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Thermodynamics

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  • Sidhant S

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    Chapter 5 ENTROPY The first law of thermodynamics deals with the property energy and the conservation of energy. The second law introduced in the previous chapter, leads to the definition of a new property called entropy. Entropy is defined in terms of a calculus operation, and no direct physical picture of it can be given. In this chapter, Clausius inequality, which forms the basis for the definition of entropy will be discussed first. It will be followed by the discussion of entropy changes that take place during various processes for different working fluids. Finally, the reversible steady-flow work and the isentropic efficiencies of various engineering devices such as turbine and compressors will be discussed. 5.1 The Clausius Inequality Consider two heat engines operating between two reservoirs kept at temperature TH and T as shown in the Figure 5.1. Of the two heat engines, one is reversible and the other is irreversible. Reservoir at T -o -o Rev Irrev w net irrev Irrev Reservoir at T For the reversible heat engine it has already been proved that rev
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    As discussed earlier, the work output from the irreversible engine should be less than that of the reversible engine for the same heat input QH. Therefore QL,Irrev will be greater than QL,Rev Let us define QL,Irrev— QL,Rev then Irrev L,lrev L,rev By combining this result with that of a reversible engine we get ... (5.1) Irrev This is known as Clausius inequality. 5.2 Entropy Clausius inequality forms the basis for the definition of a new property known as entropy. Consider a system taken from state 1 to state 2 along a reversible path A as shown in Figure 5.2. Let the system be brought back to the initial state 1 from state 2 along a reversible path B. Now the system has completed cycle. Applying Clausius one inequality we get Path A 1 2 C
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    2 dQ 1 1 ...(5.2) Instead of taking the system from state2 to statel along B, consider another reversible path C. Then for this cycle I-A-2-C-1, applying Clausius inequality : ...(5.3) 2 dQ 1 dQ 1 Comparing 5.2 & 5.3 Hence, it can be concluded that the quantity is a point function, independent of the path followed. Therefore it is a property of the system. Using the symbol S for entropy we can write upon integration we get s For a reversible process. s 1 ...(5.4) ... (5.5) 5.3 Entropy change for an irreversible process The relationship between the entropy change and heat transfer across the boundary during an irreversible processes can be illustrated with a simple cycle composed of two processes, one of which is internally reversible and the other is irreversible, as shown in Figure 5.3. The Clausius inequality applied to this irreversible cycle can be written as Since the process B is internally reversible, this process can be reversed, and therefore or ...(5.6) As defined in equation 5.5, since the process B being reversible the integral on the left hand side can be expressed as ...(5.7) 5.4 Temperature - Entropy diagram
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    In a T-s diagram consider a strip of thickness ds with mean height T as shown in Figure 5.4. Then Tds gives the area of the strip. For a reversible process the elemental heat transfer dQ Tds Area of the strip To get the total heat transfer the above equation should be integrated between the limits I and 2, so that, we get ...(5.8) This is equivalent to the area under a curve representing the process in a T-S diagram as shown in the Fig 5.4. Note: • For an isothermal process S 2 — S/ • For reversible adiabatic process S2 —S/ — 0. 5.5 Change in Entropy a) b) Solids and Liquids Change in entropy Where dq du + pdv For solids and liquids pdv 0 Where c- is the specific heat ...(5.9) For ideal gases change in entropy Substituting du We get Upon integration Also cvdT ...(5.10a)
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    Substituting and We get Upon integration cpdT ...(5.10b) 5.6 Principle of Increase in Entropy Applying Clausius inequality, For an isolated system undergoing a process ...(5.11) Consider a system interacting with its surroundings. Let the system and its surroundings are included in a boundary forming an isolated system. Since all the reactions are taking place within the combined system, we can express or ...(5.12) Whenever a process occurs entropy of the universe (System plus surroundings) will increase if it is irreversible and remain constant if it is reversible. Since all the processes in practice are irreversible, entropy of universe always increases (As) universe This is known as principle of increase of entropy. 5.7 Adiabatic Efficiency of Compressors and Turbines ...(5.13) In steady flow compressors and turbines reversible adiabatic process is assumed to be the ideal process. But due to the irreversibilities caused by friction between the flowing fluid and impellers, the process is not reversible though it is adiabatic. Percentage deviation of this process from the ideal process is expressed in terms of adiabatic efficiency. (a) Compressors : Since compressors are work consuming devices actual work required is more than ideal work. ...(5.14) For compressors handling ideal gases ...(5.15) (b) Turbines : In turbine due to irreversibilities the actual work output is less than the isentropic work. ...(5.16) For turbines handling ideal gases ...(5.17)
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    Prob : 5.1 System Known rejected Isothermal Process : To find . As Diagram Analysis Comment . Solved Problems A body at 2000C undergoes an reversible isothermal process. The heat energy removed in the process is 7875 J. Determine the change in the entropy of the body. Closed system 2000c 473 K 7875 J Q for an isothermal process 16.65 J/K. Entropy decreases as heat is removed from the system. Prob : 5.2 A mass of 5 kg of liquid water is cooled from 1000C to 200C. Determine the change in entropy. System Known Process To find Closed system Mass of water 5kg 1000c 200c 373 K 293 K Constant pressure Change in entropy
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    Diagrams : Assumptions Analysis Comment : 1) The process is reversible 2) The specific heat of liquid water is constant S2-Sl=m For this problem 4.186 5 5.053 Entropy decreases as heat is removed from the system. Prob : 5.3 System Known To find Diagram Analysis Air is compressed isothermally from 100 kPa to 800 kPa. Determine the change in specific entropy of the air. Closed/Open 100 kPa 800 kPa AS - change in Specific entropy AS R In [Since the process is isothermal] 0.287 x In 0.597 kJ/kgK.
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    Prob : 5.4 System Known Process : To find Diagram Analysis Comment Prob : 5.5 System Process : Known A mass of 5 kg of air is compressed from 90 kPa, 320C to 600 kPa in a polytropic process, pVl constant. Determine the change entropy. Closed / Open m 90 kPa 320C - 305 K 600 kPa 5 kg P VI 3 Constant AS - Change in entropy Where T2 m 305 473 K 5 0.517 kJ/K. For air the ratio of C and cv is equal to 1.4. Therefore the polytropic index n — 1.3(
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    To find Analysis Change in entropy S2-Sl=m By applying the state equation. Since V 2 Also R c c 0.286 kJ/kgK Substituting these values we get 5 3.922 10/1< Though this process is adiabatic it is not isentropic since the process of stirring is Comment . an irreversible process. Prob : 5.6 An insulated rigid vessel is divided into two chambers of equal volumes. One chamber contains air at 500 K and 2 MPa. The other chamber is evacuated. If the two chambers are connected d, what would be the entropy change ? Closed system System Unresisted expansion Process : Tl 500K Known 2 x 103 kPa To find . Entropy change Diagrams : Analysis After expansion air will occupy the entire volume of the container. V 2 - 2Vl
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    Also W Qi2 0 since it is an unresisted expansion 0 since the vessel is insulated Applying the first law of thermodynamics Q12 Therefore Au For air i.e. Hence s 2 2 1 C In + Rin 0.287 In 0.199 kJ/kgK Comment Prob . •5.7 Though the process is adiabatic entropy increases as the process involving unresisted expansion is an irreversible process. It also proves the fact that An adiabatic chamber is partitioned into two equal compartments. On one side there is oxygen at 860 kPa and 140C. On the other side also, there is oxygen, but at 100 kPa and 140C. The chamber is insulated and has a volume of 7500 cc. The partition is abruptly removed. Determine the final pressure and the change in entropy of the universe. System Process : Known Diagrams : Analysis : Closed Adiabatic Mixing Subsystem I Fluid Initial pressure Initial Temperature Initial volume Subsystem Il Oxygen 850 kPa 140c Oxygen 100 kPa 140C Here the energy interaction is taking place only between the two fluids and therefore the energy lost by one fluid should be equal to the energy gained by the other fluid. Taking tF as the final temperature we get
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    Since the same fluid is stored in both the systems at the same temperature C t 1 C 2 and 140C t 2 Therefore the final temperature will also be 140C After removing partition total mass of oxygen is occupying the entire 7500cc at 140C. Hence the final pressure can be computed as given below : 0.0427 kg 0.00503 kg To find the final pressure ml + m2 475 kPa ASI -k AS 2 AS system AS Prob . • 5.8 System Process : Known surroundings AS 8.596 universe Two vessels, A and B each of volume 3 m may be connected by a tube of negligible volume. Vessel A contains air at 0.7 MPa, 950C while vessel B contains air at 0.35 MPa, 2050C. Find the change of entropy when A is connected to B by working from the first principles and assuming the mixing to be complete and adiabatic. Closed Adiabatic mixing Properties Fluid pressure volume Temperature Subsystem A Air 0.7 MPa 3 950c Subsystem B Air 0.35 MPa 3 2050c
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    Diagrams : Analysis Since the energy interaction is taking place only between the two fluids energy lost by one fluid is equal to the energy gained by the other fluid. Taking t2 as the final temperature after mixing maca (t2 tia) mbCb(tlb t2) Since in both A and B the same fluid is stored, Ca — C Also ma 19.9 kg 19.9 (t2 2.6 (t2 b 95) 95) 2.6t2 + t2 Entropy change ASA ASB 7.65 kg 7.65 (205 - t2) (205 - t2) 205 + 2.6 x 95 125.60c ASA + ASB
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    AS sys AS 5.08 + 0.525 5.605 surr AS 5.605 universe Final pressure Prob : 5.9 525 kPa Air enters a turbine at 4000C, 30 bar and velocity 160 m/s. It leaves the turbine at 2 bar, 1200C and velocity 100 m/s. At steady state it develops 200 kJ of work per kg of air. Heat transfer occurs between the surroundings and the turbine at an average temperature of 350 K. Determine the rate of entropy generation. System Process : Known Diagram To find Analysis Open Steady flow Properties Pressure Velocity Temperature Ambient temperature Work output Inlet 30 bar 160 m/s 4000C 350 K 200 kJ/kg Outlet 2 bar 100 m/s 1200c Rate of entropy generation For unit mass where C In 1.005 In 0.236 kJ/kgK
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    (AS)sur +89.2 kJ/kg 0.255 kJ/kgK 0.019 kJ/kgK. Prob . • 5.10 A turbine operating at steady state receives air at a pressure of PI = 3.0 bar and temperature of 390 K. Air exits the turbine at a pressure of = 1.0 bar. The work developed is measured as 74 kJ/kg of air flowing through the turbine. The turbine operates adiabatically, and changes in kinetic and potential energy between inlet and exit can be neglected. Using ideal gas model for air, determine the turbine System Process Known Diagrams Analysis efficiency. Open Steady flow 3.0 bar 390 K for an ideal gas 1.0 bar 74 kJ/kg Where - 284.9K 73.63 K Hence 74 0.7 (or 70%). Prob . • 5.11 A closed system is taken through a cycle consisting of four reversible processes. Details of the processes are listed below. Determine the power developed if the system is executing 100 cycles per minutes. Process Temperature (K)
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    1-2 2-3 3-4 4-1 : Closed System Process : Known To find Diagrams : Analysis 1-2 2-3 3-4 4-1 + 1000 Initial 300 1000 1000 300 Final 1000 1000 300 300 The system is executing cyclic process. . Heat transfer in process 12, 23 and 34 and temperature change in all the No of cycles per minute. Power developed. process. To find the power developed w net per cycle must be known. From I Law w net Qnet which can be computed from the following table Process Temperature (K) 1 ooo Initial 300 1 ooo 1 ooo 300 For a cyclic process where is any property (i.e. , ) AS 12 -k AS23 -k AS34 -k AS 41 Final 1 ooo 1 ooo 300 300 EAS 1
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    Since the process 4-1 is isothermal 1 Q41 Therefore wnet and power developed 300 Id (212 + Q 2 + Q 4 + o + 1000 +0 - 300 700 kJ per cycle Qnet 700 Id 700 1166.7 kW Prob : 5.12 Two kilogram of air is heated from 2000C to 5000C at constant pressure. Determine the change in entropy. System Working fluid Process Known Diagram To find Analysis Open/closed • Air : Constant pressure heating . 1) 2000C 5000c • Change in entropy AS • AS 0.987 kJ/K Prob : 5.13 A Carnot engine operated between 40C and 2800C. If the engine produces 300 kJ of work, determine the entropy change during heat addition and heat rejection. System Process Known Diagram Open/closed The working fluid is executing Carnot cycle 2) 3) w 2800c 40C 300 Id
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    To find Analysis Comment Prob . 1) AS during heat addition 2) AS during heat rejection 1) In carnot engine heat is added at constant temperature Therefore Where AS 0.499 Therefore 601.1 Id AS 1.087 kJ/K 2) In carnot engine heat rejection is also taking place at constant temperature Therefore AS Where Qout AS 601.1 300 301.1 Id 1.087 kJ/K In a carnot change two isothermal process and two isentropic process. Therefore AS during heat addition must be equal to AS during heating rejection so that which obeys Clausius Inequality. • 5.14 Air flows through a perfectly insulated duct. At one section A the pressure and temperature are respectively 2 bar 2000C and at another section B further along the duct the corresponding values are 1.5 bar and 1500C. Which way the air flowing? System Process Known To find Diagram Open Steady flow process 2 bar 2000c 1.5 bar 1500c • To know flow direction
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    Analysis Prob 5.15 : System Process Known To find Diagram Analysis This problem cannot be solved by simple application of first law of thermodynamics. Because there is nothing to tell us whether the fluid is expanding from A to B or being compressed from B to A. However, since the duct is insulated the inference is that there is no heat transfer to or from the environment and therefore there is no change of entropy in the environment. But in any real process change of entropy of the system plus the surroundings must be positive. In otherwords ASAB > 0 Thus SA > SB and the flow is from B to A. Even though entropy cannot be measured directly it can still be used to find the sense of flow in a well insulated duct given two salient states as above. A certain fluid undergoes expansion in a nozzle reversibly and adiabatically from 500 kPa, 500 K to 100 kPa. What is the exit velocity? Take Y = 1.4 and R = 0.287 . Open Reversible adiabatic expansion 1) Inlet pressure 2) Inlet temperature 3) Exit pressure 4) The ratio of Specific heats 5) Characteristic Gas constant Exit velocity 500 kPa 500 K 100 kPa 1.4 0.287 Applying Steady Flow Energy Equation
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    Therefore where C and T 2 are unknowns. Tofind cp Tofind T2 Prob 5.16 : System Process Known p c c Substituting y and R we get It is stated in the problem that the process of expansion is reversible. Therefore Also the process is given as adiabatic. That is (or) s s 315.8 K Substituting numerical values for T 2 and C , we get Show from the first principle that, for a perfect gas with constant specific heat capacities expanding polytropically (pvn = constant) in a non-flow process, the change of entropy can be expressed by Gaseous methane is compressed polytropically by a piston from 250C and 0.8 bar to a pressure of 5.0 bar. Assuming an index of compression of 1.2, calculate the change of entropy and workdone, per unit mass of gas. The relative molecular weight of methane is 16 and y = 1.3. . Closed . Polytropic (pVn C) 2) 298 K 1 80 kPa 2 500 kPa 1.2
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    M W To find Analysis 5) 1) 2) 16 1.3 Work done AS — Change in entropy To prove S 2 — S 1 From First Law of Thermodynamics Q12 IW2+AU In differential form for a polytropic process Therefore Upon integration we get From the process relation Substituting for we get We know that R c c 1) Substituting for cv we get (2) Workdone 404.45 K Substituting numerical values (3) Change in entropy
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    Comment Prob 5.17 : System Process Known To find Analysis Prob 5.18 : The negative sign in work indicates that work is given into the system. The negative sign in entropy change indicates that there is a heat rejection by the system to the ambient during compression. A closed system undergoes the internally reversible process as shown below : Compute the heat transfer. Closed Defined by a straight line on a T-S diagram. T T s 1 s 2 200 K 600 K 1 kJ/K 3 kJ/K Heat transfer Q Area under the curve representing the process in a T-S diagram 800 Id In a refrigerant condenser superheated vapour of ammonia enters steadily at 1.4 MPa, 700c. It leaves the condenser at 200c. At 1.4 MPa condensation begins and ends at 36.280C. Cooling water enters the condenser at IOOC and leaves 150C. Determine (a) the amount of heat rejected per kg of ammonia vapour condensed for the given inlet and exit conditions. (b) mass of water to be supplied for each kg of ammonia vapour condensed (c) the change in specific entropy of ammonia (d) the entropy generation per kg of ammonia Take cp = 2.9 kJ/kgK, cp . = 4.4 kJ/kgK and latent heat of liquid vapour evaporation of ammonia at 1.4 MPa is 1118 kJ/kg. Also represent the process in a T-s diagram.
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    System Process Known To find Diagrams Analysis Open Steady flow process 1 1 2 1 2 (a) 700c 1.4 MPa 200c 1 ooc 150c the amount of heat rejected per kg of ammonia vapour condensed for the given inlet and exit conditions. (b) mass of water to be supplied for each kg of ammonia vapour condensed (c) the change in specific entropy of ammonia (d) the entropy generation per kg of ammonia (a) Heat rejected per kg of ammonia Q12 QI 2 a + Q 2 2a -2b 2.9 (70 - 36.28) + 1118 + 4.4 (36.28 1287.42 kJ/kg (b) Water flow rate required per kg of ammonia 61.51 (c) Change in Specific entropy of ammonia 20) (d) ASI 2a + AS 2a -2b 2b -2 4.153 AS AS -k AS universe Water ammonia
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    where AS mc In Water 61.51 x 4.186 x In 4.509 Comment Prob 5.19 : System Process Known To prove Proof Prob 5.20 : Substituting the values we get 4.509 + 4.153) AS universe 0.356 As heat is removed from ammonia its entropy decreases whereas entropy of water increases as it receives heat. But total entropy change will be positive as heat is transferred through finite temperature difference. The specific heats of a gas are given by Cp = a + kT and C = b + kT, where a, b and k are constants and T is in K. Show that for an isentropic expansion of this gas = constant Closed Isentropic 1) Cp=a+kT 2) Cv=b+kT constant for an isentropic process For a gas (or) R — a — b For an isentropic process ds = 0 (or) Substituting for cv and R Upon integration binT + KT + (a - b) Inv Taking antilog Tev constant constant Calculate the entropy change of the universe as a result of the following process :
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    (a) A metal block of 0.8 kg mass and specific heat capacity 250 J/kgK is placed in a lake at 80C (b) The same block, at 80C, is dropped from a height of 100 m into the lake. (c) Two such blocks, at 1000C and OOC, are joined together. Case (a) System Process Known To find Diagram Analysis : A metal block Cooling the metal block by dipping it in a lake. 1) Initial temperature of the block (T + 273 2) Final temperature of the block (T2) — 8 + 273 3) Mass of the metal block (m) 4) Specific heat capacity of the metal block (C) Entropy change of the universe AS 373 K 281 K 0.8 Kg 250 universe Where Where Q sur surroundings system Qsys 0.8 x 250 (281 18400 J 373) Substituting the values we get AS universe 56.6 + 65.48 8.84 J/K Comment : As discussed earlier the entropy change of the universe is positive. The reason is the irreversible heat transfer through finite temperature difference from the metal block to the lake.
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    Case (b) System Process Known Diagrams Analysis Comment Case (c) Systems Process Known A metal block Falling of the metal block into the lake and reaching equilibrium 1) Initial Temperature 2) Final Temperature 3) Initial height 4) mass of the metal block (m) 5) Specific heat capacity of the metal block (C) AS AS -k AS universe surroundings system 281 K 281 K 100 m 0.8 kg 250 J/kgK 0, as the system is at the same temperature at both the Where AS system initial and final state. surroundings system mgh 0.8 x 9.81 x 100 784.8 J Increase in entropy of the universe indicates that there is a irreversibility or degradation of energy. That is the high grade potential energy is converted low grade energy during this process. Two metal blocks Two metal blocks are brought in thermal contact and allowed equilibrium. Initial temperatures of the blocks Tla -373 K to reach
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    To find Diagrams Analysis Tofind T2 Comment Tlb 273 K Entropy change of the universe AS AS -k AS Where universe Tla) b Tlb) In this process also the heat transfer through finite temperature difference makes the process irreversible which in turn results in increase in entropy of the universe. Prob 5.21 : Each of three identical bodies satisfies the equation U = CT, where C is the heat capacity of each of the bodies. Their initial temperatures are 200 K, 250 K and 540 K. If C = 8.4 kJ/K, what is the maximum amount of work that can be extracted in a process in which these are brought to a final common temperature ? bodies System Process Known To find Diagram Three identical bodies Extracting work with heat transfer among the three bodies so that they reach a common temperature. Initial temperature of the three bodies - 540 K - 250 K TIC - 200 K Heat capacity of all the three bodies = 8.4 T The maximum amount of work that can be extracted.
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    Analysis Let us assume that the final temperature is greater than 250 K, so that heat is transferred from body (1) to bodies (2) and (3). The net work obtained (AU)I (AU)2 (AU)3 C [(540 - - (T2 250) - (T2 200)] 8.4 [990 — 3T2] This work will be maximum if the process under consideration is reversible. Therefore ASI -k AS 2 -k AS 3 — O Therefore T2 - 300 K This is the condition for the process to be reversible. Hence the maximum work that can be obtained is Prob 5.22 : System Process temperature. Known To find w max 8.4 (990 - 3 x 300) 8.4 (90) 756 Id A resistor of 50 ohm resistance carries a constant current of 5A. The ambient temperature and the temperature of the resistor remain to be 270C over a period of 10 seconds. What is the entropy change of the resistor, ambient and the universe ? A resistor Passing of the electrical current through a resistor at constant 1) 2) 3) Initial and final temperature of the resistor Ambient Temperature Duration (i) 300 K 300 K 10 seconds
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    Analysis Comment 1) AS resistor 2) AS ambient 3) AS universe AS universe AS resistor ambient 0+41.7 41.7 When current passes through a resistor it is converted into heat. As the resistor is to be maintained at the same temperature the heat is dissipated into the ambient and hence the process is irreversible resulting in increase of entropy of the universe. Prob 5.23 : System Process Known To find Diagram A closed system is assumed to have a heat capacity at constant volume where a = 0.002 and T is the temperature in K. The system is originally at 600 K and a thermal reservoir at 300 K is available. What is the maximum amount of work that can be recovered as the system is cooled down to the temperature of the reservoir ? A closed system Obtaining work with the help of the heat transfer from the system to the reservoir. 0.002T 600 K 300 K max
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    Exercises Analysis If the work is to be maximum, the process must be reversible. Therefore universe AS -k AS reservoir system Neglecting higher order terms, Ans : True Ans : True Therefore AS reservoir Thus maximum work AS system 1.731 911 519.3 391.7 Id 1. 2. 3. 4. 0 for processes Ans : Reversible Entropy is a point function (True / False) Entropy change of universe can never be negative (True / False) All the isentropic processes are reversible adiabatic processes (True / False)
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    5. 6. 7. 8. 9. 10. ll. 12. 13. 14. 15. 16. 17. Ans : False What is the difference between adiabatic and isentropic process? A system is losing 500 kJ of heat at a constant temperature of 500 K. What is the change in entropy ? Area under any curve in T-s diagram represents Ans : heat p — constant lines are steeper than v constant lines in T-S diagram (True/False) During throttling process entropy Ans : False (Increases / Decreases) for an ideal gas. Ans : Increases Find the entropy change of the universe when 1000 kJ of heat is transferred from 800 K to 500 K. Give the expression for change in entropy during isothermal processes and polytropic processes. Calculate the change in entropy per kg of air when it expands isothermally from 3bar. A closed system undergoes an adiabatic process. Work done on the system is The entropy change of the system a) is positive b) is negative c) can be positive or negative 6 bar to 15 kJ/kg. Ans : positive Give the interpretation of entropy from microscopic point of view. 3 A quantity of gas has an initial pressure, volume and temperature of 140 kPa, 0.14 m 1.25 and 250C respectively. It is compressed to a pressure of I .4 MPa according to the law PV C. Determine the change in entropy Take c 1.041 kJ/ kgK, - 0.743 kJ/kgK. p Ans: 0.207 kJ/kgK 1 kg of air has a volume of 56 litres and a temperature of 1900C. The air then receives heat at constant pressure until its temperature becomes 5000C. From this state the air rejects heat at constant volume until its pressure is reduced to 700 kPa. Determine the change of entropy during each process, stating whether it is an increase or decrease. Ans : 0.516 kJ/kgK, an increase — 0.88 kJ/kgK, an decrease A quantity of gas has an initial pressure, volume and temperature of 1.1 bar, and 180C respectively. It is compressed isothermally to a pressure of Determine the change of entropy. Take R 0.3 kJ/kgK. 0.16 m3 6.9 bar.
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    18. 19. 20. 21. 22. 23. Ans : -0.111 kJ/K A reversible heat engine shown in figure below operates between three constant temperature reservoirs at 600 K, 400 K and 300 K. It receives 2500 kJ of energy of heat from the reservoir at 600 K and does 1000 kJ of work. Determine the heat interactions with the other two reservoirs. Ans : Q2 = 1008, Q3 = 4926 A block of copper with a mass of 1.5 kg is initially at 700 K. It is allowed to cool by means of heat transfer to the surrounding air at 300 K. Determine the change in entropy of the copper and change in entropy of the universe after copper reaches thermal equilibrium. Assume specific heat of copper is 0.39 kJ/kgK. Ans: 0.4967 kJ/K, +0.2843 kJ/K Using the principle of increase in entropy prove that the heat transfer is always from a high-temperature body to a low temperature body. Nitrogen at 420 K and 1.4 MPa is expanded reversibly and adiabatically in a nozzle to exit pressure of 700 kPa. Determine the temperature and velocity of the nitrogen at the exit of the nozzle. Take 1.40. A vessel is divided into two temperature by means of a membrane as shown in the figure given below. What will be the final state of air and change in entropy of the universe if the membrane is removed. Ans : 750.14 kPa, 65.110C = 0.373 A given gaseous system undergoes an isentropic process from statel to state 2. a) Combine the two relations pv RT and pvY C and show that b) Integrate the two expressions, using pvY C and show that is y times by comparison.
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    24. 25 26 27 28 29 30 During the isentropic process of 1.36 kg/s of air, the temperature increases from 4.440C to 115.60C. For a non-flow process and for a steady flow process find a) Change in internal energy b) Work done c) Change in enthalpy d) Change in entropy and e) Heat transfer Air at 5 bar, IOOOC, expands reversibly in a piston-cylinder arrangement. It expands to 2 bar in an isothermal process. Calculate (a) heat transfer per unit mass of air (b) change in specific internal energy (c) change in specific entropy 1.3 One kg of air at 1 bar, 200C, is compressed according to the law pv constant until the pressure is 5 bar. Calculate the change in entropy and sketch the process on a T-S diagram indicating the area representing the heat flow. 1 kg of air at Ibar, 250C, changes its state to 6 bar and a volume of 1 m . Calculate the change of entropy and sketch the initial and final state points on the p-v and T-S fields. 0.5 m ethane (C2H4) at 7 bar, 2600C expand isentropically in a cylinder behind a piston to 1 bar, 1000C. Calculate the workdone in expansion assuming ethane as perfect gas. The 1.35 same mass is now recompressed back to 7 bar according to the law pv constant. Calculate the final temperature and the heat transfer. Calculate also the change in entropy and sketch both process on the p-v and T-S fields. ethane. Take cp for A mass m of water at Tl is mixed with equal mass of water at T 2 at the same pressure in an insulated mixing chamber. Show that the entropy change of the Universe is given as Consider a closed system consisting of air as working fluid in a piston cylinder arrangement as shown in the Figure. The weight placed on the piston is such that the air occupies a volume of 1.5 litre when there exist thermodynamic equilibrium between the system and its surroundings. The corresponding pressure and temperature are 2 bar, 300C. Heat is added until the pressure
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    increases to 5 bar. Volume of air when the piston touches the stop is 3 litres. Find the following a) b) c) d) Final temperature Workdone Heat transformed Change in entropy 31 32. An ideal vapour absorption refrigeration system may be considered as the combination of the reversible heat engine operating a reversible refrigerator as given in the following diagram. Obtain the COP of the refrigeration system which is defined as the ratio of Qe to Vapour absorption heat transformer is a novel device used for upgrading a portion of waste heat from low temperature to high temperature. An ideal vapour absorption heat transformer may be considered as the combination of the reversible heat engine operating a reversible heat pump as given in the following diagram. Obtain the COP of the vapour absorption heat transformer which is defined as the ratio of Qa to (Qg + Q).

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