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- 1. THERMODYNAMICS Part 1By. Engr. Yuri G. MellizaTerms & DefinitionProperties of FluidsForms of EnergyLaw of Conservation of MassLaw of Conservation of Energy(First Law of Thermodynamics)Ideal GasPure SubstanceProcesses of FluidsZeroth Law of Thermodynamics
- 2. Thermodynamics is a science that deals withenergy transformation or conversion of oneform of energy to another formTherme – “Heat”Dynamis – “Strength”System: A portion in the universe, an Atom, aGalaxy, a certain quantity of matter, or a certainvolume in space in which one wishes to study. It is a region enclosed by a specified boundarythat may Imaginary, Fixed or Moving.
- 3. SystemSurrounding orEnvironment
- 4. Open System: A system open to matter flow.Example: Internal Combustion Engine (ICE)
- 5. Closed System: A system close to matter flow.Example: Piston - in - cylinder
- 6. Working Substance (Working Fluid): A fluid(Liquid or Gas) responsible for thetransformation of energy.Example: air in an air compressor Air and fuel mixture in an internal combustionengine
- 7. Pure Substance: A substance that ishomogeneous in nature and is homogeneous.Example : WaterPhases of a SubstanceA phase refers to a quantity of matter that ishomogeneous throughout in both chemical compositionand physical structure. Solid Liquid Gas or Vapor
- 8. Specific Terms To Characterized PhaseTransitionSOLIDIFYING OR FREEZING - Liquid toSolidMELTING - Solid to LiquidVAPORIZATION - Liquid to VaporCONDENSATION - Vapor to LiquidSUBLIMATION - a change from soliddirectly to vapor phase without passing theliquid phase.
- 9. Mass : It is the absolute quantity of matter init.m - mass in kgAcceleration : it is the rate of change ofvelocity with respect to time t.a = dv/dt m/sec2Velocity: It is the distance per unit time.v = d/t m/sec
- 10. Force - it is the mass multiplied by the acceleration.F = ma/1000 KN1 kg-m/sec2 = Newton (N)1000 N = 1 Kilo Newton (KN)Newton - is the force required to accelerate 1 kg mass at the rate of 1 m/secper second.1 N = 1 kg-m/sec2From Newton`s Law Of Gravitation: The force of attraction between twomasses m1 and m2 is given by the equation:Fg = Gm1m2/r2 NewtonWhere: m1 and m2 - masses in kgr - distance apart in metersG - Gravitational constant in N-m2/kg2G = 6.670 x 10 -11 N-m2/kg2WEIGHT - is the force due to gravity.W = mg/1000 KNWhere: g - gravitational acceleration at sea level, m/sec2g = 9.81 m/sec2
- 11. 3mkgVmρPROPERTIES OF FLUIDSWhere: - density in kg/m3m - mass in kgV – volume in m3Specific Volume ( ) - it is the volume per unit mass or thereciprocal of its density.kgmmV3υkgm3ρ1υDensity ( ) - it is the mass per unit volume.
- 12. Specific Weight ( ) - it is the weight per unit volume.33mKNmKN1000VmgγVWγWhere: - specific weight in KN/m3m – mass in kgV – volume in m3g – gravitationalAt standard condition:g = 9.81 m/sec2
- 13. Specific Gravity Or Relative Density (S):FOR LIQUIDS: Its specific gravity or relative density is equal tothe ratio of itsdensity to that of water at standard temperature and pressure.wLwLLγγρρSFOR GASES: Its specific gravity or relative density is equal to theratio of itsdensity to that of either air or hydrogen at some specified temperature andpressureahGGρρSWhere at standard condition:w = 1000 kg/m3w = 9.81 KN/m3
- 14. Temperature: It is the measure of the intensity of heat in a body.Fahrenheit Scale:Boiling Point = 212 FFreezing Point = 32 FCentigrade or Celsius Scale:Boiling Point = 100 CFreezing Point = 0 CAbsolute Scale:R = F + 460 (Rankine)K = C + 273 (Kelvin)32F8.1F8.132FCConversion
- 15. Pressure: It is the normal component of a force per unit area.KPaor2mKNAFPWhere: P – pressure in KN/m2 or KPaF – normal force in KNA – area in m21 KN/m2 = 1 KPa (KiloPascal)1000 N = 1 KNIf a force dF acts on an infinitesimal area dA, the intensity of Pressure is;KPaor2mKNdAdFP
- 16. Pascal’s Law: At any point in a homogeneous fluid at rest the pressures arethe same in all directions:yxzABCP1A1P2A2P3A3
- 17. Fx = 0From Figure:P1A1 - P3 A3sin = 0P1A1 = P3A3sin Eq.1P2A2 - P3A3cos = 0P2A2 = P3A3 cos Eq.2sin = A1/A3A1 = A3sin Eq.3cos = A2/A3A2 = A3cos Eq.4substituting eq. 3 to eq. 1 and eq.4 to eq.2P1 = P2 = P3
- 18. Atmospheric Pressure (Pa):It is the average pressure exerted bythe atmosphere.At sea levelPa = 101.325 KPa= 0.101325 MPa= 1.01325 Bar= 760 mm Hg= 10.33 m of water= 1.033 kg/cm2= 14.7 lb/in2Pa = 29.921 in Hg= 33.88 ft. of water100 KPa = 1 Bar1000 KPa = 1 MPa
- 19. Absolute and Gauge PressureAbsolute Pressure: It is the pressure measured referred toabsolute zero using absolute zero as the base.Gauge Pressure: it is the pressure measured referred to theexisting atmospheric pressure and using atmospheric pressure asthe base.Pgauge – if it is above atmosphericPvacuum – negative gauge or vacuum if it is belowatmosphericBarometer: An instrument used to determine the absolutepressure exerted by the atmosphere
- 20. Atmospheic pressure (Pa)Absolute ZeroPvacuumPgaugePabsolutePabsolutePabs = Pgauge + PaPabs = Pvacuum - Pa
- 21. VARIATION OF PRESSUREPA(P + dP)AWdhF = 0(P + dP)A - PA - W = 0PA + dPA - PA - W = 0dPA - W = 0 or dPA = W Eq. 1but : W = dVdPA = - dV
- 22. where negative sign is used because distance h is measured upward and Wis acting downward.dV = Adh then dPA = - Adh, thereforedP = - dh(Note: h is positive when measured upward and negative if measureddownward)
- 23. MANOMETERSManometer is an instrument used in measuring gage pressure in length ofsome liquid column.1. Open Type Manometer : It has an atmospheric surface and is capable inmeasuring gage pressure.2. Differential Type Manometer : It has no atmospheric surface and iscapable in measuring differences of pressure.Open TypeOpen endManometer Fluid
- 24. Differential TypeFluid AFluid BFluid C
- 25. ENERGY FORMSWork: It is the force multiplied by the displacement in the directionof the force.W =∫Fdx KJ-W - indicates that work is done on the system+W - indicates that work is done by the system.Heat: It is a form of energy that crosses a systems boundary, becauseof a temperature difference between the system and the surrounding.Q - Heat KJ+Q - indicates that heat is added to the system-Q - indicates that heat is rejected from the system.Internal Energy: It is the energy acquired due to the overall molecularinteraction, or the total energy that a molecule has.U = mu KJU - total internal energy KJu - specific internal energy KJ/kgU- change of internal energy
- 26. Flow Energy Or Flow Work: It is the energy required in pushing a fluidusually into the system or out from the system.System orControl VolumeP1P2A1A2L1L2Ef1 = F1L1F1 = P1A1Ef1 = P1A1L1A1L1 = V1Ef1 = P1V1Ef2 = F2L2F2 = P2A2Ef2 = P2A2L2A2L2 = V2Ef2 = P2V2Ef = Ef2 – Ef1Ef = P2V2 – P1V1Ef = PVPV = P2V2 - P1V1 KJP = P2 2 - P1 1 m3/kg
- 27. Where: P – pressure in KPaV – volume in m3- specific volume in m3/kgEf = PV – Flow energy or flow workKinetic Energy: It is the energy acquired due to the motion of a body or asystem.1 2m mFd xdxFdKEdxFKE
- 28. KJ/kg2(1000)ΔKEKJ2(1000)mΔKEdtdvdx1000mKEdtdv1000m1000maFFdxKEvvvv212221222121000mΔKEdvv1000mΔKEdtdxdv1000mΔKEvv21222121Where:m – mass , kgv – velocity , m/seckgKJ10002vvKEKJ10002vvmKE21222122
- 29. Potential Energy: It is the energy required by virtue of its configuration orelevation.mmdZReference DatumkgKJ1000ZZgPEKJ1000ZZmgPEdZ1000mgPEdZWPE1212Where:W – WorkQ – HeatU – Internal EnergyPV – Flow Energy or flow workKE – Kinetic EnergyPE – Potential EnergyNote:+Z – if measured upward- Z –if measured downward
- 30. Law of Conservation of MassMass is indestructible: In applying this law we must except nuclear processesduring which mass is converted into energy.The verbal form of the law is:Mass Entering - Mass Leaving = Change of Mass stored in the systemIn equation Form:m1 - m2 = m1 2m1 m2m = 0ab cdFor a steady-state, steady-flow system m = 0, thereforem1 - m2 = 0 or m1 = m2
- 31. For one dimensional flow, where 1 = 2 =Let m1 = m2 = mContinuity Equation:υAvAvρmWhere:m - mass flw rate in kg/sec- density in kg/m3- specific volume inm3/kgA - cross sectional area in m2v - velocity in m/sec
- 32. Zeroth Law of ThermodynamicsIf two bodies are in thermal equilibrium with a third body, they are in thermalequilibrium with each other, and hence their temperatures are equal.Specific Heat or Heat Capacity: It the amount of heat required to raisethe temperature of a 1 kg mass of a substance 1 C or 1 K.tmCTmCQm;gConsiderinCdtCdTdQC;constantForK-kgKJorC-kgKJdtdQdTdQC
- 33. SENSIBLE HEAT: The amount of heat per unit mass that must betransferred (added or remove) when a substance undergoes a change intemperaturewithout a change in phase.Q = mC( t) = mC( T)where: m - mass , kgC - heat capacity or specific heat, KJ/kg- C orKJ/kg- Kt - temperature in CT - temperature in KHEAT OF TRANSFORMATION: The amount of heat per unitmass that must be transferred when a substance completely undergoes aphase changewithout a change in temperature.Q = mL
- 34. A. Heat of Vaporization: Amount of heat that must be added tovaporize a liquid or thatmust be removed to condense a gas.Q = mLwhere L - latent heat of vaporization, KJ/kgB. Heat of Fusion : Amount of heat that must be added to melt a solid orthat must beremoved to freeze a liquid.Q = mLwhere L - latent heat of fusion, KJ/kg
- 35. THE FIRST LAW OF THERMODYNAMICS(The Law of Conservation of (Energy)“Energy can neither be created nor destroyed but can onlybe converted from one form to another.”Verbal Form:Energy Entering – Energy Leaving = Change of Energystored in the systemEquation Form:E1 – E2 = Es1. First Corollary of the First Law: Application of first Law to a Closed SystemUQWFor a Closed System (Non FlowSystem),PV, KE and PE are negligible, thereforethe changeof stored energy Es = UQ – W = U 1Q = U + W 2
- 36. By differentiation:dQ = dU + dW 3where:dQ Q2 – Q1dW W2 – W1Work of a Closed System (NonFlow)PVW = PdVPdV5Eq.dVPUQ4Eq.dVPdUdQ3Eq.FromdVPdWdVPWdVAdxdxPAWPAFdxFW
- 37. 2. Second Corollary of the First Law: Application of First Law to an Open SystemSystem orControl volumeDatum LineQW12U1 + P1V1 + KE1 + PE1U2 + P2V2 + KE2 + PE2For an Open system (Steady state, Steady Flow system)Es = 0, thereforeE1 – E2 = 0 orE1 = E2 orEnergy Entering = Energy LeavingZ1Z2
- 38. U1 + P1V1 + KE1 + PE1 + Q = U2 + P2V2 + KE2 + PE2 + W 1Q = (U2 – U1) + (P2V2 – P1V1) + (KE2 – KE1) + (PE2 – PE1) + W 2Q = U + (PV) + KE + PE + W 3By differentiationdQ = dU + d(PV) + dKE + dPE + dW 4But dQ Q2 – Q1 and dW W2 – W1Enthalpy (h)h = U + PVdh = dU + d(PV) 5dh = dU + PdV + VdP 6But: dQ = dU + PdVdh = dQ + VdP 7From Eq. 3Q = h + KE + PE + W 8dQ = dh + dKE + dPE + dW 9dQ = dU + PdV + VdP + dKE + dPE + dW 10dQ = dQ + VdP + dKE + dPE + dW0 = VdP + dKE + dPE + dWdW = -VdP - dKE - dPE 11By IntegrationW = - VdP - KE - PE 12
- 39. If KE = 0 and PE = 0Q = h + W 13W = Q - h 14W = - VdP 15PEKEhVdP-WPEKEhQWWPEKEhQSYSTEMOPENanFor.BPdVWdWdUdQWUQSYSTEMCLOSEDaFor.ASUMMARY
- 40. IDEAL OR PERFECT GASPrepared By: Engr Yuri G. Melliza
- 41. 1. Ideal Gas Equation of StatePV = mRTP = RT2T2V2P1T1V1PCTPVRTPρWhere: P – absolute pressure in KPaV – volume in m3m – mass in kgR – Gas Constant in KJ/kg- KT – absolute temperature in KIDEAL OR PERFECT GAS
- 42. 2. Gas ConstantK-mkgKJ8.3143RK-kgKJMRRWhere:R- Gas Constant in KJ/kg-KKmkgKJconstantgasuniversalRM – Molecular weight kg/kgm3. Boyle’s LawIf the temperature of a certain quantity ofgas is held constant the volume V is inver-sely proportional to the absolute pressure P.
- 43. C2V2P1V1PCPVP1CVPVα4.Charle’s LawA. At Constant Pressure (P = C)If the pressure of a certain quantity ofgas is held constant, the volume V is directlyproportional to the temperature T during a qua-sistatic change of state
- 44. 2211TVTVCTVT;CV;TαVB. At Constant Volume (V = C)If the volume of a certain quantity of gas isheld constant, the pressure P varies directlyas the absolute temperature T.2211TPTPCTP;TCPTαP ;
- 45. 5. Avogadro’s LawAll gases at the same temperature andpressure have the same number of moleculesper unit of volume, and it follows that thespecific weight is directly proportional toits molecular weight M.M6.Specific HeatSpecific Heat or Heat Capacity is the amountof heat required to raise the temperature ofa 1 kg mass 1 C or 1 KA. SPECIFIC HEAT AT CONSTANT PRESSURE (Cp)From: dh = dU + PdV + VdPbut dU + VdP = dQ ; thereforedh = dQ + VdP 1
- 46. but at P = C ; dP = O; thereforedh = dQ 2and by integrationQ = h 3considering m,h = m(h2 - h1) 4Q = h = m (h2 - h1) 5From the definition of specific heat, C = dQ/TCp = dQ /dt 6Cp = dh/dT, thendQ = CpdT 7and by considering m,dQ = mCpdT 8then by integrationQ = m Cp T 9but T = (T2 - T1)Q = m Cp (T2 - T1) 10
- 47. B SPECIFIC HEAT AT CONSTANT VOLUME (Cv)At V = C, dV = O, and from dQ = dU + PdVdV = 0, thereforedQ = dU 11then by integrationQ = U 12then the specific heat at constant volumeCv is;Cv = dQ/dT = dU/dT 13dQ = CvdT 14and by considering m,dQ = mCvdT 15and by integrationQ = m U 16Q = mCv T 17Q = m(U2 - U1) 18Q = m Cv(T2 - T1) 19
- 48. From:h = U + P and P = RTh = U + RT 20and by differentiation,dh = dU + Rdt 21but dh =CpdT and dU =CvdT, thereforeCpdT = CvdT + RdT 22and by dividing both sides of theequation by dT,Cp = Cv + R 23
- 49. 7. Ratio Of Specific Heatsk = Cp/Cv 24k = dh/du 25k = h/ U 26From eq. 32,Cp = kCv 27substituting eq. 27 to eq. 24Cv = R/k-1 28From eq. 24,Cv = Cp/k 29substituting eq. 29 to eq. 24Cp = Rk/k-1 30
- 50. 8. Entropy Change ( S)Entropy is that property of a substance thatdetermines the amount of randomness and disorderof a substance. If during a process, an amount ofheat is taken and is by divided by the absolutetemperature at which it is taken, the result iscalled the ENTROPY CHANGE.dS = dQ/T 31and by integrationS = ∫dQ/T 32and from eq. 39dQ = TdS 33
- 51. 2211221122112211222111MMLAWSAVOGADRO.4CTVTVCPAtb.CTPTPCVAta.LAWCHARLES.3CVPVPC)T(LAWBOYLES2.mRTPVCTVPTVPStateofEquation.1SUMMARYTdQSCHANGEENTROPY.8CvCpkHEATSPECIFICOFRATIO.7RCvCp1-kRCv;1-kRkCpHEATPECIFICS6.kgkgR8.3143MK-kgKJM8.3143RCONSTANTGAS.5mol
- 52. GAS MIXTURE Total Mass of a mixtureinnmmx iiimm Mass Fraction Total Moles of a mixturenny ii Mole FractionWhere:m – total mass of a mixturemi – mass of a componentn – total moles of a mixtureni – moles of a componentxi – mass fraction of a componentyi - mole fraction of a component
- 53. Equation of StateMass BasisA. For the mixtureiiiii TRmVPmRTPVTRnPVB. For the componentsiiiii TRnVPMole BasisA. For the mixtureB. For the componentsWhere:R – Gas constant of a mixturein KJ/kg- K- universal gas constant inKJ/kgm- KR
- 54. AMAGAT’S LAWThe total volume of a mixture V is equal to the volume occupied by eachcomponent at the mixture pressure P and temperature T.1n1V12n2V23n3V3P,TP = P1 = P2 = P3T = T1 = T2 = T3
- 55. For the components:TRPVn;TRPVn;TRPVn 332211The mole fraction:VVyiTRPVTRPVynnyiiiii321321321321VVVVPTRTRPVTRPVTRPVTRPVTRPVTRPVTRPVTRPVnnnnThe total moles n:
- 56. DALTON’S LAWThe total pressure of a mixture P is equal to the sum of the partial pressurethat each gas would exert at mixture volume V and temperature T.1n1P12n2P23n3P3MIXTUREnPT1 = T2 = T3 = TV1 = V2 = V3 = VFor the mixtureFor the componentsTRVPnTRVPnTRVPn332211TRPVn
- 57. 321321321321PPPPVTRTRVPTRVPTRVPTRPVTRVPTRVPTRVPTRPVnnnnThe total moles n: The mole fraction:PPyiTRPVTRVPynnyiiiii
- 58. Molecular Weight of a mixtureRRMMyM iiMRRRxR ii Gas Constant of a mixture Specific Heat of a mixtureRCCCxCCxCvpviivpiip Ratio of Specific HeatuhCCkvp
- 59. Gravimetric and Volumetric AnalysisGravimetric analysis gives the mass fractions of the componentsin the mixture. Volumetric analysis gives the volumetric or molal fractionsof the components in the mixture.iiiiiiiiiiMxMxyMyMyx
- 60. PROPERTIES OF PURE SUBSTANCEa - sub-cooled liquidb - saturated liquidc - saturated mixtured - saturated vapore - superheated vaporConsidering that the system is heated at constantpressure where P = 101.325 KPa, the 100 C is thesaturation temperature corresponding to 101.325 KPa,and 101.325 KPa is the saturation pressure correspon-ding 100 C.P P P P PQ30°C100°C100°C 100°CT 100°C(a) (b) (c) (d) (e)Q Q Q Q
- 61. Saturation Temperature (tsat) - is the highest temperature at a given pressure in whichvaporization takes place.Saturation Pressure (Psat) - is the pressure corresponding to the temperature.Sub-cooled Liquid - is one whose temperature is less than the saturation temperaturecorresponding to the pressure.Compressed Liquid - is one whose pressure is greater than the saturation pressurecorresponding to the temperature.Saturated Liquid - a liquid at the saturation temperatureSaturated Vapor - a vapor at the saturation temperatureSaturated Mixture - a mixture of liquid and vapor at the saturation temperature.Superheated Vapor - a vapor whose temperature is greater than the saturation temperature.ab c deT FSaturated VaporSaturated Vapor30°C100°Ct 100°CSaturated MixtureP = CCritical PointT- Diagram
- 62. ab c deTSFSaturated VaporSaturated Vapor30°C100°Ct 100°CSaturated MixtureP = CCritical PointT-S DiagramF(critical point)- at the critical point the temperature and pressure is unique.For Steam: At Critical Point, P = 22.09 MPa; t = 374.136 C
- 63. ab c deTSFSaturated VaporSaturated VaportatsatteSaturated MixtureP = CCritical PointT-S Diagramtsat - saturation temperature corresponding the pressure Pta - sub-cooled temperature which is less than tsatte - superheated vapor temperature that is greater than tsat
- 64. h-S (Enthalpy-Entropy Diagram)hSt = C (constant temperature curve)P = C (constant pressure curve)FIIIIIII - subcooled or compressed liquid regionII - saturated mixture regionIII - superheated vapor region
- 65. Quality (x):LvvLvvmmmmmmmmxWhere:mv – mass of vapormL – mass of liquidm – total massx- qualityThe properties at saturated liquid, saturated vapor, superheatedvapor and sub-cooled or compressed liquid can be determinedfrom tables. But for the properties at saturated mixture (liquidand vapor) they can be determined by the equationrc = rf + x(rfg)rfg = rg – rfWhere: r stands for any property ( , U, h and S)rg – property at saturated vapor (from table)rf – property at saturated liquidNote: The properties at siub-cooled or compressed liquid isapproximately equal to the properties at saturated liquidcorresponding the sub-cooled temperature.
- 66. Throttling CalorimeterMain Steam LineP1 – steam linepressureTo main steam lineP2 -Calorimeter pressureh1 = h2h1 = hf1 + x1(hfg1)Where:1 – main steam line2 - calorimeterthermometer
- 67. P1P212TSh = CT-S Diagram Throttling ProcessP1 – steam line pressureP2 – calorimeter pressure
- 68. 1. Isobaric Process ( P = C): An Isobaric Process is an internallyreversible constant pressure process.A. Closed System:(Nonflow)PV21PdVQ = U + W 1 any substanceW = PdV 2 any substanceU = m(U2 - U1) 3 any substanceW = P(V2 - V1) 4 any substanceQ = h = m(h2-h1) 5 any substanceTS12dSTP = CPROCESSES OF FLUIDS
- 69. For Ideal Gas:PV = mRTW =mR(T2-T1) 5U = mCv(T2-T1) 6Q = h = mCP (T2-T1) 7Entropy ChangeS = dQ/T 8 any substancedQ = dhFor Ideal Gasdh = mCPdTS = dQ/TS = mCP dT/TS = mCP ln(T2/T1) 9B. Open System:Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance- VdP = 0
- 70. Q = h 12W = - KE - PE 13If KE = 0 and PE = 0W = 0 14Q = mCP(T2-T1) 15 Ideal Gas2. Isometric Process (V = C): An Isometric process is internallyreversible constant volume process.A. Closed System: (Nonflow)PV12TSTdS12V = C
- 71. Q = U + W 1 any substanceW = PdV at V = C; dV = 0W = 0Q = U = m(U2 - U1) 2 any substanceh = m(h2-h1) 3 any substanceFor Ideal Gas:Q = U = mCv(T2-T1) 4h = mCP(T2-T1) 5Entropy Change:S = dQ/T 6 any substancedQ = dUdU = mCvdT for ideal gasS = dU/T = mCv dT/TS = mCv ln(T2/T1) 6
- 72. B. Open System:Q = h + KE + PE + W 7 any substanceW = - VdP - KE - PE 8 any substance- VdP = -V(P2-P1) 9 any substanceQ = U = m(U2 - U1) 10 any substanceh = m(h2-h1) 11 any substanceFor Ideal Gas:- VdP = -V(P2-P1) = mR(T1-T2)Q = U = mCv(T2-T1) 12h = mCP(T2-T1) 13If KE = 0 and PE = 0Q = h + W 14 any substanceW = - VdP 15W = - VdP = -V(P2-P1) 16 any substanceW = mR(T1-T2) 16 ideal gash = mCP(T2-T1) 17 ideal gas
- 73. 3. Isothermal Process(T = C): An Isothermal process is a reversibleconstant temperature process.A. Closed System (Nonflow)dSTST1 2PV12PdVPV = C orT = CQ = U + W 1 any substanceW = PdV 2 any substanceU = m(U2 - U1) 3 any substanceFor Ideal Gas:dU = mCv dT; at T = C ; dT = 0Q = W 4
- 74. W = PdV ; at PV = C ;P1V1 = P2V2 = C; P = C/VSubstituting P = C/V to W = PdVW = P1V1 ln(V2/V1) 5Where (V2/V1) = P1/P2W = P1V1 ln(P1/P2) 6P1V1 = mRT1Entropy Change:dS = dQ/T 7S = dQ/TdQ = TdS ;at T = CQ = T(S2-S1)(S2-S1) = S = Q/T 8S = Q/T = W/T 9 For Ideal Gas
- 75. B. Open System (Steady Flow)Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance- VdP = -V(P2-P1) 12 any substanceh = m(h2-h1) 13 any substanceFor Ideal Gas:- VdP = -P1V1ln(P2/P1) 14- VdP = P1V1ln(P1/P2) 15P1/P2 = V2/V1 16dh = CPdT; at T = C; dT = 0h = 0 16If KE = 0 and PE = 0Q = h + W 17 any substanceW = - VdP = P1V1ln(P1/P2) 18For Ideal Gash = 0 19Q = W = - VdP = P1V1ln(P1/P2) 20
- 76. 4. Isentropic Process (S = C): An Isentropic Process is an internally“Reversible Adiabatic” process in which the entropy remains constantwhere S = C and PVk = C for an ideal or perfect gas.For Ideal Gas121112122221kkkk22k1111kVVPPTTVPVPandTVPTVPCPVandCTPVUsing
- 77. A. Closed System (Nonflow)TS12PV12dVPS = C orPVk = CQ = U + W 1 any substanceW = PdV 2 any substanceU = m(U2 - U1) 3 any substanceQ = 0 4W = - U = U = -m(U2 - U1) 5
- 78. For Ideal GasU = mCV(T2-T1) 6From PVk = C, P =C/Vk, and substituting P=C/Vkto W = ∫PdV, then by integration,11111112111121kkVPkk121122PPkPdVPPkmRTk-1T-TmRPdVkVP-VPPdVW 789Q = 0
- 79. Entropy ChangeS = 0S1 = S2B. Open System (Steady Flow)Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substanceh = m(h2-h1) 12 any substanceQ = 0W = - h - KE - PE 13From PVk = C ,V =[C/P]1/k, substituting V to-∫VdP, then by integration,
- 80. 11111112111121kkkk121122PPkVkPVdPPPkkmRTk-1T-TkmRVdPkVP-VPkVdPPdVkVdP141516If KE = 0 and PE = 00 = h + W 17 any substanceW = - VdP = - h 18 any substanceh = m(h2-h1) 19 any substanceQ = 0
- 81. 12Pkkkk121122T-TmChWPPkVkPWPPkkmRTk-1T-TkmRWkVP-VPkPdVkVdPW1111111211112120222123
- 82. PVdPVArea = - VdPS = C
- 83. 121112122221nnnn22n1111nVVPPTTVPVPandTVPTVPCPVandCTPVUsing5. Polytropic Process ( PVn = C): A Polytropic Process is aninternally reversible process of an Ideal or Perfect Gas in whichPVn = C, where n stands for any constant.
- 84. A. Closed System: (Nonflow)Q = U + W 1W = PdV 2U = m(U2 - U1) 3Q = mCn(T2-T1) 4U = m(U2 - U1) 5PV12dVPPVn = CTS21dSTPVn = CK-kgKJorC-kgKJheatspecificpolytropicCn1nkCCnvn
- 85. From PVn = C, P =C/Vn, and substitutingP =C/Vn to W = ∫PdV, then by integration,11111112111121nnVPnn121122PPnPdVWPPnmRTn-1T-TmRPdVWnVP-VPPdVWEntropy ChangedS = dQ/TdQ = mCndTS = mCnln(T2/T1)68910
- 86. B. Open System (Steady Flow)Q = h + KE + PE + W 11W = - VdP - KE - PE 12h = m(h2-h1) 13Q = mCn(T2-T1) 14dQ = mCn dTW = Q - h - KE - PE 15From PVn = C ,V =[C/P]1/n, substituting V to-∫VdP, then by integration,nVP-VPnVdPPdVnVdP1122116
- 87. 1111112111121nnnn12PPnVnPVdPPPnnmRTn-1T-TnmRVdPIf KE = 0 and PE = 0Q = h + W 19 any substanceW = - VdP = Q - h 20 any substanceh = m(h2-h1) 21 any substanceh = mCp(T2-T1)Q = mCn(T2-T1) 221718
- 88. 1111112111121nnnn12PPnVnPWPPnnmRTn-1T-TnmRW2423
- 89. 6. Isoenthalpic or Throttling Process: It is a steady - state, steadyflow process in which Q = 0; PE = 0; KE = 0; W = 0 and theenthalpy remains constant.h1 = h2 or h = CThrottling valveMain steam linethermometerPressure GaugePressure GaugeTo main steam lineThrottling Calorimeter
- 90. Irreversible or Paddle WorkmWQUWpQ = U + W - Wpwhere: Wp - irreversible or paddle work
- 91. THERMODYNAMICS Part 2By. Engr. Yuri G. Melliza2nd Law of ThermodynamicsCarnot CyclesSteam CyclesFuels and CombustionICE Cycles
- 92. 2nd Law of Thermodynamics• Second Law of Thermodynamics• Kelvin – Planck Statement• Carnot engine• Carnot Refrigerator• Sample Problems
- 93. Second Law ofThermodynamics:Whenever energy is transferred, thelevel of energy cannot be conservedand some energy must be permanentlyreduced to a lower level.When this is combined with the first law ofthermodynamics, the law of energyconservation, the statement becomes:
- 94. Second Law ofThermodynamics:Whenever energy is transferred, energymust be conserved, but the level ofenergy cannot be conserved and someenergy must be permanently reducedto a lower level.
- 95. Kelvin-Planck statement of the SecondLaw:No cyclic process is possible whose sole resultis the flow of heat from a single heat reservoirand the performance of an equivalent amountof work.For a system undergoing a cycle:The net heat is equal to the net work.QWdWdQ Where:W - net workQ - net heat
- 96. CARNOT CYCLENicolas Leonard Sadi Carnot1796-18321.Carnot EngineProcesses:1 to 2 - Heat Addition (T = C)2 to 3 - Expansion (S = C)3 to 4 - Heat Rejection (T = C)4 to 1 - Compression (S = C)
- 97. PV2134T = CS = CS = CT = C
- 98. TS2134T HT LQ AQ R
- 99. Heat Added (T = C)QA = TH( S) 1Heat Rejected (T = C)QR = TL( S) 2S = S2 - S1 = S3 – S4 3Net WorkW = Q = QA - QR 4W = (TH - TL)( S) 5
- 100. %xQQe%xQQQe%xQWeARARAA1001100100 678
- 101. Substituting eq.1 and eq. 5 to eq 6%xTTe%xTTTeHLHLH1001100 910
- 102. THTLWQAQRECarnot Engine
- 103. 2. Carnot Refrigerator:Reversed Carnot CycleProcesses:1 to 2 - Compression (S =C)2 to 3 - Heat Rejection (T = C)3 to 4 - Expansion (S = C)4 to 1 - Heat Addition (T = C)
- 104. QRQA1243STTHTL
- 105. Heat Added (T = C)QA = TL( S) 1Heat Rejected (T = C)QR = TH( S) 2S = S1 - S4 = S2 - S33Net WorkW = Q 4W = QR - QA 5W = (TH - TL)( S) 6
- 106. Coefficient of Performance(COP)LHLATTTCOPWQCOP 78
- 107. 1HLTTCOP 9Tons of Refrigeration211 KJ/min = 1 TR3. Carnot Heat Pump:A heat pumpuses the same components astherefrigerator but its purposeisto reject heat at high energylevel.
- 108. Performance Factor (PF)ARRRQQQPFWQPF 1011
- 109. 111COPPFTTPFQQPFTTTPFLHARLHH12131415
- 110. THTLWQAQRRCarnot Refrigerator
- 111. A Carnot engine operating between 775 K and305 K produces 54 KJ of work. Determine thechange of entropy during heat addition.TH = 775 K ; TL = 305 KW = 54 KJ
- 112. THTLWQAQRE
- 113. KKJ0.01577589.04TQS-S)S-(STQKJ89.040.60654eWQQWe0.606775305775TTTeHA1212HAAAHLH
- 114. A Carnot heat engine rejects 230 KJ ofheat at 25 C. The net cycle work is 375 KJ.Determine the cycle thermal efficiency andthe cycle high temperature .Given:QR = 230 KJTL = 25 + 273 = 298 KW = 375 KJ
- 115. TL = 298 KTHWEQR = 230 KJQAK87.783772.0605)S-(SQTKKJ/-0.772)S-(SKKJ/772.0)SS()SS(298230)SS(SS)SS(TQ)SS(TQ62.0605375QAWeKJ605QA)230375(QWQQQW12AH123434123434LR12HARARA
- 116. A Carnot engine operates between temperaturereservoirs of 817 C and 25 C and rejects 25 KW tothe low temperature reservoir. The Carnot enginedrives the compressor of an ideal vapor compres-sion refrigerator, which operates within pressurelimits of 190 KPa and 1200 Kpa. The refrigerant isammonia. Determine the COP and the refrigerantflow rate.(4; 14.64 kg/min)TH = 817 + 273 = 1090 KTL = 25 + 273 = 298 KQR = 25 KW
- 117. Internal Combustion Engine Cycles1. Air Standard Otto Cycle (Spark Ignition Engine Cycle)Processes1 to 2 - Isentropic Compression (S = C)2 to 3 - Constant Volume Heat Addition ( V = C)3 to 4 - Isentropic Expansion (S =C)4 to 1 - Constant Volume Heat Rejection (V = C)PPmVVDCVDW1423S = CS = C3QATS124V = CV = CQR
- 118. Compression Ratio3421VVVVrwhere:r - compression ratioV1 = V4 and V2 = V31Heat Added (V = C)QA = mCV(T3 - T2) 2Heat RejectedQR = mCV(T4 - T1) 3Net Cycle WorkW = QA - QRW = mCV[(T3 - T2) - (T4 - T1)] 4
- 119. Thermal Efficiency100%r11e100%TTTT1e100%xQQ1e100%QQQe100%xQWe1-k2314ARARAAxxx56789
- 120. Mean Effective PressureKPaVWPDmwhere:W - net work, KJ, KJ/kg, KWVD - Displacement Volume, m3, m3/kg, m3/secVD = V1 - V2 m3VD = 1 - 2 m3/kg10Percent Clearance100%xVVCD2V2 = CVDV1 = Vd + CVDCC1VVr2111 12
- 121. 2. Diesel Cycle: (Compression Ignition Engine Cycle)Processes1 to 2 - Isentropic Compression (S = C)2 to 3 - Constant Pressure Heat Addition (P = C)3 to 4 - Isentropic Expansion (S = C)4 to 1 - Constant Volume Heat Rejection (V = C)PVTS2 341S = CS = CVDCVD1234V = CP = CQRQA
- 122. Heat Added (P = C)QA = mCP(T3 - T2) 3QA = mkCV(T3 - T2) 4Heat Rejected (V = C)QR = mCV(T4 - T1) 5Net Cycle WorkW = QA - QRW = mCV[k(T3 - T2) - (T4 - T1)] 6Compression Ratio3421VVVVr 1Cut - Off Ratio23c VVr 2
- 123. Thermal Efficiency100%1)k(r1)(rr11e100%TTTT1e100%xQQ1e100%QQQe100%xQWeckc1-k2314ARARAAxxkx7891011
- 124. where:W - net work, KJ, KJ/kg, KWVD - Displacement Volume, m3, m3/kg, m3/secVD = V1 - V2 m3VD = 1 - 2 m3/kgKPaVWPDmMean Effective Pressure12
- 125. 3. Air Standard Dual CycleProcesses:1 to 2 - Compression (S = C)2 to 3 -Heat Addition (V = C)3 to 4 - Heat Addition (P = C)4 to 5 -Expansion (S = C)5 to 1 _ Heat Rejection (V = C)1122334455S = CS = CP = CV = CV = CQA1QA2QRPVTSVDCVD
- 126. Comprssion Ratio453521VVVVVVrV5 = V1 ; V2 = V31Pressure Ratio2P4P23p PPr 32434c VVrVV2Cut-Off Ratio
- 127. Heat Added (V = C)QA1 = mCV(T3 - T2) 4Heat Added (P =C)QA2 = mCP(T4 - T3) 5QA2 = mkCV(T4 - T3) 6Heat Rejected (V = C)QR = mCV(T5 - T1) 7Net Cycle WorkW = QA - QR 8QA = QA1 + QA2 9QA = mCV[(T3 - T2) +k (T4 - T3)] 10W = mCV[(T3 - T2) + k(T4 - T3) - (T5 - T1) ]100%xQQ1e100%QQQe;100%xQWeARARAAx 1213Thermal Efficiency11
- 128. 100%1-(rkr1)-(r1)-r(rr11e100%)]T-(Tk)T-[(T)T-(T1ecppkcp1-k342315xxMean Effective PressurePm = W/VD KPawhere: VD = V1 - V2 m3 ; W in KJV1 = V5 ; V2 = V3VD = 1 - 2 m3/kg ; W in KJ/kg1 = 5 ; 2 = 3For Cold Air Standard: K = 1.4For Hot Air Standard: K = 1.3
- 129. Vapor Power CycleRANKINE CYCLEProcesses:1 to 2 - Expansion (S = C)2 to 3 - Heat Rejection (P = C)3 to 4 - Compression or Pumping (S = C)4 to 1 - Heat Addition (P = C)Boiler or SteamGeneratorTurbineCondenserPumpWPQAQRWt1234
- 130. Major Components of a Rankine Cycle1. Steam Generator or Boiler: The working substance absorbs heatfrom products of combustion or other sources of heat at constantpressure which in turn changes the state of the working substance(water or steam) from sub-cooled liquid and finally to superheatedvapor whence at this point it enters the turbine.2. Steam Turbine: A steady state, steady flow device where steamexpands isentropically to a lower pressure converting some formsof energy (h, KE, PE) to mechanical work that finally be convertedinto electrical energy if the turbine is used to drive an electric gene-rator.3. Condenser: Steam exiting from the turbine enters this device to re-ject heat to the cooling medium and changes its state to that of thesaturated liquid at the condenser pressure which occurred at a cons-tant pressure process.
- 131. 4. Pump: It is also a steady state, steady flow machine where thecondensate leaving the condenser at lower pressure be pumpedback to the boiler in an isentropic process in order to raise thepressure of the condensate to that of the boiler pressure.hS ST34213412P1P2P1P24’2’2’4’
- 132. Turbine Worka) Ideal CycleWt = (h1 - h2) KJ/kgWt = ms(h1 - h2) KWb) Actual CycleWt’ = (h1 - h2’) KJ/kgWt’ = ms(h1 - h2’) KWwhere: ms - steam flow rate in kg/secTurbine Efficiency100%xhhhhη100%xWWtη2121ttt
- 133. Pump Worka) Ideal CycleWP = (h4 - h3) KJ/kgWP = ms(h4 - h3) KWb) Actual CycleWP’ = (h4’ - h3) KJ/kgWP’ = ms(h4’ - h3) KWPump Efficiency100%xhhhhη100%xWWη3434pppp
- 134. Heat Rejecteda) Ideal CycleQR = (h2 - h3) KJ/kgQR = ms(h2 - h3) KWQR = ms(h2 - h3) KW = mwCpw(two - twi) KWb) Actual CycleQR = (h2’ - h3) KJ/kgQR = ms(h2’ - h3) KW = mwCpw(two - twi) KWWhere: mw - cooling water flow rate in kg/sectwi - inlet temperature of cooling water in Ctwo - outlet temperature of cooling water in CCpw - specific heat of water in KJ/kg- C or KJ/kg- KCpw = 4.187 KJ/kg- C or KJ/kg- K
- 135. Heat Added:a) Ideal CycleQA = (h1 - h4) KJ/kgQA = ms (h1 - h4) KWb) Actual CycleQA = (h1 - h4’) KJ/kgQA = ms (h1 - h4’) KWSteam Generator or boiler Efficiency100%x(HV)m)h(hmη100%xQQηf41sBSABWhere: QA - heat absorbed by boiler in KWQS - heat supplied in KWmf - fuel consumption in kg/secHV - heating value of fuel in KJ/kg
- 136. Steam RateKW-seckgProducedKWrateFlowSteamSRHeat RateKW-secKJProducedKWSuppliedHeatHRReheat CycleA steam power plant operating on a reheat cycle improves the thermalefficiency of a simple Rankine cycle plant. After partial expansion ofthe steam in the turbine, the steam flows back to a section in the boilerwhich is the re-heater and it will be reheated almost the same to itsinitial temperature and expands finally in the turbine to the con-denser pressure.
- 137. ReheaterQAWPQRWt1 kg1 2 3456Regenerative CycleIn a regenerative cycle, after partial expansion of the steam in theturbine, some part of it is extracted for feed-water heating in an open orclose type feed-water heater. The bled steam heats the condensate fromthe condenser or drains from the previous heater causing a decrease inheat absorbed by steam in the boiler which result to an increase inthermal efficiency of the cycle.
- 138. QAWP1QRWt1 kg1234567WP2mReheat-Regenerative CycleFor a reheat - regenerative cycle power plant, part of the steam is re-heated in the re-heater and some portion is bled for feed-water heatingto an open or closed type heaters after its partial expansion in theturbine. It will result to a further increase in thermal efficiency of theplant.
- 139. QAWP1QRWt1 kg1245678WP2m231-m1-mFor a 1 kg basis of circulating steam, m is the fraction of steamextracted for feed-water heating as shown on the schematic diagramabove, where the reheat and bled steam pressure are the same.
- 140. FUELSandCOMBUSTIONBy. Engr. Yuri G. Melliza
- 141. FUELS AND COMBUSTION Fuels and Combustion Types of Fuels Complete/Incomplete Combustion Oxidation of Carbon Oxidation of Hydrogen Oxidation of Sulfur Air composition Combustion with Air Theoretical Air Hydrocarbon fuels Combustion of Hydrocarbon Fuel
- 142. Fuels and CombustionFuel: Substance composed ofchemical elements which in rapidchemical union with oxygenproduced combustion.
- 143. Combustion:Is that rapid chemical union withoxygen of an element, whoseexothermic heat of reaction issufficiently great and whose rateof reaction is suffi-ciently fastwhereby useful quantities of heatare liberated at elevatedtemperature.
- 144. TYPES OF FUELS Solid Fuelsex: Wood, coal, charcoal Liquid Fuelsex: gasoline, diesel, kerosene Gaseous Fuelsex: LPG, Natural Gas, Methane Nuclear Fuelsex: UraniumCombustible Elements1. Carbon (C) 3. Sulfur (S)2. Hydrogen (H2)
- 145. Complete Combustion: Occurs when allthe combustible elements has been fullyoxidized.Ex:C + O2 CO2Incomplete Combustion: Occurs whensome of the combustible elements has notbeen fully oxidized.Ex:C + O2 CO
- 146. Common Combustion GasesGAS MOLECULARWeight (M)C 12H 1H2 2O 16O2 32N 14N2 28S 32
- 147. THE COMBUSTION CHEMISTRYOxidation of Carbon118344361232)1(121(16)1(12)BasisMass111BasisMoleCOOC 22
- 148. Oxidation of Hydrogen981181622)1(16(32)1(2)BasisMass11BasisMoleOHOH2121222 21
- 149. Oxidation of Sulfur21164323232)1(32(32)1(32)BasisMass111BasisMoleOSOS122
- 150. Composition of AIRa. Percentages by Volume (bymole)O2 = 21%N2 = 79%b. Percentages by MassO2 = 23%N2 = 77%7632179.22OofMoleNofMoles
- 151. Combustion with AirA. Combustion of Carbon with airC + O2 + 3.76N2 CO2 + 3.76N2Mole Basis:1 + 1 + 3.76 1+ 3.76Mass Basis:1(12) + 1(32) + 3.76(28) 1(44) +3.76(28)12 + 32 + 3.76(28) 44 + 3.76(28)3 + 8 + 3.76(7) 11+ 3.76(7)
- 152. kg of air per kg of Carbon:Cofkgairofkg11.44=33.76(7)+8=Cofkgairofkg
- 153. B. Combustion of Hydrogen with airH2 + ½ O2 + ½ (3.76)N2 H2O +½(3.76)N2Mole Basis:1 + ½ + ½(3.76) 1 + ½(3.76)Mass Basis:1(2) + ½ (32) + ½(3.76)(28) 1(18) +½ (3.76)(28)2 + 16 + 3.76(14) 18 + 3.76(14)1 + 8 + 3.76(7) 9 + 3.76(7)
- 154. kg of air per kg of Hydrogen:22 Hofkgairofkg34.32=13.76(7)+8=Hofkgairofkg
- 155. C. Combustion of Sulfur with airS + O2 + 3.76N2 SO2 + 3.76N2Mole Basis:1 + 1 + 3.76 1 + 3.76Mass Basis:1(32) + 1(32) + 3.76(28) 1(64) +3.76(28)32 + 32 + 105.28 64 + 105.28
- 156. kg of air per kg of Sulfur:Sofkgairofkg4.29=32105.2832=Sofkgairofkg
- 157. Theoretical AirIt is the minimum amount of air required tooxidize the reactants or the combustibleelements found in the fuel. With theoreticalair no O2 is found in products.
- 158. Excess AirIt is an amount of air in excess of theTheoretical requirements in order toinfluence complete combustion. Withexcess air O2 is present in theproducts.
- 159. HYDROCARBON FUELSFuels containing the element s Carbonand Hydrogen.Chemical Formula: CnHm
- 160. Family Formula Structure SaturatedParaffin CnH2n+2 Chain YesOlefin CnH2n Chain NoDiolefin CnH2n-2 Chain NoNaphthene CnH2n Ring YesAromaticBenzene CnH2n-6 Ring NoNaphthalene CnH2n-12 Ring NoAlcohols Note: Alcohols are not purehydrocarbon, because one of itshydrogen atom is replace by anOH radical. Sometimes it is usedas fuel in an ICE.Methanol CH3OHEthanol C2H5OH
- 161. Saturated Hydrocarbon: All the carbon atoms arejoined by a single bond.Unsaturated Hydrocarbon: It has two or moreadjacent Carbon atoms joined by a double ortriple bond.Isomers: Two hydrocarbons with the samenumber of carbon and hydrogen atoms but atdifferent structures.
- 162. H H H HH C C C C HH H H HChain structure SaturatedH HH C C=C C HH H H HChain Structure UnsaturatedRing structure SaturatedH HH C HC CH C HH H
- 163. Theoretical Air: It is the minimum or theoretical amountof air required to oxidized the reactants. With theoreticalair no O2 is found in the products.Excess Air: It is an amount of air in excess of the theo-retical air required to influence complete combustion.With excess air O2 is found in the products.Combustion of HydrocarbonFuel(CnHm)A. Combustion with 100% theoretical airCnHm + aO2 + a(3.76)N2 bCO2 + cH2O +a(3.76)N2fuelairt kgkgm12n)a(3.76)(28a(32)FA
- 164. Combustion of Hydrocarbon FuelFormula: (CnHm)A. Combustion with 100% theoretical airCnHm + aO2 + a(3.76)N2 bCO2 + cH2O +a(3.76)N2fuelairt kgkgm12n)a(3.76)(28a(32)FA
- 165. fuelaira kgkgm12n)a(3.76)(28a(32)e)(1FAB. Combustion with excess air eCnHm +(1+e) aO2 + (1+e)a(3.76)N2 bCO2+ cH2O + dO2 + (1+e)a(3.76)N2Actual Air – Fuel Ratiofuelairta kgkgFAe)(1FAWhere: e – excess air in decimalNote: Sometimes excess air is expressible in terms oftheoretical air.Example: 25% excess air = 125% theoretical air
- 166. Orsat Analysis: Orsat analysis gives thevolumetric or molal analysis of thePRODUCTS on a DRY BASIS, (no amount ofH2O given).Proximate Analysis: Proximate analysisgives the amount of FixedCarbon, Volatiles, Ash and Moisture, inpercent by mass. Volatiles are thosecompoundsthat evaporates at low temperature whenthe solid fuel is heated.
- 167. ULTIMATE ANALYSIS: Ultimate analysisgives the amount of C, H, O, N, S inpercentages by mass, and sometimes theamount of moisture and ash are given.
- 168. SOLID FUELSComponents of Solid Fuels:1. Carbon (C)2. Hydrogen (H2)3. Oxygen (O2)4. Nitrogen (N2)5. Sulfur (S)6. Moisture (M)7. Ash (A)
- 169. A.Combustion with 100% theoretical airaC + bH2 + cO2 + dN2 + eS + fH2O + gO2 +g(3.76)N2 hCO2 + iH2O + jSO2 + kN2B.Combustion with excess air x:aC + bH2 + cO2 + dN2 + eS + fH2O +(1+x)gO2 +(1+x)g(3.76)N2 hCO2 +iH2O + jSO2 + lO2 + mN2WHERE: a, b, c, d, e, f, g, h, I, j, k, x are thenumber of moles of the elements.x – excess air in decimal
- 170. fuelkgairkg18f32e28d32c2b12a3.76(28)g32gFAtTheoretical air-fuel ratio:Actual air-fuel ratio:fuelkgairkg18f32e28d32c2b12a3.76(28)g32gx)(1aFA
- 171. MASS FLOW RATE OF FLUE GAS (Products)Air+Fuel ProductsA. Without considering Ash loss1FAmm FgB. Considering Ash losslossAsh1FAmm Fg
- 172. Heating ValueHeating Value - is the energy released byfuel when it is completely burned and theproducts of combustion are cooled to theoriginal fuel temperature.Higher Heating Value (HHV) - is theheating value obtained when the water inthe products is liquid.Lower Heating Value (LHV) - is theheating value obtained when the waterin the products is vapor.
- 173. For Solid Fuels with the presence ofFuel’s ULTIMATE ANALYSISkgKJS93048OH212,144C820,33HHV 22where: C, H2, O2, and S are in decimals from theultimate analysis
- 174. HHV = 31 405C + 141 647H KJ/kgHHV = 43 385 + 93(Be - 10) KJ/kgFor Liquid Fuelswhere: Be - degrees Baume For Coal and Oils with the absence ofUltimate AnalysisfuelofkgairofKg3041HHVFAt
- 175. For GasolinekgKJ)API(93639,38LHVkgKJ)API(93160,41HHVkgKJ)API(93035,39LHVkgKJ)API(93943,41HHVFor Kerosene
- 176. For Fuel OilsInstitutePetroleumAmericanAPIkgKJ)API(6.139105,38LHVkgKJ)API(6.139130,41HHV
- 177. For Fuel Oils (From Bureau of StandardFormula)).t(.St@S 561500070API131.5141.5SHHV = 51,716 – 8,793.8 (S)2 KJ/kgLHV = HHV - QL KJ/kgQL = 2442.7(9H2) KJ/kgH2 = 0.26 - 0.15(S) kg of H2/ kg offuel
- 178. WhereS - specific gravity of fuel oil at 15.56 CH2 - hydrogen content of fuel oilQL - heat required to evaporate and superheat the watervapor formed bythe combustion of hydrogen in the fuelS @ t - specific gravity of fuel oil at any temperature tOxygen Bomb Calorimeter - instrument used in mea-suring heating value of solid and liquid fuels.Gas Calorimeter - instrument used for measuringheating value of gaseous fuels.
- 179. Properties of Fuels and Lubricantsa) Viscosity - a measure of the resistance to flow that alubricant offers when it is subjected to shear stress.b) Absolute Viscosity - viscosity which is determined bydirect measurement of shear resistance.c) Kinematics Viscosity - the ratio of the absolute viscosityto the densityd) Viscosity Index - the rate at which viscosity changes withtemperature.e) Flash Point - the temperature at which the vapor abovea volatile liquid forms a combustible mixture with air.f) Fire Point - The temperature at which oil gives off vaporthat burns continuously when ignited.
- 180. g) Pour Point - the temperature at which oil will no longer pour freely.h) Dropping Point - the temperature at which greasemelts.i) Condradson Number(carbon residue) - the percentage amount by mass of thecarbonaceous residue remaining after destructive distillation.j) Octane Number - a number that provides a measure of the ability of a fuel toresist knocking when it is burnt in a gasoline engine. It is the percentage by volumeof iso-octane in a blend with normal heptane that matches the knocking behavior ofthe fuel.
- 181. k) Cetane Number - a number that provides a measure of the ignitioncharacteristics of a diesel fuel when it is burnt in a standard diesel engine. It is thepercentage of cetane in the standard fuel.Prepared By: ENGR YURI G. MELLIZA, RME

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