Claus borgnakke, richard e sonntag fundamentals of thermodynamics 7th edition solution manual

SOLUTION MANUAL CHAPTER www.elsolucionario.org www.elsolucionario.org Borgnakke and Sonntag CONTENT SUBSECTION Concept Problems Properties and Units Force and Energy Specific Volume Pressure Manometers and Barometers Temperature Review problems PROB NO 1-18 19-22 23-34 35-40 41-56 57-77 78-83 84-89 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 www.elsolucionario.org 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 www.elsolucionario.org Borgnakke and Sonntag 2.a Make a control volume around the turbine in the steam power plant in Fig 1.1 and list the flows of mass and energy that are there Solution: We see hot high pressure steam flowing in at state from the steam drum through a flow control (not shown) The steam leaves at a lower pressure to the condenser (heat exchanger) at state A rotating shaft gives a rate of energy (power) to the electric generator set W T 2.b Take a control volume around your kitchen refrigerator and indicate where the components shown in Figure 1.6 are located and show all flows of energy transfer Solution: The valve and the cold line, the evaporator, is inside close to the inside wall and usually a small blower distributes cold air from the freezer box to the refrigerator room Q leak The black grille in the back or at the bottom is the condenser that gives heat to the room air Q W The compressor sits at the bottom cb 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 2.c Why people float high in the water when swimming in the Dead Sea as compared with swimming in a fresh water lake? As the dead sea is very salty its density is higher than fresh water density The buoyancy effect gives a force up that equals the weight of the displaced water Since density is higher the displaced volume is smaller for the same force www.elsolucionario.org 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 www.elsolucionario.org Borgnakke and Sonntag 2.d Density of liquid water is ρ = 1008 – T/2 [kg/m3] with T in oC If the temperature increases, what happens to the density and specific volume? Solution: The density is seen to decrease as the temperature increases ∆ρ = – ∆T/2 Since the specific volume is the inverse of the density v = 1/ρ it will increase 2.e A car tire gauge indicates 195 kPa; what is the air pressure inside? The pressure you read on the gauge is a gauge pressure, ∆P, so the absolute pressure is found as ` P = Po + ∆P = 101 + 195 = 296 kPa 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 2.f Can I always neglect ∆P in the fluid above location A in figure 2.12? What does that depend on? If the fluid density above A is low relative to the manometer fluid then you neglect the pressure variation above position A, say the fluid is a gas like air and the manometer fluid is like liquid water However, if the fluid above A has a density of the same order of magnitude as the manometer fluid then the pressure variation with elevation is as large as in the manometer fluid and it must be accounted for 2.g A U tube manometer has the left branch connected to a box with a pressure of 110 kPa and the right branch open Which side has a higher column of fluid? Solution: Box Since the left branch fluid surface feels 110 kPa and the right branch surface is at 100 kPa you must go further down to match the 110 kPa The right branch has a higher column of fluid Po www.elsolucionario.org H cb 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 www.elsolucionario.org 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 2.1 Make a control volume around the whole power plant in Fig 1.2 and with the help of Fig 1.1 list what flows of mass and energy are in or out and any storage of energy Make sure you know what is inside and what is outside your chosen C.V Solution: Smoke Boiler building stack Coal conveyor system Storage gypsum cb flue gas Coal storage Turbine house Dock Flue gas www.elsolucionario.org Combustion air Underground Welectrical power cable District heating Cold return Hot water m m m Storage for later transport out: m Coal Gypsum, fly ash, slag m 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 www.elsolucionario.org Borgnakke and Sonntag 17.74 A sharp-edged orifice is used to measure the flow of air in a pipe The pipe diameter is 100 mm and the diameter of the orifice is 25 mm Upstream of the orifice, the absolute pressure is 150 kPa and the temperature is 35°C The pressure drop across the orifice is 15 kPa, and the coefficient of discharge is 0.62 Determine the mass flow rate in the pipeline 0.4 15 k-1∆P ∆T = Ti k  P = 308.15 × 1.4 × 150 = 8.8 K   i vi = RTi/Pi = 0.5896 m3/kg Pe = 135 kPa, Te = 299.35 K, ve = 0.6364 m3/kg mi = me ⇒ Vi / Ve = (De/Di)2 vi/ve = 0.0579 hi - he = V2e(1 - 0.05792)/2 = Cp(Ti - Te) Ve s = × 1000 × 1.004 × 8.8/(1 - 0.0579)2 = 133.1 m/s m = CD AV/v = 0.62 (π/4) (0.025)2 133.1 / 0.6364 = 0.06365 kg/s 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 17.75 A critical nozzle is used for the accurate measurement of the flow rate of air Exhaust from a car engine is diluted with air so its temperature is 50°C at a total pressure of 100 kPa It flows through the nozzle with throat area of 700 mm2 by suction from a blower Find the needed suction pressure that will lead to critical flow in the nozzle and the mass flow rate P* = 0.5283 Po = 52.83 kPa, T* = 0.8333 To = 269.3 K v* = RT*/P* = 0.287 × 269.3/52.83 = 1.463 m3/kg c* = kRT* = 1.4 × 1000 × 0.287 × 269.3 = 328.9 m/s m = Ac*/v* = 700 × 10-6 × 328.9/1.463 = 0.157 kg/s www.elsolucionario.org 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 www.elsolucionario.org Borgnakke and Sonntag 17.76 Air is expanded in a nozzle from 700 kPa, 200°C, to 150 kPa in a nozzle having an efficiency of 90% The mass flow rate is kg/s Determine the exit area of the nozzle, the exit velocity, and the increase of entropy per kilogram of air Compare these results with those of a reversible adiabatic nozzle T2s = T1(P2/P1)(k-1)/k = 473.2 (150/700)0.286 = 304.6 K V2s2 = × 1000 × 1.004(473.2 - 304.6) = 338400 J/kg V22 = 0.9 × 338400 ⇒ V2 = 552 m/s h2 + V22/2 = h1 ⇒ T2 = T1 - V22/2Cp T2 = 473.2 - 5522/(2 × 1000 × 1.004) = 321.4 K ; v2 = 0.287 × 321.4/150 = 0.6149 m3/kg A2 = × 0.6149/552 = 0.00446 m2 = 4460 mm2 321.4 150 s2 - s1 = 1.0035 ln473.2 - 0.287 ln700 = 0.0539 kJ/kg 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 17.77 Steam at a pressure of MPa and temperature of 400°C expands in a nozzle to a pressure of 200 kPa The nozzle efficiency is 90% and the mass flow rate is 10 kg/s Determine the nozzle exit area and the exit velocity First the ideal reversible adiabatic nozzle s2s= s1= 7.4651 kJ/kg K, h1= 3263.9 kJ/kg ⇒ T2s = 190.4°C ; h2s = 2851 kJ/kg Now the actual nozzle can be calculated h1 - h2ac = ηD(h1 - h2s) = 0.9(3263.9 - 2851) = 371.6 kJ/kg h2ac = 2892.3 kJ/kg, T2 = 210.9°C, v2 = 1.1062 m3/kg V2 = 2000(3263.9 - 2892.3) = 862 m/s A2 = mv2/V2 = 10 × 1.1062/862 = 0.01283 m2 www.elsolucionario.org 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 www.elsolucionario.org Borgnakke and Sonntag 17.78 Steam at 800 kPa, 350°C flows through a convergent-divergent nozzle that has a throat area of 350 mm2 The pressure at the exit plane is 150 kPa and the exit velocity is 800 m/s The flow from the nozzle entrance to the throat is reversible and adiabatic Determine the exit area of the nozzle, the overall nozzle efficiency, and the entropy generation in the process ho1 = 3161.7 kJ/kg, so1 = 7.4089 kJ/kg K P*/Po1 = (2/(k+1))k/(k-1) = 0.54099 ⇒ P* = 432.7 kPa At *: (P*,s* = so1) ⇒ h* = 2999.3 kJ/kg, v* = 0.5687 m3/kg ∆ h = V2/2 ⇒ V* = 2000(3161.7-2999.3) = 569.9 m/s m = AV*/v* = 350 × 10-6 × 569.9/0.5687 = 0.3507 kg/s he = ho1 - V2e/2 = 3161.7 - 8002/2 × 1000 = 2841.7 kJ/kg Exit: Pe, he: ve = 1.395 m3/kg, se = 7.576 kJ/kg K Ae = mve/Ve = 0.3507 × 1.395/800 = 6.115 × 10-4 m2 sgen = se - so1 = 7.576 - 7.4089 = 0.167 kJ/kg 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 17.79 A convergent nozzle with exit diameter of cm has an air inlet flow of 20°C, 101 kPa (stagnation conditions) The nozzle has an isentropic efficiency of 95% and the pressure drop is measured to 50 cm water column Find the mass flow rate assuming compressible adiabatic flow Repeat calculation for incompressible flow Convert ∆P to kPa: ∆P = 50 cm H2O = 0.5 × 9.8064 = 4.903 kPa T0 = 20°C = 293.15 K Assume inlet Vi = P0 = 101 kPa Pe = P0 - ∆P = 101 - 4.903 = 96.097 kPa Pe k-1 96.097 Te = T0 (P ) k = 293.15 ×( 101 )0.2857 = 289.01 V2e/2 = hi - he = Cp (Ti - Te) = 1.004 × (293.15 - 289.01) = 4.1545 kJ/kg = 4254.5 J/kg Ve 2ac/2 = η Ve2s/2 = 0.95 × 4154.5 = 3946.78 => Ve = 91.15 m/s ⇒ Ve ac = 88.85 m/s www.elsolucionario.org Te ac = Ti - Ve 2ac/2 Pe ρe ac = RT = p Cp 3.9468 = 293.15 - 1.0035 = 289.2 K 96.097 = 1.158 kg/m3 0.287 × 289.2 π m = ρAV = 1.158 × × 0.022 × 88.85 = 0.0323 kg/s 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 www.elsolucionario.org Borgnakke and Sonntag 17.80 The coefficient of discharge of a sharp-edged orifice is determined at one set of conditions by use of an accurately calibrated gasometer The orifice has a diameter of 20 mm and the pipe diameter is 50 mm The absolute upstream pressure is 200 kPa and the pressure drop across the orifice is 82 mm of mercury The temperature of the air entering the orifice is 25°C and the mass flow rate measured with the gasometer is 2.4 kg/min What is the coefficient of discharge of the orifice at these conditions? ∆P = 82 × 101.325/760 = 10.93 kPa 0.4 k-1 ∆T = Ti  k  ∆P/Pi = 298.15 × 1.4 × 10.93/200 = 4.66   vi = RTi/Pi = 0.4278 m3/kg, ve = RTe/Pe = 0.4455 m3/kg Vi = VeAevi/Aive = 0.1536 Ve (V2e - V2i )/2 = V2e(1 - 0.15362)/2 = hi - he = Cp∆T Ve = × 1000 × 1.004 × 4.66/(1 - 0.15362) = 97.9 m/s π m = AeVe/ve = × 0.022 × 97.9/0.4455 = 0.069 kg/s CD = 2.4/60 × 0.069 = 0.58 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 17.81 A convergent nozzle is used to measure the flow of air to an engine The atmosphere is at 100 kPa, 25°C The nozzle used has a minimum area of 2000 mm2 and the coefficient of discharge is 0.95 A pressure difference across the nozzle is measured to 2.5 kPa Find the mass flow rate assuming incompressible flow Also find the mass flow rate assuming compressible adiabatic flow Assume Vi ≅ 0, vi = RTi/Pi = 0.287 × 298.15/100 = 0.8557 m3/kg Incompressible flow: Ve,s2/2 = hi - he,s = vi(Pi - Pe) = 2.1393 kJ/kg Ve,s = × 1000 × 2.1393 = 65.41 m/s ms = AVe,s/vi = 2000 × 10-6 × 65.41/0.8557 = 0.153 kg/s ma = CDms = 0.1454 kg/s Compressible flow: Te,s = Ti (Pe/Pi)(k-1)/k = 298.15(97.5/100)0.2857 = 296 K ∆h = Cp∆T = 1.0035 × 2.15 = 2.1575 = Ve,s2/2 × 1000 × 2.1575 = 65.69 m/s www.elsolucionario.org Ve,s = ve,s = 0.287 × 296/97.5 = 0.8713 m3/kg ms = AVe,s/ve,s = 2000 × 10-6 × 65.69/0.8713 = 0.1508 kg/s ma = CDms = 0.1433 kg/s 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 www.elsolucionario.org Borgnakke and Sonntag Review 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 17.82 Atmospheric air is at 20°C, 100 kPa with zero velocity An adiabatic reversible compressor takes atmospheric air in through a pipe with cross-sectional area of 0.1 m2 at a rate of kg/s It is compressed up to a measured stagnation pressure of 500 kPa and leaves through a pipe with cross-sectional area of 0.01 m2 What are the required compressor work and the air velocity, static pressure, and temperature in the exit pipeline? C.V compressor out to standing air and exit to stagnation point m ho1 + W c = m(h + V2/2)ex = mho,ex mso1 = mso,ex ⇒ Pr,o,ex = Pr,o1 (Pst,ex/Po1) = 1.028(500/100) = 5.14 ⇒ To,ex = 463 K, ho,ex = 465.38 kJ/kg, ho1 = 209.45 kJ/kg Wc = m(ho,ex - ho1) = 1(465.38 - 209.45) = 255.9 kW Pex = Po,ex(Tex/To,ex)k/(k-1) Tex = To,ex - Vex /2Cp m = kg/s = (ρAV)ex = PexAVex/RTex Now select unknown amongst Pex, Tex, Vex and write the continuity eq m www.elsolucionario.org and solve the nonlinear equation Say, use Tex then Vex = 2Cp(To,ex - Tex) m = kg/s = Po,ex(Tex/To,ex)k/k-1A 2Cp(To,ex - Tex)/RTex solve for Tex/To,ex (close to 1) Tex = 462.6 K ⇒ Vex= 28.3 m/s, Pex = 498.6 kPa 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 www.elsolucionario.org Borgnakke and Sonntag 17.83 The nozzle in Problem 17.46 will have a throat area of 0.001272 m2 and an exit area 2.896 times as large Suppose the back pressure is raised to 1.4 MPa and that the flow remains isentropic except for a normal shock wave Verify that the shock mach number (Mx) is close to and find the exit mach number, the temperature and the mass flow rate through the nozzle (a) From Table A.12: ME = 2.6 PE = 2.0 × 0.05012 = 0.1002 MPa T* = 423.15 × 0.8333 = 352.7 K P* = 2.0 × 0.5283 = 1.057 MPa c* = 1.4 × 1000 × 0.287 × 352.7 = 376.5 m/s v* = 0.287 × 352.7/1057 = 0.0958 m3/kg A* = × 0.0958/376.5 = 1.272 × 10-3m2 AE = 1.272 × 10-3 × 2.896 = 3.68 10-3m2 TE = 423.15 × 0.42517 = 179.9 K Assume Mx = then My = 0.57735, Poy/Pox = 0.72088, AE/A*x = 2.896 Ax/A*x = 1.6875, Ax/Ay* = 1.2225, AE/Ay* = 2.896 × 1.2225/1.6875 = 2.098 ⇒ ME = 0.293, PE/Poy = 0.94171 PE = 0.94171 × 0.72088 × 2.0 = 1.357 MPa, OK close to the 1.4 MPa TE = 0.98298 × 423.15 = 416 K, m = kg/s 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 17.84 At what Mach number will the normal shock occur in the nozzle of Problem 17.53 if the back pressure is 1.4 MPa? (trial and error on Mx) Relate the inlet and exit conditions to the shock conditions with reversible flow before and after the shock It becomes trial and error Assume Mx = 1.8 ⇒ My = 0.6165 ; Poy/Pox = 0.8127 AE/A*x = A2/A* = 0.002435/0.001516 = 1.6062 Ax/A*x = 1.439 ; Ax/A*y = 1.1694 AE/Ay* = (AE/A*x)(Ax/Ay*)/(Ax/A*x) = ⇒ ME = 0.5189 ; PE/Poy = 0.8323 1.6062(1.1694) = 1.3053 1.439 PE = (PE/Poy)(Poy/Pox)Pox = 0.8323 × 0.8127 × 2000 = 1353 kPa < 1.4 MPa So select the mach number a little less Mx = 1.7 ⇒ My = 0.64055 ; Poy/Pox = 0.85573 Ax/A*x = 1.3376 ; Ax/Ay* = 1.1446 AE/Ay* = (AE/A*x)(Ax/Ay*)/(Ax/A*x) = ⇒ ME = 0.482 ; PE/Poy = 0.853 1.6062(1.1446) = 1.3744 1.3376 www.elsolucionario.org PE = (PE/Poy)(Poy/Pox)Pox = 0.853 × 0.85573 × 2000 = 1459.9 kPa Now interpolate between the two Mx = 1.756 and we check ⇒ My = 0.6266 ; Poy/Pox = 0.832 Ax/A*x = 1.3926 ; Ax/Ay* = 1.1586 AE/Ay* = 1.6062 × 1.1586/1.3926 = 1.3363 ⇒ ME = 0.5 ; PE/Poy = 0.843 PE = 0.843 × 0.832 × 2000 = 1402.7 kPa 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 www.elsolucionario.org Borgnakke and Sonntag Solution using the Pr or vr functions 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 17.44 A jet plane travels through the air with a speed of 1000 km/h at an altitude of km, where the pressure is 40 kPa and the temperature is −12°C Consider the inlet diffuser of the engine where air leaves with a velocity of 100 m/s Determine the pressure and temperature leaving the diffuser, and the ratio of inlet to exit area of the diffuser, assuming the flow to be reversible and adiabatic RT 0.287 × 261.15 V = 1000 km/h = 277.8 m/s, v1 = P = = 1.874 m3/kg 40 h1 = 261.48 kJ/kg, Pr1 = 0.6862 ho1 = 261.48 + 277.82/2000 = 300.07 kJ/kg ⇒ To1 = 299.7 K, Pro1 = 1.1107 The ratio of the pressures equals the ratio of the Pr functions when s = constant Po1 = P Pro1/ Pr1 = 40 × 1.1107/0.6862 = 64.74 kPa h2 = 300.07 - 1002/2000 = 295.07 ⇒ T2 = 294.7 K, Pr2 = 1.0462 P2 = 64.74 × 1.0462/1.1107 = 61 kPa v2 = RT2/P2 = 0.287 × 294.7/61 = 1.386 m3/kg www.elsolucionario.org A1/A2 = (v1/v2)(V2/V1) = (1.874/1.386)(100/277.8) = 0.487 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 www.elsolucionario.org Borgnakke and Sonntag 17.73 Repeat Problem 17.44 assuming a diffuser efficiency of 80% RT 0.287 × 261.15 = 1.874 m3/kg V = 1000 km/h = 277.8 m/s, v1 = P = 40 h1 = 261.48 kJ/kg, Pr1 = 0.6862 ho1 = 261.48 + 277.82/2000 = 300.07 kJ/kg ⇒ To1 = 299.7 K, Pro1 = 1.1107 Same as problem 17.44, except ηD = 0.80 We thus have from 17.44 h3 - h1 h h3 - 261.48 ho1 - h1 = 300.07 - 261.48 = 0.8 ⇒ h3 = 292.35 kJ/kg, Pr3 = 1.0129 Po2 = P3 = 40 × 1.0129/0.6862 = 59.04 kPa 01 02 s Pro2 = Pro1 = 1.1107 h2 = 300.07 - 1002/2000 = 295.07 kJ/kg ⇒ T2 = 294.7 K, Pr2 = 1.0462 P2 = Po2 Pr2 / Pro2 = 59.04 × 1.0462/1.1107 = 55.6 kPa v2 = RT2/P2 = 0.287 × 294.7/55.6 = 1.521 m3/kg A1/A2 = (v1/v2)(V2/V1) = (1.874/1.521) (100/277.8) = 0.444 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 ... the temperature increases, what happens to the density and specific volume? Solution: The density is seen to decrease as the temperature increases ∆ρ = – ∆T/2 Since the specific volume is the... shown) The steam leaves at a lower pressure to the condenser (heat exchanger) at state A rotating shaft gives a rate of energy (power) to the electric generator set W T 2.b Take a control volume around... energy transfers that enter or leave the C.V Solution: Hot steam from generator c Electric power gen cb WT Welectrical The electrical power also leaves the C.V to be used for lights, instruments