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- 1. Mid-Term ReviewTopics1) Energy Analysis of Closed System (Lecture 4, Chap. 4of textbook).2) Mass and Energy Analysis Of Control Volumes(Lecture 5, Chap. 5 of textbook).3) Second Law of Thermodynamics (Lecture 6, Chap. 6of textbook).1
- 2. 21) Energy Analysis of Closed SystemKnow how to interpret condition at final or initialstate, given the nature of the process; e.g. constantpressure process, isothermal process, process inside aclosed rigid tank, process inside a piston-cylinder device.Know how to obtain boundary work wb for the constantpressure process, the isothermal process and thepolytropic process.Know and understand the various forms of the energybalance equation for a closed system.
- 3. 3Examples (Energy Analysis of Closed System)1. An insulated piston–cylinder device contains 5 L of saturatedliquid water at a constant pressure of 175 kPa. Water is stirred bya paddle wheel while a current of 8 A flows for 45 min through aresistor placed in the water.If one-half of the liquid is evaporated during this constantpressure process and the paddle-wheel work amounts to 400kJ, determine the voltage of the source. Also, show the process ona P-v diagram.
- 4. 4SolutionWe take the contents of the cylinder as the system. This is aclosed system since no mass enters or leaves. The energybalance for this stationary closed system can be expressed assince ΔU + Wb = ΔH during a constant pressure quasi-equilibrium process.
- 5. 5The properties of water areSubstituting,
- 6. 62. A mass of 5 kg of saturated liquid–vapor mixture of water iscontained in a piston–cylinder device at 150 kPa. Initially, half ofthe water is in the liquid phase and the rest is in the vapor phase.Heat is now transferred to the water, and the piston, which isresting on a set of stops, starts moving when the pressure insidereaches 300 kPa. Heat transfer continues until the total volumeincreases by 20 percent. Determine (a) the initial and finaltemperatures, (b) the mass of liquid water when the piston firststarts moving, and (c) the work done during this process.Also, show the process on a P-v diagram.
- 7. 7SolutionInitially the system is a saturated mixture at 150 kPa pressure, andthus the initial temperature isTi = Tsat@150 kPa = 111.35oCThe specific volumes at 150 kPa are vf = 0.001053 m3/kg and vg =1.1594 m3/kg; thus total initial volume isV1 = mf vf + mg vg= (2.5 kg)( 0.001053 m3/kg) + (2.5 kg)(1.1594 m3/kg)= 2.901 m3Then the total and specific volumes at the final state areV3 = 1.2 V1 = 1.2 2.901 = 3.481 m3v3 = V3/m = 3.481 m3 / 5 kg = 0.69627 m3/kThus at the final state, P3 = 300 kPa, v3 = 0.69627 m3/kg.
- 8. 8At 300 kPa, vg = 0.60582 m3/kg. Hence since v3 > vg@300 KPa, state3 is superheated vapour. Interpolating from Table A-6(b) When the piston first starts moving, P2 = 300 kPa and V2 = V1= 2.901 m3. The specific volume at this state isV2 = V2/m = 2.901 m3 / 5 kg = 0.5802 m3/kgwhich is less than vg = 0.60582 m3/kg at 300 kPa. Thus the qualityat state 2 isThus mass of the liquid at state 2 = (5 kg)(1 0.9576) = 0.212 kg.33150 0.69627 0.63402 0.69627 0.63402150 200 150 187.8 C200 150 0.71643 0.63402 0.71643 0.63402TT 2 @300 kPa2@300 kPa0.5802 0.0010730.95760.60582 0.001073ffgv vxv
- 9. 9(c) No work is done during process 1-2 since V1 = V2. Thepressure remains constant during process 2-3 and the work doneduring this process is3 32 3 2 300 kPa 3.481 2.901 m 206.4 kPa m 174 kJbW P V V
- 10. 10ExampleSteam enters a turbine operating at steady state with a mass flowrate of 4600 kg/h. The turbine develops a power output of 1000kW. At the inlet, the pressure is 60 bar, the temperature is400C, and the velocity is 10 m/s. At the exit, the pressure is 0.1bar, the quality is 0.9 (90%), and the velocity is 50 m/s.Calculate the rate of heat transfer between the turbine andsurroundings, in kW.
- 11. 11(assuming net heat transfer out ofthe system and pe = 0.)
- 12. 12From the tables and using the usual procedures we obtainh1 = 3178.3 kJ/kg and h2 = 2344.7 kJ/kg. AlsoSubstituting,Thus heat is transferred out of the system at a rate of 63.6 kW.
- 13. 13Expansion or compression process wherepressure is not constant - Examples
- 14. 14FIGURE 4–10 Schematic and P-V diagram for Example 4–4.
- 15. 15
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- 17. 17
- 18. 18.
- 19. 19
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- 21. 21
- 22. 22A waterheater insteadyoperation.MassbalanceEnergybalance2) Mass and Energy Analysis OfControl Volumes
- 23. 23Turbines andCompressorsTurbine drives the electric generator In steam,gas, or hydroelectric power plants.As the fluid passes through the turbine, work isdone against the blades, which are attached tothe shaft. As a result, the shaft rotates, and theturbine produces work.Compressors, as well as pumps and fans, aredevices used to increase the pressure of afluid. Work is supplied to these devices froman external source through a rotating shaft.A fan increases the pressure of a gas slightlyand is mainly used to mobilize a gas.A compressor is capable of compressing thegas to very high pressures.Pumps work very much like compressorsexcept that they handle liquids instead ofgases.Energy balance for thecompressor in this figure:
- 24. 24Example 1The power output of anadiabatic steam turbine is 5MW, and the inlet and theexit conditions of the steamare as indicated in thediagram.(a) Compare the magnitudesof h, ke, and pe.(b) Determine the work doneper unit mass of the steamflowing through the turbine.(c) Calculate the mass flowrate of the steam.
- 25. 25SolutionWe take the turbine as the system. This is a control volume sincemass crosses the system boundary during the process. Weobserve that there is only one inlet and one exit and thus(a) At the inlet, steam is in a superheated vapor state, and itsenthalpy isAt the turbine exit, we have a saturated liquid–vapor mixture at15 kPa. The enthalpy at this state is.21 mm
- 26. 26Then
- 27. 27Single inlet and outlet, subs. 1 inlet, 2 outlet
- 28. 28
- 29. 29A steam turbine has an inlet of 2 kg/s steam at 1000 kPa, 350 Cand velocity of 15 m/s. The exit is at 100 kPa, x = 1 and very lowvelocity. Determine the power output of the turbine in kW.SolutionProperties of waterExample 2
- 30. 30The energy balance equation for the steady flow turbine isNeglecting change in PE and given that V2 0, we then haveOrSubstituting values, we have
- 31. 31At 2 kg/s steam flow rate, the power output
- 32. 32A well-insulated turbine operating at steady conditions develops 23MW of power for a steam flow rate of 40 kg/s. The steam enters at360°C with a velocity of 35 m /s and exits as saturated vapor at 0.06bar with a velocity of 120 m /s. Neglecting potential energyeffects, determine the inlet pressure, in bar.SolutionProperties of waterExample 3
- 33. 33Neglecting change in PE and heat transfer (as turbine is wellinsulated), the energy balance equation for the steady flow processon a unit mass basis is,ThusSubstituting values,= 3147.6 kJ/kg
- 34. 34HenceChecking Table A-4, we find hg@360 C = 2481.6 kJ/kg. Since h1 ishigher, we have superheated vapour state at 1.(Note: Steam at turbine inlet is always superheated.)Interpolating from Table A-6 (see below), we find P1 = 2.6 MPa or26 bar.
- 35. 35Refrigerant-134a enters an adiabatic compressor as saturated vaporat 24°C and leaves at 0.8 MPa and 60°C. The mass flow rate ofthe refrigerant is 1.2 kg/s. Determine (a) the power input to thecompressor and (b) the volume flow rate of the refrigerant at thecompressor inlet.SolutionFrom the refrigerant tablesExample 4
- 36. 36The energy balance for this steady-flow system can be expressed inthe rate form as
- 37. 37Air enters the compressor of a gas-turbine plant at ambientconditions of 100 kPa and 25 C with a low velocity and exits at1 MPa and 347 C with a velocity of 90 m/s. The compressor iscooled at a rate of 1500 kJ/min, and the power input to thecompressor is 250 kW. Determine the mass flow rate of airthrough the compressor.Example 5
- 38. 38SolutionThe energy balance for this steady-flow system can be expressedin the rate form asSubstituting, the mass flow rate is determined to be
- 39. 39Throttling valvesThrottling valves are any kind of flow-restricting devices thatcause a significant pressure drop in the fluid.What is the difference between a turbine and a throttlingvalve?The pressure drop in the fluid is often accompanied by a largedrop in temperature, and for that reason throttling devices arecommonly used in refrigeration and air-conditioningapplications.The temperature of an ideal gas does notchange during a throttling (h =constant) process since h = h(T).During a throttling process, the enthalpy of afluid remains constant. But internal and flowenergies may be converted to each other.Energybalance
- 40. 40Example – Expansion of R-134a in a RefrigeratorRefrigerant-134a enters the capillary tube of a refrigerator assaturated liquid at 0.8 MPa and is throttled to a pressure of 0.12MPa. Determine the quality of the refrigerant at the final stateand the temperature drop during this process.
- 41. 41SolutionA capillary tube is a simple flow-restricting device that iscommonly used in refrigeration applications to cause a largepressure drop in the refrigerant. Flow through a capillary tube is athrottling process; thus, the enthalpy of the refrigerant remainsconstant.Since hf < h2 < hg, the refrigerant exists as a saturated mixture atthe exit. The quality of this mixture is
- 42. 42Since the exit state is a saturated mixture at 0.12 MPa, the exittemperature must be the saturation temperature at thispressure, which is 22.32°C. Then the temperature change for thisprocess is
- 43. Part of the heat received by a heatengine is converted to work, while therest is rejected to a sink.The devices that convert heat to work.1. They receive heat from a high-temperature source (solar energy, oilfurnace, nuclear reactor, etc.).2. They convert part of this heat towork (usually in the form of arotating shaft.)3. They reject the remaining wasteheat to a low-temperature sink (theatmosphere, rivers, etc.).4. They operate on a cycle.Heat engines and other cyclic devicesusually involve a fluid to and from whichheat is transferred while undergoing acycle. This fluid is called the workingfluid.Work can be converted to heat directlyand completely, but converting heat towork requires the use of heat engines.433) Second Law of ThermodynamicsHeat Engines
- 44. 44Thermal efficiencySome heat engines perform betterthan others (convert more of theheat they receive to work).Schematic of aheat engine.Even the mostefficient heatengines rejectalmost one-halfof the energythey receive aswaste heat.
- 45. 45Example 1Heat is transferred to a heatengine from a furnace at a rateof 80 MW. If the rate of wasteheat rejection to a nearby riveris 50 MW, determine the netpower output and the thermalefficiency for this heat engine.
- 46. 46SolutionA schematic of the heat engine is shown in the previous slide.The furnace serves as the high-temperature reservoir for thisheat engine and the river as the low-temperature reservoir. Thegiven quantities can be expressed as
- 47. 47Example 2A car engine produces 136 hp on theoutput shaft. The thermal efficiency ofthe engine is 30%. Find the rate of heatrejected to the ambient and the rate offuel consumption if the engine uses afuel with a heating value of 35,000kJ/kg.Note: 1 hp = 745.7 W = 0.7457 kW
- 48. 48Solutionskg.kJ/kgkW.HVnconsumptiofuelofRatekW.hp..kWhp..009703500023387236331713634532338345330136HLtHHtQQ.WQQW
- 49. 49An automobile engine consumes fuel at a rate of 28 L/h and delivers60 kW of power to the wheels. If the fuel has a heating value of44,000 kJ/kg and a density of 0.8 kg/L, determine the thermalefficiency of this engine.Example 3
- 50. 50SolutionThe mass consumption rate of the fuel isThe rate of heat supply to the car isThe thermal efficiency of the car is then
- 51. 51REFRIGERATORS AND HEAT PUMPS• The transfer of heat from a low-temperature medium to a high-temperature one requires specialdevices called refrigerators.• Refrigerators, like heat engines, arecyclic devices.• The working fluid used in therefrigeration cycle is called arefrigerant.• The most frequently usedrefrigeration cycle is the vapor-compression refrigeration cycle.Basic components of arefrigeration system andtypical operating conditions.In a household refrigerator, the freezer compartmentwhere heat is absorbed by the refrigerant serves as theevaporator, and the coils usually behind the refrigeratorwhere heat is dissipated to the kitchen air serve as thecondenser.
- 52. 52Coefficient of PerformanceThe objective of a refrigerator is toremove QL from the cooled space.The efficiency of a refrigerator is expressed interms of the coefficient of performance (COP).The objective of a refrigerator is to remove heat(QL) from the refrigerated space.
- 53. 53HeatPumpsThe objective ofa heat pump istosupply heat QHinto thewarmer space.The worksupplied to aheat pump isused to extractenergy from thecold outdoorsand carry it intothe warmindoors.for fixed values of QL and QH
- 54. 54The food compartment of arefrigerator, shown at right, ismaintained at 4°C by removingheat from it at a rate of 360kJ/min. If the required powerinput to the refrigerator is 2kW, determine(a) the COP of the refrigerator and(b) the rate of heat rejection tothe room that houses therefrigerator.Example 1
- 55. 55Solution(a) The coefficient of performance of the refrigerator isThat is, 3 kJ of heat is removed from the refrigerated space for eachkJ of work supplied.(b) The rate at which heat is rejected to the room that houses therefrigerator is determined from the conservation of energy relation:
- 56. 56A heat pump is used to meet theheating requirements of a house andmaintain it at 20°C. On a day when theoutdoor air temperature drops to 2°C,the house is estimated to lose heat ata rate of 80,000 kJ/h. If the heat pumpunder these conditions has a COP of2.5, determine(a) the power consumed by the heatpump and(b) the rate at which heat is absorbedfrom the cold outdoor air.Example 2
- 57. 57(a) The power consumed by this heat pump is determined from thedefinition of the coefficient of performance to be(b) The house is losing heat at a rate of 80,000 kJ/h. If the house isto be maintained at a constant temperature of 20°C, the heat pumpmust deliver heat to the house at the same rate, that is, at a rate of80,000 kJ/h. Then the rate of heat transfer from the outdoorbecomes
- 58. 58When a man returns to his well-sealed house, he finds that thehouse is at 32°C. He turns on the air conditioner, which cools theentire house to 20°C in 15 min. If the COP of the air-conditioningsystem is 2.5, determine the power drawn by the air conditioner.Assume the entire mass within the house is equivalent to 800 kg ofair for which cv = 0.72 kJ/kg · °C and cp = 1.0 kJ/kg · °C.SolutionWe first have to find the amount of heat that need to be removed.Since the house is well-sealed, we can assume a constant volumecooling process so thatThis heat is removed in 15 minutes. Thus the average rate of heatremoval from the house isExample 3
- 59. 59Using the definition of the coefficient of performance, the powerinput to the air-conditioner is determined to be
- 60. 60Refrigerant-134a enters the condenser of a residentialheat pump at 800 kPa and 35°C at a rate of 0.018 kg/s andleaves at 800 kPa as a saturated liquid. If the compressor consumes1.2 kW of power, determine(a) the COP of the heat pump and(b) the rate of heat absorption from the outside air.Example 4
- 61. 61Solution(a) The enthalpies of R-134a at the condenser inlet and exit areAn energy balance on the condenser gives the heat rejected inthe condenserThe COP of the heat pump is
- 62. 62(b) The rate of heat absorbed from the outside air
- 63. 63THE CARNOT HEAT ENGINEThe Carnotheat engineis the mostefficient ofall heatenginesoperatingbetween thesame high-and low-temperaturereservoirs.No heat engine can have a higher efficiencythan a reversible heat engine operatingbetween the same high- and low-temperature reservoirs.Any heatengineCarnot heatengine
- 64. 64A Carnot heat engine receives 500 kJof heat per cyclefrom a high-temperature source at652°C and rejects heat to a low-temperature sink at 30°C. Determine(a) the thermal efficiency of thisCarnot engine and(b) the amount of heat rejected tothe sink per cycle.Example
- 65. 65(a) The Carnot heat engine is a reversible heat engine, and so itsefficiency is given by(b) The amount of heat rejected QL by this reversible heat engineisSolution
- 66. 66THE CARNOT REFRIGERATORAND HEAT PUMPNo refrigerator can have a higher COPthan a reversible refrigerator operatingbetween the same temperature limits.Any refrigerator or heat pumpCarnot refrigerator or heat pump
- 67. 67Example 1An inventor claims to have developed a refrigerator thatmaintains the refrigerated space at 2°C while operating in a roomwhere the temperature is 24°C and that has a COP of 13.5. Is thisclaim reasonable?SolutionThe maximum COP is the COP of a reversible refrigeratoroperating between these temperature limits.Since the COPR,rev is lower than the claimed COP, the claim cannotbe justified.5121227324273111.//COPCOP max,Rrev,RLH TT
- 68. 68Example 2A heat pump is to be used toheat a house during winter.The house is to bemaintained at 21°C at alltimes. The house isestimated to be losing heatat a rate of 135,000 kJ/hwhen the outsidetemperature drops to 5°C.Determine the minimumpower required to drive thisheat pump.
- 69. 69SolutionThe heat pump must supply heat to the house at a rate of QH =135,000 kJ/h = 37.5 kW. The power requirements are minimumwhen a reversible heat pump is used to do the job. The COP of areversible heat pump operating between the house and theoutside air isThe required power input to this reversible heat pump is then3112127352731111.//COP rev,HPHL TTkW..kW.COPHP323311537H,innetQW
- 70. 70THE CARNOTCYCLEReversible Isothermal Expansion (process 1-2, TH = constant)Reversible Adiabatic Expansion (process 2-3, temperature drops from TH to TL)Reversible Isothermal Compression (process 3-4, TL = constant)Reversible Adiabatic Compression (process 4-1, temperature rises from TL to TH)Most efficientcycles arereversiblecycles. Mostfamiliarreversiblecycle is theCarnot cycleExecution of theCarnot cycle in aclosed system.
- 71. 71P-V diagram of the Carnot cycle. P-V diagram of the reversedCarnot cycle.The Reversed Carnot CycleThe Carnot heat-engine cycle is a totally reversible cycle.Therefore, all the processes that comprise it can be reversed, in whichcase it becomes the Carnot refrigeration cycle.
- 72. 72THE CARNOT PRINCIPLES1. The efficiency of anirreversible heat engineis always less than theefficiency of areversible oneoperating between thesame two reservoirs.2. The efficiencies of allreversible heat enginesoperating between thesame two reservoirs arethe same.
- 73. 73Proof of the First Carnot principle.Let reversible HE run as refrigerator (dashedline). If heat from refrigerator is supplied directlyto irrev. HE, can eliminate HT reservoir. But whatdo we have then?A HE operating from a single reservoir andproducing work amounting to Wirrev WrevIf Wirrev > Wrev, net work is produced and thisviolate the 2nd Law. Hence Wirrev < Wrev.

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