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1 Previous lecture ds relations (derive from steady energy balance) Gibb s equations Entropy change in liquid and solid Equations of & v, & P, and P & for steady isentropic process of ideal gas Isentropic efficiencies of power cycle devices, such as turbine, compressor and pump, and nozzle oday lecture Summary of -ds relations and error in using formula An example of isentropic efficiency in nozzle Entropy balance

2 Objectives Able to use entropy balance for analyzing entropy generation due to - Heat transfer -Wor -Mass flow

3 ds du Pdv d ds = Cv + R = + ( ) d v s s = Cv ( ) + Rln v dv v v s s = Cvav, ln + Rln v ds dh vdp = ( ) d s s = Cp ( ) Rln P P Δs=0 P P v = v d RdP ds = C p P v = v s s = Cpav, ln Rln P P Δs=0 P = P 3

4 Entropy change in liquid or solid d s s = C + R v v v ( ) ln p ( ) d s s = C Rln P P For liquid and solid substance d = s s C 4

5 Error in using formula P P v = v v = v P = P ( ) Δ h = C Δ u = C ( ) pave, vave, 5

6 Caution! 6

7 Isentropic efficiency in nozzle P P Actual KE at nozzle exit η N = Isentropic KE at nozzle exit Va / η N = V / s Neglect inlet velocity h h V V η N = = h h V V a a s s Consider inlet velocity 7

8 Example (Neglect inlet velocity) Nitrogen expands in a nozzle from a temperature of 500 K while its pressure decreases by factor of two. What is the exit velocity of the nitrogen when the nozzle isentropic efficiency is 95 percent?. Assume - constant specific heats - an isentropic process and then apply the nozzle isentropic efficiency to find the actual exit velocity. 8

9 Example (cont.) For the isentropic case, Q net = 0. Assume steady-state, steady-flow, no wor is done. Neglect the inlet inetic energy and changes in potential energies. hen for one entrance, one exit, the first law reduces to when v << v he conservation of mass gives E in = Eout V s mh = m ( hs + ) m = m = m he conservation of energy reduces to V = ( h h ) s s Using the ideal gas assumption with constant specific heats, the isentropic exit velocity is V = C ( ) s p s 9

10 Example (cont.) he isentropic temperature at state is found from the isentropic relation P s = P P = 500 K = 40 K ( ) P V = C ( ) s p s =.005 ( K) = 44.8 g.k J/g s 3 J 0 m /s m he nozzle exit velocity is obtained from the nozzle isentropic efficiency as N V = V a s / η Va Vs ηn / m m = = = 4.8 s s 0

11 Example (Consider inlet velocity) P /P = 0.5 η N =0.9 h h V V η N = = h h V V V V η = V V ( ) a a s s s N a =500K E in = E out V V s m h+ = m hs + m = const V s V h hs = V s V Cpave, ( s) = P = P s s V = V V η + V ( ) a s N P s s = P is also an average value

12 ( )( ) s = 500 K 0.5 = 40K V s V Cpave, ( s) = V = C + V ( ) s p, ave s V = C ( ) s pave, s Example (cont.) if V << V V = V V η + V ( ) a s N V = η ( V ) a s N

13 Entropy Balance - Property of entropy is a measure of molecular disorder or randomness of a system -From nd law of hermodynamics, entropy can be created but it can not be destroyed 3

Entropy Balance he principle of increase of entropy for any system is expressed as an entropy balance given by S in S out + S gen = Δ S system he entropy balance relation can be stated as:

14 Entropy Balance he principle of increase of entropy for any system is expressed as an entropy balance given by S in S out + S gen = Δ S system he entropy balance relation can be stated as: the entropy change of a system during a process is equal to the net entropy transfer through the system boundary and the entropy generated within the system as a result of irreversibilities. 4

15 Entropy change of a system he entropy change of a system is the result of the process occurring within the system. Entropy change = Entropy at final state Entropy at initial state Initial Final 5

16 Mechanisms of Entropy ransfer, S in and S out - Entropy can be transferred to or from a system by two mechanisms: heat transfer and mass flow. - Entropy transfer occurs at the system boundary as it crosses the boundary, and it represents the entropy gained or lost by a system during the process. - he only form of entropy interaction associated with a closed system is heat transfer, and thus the entropy transfer for an adiabatic closed 6 system is zero.

Heat transfer he ratio of the heat transfer Q at a location to the absolute temperature at that location is called the entropy flow or entropy transfer and is given as Q Entropy transfer by heat

17 Heat transfer he ratio of the heat transfer Q at a location to the absolute temperature at that location is called the entropy flow or entropy transfer and is given as Q Entropy transfer by heat transfer: S heat = ( =constant) -Q/represents the entropy transfer accompanied by heat transfer - Direction of entropy transfer is the same as the direction of heat transfer since the absolute temperature is always a positive quantity. - When the temperature is not constant, the entropy transfer for process - can be determined by 7

18 Wor - Wor is entropy-free, and no entropy is transferred by wor. - Energy is transferred by both wor and heat ( st law of hermodynamics), whereas entropy is transferred only by heat and mass. Entropy transfer by wor: = 0 S wor 8

19 Mass flow - Mass contains entropy and energy - Entropy and energy contents of a system are proportional to the mass. - When a mass in the amount m enters or leaves a system, entropy in the amount of ms enters or leaves, where s is the specific entropy of the mass. Entropy transfer by mass: Smass = ms 9

Entropy Generation, S gen Irreversibilities such as friction,

temperature difference, unrestrained expansion,

entropy of a system to increase, and entropy generation is a

20 Entropy Generation, S gen Irreversibilities such as friction, mixing, chemical reactions, heat transfer through a finite temperature difference, unrestrained expansion, nonquasi-equilibrium expansion, or compression always cause the entropy of a system to increase, and entropy generation is a measure of the entropy created by such effects during a process. Friction loss Heat transfer Mixing 0

21 Entropy Generation, S gen - For a reversible process, the entropy generation is zero and the entropy change of a system is equal to the entropy transfer. - Entropy transfer by heat is zero for an adiabatic system - Entropy transfer by mass is zero for a closed system. he entropy balance for any system undergoing any process can be ex-pressed in the general form as

22 Rate of entropy transfer he entropy balance for any system undergoing any process can be ex-pressed in the general rate form, as S S + S = in out gen where the rates of entropy transfer by heat transferred at a rate of Q and mass flowing at a rate of m are S = Q / and S = ms heat ds dt sys mass he entropy balance can also be expressed on a unit-mass basis as ( s s ) + s = Δs ( J / g K) in out gen system

Isolated system: Entropy Generation, S gen - he term S gen is the entropy generation within the system boundary only, and not the entropy generation that may occur outside the system boundary during

23 Isolated system: Entropy Generation, S gen - he term S gen is the entropy generation within the system boundary only, and not the entropy generation that may occur outside the system boundary during the process as a result of external irreversibilities. - S gen = 0 for the internally reversible process, but not necessarily zero for the totally reversible process. - he total entropy generated during any process is obtained by applying the entropy balance to an Isolated System that contains the system itself and its immediate surroundings. Isolated system S gen, outside = 0 3

24 Closed system: Entropy Generation, S gen Closed Systems aing the positive direction of heat transfer to the system to be positive, the general entropy balance for the closed system is Q + Sgen = ΔSsystem = S S ( J / K) For an adiabatic process (Q = 0), this reduces to Adiabatic closed system: S = ΔS gen adiabatic system A general closed system and its surroundings (an isolated system) can be treated as an adiabatic system, and the entropy balance becomes System + surroundings: S = ΔS = ΔS + ΔS gen system surroundings 4

Q i i e e gen CV + ms ms + S = ( S S ) ( J/ K) In the rate form we have Q Control volume :Entropy Generation, S gen Control Volumes he entropy balance for control volumes differs from that for closed

25 Q i i e e gen CV + ms ms + S = ( S S ) ( J/ K) In the rate form we have Q Control volume :Entropy Generation, S gen Control Volumes he entropy balance for control volumes differs from that for closed systems in that the entropy exchange due to mass flow must be included. i i e e gen CV + ms ms + S = ΔS ( W/ K) - Rate of entropy change within the control volume during a process is equal to the sum of the rate of entropy transfer through the control volume boundary by heat transfer, the net rate of entropy transfer into the control volume by mass flow, and the rate of entropy generation within the boundaries of the control volume as a result of irreversibilities. 5

26 Control volume :Entropy Generation, S gen Q ms i i ms e e Sgen SCV + + =Δ For a general steady-flow process, by setting Δ S CV = 0 the entropy balance simplifies to Q Sgen = me se ms i i For a single-stream (one inlet and one exit), steady-flow device, the entropy balance becomes S gen = m ( se si ) Q Heat transfer 6

27 Adiabatic process: Entropy Generation, S gen For an adiabatic single-stream device, the entropy balance becomes S = m ( s s ) gen e i his states that the specific entropy of the fluid must increase as it flows through an adiabatic device since S gen 0. If the flow through the device is reversible and adiabatic, then the entropy will remain constant regardless of the changes in other properties. herefore, for steady-flow, single-stream, adiabatic and reversible process: s e = s i 7

28 Example An inventor claims to have developed a water mixing device in which 0 g/s of water at 5 o C and 0. MPa and 0.5 g/s of water at 00 o C, 0. MPa, are mixed to produce 0.5 g/s of water as a saturated liquid at 0. MPa. If the surroundings to this device are at 0 o C, is this process possible? If not, what temperature must the surroundings have for the process to be possible? Assume steady-flow 8

29 Example (cont.) If there is a heat transfer from the surroundings to the mixing chamber. Assume there is no wor done during the mixing process, and neglect inetic and potential energy changes. hen for two entrances and one exit, the first law becomes P 0.MPa g J h h = = o = 5 C J s = g K P 0.MPa g J h = = = o 00 C J s = 7.36 g K 9

30 3 P3 = 0.MPa g Sat. liquid net J h = 47.5 J s = g K Q = m h mh m h 3 3 g J g J g J = s g s g s g = J s So, J/s of heat energy must be transferred from the surroundings to this mixing process, or Q = Q Example (cont.) net, surr net, CV For the process to be possible, the second law must be satisfied. Write the second law for the isolated system, Q ms ms S ΔS + + = i i e e gen CV 30

31 Example (cont.) For steady-flow Δ S CV =. 0 Solving for entropy Q Sgen = mese ms i i generation, we have Q cv = ms 3 3 ms ms g J g J = s g K s g K g J J / s s g K (0 + 73) K J = 0.49 K s Since S gen must be 0 to satisfy the second law, this process is impossible, and the inventor's claim is false. surr o find the minimum value of the surrounding temperature to mae this mixing process possible, set = 0 and solve for surr. S gen 3

32 S m s ms gen = e e i i = surr = Q Q cv ms ms ms 3 3 Example (cont.) J / s = g J g J g J s g K s g K s g K = 35.75K One way to thin about this process is as follows: Heat is transferred from the surroundings at K (4.75 o C) in the amount of J/s to increase the water temperature to approximately 4.75 o C before the water is mixed with the superheated steam. Recall that the surroundings must be at a temperature greater than the water for the heat transfer to tae place from the surroundings to the water. 3

Chapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn

Chapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Chapter 7 Entropy by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics: An Engineering Approach, 5th ed.,