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EGR 334 Thermodynamics Chapter 6: Sections 6-8 Lecture 25: Entropy and closed system analysis Quiz Today? Today’s main concepts: • Heat transfer of an internally reversible process can be represented as an area on a T-s diagram. • Learn how to evaluate the entropy balance for a closed system Reading Assignment: Read Chapter 6, Sections 9-10 Homework Assignment: Problems from Chap 6: 36, 38, 59, 66 3 Recall from last time: Energy Balance: E sys Q sys W sys dE sys Energy Rate Balance: d E sys dt W Q dt Q sys W sys Entropy Balance: S sys Q T Q g en T Entropy Rate Balance: d S sys dt Q T d S sys dt gen g en Entropy Balance for Closed Systems Q S T b change of entropy entropy transfer entropy production where the subscript b indicates the integral is evaluated at the system boundary. • Unlike mass and energy balances, the entropy balance doesn’t represent a conserved quantity. : = 0 (no irreversibilities present within the system) > 0 (irreversibilities present within the system) < 0 (impossible) Change of Entropy of the System Q S T b As seen in the previous lecture, ∆S = S2-S1, represents a difference of state properties and may be evaluated by -- looking up s values on substance property tables. -- applying Tds relationships for ideal gas • If two or more properties of the end states of a process are known, then the Change of Entropy per unit mass, ∆s, is completely defined and discernible. Entropy Transfer • Consider the Entropy Transfer term of the Entropy Balance. On a differential basis it can expressed • This expression 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 ►In an internally reversible, adiabatic process (no heat transfer), entropy remains constant. Such a constantentropy process is called an isentropic process. ► Rearranging the differential expression gives which can be integrated from state 1 to state 2, 2 Q int rev T dS 1 Entropy and Heat Transfer Consider how this integral would represented on a T-S diagram: 2 Q in t rev T d S A rea u n d er T -S 1 An energy transfer by heat to a closed system during an internally reversible process is represented by an area on a temperature-entropy diagram: 6.6&7 : Entropy and Closed Systems 9 For the Carnot cycle: What does the area represent? Carnot Work: W PdV What does the area represent? Carnot Heat Transfer: Q In the Carnot Cycle, only the constant temperature processes contribute to Heat (and Entropy) transfer. TdS 6.6&7 : Entropy and Closed Systems 10 For the Carnot cycle: 3 Q 23 TdS 2 Q dS T in t Q TdS 0 1 Q 41 rev. TdS 4 0 Q Q1 2 Q 2 3 Q 3 4 Q 4 1 (note that Q41 is negative) Entropy Balance for Closed Systems That 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 accounts for entropy produced (or generated) within the system by action of irreversibilities. Q S T int rev. 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 A general approach to analyze a Closed System with entropy balance. • Step 1: Identify properties at each state including T, p, v, u, x, and s. • Step 2: Apply 1st law and attempt to evaluate ∆U, Q, and W for each process. • Step 3: Write the 2nd Law (Entropy balance) and attempt to evaluate ∆S and ΣQ/T for each process. Usually σ will be determined from σ = ∆S - ΣQ/T • Step 4: Explain the significance of σ. Example: One kg of water vapor contained within a pistoncylinder 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. Since the expansion occurs adiabatically, Boundary Q and entropy balance reduces to give 0 S 2 S1 1 2 Q T b → Find property values: using m = 1 kg and Table A-4 m(s2 – s1) = get s1 = 7.7938 kJ/kg∙K. Example 1 continued: (a) At p = 1 bar and T = 200 deg C. Table A-4 gives, s2 = 7.8343 kJ/kg∙K. then m(s2 – s1) = (1 kg)(7.8343 – 7.7938) kJ/kg∙K = 0.0405 kJ/K (Since s is positive, irreversibilities are present within the system during expansion ) (b) At p = 1 bar and T = 100 deg C. Table A-4 gives, s2 = 7.3614 kJ/kg∙K. then m(s2 – s1) = (1 kg)(7.3614 – 7.7938) kJ/kg∙K = –0.4324 kJ/K (Since s is negative, expansion (b) is impossible. It cannot occur adiabatically.) Example 1 continued: Just a little more analysis of part b) The result of part b was that is negative. The process cannot occur adiabatically. <0 = <0 + ≥0 ► Since cannot be negative and ► For expansion in (part b) S is negative, then ► By inspection the integral must be negative and so heat transfer from the system must occur in expansion (b). Entropy Rate Balance for Closed Systems Some problems are presented in the form of a closed system entropy rate balance given by where dS the time rate of change of the entropy of the system the time rate of entropy transfer through the portion of the boundary whose temperature is Tj dt Q j Tj time rate of entropy production due to irreversibilities within the system Example 2: An inventor claims that the device shown generates electricity at a rate of 100 kJ/s while receiving a heat transfer of energy at a rate of 250 kJ/s at a temperature of 500 K, receiving a second heat transfer at a rate of 350 kJ/s at 700 K, and discharging energy by heat transfer at a rate of 500 kJ/s at a temperature of 1000 K. Each heat transfer is positive in the direction of the accompanying arrow. For operation at steady state, evaluate this claim. Q 1 250 kJ/s T1 = 500 K Q 2 350 kJ/s + – T2 = 700 K T3 = 1000 K Q 3 500 kJ/s Example 2 continued: Applying an energy rate balance dE at steady state dt Solving 0 0 Q 1 Q 2 Q 3 W e W e 250 kJ/s 350 kJ/s 500 kJ/s 100 kJ/s The claim is in accord with the first law of thermodynamics. Applying an entropy rate balance at steady state Solving 250 kJ/s 500 K 350 kJ/s 700 K 0 . 5 0 . 5 0 . 5 kJ/s K dS 0 0 dt Q 1 T1 Q 2 T2 Q 3 T3 500 kJ/s 1000 K 0.5 kJ/s K ∙ negative, the claim is not in accord with the 2nd Law of Since σ is Thermodynamics and is therefore denied. 19 Example 3 (6.14): One kilogram of water contained in a piston-cylinder assembly, initially at 160°C, 150 kPa undergoes an isothermal compression process to saturated liquid. For the process, W= –471.5 kJ. Determine for the process. (a) (b) (c) (d) Sketch the process on a T-s diagram. The heat transfer, in kJ The change in entropy in kJ/K The entropy generated in the process. State 1: T1= 160 oC p1 = 150 kPa = 1.50 bar super heated vapor State 2: T2= T1 =160 oC sat. liquid. T 2 1 s 20 Example 3 (6.14): One kilogram of water contained in a piston-cylinder assembly, initially at 160°C,150 kPa undergoes an isothermal expansion from compression process to saturated liquid. For the process, W= –471.5 kJ. b) find the heat transfer State 1 2 Look up the rest of the state properties. Then apply the 1st Law. U Q W Q m u 2 u 1 W T (°C) 160 160 p (kPa) 1.50 150 617.8 SH 0 u (kJ/kg 2595.2 674.86 s (kJ/kg K) 7.4665 1.9427 x Q 1kg 674.86 2595.2 kJ / kg ( 471.5 kJ ) 2391.84 kJ / K (c) To find the change in entropy S m ( s 2 s1 ) S 1kg 1 .9 4 2 7 7 .4 6 6 5 kJ / kg K 5 .5 2 3 8 kJ kg K 21 Example 3 (6.14): One kilogram of water contained in a piston-cylinder assembly, initially at 160°C,150 kPa undergoes an isothermal compression process to saturated liquid. For the process, W= –471.5 kJ. Determine for the process. (d) To find the entropy generation S where Q T S m s 2 s1 5.5238 kJ / kg K Q T 2391.84 kJ (160 273) K S Q State 1 2 T (°C) 160 160 p (kPa) 150 617.8 x SH 0 u (kJ/kg 2595.2 674.86 s (kJ/kg K) 7.4665 1.9427 5.5239 kJ / K 5.5238 ( 5.5239) 0.0001 0 kJ / K T What does this mean? Process had no irreversibilities 6.11 : Isentropic Processes States may be given as having the same entropy (two-phase, saturated vapor, superheated vapor) Any process where the entropy does not change is called isentropic. 22 6.11 : Isentropic Processes 23 Consider isentropic process for an ideal gas Starting with s and Next with c V T dT T cV s R k 1 T1 T 2 s cP T dT T and R cP k 1 v R ln 2 T v1 dT T2 ln k 1 T1 1 thus v R ln 2 v1 s k 1 kR k 1 p2 R ln T p 1 dT T2 p2 ln ln k 1 T1 p 1 v2 ln v 1 v2 v 1 kR p2 R ln p 1 k 1 k 1 thus and An isentropic process is a p 2 v1 type of polytropic process p v 1 2 T2 p 2 T p 1 1 ( k 1) k k p 1 v1 p 2 v 2 k k 24 Example (6.27): Air in a piston-cylinder assembly and modeled as an ideal gas undergoes two internally reversible processes in series from State 1 where T1 = 290 K, p1 = 1 bar. Process 1 – 2 : Compression to p2 = 5 bar during which pV1.19 = constant Process 2 – 3 : Isentropic expansion to p3 = 1 bar. (a) Sketch the two processes on T-s coordinates (b) Determine the temperature at State 2 in K Air (c) Determine the net work in kJ 5 bar 2 TT 1 3 SS 290 K 1 bar 25 Example (6.27): Air in a piston-cylinder assembly and modeled as an ideal gas undergoes two internally reversible processes in series from state 1 where T1 = 290 K, p1 = 1 bar. Process 1 – 2 : Compression to p2 = 5 bar during which pV1.19 = constant Process 2 – 3 : Isentropic expansion to p3 = 1 bar. (a) Sketch the two processes on T-s coordinates (b) Determine the temperature at state 2 in K (c) Determine the net work in kJ p1V1 p 2V 2 Process 1-2: N N ideal gas: T1 T2 polytropic: p1V1 p 2V 2 V2 V1 p 1 p2 1/ N V2 p1 p2 p2 T 2 T1 p1 1/ N n 1 p1T 2 p1T 2 V1 p 2T2 1 .1 9 1 1 .1 9 p 2T2 n 5 290 K 1 375 K 26 Example (6.27): Process 1 – 2 : Compression to p2 = 5 bar during which pV1.19 = constant Process 2 – 3 : Isentropic expansion to p3 = 1 bar. Find the state properties State 1: p1 = 1 bar T1 = 290 K State 2: p2 = 5 bar T2 = 375 K State 1 2 T (K) 290 375 1 5 u (kJ/kg) 206.91 268.075 s °(kJ/kg K) 1.66802 1.92657 p (bar) State 3: p3 = 1 bar s3 = s2 From Table A-22: at T = 290 K: u = 206.91 so = 1.66802 From Table A-22: at T = 375 K: u = 268.075 so = 1.92657 3 1 27 Example (6.27): Process 2 – 3 : Isentropic expansion to p3 = 1 bar. State State 11 22 33 TT (K) (K) 290 290 375 375 236.83 11 55 11 uu (kJ/kg) (kJ/kg) 206.91 206.91 268.075 268.075 168.86 ss °(kJ/kg °(kJ/kg K) K) 1.66802 1.66802 1.92657 1.92657 1.46466 pp (bar) (bar) At State 3: p3 s 3 s 2 s T3 s T 2 R ln p 2 p 0 s T3 s T 2 R ln 3 p2 p3 s T3 s T 2 R ln p 2 1b a r s T3 (1 .9 2 6 5 7 kJ / kg K ) (0 .2 8 7 0 kJ / kg K ) ln 1 .4 6 4 6 6 kJ / kg K 5b a r From Table A-22: for so=1.46466 find T and u T3 = 236.82 K and u3 = 168.86 kJ/kg 28 Example (6.27): (c) Determine the net work in kJ Process ΔU 1–2 +61.2 Q W State 1 2 3 -67.2 -128.4 T (K) 290 375 636.82 1 5 1 u (kJ/kg) 206.91 268.075 168.86 s °(kJ/kg K) 1.66802 1.92657 1.46466 2–3 p (bar) Process 1 – 2: U 12 Q12 W 12 where U 12 m ( u 2 u 1 ) U 12 m 268.08 206.91 W 12 pdV W 12 0.287 kJ m 61.17 kJ / kg Q12 m U 12 m p 2V 2 p1V1 1 n m R T 2 T1 1 n / kg K (375 290) K 1 1.19 128.39 kJ / kg W 12 m 61.17 ( 128.39) 67.22 kJ / kg 29 Example (6.27): Process ΔU Q W 1–2 +61.2 -67.2 -128.4 2–3 -99.2 0 +99.2 Process 2-3: State 1 2 3 T (K) 290 375 236.82 1 5 1 u (kJ/kg) 206.91 268.075 168.86 s °(kJ/kg K) 1.66802 1.92657 1.46468 P (bar) U 23 Q 23 W 23 3 where Q 23 T ds 0 U 23 m ( u 3 u 2 ) U 23 2 168.86 268.08 99.22 kJ / kg m W 23 m Q 23 U 23 m 0 ( 99.22) 99.22 kg / kg m Net Work over both processes: W net m W12 m W 23 m 128 . 39 kJ kg 99 . 22 kJ kg 29 . 17 kJ kg 6.8 : Directionality of Processes 30 Second Law statement: “It is impossible for a system to operate such that entropy is destroyed.” This can be seen in the entropy balance, but first look at the energy balance. E E isolated system system E 0 en viro n m en t 0 Energy is neither created nor destroyed. S Isolated System S system Q T int S rev. en viro n m en t 31 End of Slides for Lecture 25