Claus borgnakke, richard e sonntag fundamentals of thermodynamics, instructor solution manual wiley (2012)

Borgnakke Sonntag Fundamentals of Thermodynamics SOLUTION MANUAL CHAPTER 8e Updated June 2013 www.elsolucionario.org Borgnakke and Sonntag CONTENT CHAPTER SUBSECTION Concept Problems Properties, Units and Force Specific Volume Pressure Manometers and Barometers Energy and Temperature Review problems PROB NO 1-21 22-37 38-44 45-61 62-83 84-95 96-101 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Borgnakke and Sonntag In-Text Concept Questions Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful www.elsolucionario.org Borgnakke and Sonntag 1.a Make a control volume around the turbine in the steam power plant in Fig 1.2 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 WT 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 1.b Take a control volume around your kitchen refrigerator and indicate where the components shown in Figure 1.3 are located and show all flows of energy transfers 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 www.elsolucionario.org Borgnakke and Sonntag 1.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 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 1.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 1.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 www.elsolucionario.org Borgnakke and Sonntag 1.f Can I always neglect ∆P in the fluid above location A in figure 1.13? 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 1.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 H Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Borgnakke and Sonntag Concept Problems Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Borgnakke and Sonntag Flow Through Nozzles, Shocks 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 15.92E Steam flowing at 50 ft/s 200 psia, 600 F expands to 150 psia in a converging nozzle Find the exit velocity and area ratio Ae / Ai Solve the problem with the steam tables Inlet state: vi = 3.058 ft3/lbm, hi = 1322.05 Btu/lbm, si = 1.6767 Btu/lbm-R Exit state: (Pe,se = si) ve = 3.8185 ft3/lbm, he = 1290.69 Btu/lbm Energy Eq.: Ve = V2i / + hi = V2e / + he ; V2e = V2i + 2(hi − he) 50 × 50 + × 25037 × (1322.05 − 1290.69) = 1254 ft/s Recall conversion Btu/lbm = 25 037 ft2/s2 (= 32.174 × 778.1693) Same mass flow rate so 3.8185 50 Ae/Ai = (ve/vi)(Vi/Ve) = 3.058 × 1254 = 0.0498 If we solved as ideal gas with constant specific heat we get (k = 1.327) (k-1)/k Te = Ti (Pe/Pi) Ve = V2i + 2Cp(Ti − Te) = = 1274.9 ft/s 0.2464 = 1059.7 (150/200) = 987.2 R 50 ×50 + ×0.447 ×25037(1059.7 − 987.2) 1/k 50 2000.7536 × 1274.9 Ae/Ai = (ve/vi)(Vi/Ve) = (Pi/Pe) (Vi/Ve) = 150   = 0.0487 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 15.93E A convergent nozzle has a minimum area of ft2 and receives air at 25 lbf/in.2, 1800 R flowing with 330 ft/s What is the back pressure that will produce the maximum flow rate and find that flow rate? P* k k-1 = ( Po k+1) = 0.528 Critical Pressure Ratio E E A A A A E Find Po: Cp = (463.445 - 449.794)/50 = 0.273 Btu/lbm-R from table F.5 ⇒ T0 = Ti + V2/2Cp h0 = h1 + V21/2 A E AE A A 3302/2 T0 = 1800 + = 1807.97 => T* = 0.8333 To = 1506.6 R 25 037 × 0.273 Recall conversion Btu/lbm = 25 037 ft2/s2 (= 32.174 × 778.1693) A A E EA A E E A E A A E A A A P0 = Pi (T0/Ti)k/(k-1) = 25 × (1807.97/1800)3.5 = 25.39 lbf/in.2 E A E A A P* = 0.528 Po = 0.528 × 25.39 = 13.406 lbf/in2 E A A E A P* 13.406 × 144 ρ = * = 53.34 × 1506.6 = 0.024 lbm/ft RT * A A E A E A A EA A A A A E A E E V=c= A EA AE kRT* = A AE A 1.4 × 53.34 × 1506.6 × 32.174 = 1902.6 ft/s EA m = ρAV = 0.024 × × 1902.6 = 45.66 lbm/s E A A 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 15.94E A jet plane travels through the air with a speed of 600 mi/h at an altitude of 20000 ft, where the pressure is 5.75 lbf/in.2 and the temperature is 25 F Consider the diffuser of the engine where air leaves at with a velocity of 300 ft/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 E A A V = 600 mi/h = 880 ft/s v1 = 53.34 × 484.67/(5.75 × 144) = 31.223 ft3/lbm, h1 = 115.91 Btu/lbm, E A A ho1 = 115.91 + 8802/(2 × 25 037) = 131.38 Btu/lbm E A A Recall conversion Btu/lbm = 25 037 ft2/s2 (= 32.174 × 778.1693) Table F.5 ⇒ To1 = 549.2 R, E A E A A A Po1 = P1 (To1/T1)k/(k-1) = 5.75 × (549.2/484.67)3.5 = 8.9 lbf/in.2 E A E A A A E A h2 = 131.38 - 3002/(2 × 32.174 × 778) = 129.58 Btu/lbm T2 = 542 R, => E A A P2 = Po1 (T2/To1)k/(k-1) = 8.9 × (542/549.2)3.5 = 8.5 lbf/in.2 E A E A A A A E A v2 = 53.34 × 542/(8.5 × 144) = 23.62 ft3/lbm A1/A2 = (v1/v2)(V2/V1) = (31.223/23.62)(300/880) = 0.45 E A A 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 15.95E An air flow at 90 psia, 1100 R, M = 0.3 flows into a convergent-divergent nozzle with M = at the throat Assume a reversible flow with an exit area twice the throat area and find the exit pressure and temperature for subsonic exit flow to exist To find these properties we need the stagnation properties from the inlet state From Table A.12: Mi = 0.3: Pi/Po = 0.93947, Ti/To = 0.98232 Po = 90 / 0.93947 = 95.8 psia, To = 1100 / 0.98232 = 1119.8 R This flow is case c in Figure 15.13 From Table A.12: AE/A* = E A A PE/Po = 0.9360, TE/To = 0.98127 PE = 0.9360 Po = 0.936 × 95.8 = 89.67 psia TE = 0.98127 To = 0.98127 × 1119.8 = 1099 R 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 15.96E Air is expanded in a nozzle from 300 lbf/in.2, 1100 R to 30 lbf/in.2 The mass flow rate through the nozzle is 10 lbm/s Assume the flow is reversible and adiabatic and determine the throat and exit areas for the nozzle E A A Mach #  k P* = Po k+1k-1   = 300 × 0.5283 = 158.5 lbf/in.2 T* = To× 2/(k+1) = 1100 × 0.8333 = 916.6 R E A E A A E A E A Velocity A Area A v* = RT*/P* = 53.34 × 916.6/(158.5 × 144) = 2.1421 ft3/lbm E Density A E A E A A E A A A E A A P 300 psia 30 psia The critical speed of sound is c* = kRT* = E A A A EA A 1.4 × 32.174 × 53.34 × 916.6 = 1484 ft/s EA A* = mv*/c* = 10 × 2.1421/1484 = 0.0144 ft2 P2/Po = 30/300 = 0.1 Table A.11 ⇒ M2* = 1.701 = V2/c* E E A E A A A A E A A E A A A E A AE E A A We used the column in Table A.12 with mach no based on throat speed of sound V2 = 1.701 × 1484 = 2524 ft/s T2 = 916.6 × 0.5176 = 474.4 R v2 = RT2/P2 = 53.34 × 474.4/(30 × 144) = 5.8579 ft3/lbm E A A2 = mv2/V2 = 10 × 5.8579 / 2524 = 0.0232 ft2 A E A A A E A 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 15.97E A 50-ft3 uninsulated tank contains air at 150 lbf/in.2, 1000 R The tank is now discharged through a small convergent nozzle to the atmosphere at 14.7 lbf/in.2 while heat transfer from some source keeps the air temperature in the tank at 1000 R The nozzle has an exit area of × 10−4 ft2 a Find the initial mass flow rate out of the tank b Find the mass flow rate when half the mass has been discharged c Find the mass of air in the tank and the mass flow rate out of the tank when the nozzle flow changes to become subsonic E A A E A A E A A AIR P e cb PB/Po = 14.7/150 = 0.098 < (P*/Po)crit = 0.5283 E A A a The flow is choked, max possible flow rate ME =1 ; PE = 0.5283 × 150 = 79.245 lbf/in.2 A E A TE = T* = 0.8333 × 1000 = 833.3 R E A A VE = c = A kRT* = EA 1.4 × 53.34 × 833.3 × 32.174 = 1415 ft/s A EA vE = RT /PE = 53.34 × 833.3/(79.245 × 144) = 3.895 ft3/lbm * E A E A A A Mass flow rate is : m1 = AVE/vE = × 10-4 × 1415/3.895 = 0.0727 lbm/s E E A A A A b m1 = P1V/RT1 = 150 × 50 × 144/53.34 × 1000 = 20.247 lbm m2 = m1/2 = 10.124 lbm, P2 = P1/2 = 75 lbf/in.2 ; T2 = T1 E A A PB/P2= 14.7/75 = 0.196 < (P*/Po)crit E A A The flow is choked and the velocity is the same as in a) PE = 0.5283 × 75 = 39.623 lbf/in.2 ; ME =1 E A A × 10-4 × 1415 × 39.623 × 144 m2 = AVEPE/RTE = = 0.0303 lbm/s 53.34 × 1000 E A A A A E E 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 c Flow changes to subsonic when the pressure ratio reaches critical PB/Po = 0.5283 P3 = 27.825 lbf/in.2 E A A m3 = m1P3/P1 = 3.756 lbm ; T3 = T1 ⇒ VE = 1415 ft/s × 10-4 × 1415 × 27.825 × 144 m3 = AVEPE/RTE = = 0.02125 lbm/s 53.34 × 1000 E A A A A E E 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 15.98E A flow of helium flows at 75 psia, 900 R with 330 ft/s into a convergent-divergent nozzle Find the throat pressure and temperature for reversible flow and M = at the throat We need to find the stagnation properties first ( k = 1.667 ) T0 = T1 + V21/2Cp = 900 + 3302/(2 × 25037 × 1.24) = 901.754 R E A AE A A Recall conversion Btu/lbm = 25 037 ft2/s2 (= 32.174 × 778.1693) E A E A A A 2.5 P0 = P1 (T0/T1)k/(k-1) = 75 (901.754/900) E E A A A A = 75.366 psia From the analysis we get Eqs.15.37-38  k/(k-1)  2.5 P* = P0 k + 1 = 75.366 1.667 + 1 = 36.7 psia     2 T* = T0 k + = 901.754 × 1.667 + = 676.2 R E E E A A A A A A E A A A A E A A E 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 15.99E The products of combustion enter a nozzle of a jet engine at a total pressure of 18 lbf/in.2, and a total temperature of 1200 F The atmospheric pressure is 6.75 lbf/in.2 The nozzle is convergent, and the mass flow rate is 50 lbm/s Assume the flow is adiabatic Determine the exit area of the nozzle E A A E A A Pcrit = P2 = 18 × 0.5283 = 9.5 lbf/in.2 > Pamb The flow is then choked T2 = 1660 × 0.8333 = 1382 R E A V2 = c2 = kRT = A EA A A 1.4 × 32.174 × 53.34 × 1382 = 1822 ft/s EA v2 = 53.34 × 1382/9.5 × 144 = 53.9 ft /lbm A E A A2 = m v2/ V2 = 50 × 53.9/1822 = 1.479 ft2 E A A A E A 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 15.100E A normal shock in air has upstream total pressure of 75 psia, stagnation temperature of 900 R and Mx = 1.4 Find the downstream stagnation pressure From the normal shock relations in Section 15.8 found in Table A.13 we get Mx = 1.4: Po y/Po x = 0.95819 Po y = 0.95819 Po x = 0.95819 × 75 = 71.86 psi Remark: The stagnation temperature would be unchanged (energy equation) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful www.elsolucionario.org Borgnakke and Sonntag Nozzles, Diffusers and Orifices 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 15.101E Air enters a diffuser with a velocity of 600 ft/s, a static pressure of 10 lbf/in.2, and a temperature of 20 F The velocity leaving the diffuser is 200 ft/s and the static pressure at the diffuser exit is 11.7 lbf/in.2 Determine the static temperature at the diffuser exit and the diffuser efficiency Compare the stagnation pressures at the inlet and the exit V21 To1 = T1 + 2g C = 480 + 6002/(2 × 25 037 × 0.24) = 510 R E A A E A A AE E A E A A c p Recall conversion Btu/lbm = 25 037 ft2/s2 (= 32.174 × 778.1693) To1 - T1 k-1 Po1 - P1 = k ⇒ Po1 - P1 = 2.1875 ⇒ Po1 = 12.2 lbf/in.2 T P E A A A A A To2 = To1 E A A E ⇒ T2 = To2 - V22/2Cp = 510 - 2002/(2 × 25 037 × 0.24) = 506.7 R E A To1 - T1 T1 To2 - T2 T2 = AE A A k-1 Po1 - P1 ⇒ Po1 - P1 = 2.1875 ⇒ Po1 = 12.2 lbf/in.2 P1 k A A E A A E k-1 Po2 - P2 = k P2 A ⇒ Po2 - P2 = 0.267 ⇒ Po2 = 11.97 lbf/in.2 A A E E A Tex,s = T1 (Po2/P1)(k-1)/k = 480 R × 1.0528 = 505.3 R E A A Tex,s - T1 505.3 - 480 ηD = T - T = 51 - 480 = 0.844 o1 A A E 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 A E www.elsolucionario.org Borgnakke and Sonntag 15.102E Repeat Problem 15.94 assuming a diffuser efficiency of 80% From solution to 15.94 h1 = 115.91 Btu/lbm, v1 = 31.223 ft3/lbm h 01 E A A ho1 = 115.91 + 8802/(2 × 25 037) = 131.38 Btu/lbm Recall conversion Btu/lbm = 25 037 ft2/s2 (= 32.174 × 778.1693) Table F.5 ⇒ To1 = 549.2 R, 02 E A A E A ηD = (h3 - h1)/(ho1 - h1) = 0.8 A A E A s ⇒ h3 = 128.29 Btu/lbm, T3 = 536.29 R Po2 = P3 = P1 (Τ3/Τ1)k/(k-1) = 5.75 × (536.29/484.67)3.5 = 8.194 lbf/in.2 E A E A A A A E A To2 = To1 = 549.2 R h2 = 131.38 - 3002/(2 × 25 037) = 129.58 Btu/lbm E A A T2 = 542 R, => P2 = Po2 (T2/To1)k/(k-1) = 8.194 × (542/549.2)3.5 = 7.824 lbf/in.2 E A ⇒ v2 = E A A A A E A 53.34 × 542 = 25.66 ft3/lbm 7.824 × 144 E A A A A E A1/A2 = v1V2/v2V1 = 31.223 × 300/(25.66 × 880) = 0.415 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 15.103E A convergent nozzle with exit diameter of in has an air inlet flow of 68 F, 14.7 lbf/in.2 (stagnation conditions) The nozzle has an isentropic efficiency of 95% and the pressure drop is measured to 20 in water column Find the mass flow rate assuming compressible adiabatic flow Repeat calculation for incompressible flow E A A Convert ∆P to lbf/in2 E A ∆P = 20 in H2O = 20 × 0.03613 = 0.7226 lbf/in2 E A T0 = 68 F = 527.7 R P0 = 14.7 lbf/in2 Assume inlet Vi = Pe = P0 - ∆P = 14.7 - 0.7226 = 13.977 lbf/in2 E A A Pe k-1 13.977 Te = T0 (P ) k = 527.7 ×( 14.7 )0.2857 = 520.15 R E A E A A A A A E V2e/2 = hi - he = Cp (Ti - Te) = 0.24 × (527.7 - 520.15) = 1.812 Btu/lbm A AE Ve 2ac/2 = η V2e/2 = 0.95 × 1.812 = 1.7214 Btu/lbm A AE E A ⇒ Ve ac = AE × 25 037 × 1.7214 = 293.6 ft/s A EA Recall conversion Btu/lbm = 25 037 ft2/s2 (= 32.174 × 778.1693) E A Ve 2ac/2 A Te ac = Ti - = 527.7 E A A 1.7214 = 520.53 R 0.24 E Cp A Pe ρe ac = RT AE E A A A E 13.977 × 144 = 0.07249 lbm/ft3 53.34 × 520.53 = A e ac A E A E π m = ρAV = 0.07249 × × ( )2 × 293.6 = 0.116 lbm/s 12 P0 14.7 × 144 Incompressible: ρi = RT = = 0.0752 lbm/ft3 53.34 × 527.7 E E A A A A A A E A A E A A E A E V2e/2 = vi (Pi - Pe) = A AE ∆P 0.7226 × 144 = = 1.7785 Btu/lbm ρi 0.0752 × 778 A A E Ve 2ac/2 = η V2e/2 = 0.95 × 1.7785 = 1.6896 Btu/lbm A AE E ⇒ Ve ac = A A AE × 25 037 × 1.6896 = 290.84 ft/s EA π m = ρAV = 0.0752 × × (12)2 × 290.84 = 0.119 lbm/s E E A A A A E A A A A E 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 E ... 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... 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 WT Excerpts from this work may be... Borgnakke and Sonntag 1.17 What is the lowest temperature in degrees Celsuis? In degrees Kelvin? Solution: The lowest temperature is absolute zero which is at zero degrees Kelvin at which point the