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- 1. 1. Fundamental Concepts & Definition1. Explain dependent properties and thermodynamic equilibrium.2. Show that the negative absolute temperature scale is impossible3. Explain macroscopic and microscopic point of view.4. Explain briefly the followings: i) System boundary & surrounding. ii) Open cycle & closed cycle. iii) Quasi-Static Process.5. (i)Explain quasi-static process with p-V diagram. (ii)Explain thermodynamic equilibrium.6. Show that the thermodynamic temperature scale is independent of working substance.7. Explain thermodynamic temperature scale.8. Explain quasi-static process with p-V diagram9. Explain point function and path function.10. Write down Vander walls equation of state. How does it differ from ideal gas equation?2. First-Law of Thermodynamics1. Write steady flow energy equation and reduce the same for Turbine, Nozzle,Automobile radiatorand throttling device. Clearly sWe the asmn#oris made in each cam.2. Explain Enthalpy of reaction.3. What are the requirements of a steady flow process ?4. Derive the General energy equation for steady flow system and simplify when applied for thefollowing systems.(i)Centrifugal water pump (ii) Reciprocating air compressor(iii)Steam nozzle (iv) Steam turbine(v) Gas turbine
- 2. 5. A turbine, operating under steady flow conditions, receives 4500 Kg. of steam per hour. Thesteam enters the turbine at a velocity of 2800 m/min, an elevation of 5.5 m and a specific enthalpyof 2800 KJ/Kg. It leaves the turbine at a velocity of 5600 m/min, an elevation of 1.5 in and a specificenthalpy of 2300 KJ/Kg. Heat losses from the turbine to the surroundings amount to 16,000 KJ/hr.Determine the power output of the turbine.6. For isothermal flow and non-flow steady processes, Prove that, Jpdv = - Jvdp .Also state theassumptions made.7. Discuss perpetual motion machines of the first and second kind.8. What are the limitations of the first law of Thermodynamics.9. An air standard dual cycle has a compression ratio of 16 bar, and compression begins at 1 bar, 50 0C.The maximum pressure is 70 bar. The heat transferred to air at a constant pressure is equal to that atconstant volume.Calculate:- (1) The pressure and temperature at the cardinal points of the cycle. (2) The cycle efficiency. (3) The m.e.p.of the cycle.Take:- Cv=0.718 Kj/Kg K Cp=1.005 Kj/Kg K R=0.278 Kj/Kg K10. Define & explain the first law of thermodynamics for a closed system undergoing a (i) cyclicprocess & (ii) non-cyclic process.11. Explain Phase change processes for pure substance with the help of PT and PV diagrams 3 012. An air receiver of volume 6 m contains air at 16 bar and 40.5 C. A valve is opened and some air isallowed to blow out to atmosphere. The pressure of the air in the receiver drops rapidly to 13 bar whenthe valve is then closed. Calculate the mass of air which has left the receiver.13. For unsteady flow, derive the energy equation for filling process by Control Volume analysis.14. Define the specific heat and explain the specific heat at constant volume & constant pressureprocess.15. Derive the steady flow energy equation for a single stream entering and leaving a controlvolume and explain various terms in it.16. 12 kg of a fluid per minute goes through a reversible steady flow process. The properties of fluid atthe inlet are p1 = 1.4 bar, ρ1 = 25 kg/m3, V1 = 120 m/s & u1 = 920 kJ/kg and at the exit are p2 = 5.6 bar, ρ2 =5 kg/m3, V2 = 180 m/s and u2 = 720 kJ/kg. During the passage, the fluid rejects 60 kJ/s and raises through60m. Determine i) the change in enthalpy ii) work done during the process.
- 3. 17. Define first law of thermodynamics. Also define internal energy of the system and show thatthe internal energy is a property of the system.18. An air receiver of volume 6 m3 contains air at 15 bar and 40.50C. A valve is opened and some air isallowed to blow out to atmosphere. The pressure of the air in the receiver drops rapidly to 12 bar whenthe valve is then closed. Calculate the mass of air which has left the receiver.19. Analyze the general equation of the first law of thermodynamics applied to open system.20. Show that the work transferred by steady flow process is given by Wsf=-fvdp21. A steam at 6.85 bar and 205°C enters in an insulated nozzle with velocity of 50 m/sec. It leavesat a pressure of 1.35 bar and a velocity of 510 m/sec. Determine the final enthalpy steam.22. Explain flow work, shaft work and paddle-wheel work transfer.23. Steam enters a nozzle at a pressure of 7 bar and 20 °C and leaves at a pressure of 2 bar. Theinitial velocity of steam at the entrance is 50m/s and exit velocity from the nozzle is 750 m/s.The mass flow rate through the nozzle is 1500 kg/h. The heat loss from the nozzle is, 12000kJ/h. Determine the final enthalpy of the steam and nozzle area if the specific volume is 1.25m3/kg. Take initial enthalpy 2850kJ/kg.24. A fluid undergoes a reversible adiabatic compression from 1 MPa and 0.3 m3 to0.05 m3 according to the law pvl.3 constant. Determine the change in enthalpy,entropy, heattransfer and work transfer.25. Explain PVT behavior of pure substances with the help of PT and PV diagrams.26. A centrifugal air :compressor delivers 12 kg of air per minute. The inlet and outletconditions of air are:C1=12 m/sec, pi=1 bar, v,=0.5 m3/kg andC2=90 m/sec, P2=8 bar, v2=0.14 m3lkg. The increase in enthalpy of air passing through thecompressor is 150kJ/kg and heat loss to the surroundings is 700 kJ/min.Find: (1) Motor power required to drive the compressor. (2) Ratio of inlet to outlet pipe diameter. Assume that inlet & discharge lines are at the same level.3. Second law of thermodynamics 1. State and prove carrots theorem. 2. A heat engine is operated between 700 °C and 30 °C. It drives a heat pump which works between100 °C and 30 °C. Efficiency and COP of the heat engine and the heat pump are half of that ofcorresponding Carnot values. Calculate amount of heat rejected by heat pump at 100 °C when 100 kJis absorbed by heat engine at700°C.
- 4. 3.10 kg/s of chilled water enters a tall building with velocity of 50 m/s at an elevation of 30 m fromground. The water leaves the system with velocity of 10 m/s at an elevation of 60 m/s. Thetemperature of water entering in and leaving out are 7 °C and 12 °C respectively. The rate of work doneby pump in the line is 35 kW. Calculate rate of heat removed by water.4. What are the limitations of Carrot cycle ? Give comparison of Carrot and Rankine cycle.5. Write comparison of First and Second law of thermodynamics.6. Two Carrot engines work in series between the source and sink temperatures of 550 ° K and 350 °K. If both engines develop equal power determine the intermediate temperature.7. Explain reversible and irreversible process. Give some examples.8. Prove that no engine working in cycle between, the two reservoirs can be more efficient than areversible engine working between the same reservoirs.9. Write the different statements of Second law of thermodynamics & prove their equivalence with neatfigures.10. In a water cooled compressor 0.4 kg/sec of air is compressed. A shaft input of 48 kW is required torun the compressor. Heat lost to the cooling water is 30% of input and 10% of the input is lost inbearings and other frictional effects. Air enters the compressor at 1 bar and 23 0C. Neglecting thechanges in KE & PE, determine the exit air temperature. Take C p = 1kJ/kg0C air.11. Show that heat transfer through a finite temperature difference is irreversible.12. Explain reversible and irreversible process with suitable examples. Write causes ofirreversibility.0413. State the limitation of the first law of thermodynamics. How the second law ofthermodynamics has helped make them clear? Discuss perpetual motion machines of the firstand second kind.14. In a water cooled compressor 0.5 kg of air is compressed/sec. A shaft input of 60 kW isrequired to run the compressor. Heat lost to the cooling water is 30% of input and 10% of theinput is lost in bearings and other frictional effects. Air enters the compressor at 1 bar and 200C.Neglecting the changes in KE & PE, determine the exit air temperature. Take Cp = 1 kJ/kg 0Cfor air.15. There are two ways of increasing the efficiency of a Carnot heat engine:i) Lowering the temperature T2 of the low temperature reservoir by ∆T, while keeping thetemperature T1 of the high temperature reservoir a constant.
- 5. ii) Increasing the temperature T1 by ∆T, while keeping the temperature T2 a constant. Which ismore effective? Prove your answer.16. A reversible refrigerator operates between 35 °C and -12 °C. If the heat rejected to 35 °Creservoir is 1.3 kW, determine the rate at which to heat is leaking into the refrigerator.17. Prove the equivalence of Clausius and Kelvin statements.18. Three carnot Engines Ei,E2,E3 operate between temperatures 1000 K and 300 K. Makecalculations for the intermediate temperatures if the work produced by the engines are in theratio of 4:3:2.19. State and prove camot theorem for heat engine. Also write the statements of carnot theoremin the context of refrigerator or heat pump.20. A heat pump operates between two identical systems; both being attemperature T, to start with. Due to operation of pump, one of the systems getscooled down to temperature T2.Show that for this operation of pump, theminimum work required by the heat pump is T2 Wmin=C T-+T2-2T,I zWhere C is the specific heat of the two identical systems21. Prove that all reversible engines working between the two constant temperature reservoirs havethe same efficiency.22. Carnet cycle is not practical. Justify.23. A heat engine receives 999 kW of heat at constant temperature of 286 °C. The heat isrejected at 6 °C. The possible heat rejected are (a) 850 kW (b) 490kWand (c) 400 kW.24. To get maximum work from two. finite bodies at temperature TI and T2, derivethe relevant equation.25. Gas enters a nozzle at 15 bar and 1500 K with a velocity of 30 m/s. The pressure at the exit ofnozzle is 5 bar. If the nozzle efficiency is 90%, calculate the actual exit velocity. Neglectchanges in PE and heat exchange between nozzle and surrounding. Take Cp=1.005 kJ/kgK.26. Prove that the COP of a reversible refrigerator operating between two giventemperatures is the maximum.27. A household refrigerator is maintained at a temperature 2°C. Every time door is opened,warm material is placed inside, introducing an average of 420 KJ, but making only a smallchange in the temperature of the refrigerator. The door is opened 20 times a day and therefrigerator operates at 15% of the ideal COP. The cost of work is 1.50 per kWh. What is themonthly bill for this refrigerator? The atmosphere is at 30°C.
- 6. 28. Why the second law is called a directional law of nature.29. A heat engine is supplied with 420KJ/cycle of heat at a constant fixed temperature of327°C and heat rejection takes place at 27°C.here are no other heat transfers. The followingresults were reported.1.210KJ/cycle heat is rejected 2. 105 K.J/cycle heat is rejected3. 315 KJ/cycle heat is rejected.Compute the cyclic integral of dQ/T. From these classifywhich of the results report a reversible cycle or irreversible cycle or impossible cycle.30. Why the second law is called the law of degradation of energy?31. An engineer claims to have developed an engine which develops 3.4 kW whileconsuming 0.44 Kg of fuel of calorific value 41870 kJ / kg in one hour. The maximumand minimum temperatures recorded in the cycle are 1400 °C & 350 °C respectively. Isthe claim of the engineer genuine?4. Entropy 1. Derive inequality of clausius. 2. 2 kg. of ice at -10 °C is exposed to atmosphere at 37 °C. The ice melts and comes into thermalequilibrium with atmosphere. Calculate entropy change of the universe. Take Cp of ice = 2 kJ/kg °Kand latent heat of fusion for ice = 335 kJ/kg.3. What are the characteristics of Entropy ? Prove that entropy is a property of a system.4. Derive an expression for the change in entropy of the universe5. 2 Kg of water at 94 ° C are mixed with 3 Kg of water at 10 ° C in an isolated system. Calculate thechange of entropy due to mixing process.6. State and prove Clausius’ theorem.7. State and explain the Clausius inequality and prove it.8. Prove that entropy is a point function and hence a thermodynamic property.9. 5 kg of water at 30°C is mixed with 1 kg of ice at 0°C.The process of mixingis adiabatic and the system is open to atmo sphere. Make calculations forthe temperature of mixture and the change of entropy for the spontaneousmixing process. Take specific heat of water= 4.187 kJ/kg k and latent heat ofice=335 kJ/kg.10. 2 kg of water 94°C is mixed with 4 kg of water at 15°C in an isolated systemCalculate the changes in entropy due to the mixing process11.Explain principle of increase of entropy.12. Show that the internal energy is a property of the system.
- 7. 5. Availability and Irreversibility1. Derive the expression for available energy from a finite energy source (heat capacity ,Cp) attemperature T when atmosphere temperature is T o.2. Show that due to friction kinetic energy is converted into internal energy and this conversion isirreversible.3. Show that there is decrease in available energy when heat is transferred through infinte temperaturedifference.4. A copper block having heat capacity of 150 kJ/°K and temperature of 100 °C is brought incontact with an aluminium block having heat capacity 200 kJ/°K and temperature of 50 °Cadiabatically. Calculate irreversibility of the process. Ambient temperature is 37 °C. State how thesetwo blocks can be brought into thermal equilibrium reversibly.5.Explain the concept of decrease in Available energy when heat is transferred through a finitetemperature difference with the aid of Temperature - Entropy diagram.6. Derive an expression for availability of non-flow process.7. A system at 500 ° K receives 7200 KJ/min from a source at 1000 ° K. The temperature of atmosphere is3000K. Assuming that the temperature of system and source remain constant during heat transfer, findout,(i)The decrease, in Available energy after heat transfer(ii)The entropy produced during heat transfer.8. Deduce the expression for available energy from a finite energy source at temperature T whenthe environmental temperature is To.9. Calculate the decrease in available energy when 25 Kg of water at 95OC mix with 35 Kg ofwater at 35OC, the pressure being taken as constant and the temperature of the surrounding being15OC. (Cp of water =4.2 KJ/Kg K )10. Define available and unavailable energy. Derive an expression for availability of a non-flow process. 11. A lump of 800 kg of steel at 1250 K is to be cooled to 500 K.If it is desired to use thesteel as source of energy, calculate the available and unavailable energies. Takespecific heat of steel as 0.5 kJ/kgK and ambient temperature 300 K.12. What is irreversibility? State various types of irreversibility and explain them.
- 8. 6. General Thermodynamic Relationships1. What is the importance of general Thermodynamic relation ? Define co-efficient of volumeexpansion and isothermal compressibility.2. State and derive the Maxwell’s relations k 13. Derive the equation Tds = CV dp + C P dv β βV4. What is the importance of thermodynamic relations ? Define Helmholtz and Gibbs free energy.5. Explain thermodynamic equilibrium.6. Explain the Joule-Kelvin effect with various diagrams.7. Derive Clapeyron equation and state its practical utility.7. Properties of Gases and Mixtures1. What is Joule-Thomson coefficient? Explain and show that it is zero for an ideal gas.2. Define the Perfect and real gases.3. Explain the Dalton’s law of Partial Pressure & Partial Volume.

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