1Chapter 2: Energy and the 1stLaw of ThermodynamicsThe Study of Energy in Closed SystemsWork & Power∫•=⇔•=→2121sssdFWsdFdWrrrr• Work is the (dot) product of force and displacementFds• Power is the rate of change of workvFdtsdFdtdWprrrr•=•==• For rotational systemsω•τ==rrdtdWp2Energy: The Property• Common Units– J(N·m) or kJ– ft·lbf–Btu)(212122VVmKE −⋅=∆)(12zzgmPE−⋅=∆• Kinetic Energy• Potential EnergyConservation of EnergydtvdmmgFmaFrr=−⇒=∑j• By the Newton’s 2ndLawFds• Multiplying (dot-product) both sides by ds and integrating yields-mgkik()4342143421KE2122PE122121)(∆∆→−+−= vvmzzmgWWhere is the work done on the system by force F:∫•=→2121sssdFWrr3• Work can either be done on the system or by the system on the surrounding environment.• Work Sign Convention– W > 0 : Work done by the system– W < 0 : Work done on the system– Time rate of work is Power:Work Energy Transfer•WConservation of Energy:MechanicsChange in amount of energy contained within the system during some time interval+Net amount of energy transferred out across the system boundary by work during the time interval0WPEKE=+∆+∆=04Inclined Surface ExampleZ=5’V1=2 ft/secV2=?1. Find V2when the block reaches the bottom.2. Find the work done on the system by friction when m stops.m=32.2 lbmExpansion Work• Infinitesimal work done by system: dW=Fdx• Force F=pA⇒ dW=pdV• Conclusion: Work done by system during expansion is the area under the p-v curve.• In general, W is path dependent thus not a property.dVpWVV∫⋅=215Polytropic Expansion WorkdVpWVV∫⋅=21cpVn=Polytropic:Where c and n are positive constant.Work:Shaft Work• This work is done or by system through a shaft:dtdWttω⋅τ=θ⋅τ=∫∫θθ2121ω⋅τ== WPower&W>0W<0τ6Other Examples WorkWork is process (path) dependent,and is NOT a property of the system• Elongation of a solid bar• Stretching of a Liquid Film • Rotating Shaft•Electric• Polarization/MagnetizationInternal Energy & Energy Changes• Total Energy: An extensive property of a system– Kinetic Energy (Mechanical)– Potential Energy (Mechanical)– Internal Energy: U or u • Represents all other forms of energy• Includes all microscopic forms of energyEKE PE U∆=∆ +∆ +∆7Energy Transfer by Heat• One form of enrgy transfer is by means of work interaction.• Energy can also be transferred by means of heat or thermal interaction.• A heat interaction is defined as any process in which energy is transferred to/from a system by means other than work. The amount of energy transferred is the heat denoted by Q.Heat Transfer Sign Convention• Heat Transfer– Q > 0 : Heat transfer into the system– Q < 0 : Heat transfer out of the system– Rate of heat transfer:•Q8Heat Transfer• Conduction• Radiation• ConvectionxdTQAdxκ•=−4beQATεσ•=−()bfcQhATT•=−Photo courtesy of Mike BensonConservation of Energy:The 1stLaw of ThermodynamicsKE PE U Q W∆+∆+∆=−Change in amount of energy contained within the system during some time interval=Net amount of energy transferred in across the system boundary by heat transfer during the time interval-Net amount of energy transferred out across the system boundary by work during the time interval9Alternative Forms of the Energy Balance• Differential Form:dE Q Wδδ=−• Time Rate Form:dEQWdt••=−Example Problem10Example ProblemExample Problem11Example 2.3m=100 lbmPatm=14.7 psiConstant gas pressureFind the heater input energyA=1 ft2mair=0.6 lbm∆v=1.6 ft3∆uair=18 Btu/lbFirst law for Cycles• First law for an arbitrary system: ∆E=Q-W• For a cycle, ∆Ecycle=0 since it begins and ends at the same state. Thuscycle cycleQW=12Power Cycle• A Power system receives heat Qinfrom a hot reservoir and rejects heat Qoutto a cold reservoir.• Net cycle heat Qcycle=Qin-Qout• Thus Wcycle=Qin-Qout• Thermal efficiency:inoutinoutinincycleQQQQQQW−=−==η 1Refrigeration Cycle• A Refrigeration cycle extracts heat from a cold reservoir and rejects heat Qoutto a hot reservoir.• Net cycle heat Qcycle=Qin-Qout• Thus Wcycle+Qcycle=0• Coefficient of Performance (COP):1>−==βinoutincycleinQQQWQ13Heat Pump Cycle• Heat pump is similar to refrigeration except that Qoutis the output of interest .• Coefficient of Performance (COP):1>−==γinoutoutcycleoutQQQWQSummary of Cycle AnalysisPower Cycles Refrigeration & Heat Pump Cyclescycle cycle cycleEQW∆= −cycle cycleQW=cycleinWQη=incycleQWβ=outcycleQWγ=14Example ProblemsExample Problem 2.7715Refrigeration ExampleHeat Pump