Baixe Exercícios de Termodinâmica Resolvidos Cap 3 e outras Notas de estudo em PDF para Termodinâmica, somente na Docsity! SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 3 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG • BORGNAKKE • VAN WYLEN Sonntag, Borgnakke and Wylen CHAPTER 3 SUBSECTION PROB NO. Concept-Study Guide Problems 128-132 Phase diagrams 133-134 General Tables 135-145 Ideal Gas 146-148 Compressibility Factor 149, 157, 158 Review Problems 150-156 Correspondence table The correspondence between the problem set in this sixth edition versus the problem set in the 5'th edition text. Problems that are new are marked new and the SI number refers to the corresponding SI unit problem. New 5th Ed. SI New 5th Ed. SI 128 new 5 143 77E 53 129 new 7 144 new 62 130 new 9 145 79E 58 131 new 11 146 62E 69 132 new 17 147 new 65 133 new 23 148 69E c+d 70E d - 134 61E 27 149 72E 81 135 68E a-c 30 150 64E 113 136 68E d-f 30 151 new 74 137 new 40 152 81E 49 138 70E 36 153 new 99 139 73E 47 154 71E 95 140 74E 41 155 80E 61 141 new 44 156 83E 106 142 76E 51 157 65E 89 158 66E - Sonntag, Borgnakke and Wylen 3.132E Calculate the ideal gas constant for argon and hydrogen based on Table F.1 and verify the value with Table F.4 The gas constant for a substance can be found from the universal gas constant from table A.1 and the molecular weight from Table F.1 Argon: R = R _ M = 1.98589 39.948 = 0.04971 Btu lbm R = 38.683 lbf-ft lbm R Hydrogen: R = R _ M = 1.98589 2.016 = 0.98506 Btu lbm R = 766.5 lbf-ft lbm R Recall from Table A.1: 1 Btu = 778.1693 lbf-ft Sonntag, Borgnakke and Wylen Phase Diagrams 3.133E Water at 80 F can exist in different phases dependent on the pressure. Give the approximate pressure range in lbf/in2 for water being in each one of the three phases, vapor, liquid or solid. Solution: The phases can be seen in Fig. 3.7, a sketch of which is shown to the right. T = 80 F = 540 R = 300 K From Fig. 3.7: PVL ≈ 4 × 10−3 MPa = 4 kPa = 0.58 psia, PLS = 103 MPa = 145 038 psia ln P T V L S CR.P. S 0 < P < 0.58 psia VAPOR 0.58 psia < P < 145 038 psia LIQUID P > 145 038 psia SOLID(ICE) Sonntag, Borgnakke and Wylen 3.134E A substance is at 300 lbf/in.2, 65 F in a rigid tank. Using only the critical properties can the phase of the mass be determined if the substance is nitrogen, water or propane? Solution: Find state relative to the critical point properties, Table F.1 a) Nitrogen 492 lbf/in.2 227.2 R b) Water 3208 lbf/in.2 1165.1 R c) Propane 616 lbf/in.2 665.6 R P < Pc for all and T = 65 F = 65 + 459.67 = 525 R a) N2 T >> Tc Yes gas and P < Pc b) H2O T << Tc P << Pc so you cannot say c) C3H8 T < Tc P < Pc you cannot say ln P T Vapor Liquid Cr.P. a c b Sonntag, Borgnakke and Wylen 3.137E Give the phase and the missing property of P, T, v and x. a. R-134a T = -10 F, P = 18 psia b. R-134a P = 50 psia, v = 1.3 ft3/lbm c. NH3 T = 120 F, v = 0.9 ft3/lbm d. NH3 T = 200 F, v = 11 ft3/lbm Solution: a. Look in Table F.10.1 at –10 F: P > Psat = 16.76 psia This state is compressed liquid so x is undefined and v = vf = 0.01173 ft3/lbm b. Look in Table F.10.1 close to 50 psia there we see v > vg = 0.95 ft3/lbm so superheated vapor Look then in Table F.10.2 under 50 psia which is not printed so we must interpolate between the 40 and 60 psia sections. (60 psia, 1.3 ft3/lbm) : T = 300 F (40 psia, 1.3 ft3/lbm) : T = 66.6 F Linear interpolation between these gives T = 183 F for a better accuracy we must use the computer software. c. Look in Table F.8.1 at 120 F: v < vg = 1.0456 ft3/lbm so two-phase P = Psat = 286.5 psia x = v - vf vfg = 0.9 - 0.02836 1.0172 = 0.8569 d. Look in Table F.8.1 at 200 F: v > vg = 0.3388 ft3/lbm so sup. vapor Look in Table F.8.2 start anywhere say at 15 psia, 200 F there we see v = 27.6 ft3/lbm so P larger We can bracket the state between 35 and 40 psia so we get P = 35 + 5 11 – 11.74 10.2562 – 11.74 = 37.494 psia Sonntag, Borgnakke and Wylen 3.138E Give the phase and the specific volume. Solution: a. R-22 T = -10 F, P = 30 lbf/in.2 Table C.10.1 P < Psat = 31.2 psia ⇒ sup.vap. v ≅ 1.7439 + -10+11.71 11.71 (1.7997 – 1.7439) = 1.752 ft 3/lbm b. R-22 T = -10 F, P = 40 lbf/in.2 Table C.10.1 Psat = 31.2 psia P > Psat ⇒ compresssed Liquid v ≅ vf = 0.01178 ft 3/lbm c. H2O T = 280 F, P = 35 lbf/in. 2 Table C.8.1 P < Psat = 49.2 psia ⇒ sup.vap v ≅ 21.734 + ( 10.711 – 21.734) ×(15/20) = 1.0669 ft3/lbm d. NH3 T = 60 F, P = 15 lbf/in. 2 Table C.9.1 Psat = 107.6 psia P < Psat ⇒ sup.vap v ≅ 21.564 ft 3/lbm Sonntag, Borgnakke and Wylen 3.139E A water storage tank contains liquid and vapor in equilibrium at 220 F. The distance from the bottom of the tank to the liquid level is 25 ft. What is the absolute pressure at the bottom of the tank? Solution: Table F.7.1: vf = 0.01677 ft 3/lbm ∆P = g l vf = 32.174 × 25 32.174 × 0.01677 × 144 = 10.35 lbf/in 2 Since we have a two-phase mixture the vapor pressure is the saturated Psat so P = Psat + ∆P = 17.189 + 10.35 = 27.54 lbf/in 2 H Sonntag, Borgnakke and Wylen 3.142E Saturated water vapor at 200 F has its pressure decreased to increase the volume by 10%, keeping the temperature constant. To what pressure should it be expanded? Solution: v = 1.1 × vg = 1.1 × 33.63 = 36.993 ft 3/lbm Interpolate between sat. at 200 F and sup. vapor in Table F.7.2 at 200 F, 10 lbf/in2 P ≅ 10.54 lbf/in2 P C.P. v T C.P. v T P = 10 lbf/in 200 Fo 10 lbf/in2 2 Sonntag, Borgnakke and Wylen 3.143E A boiler feed pump delivers 100 ft3/min of water at 400 F, 3000 lbf/in.2. What is the mass flowrate (lbm/s)? What would be the percent error if the properties of saturated liquid at 400 F were used in the calculation? What if the properties of saturated liquid at 3000 lbf/in.2 were used? Solution: Table F.7.3: v = 0.0183 ft3/lbm (interpolate 2000-8000 psia) m . = V . v = 100 60 × 0.018334 = 91.07 lbm/s vf (400 F) = 0.01864 ⇒ m . = 89.41 lbm/s error 1.8% vf (3000 lbf/in2) = 0.03475 ft 3/lbm ⇒ m . = 47.96 lbm/s error 47% P C.P. v T C.P. v P = 3000 psia 400 F 3000 400 247 695 Sonntag, Borgnakke and Wylen 3.144E A pressure cooker has the lid screwed on tight. A small opening with A = 0.0075 in2 is covered with a petcock that can be lifted to let steam escape. How much mass should the petcock have to allow boiling at 250 F with an outside atmosphere at 15 psia? Solution: Table F.7.1 at 250 F: Psat = 29.823 psia Fnet = (Psat – Patm) A = (29.823 - 15) psia × 0.0075 in 2 = 0.111 lbf Fnet = mpetcock g mpetcock = Fnet/g = 0.111 lbf 32.174 ft/s2 = 0.111 × 32.174 lbm ft/s2 32.174 ft/s2 = 0.111 lbm Some petcocks are held down by a spring, the problem deals with one that stays on due to its weight. Sonntag, Borgnakke and Wylen 3.147E A spherical helium balloon of 30 ft in diameter is at ambient T and P, 60 F and 14.69 psia. How much helium does it contain? It can lift a total mass that equals the mass of displaced atmospheric air. How much mass of the balloon fabric and cage can then be lifted? V = π 6 D 3 = π 6 30 3 = 14 137 ft3 mHe = ρV = V v = PV RT = 14.69 × 14 137 × 144 386.0 × 520 = 148.99 lbm mair = PV RT = 14.69 × 14 137 × 144 53.34 × 520 = 1078 lbm mlift = mair – mHe = 1078 - 149 = 929 lbm Sonntag, Borgnakke and Wylen 3.148E Give the phase and the specific volume for each of the following. Solution: a. CO2 T = 510 F P = 75 lbf/in. 2 Table F.4 superheated vapor ideal gas v = RT/P = 35.1 × (510 + 459.7) 75 × 144 = 3.152 ft 3/lbm b. Air T = 68 F P = 2 atm Table F.4 superheated vapor ideal gas v = RT/P = 53.34 × (68 + 459.7) 2 × 14.6 × 144 = 6.6504 ft 3/lbm c. Ar T = 300 F, P = 30 lbf/in.2 Table F.4 Ideal gas: v = RT/P = 38.68 (300 + 459.7) / (30 × 144) = 6.802 ft3/lbm Sonntag, Borgnakke and Wylen Review Problems 3.149E What is the percent error in specific volume if the ideal gas model is used to represent the behavior of superheated ammonia at 100 F, 80 lbf/in.2? What if the generalized compressibility chart, Fig. D.1, is used instead? Solution: Ammonia Table F.8.2: v = 4.186 ft3/lbm Ideal gas v = RT P = 90.72 × 559.7 80 × 144 = 4.4076 ft 3/lbm 5.3% error Generalized compressibility chart and Table D.4 Tr = 559.7/729.9 = 0.767, Pr = 80/1646 = 0.0486 => Z ≅ 0.96 v = ZRT/P = 0.96 × 4.4076 = 4.231 ft3/lbm 1.0% error Sonntag, Borgnakke and Wylen 3.152E Two tanks are connected together as shown in Fig. P3.49, both containing water. Tank A is at 30 lbf/in.2, v = 8 ft3/lbm, V = 40 ft3 and tank B contains 8 lbm at 80 lbf/in. 2, 750 F. The valve is now opened and the two come to a uniform state. Find the final specific volume. Solution: Control volume both tanks. Constant total volume and mass process. A B sup. vapor State A1: (P, v) two-phase, mA = VA/vA = 40/8 = 5 lbm State B1: (P, T) Table F.7.2: vB = (8.561 + 9.322)/2 = 8.9415 ⇒ VB = mBvB = 8 × 8.9415 = 71.532 ft 3 Final state: mtot = mA + mB = 5 + 8 = 13 lbm Vtot = VA + VB = 111.532 ft 3 v2 = Vtot/mtot = 111.532/13 = 8.579 ft 3/lbm Sonntag, Borgnakke and Wylen 3.153E A 35 ft3 rigid tank has air at 225 psia and ambient 600 R connected by a valve to a piston cylinder. The piston of area 1 ft2 requires 40 psia below it to float, Fig. P3.99. The valve is opened and the piston moves slowly 7 ft up and the valve is closed. During the process air temperature remains at 600 R. What is the final pressure in the tank? mA = PAVA RTA = 225 × 35 × 144 53.34 × 600 = 35.433 lbm mB2 - mB1 = ∆VA vB = ∆VBPB RT = 1 × 7 × 40 × 144 53.34 × 600 = 1.26 lbm mA2 = mA – (mB2 - mB1) = 35.433 – 1.26 = 34.173 lbm PA2 = mA2RT VA = 34.173 × 53.34 × 600 35 × 144 = 217 psia Sonntag, Borgnakke and Wylen 3.154E Give the phase and the missing properties of P, T, v and x. These may be a little more difficult if the appendix tables are used instead of the software. Solution: a. R-22 at T = 50 F, v = 0.6 ft3/lbm: Table F.9.1 v > vg sup. vap. F.9.2 interpolate between sat. and sup. vap at 50 F. P ≅ 98.73 + (0.6 - 0.5561)(80 -98.73)/(0.708 - 0.5561) = 93.3 lbf/in2 b. H2O v = 2 ft3/lbm, x = 0.5: Table F.7.1 since vf is so small we find it approximately where vg = 4 ft3/lbm. vf + vg = 4.3293 at 330 F, vf + vg = 3.80997 at 340 F. linear interpolation T ≅ 336 F, P ≅ 113 lbf/in2 c. H2O T = 150 F, v = 0.01632 ft3/lbm: Table F.7.1, v < vf compr. liquid P ≅ 500 lbf/in2 d. NH3 T = 80 F, P = 13 lbf/in. 2 Table F.8.1 P < Psat sup. vap. interpolate between 10 and 15 psia: v = 26.97 ft3/lbm v is not linear in P (more like 1/P) so computer table is more accurate. e. R-134a v = 0.08 ft3/lbm, x = 0.5: Table F.10.1 since vf is so small we find it approximately where vg = 0.16 ft3/lbm. vf + vg = 0.1729 at 150 F, vf + vg = 0.1505 at 160 F. linear interpolation T ≅ 156 F, P ≅ 300 lbf/in2 Sonntag, Borgnakke and Wylen Compressiblity Factor 3.157E A substance is at 70 F, 300 lbf/in.2 in a 10 ft3 tank. Estimate the mass from the compressibility chart if the substance is a) air, b) butane or c) propane. Solution: Use Fig. D.1 for compressibility Z and table F.1 for critical properties m = PV ZRT = 300 ×144 ×10 530 ZR = 815.09 ZR Air use nitrogen Pc = 492 lbf/in. 2; Tc = 227.2 R Pr = 0.61; Tr = 2.33; Z = 0.98 m = PV ZRT = 815.09 ZR = 815.09 0.98× 55.15 = 15.08 lbm Butane Pc = 551 lbf/in. 2; Tc = 765.4 R Pr = 0.544; Tr = 0.692; Z = 0.09 m = PV ZRT = 815.09 ZR = 815.09 0.09× 26.58 = 340.7 lbm Propane Pc = 616 lbf/in. 2; Tc = 665.6 R Pr = 0.487; Tr = 0.796; Z = 0.08 m = PV ZRT = 815.09 ZR = 815.09 0.08× 35.04 = 290.8 lbm ln Pr Z T = 2.0r a bc T = 0.7r T = 0.7r 0.1 1 Sonntag, Borgnakke and Wylen 3.158E Determine the mass of an ethane gas stored in a 25 ft3 tank at 250 F, 440 lbf/in.2 using the compressibility chart. Estimate the error (%) if the ideal gas model is used. Solution Table F.1: Tr = ( 250 + 460 ) / 549.7 = 1.29 and Pr = 440/708 = 0.621 Figure D.1 ⇒ Z = 0.9 m = PV/ZRT = 440 × 144 × 25 / (51.38 × 710 × 0.9) = 48.25 lbm Ideal gas Z = 1 ⇒ m = 43.21 lbm 10% error