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Mechanical Engg. Dept D.N.Thatoi, S.N.Das, A.K.Mishra ITER FLUID MECHANICS AND THERMODYNAMICS ASSIGNMENT – 5 SECOND LAW OF THERMODYNAMICS AND ENTROPY Introductory Notes Thermal Reservoir is a large system to which a finite amount of heat can be added or extracted without changing its temperature. A source is a thermal reservoir at higher temperature from which energy in the form of heat is withdrawn isothermally. A sink is a thermal reservoir at lower temperature to which energy as heat can be added isothermally. Thermal Efficiency of a Heat Engine: Performance of a heat engine is called its thermal efficiency. It is a measure of the degree of useful W net , out Q i n−Q out Q out utilization of the heat input into a heat engine. So, th = = =1− Qi n Qi n Qi n Second Law of Thermodynamics : Kelvin-Planck Statement “It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.” Clausius Statement – “It is impossible to have a device which working cyclically will produce no other effect than to transfer heat from a low temperature body to a high temperature body.” The two statements of the Second Law are equivalent and violation of one leads/implies the violation of the other statement. Factors that cause irreversibility: – Friction (solid friction or viscosity) – heat transfer across a finite temperature difference – electric resistance – expansion and compression due to finite pressure difference – spontaneous mixing of two fluids – unrestrained expansion of a gas – elastic deformation of solids Carnot Cycle: P-v diagram of Carnot Cycle T-s diagram of Carnot Cycle Mechanical Engg. Dept D.N.Thatoi, S.N.Das, A.K.Mishra Process 1-2 → Reversible isothermal expansion at TH Process 2-3 → Reversible adiabatic expansion where temperature changes from TH to TL Process 3-4 → Reversible isothermal compression at TL Process 4-1 → Reversible adiabatic compression where temperature changes from TL to TH TL Thermal efficiency of a Carnot cycle rev =1− This formula is applicable for a reversible cycle TH only. QL Note: The efficiency of a heat engine (reversible or irreversible) is 1− . But for a reversible QH QL T QL T L heat engine rev =1− =1− L , since for reversible engines, = . Here, temperatures QH TH QH T H must be taken in Kelvin. Remember that efficiency of a reversible heat engine (say Carnot Engine) is never 100%. Carnot Principle: I. No heat engine operating between the two thermal reservoirs, with fixed temperatures, can be more efficient than a reversible engine operating between the same two thermal reservoirs. II. The efficiencies of all reversible engines operating between the same two reservoirs are the same. Clausius Inequality: ∮ Q ≤0 T The limiting case of equality is for reversible processes i.e ∮ Q T i nt. rev. =0 Entropy: dS = Q T i nt.rev. Here, dS is the infinitesimal change in entropy, δQ is the infinitesimal amount of heat interaction across the boundary where the temperature is T in Kelvin. S =S 2−S 1=∫1 2 Q T i nt.rev. kJ /K Entropy is an extensive property. Change in Entropy during an internally reversible isothermal heat transfer : 1 2 S = ∫1 Qi nt.rev. =1 Q 2 / T kJ / K This expression is useful to determine the entropy changes T of a thermal reservoir that can absorb or supply heat indefinitely at a constant temperature. For an isentropic process, ΔS = 0. The performance of devices like pumps, turbines, nozzles, compressors and diffusers etc. are compared with an idealized isentropic version of their operations. [Note: A reversible, adiabatic process is necessarily isentropic but an isentropic process is not necessarily a reversible, adiabatic process. The entropy increase of a substance during a process as a result of irreversibilities may be compensated by a decrease in entropy as a result of heat losses.] 1Q2 int.rev. = T ΔS Combining the First and Second law for a closed system: (for both reversible and irreversible process) TdS =dU + PdV. Using H = U + PV, we can also express this as TdS = dH – VdP Mechanical Engg. Dept D.N.Thatoi, S.N.Das, A.K.Mishra du dv ds= P T T So, dh dP ds= −v T T Entropy Change of Solids and Liquids: Approximating liquids and solids as incompressible substances, Cp = Cv = C. If we also assume that the heat capacity of the substance does not change much with temperature, s 2−s1=C ln T2 T1 . For an isentropic process with incompressible substances, ΔS = 0 and hence, T2 = T1. Therefore, for a truly incompressible substance, an isentropic process is also isothermal. Entropy change for Ideal Gases: (taking constant specific heats) Using du = Cv dT, dh = Cp dT and Pv = RT: T v s 2−s1=C v ln 2 R ln 2 kJ /kg.K T1 v1 =C p ln T2 T1 P − R ln 2 kJ /kg.K P1 Principle of Entropy Increase: “For an isolated system, the entropy never decreases. It increases for any irreversible process and remains the same for reversible processes only.” S 2−S 1= S system=∫1 2 Q T S gen Assignment Problems 1. An inventor claims that he has built an engine working between temperature limits of 1000 K and 400 K and having an efficiency of 70%. Is his claim valid? 2. Among the following processes, pick those which are internally reversible: • A process with friction • Throttling process • Frictionless adiabatic process • A process involving mixing • Frictionless constant pressure process • Free expansion process 3. A reversible engine working between temperature limits of 800 K and 300 K receives 1000 kW of heat from the high temperature reservoir. What power is developed by this engine? (625kW) 4. A Carnot refrigerator takes heat at the rate of 150 kW from a space maintained at 300 K. If the refrigerator rejects heat at 320 K, what is the power required to drive the refrigerator? (10 kW) 5. A cycle is constructed with three frictionless processes and one Q process with friction. What will be the value of ∮ Q T for this W cycle – positive, negative or zero ? 6. Prove that the cyclic device shown(Figure 1) can not be reversible. 7. An inventor claims that he has developed an engine that works between temperature limits of 500 K and 1000 K and develops Figure 1 Mechanical Engg. Dept D.N.Thatoi, S.N.Das, A.K.Mishra a power of 20 kW while consuming heat at the rate of 50 KW. Is this engine possible? If possible, is it reversible or irreversible? 8. Working between the same temperature limits, show that coefficient of performance of a heat pump is greater than that of a refrigerator by unity. 9. Two Carnot engines A and B are connected in series between two thermal reservoirs maintained at 1000 K and 100 K respectively. Engine A receives 1680 kJ of heat from the high temperature reservoir and rejects heat to the Carnot engine B. Engine B takes in heat rejected by engine A and rejects heat to the low temperature reservoir. If engines A and B have equal thermal efficiencies, determine : (i ) the heat rejected by engine B (ii ) the temperature at which engine A rejects heat (iii ) the work done. (168 kJ, 316.23 K, 1512 kJ) 10. Two reversible engines are arranged in series as detailed below (Figure 2). The first engine receives energy from a reservoir at TH and rejects energy to a reservoir at temperature T. The second engine receives the energy rejected by the first engine from the reservoir at T and rejects energy to a reservoir at temperature TL . Here, TH>T > TL . Derive an expression for the temperature T in terms of TH and TL, if the net work of the two engines are equal. Figure 2 11. A reversible engine operates as shown in Figure 3. It has two sources and a sink. One source is at 1200 K and supplies heat at the rate of 800 kW, while the other source is at 1000 K and supplies heat at the rate of 500 kW. The engine rejects heat at the rate of 400 kW to a sink at an unknown temperature. Determine the sink temperature, the power developed and the engine efficiency. (343 K, 900 kW, 0.6923) Mechanical Engg. Dept D.N.Thatoi, S.N.Das, A.K.Mishra Figure 3 12. A heat engine operating between two reservoirs at 1000 K and 300 K is used to drive a heat pump which extracts heat from a reservoir at 250 K at a rate twice that at which the engine rejects heat. If the efficiency of the engine is 80% of the maximum possible and COP of the heat pump is 1/3 of the maximum possible, what is the temperature of the reservoir to which the heat pump rejects heat? What is the rate of heat rejection from the heat pump if the rate of heat supply to the engine is 50 kW? (303 K, 72 kW) 13. A compressor takes in 500 kg/minute of air at 0.98 bar and 68oC. The diameters of inlet and delivery pipes are 450 mm and 200 mm. The power input is 1000 kW. Determine the rate and direction of heat flow. (Cp = 1.005kJ/kg OC) (-580.3 kW) 14. An inventor claims that he has developed a heat engine which absorbs 1200 kJ and 800kJ from reservoir at 800 K and 600 K respectively and rejects 600 kJ and 200 kJ as heat to reservoir at 400 K and 300 K respectively. It delivers 1200 kJ work. Determine whether the heat engine is theoretically possible. 15. A system at 27OC is compressed reversibly and isothermally such that it receives 30 kJ of work. Calculate its entropy change. (-0.1 kJ/K) 16. During a reversible isothermal process occurring at 400 K, the entropy change was found to be 0.2 kJ/K. Find the work done during this process. (80 kJ) 17. A system undergoes a process during which its entropy changes by -0.2 kJ/K. The system also transfers 90 kJ of heat to the surroundings at 300 K during this process. What is the change in entropy of the universe due to this process? (0.1 kJ/K ) 18. A system performs an irreversible process at 200 K and receives 200 J of heat from the surroundings. Then which of the following relations is correct for the entropy changes (ΔS) of the system (a ) ΔS = 1 J/K (b ) ΔS > 1 J/K (c ) ΔS < 1 J/K (d ) ΔS = 0 J/ K 19. If an ideal gas undergoes a reversible adiabatic process and changes from (P1, T1) to (P2, T2), show that the change in entropy is equal to zero. 20. Two identical bodies of constant heat capacity (Cp) are at the same initial temperature T1. A refrigerator operates between these two bodies until one body is cooled to temperature T2. If the bodies remain at constant pressure and undergo no change of phase, show that the minimum amount of work needed to do this is Wmin = Cp [ T2 1 T2 T 2−2 T 1 ] 21. A system maintained at constant volume is initially at temperature T1 and a heat reservoir at the lower temperature T0 is available. Show that the maximum recoverable work as the system Mechanical Engg. Dept D.N.Thatoi, S.N.Das, A.K.Mishra is cooled to T0 is [ W max =C v T 1 −T 0 −T 0 ln ] T1 T0 . 22. A rigid vessel contains air at 2 bar, 100oC while the surroundings are at 30oC. The air in the vessel gets cooled by heat transfer to the surroundings and reaches thermal equilibrium with the surroundings. For 5 kg of this air, determine the : (i ) Entropy change of air. (-0.746 kJ/K) (ii ) Entropy change of surroundings. (0.829 kJ/K ) (iii ) Entropy change of the universe. (0.083 kJ/K ) 23. One kg of superheated steam at 0.2 M Pa and 200oC contained in a piston-cylinder assembly is kept at ambient condition of 300 K till steam is condensed to saturated liquid at constant pressure. Calculate the change in the entropy of the universe associated with this process. (1.9089 kJ/K) 24. Steam at 0.8 MPa, 2500C and flowing at the rate of 1 kg/s passes into a pipe carrying wet steam at 0.8 MPa, 0.95 dryness fraction. After adiabatic mixing, the flow rate is 2.3 kg/s. Determine the condition of steam after mixing. The mixture is expanded in a frictionless nozzle isentropically to the pressure of 0.4 MPa. Determine the quality of steam leaving the nozzle. (178.965oC, 6.709 kJ/kg-K; 0.96)

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an assignment on second law of thermodynamics and entropy

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