# Chapter 1. Review Of Thermodynamics And Heat Transfer

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AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter1Chapter 1. Review of Thermodynamics and Heat Transfer1.1 Introduction1.2 Basic concepts of Heat Transfer1.3 First Law of Thermodynamics1.4 Second Law of Thermodynamics1.5 Ideal GasReadings: M.J. Moran and H.N. Shapiro, Fundamentals of Engineering Thermodynamics,3rd ed.,John Wiley & Sons, Inc., or Other thermodynamics texts1.1 Introduction1.1.1 ThermodynamicsThermodynamics is the science devoted to the study of energy, its transformations, and itsrelation to the status of matter.energy the first law (conservation of energy)entropy the second law (quality of energy)every naturally occurring transformation of energy is accompaniedsomewhere by a loss in the availability of energy for future performance ofwork1.1.2 System and SurroundingsSystem: an object, any quantity of matter, any region of space, etc. selected for studySurroundings: the restBasic system types: Closed system (control mass) and Open system (control volume)1.1.3 Property, State, Process and EquilibriumA property is the observable characteristic of a system such as temperature, pressure anddensity.A state is the condition of the system defined by properties.A process is transformation of the system from one state to another.A thermodynamic cycle is a process that begins and ends at the same state.1

AE 310 Fundamentals of Heating, Ventilating, and e surface for a substance that expands on freezing2

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter11.1.4 p-v-T RelationshipsThermodynamic properties: tables, graphs, equations and programs.1.2 Basic ConceptsThe engineering discipline of heat transfer is concerned with methods of calculating ratesof heat transfer. These methods are used by engineers to design components and systems inwhich heat transfer occurs.(1) Temperature: Degree of molecular movementS.I. oC, KI-P: oF, oR(2) Energy: E (Btu, J)Capacity to do work. Examples: Thermal Light Mechanical Electrical Chemical1 Btu 1055 JWork (W) is an action of a force on a moving system.(3) Heat (Thermal energy): Q (Btu, J)Heat is energy transferred across the system boundary by temperature difference ( T).3

AE 310 Fundamentals of Heating, Ventilating, and Air-Conditioning(4) Specific heat: Cp (Btu/lboF, J/kg K)Heat needed to raise temperature of 1 lb (1 kg) material for 1 oF (1 oC).(5) Heat capacity: (Btu/ft3 oF) and thermal mass (Btu/oF)Ability to store energyHeat capacity density x specific heatThermal mass density x volume x specific heat ρ Cp ρ V Cp(6) Energy change (heat transfer): Q (Btu)Q ρ V Cp Twhere ρ - density, V - volume, T - temperature difference.& (Btu/h)(7) Heat flow (heat transfer rate or energy change rate): QEnergy transferred per unit time.4Chapter1

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter11.3 First Law of ThermodynamicsThe first law of thermodynamics expresses principle of conservation of energy.1KE mv 2Kinetic (motion of the system)Forms of energy:2PE mgzPotential (position of the system)U U(T)Internal (stored in matter)System total energy (E) sum of all forms of energyEnergy transfer mechanisms are work (W) and heat (Q), which are not properties of the system.Conventions:Work done by a system is positive.Heat transfer to a system is positive.1.3.1 First Law for the Closed SystemHeat (Q) – Work (W) Change in Total Energy ( E)Closed system with process between states “1” and “2”: v 2 v12 g(z 2 z 1 ) Q1 2 W1 2 m (u 2 u 1 ) 2 2g CgC Example 1Constant volume heating of a cylinder filled with a gas.1st Law:Q1-2 m u m c V Tweightp const.V const.QQExample 1: Constant volume heatingExample 2: Constant pressure heating5

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter1Example 2Constant pressure heating of a cylinder filled with a gas.1st Law:where i u pv is enthalpy.Enthalpy is a property that combines u and pv work that are only forms of energychange in many processes.cP cV Rwhere R - specific ideal gas constant, and cP cV because constant input does work.1.3.2 First Law for the Open System (Control Volume Formulation)& ) – Net Rate of Work ( W& ) Net Rate of Energy FlowHeat Transfer ( Q(across the system boundary)dE cvdtEoutEin-- &Q &W v2gz v2gz dE cv& W&&& Qmimi CVCV 2g C g C in 2g C g C dtout 6

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter1Special case of the 1st law is for steady-state flow that has constant flow across theboundary and no mass or energy change in CV.For a typical HVAC system two assumptions are valid:& cvdEdm1) Steady-state ( cv 0, 0) flow across the system boundarydtdt2) KE and PE terms usually smallTherefore, for the typical HVAC system:& W&&&QCVCV m i - m ioutinExample 3A house/building is a thermal system and its envelope is the boundary. Let us consider someenergy transfer in a single family house.& infiltration 150 Btu/hQ& conduction 200 Btu/hQ& occupant 300 Btu/hQ& ac 350 Btu/hQ& solar 500 Btu/hQ& ground 100 Btu/hQThe thermal mass of the house is assumed to be 700 Btu/oF.(a) Is the system in equilibrium?7

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter1& in Q& out, the system is in equilibrium. The heat flows are steady state. The temperatureSince Qwill not change.(b) What will happen if the A/C is shut off?8

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter11.4 Second Law of ThermodynamicsAll processes obey the 1st Law of Thermodynamics. However, some 1st law processesnever occur. For example, heat transfer from cold reservoir to hot reservoir or flow from lowpressure to high pressure. The 2nd Law of Thermodynamics defines:direction of change for processesfinal equilibrium for spontaneous processescriterion for theoretical performance limits of cyclesquality of energyEnergy changes and transfer involves both conservation principle and degradation inquality. Therefore, the thermal efficiency of all heat engines must be less than 100% due todissipative effects. Processes occurring in a system such as heat engine are irreversible sinceeither the system or its surroundings cannot be returned to their initial states. A reversibleprocess is an idealization.Heat engines (heat pumps) are closed systems, which operates continuously, orcyclically, and produce (use) work while exchanging heat across its boundaries.1.4.1 Heat EngineWork produced while heat extracted from high temperature (TH) reservoir and rejected tolow temperature reservoir (TL).TH&QH&W&QLTL9

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter11.4.2 Heat PumpWork used to extract heat from low temperature reservoir (TL) and reject to hightemperature (TH) reservoir.TH&QH&W&QLTL1.4.3 Performance evaluation of cyclesPerformance evaluation of cycles: comparisons with the ideal Carnot heat engine that isa totally reversible heat engine or pump.& T of reservoir (absolute scale)Q(1) “Efficiency” (η) of a Heat Engineη & Q&&QW QL H 1 L ,&&&QQQHHH100 η 1

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningCarnot cycle:η Carnot 1 TL,THChapter1η Real Process η Carnotη Real Process 1Real irreversibilities: Friction, electrical resistance, heat transfer across finite temperature difference, (2) “Coefficient of Performance” (COP) of a Heat PumpHeating conditions (heat pump):Carnot cycle:COPHeat, Carnot THTH TLCooling conditions (refrigeration):Carnot cycle:COPCool, Carnot TLTH TLFor energy exchange with the same two thermal reservoirs:Carnot efficiency of a heat pump:η pump COPActual&QCarnot11

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter11.5 Ideal Gaspv RuTMwherepvMRuT gas pressure, psi, Pa specific volume, ft3/lbm, m3/kg, v 1/ρ molecular mass, lbm/mol, kg/mol universal gas constant, Ru 1545.32 ft-lbf/(mol-oR) 8314 J/(mol-K) temperature, oR, KDefine:R RuMwhereR specific gas constantair: Ra 53.34 ft-lbf/(lbm-oR) 287 J/(kg-K)water: Rv 85.76 ft-lbf/(lbm-oR) 462 J/(kg-K)Then:pv RT12

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter2Chapter 2.Moist Air Properties and Air-Conditioning Processes2.1 Moist Air and Its Properties2.2 Methods of Measurements and Analysis2.3 Typical Air Conditioning Processes2.4 Characteristics of “Real” Systems2.5 Psychrometric Analysis of Complete SystemsReadings: McQuiston & Parker (M&P) Ch 3 Texts on moist air in most thermodynamics books2.1 Moist Air and Its Properties2.1.1 Air Composition (two components)Dry air composition (volume fraction):NitrogenOxygenArgonCarbon dioxide78.084%20.948%0.934%0.031%Minor gases0.003%Water VaporDust, Fog, Microbe2.1.2 Ideal Gas LawFor dry airRa Ru 1545.32 53.352(ft lbf)/(lbm R) Ma28.965For water vaporRv Ru 1545.32 85.78ft lbf/(lbm R) Mv18.02The following data for US STANDARD ATSMOPHERE are from ASHARE Handbook ofFundamentals (Chapter6). Several assumptions are introduced such as:The atmosphere consists of dry air that behaves as an ideal gas.(see pg.50 in thetextbook)1

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningReference p and T are functions of altitude. Formula for p as function of elevation Z:p[psi] 14.696 (1-6.8753 10-6 Z[ft])5.2559p[bar] 1.013 (1-2.256 10-6 Z[m])5.25592Chapter2

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter2ExampleSea level: p 14.696 psi, T 59oF3.1.3 Fundamental Parameters(1) PressureThe air layer above the earth forms atmospheric pressure. Atmospheric pressure: sea level 14.692 psi elevation of 6600 ft, 11.513 psi(note: pressure inch mercury)a,b ---Table 3-2 in M&PPartial pressure (Dalton’s Law)Component/partial pressure:p p1 p2 p3 .Gibbs Dalton’s Law for Moist air:pa—dry air component (constant)pv—vapor component (change with moisture content)Note: When applying ideal gas law to each component of a mixture (e.g., moist air), should usepartial pressure for the component.For component i:Pivi RiTiWhere Pi is partial pressure for component i.ExampleOne lbm H2O vapor in 100 lbm dry air at standard pressure.(a) What is pv vapor pressure? (b) What is saturation T at this pv?3

AE 310 Fundamentals of Heating, Ventilating, and Air-Conditioning(a)Chapter2,(b) Saturation: State of maximum concentration for mixture components.(2) TemperatureTemperature is the macro results of molecular kinetics.0th law of thermodynamicsTa TbTb Tc Ta TcTable 2.1 Relationship between Different Temperature ScalesRelationshipbetweenKelvin ( K)Kelvin (K)Celsius( C)Fahrenheit( F)Rankine (R)ExampleTemperature inWater BoilingIce PointAbsolute ZerooC K 27315.9F o C 3255C ( oF 32)9oC o5.R 273159F R-459.67--Kelvin373.15K273.15K0KFahrenheit212 F32 F-459.67 F4Rankine617.67 R491.67 R0 R(R)5R9--9K5Celsius100 C0 C-273.15 CRankineK o-oR Fahrenheit( F)K oC 27315.-Celsius( C)

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter2(3) Humidity Ratio, Wmv( Kg / Kg drry air )mai.e. 1 kg dry air w kg water vapor (1 W) kg moist airDefinition: W where P is the atmospheric pressure. Because pv P , thus W pv(4) Relative Humidity φThermodynamic fluid statesPPPvaporliquidliquidliquidT TsatsubcooledT TsatsaturatedT Tsatquality vaporPvaporT Tsatsaturated vaporPPvaporvaporT Tsatsuperheated vapor5T Tsatgas

AE 310 Fundamentals of Heating, Ventilating, and Air-Conditioningφ pvp 100% v 100%pv , spsChapter2pv , s p spv partial pressure of the water vapor in the airps partial pressure of the water vapor in a saturated mixture under the same temperatureDry air: φ 0%Saturated air: φ 100%Difference between W and φ:pvMoist air:W 0.622P pvpsSaturated air:Ws 0.622P psp P psP psW v φ Wsp s P pvP pvW P pvφ 100%Ws P psSince P pv and P psFurtherExampleDetermine the humidity ratio of moist air at a temperature of 24 C and a relative humidity of50% at a standard pressure 1atmGiven: T, φFind: WSolution:(5) Dewpoint Temperature, TdTd the saturated temperature of a given mixture at the same pressure and humidity ratio.6

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningExampleFind Td of the air in the above example.Solution:At the dew point temperature: air mixture φ 100%(6) EnthalpyEnthalpy of the moist air enthalpy of the dry air enthalpy of the water vaporEnthalpy is energy per unit mass.i ia W ivia Cp,a Tiv ig Cp,v Twhere specific heat of dry air kJ/(kgoC), Btu/(lboF)Cp,a specific heat of water vapor kJ/(kgoC), Btu/(lboF)Cp,v the enthalpy of saturated water vapor at 0oC or 0oF.igig 2501.3 kJ/kg at 0oC; ig 1061.2 Btu/lb at 0oF.Therefore, we havei Cp,aT W(ig Cp,v T)ExampleFind the enthalpy of the air mixture in the above example.Solution:7Chapter2

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter22.2 Methods of Measurement and Analysis2.2.1 Measurement of air temperature Liquid-in-glass thermometersThermocouples2.2.2 Measurement of pressure Absolute pressure (vacuum tube with mercury)Differential pressureManometersPressure transducerPP2.2.3 Measurement of other parameters of moist airTo determine state of moist air, one property in addition to the pressure and temperaturemust be known. It can be v, I, φ, or W. However, none of them can be directly measured. As analternative, we seek an indirect measuring technique. In this section, the method used todetermine air humidity will be introduced. Adiabatic saturation deviceFirst we look at a special process:ia,1W1iv,1T1ia,2W2iv,2T2waterAdiabatic saturationdeviceThe equation that describes above process is:1st LawThe process is adiabatic, and only flow work is present.:where m& 1 m& a W1 m& a , m& 2 m& a W2 m& a , and form mass balance8

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter2The equation that describe above process becomes:dry air water vapor water vapor added dry air water vapor(in)(in)(added )(out )(out )so we havewhere ifg is enthalpy difference between liquid water and saturated vapor at the temperature T2W2 0.622pv ,2P pv ,2Then the state of moist are can be determined.T2 is called wet-bulb temperature.ExampleFind W and φ of the above adiabatic saturation device.Given: P 1.01325 105Pa, T1 30 C, T2 26 CFind: W1 , φ1Solution:Since the state 2 is in saturation, from the Table A-1b (McQuiston & Paker, p587), we can find:for T2 26 C, pv,2 ps 0.03363 105 (φ 100%), ifg 2439.1 kJ/kg, iw 109.07 kJ/kgfor T1 30 C, iv,1 2555.3 kJ/kg1) Find W19

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter22) Find φ1φ p v ,1p s1 100%pv,1 can be found viaFrom Table A-1b The PsychrometerPsychrometer TWB used in place of T2 for practical humidity measurement.In wet-bulb, heat transfer from air bulbTdryp TDB TWB Moist air stateTWBKey issues to measure TWB TWet bulb unshieldedWet bulb well ventilated (V 100 fpm)For thermocouples V could be lower.Then the accuracy is in order of 0.27oC (0.5oF).10

AE 310 Fundamentals of Heating, Ventilating, and Air-Conditioning2.2.4 The psychrometric chart(1) Selection of the CoordinatesHorizontal coordinate Enthalpy (155 )Vertical coordinate Humidity Ratio(2) Dry Bulb Temperaturei 1.01T W (2501.3 1.86T ) (SI Unit)T const i WIsothermal lines are not parallel.(3) Relative HumidityW 0.622φ pv,sP φ pv ,sφ Under a certain P, W f ( pv ,s ) pv ,s f (T )From P587 Table A-1b. find W-T relationshipWApproximately φ 100% (equal division)Ws11W P pv 100%Ws P p sChapter2

AE 310 Fundamentals of Heating, Ventilating, and Air-Conditioning(4) Specific Volume11 ρ ρa ρvppρa ρv a vRa T RvTTv p a pv Ra Rvv (5) Wet-Bulb Temperature12Chapter2

AE 310 Fundamentals of Heating, Ventilating, and Air-Conditioningi (Cp,a W Cp,v) Twet W x 2501Chapter2when Twet const, linear relationshipi (Cp,a W Cp,v) Twet W x 2501 W iw when φ 100%(6) Sensible Heat /Total HeatEnthalpy/Humidity ratioSee figure: Primary moist air parameter on the psychrometric chart.For specific pressure:Repeat the previousproblem by using the psychrometricTDBW, Tchart.DPGiven: Tdry 30 o C , Twet 26 o CFind: Primary moist air parameter on psychrometric chartSolution:From the psychrometric chart (SI Unit)i, TWBW 0.01971φ 74%φi 81kJ / Kgv 0.885 m / Kg313Tdewpo intv 24.7o C

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningExampleRepeat the previous problem by using the psychrometric chart.Given: Tdry 30 o C , Twet 26 o CFind: Primary moist air parameter on psychrometric chartSolution:14Chapter2

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter22.3 Typical Air-Conditioning ProcessTypical Air-Conditioning processes are: Sensible Heating / CoolingCooling and dehumidificationHeating and humidificationAdiabatic HumidificationAdiabatic Mixing of AirGoverning equations:STATE is a point, and PROCESS (sequence of states) is a line on the Chart.Process may involve: Sensible Heat (change TDB, constant W)Latent Heat (constant TDB, change W)Both2.3.1 Sensible heating and coolingQ&m& aW1T1i1On the psychrometric chartEnergy conservation (1stLaw)q i2 i1 C p (T2 T1 ). iheatingW1 W21221cooling15

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter2ExampleDetermine the energy (heat flux) required for sensible heating of air at 15 C and 50% RH to32 C. Also find φ2.Given: State 1: 15 C, RH 50%,Find: q& , φ2State 2: 32 CSolution:Heat Flux:Relative humidity:From the psychrometric chart (the process is a horizontal line):Heat flux isRelative humidity:How to check if W is constant, i.e. no latent heat?16

AE 310 Fundamentals of Heating, Ventilating, and Air-Conditioning2.3.2 Cooling and dehumidificationQ&m& aW1T1i1m& wiwMoisture is removed as saturated liquid.A1cooling2Bwhere iw water enthalpym& a (W1 W2 )iW is normally smallSensible heat:Latent heatq& sensible C p (T2 T1 )q& latent (W2 W1 )i fg17Chapter2

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter2ExampleAir at 60% RH, Tdry 30 C, Cooled to 18 C. Determine φnew, qsensible, qlatentSolution:φnew 100%, from the psychrometric chart, we can findBy using the formulaSensible Heat Factor (SHF) isQ& SQ&Defines process slope on the chart. Use protractor (semicircular scale) in the upper left handcorner to read the sensible heat factor.18

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter22.3.3 Heating and humidificationQm& aW1T1i1mw, iw2A1enthalpyhumidity ratio- Look at the semicircular scale in the psychrometric chart.Defines the process slope.For adiabatic humidification, thenQ& 0i2 - i1 W iwW2 - W1 WSince iw of the water is rather small, i2 i119

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningChapter2ExampleIn Phoenix, it is possible to use evaporative cooling in summer. In a room of 50 m3, the airtemperature is 45oC and relative humidity is 20%. Comfort standard allows the relative humidityto be increased to 60% by evaporative cooling. Determine the dry bulb temperature and waterneeded if there is no internal heat source and no air infiltration. Assume local pressure is 101325Pa.Solution:This is an adiabatic humidification process. The air process in a psychrometric chart is isoenthalpy. For the psychrometric chart, we can determine the starting humidity ratio and endinghumidity ratio as2.3.4 Other Humidity ProcessSteamHot waterSuper-heated steamAdiabatic20

AE 310 Fundamentals of Heating, Ventilating, and Air-ConditioningFor humidification: i w i WAdiabatic humidification:i w i f at TWB, process i const.Hot water:i w i g at TDB, process TDB Saturated steam:i w i g at TDB, process TDB const.Super-heated steam

1.4 Second Law of Thermodynamics 1.5 Ideal Gas Readings: M.J. Moran and H.N. Shapiro, Fundamentals of Engineering Thermodynamics,3rd ed., John Wiley & Sons, Inc., or Other thermodynamics texts 1.1 Introduction 1.1.1 Thermodynamics Thermodynamics is the science devoted to the study of energy, its transformations, and its

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