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SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 5 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and van Wylen CONTENT SUBSECTION Correspondence table Concept-Study Guide Problems Kinetic and potential energy Properties (u,h) from general tables Energy equation: simple process Energy eqaution: multistep process Energy equation: solids and liquids Properties (u, h, Cv, Cp), ideal gas Energy equation: ideal gas Energy equation: polytropic process Energy equation in rate form Review Problems PROB NO. 1-19 20-27 28-34 35-60 61-73 74-81 82-88 89-102 103-115 116-125 126-138 Sonntag, Borgnakke and van Wylen CHAPTER 5 CORRESPONDENCE TABLE The correspondence between this problem set and 5th edition chapter 5 problem set. Study guide problems 5.1-5.19 are all new New 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 5th 1 4 2mod 3 new 5 new new 6 mod new 7 mod new 8 mod 9 mod new 10 mod new 12 14 11 new 13 15 21 new new new 26 41 new New 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 5th 28 new 17 new 27 51 53 40 37 44 42 new 38 39 20 23 mod 43 24 45 new new 49 mod 55 36 new 58 60 new 59 61 New 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 5th new new new new new 67 mod new 68 mod 62 72 mod 63 new new 79 new 64 new 65 new new new 69 new new 74 76 new 66 new 46 New 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 5th new 84 77 30 54 82 new 89 87 new 90 new 86 new new new 22 29 57 35 31 32 48 56 18 new 83 new 85 Sonntag, Borgnakke and van Wylen The english unit problem set corresponds to the 5th edition as New 139 140 141 142 143 144 145 146 147 148 149 150 5th new new new new new new new 102 103 104 mod 105 mod 104 mod New 151 152 153 154 155 156 157 158 159 160 161 162 5th 107 108 106 new 112 115 111 110 109 113 114 118 New 163 164 165 166 167 168 169 170 171 172 173 174 5th 124 119 new 120 new 122 121 new 125 130 129 123 New 175 176 177 178 179 180 181 182 5th 127 new 131 132 135 new 136 134 Sonntag, Borgnakke and van Wylen Concept-Study Guide Problems 5.1 What is 1 cal in SI units and what is the name given to 1 N-m? Look in the conversion factor table A.1 under energy: 1 cal (Int.) =4.1868 J =4.1868 Nm =4.1868 kg m2/s2 This was historically defined as the heat transfer needed to bring 1 g of liquid water from 14.5oC to 15.5oC, notice the value of the heat capacity of water in Table A.4 1 N-m =1 J or Force times displacement =energy =Joule 5.2 In a complete cycle what is the net change in energy and in volume? For a complete cycle the substance has no change in energy and therefore no storage, so the net change in energy is zero. For a complete cycle the substance returns to its beginning state, so it has no change in specific volume and therefore no change in total volume. 5.3 Why do we write E or E2 E1 whereas we write 1Q2 and 1W2? E or E2 E1 is the change from state 1 to state 2 and depends only on states 1 and 2 not upon the process between 1 and 2. 1Q2 and 1W2 are amounts of energy transferred during the process between 1 and 2 and depend on the process path. Sonntag, Borgnakke and van Wylen 5.4 When you wind a spring up in a toy or stretch a rubber band what happens in terms of work, energy and heat transfer? Later when they are released, what happens then? In both processes work is put into the device and the energy is stored as potential energy. If the spring or rubber is inelastic some of the work input goes into internal energy (it becomes warmer) and not its potential energy and being warmer than the ambient air it cools slowly to ambient temperature. When the spring or rubber band is released the potential energy is transferred back into work given to the system connected to the end of the spring or rubber band. If nothing is connected the energy goes into kinetic energy and the motion is then dampened as the energy is transformed into internal energy. 5.5 Explain in words what happens with the energy terms for the stone in Example 5.2. What would happen if it were a bouncing ball falling to a hard surface? In the beginning all the energy is potential energy associated with the gravitational force. As the stone falls the potential energy is turned into kinetic energy and in the impact the kinetic energy is turned into internal energy of the stone and the water. Finally the higher temperature of the stone and water causes a heat transfer to the ambient until ambient temperature is reached. With a hard ball instead of the stone the impact would be close to elastic transforming the kinetic energy into potential energy (the material acts as a spring) that is then turned into kinetic energy again as the ball bounces back up. Then the ball rises up transforming the kinetic energy into potential energy (mgZ) until zero velocity is reached and it starts to fall down again. The collision with the floor is not perfectly elastic so the ball does not rise exactly up to the original height loosing a little energy into internal energy (higher temperature due to internal friction) with every bounce and finally the motion will die out. All the energy eventually is lost by heat transfer to the ambient or sits in lasting deformation (internal energy) of the substance. Sonntag, Borgnakke and van Wylen 5.6 Make a list of at least 5 systems that store energy, explaining which form of energy. A spring that is compressed. Potential energy (1/2)kx2 A battery that is charged. Electrical potential energy. V Amp h A raised mass (could be water pumped up higher) Potential energy mgH A cylinder with compressed air. Potential (internal) energy like a spring. A tank with hot water. Internal energy mu A fly-wheel. Kinetic energy (rotation) (1/2)I2 A mass in motion. Kinetic energy (1/2)mV2 5.7 A 1200 kg car is accelerated from 30 to 50 km/h in 5 s. How much work is that? If you continue from 50 to 70 km/h in 5 s is that the same? The work input is the increase in kinetic energy. E2 E1 =(1/2)m[V2 -V1] =1W2 km2 =0.5 1200 kg [502 302] h 1000 m2 =600 [ 2500 900 ] kg 3600 s =74 074 J =74.1 kJ The second set of conditions does not become the same 2 2 1000 m2 E2 E1 =(1/2)m[V2 -V1] =600 [ 702 502 ] kg 3600 s =111 kJ 2 2 Sonntag, Borgnakke and van Wylen 5.8 A crane use 2 kW to raise a 100 kg box 20 m. How much time does it take? L Power =W =FV =mgV =mg t mgL 100 kg 9.807 m/s2 20 m =9.81 s t= 2000 W W 5.9 Saturated water vapor has a maximum for u and h at around 235oC. Is it similar for other substances? Look at the various substances listed in appendix B. Everyone has a maximum u and h somewhere along the saturated vapor line at different T for each substance. This means the constant u and h curves are different from the constant T curves and some of them cross over the saturated vapor line twice, see sketch below. P C.P. T u=C T v C.P. P=C u=C v Constant h lines are similar to the constant u line shown. Notice the constant u(h) line becomes parallel to the constant T lines in the superheated vapor region for low P where it is an ideal gas. In the T-v diagram the constant u (h) line becomes horizontal. Sonntag, Borgnakke and van Wylen 5.10 A pot of water is boiling on a stove supplying 325 W to the water. What is the rate of mass (kg/s) vaporizing assuming a constant pressure process? To answer this we must assume all the power goes into the water and that the process takes place at atmospheric pressure 101 kPa, so T =100oC. Energy equation dQ =dE +dW =dU +PdV =dH =hfg dm dQ dm =hfg dt dt .325 W dm Q =h =2257 kJ/kg =0.144 g/s dt fg The volume rate of increase is dV dm 3 dt =dt vfg =0.144 g/s 1.67185 m /kg =0.24 10-3 m3/s =0.24 L/s 5.11 A constant mass goes through a process where 100 W of heat transfer comes in and 100 W of work leaves. Does the mass change state? Yes it does. As work leaves a control mass its volume must go up, v increases As heat transfer comes in at a rate equal to the work out means u is constant if there are no changes in kinetic or potential energy. Sonntag, Borgnakke and van Wylen 5.12 I have 2 kg of liquid water at 20oC, 100 kPa. I now add 20 kJ of energy at a constant pressure. How hot does it get if it is heated? How fast does it move if it is pushed by a constant horizontal force? How high does it go if it is raised straight up? a) Heat at 100 kPa. Energy equation: E2 E1 =1Q2 1W2 =1Q2 P(V2 V1) =H2 H1= m(h2 h1) h2 =h1 +1Q2/m =83.94 +20/2 =94.04 kJ/kg Back interpolate in Table B.1.1: T2 =22.5oC (We could also have used T =1Q2/mC =20 / (2*4.18) =2.4oC) b) Push at constant P. It gains kinetic energy. 0.5 m V2 =1W2 V2 =2 1W2/m =2 20 1000 J/2 kg =141.4 m/s c) Raised in gravitational field m g Z2 =1W2 Z2 =1W2/m g =20 000 J =1019 m 2 kg 9.807 m/s2 2 Sonntag, Borgnakke and van Wylen 5.13 Water is heated from 100 kPa, 20oC to 1000 kPa, 200oC. In one case pressure is raised at T =C, then T is raised at P =C. In a second case the opposite order is done. Does that make a difference for 1Q2 and 1W2? Yes it does. Both 1Q2 and 1W2 are process dependent. We can illustrate the work term in a P-v diagram. P Cr.P. S 1000 100 L a 1 20 200 2 V T P T C.P. 1553 kPa 1000 a 1000 100 1 2 20 C 200 C 200 20 a 1 2 b v 100 b 180 C v In one case the process proceeds from 1 to state "a" along constant T then from "a" to state 2 along constant P. The other case proceeds from 1 to state "b" along constant P and then from "b" to state 2 along constant T. Sonntag, Borgnakke and van Wylen 5.14 Two kg water at 120oC with a quality of 25% has its temperature raised 20oC in a constant volume process. What are the new quality and specific internal energy? Solution: State 1 from Table B.1.1 at 120oC v =vf +x vfg =0.001060 +0.25 0.8908 =0.22376 m3/kg State 2 has same v at 140oC also from Table B.1.1 v -vf 0.22376 -0.00108 x= v =0.4385 0.50777 fg u =uf +x ufg =588.72 +0.4385 1961.3 =1448.8 kJ/kg P C.P. 140 C 120 C T C.P. 361.3 198.5 T v 140 120 v Sonntag, Borgnakke and van Wylen 5.15 Two kg water at 200 kPa with a quality of 25% has its temperature raised 20oC in a constant pressure process. What is the change in enthalpy? Solution: State 1 from Table B.1.2 at 200 kPa h =hf +x hfg =504.68 +0.25 2201.96 =1055.2 kJ/kg State 2 has same P from Table B.1.2 at 200 kPa T =T +20 =120.23 +20 =140.23oC 2 sat so state 2 is superheated vapor (x =undefined) from Table B.1.3 20 h2 =2706.63 +(2768.8 2706.63)150 -120.23 =2748.4 kJ/kg h2 h1 =2748.4 1055.2 =1693.2 kJ/kg P C.P. 140 C T C.P. 200 kPa 200 120.2 C T v 140 120 v 5.16 You heat a gas 10 K at P =C. Which one in table A.5 requires most energy? Why? A constant pressure process in a control mass gives (recall Eq.5.29) 1q2 =u2 -u1 +1w2 =h2 -h1 Cp T The one with the highest specific heat is hydrogen, H2. The hydrogen has the smallest mass but the same kinetic energy per mol as other molecules and thus the most energy per unit mass is needed to increase the temperature. Sonntag, Borgnakke and van Wylen 5.17 Air is heated from 300 to 350 K at V =C. Find 1q2? What if from 1300 to 1350 K? Process: V =C Energy Eq.:1W2 =1q2 =u2 -u1 u2 -u1 =1q2 0 Read the u-values from Table A.7.1 a) 1q2 =u2 -u1 =250.32 214.36 =36.0 kJ/kg b) 1q2 =u2 -u1 =1067.94 1022.75 =45.2 kJ/kg case a) Cv 36/50 =0.72 kJ/kg K ,see A.5 case b) Cv 45.2/50 =0.904 kJ/kg K (25 %higher) 5.18 A mass of 3 kg nitrogen gas at 2000 K, V =C, cools with 500 W. What is dT/dt? Process: V=C 1W2= 0 .dE dU dU dT .dt =m dt =mCv dt =Q W =Q =500 W dt du u u2100 -u1900 1819.08 -1621.66 Cv 2000 =dT =0.987 kJ/kg K 200 T 2100-1900 .dT Q -500 W K =mC =0.17 dt s v 3 0.987 kJ/K Remark: Specific heat from Table A.5 has Cv 300 =0.745 kJ/kg K which is nearly 25% lower and thus would over-estimate the rate with 25%.Sonntag, Borgnakke and van Wylen 5.19 A drag force on a car, with frontal area A =2 m2, driving at 80 km/h in air at 20oC is Fd =0.225 A airV2. How much power is needed and what is the traction force? W =FV km 1000 V =80 h =80 3600 ms-1 =22.22 ms-1 P 101 AIR =RT =1.20 kg/m3 0.287 293 Fd =0.225 AV2 =0.225 2 1.2 22.222 =266.61 N .W =FV =266.61 N 22.22 m/s =5924 W =5.92 kW Sonntag, Borgnakke and van Wylen Kinetic and Potential Energy 5.20 A hydraulic hoist raises a 1750 kg car 1.8 m in an auto repair shop. The hydraulic pump has a constant pressure of 800 kPa on its piston. What is the increase in potential energy of the car and how much volume should the pump displace to deliver that amount of work? Solution: C.V. Car. No change in kinetic or internal energy of the car, neglect hoist mass. E2 E1 =PE2 -PE1 =mg (Z2 Z1) =1750 9.80665 1.8 =30 891 J The increase in potential energy is work into car from pump at constant P. W =E2 E1 =P dV =P V V =E2 E1 30891 =800 1000 =0.0386 m3 P Sonntag, Borgnakke and van Wylen 5.21 A piston motion moves a 25 kg hammerhead vertically down 1 m from rest to a velocity of 50 m/s in a stamping machine. What is the change in total energy of the hammerhead? Solution: C.V. Hammerhead The hammerhead does not change internal energy (i.e. same P, T), but it does have a change in kinetic and potential energy. E2 E1 =m(u2 u1) +m[(1/2)V2 2 0] +mg (h2 -0) =0 +25 (1/2) 502 +25 9.80665 (-1) =31250 245.17 =31005 J =31 kJ Sonntag, Borgnakke and van Wylen 5.22 Airplane takeoff from an aircraft carrier is assisted by a steam driven piston/cylinder device with an average pressure of 1250 kPa. A 17500 kg airplane should be accelerated from zero to a speed of 30 m/s with 30% of the energy coming from the steam piston. Find the needed piston displacement volume. Solution: C.V. Airplane. No change in internal or potential energy; only kinetic energy is changed. E2 E1 =m (1/2) (V2 -0) =17500 (1/2) 302 =7875 000 J =7875 kJ The work supplied by the piston is 30% of the energy increase. 2 W =P dV =Pavg V =0.30 (E2 E1) =0.30 7875 =2362.5 kJ W 2362.5 V =P =1250 =1.89 m3 avg Sonntag, Borgnakke and van Wylen 5.23 Solve Problem 5.22, but assume the steam pressure in the cylinder starts at 1000 kPa, dropping linearly with volume to reach 100 kPa at the end of the process. Solution: C.V. Airplane. P E2 E1 =m (1/2) (V22 -0) =3500 (1/2) 302 =1575000 J =1575 kJ W =0.25(E2 E1) =0.25 1575 =393.75 kJ W =P dV =(1/2)(Pbeg +Pend) V W 2362.5 V =P =1/2(1000 +100) =4.29 m3 avg 1000 1 2 V 100 W Sonntag, Borgnakke and van Wylen 5.24 A 1200 kg car accelerates from zero to 100 km/h over a distance of 400 m. The road at the end of the 400 m is at 10 m higher elevation. What is the total increase in the car kinetic and potential energy? Solution: KE =m (V2 -V1) V2 =100 km/h =27.78 m/s 100 1000 m/s 3600 2 2 KE =1200 kg (27.782 02) (m/s)2 =463 037 J =463 kJ PE =mg(Z2 Z1) =1200 kg 9.807 m/s2 ( 10 -0 ) m =117684 J =117.7 kJ Sonntag, Borgnakke and van Wylen 5.25 A 25 kg piston is above a gas in a long vertical cylinder. Now the piston is released from rest and accelerates up in the cylinder reaching the end 5 m higher at a velocity of 25 m/s. The gas pressure drops during the process so the average is 600 kPa with an outside atmosphere at 100 kPa. Neglect the change in gas kinetic and potential energy, and find the needed change in the gas volume. Solution: C.V. Piston (E2 E1)PIST. m(u2 u1) +m[(1/2)V2 2 0] +mg (h2 0) =0 +25 (1/2) 252 +25 9.80665 5 =7812.5 +1225.8 =9038.3 J =9.038 kJ Energy equation for the piston is: E2 E1 =Wgas -Watm =Pavg Vgas Po Vgas (remark Vatm =Vgas so the two work terms are of opposite sign) Vgas =9.038/(600 100) =0.018 m3 V Po g P H Pavg 1 2 V Sonntag, Borgnakke and van Wylen 5.26 The rolling resistance of a car depends on its weight as: F =0.006 mg. How far will a car of 1200 kg roll if the gear is put in neutral when it drives at 90 km/h on a level road without air resistance? Solution: The car decreases its kinetic energy to zero due to the force (constant) acting over the distance. m (1/2V2 1/2V1) =1W2 =F dx =FL V2 =0, 2 2 2 km 90 1000 V1 =90 h =3600 ms-1 =25 ms-1 -1/2 mV1 =FL =0.006 mgL 0.5 V1 0.5 252 m2/s2 =5311 m L =0.0006g =0.006 9.807 m/s2 Remark: Over 5 km! The air resistance is much higher than the rolling resistance so this is not a realistic number by itself. 2 Sonntag, Borgnakke and van Wylen 5.27 A mass of 5 kg is tied to an elastic cord, 5 m long, and dropped from a tall bridge. Assume the cord, once straight, acts as a spring with k =100 N/m. Find the velocity of the mass when the cord is straight (5 m down). At what level does the mass come to rest after bouncing up and down? Solution: Let us assume we can neglect the cord mass and motion. 1: V1 =0, 3: V3 =0, 1 2: Z1= 0 2 :V2, Z2= 5m Z3= L ,Fup =mg =ks L 2 2 mV1 +mg Z1 =V2 +mgZ2 Divide by mass and left hand side is zero so V2 +g Z2 =0 V2 =(-2g Z2)1/2 =( -2 9.807 (-5)) 1/2 =9.9 m/s State 3: m is at rest so Fup =Fdown 2 ks L =mg mg 5 9.807 kg ms-2 L =k =100 =0.49 m s Nm-1 L =Lo +L =5 +0.49 =5.49 m So: Z2 =L =5.49 m BRIDGE m V Sonntag, Borgnakke and van Wylen Properties (u, h) from General Tables 5.28 Find the missing properties. a. H2O T =250 C, v =0.02 m3/kg b. c. d. N2 H2O R-134a Solution: T =120 K, P =0.8 MPa T =2 C, P =100 kPa P =200 kPa, v =0.12 m3/kg vf v vg P=?u=?x=?h=?u=?v=?u=?T=?P =Psat =3973 kPa a) Table B.1.1 at 250 C: x =(v -vf)/ vfg =(0.02 0.001251)/0.04887 =0.38365 u =uf +x ufg =1080.37 +0.38365 1522.0 =1664.28 kJ/kg b) Table B.6.1 Table B.6.2: P is lower than Psat so it is super heated vapor and we find the state in Table B.6.2 h =114.02 kJ/kg =x =undefined c) Table B.1.1 :T Ttriple point =B.1.5: P Psat so compressed solid u ui =337.62 kJ/kg v vi =1.09 10-3 m3/kg approximate compressed solid with saturated solid properties at same T. d) Table B.5.1 v vg superheated vapor =Table B.5.2. T 32.5 C =30 +(40 30) (0.12 0.11889)/(0.12335 -0.11889) u =403.1 +(411.04 403.1) 0.24888 =405.07 kJ/kg P L S T c v a P C.P. b d a c T v c a T C.P. b P=C d C.P. V d b v Sonntag, Borgnakke and van Wylen 5.29 Find the missing properties of T, P, v, u, h and x if applicable and plot the location of the three states as points in the T-v and the P-v diagrams a. Water at 5000 kPa, u =800 kJ/kg b. Water at 5000 kPa, v =0.06 m3/kg c. R-134a at 35oC, v =0.01 m3/kg Solution: a) Look in Table B.1.2 at 5000 kPa u uf =1147.78 Table B.1.4: compressed liquid between 180 oC and 200 oC 800 -759.62 T =180 +(200 -180) 848.08 -759.62 =180 +20*0.4567 =189.1 C v =0.001124 +0.4567 (0.001153 -0.001124) =0.001137 b) Look in Table B.1.2 at 5000 kPa v vg =0.03944 =superheated vapor Table B.1.3: between 400 oC and 450 oC. T =400 +50*(0.06 -0.05781)/(0.0633 -0.05781) =400 +50*0.3989 =419.95 oC h =3195.64 +0.3989 *(3316.15 -3195.64) =3243.71 c) B.5.1: v f v vg =2-phase, P =Psat =887.6 kPa, x =(v -vf ) / vfg =(0.01 -0.000857)/0.02224 =0.4111 u =uf +x ufg =248.34 +0.4111*148.68 =309.46 kJ/kg P C.P. T C.P. P =const. a T c v b a c v b States shown are placed relative to the two-phase region, not to each other. Sonntag, Borgnakke and van Wylen 5.30 Find the missing properties and give the phase of the ammonia, NH3. a. T =65oC, P =600 kPa b. T =20oC, P =100 kPa c. T =50oC, v =0.1185 m3/kg Solution: a) Table B.2.1 P Psat u=?v=?u=?v=?x=?u=?P=?x=?superheated vapor Table B.2.2: v =0.5 0.25981 +0.5 0.26888 =0.2645 m3/kg u =0.5 1425.7 +0.5 1444.3 =1435 kJ/kg b) Table B.2.1: P Psat =x =undefined, superheated vapor, from B.2.2: v =1.4153 m3/kg ;u =1374.5 kJ/kg c) Sup. vap. ( v vg) Table B.2.2. P =1200 kPa, x =undefined u =1383 kJ/kg P C.P. T c C.P. c T b v States shown are placed relative to the two-phase region, not to each other. 1200 kPa 600 kPa a b v a Sonntag, Borgnakke and van Wylen 5.31 Find the phase and missing properties of P, T, v, u, and x. a. Water at 5000 kPa, u =1000 kJ/kg (Table B.1 reference) b. R-134a at 20oC, u =300 kJ/kg c. Nitrogen at 250 K, 200 kPa Show also the three states as labeled dots in a T-v diagram with correct position relative to the two-phase region. Solution: a) Compressed liquid: B.1.4 interpolate between 220oC and 240oC. T =233.3oC, v =0.001213 m3/kg, x =undefined b) Table B.5.1: u ug =two-phase liquid and vapor x =(u -uf)/ufg =(300 -227.03)/162.16 =0.449988 =0.45 v =0.000817 +0.45*0.03524 =0.01667 m3/kg c) Table B.6.1: T Tsat (200 kPa) so superheated vapor in Table B.6.2 x =undefined v =0.5(0.35546 +0.38535) =0.3704 m3/kg, u =0.5(177.23 +192.14) =184.7 kJ/kg P C.P. T a b c T v b C.P. P =const. c States shown are placed relative to the two-phase region, not to each other. a v Sonntag, Borgnakke and van Wylen 5.32 Find the missing properties and give the phase of the substance a. b. c. d. e. H2O H2O N2 NH3 N2 T =120 C, v =0.5 m3/kg T =100 C, P =10 MPa T =200 K, P =200 kPa T =100 C, v =0.1 m3/kg T =100 K, x =0.75 u=?P=?x=?u=?x=?v=?v=?u=?P=?x=?v=?u=?Solution: a) Table B.1.1: vf v vg =L+V mixture, P =198.5 kPa, x =(0.5 -0.00106)/0.8908 =0.56, u =503.48 +0.56 2025.76 =1637.9 kJ/kg b) Table B.1.4: compressed liquid, v =0.001039 m3/kg, u =416.1 kJ/kg c) Table B.6.2: 200 K, 200 kPa u =147.37 kJ/kg v =0.29551 m3/kg ;d) Table B.2.1: v vg =superheated vapor, x =undefined 0.1 -0.10539 B.2.2: P =1600 +400 0.08248-0.10539 =1694 kPa e) Table B.6.1: 100 K, x =0.75 v =0.001452 +0.75 0.02975 =0.023765 m3/kg u =74.33 +0.75 137.5 =28.8 kJ/kg P C.P. T c a e v d T b C.P. a e v c States shown are placed relative to the two-phase region, not to each other. b P =const. d Sonntag, Borgnakke and van Wylen 5.33 Find the missing properties among (T, P, v, u, h and x if applicable) and give the phase of the substance and indicate the states relative to the two-phase region in both a T-v and a P-v diagram. a. R-12 P =500 kPa, h =230 kJ/kg b. c. Solution: a) Table B.3.2: h hg =superheated vapor, look in section 500 kPa and interpolate T =68.06 C, v =0.04387 m3/kg, u =208.07 kJ/kg R-22 R-134a T =10oC, u =200 kJ/kg T =40oC, h =400 kJ/kg b) Table B.4.1: u ug =L+V mixture, P =680.7 kPa u -uf 200 -55.92 x =u =173.87 =0.8287, fg v =0.0008 +0.8287 0.03391 =0.0289 m3/kg, h =56.46 +0.8287 196.96 =219.7 kJ/kg c) Table B.5.1: h hg =two-phase L +V, look in B.5.1 at 40 C: h -hf 400 -256.5 x =h =163.3 =0.87875 fg P =Psat =1017 kPa, v =0.000 873 +0.87875 0.01915 =0.0177 m3/kg u =255.7 +0.87875 143.8 =382.1 kJ/kg P C.P. T C.P. P=C a b, c T v b, c a States shown are placed relative to the two-phase region, not to each other. v Sonntag, Borgnakke and van Wylen 5.34 Saturated liquid water at 20oC is compressed to a higher pressure with constant temperature. Find the changes in u and h from the initial state when the final pressure is a) 500 kPa, b) 2000 kPa, c) 20 000 kPa Solution: State 1 is located in Table B.1.1 and the states a-c are from Table B.1.4 State 1 a b c u [kJ/kg] 83.94 83.91 83.82 82.75 h [kJ/kg] 83.94 84.41 85.82 102.61 u =u -u1 -0.03 -0.12 -1.19 h =h -h1 0.47 1.88 18.67 (Pv) 0.5 2 20 For these states u stays nearly constant, dropping slightly as P goes up. h varies with Pv changes. P c b a 1 T =20 C v v o T c,b,a,1 P L T cb C.P. V c b a S 1 v Sonntag, Borgnakke and van Wylen Energy Equation: Simple Process 5.35 A 100-L rigid tank contains nitrogen (N2) at 900 K, 3 MPa. The tank is now cooled to 100 K. What are the work and heat transfer for this process? Solution: C.V.:Nitrogen in tank. Energy Eq.5.11: m2 =m1 ;/ 1W2 =0 m(u2 -u1) =1Q2 -1W2 Process: V =constant, v2 =v1 =V/m Table B.6.2: State 1: v1 =0.0900 m3/kg =m =V/v1 =1.111 kg u1 =691.7 kJ/kg State 2: 100 K, v2 =v1 =V/m, look in Table B.6.2 at 100 K 200 kPa: v =0.1425 m3/kg; u =71.7 kJ/kg 400 kPa: v =0.0681 m3/kg; u =69.3 kJ/kg so a linear interpolation gives: P2 =200 +200 (0.09 0.1425)/(0.0681 0.1425) =341 kPa 0.09 0.1425 u2 =71.7 +(69.3 71.7) 0.0681 0.1425 =70.0 kJ/kg, 1Q2 =m(u2 -u1) =1.111 (70.0 691.7) =690.7 kJ Sonntag, Borgnakke and van Wylen 5.36 A rigid container has 0.75 kg water at 300oC, 1200 kPa. The water is now cooled to a final pressure of 300 kPa. Find the final temperature, the work and the heat transfer in the process. Solution: C.V. Water. Constant mass so this is a control mass Energy Eq.:Process eq.:o U2 -U1 =1Q2 -1W2 V =constant. (rigid) 1W2 =P dV =0 P 1200 1 State 1: 300 C, 1200 kPa =superheated vapor Table B.1.3 v =0.21382 m3/kg, u =2789.22 kJ/kg 300 2 v State 2: 300 kPa and v2 =v1 T2 =Tsat =133.55oC from Table B.1.2 v2 vg two-phase v2 -vf 0.21382 -0.001073 =0.35179 x2 =v 0.60475 fg u2 =uf +x2 ufg =561.13 +x2 1982.43 =1258.5 kJ/kg 1Q2 =m(u2 -u1) +1W2 =m(u2 -u1) =0.75 (1258.5 -2789.22) =1148 kJ Sonntag, Borgnakke and van Wylen 5.37 A cylinder fitted with a frictionless piston contains 2 kg of superheated refrigerant R134a vapor at 350 kPa, 100oC. The cylinder is now cooled so the R-134a remains at constant pressure until it reaches a quality of 75%.Calculate the heat transfer in the process. Solution: C.V.:R-134a Energy Eq.5.11 m2 =m1 =m; m(u2 -u1) =1Q2 -1W2 Process: P =const. 1W2 =PdV =PV =P(V2 -V1) =Pm(v2 -v1) P T 1 2 V State 1: Table B.5.2 State 2: Table B.5.1 V 2 1 h1 =(490.48 +489.52)/2 =490 kJ/kg h2 =206.75 +0.75 194.57 =352.7 kJ/kg (350.9 kPa) 1Q2 =m(u2 -u1) +1W2 =m(u2 -u1) +Pm(v2 -v1) =m(h2 -h1) 1Q2 =2 (352.7 -490) =274.6 kJ Sonntag, Borgnakke and van Wylen 5.38 Ammonia at 0 C, quality 60% is contained in a rigid 200-L tank. The tank and ammonia is now heated to a final pressure of 1 MPa. Determine the heat transfer for the process. Solution: C.V.:NH3 P 2 1 V Continuity Eq.:Energy Eq.5.11: m2 =m1 =m ;m(u2 -u1) =1Q2 -1W2 v2 =v1 & 1W2 =0 Process: Constant volume State 1: Table B.2.1 two-phase state. v1 =0.001566 +x1 0.28783 =0.17426 m3/kg u1 =179.69 +0.6 1138.3 =862.67 kJ/kg m =V/v1 =0.2/0.17426 =1.148 kg State 2: P2 ,v2 =v1 superheated vapor Table B.2.2 T2 100 C, u2 1490.5 kJ/kg So solve for heat transfer in the energy equation 1Q2 =m(u2 -u1) =1.148(1490.5 -862.67) =720.75 kJ Sonntag, Borgnakke and van Wylen 5.39 Water in a 150-L closed, rigid tank is at 100 C, 90% quality. The tank is then cooled to -10 C. Calculate the heat transfer during the process. Solution: C.V.:Water in tank. Energy Eq.5.11: m2 =m1 ;m(u2 -u1) =1Q2 -1W2 Process: V =constant, v2 =v1, 1W2 =0 State 1: Two-phase L +V look in Table B.1.1 v1 =0.001044 +0.9 1.6719 =1.5057 m3/kg u1 =418.94 +0.9 2087.6 =2297.8 kJ/kg State 2: T2, v2 =v1 mix of saturated solid +vapor Table B.1.5 =x2 =0.003224 v2 =1.5057 =0.0010891 +x2 466.7 m =V/v1 =0.15/1.5057 =0.09962 kg 1Q2 =m(u2 -u1) =0.09962(-345.34 -2297.8) =263.3 kJ u2 =354.09 +0.003224 2715.5 =345.34 kJ/kg P C.P. T C.P. P =const. 1 1 T 2 v 2 v P L T S 2 1 C.P. V L+V S+V v Sonntag, Borgnakke and van Wylen 5.40 A piston/cylinder contains 1 kg water at 20oC with volume 0.1 m3. By mistake someone locks the piston preventing it from moving while we heat the water to saturated vapor. Find the final temperature and the amount of heat transfer in the process. Solution: C.V. Water. This is a control mass Energy Eq.:m (u2 -u1 ) =1Q2 -1W2 Process :State 1: 1W2 =0 T, v1 =V1/m =0.1 m3/kg vf so two-phase V =constant v1 -vf 0.1-0.001002 x1 =v =57.7887 =0.0017131 fg u1 =uf +x1 ufg =83.94 +x1 2318.98 =87.913 kJ/kg State 2: v2 =v1 =0.1 & x2 =1 found in Table B.1.1 between 210 C and 215 C 0.1-0.10441 T2 =210 +5 0.09479-0.10441 =210 +5 0.4584 =212.3 C u2 =2599.44 +0.4584 (2601.06 2599.44) =2600.2 kJ/kg From the energy equation 1Q2 =m(u2 -u1) =1( 2600.2 87.913) =2512.3 kJ P 2 1 V T 2 1 V Sonntag, Borgnakke and van Wylen 5.41 A test cylinder with constant volume of 0.1 L contains water at the critical point. It now cools down to room temperature of 20 C. Calculate the heat transfer from the water. Solution: C.V.:Water P m2 =m1 =m ;1 Energy Eq.5.11: m(u2 -u1) =1Q2 -1W2 2 v Process: Constant volume v2 =v1 Properties from Table B.1.1 State 1: v1 =vc =0.003155 m3/kg, u1 =2029.6 kJ/kg m =V/v1 =0.0317 kg State 2: T2, v2 =v1 =0.001002 +x2 57.79 x2 =3.7 10-5, u2 =83.95 +x2 2319 =84.04 kJ/kg Constant volume =/ 1W2 =0 1Q2 =m(u2 -u1) =0.0317(84.04 -2029.6) =61.7 kJ Sonntag, Borgnakke and van Wylen 5.42 A 10-L rigid tank contains R-22 at -10 C, 80% quality. A 10-A electric current (from a 6-V battery) is passed through a resistor inside the tank for 10 min, after which the R-22 temperature is 40 C. What was the heat transfer to or from the tank during this process? Solution: C.V. R-22 in tank. Control mass at constant V. Continuity Eq.:m2 =m1 =m ;Energy Eq.:Process: m(u2 -u1) =1Q2 -1W2 1 V Constant V v2 =v1 =no boundary work, but electrical work State 1 from table B.4.1 v1 =0.000759 +0.8 0.06458 =0.05242 m3/kg u1 =32.74 +0.8 190.25 =184.9 kJ/kg m =V/v =0.010/0.05242 =0.1908 kg State 2: Table B.4.2 at 40 C and v2 =v1 =0.05242 m3/kg =sup.vapor, so use linear interpolation to get P2 =500 +100 (0.05242 0.05636)/(0.04628 0.05636) =535 kPa, u2 =250.51 +0.35 (249.48 250.51) =250.2 kJ/kg 1W2 elec =power t =Amp volts t =P 2 10 6 10 60 =36 kJ 1000 1Q2 =m(u2 u1) +1W2 =0.1908 ( 250.2 184.9) 36 =23.5 kJ Sonntag, Borgnakke and van Wylen 5.43 A piston/cylinder contains 50 kg of water at 200 kPa with a volume of 0.1 m3. Stops in the cylinder are placed to restrict the enclosed volume to a maximum of 0.5 m3. The water is now heated until the piston reaches the stops. Find the necessary heat transfer. Solution: C.V. H2O m =constant Energy Eq.5.11: m(e2 e1) =m(u2 u1) =1Q2 -1W2 Process :P =constant (forces on piston constant) 1W2 =P dV =P1 (V2 V1) P 1 0.1 Properties from Table B.1.1 2 0.5 V State 1 :v1 =0.1/50 =0.002 m3/kg =2-phase as v1 vg v1 vf 0.002 0.001061 x= 0.001061 0.88467 vfg =h =504.68 +0.001061 2201.96 =507.02 kJ/kg State 2 :v2= 0.5/50 =0.01 m3/kg also 2-phase same P v2 vf 0.01 0.001061 =0.01010 x2 =v 0.88467 fg h2 =504.68 +0.01010 2201.96 =526.92 kJ/kg Find the heat transfer from the energy equation as 1Q2 =m(u2 u1) +1W2 =m(h2 h1) 1Q2 =50 kg (526.92 507.02) kJ/kg =995 kJ [ Notice that 1W2 =P1 (V2 V1) =200 (0.5 0.1) =80 kJ ] Sonntag, Borgnakke and van Wylen 5.44 A constant pressure piston/cylinder assembly contains 0.2 kg water as saturated vapor at 400 kPa. It is now cooled so the water occupies half the original volume. Find the heat transfer in the process. Solution: C.V. Water. This is a control mass. Energy Eq.5.11: m(u2 u1) =1Q2 1W2 Process: P =constant =1W2 =Pm(v2 v1) So solve for the heat transfer: 1Q2 =m(u2 -u1) +1W2 =m(u2 -u1) +Pm(v2 -v1) =m(h2 -h1) State 1: Table B.1.2 v1 =0.46246 m3/kg; h1 =2738.53 kJ/kg State 2: v2 =v1 / 2 =0.23123 =vf +x vfg from Table B.1.2 x2 =(v2 vf) / vfg =(0.23123 0.001084) / 0.46138 =0.4988 h2 =hf +x2 hfg =604.73 +0.4988 2133.81 =1669.07 kJ/kg 1Q2 =0.2 (1669.07 2738.53) =213.9 KJ Sonntag, Borgnakke and van Wylen 5.45 Two kg water at 120oC with a quality of 25% has its temperature raised 20oC in a constant volume process as in Fig. P5.45. What are the heat transfer and work in the process? Solution: C.V. Water. This is a control mass Energy Eq.:m (u2 -u1 ) =1Q2 -1W2 Process :V =constant 1W2 =P dV =0 State 1: T, x1 from Table B.1.1 v1 =vf +x1 vfg =0.00106 +0.25 0.8908 =0.22376 m3/kg u1 =uf +x1 ufg =503.48 +0.25 2025.76 =1009.92 kJ/kg State 2: T2, v2 =v1 vg2 =0.50885 m3/kg so two-phase v2 -vf2 0.22376 -0.00108 =0.43855 x2 =v 0.50777 fg2 u2 =uf2 +x2 ufg2 =588.72 +x2 1961.3 =1448.84 kJ/kg From the energy equation 1Q2 =m(u2 -u1) =2 ( 1448.84 1009.92 ) =877.8 kJ P C.P. 140 C 120 C T C.P. 361.3 198.5 T v 140 120 v Sonntag, Borgnakke and van Wylen 5.46 A 25 kg mass moves with 25 m/s. Now a brake system brings the mass to a complete stop with a constant deceleration over a period of 5 seconds. The brake energy is absorbed by 0.5 kg water initially at 20oC, 100 kPa. Assume the mass is at constant P and T. Find the energy the brake removes from the mass and the temperature increase of the water, assuming P =C. Solution: C.V. The mass in motion. E2 -E1= E =0.5 mV =0.5 25 25 /1000 =7.8125 kJ C.V. The mass of water. m(u2 -u1) H2O =E =7.8125 kJ Assume u2 =uf =u2 -u1 =7.8125 / 0.5 =15.63 T2 23.7oC, T =3.7oC u2 =u1 +15.63 =83.94 +15.63 =99.565 kJ/kg then from Table B.1.1: 2 2 We could have used u2 -u1 =CT with C from Table A.4: C =4.18 kJ/kg K giving T =15.63/4.18 =3.7oC. Sonntag, Borgnakke and van Wylen 5.47 An insulated cylinder fitted with a piston contains R-12 at 25 C with a quality of 90% and a volume of 45 L. The piston is allowed to move, and the R-12 expands until it exists as saturated vapor. During this process the R-12 does 7.0 kJ of work against the piston. Determine the final temperature, assuming the process is adiabatic. Solution: Take CV as the R-12. Continuity Eq.:m2 =m1 =m ;Energy Eq.5.11: State 1: (T, x) m(u2 -u1) =1Q2 -1W2 Tabel B.3.1 =v1 =0.000763 +0.9 0.02609 =0.024244 m3/kg m =V1/v1 =0.045/0.024244 =1.856 kg u1 =59.21 +0.9 121.03 =168.137 kJ/kg State 2: (x =1, ) We need one property information. Apply now the energy equation with known work and adiabatic so / 1Q2 =0 =m(u2 -u1) +1W2 =1.856 (u2 -168.137) +7.0 =u2 =164.365 kJ/kg =ug at T2 Table B.3.1 gives ug at different temperatures: T2 -15 C P T 1 1 2 v 2 v Sonntag, Borgnakke and van Wylen 5.48 A water-filled reactor with volume of 1 m3 is at 20 MPa, 360 C and placed inside a containment room as shown in Fig. P5.48. The room is well insulated and initially evacuated. Due to a failure, the reactor ruptures and the water fills the containment room. Find the minimum room volume so the final pressure does not exceed 200 kPa. Solution: Solution: C.V.:Containment room and reactor. Mass: m2 =m1 =Vreactor/v1 =1/0.001823 =548.5 kg Energy: m(u2 -u1) =1Q2 -1W2 =0 -0 =0 v1 =0.001823 m3/kg, u1 =1702.8 kJ/kg Energy equation then gives u2 =u1 =1702.8 kJ/kg State 1: Table B.1.4 State 2: P2 =200 kPa, u2 ug =Two-phase Table B.1.2 x2 =(u2 -uf)/ ufg =(1702.8 504.47)/2025.02 =0.59176 v2 =0.001061 +0.59176 0.88467 =0.52457 m3/kg V2 =m2 v2 =548.5 0.52457 =287.7 m3 P 1 200 2 v T 1 2 200 kPa u =const v P L T 1 C.P. 200 kPa 2 v Sonntag, Borgnakke and van Wylen 5.49 A piston/cylinder arrangement contains water of quality x =0.7 in the initial volume of 0.1 m3, where the piston applies a constant pressure of 200 kPa. The system is now heated to a final temperature of 200 C. Determine the work and the heat transfer in the process. Take CV as the water. Continuity Eq.:Energy Eq.5.11: Process: P =constant m2 =m1 =m ;m(u2 -u1) =1Q2 -1W2 1W2 =PdV =Pm(v2 -v1) State 1: Table B.1.2 T1 =Tsat at 200 kPa =120.23 C v1 =vf +xvfg =0.001061 +0.7 0.88467 =0.62033 m3 h1 =hf +xhfg =504.68 +0.7 2201.96 =2046.05 kJ/kg Total mass can be determined from the initial condition, m =V1/v1 =0.1/0.62033 =0.1612 kg T2 =200 C, P2 =200 kPa (Table B.1.3) gives v2 =1.08034 m3/kg h2 =2870.46 kJ/kg (Table B.1.3) V2 =mv2 =0.1612 kg 1.08034 m3/kg =0.174 m3 Substitute the work into the energy equation 1Q2 =U2 -U1 +1W2 =m ( u2 u1 +Pv2 Pv1) =m(h2 -h1) 1Q2= 0.1612 kg (2870.46-2046.05) kJ/kg =132.9 kJ (heat added to system). P 1 2 T 2 1 V V Sonntag, Borgnakke and van Wylen 5.50 A piston/cylinder arrangement has the piston loaded with outside atmospheric pressure and the piston mass to a pressure of 150 kPa, shown in Fig. P5.50. It contains water at -2 C, which is then heated until the water becomes saturated vapor. Find the final temperature and specific work and heat transfer for the process. Solution: C.V. Water in the piston cylinder. Continuity: m2 =m1, Energy Eq. per unit mass: Process: P =constant =P1, u2 -u1 =1q2 -1w2 2 =1w2 =P dv =P1(v2 -v1) 1 State 1: T1 ,P1 =Table B.1.5 compressed solid, take as saturated solid. v1 =1.09 10-3 m3/kg, v2 =vg(P2) =1.1593 m3/kg, From the process equation -3 1w2 =P1(v2 -v1) =150(1.1593 -1.09 10 ) =173.7 kJ/kg From the energy equation 1q2 =u2 -u1 +1w2 =2519.7 -(-337.62) +173.7 =3031 kJ/kg u1 =337.62 kJ/kg State 2: x =1, P2 =P1 =150 kPa due to process =Table B.1.2 T2 =111.4 C ;u2 =2519.7 kJ/kg P L S T 1 L+V S+V v 2 C.P. V P C.P. 1 T C.P. P=C 2 v 2 v 1 Sonntag, Borgnakke and van Wylen 5.51 A piston/cylinder assembly contains 1 kg of liquid water at 20oC and 300 kPa. There is a linear spring mounted on the piston such that when the water is heated the pressure reaches 1 MPa with a volume of 0.1 m3. Find the final temperature and the heat transfer in the process. Solution: Take CV as the water. m2 =m1 =m ;m(u2 -u1) =1Q2 -1W2 u1 =uf =83.94 kJ/kg so T2 =Tsat =179.9 C State 1: Compressed liquid, take saturated liquid at same temperature. v1 =vf(20) =0.001002 m3/kg, Two phase as v2 vg State 2: v2 =V2/m =0.1/1 =0.1 m3/kg and P =1000 kPa x2 =(v2 -vf) /vfg =(0.1 -0.001127)/0.19332 =0.51145 u2 =uf +x2 ufg =780.08 +0.51147 1806.32 =1703.96 kJ/kg Work is done while piston moves at linearly varying pressure, so we get 1W2 =P dV =area =Pavg (V2 -V1) =0.5 (300 +1000)(0.1 -0.001) =64.35 kJ Heat transfer is found from the energy equation 1Q2 =m(u2 -u1) +1W2 =1 (1703.96 -83.94) +64.35 =1684 kJ P P2 P 1 1 cb 2 v Sonntag, Borgnakke and van Wylen 5.52 A closed steel bottle contains ammonia at -20 C, x =20% and the volume is 0.05 m3. It has a safety valve that opens at a pressure of 1.4 MPa. By accident, the bottle is heated until the safety valve opens. Find the temperature and heat transfer when the valve first opens. Solution: C.V.:NH3 :m2 =m1 =m ;Energy Eq.5.11: m(u2 -u1) =1Q2 -1W2 P 2 Process: constant volume process 1W2 =0 State 1: (T, x) Table B.2.1 v1 =0.001504 +0.2 0.62184 =0.1259 m3/kg =m =V/v1 =0.05/0.1259 =0.397 kg u1 =88.76 +0.2 1210.7 =330.9 kJ/kg State 2: P2 ,v2 =v1 =superheated vapor, interpolate in Table B.2.2: 1 V T 110 C =100 +20(0.1259 0.12172)/(0.12986 0.12172), u2 =1481 +(1520.7 1481) 0.51 =1501.25 kJ/kg 1Q2 =m(u2 -u1) =0.397(1501.25 330.9) =464.6 kJ Sonntag, Borgnakke and van Wylen 5.53 Two kg water at 200 kPa with a quality of 25% has its temperature raised 20oC in a constant pressure process. What are the heat transfer and work in the process? C.V. Water. This is a control mass Energy Eq.:m (u2 -u1 ) =1Q2 -1W2 Process :P =constant 1W2 =P dV =mP (v2 -v1) State 1: Two-phase given P,x so use Table B.1.2 v1 =0.001061 +0.25 0.88467 =0.22223 m3/kg u1 =504047 +0.25 2025.02 =1010.725 kJ/kg T =T +20 =120.23 +20 =140.23 State 2 is superheated vapor 20 v2 =0.88573 +150-120.23 (0.95964 0.88573 ) =0.9354 m3/kg 20 u2 =2529.49 +150-120.23 (2576.87- 2529.49) =2561.32 kJ/kg From the process equation we get 1W2 =mP (v2 -v1) =2 200 ( 0.9354 -0.22223) =285.3 kJ From the energy equation 1Q2 =m (u2 -u1) +1W2 =2 ( 2561.32 1010.725 ) +285.3 =3101.2 +285.27 =3386.5 kJ P 1 2 T 2 1 V V Sonntag, Borgnakke and van Wylen 5.54 Two kilograms of nitrogen at 100 K, x =0.5 is heated in a constant pressure process to 300 K in a piston/cylinder arrangement. Find the initial and final volumes and the total heat transfer required. Solution: Take CV as the nitrogen. Continuity Eq.:m2 =m1 =m ;Energy Eq.5.11: Process: P =constant State 1: Table B.6.1 v1 =0.001452 +0.5 0.02975 =0.01633 m3/kg, h1 =73.20 +0.5 160.68 =7.14 kJ/kg State 2: (P =779.2 kPa ,300 K) =sup. vapor interpolate in Table B.6.2 v2 =0.14824 +(0.11115-0.14824) 179.2/200 =0.115 m3/kg, V2 =0.23 m3 h2 =310.06 +(309.62-310.06) 179.2/200 =309.66 kJ/kg Now solve for the heat transfer from the energy equation 1Q2 =m(u2 -u1) +1W2 =m(h2 -h1) =2 (309.66 -7.14) =605 kJ m(u2 -u1) =1Q2 -1W2 1W2 =PdV =Pm(v2 -v1) V1 =0.0327 m3 P 1 2 T 2 1 V V Sonntag, Borgnakke and van Wylen 5.55 A 1-L capsule of water at 700 kPa, 150 C is placed in a larger insulated and otherwise evacuated vessel. The capsule breaks and its contents fill the entire volume. If the final pressure should not exceed 125 kPa, what should the vessel volume be? Solution: C.V. Larger vessel. Continuity: m2 =m1 =m =V/v1 =0.916 kg Process: expansion with 1Q2 =0 ,1W2 =0 / / Energy: m(u2 -u1) =1Q2 -1W2 =0 u2 =u1 / u1 uf =631.66 kJ/kg 631.66 444.16 =0.09061 2069.3 State 1: v1 vf =0.001091 m3/kg; State 2: P2 ,u2 x2 =v2 =0.001048 +0.09061 1.37385 =0.1255 m3/kg V2 =mv2 =0.916 0.1255 =0.115 m3 =115 L P 1 200 2 v T 1 2 200 kPa u =const v P L T 1 C.P. 200 kPa 2 v Sonntag, Borgnakke and van Wylen 5.56 Superheated refrigerant R-134a at 20 C, 0.5 MPa is cooled in a piston/cylinder arrangement at constant temperature to a final two-phase state with quality of 50%.The refrigerant mass is 5 kg, and during this process 500 kJ of heat is removed. Find the initial and final volumes and the necessary work. Solution: C.V. R-134a, this is a control mass. Continuity: m2 =m1 =m ;Energy Eq.5.11: m(u2 -u1) =1Q2 -1W2 =500 -1W2 v1 =0.04226 m3/kg ;u1 =390.52 kJ/kg State 1: T1 ,P1 Table B.5.2, V1 =mv1 =0.211 m3 State 2: T2 ,x2 Table B.5.1 u2 =227.03 +0.5 162.16 =308.11 kJ/kg, v2 =0.000817 +0.5 0.03524 =0.018437 m3/kg =V2 =mv2 =0.0922 m3 1W2 =500 -m(u2 -u1) =500 -5 (308.11 -390.52) =87.9 kJ P 2 1 T 2 1 v v Sonntag, Borgnakke and van Wylen 5.57 A cylinder having a piston restrained by a linear spring (of spring constant 15 kN/m) contains 0.5 kg of saturated vapor water at 120 C, as shown in Fig. P5.57. Heat is transferred to the water, causing the piston to rise. If the piston cross-sectional area is 0.05 m2, and the pressure varies linearly with volume until a final pressure of 500 kPa is reached. Find the final temperature in the cylinder and the heat transfer for the process. Solution: C.V. Water in cylinder. Continuity: m2 =m1 =m ;Energy Eq.5.11: m(u2 -u1) =1Q2 -1W2 v1 =0.89186 m3/kg, 15 0.5 u1 =2529.2 kJ/kg ksm State 1: (T, x) Table B.1.1 =Process: State 2: P2 =P1 +(v -v ) =198.5 +(v -0.89186) Ap2 2 1 (0.05)2 2 P2 =500 kPa and on the process curve (see above equation). v2 =0.89186 +(500 -198.5) (0.052/7.5) =0.9924 m3/kg =T2 =803 C; u2 =3668 kJ/kg (P, v) Table B.1.3 P1 +P2 W12 =PdV =2 m(v2 -v1) 198.5 +500 =0.5 (0.9924 -0.89186) =17.56 kJ 2 1Q2 =m(u2 -u1) +1W2 =0.5 (3668 -2529.2) +17.56 =587 kJ P 2 1 ksm A2 p v T 2 1 v Sonntag, Borgnakke and van Wylen 5.58 A rigid tank is divided into two rooms by a membrane, both containing water, shown in Fig. P5.58. Room A is at 200 kPa, v =0.5 m3/kg, VA =1 m3, and room B contains 3.5 kg at 0.5 MPa, 400 C. The membrane now ruptures and heat transfer takes place so the water comes to a uniform state at 100 C. Find the heat transfer during the process. Solution: C.V.:Both rooms A and B in tank. A B Continuity Eq.:Energy Eq.:m2 =mA1 +mB1 ;m2u2 -mA1uA1 -mB1uB1 =1Q2 -1W2 mA1 =VA/vA1 =1/0.5 =2 kg State 1A: (P, v) Table B.1.2, v vf 0.5 -0.001061 =0.564 xA1 =v 0.88467 fg uA1 =uf +x ufg =504.47 +0.564 2025.02 =1646.6 kJ/kg State 1B: Table B.1.3, vB1 =0.6173, uB1 =2963.2, VB =mB1vB1 =2.16 m3 Process constant total volume: m2 =mA1 +mB1 =5.5 kg State 2: T2 ,v2 Table B.1.1 x2 =Vtot =VA +VB =3.16 m3 and 1W2 =0 / =v2 =Vtot/m2 =0.5746 m3/kg two-phase as v2 vg v2 vf 0.5746 0.001044 =0.343 ,1.67185 vfg u2 =uf +x ufg =418.91 +0.343 2087.58= 1134.95 kJ/kg Heat transfer is from the energy equation 1Q2 =m2u2 -mA1uA1 -mB1uB1 =7421 kJ Sonntag, Borgnakke and van Wylen 5.59 A 10-m high open cylinder, Acyl =0.1 m2, contains 20 C water above and 2 kg of 20 C water below a 198.5-kg thin insulated floating piston, shown in Fig. P5.59. Assume standard g, Po. Now heat is added to the water below the piston so that it expands, pushing the piston up, causing the water on top to spill over the edge. This process continues until the piston reaches the top of the cylinder. Find the final state of the water below the piston (T, P, v) and the heat added during the process. Solution: C.V. Water below the piston. Piston force balance at initial state: F =F =PAA =mpg +mBg +P0A State 1A,B: Comp. Liq. v vf =0.001002 m3/kg; VA1 =mAvA1 =0.002 m3; mass above the piston u1A =83.95 kJ/kg mtot =Vtot/v =1/0.001002 =998 kg (198.5+996)*9.807 =218.5 kPa 0.1*1000 mB1 =mtot -mA =996 kg PA1 =P0 +(mp +mB)g/A =101.325 +State 2A: mpg PA2 =P0 +A =120.8 kPa ;vA2 =Vtot/ mA= 0.5 m3/kg xA2 =(0.5 -0.001047)/1.4183 =0.352 ;T2 =105 C uA2 =440.0 +0.352 2072.34 =1169.5 kJ/kg Continuity eq. in A: Process: mA2 =mA1 P Energy: mA(u2 -u1) =1Q2 -1W2 P linear in V as mB is linear with V 1 1W2 =PdV =2(218.5 +120.82)(1 -0.002) =169.32 kJ 1Q2 =mA(u2 -u1) +1W2 =2170.1 +169.3 =2340.4 kJ 1 2 V W cb Sonntag, Borgnakke and van Wylen 5.60 Assume the same setup as in Problem 5.48, but the room has a volume of 100 m3. Show that the final state is two-phase and find the final pressure by trial and error. C.V.:Containment room and reactor. Mass: m2 =m1 =Vreactor/v1 =1/0.001823 =548.5 kg Energy: m(u2 -u1) =1Q2 -1W2 =0 -0 =0 u2 =u1 =1702.8 kJ/kg v2 =Vroom/m2 =0.1823 m3/kg Total volume and mass =State 2: u2 ,v2 Table B.1.1 see Figure. Note that in the vicinity of v =0.1823 m3/kg crossing the saturated vapor line the internal energy is about 2585 kJ/kg. However, at the actual state 2, u =1702.8 kJ/kg. Therefore state 2 must be in the two-phase region. T Trial & error v =vf +xvfg ;u =uf +xufg 1060 kPa 1060 kPa u=2585 v2 -vf u2 =1702.8 =uf +v ufg fg Compute RHS for a guessed pressure P2: sat vap 0.184 v P2 =600 kPa: RHS =669.88 +P2 =550 kPa: RHS =655.30 +0.1823-0.001101 1897.52 =1762.9 0.31457 0.1823-0.001097 1909.17 =1668.1 0.34159 P2 568.5 kPa too large too small Linear interpolation to match u =1702.8 gives Sonntag, Borgnakke and van Wylen Energy Equation: Multistep Solution 5.61 10 kg of water in a piston cylinder arrangement exists as saturated liquid/vapor at 100 kPa, with a quality of 50%.It is now heated so the volume triples. The mass of the piston is such that a cylinder pressure of 200 kPa will float it, as in Fig. 4.68. Find the final temperature and the heat transfer in the process. Solution: Take CV as the water. m2 =m1 =m ;m(u2 -u1) =1Q2 -1W2 Process: v =constant until P =Plift ,then P is constant. State 1: Two-phase so look in Table B.1.2 at 100 kPa u1 =417.33 +0.5 2088.72 =1461.7 kJ/kg, v1 =0.001043 +0.5 1.69296 =0.8475 m3/kg State 2: v2, P2 Plift =v2 =3 0.8475 =2.5425 m3/kg ;Interpolate: T2 =829 C, u2 =3718.76 kJ/kg =V2 =mv2 =25.425 m3 1W2 =Plift(V2 -V1) =200 10 (2.5425 -0.8475) =3390 kJ 1Q2 =m(u2 -u1) +1W2 =10 (3718.76 -1461.7) +3390 =25 961 kJ Po cb P 2 P2 H2O P1 1 cb V Sonntag, Borgnakke and van Wylen 5.62 Two tanks are connected by a valve and line as shown in Fig. P5.62. The volumes are both 1 m3 with R-134a at 20 C, quality 15% in A and tank B is evacuated. The valve is opened and saturated vapor flows from A into B until the pressures become equal. The process occurs slowly enough that all temperatures stay at 20 C during the process. Find the total heat transfer to the R-134a during the process. Solution: C.V.:A +B State 1A: vA1 =0.000817 +0.15 0.03524 =0.006103 m3/kg uA1 =227.03 +0.15 162.16 =251.35 kJ/kg mA1 =VA/vA1 =163.854 kg Process: Constant temperature and constant total volume. m2 =mA1 ;V2 =VA +VB =2 m3 ;v2 =V2/m2 =0.012206 m3/kg 1W2 =P dV =0 State 2: T2 ,v2 x2 =(0.012206 0.000817)/0.03524 =0.3232 u2 =227.03 +0.3232 162.16 =279.44 kJ/kg 1Q2 =m2u2 -mA1uA1 -mB1uB1 +1W2 =m2(u2 -uA1) =163.854 (279.44 -251.35) =4603 kJ A B Sonntag, Borgnakke and van Wylen 5.63 Consider the same system as in the previous problem. Let the valve be opened and transfer enough heat to both tanks so all the liquid disappears. Find the necessary heat transfer. Solution: C.V. A +B, so this is a control mass. State 1A: vA1 =0.000817 +0.15 0.03524 =0.006 103 m3/kg uA1 =227.03 +0.15 162.16 =251.35 kJ/kg mA1 =VA/vA1 =163.854 kg Process: Constant temperature and total volume. m2 =mA1 ;V2 =VA +VB =2 m3 ;v2 =V2/m2 =0.012 206 m3/kg State 2: x2 =100%,v2 =0.012206 T2 =55 +5 (0.012206 0.01316)/(0.01146 0.01316) =57.8 C u2 =406.01 +0.56 (407.85 406.01) =407.04 kJ/kg 1Q2 =m2(u2 -uA1) =163.854 (407.04 -251.35) =25 510 kJ A B Sonntag, Borgnakke and van Wylen 5.64 A vertical cylinder fitted with a piston contains 5 kg of R-22 at 10 C, shown in Fig. P5.64. Heat is transferred to the system, causing the piston to rise until it reaches a set of stops at which point the volume has doubled. Additional heat is transferred until the