2013-3-6 Entropy Facts For problem solving,specific entropy values are provided in Tables A-2 through A-18.Values for specific entropy are obtained from these tables sing thesm procedur res as for specific volume,internal energy,and enthalpy,including use of s=(1-x+8 =sr+x(52-s) (Eq.6.4) for two-phase liquid-vapor mixtures,and s(T.p)~s(T) (E4.6.5) for liquid water,each of which is similar in form to expressions introduced in Chap.3 for evaluating v,"and I. Entropy Facts For problem solving,states often are shown on property diagrams having specific entropy as a coordinate:the temperature-entropy and enthalpy-entropy(Mollier)diagrams shown here 6
2013-3-6 6 Entropy Facts ►For problem solving, specific entropy values are provided in Tables A-2 through A-18. Values for specific entropy are obtained from these tables using the same procedures as for specific volume, internal energy, and enthalpy, including use of (Eq. 6.4) for two-phase liquid-vapor mixtures, and (Eq. 6.5) for liquid water, each of which is similar in form to expressions introduced in Chap. 3 for evaluating v, u, and h. Entropy Facts ►For problem solving, states often are shown on property diagrams having specific entropy as a coordinate: the temperature-entropy and enthalpy-entropy (Mollier) diagrams shown here
2013-3-6 Entropy and Heat Transfer By inspection of Eq.6.2a,the defining equation for entropy change on a differential basis is s-(9 (Eq.6.2b) Equation 6.2b indicates that when a closed system undergoing an internally reversible process receives energyby heat increase in entropy.Conversely,when energy is removed by heat transfer,the entropy of the system decreases. From these considerations,we say that entropy transfer accompanies heat transfer.The direction of the entropy Entropy and Heat Transfer In an internally reversible.adiabatic process (no heat transfer),entropy remains constant.Such a constant entropy process is called an isentropic process. On rearrangement,Eq.6.2b gives (80)m=Tds Integrating from state 1 to state 2, 0n-Tds (E4.6.23) 7
2013-3-6 7 Entropy and Heat Transfer ►By inspection of Eq. 6.2a, the defining equation for entropy change on a differential basis is (Eq. 6.2b) ►Equation 6.2b indicates that when a closed system undergoing an internally reversible process receives energy by heat transfer, the system experiences an increase in entropy. Conversely, when energy is removed by heat transfer, the entropy of the system decreases. From these considerations, we say that entropy transfer accompanies heat transfer. The direction of the entropy transfer is the same as the heat transfer. Entropy and Heat Transfer Integrating from state 1 to state 2, (Eq. 6.23) ►On rearrangement, Eq. 6.2b gives ►In an internally reversible, adiabatic process (no heat transfer), entropy remains constant. Such a constantentropy process is called an isentropic process
2013-3-6 Entropy and Heat Transfer From this it follows that an energy transfer by e=j片rds heat to a closed system T (6m=Td during an internally reversible process is represented by an area on a temperature-entropy diagram: Entropy Balance for Closed Systems The entropy balance for closed systems can be developed using the Clausius inequality expressed as Eq. 5.13 and the defining equation for entropy change,Eg. 6.2a The result is -=9》+。 (Eq.6.24 is evaluated at the system boundary. In accord with the interpretation ofin the Clausius inequality,Eq.5.14,the value of o in Eq.6.24 adheres to the following interpretation nirrbliiespresthn the system) ties present within the system) <0(impossible) 8
2013-3-6 8 Entropy and Heat Transfer From this it follows that an energy transfer by heat to a closed system during an internally reversible process is represented by an area on a temperature-entropy diagram: Entropy Balance for Closed Systems ►The entropy balance for closed systems can be developed using the Clausius inequality expressed as Eq. 5.13 and the defining equation for entropy change, Eq. 6.2a. The result is (Eq. 6.24) ►In accord with the interpretation of σcycle in the Clausius inequality, Eq. 5.14, the value of σ in Eq. 6.24 adheres to the following interpretation = 0 (no irreversibilities present within the system) > 0 (irreversibilities present within the system) < 0 (impossible) σ: where the subscript b indicates the integral is evaluated at the system boundary
2013-3-6 Entropy Balance for Closed Systems That o has a value of zero when there are no internal irreversibilities and is positive when irreversibilities are present within the system leads to the interpretation that o accounts for entropy produced(or generated)within the system by action of irreversibilities. Expressed in words,the entropy balance is within the system across the system boundar the system during some time inte Entropy Balance for Closed Systems Example:One kg of water vapor contained within a piston-cylinder assembly,initially at 5 bar,400C,undergoes an adiabatic expansion to a state where pressure is 1 bar Water and the temperature is (a)200C,(b)100C. Using the entropy balance,determine the Boundary nature of the process in each case. Since the expansion occurs adiabatically,Eq.6.24 reduces to give --Aa一ma-=。0 where m=1 kg and Table A-4 gives s =7.7938 k./kg-K. 9
2013-3-6 9 Entropy Balance for Closed Systems ►That σ has a value of zero when there are no internal irreversibilities and is positive when irreversibilities are present within the system leads to the interpretation that σ accounts for entropy produced (or generated) within the system by action of irreversibilities. ►Expressed in words, the entropy balance is change in the amount of entropy contained within the system during some time interval net amount of entropy transferred in across the system boundary accompanying heat transfer during some time interval amount of entropy produced within the system during some time interval + Entropy Balance for Closed Systems ►Since the expansion occurs adiabatically, Eq. 6.24 reduces to give Example: One kg of water vapor contained within a piston-cylinder assembly, initially at 5 bar, 400oC, undergoes an adiabatic expansion to a state where pressure is 1 bar and the temperature is (a) 200oC, (b) 100oC. Using the entropy balance, determine the nature of the process in each case. σ δ ⎟ + ⎠ ⎞ ⎜ ⎝ ⎛ − = ∫ b 2 1 T Q S S 1 2 0 → m(s2 – s1) = σ (1) where m = 1 kg and Table A-4 gives s1 = 7.7938 kJ/kg·K. Boundary
2013-3-6 Entropy Balance for Closed Systems (a)Table A-4 gives,s2=7.8343 kJ/kg-K.Thus Eq.(1)gives o=(1kg)7.8343-7.7938)kJ/kgK=0.0405kJ/K Sinceo is positive,irreversibilities are present within the system during expansion(a). (b)Table A-4 gives,s2=7.3614 k.J/kg-K.Thus Eq (1)gives c=(1kg)(7.3614-7.7938)kJ/kgK=-0.4324kJ/K Since o is negative,expansion(b)is impossible:it cannot occur adiabatically. Entropy Balance for Closed Systems More about expansion(b):Considering Eq.6.24 s-5=9)+。 <0=<0+20 Since o cannot be negative and For expansion(b)AS is negative,then By inspection the integral must be negative and so heat transfer from the system must occur in expansion(b). 10
2013-3-6 10 Entropy Balance for Closed Systems (a) Table A-4 gives, s2 = 7.8343 kJ/kg·K. Thus Eq. (1) gives σ = (1 kg)(7.8343 – 7.7938) kJ/kg·K = 0.0405 kJ/K Since σ is positive, irreversibilities are present within the system during expansion (a). (b) Table A-4 gives, s2 = 7.3614 kJ/kg·K. Thus Eq. (1) gives σ = (1 kg)(7.3614 – 7.7938) kJ/kg·K = –0.4324 kJ/K Since σ is negative, expansion (b) is impossible: it cannot occur adiabatically. Entropy Balance for Closed Systems ► Since σ cannot be negative and ► For expansion (b) ΔS is negative, then ► By inspection the integral must be negative and so heat transfer from the system must occur in expansion (b). More about expansion (b): Considering Eq. 6.24 < 0 = < 0 + ≥ 0