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Chapter 1 Solutions P1.1 This example shows how the First Law can be applied to individual processes and how these can make up a cycle. A cycle of events is shown in Fig A.10. It is made up of four processes, and the heat and work associated with those processes is as given. Fig A.10: Cycle of events made up of four processes. Process 12: Q = +10J W = -18J Process 23: Q = +100J W = 0J Process 34: Q = -20J W = +70J Process 41: Q = -10J W = +28J Calculate the values of U - U , the net work, the net heat transfer and the heat supplied for the cycle. n 1 Solution This problem can be solved by applying the First Law to each of the processes in turn, when the change in internal energy is dU QW. Process 12: dU U U Q W 101828J 12 2 1 12 12 Process 23: dU U U Q W 1000100J 23 3 2 23 23 Process 34: dU U U Q W 207090J 34 4 3 34 34 Process 41: dU U U Q W 102838J 41 1 4 41 41 The change in internal energy relative to the internal energy at point 1, U , is given by 1 dU U U U U U U ......U U 1n n 1 n n1 n1 n2 2 1 dU dU ......dU n1,n n2,n1 12 Hence: dU U U 28J 12 2 1 dU dU dU 28100128J 13 12 23 dU dU dU dU 28100(90)38J 14 12 23 34 dU dU dU dU dU 28100(90)(38)0J 11 12 23 34 41 The result dU = 0 confirms that the four processes constitute a cycle, because the net change of state 11 is zero. The net work done in the cycle is W W W W W 12 23 34 41 cycle 180702880J The positive sign indicates that the system does work on the surroundings. The net heat supplied in the cycle is Solutions Chapter 1 Page 1  D E Winterbone Current edition: 13/02/2015 Chapter 1 Solutions QQ Q Q Q 12 23 34 41 cycle 10100201080J Hence the net work done and net heat supplied are both 80J, as would be expected because the state of the system has not changed between both ends of the cycle. It is also possible to differentiate between the heat supplied to the system, Q > 0, and heat rejected from the system, Q < 0. In this case the total heat supplied is QQ Q 12 23 cycle 10100110J while the heat rejected is the sum of the negative heat transfer terms, viz.: QQ Q 34 41 . cycle 201030J It should also be noted that the work done, W, is the difference between the heat supplied and that rejected. This is an important point when considering the conversion of heat into work, which is dealt with by the Second Law of Thermodynamics. __________________________________________________________________________________ P1.2 There are two ways to solve this problem: 1. A simple one based on the p-V diagram 2. A more general one based on the p-V relationship Work done = Area under lines. Assuming a linear spring, then a linear (straight) line relationship joins the points. 1 105 Hence, Work done by spring = Area abca = W  0.54 100kJ s 2 103 Work done by gas = Area adeca = Area abca + Area adbca 105 = 1001.00.5 150kJ 103 More general method is to evaluate the work as W pdV . First find the relationship for the spring. If linear p kV k,and p kV k g1 1 g2 2 p p k g2 g1 8bar/m3, and k7bar Thus V V 2 1 giving p 8V 7. g Solutions Chapter 1 Page 2  D E Winterbone Current edition: 13/02/2015 Chapter 1 Solutions δW  p dV, g g 2 8V2 2 giving W 8V 7dV  7V Work done by gas, g  2  1 1 42.25171.51102 150kJ δW p dV s s Work done by spring, 2 V2 2 W 8V 8dV 8 V 100kJ s  2  1  1 The benefit of the latter approach is that it can be applied in the case of a non-linear spring: it is more difficult to use the simpler approach. _________________________________________________________________________________________ P1.3 p 1.5 Pressure of gas is proportional to diameter, i.e.pd, giving p kd , and hence k 1  5bar/m 1 1 d 0.3 1 d3 6V 13 VolumeV  , thus d   6    Work done during process, W pdV 3d2 d2 Working in terms of d dV  dd  dd 6 2 2 d2 k2 kd 4 d4 W kd dd  d3.dd   2  1  1 2 2 1 2  4 4  5  0.3340.34105 738J 42  This problem can also be solved in terms of V; however, it cannot be solved using a linear approximation. You might have been close to the correct solution for P1.3, but it does not work for P1.4. _________________________________________________________________________________________ P1.4 The problem is the same as P1.3, but the final diameter is 1m. 2 d2 k2 kd 4 d4 W kd dd  d3.dd   2  1  The work done is 1 2 2 1 2  4 4  5  1.040.34105 194759J 42  Using the WRONG APPROACH, assuming a linear relationship, gives W 1.55d 3d3105 165580J 2 6 2 1 _________________________________________________________________________________________ P1.5 t at 20bar = 212.4C s Initial conditions: u 2600kJ /kg;h 2799kJ /kg;v 0.0995m3 /kg g g g p20bar;t 500C Final conditions: u 3116kJ /kg;h 3467kJ /kg;v 0.1756m3 /kg g g g To evaluate the energy added use 1st Law for closed system QdU W dU  pdV Solutions Chapter 1 Page 3  D E Winterbone Current edition: 13/02/2015 Chapter 1 Solutions  20105  Qm31162600 0.17560.09957 Hence energy added,  103  3516152.062004kJ Alternative method Constant pressure process, hence enthalpy can be used Qdhmh h 3346727992004kJ 2 1 20105 W  0.17560.99573456kJ 103 _________________________________________________________________________________________ P1.6 This is a constant volume (isochoric) process. Hence v 0.00317m3/kg at critical point, and v 0.00317m3/kg. 1 1 But v  xv 1xv v xv 2 g f f fg At p 27.5bar, interpolation v 0.072788m3/kg, and t 229C 2 g 2 Using the saturated water tables, v = 0.001209m3/kg f 0.003170.001209 Hence, x 0.02742.74% 0.0727880.001209 _________________________________________________________________________________________ P1.7 Initial and final conditions: p 3.5bar;m 1.0kg;x 1.0;u 2549kJ /kg;u 2549kJ /kg 1A 1A 1A g1A 1A p 7.0bar;m 2.0kg;x 0.8;u 2573kJ /kg;u 696kJ /kg;u 2197.6kJ /kg 1B 1B 1B g1B f1B 1B p 5bar;m 3.0kg 2 2 Total U 25494395.26944.2kJ /kg 1 Volumes v 0.5241m3 /kg;V 0.5241m3 /kg g1A 1A v 0.2728m3 /kg;v 0.1107102m3/kg(interp); V 0.21846m3/kg g1B f1B 1B Total volume, V 0.524120.218460.9610m3/kg 1 Conditions at 2 V V 0.9610m3;v  2 0.3203m3 /kg 2 2 m 2 At 5bar, v 0.3748m3 /kg;v 01093102m3 /kg g f v v 0.3192 x  2 f2  0.8541 2 v v 0.3737 g2 f2 Thus U 30.854125620.14596396846kJ 2 Total heat transfer, QU U 98kJ 2 1 _________________________________________________________________________________________ P1.8 Solutions Chapter 1 Page 4  D E Winterbone Current edition: 13/02/2015 Chapter 1 Solutions m5kg;p 14bar;x 0.8 1 1 v 0.1408m3 /kg;v 0.001149m3 /kg g1 f1 u 2593kJ /kg;u 828kJ /kg g1 f1 v  xv 1x v 0.1129m3 /kg:V 50.11290.5645m3 1 1 g1 1 f1 1 u  xu 1x u 2240kJ /kg:U 5224011200kJ 1 1 g1 1 f1 1 Spring equation, p = kV, because p = 0 when V = 0. When volume is 150% of initial value, p 1.5p 21bar:V 1.50.56450.8468m3 2 1 2 Work done, W  pV kVdV p p pV W  pdV kVdV  1 VdV  1 V2 V2 1 11.25 V 2V  2 1  2 Hence, 1 1 140.5645 105  1.25 494kJ 2 103 Conditions at 2: V 0.8468m3;v 0.16936m3/kg: p 21bar 2 2 2 Based on steam tables condition 2 is in the superheat region: volumes listed below p = 20bar 0.1634 0.1756 p = 21bar 0.1578 0.1696 (interpolated) p = 30bar 0.1078 0.1161 Hence t = 500C 2 u 31160.1310831163115kJ /kg 2 U 5311515576kJ 2 Thus Heat transfer from 1st Law QdU W 15576112004944870kJ _________________________________________________________________________________________ P1.9 t 20C;p 5.673bar;h 195.78kJ /kg;h 54.87kJ /kg A s g f Tank A: 1kg Freon-12 105 u h  p v 195.785.673 0.0308178.3kJ /kg g g g g 103 Cylinder: isobaric expansion at 2bar Heat transferred so temperature constant at 20°C: calculate heat transfer. V 0m3;V ? Flow across valve is isenthalpic: p1 p2 Initial energy, U 178.3kJ; initial volume, V 0.0308m3 1 1 Final volume, V 0.0969m3 2 2105 Hence work done, W  pdV  0.09690.030813.22kJ 103 2105 Final energy, U 202.27 0.0969182.9kJ 2 103 Hence, heat transfer, QdU W 182.9178.313.2217.8kJ ___________________________________________________________________________________ Solutions Chapter 1 Page 5  D E Winterbone Current edition: 13/02/2015 Chapter 1 Solutions Problems P1.10 to P1.14 require the USFEE These examples demonstrate that it is possible to obtain solutions for problems that have flows across the system boundaries by closed system methods. P1.10 An insulated bottle is initially evacuated, i.e. it contains a vacuum, and then the stopper is removed allowing atmospheric air to fill the bottle. Evaluate the final conditions in a bottle when the air has just filled it. Comment: This is a relatively complex problem in unsteady gas dynamics if the processes between the removal of the stopper and the quiescent end state are considered in detail. In the flow processes pressure waves will travel into the bottle and be reflected; these will cause the atmospheric air to start flowing into the bottle. The waves will ultimately die out due to the flow interactions at the neck of the bottle and fluid friction, and finally a steady state will be reached. The great strength of the approaches of thermodynamics are that they allow the final state to be evaluated without any knowledge of gas dynamics. The bottle is shown in Fig A. 11; it has a volume of V. At t = 0, referred to as state 1, there is no gas in the bottle and its pressure is zero. Hence, p = 0 and m = 0. Gas is admitted to the bottle until the 1 1 pressure of the gas in the bottle is atmospheric pressure; p = p . It is possible to treat this as a closed 2 atm system problem if the system boundaries are drawn in such a way that no flow occurs across them during the filling process. This has been done by drawing a boundary around the gas which flows into the bottle during the process: in this case it could be an actual boundary such as an extremely flexible balloon which has been filled sufficiently to hold all the air required to fill the bottle. Fig A.11: Bottle being filled with atmospheric air. The bottle is initially evacuated: m = 0 at t = 0. a Considering the total system at time t = 0, i.e. system a + system b. The First Law can be applied to the combined system, giving QdUW. In this case Q = 0, because the bottle is insulated. The total work done during the time interval is the pdV work done on the gas in system b, because system a does not change volume and no stirring work is done. Hence dU U U W, 2 1 where U U U 0m u m u 1 1a 1b in in in in and U U U m u 0m u 2 2a 2b in 2 in 2. Rearranging the equation gives U U W 2 1 where Wp V m p v . atm in in atm in Solutions Chapter 1 Page 6  D E Winterbone Current edition: 13/02/2015 Chapter 1 Solutions Thus U m u m u m p v m u  p v , 2 in 2 in 1 in atm in in in atm in giving the surprising result that the specific internal energy of the gas in the bottle is not the same as that of the atmosphere, but is u u  p v h . 2 in atm in in i.e. the internal energy of the gas in the bottle is equal to the enthalpy of the gas that was forced into the bottle. The reason for this is that the atmosphere did work on pushing the gas from system b into system a. ________________________________________________________________________________ P1.11 Filling a bottle which already contains some fluid. The analysis adopted above may be applied to a bottle which is not initially evacuated, i.e. p  0. a1 This is shown in Fig P1.11. The approach to solving this problem is similar to that adopted above. The same technique is used to make the problem a closed system one. As before, Q = 0, because the bottle is insulated. The total work done during the time interval is the pdV work done on the gas in system b, because system a does not change volume and no stirring work is done. The mass of air in the bottle at the beginning of the process will be denoted by m , and its 1 specific internal energy will be u . Hence, again 1 dU U U W, 2 1 but now U U U mu m u 1 1a 1b 1 1 in in and U U U m u 0m u 2 2a 2b in 2 in 2. Fig P1.11(a): Filling of a bottle which is not initially evacuated Rearranging the equation gives U U W 2 1 where Wp V m p v . atm in in atm in Thus U m u mu m u m p v mu m u  p v , 2 in 2 1 1 in in in atm in 1 1 in in atm in where m m m. Substituting this value in the equation for u gives in 2 1 2 m u  p v m h m m h m u mu in in in in in in 2 1 in 2 2 1 1 which can be rearranged to give m h u mh u . 2 in 2 1 in 1 This equation has two unknowns, m and u , and, in general, has to be solved iteratively, i.e. by trial 2 2 R R and error. In the special case of a perfect gas, where pv RT, with c  and c  , it is v 1 p 1 possible to get the following solution. Solutions Chapter 1 Page 7  D E Winterbone Current edition: 13/02/2015 Chapter 1 Solutions Using the equation m h u mh u  2 in 2 1 in 1 and substituting the following values for the initial and final conditions RT pV u c T  1 , m  1 1 , 1 v 1 1 1 RT 1 RT pV u c T  2 , m  2 2 2 v 2 1 2 RT 2 and the value for the inflowing enthalpy RT h  a in 1 gives pV RT RT  pV RT RT  a 1  a  2  1 1  a  1 . RT 1 1 RT 1 1 1 1 This equation can be rearranged to give T T  a 2 p T  1  a 11 pa  T1  If p = 0, then the result reduces to that for filling an evacuated bottle, viz., T = T . Fig P1.11(b) 1 2 a shows how the value of T varies with the ratios p /p and T /T . 2 1 a a 1 1.4 a T 1.2 /2 T ,o i tar 1 e r T1/Ta=1.0 u ta re T1/Ta=0.8 p m 0.8 T1/Ta=1.2 e T 0.6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Pressure ratio, p1/pa Fig P1.11(b): Variation of temperature ratio, T /T , with pressure ratio, p /p , for different levels of 2 a 1 a temperature ratio, T /T . 1 a In the case of gases which are not perfect it is not possible to derive a simple equation for the variation of the final temperature with initial conditions, and an iteration must be performed to evaluate it. _________________________________________________________________________________ P1.12 Discharge from a bottle The two questions above have dealt with filling a bottle, and shown how this can be considered to be a closed system problem. A similar approach can be applied to a bottle discharging to the surroundings. The system is shown in Fig P1.12. Solutions Chapter 1 Page 8  D E Winterbone Current edition: 13/02/2015 Chapter 1 Solutions Fig P1.12: Bottle discharging to atmosphere. The initial mass in the bottle (system a) is m with a specific internal energy of u , while the final mass 1 1 in the bottle is m with a specific internal energy of u . The mass flow out of the cylinder, into system 2 2 b, is m , and the mean specific internal energy of the fluid passing from system a to system b is u . exh exh If the systems are insulated, then Q = 0. During the discharge process work is done by the gas on the surroundings as the fluid "inflates" system b. This work is Wm p v . exh exh exh Applying the First Law to the process QWdEU U , 2 1 which gives on substitution 0m p v mu m u mu, exh exh exh 2 2 exh exh 1 1 which can be rearranged to give m m p v u m u mu m m h . 1 2 exh exh exh 2 2 1 1 1 2 exh Hence m h u mh u , 2 exh 2 1 exh 1 h u  giving m m exh 1 . 2 1h u  exh 2 The final mass has to be evaluated by trial and error in most cases because the value of h is an exh average value during the exhausting process. __________________________________________________________________________________________ Solutions Chapter 1 Page 9  D E Winterbone Current edition: 13/02/2015 Solutions Manual Chapter 2 Pz.l A mass of 10 kg of water at 0"C is brought into contact with a large heat reservoir at 100"C. (a) When the water has reached l00oC what has been: (r) the change of entropy of the water; (ii) the change of antropy of the reservoir; (iii) the change of entropy of the universe? ( 13. 1 lkJ/K; -lt.26kJ lK:' 1.85kJ/K) O) If the water had been heated from OoC to 100oC by first bringing it into contact with a reservoir at 30oC and then a reservoir of 100"C, what would have been the change in the entropy of the universe? (1.070kJ/K) (c) Explain how the water could have been heated to give no change in the entropy of the universe' flnfinite number of reservoirs - reversible heat transfer) P2.2, A system contains a fluid at a temperature of 70"C and I bar. It undergoes a reversible process during which the temp€rature of the system remains constant. Given that the heat transfer to the fluid during the process is 100 kJ, evaluate: (a) the increase in entropy. (b) if the system has a mass of 2.31kg. Evaluate the increase in specific enfopy of the system. (c) If a second fluid system, identical to the first one urdergoes an irreversible isothermal process from the same initial state to the same final state as above; and the heat transfer to the fluid in this irreversible process is 180 kJ; evaluate the increase in entropy ofthe fluid. (0.29 t SkJ/ K; 0.1262kJ lk gK; 0.29 I skJ/kgK) P2.3 Calculate the gain in entropy when 1 kg of water at 30"C is converted into steam at 150"C and then superheated to 300"C, with the process taking place at constant pressue. Take cp(wate )= 4.2 kJ/kg K cp(steam) :2.1 kJ/kg K hg:2600 - 1.5 t, where /: temperature in oC. (7.6533kJ/K) P2.4 A mass of a liquid, m, at temperature, 21, is mixed with an equal mass of the same liquid at tempereture, 72. The system is therrnally insulated. Show that the change of entropy of the Universe is: zm"c ,t(n.( <r'*rlz) 'lr'r' ) and prove that this is necessarily positive. [Hint Do not specifically consider the mixing of the fluids.]. P2.5 A substance has the following physical properties at a certain pressure: :76oC; Saturation temperature, t, 4 = 61.lkJ/kg cp(iquicr) = 0.2|kJlkgKi cp1,ryo-) = 0.14+0.000362 kJ/kgK, where Z==temperature in K Determine the enthalpy and entropy of superheated vapour at 150"C assuming these properties are zero for the fluid in liquid state at -40oC. ( I 06. I 0kJ/kg; 0.3 I 3a8kl&gK)