MECH 310 Problems Solution Manual Chapter 8, Fall 2011

SOLUTION MANUAL CHAPTER 8 Borgnakke and Sonntag CONTENT SUBSECTION PROB NO. In-Text concept questions a-j Concept-Study Guide problems 1-16 Inequality of Clausius 17-23 Entropy of a pure substance 24-34 Reversible processes 35-66 Entropy of a liquid or solid 67-79 Entropy of ideal gases 80-103 Polytropic processes 104-116 Entropy generation 117-157 Rates or fluxes of entropy 158-168 Review 169-185 Problem solution repeated, but using the Pr and vr functions in Table A.7.2: 97, 107, 112 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag In-Text Concept Questions Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.a Does Clausius say anything about the sign for ∫o dQ ? No. The total (net) heat transfer can be coming in like in a heat engine (Wout = QH – QL) in which case it is positive. It can also be net going out like in a refrigerator or heat pump (Win = QH – QL) in which case the sign is negative. Finally if you look at a transmission gearbox there could be no heat transfer (first approximation) in which case the integral is zero. 8.b Does the statement of Clausius require a constant T for the heat transfer as in a Carnot cycle? No. The statement for a cycle involves an integral of dQ/T so T can vary, which it does during most processes in actual devices. This just means that you cannot that easily get a closed expression for the integral. 8.c How can you change s of a substance going through a reversible process? From the definition of entropy ds = dq T for a reversible process. Thus only heat transfer gives a change in s, expansion/compression involving work does not give such a contribution. 8.d A reversible process adds heat to a substance. If T is varying does that influence the change in s? Yes. Reversible: ds = dq T So if T goes up it means that s changes less per unit of dq, and the opposite if T decreases then s changes more per unit of dq. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.e Water at 100 kPa, 150oC receives 75 kJ/kg in a reversible process by heat transfer. Which process changes s the most: constant T, constant v or constant P? ds = dq T Look at the constant property lines in a T-s diagram, Fig. 8.5. The constant v line has a higher slope than the constant P line also at positive slope. Thus both the constant P and v processes have an increase in T. As T goes up the change in s is smaller. The constant T (isothermal) process therefore changes s the most. 8.f A liquid is compressed in a reversible adiabatic process. What is the change in T? If the process is reversible then s is constant, ds = dq T = 0 Change in s for a liquid (an incompressible substance) is Eq. 8.10: ds = C T dT From this it follows that if ds = 0 then T is constant. 8.g An ideal gas goes through a constant T reversible heat addition process. How do the properties (v, u, h, s, P) change (up, down or constant)? Ideal gas: u(T), h(T) so they are both constant Eq. 8.11 gives: ds = dq/T + dsgen = dq/T + 0 > 0 so s goes up by q/T Eq. 8.21 gives: ds = (R/v) dv so v increases Eq. 8.23 gives: ds = -(R/P) dP so P decreases 2 1 P v T s 1 2 T q Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.h Carbon dioxide is compressed to a smaller volume in a polytropic process with n = 1.2. How do the properties (u, h, s, P, T) change (up, down or constant)? For carbon dioxide Table A.5 k = 1.289 so we have n < k and the process curve can be recognized in Figure 8.18. From this we see a smaller volume means moving to the left in the P-v diagram and thus also up. From P-v diagram: P up, T up From T-s diagram Since T is up then s down. As T is up so is h and u. T P 2 2 1 1 T (n = 1) n = 1.2 v s Borgnakke and Sonntag 8.i A substance has heat transfer out. Can you say anything about changes in s if the process is reversible? If it is irreversible? Reversible: ds = dq T < 0 since dq < 0 Irreversible: ds = dq T + dsgen = ? dq < 0 but dsgen > 0 You cannot say, ds depends on the magnitude of dq/T versus dsgen 8.j A substance is compressed adiabatically so P and T go up. Does that change s? If the process is reversible then s is constant, ds = dq T = 0 If the process is irreversible then s goes up, ds = dq T + dsgen = dsgen > 0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Concept Problems Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.1 When a substance has completed a cycle, v, u, h, and s are unchanged. Did anything happen? Explain. Yes. During various parts of the cycle work and heat transfer may be transferred. That happens at different P and T. The net work out equals the net heat transfer in (energy conservation) so dependent upon the sign it is a heat engine or a heat pump (refrigerator). The net effect is thus a conversion of energy from one storage location to another and it may also change nature (some Q was changed to W or the opposite) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.2 Assume a heat engine with a given QH. Can you say anything about QL if the engine is reversible? If it is irreversible? For a reversible heat engine it must be that: ∫o dQ T = 0 = QH TH − QL TL or integrals if T not constant So as TL is lower than TH then QL must be correspondingly lower than QH to obtain the net zero integral. For an irreversible heat engine we have ∫o dQ T = QH TH − QL TL < 0 This means that QL is larger than before (given QH and the T’s). The irreversible heat engine rejects more energy and thus gives less out as work. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.3 CV A is the mass inside a piston/cylinder, CV B is that plus part of the wall out to a source of 1Q2 at Ts. Write the entropy equation for the two control volumes assuming no change of state of the piston mass or walls. Fig. P8.3 P o m p Ts A m The general entropy equation for a control mass is Eq.8.37 S2 – S1 = ⌡⌠ 1 2 dQ T + 1S2 gen The left hand side is storage so that depends of what is inside the C.V. and the integral is summing the dQ/T that crosses the control volume surface while the process proceeds from 1 to 2. C.V. A: mA (s2 – s1) = ⌡⌠ 1 2 dQ TA + 1S2 gen CV A C.V. B: mA (s2 – s1) = ⌡⌠ 1 2 dQ Ts + 1S2 gen CV B In the first equation the temperature is that of mass mA which possibly changes from 1 to 2 whereas in the second equation it is the reservoir temperature Ts. The two entropy generation terms are also different the second one includes the first one plus any s generated in the walls that separate the mass mA from the reservoir and there is a Q over a finite temperature difference. When the storage effect in the walls are neglected the left hand sides of the two equations are equal. Borgnakke and Sonntag 8.4 Consider the previous set-up with the mass mA and the piston cylinder of mass mp starting out at two different temperatures. After a while the temperature becomes uniform without any external heat transfer. Write the entropy equation storage term (S2 – S1) for the total mass. In this case the storage effect must be summed up over all the mass inside the control volume. S2 – S1 = mA (s2 – s1)A + mP (s2 – s1)P ≈ mA (s2 – s1)A + mP CP ln ( T2 T1 P ) The last equation assumed a constant specific heat for the solid material of the piston, a common assumption. There is only a single temperature T2, but there are two different temperatures for state 1 T1 A and T1 P. The temperature T2 would be found from the energy equation. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.5 Water at 100oC, quality 50% in a rigid box is heated to 110oC. How do the properties (P, v, x, u and s) change? (increase, stay about the same, or decrease) A fixed mass in a rigid box give a constant v process. So P goes up (in the two-phase region P = Psat at given T) v stays constant. x goes up ( we get closer to the saturated vapor state see P-v diagram) u goes up (Q in and no work) s goes up (Q in) P T 2 2 1 1 v s Borgnakke and Sonntag 8.6 Liquid water at 20oC, 100 kPa is compressed in a piston/cylinder without any heat transfer to a pressure of 200 kPa. How do the properties (T, v, u, and s) change? (increase, stay about the same, or decrease) Adiabatic dq = 0: dq = T ds = 0; Incompressible dv = 0: dw = P dv = 0 (T, v, u, and s) they are all constant. Only the pressure and enthalpy goes up. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.7 A reversible process in a piston/cylinder is shown in Fig. P8.7. Indicate the storage change u2 - u1 and transfers 1w2 and 1q2 as positive, zero, or negative P T 2 u = C 1 2 1 s v 1w2 = ∫ P dv > 0 ; 1q2 = ∫ T ds > 0 u2 - u1 > 0 from general shape of the constant u curves. Further out in the ideal gas region the constant u curve become horizontal ( u = fct(T) only ). Borgnakke and Sonntag 8.8 A reversible process in a piston/cylinder is shown in Fig. P8.8. Indicate the storage change u2 - u1 and transfers 1w2 and 1q2 as positive, zero, or negative P v 1 2 2 1 T s 1w2 = ∫ P dv > 0 ; 1q2 = ∫ T ds = 0 u2 - u1 = 1q2 - 1w2 < 0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.9 Air at 290 K, 100 kPa in a rigid box is heated to 325 K. How do the properties (P, v, u and s) change? (increase, stay about the same, or decrease) Rigid box: v = constant, (P, u, and s) all increases. T P 2 P 2 2 P 1 1 1 s v Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.10 Air at 20oC, 100 kPa is compressed in a piston/cylinder without any heat transfer to a pressure of 200 kPa. How do the properties (T, v, u and s) change? (increase, about the same or decrease) T goes up, v goes down u goes up (work in, q = 0) s = constant T P 2 P 2 2 P 1 1 1 s v Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8.11 Carbon dioxide is compressed to a smaller volume in a polytropic process with n = 1.4. How do the properties (u, h, s, P, T) change (up, down or constant)? For carbon dioxide Table A.5 k = 1.289 so we have n > k and the process curve can be recognized in Figure 8.18. From this we see a smaller volume means moving to the left in the P-v diagram and thus also up. P up, T up. As T is up so is h and u. From the T-s diagram as n > k then we move to larger T means s is up. T P 2 2 1 T (n = 1) 1 n = 1.4 v s Borgnakke and Sonntag 8.12 Process A: Air at 300 K, 100 kPa is heated to 310 K at constant pressure. Process B: Heat air at 1300 K to 1310 K at constant 100 kPa. Use the table below to compare the property changes. Property ∆A > ∆B ∆A ≈ ∆B ∆A < ∆B a ∆ = v2 – v1 √ b ∆ = h2 – h1 √ c ∆ = s2 – s1 √ a. Ideal gas Pv = RT so v goes with absolute T ∆v = (R/P) ∆T thus the same b. Since dh = CP dT and CP increases with T c. At constant P: ds = (CP/T) dT CP is only 15% higher at 1300 K compared to 300 K (see Fig. 5.11) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.13 Why do we write ∆S or S2 – S1 whereas we write ∫dQ/T and 1S2 gen? This is similar to the terms in the continuity equation m2 – m1 versus mi, me and the energy equation E2 – E1 versus 1Q2, 1W2. The first part is the change in the storage S2 – S1 of entropy in the mass and the second part is the transfer or generation during the process from 1 to 2. The storage terms correspond to the left hand side of the balance equation for entropy. The integral ∫dQ/T represents a transfer of entropy across the control volume surface during the process from 1 to 2 and the 1S2 gen expresses the total amount of entropy being generated inside the control volume and both are on the right hand side of the balance equation. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.14 A reversible heat pump has a flux of s entering as Q . L/TL. What can you say about the exit flux of s at TH? For the entropy equation 8.3 and 8.42, the rate of storage is zero and the generation term is zero. Thus the entropy equation becomes 0 = Q . L TL – Q . H TH + 0 So: Q . L TL = Q . H TH = flux of s We have the same flux of s in as out matching the result in chapter 7. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.15 An electric baseboard heater receives 1500 W of electrical power that heats the room air which loses the same amount out through the walls and windows. Specify exactly where entropy is generated in that process. a Electrical heating wire (electrical work turned into internal energy, leaving as heat transfer). b Heat transfer from hot wire to cooler room air, i.e. in the wire coverings c Room air to walls d Inside walls and windows, heat transfer over a finite ∆T e from outside wall surface to ambient T Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.16 A 500 W electric space heater with a small fan inside heats air by blowing it over a hot electrical wire. For each control volume: a) wire at Twire only b) all the room air at Troom and c) total room plus the heater, specify the storage, entropy transfer terms and entropy generation as rates (neglect any Q . through the room walls or windows). Storage Q . /T S . gen C.V. Wire only 0 −500 W Twire 500 W Twire C.V. Room air 500 W Troom 500 W Twire 500 W ( 1 Troom − 1 Twire ) C.V. Total room 500 W Troom 0 500 W Troom Remark: Room only receives the electrical power input of 500 W. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Inequality of Clausius Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.17 Consider the steam power plant in Example 6.9 and assume an average T in the line between 1 and 2. Show that this cycle satisfies the inequality of Clausius. Solution: Show Clausius: ⌡⌠dQ T ≤ 0 For this problem we have three heat transfer terms: qb = 2831 kJ/kg, qloss = 21 kJ/kg, qc = 2173.3 kJ/kg ⌡⌠dq T = qb Tb – qloss Tavg 1-2 – qc Tc = 2831 573 − 21 568 − 2173.3 318 = –1.93 kJ/kg K < 0 OK Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.18 A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequality of Clausius. a. W . = 6 kW b. W . = 0 kW c. Carnot cycle Solution: TH = 250 + 273 = 523 K ; TL = 30 + 273 = 303 K Case a) ⌡ ⌠ dQ . T = 6000 523 – 0 303 = 11.47 kW/K > 0 Impossible b) ⌡ ⌠ dQ . T = 6000 523 – 6000 303 = –8.33 kW/K < 0 OK c) ⌡ ⌠ dQ . T = 0 = 6000 523 – Q . L 303 ⇒ Q . L = 303 523 × 6 kW = 3.476 kW W . = Q . H – Q . L = 2.529 kW Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.19 Use the inequality of Clausius to show that heat transfer from a warm space towards a colder space without work is a possible process i.e. a heat engine with no work output. Clausius: ⌡⌠dQ T ≤ 0 or ⌡ ⌠ dQ . T ≤ 0 Take C.V. as the space separating the warm and cold space. It is the same Q that crosses each surface (from energy equation) so ⌡⌠dQ T = Q Twarm – Q Tcold = Q ( 1 Twarm – 1 Tcold) < 0 OK Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.20 Use the inequality of Clausius to show that heat transfer from a cold space towards a warmer space without work is an impossible process i.e. a heat pump with no work input. Clausius: ⌡⌠dQ T ≤ 0 or ⌡ ⌠ dQ . T ≤ 0 Take C.V. as the space separating the warm and cold space. It is the same Q that crosses each surface (from energy equation) so ⌡⌠dQ T = Q Tcold – Q Twarm = Q ( 1 Tcold – 1 Twarm) > 0 Impossible! Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.21 Assume the heat engine in Problem 7.32 has a high temperature of 1200 K and a low temperature of 400 K. What does the inequality of Clausius say about each of the four cases? Solution: Cases a) ⌡ ⌠ dQ . T = 6 1200 – 4 400 = – 0.005 kW/K < 0 OK b) ⌡ ⌠ dQ . T = 6 1200 – 0 400 = 0.005 kW/K > 0 Impossible c) ⌡ ⌠ dQ . T = 6 1200 – 2 400 = 0 kW/K Possible if reversible d) ⌡ ⌠ dQ . T = 6 1200 – 6 400 = – 0.001 kW/K < 0 OK H Q W L Q H HE cb T = 400 K L T = 1200 K Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.22 Let the steam power plant in Problem 7.35 have 700oC in the boiler and 40oC during the heat rejection in the condenser. Does that satisfy the inequality of Clausius? Repeat the question for the cycle operated in reverse as a refrigerator. Solution: Q . H = 1 MW Q . L = 0.58 MW ⌡ ⌠ dQ . T = 1000 973 – 580 313 = –0.82 kW/K < 0 OK Refrigerator ⌡ ⌠ dQ . T = 580 313 – 1000 973 = 0.82 > 0 Cannot be possible WT QH QL . WP, in Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.23 Examine the heat engine given in Problem 7.54 to see if it satisfies the inequality of Clausius. Solution: QH = 325 kJ at TH = 1000 K QL = 125 kJ at TL = 400 K ⌡⌠ dQ T = 325 1000 – 125 400 = 0.0125 kJ/K > 0 Impossible H Q = 325 kJ W = 200 kJ L Q = 125 kJ T = 1000 K H HE cb T = 400 K L Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Entropy of a pure substance Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 8.24 Find the missing properties of T, P, s and x for water at: a. P = 25 kPa, s = 7.7 kJ/kgK b. P = 10 MPa, u = 3400 kJ/kg c. T = 150oC, s = 7.4 kJ/kgK Solution: a) Table B.1.2 sf < s < sg so two-phase and T = Tsat(P) = 64.97°C x = (s – sf)/sfg = 7.70 − 0.893 6.9383 = 0.981 h = 271.9 + 0.981 × 2346.3 = 2573.8 kJ/kg b) Table B.1.2 u > ug => Superheated vapor Table B.1.3, x = undefined T ≅ 682°C , s ≅ 7.1223 kJ/kg K c) Table B.1.1 s > sg so superheated vapor. Table B.1.3, x = undefined State is located between 100 and 200 kPa, so interpolate P = 100 kPa + 100 kPa × 7.40 − 7.6133 7.2795 − 7.6133 = 163.9 kPa b P v T s b a a T P c c Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has b