T (b) ∫ IA AS-10(ii). A cyclic c heat engine rece eives 600 kJ of heat fr rom a 1000 K source and a rejects 450 0 kJ to a 300 3 K sink. The quanttity resspectively (a) 2.1 kJ/K and d 70% nd 70% (c) + 0.9 kJ/K an ∫ dQ a and efficien ncy of the engine e are T (b)) –0.9 kJ/K and a 25% (d)) –2.1 kJ/K and a 25% [IAS-2001] A Applicattions off Entrop py Prin nciple IA AS-11. Wh hich one of the t followin ng statemen nts is not co orrect? [IAS-2003] (a) Change in entropy e durin ng a reversiblle adiabatic process p is zerro (b) Entropy increases with the t addition of heat ansion processs (c) Throttling is a constant entropy expa hen a gas is heated under u consta ant pressure given by (d) Change in entropy wh s2 − s1 = mC C p log e IA AS-12. T2 T1 Asssertion (A): Entropy ch hange for a reversible adiabatic a pr rocess is zero. 106 6 Entropy S K Mondal’s Chapter 4 Reason (R): There is no heat transfer in an adiabatic process. [IAS 1994] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true Entropy Generation in a Closed System IAS-13. 1600 kJ of energy is transferred from a heat reservoir at 800 K to another heat reservoir at 400 K. The amount of entropy generated during the process would be: [IAS-2000] (a) 6 kJ/k (b) 4 kJ/k (c) 2kJ/k (d) Zero IAS-14. An electric motor of 5 kW is subjected to a braking test for 1 hour. The heat generated by the frictional forces in the process is transferred to the surroundings at 20°C. The resulting entropy change will be: [IAS-1998] (a) 22.1 kJ/K (b) 30.2 kJ/K (c) 61.4 kJ/K (d) 82.1 kJ/K Entropy and Direction: The Second Law a Directional law of Nature IAS-15. M1 kg of water at T1 is isobarically and adiabatically mixed with M2 kg of water at T2 (T1 > T2). The entropy change of the universe is: [IAS-2004] (a) Necessarily positive (b) Necessarily negative (c) Always zero (d) Negative or positive but not zero IAS-16. In which one of the following processes is there an increase in entropy with no degradation of energy? [IAS-1996] (a) Polytropic expansion (b) Isothermal expansion (c) Isochoric heat addition (d) Isobaric heat addition 107 Entropy S K Mondal’s Chapter 4 Answers with Explanation (Objective) Previous 20-Years GATE Answers GATE-1. Ans. –11858 J/K, 12787 J/K. GATE-1A Ans. (b) ⎛υ ⎞ ⎛υ ⎞ ⎛2⎞ GATE-2. Ans. (b) w1− 2 = mRT In ⎜ 2 ⎟ = pυ In ⎜ 2 ⎟ = 200 × 1× In ⎜ ⎟ kJ = 138.6 kJ ⎝ 1⎠ ⎝ υ1 ⎠ ⎝ υ1 ⎠ GATE-3. Ans. (c) It is reversible process so ( ΔS )universe = 0 GATE-4. Ans. (d) GATE-5. Ans. (d) It is a case of spontaneous process i.e. irrepressibility involved that so why entropy change of the universe is positive. GATE-6. Ans. (c) TQ = 300 K, PQ = 50 kPa TP = 350 K, PP = 150 kPa T P SQ − SP = c P ln Q − R ln Q TP PP ⎛ 300 ⎞ ⎛ 50 ⎞ SQ − SP = 1.005 ln ⎜ ⎟ − 0.287 ln ⎜ ⎟ ⎝ 350 ⎠ ⎝ 150 ⎠ SQ − SP = 0.160 kJ/kg-K GATE-7. Ans. (b) If air has to flow from station P to station Q adiabatically means no entropy change in surroundings, then P Q SQ − SP ≥ 0 ⇒ c P ln ⇒ ⇒ ⇒ TQ TP − R ln PQ PP ≥0 ⎛ PQ ⎞ ⎛ 300 ⎞ 1.005 ln ⎜ ⎟ − 0.287 ln ⎜ ⎟≥0 ⎝ 350 ⎠ ⎝ 150 ⎠ ⎛ P ⎞ −0.287 ln ⎜ Q ⎟ ≥ 0.15492 ⎝ 150 ⎠ PQ ≤ 87.43 So, Ans. is PQ = 87 kPa 108 ⇒ ⇒ ⎛ P ⎞ −0.15492 − 0.287 ln ⎜ Q ⎟ ≥ 0 ⎝ 150 ⎠ ⎛ P ⎞ ln ⎜ Q ⎟ ≤ − 0.53979 ⎝ 150 ⎠ Enttropy S K Mondal’s Cha apter 4 Previious 20-Yearrs IES Answ wers ⎛ dQ ⎞ ⎟ ⎝ T ⎠R Rev IES-1. Ans. (a)) Remember ds = ⎜ IES-2. Ans. (b b) The chang ge of entropy y may be regarded as a measure m of th he rate of av vailability of hea at for transformation intoo work. IES-2a Ans. (b b) S gen = − 1000 492 + = −0.02233kkW / K (285 ( + 273) (5 + 273) So cycle is im mpossible Cy ycle See in both the case Carrnot Cycle an nd Reversiblee cycle entrop py change of the t Universe wiill be zero. Irreversible cycle entrop py change willl be positive. IES-3. Ans. (c)) IES-4. Ans. (b) Q1 Q1 Q1 − Q2 W = = = T1 T1 T1 − T2 T1 − T2 Q1 = 600kJ Thee entropy ch hange durin ng heat add dition Q1 600 = = = 1 kJ k /K T1 600 IES-4a Ans. (b) IES-5. Ans. (c)) IES-6. Ans. (b) IES-7. Ans. (b) IES-8. Ans. (c)) IES-9. Ans. (d) AB constan nt pressure heat h addition. IES-10. Ans. (c) XA consstant pressu ure heat rejeection. XB = const. temp p. heat rejecction. XC = isen ntropic heat rejection. r IES-11. Ans. (b b) IES-12. Ans. (b b) IES-13. Ans. (b b) b) IES-14. Ans. (b IES-15. Ans. (c c) 109 9 Enttropy S K Mondal’s IES-16. Ans. (a a) IES-17. ∫ Cha apter 4 dQ = 0 does not neccessarily meaans reversiblle process. If dQ = 0 . T Ans. (c) In irreversible i heat rejeection process enttropy decreases. In an irreverrsible y of the univ verse process entropy ways increasses i.e. sum m of alw systtem + su urroundings will incrreases. Con nsider the prrocess 3–4 iff it is irreeversible prrocess then also entropy will deccrease. IES-18. Ans. (c) ( In reversiible adiabaticc expansion, entropy chan nge is zero and no changee in entropy of surroundings s s. b) Entropy in ncrease in prrocess = 100 (0.4 – 0.3) = 10 kJ/kg IES-19. Ans. (b Enttropy change e of surround dings = 5 kJ/K K Thu us net entrop py increases and a the proceess is irreverrsible. IES-20. Ans. (c) ( In reversiible process entropy e chan nge is zero an nd in four figu ures it is rep presented by stra aight vertical line. Howev ver, in irreveersible processs, entropy in ncreases in all a processes (exp pansion as well w as compression). IES-21. Ans. (b b) ( T1 − T2 ) ΔS = W or ΔS = 300 = 1 kJ / k 600 − 300 IES-22. Ans. (c c) ⎛T ⎞ dQ 1 α + 2 β T IES-23. Ans. (d d) ∫ dT =∫ T = α ln ⎜ 2 ⎟ + 2 β (T2 − T1 ) T T ⎝ T1 ⎠ T1 T1 T1 T IES-24. Ans. (c c) IES-25. Ans. (d d) 110 0 Entropy S K Mondal’s Chapter 4 Q 6 = 2 800 100 Q 8 = 4 600 100 Q + Q4 η = 1− 2 Q1 + Q3 3 4 + 4 3 = 0.85 = 1− 6+8 IES-26. Ans. (d) Refer to given figure, as given Engine work developed = 100 kW = 100 × 1000 × 60 = 6 × 106 J/min. Qs = total heat supplied Thus, = 6 × 106 +3.2 × 106 = 9.2 × 106 J/min. Let reservoir at 1200 K supply Qs1 J/min. Therefore reservoir at 800 o K will supply. Qs2 = 9.2 × 106 – Qs1 Also, by data the engine is a reversible heat engine completing 600 cycles/min. and therefore entropy change after every complete cycle is zero. Thus, or Qs1 Qs2 QR + − =0 1200 800 400 Qs1 9.2 × 106 − Q 6 × 3.2 × 106 + − =0 1200 800 400 2Qs1 + 3(9.2 × 106 − Qs1 ) − 6 × 3.2 × 106 =0 2400 or Qs1 = 3 × 9.3 × 106 − 6 × 3.2 × 106 = 8.4 × 106 J/min Qs2 = 9.2 × 106 − 8.4 × 106 = 0.8 × 106 J/min 8.4 Hence ratio = =10.5 0.8 IES-27. Ans. (b) It is isobaric compression. 111 Enttropy S K Mondal’s Cha apter 4 IES-28. Ans. (d d) IES-29 Ans. (d) Previious 20 0-Yearrs IAS Answ wers IA AS-1. Ans. (b) IA AS-2. Ans. (b) +Q = 10 0 500 or Q = 5000 kJ −Q −5000 S1 = = = −5kJ / k 1000 1000 1 t thesystem m is +ive ⎤ ⎡∴Heat added to ⎢ Heat rejecteed from the syystem is -ive ⎥ ⎦ ⎣ S2 = IA AS-3. Ans. (d) 1 × ( 5 − 1) × ( 800 − 40 00 ) Work done d area1 − 2 − 3 = = 2 = 0.25 IA AS-4. Ans. (d) η = Heat ad dded areaunder curve 2 − 3 ( 5 − 1) × 800 IA AS-5. Ans. (d) That so wh hy we are usin ng p–v or T– –s diagram. IA AS-6. Ans. (a)) IA AS-7. Ans. (b b) IA AS-8. Ans. (d) IA AS-9. Ans. (b) 112 2 Entropy S K Mondal’s Chapter 4 dQ = 0 does not necessarily means reversible process. If dQ = 0 . T Q − Q2 Q 450 = 1− 2 = 1− = 0.25 = 25% IAS-10(ii). Ans. (b) η = 1 600 Q1 Q1 IAS-10(i). Ans. (a) ∫ IAS-11. Ans. (c) Throttling is a constant enthalpy expansion process. IAS-12. Ans. (b) dQ dQ 1600 1600 − = − = 2kJ / K IAS-13. Ans. (c) Entropy generated = dsat 400K − dsat 800K = 400 800 400 800 ΔQ 5 × 3600 = kJ / K = 61.4kJ / K IAS-14. Ans. (c) ΔS = T 293 IAS-15. Ans. (a) IAS-16. Ans. (b) 113 Availability, Irreversibility S K Mondal’s 5. Chapter 5 Availability, Irreversibility Theory at a Glance (For GATE, IES & PSUs) That part of the low grade energy which is available for conversion is referred to as available energy, while the part which, according to the second law, must be rejected, is known as unavailable energy. Availability The availability of a given system is defined as the maximum useful work that can be obtained in a process in which the system comes to equilibrium with the surroundings or attains a dead state. Clearly, the availability of a system depends on the condition of the system as well as those of the surroundings. Availability: • Yields the maximum work producing potential or the minimum work requirement of a process • Allows evaluation and quantitative comparison of options in a sustainability context Availability = Maximum possible work-Irreversibility W useful = W rev – I Irreversibility The actual work done by a system is always less than the idealized reversible work, and the difference between the two is called the irreversibility of the process. I = Wmax – W This is also sometimes referred to as 'degradation' or 'dissipation'. Availability and Irreversibility 1. Available Energy (A. E.) T ⎛ T0 ⎞ T ⎞ ⎛ = Wmax = Q1 ⎜1 − ⎟ = mcp ∫ ⎜1 − 0 ⎟dT T1 ⎠ T ⎠ T0 ⎝ ⎝ = (T1 − T0 ) ΔS Where To is surroundings temperature. 112 Availability, Irreversibility S K Mondal’s = u − u − T (S − S 1 2 0 2 ) 2 ) 1 change of volume is present there. = h − h − T (S − S 1 2 0 1 Chapter 5 (For closed system), it is NOT ( φ1 – φ2) because (For steady flow system), it is (A1-A2) as in steady state no change in volume in C.V. (i.e. change in availability in steady flow) 2. Decrease in Available Energy = T0 [ ΔS ′ − ΔS ] 3. Take ΔS′ and ΔS both positive quantity. Availability function 2 C A = h − T0S + + gz 2 φ = u − T0S + P0V For open system For closed system Availability = maximum useful work. For steady flow C12 Availability = A1 − A0 = ( h1 − h0 ) − T0 ( S1 − S0 ) + + gz 2 For closed system Availability = φ1 − φ0 = u1 − u0 − T0 ( S1 − S0 ) + P0 ( V1 − V0 ) 4. Unavailable Energy (U.E.) = T0 ( S1 − S2 ) 5. Increase in unavailable Energy = Loss in availability 6. Irreversibility = T0 ( ΔS )Univ 113