Tarik Al-Shemmeri. Engineering Thermodynamics (776123), страница 5
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for a constant pressure process, the work during an isobaric process is simply :W= p (V2 - V1)(31)b. for a constant volume process, dv =0; hence the work during an isochoric process is simply :W=0(32)c. for a constant temperature ,isothermal process,hencebutPV = cVcW = ∫ dV = ln 2vV1P1V1 = P2V2 ∴(33)V2 P1=V2 P2hence the work done can be written in terms of pressure ratio:∴ W = mRT1 lnP1P2d. for an adiabatic ( polytropic ) process,(34)PV n = cintegrating between states 1 and 2, the work done is derived :W =P1V1 − P2V2n−l(35)and using the ideal gas definition, the work done can be written in terms of intial and final statestemperatures:Download free ebooks at bookboon.com443. Laws of ThermodynamicsEngineering ThermodynamicsW =mR (T1 − T2 )n −1(36)3.2.5 First Law of Thermodynamics Applied to Closed Systems (Cycle)Since in a cycle the working fluid undergoing changes in its state will retain its initial conditions atany fixed point along the cycle.
Hence the energy equation applied to a cycle is:- Q−∑ W=0∑(37)where ∑ means the sum of heat or work around the cycle.Practical Application of a Closed System (Cycle) - Assume compression and expansion to beadiabatic, from first law :Qs – Qr = We - Wc(38)Please click the advertdongenergy.com/jobMay we offer you oneof the world’s greatestchallenges?In all humility.We are looking for highlyeducated engineers andfinance students.Join us at dongenergy.com/jobDownload free ebooks at bookboon.com453. Laws of ThermodynamicsEngineering ThermodynamicsWhere Qs heat suppliedQr heat rejectedWc work of compressionWe work of expansion3.2.6 First Law of Thermodynamics Applied to Open SystemsThe first law of thermodynamics is based on the conversation of energy within a system.
Opensystems are associated with those, which have a steady flow, thus the first law applied to suchsystems is known as the Steady Flow Energy Equation (SFEE):-Q − W = (∆H + ∆KE + ∆PE )()1 22= m (h2 − h1 ) + V2 − V1 + g (Z 2 − Z 1 )2(39)VariableHeat transferWork transferMass flow rateSpecific enthalpyVelocityGravitational accelerationElevation above datumSymbolQWmhVgzUnitsWWkg/sJ/kgm/s9.81 m/s2m3.2.6 Application of SFEEa. Turbines or Compressorsif the SFEE is applied to the expansion of a fluid in a turbine as shownQ - W = m (∆h + ∆ke + ∆Pe)With the following simplifications are madeDownload free ebooks at bookboon.com463.
Laws of ThermodynamicsEngineering ThermodynamicsQ=0,∆ke = 0 ,∆Pe = 0 are all neglected.∴ W = m(h1 - h 2)(40)hence for a turbine, the amount of energy produced “Work “ is equal to the enthalpychange between inlet and outlet.b. Boilers or Condencorsif the SFEE is applied to the heating or cooling ( evaporation or condensation ) of a fluid in a boileror condensorQ - W = m (∆h + ∆ke + ∆Pe)With the following simplifications are madeThere is no process work on the fluidW=0,If Kinetic energy and Potential energy changes ∆ke = 0 , ∆Pe = 0 are neglected. Then the SFEEreduces to :∴ Q = m(h1 - h 2)(41)Download free ebooks at bookboon.com473. Laws of ThermodynamicsEngineering Thermodynamicshence for a boiler or a condensor, the amount of energy supplied or exctracted from the fluid “Heat“ is equal to the enthalpy change for the fluid between inlet and outlet.c. Throttling valveConsider the flow of fluid through a small valve as shown12if the SFEE is applied between sections 1 and 2 :Q - W = m (∆h = ∆ke + ∆Pe)Q=0Assuming adiabaticPlease click the advertdongenergy.com/jobAmbitiousand curious?We are looking for young andhighly educated engineersand finance students.Join us at dongenergy.com/jobDownload free ebooks at bookboon.com483.
Laws of ThermodynamicsEngineering ThermodynamicsW=0No displacement work ( no work is inputted or extracted, ie no pump or turbineisattached )∆ke = 0 Assumed ( inlet and exit velocities are similar or slow)∆Pe = 0Assumed ( entry and exit at the same or nearly the same elevation )Hence, The SFEE, reduces to:∴m (h2 - h1) = 0divide by the mass flow m to get:∴ h2 = h1(42)hence for a control valve, the enthalpy of the fluid remains constant.d. DiffuserConsider the flow of fluid through a diffuser which is a device used in aircraft to reduce the kineticenergy of exhaust gases, as shown21if the SFEE is applied between sections 1 and 2:Q - W = m (∆h = ∆ke + ∆Pe)Q=0Assuming adiabaticW=0No displacement work ( no work is inputted or extracted, ie no pump orturbineis attached )Download free ebooks at bookboon.com493.
Laws of ThermodynamicsEngineering Thermodynamics∆Pe = 0Assumed ( entry and exit at the same or nearly the same elevation )Hence, The SFEE, reduces to:∴h2 − h1 =V22 − V122(43)3.3 The Second Law of ThermodynamicsThe second law of thermodynamics is a general principle which places constraints upon thedirection of heat transfer and the attainable efficiencies of heat engines. It's implications may bevisualized in terms of the waterfall analogy.Figure 3.2: Analogy of the 2nd Law of Thermodynamics3.3.1 Second Lay of Thermodynamics – statements:The second law, indicates that, although the net heat supplied,Q1-Q2, in a cycle is equal to thenetwork done, W, the gross heat supplied, Q1 must be greater than the network done; some heatmust always be rejected by the system.Q1 > W,or to be precise:-W = Q1 – Q2(44)Where, Q1 is the heat supplied and Q2 is the heat rejected.The ratio of network output to heat supplied is known as the thermal efficiency of the system.There are two ways in which the second law is expressed:Download free ebooks at bookboon.com503.
Laws of ThermodynamicsEngineering Thermodynamicsa) Kelvin-Planck Statement“It is impossible to construct a system which when operating in a cycle will extract heatenergy from a single source and do an equal amount of work on the surroundings”.Ie never possible to achieve 100% thermal efficiencyb) Clausius Statement“It is impossible to construct a device which when operating in a cycle has no other effectthan the transfer of heat energy from a cool to a hotter body”.Ie some work is done on or by the working fluid to the surroundings or vice versa.In 1865 Clausius observed that the amount dQ/T is proper to describe the thermodynamicphenomenon.
This amount was named reduced heat or entropy. During a process, the change inentropy is defined as :dS = Q/T(45)Definitions of Entropy :1. is a state variable whose change is defined for a reversible process at T where Q is the heatabsorbed.2. a measure of the amount of energy which is unavailable to do work.3. a measure of the disorder of a system.Please click the advertdongenergy.com/jobEverybodyis talking......about future energy supply. We are not.Stop talking and make a careermoving energy forward.Ambitious engineers andfinance students go todongenergy.com/jobDownload free ebooks at bookboon.com513.
Laws of ThermodynamicsEngineering Thermodynamics3.3.2 Change of Entropy for a Perfect Gas Undergoing a ProcessThe First Law of Thermodynamics:-Q – W = ΔUAnd the Second Law of Thermodynamics:Q = TdsTherefore,Tds – pdv = CvdT,divide by TdTdV+RTVTVs 2 − s1 = C v n 2 + Rn 2T1V1ds = CvSince,P1V1 P2V 2=T1T2thenV2 P1 .T2=V1 P2 .T1Therefore,s 2 − s1 = C v n= C p n P T T2+ Rn 1 2 T1 P2 T1 PT2− Rn 2T1 P1Since, Cv + R = CpThis can also be written in terms of volumes and pressures as:-S 2 − S1 = C v n( P2 / P1 ) + C p n(V2 / V1 )(46)3.3.3 Implications of the Second Law of ThermodynamicsSince entropy is defined as a property which remains constant during a reversible adiabatic process;it follows that a temperature-entropy diagram would indicate a process by a straight lineperpendicular to the entropy axis if the process is purely isentropic Figure (Figure 3.3).
Thefriction in an irreversible process will cause the temperature of the fluid to be higher than it wouldhave been in a frictionless (reversible) process. The entropy increased during an irreversibleprocess.Isentropic Efficiency – the entropy change in an irreversible adiabatic process leads to processefficiency. The ideal constant entropy process is termed isentropic and the ratio of the specificwork transfer in the ideal process to that in the actual process is called the isentropic efficiency.Download free ebooks at bookboon.com523. Laws of ThermodynamicsEngineering ThermodynamicsSince the majority of adiabatic machines are flow processes, isentropic efficiency is usuallyexpressed in terms of the useful work W.η ic =W12'W12(47)And for the expansion process η it =W12W12'(48)For the compression process,If changes in kinetic and potential energy are negligible, the SFEE may be used to rewrite theseexpressions in terms of specific enthalpy change and for a perfect gas, enthalpy change may beexpressed by temperature change.
Thus, for compression processes,(Figure 3.3):η ic =h2 ' − h1h2 − h1(49)for a perfect gas it becomesη icT2' − T1T2 − T1(50)The physical interpretation of this efficiency is that an irreversible compression process requiresmore work than an ideal process ( Figure 3.3) and in irreversible expansion process gives less workthan an ideal process.ReversibleT22p22’IrreversibleWp111SFigure 3.3 isentropic efficiency conceptDownload free ebooks at bookboon.com533. Laws of ThermodynamicsEngineering Thermodynamics3.4 Third LawThe entropy of a perfect crystal is zero when the temperature of a the crystal is equal to absolutezero (0 K).•At 0 K, there is no thermal motion, and if the crystal is perfect, there will be no disorder•Once the temperature begins to rise above 0, the particles begin to move and entropygradually increases as the average kinetic energy of the particles increases.•When temperature reaches the melting point of the substance (Tm), there is an abruptincrease in entropy as the substance changes from a solid to a more disordered liquid.Again the entropy increases gradually as the motion of the particles increases until the temperaturereaches the boiling point of the substance (Tb).
At this point, there is another drastic increase inentropy as the substance changes from a confined liquid particles to radom motion gas particles.Please click the advertdongenergy.com/jobdDear highly educated engineeringand finance students,if you are driven, ambitious, open-mindedand focused - we have a challenge for you.Actually, the greatest challenge in theworld. Curious? Visit dongenergy.com/jobBest wishesDONG EnergyDownload free ebooks at bookboon.com543. Laws of ThermodynamicsEngineering ThermodynamicsFigure 3.4 Temperature – Entropy relationship3.4.1 The Third Law of Thermodynamics - AnalysisIf we have sufficient heat capacity data (and the data on phase changes) we could write(51)(If there is a phase change between 0 K and T we would have to add the entropy of the phasechange.)Experimentally it appears that the entropy at absolute zero is the same for all substances.
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