SOLUION MANUAL ENGLISH UNI PROBLEMS CHAPER SONNAG BORGNAKKE VAN WYLEN FUNDAMENALS of hermodynamic Sixth Edition
CHAPER SUBSECION PROB NO. Rankine Cycle 67-8 Brayton Cycle 8-87 Otto, Dieel, Stirling and Carnot Cycle 88-0 Refrigeration Cycle 0-07 Availability and Combined Cycle 08- Review Problem -7 Correpondence Lit he correpondence between the new Englih unit problem et and the previou 5th edition chapter problem et and the current SI problem. New 5th SI New 5th SI New 5th SI 67 7 mod 8 new 7 0 8b 8 68 8 mod 85 7 0 new - 69 new 86 6 86 0 9 0 70 9 6 87 7 89 0 50 7 0 7 88 8 9 05 5 5 7 new 89 9 95 06 new 0 7 mod 90 new 97 07 new 7 7 mod 5 9 new 98 08 new 6 75 mod 7 9 0 09 55 7 76 5 mod 5 9 0 05 0 new 8 77 8 9 07 new 50 78 7 mod 55 95 09 5 79 8 mod 57 96 6-80 9 60 97 5 0 57 8 mod 66 98 6 5 60 8 7 99 7 6 6 5 60 8 new 7 00 8a 7 7 5
Rankine cycle Sonntag, Borgnakke and van Wylen
.67E A team power plant, a hown in Fig.., operating in a Rankine cycle ha aturated vapor at 600 lbf/in. leaving the boiler. he turbine exhaut to the condener operating at.5 lbf/in.. Find the pecific work and heat tranfer in each of the ideal component and the cycle efficiency. Solution: For the cycle a given: : h = 97.97 Btu/lbm, v = 0.065 ft /lbm, : h = h g = 0.06 Btu/lbm, = g =.6 Btu/lbm R C.V. Pump Reverible and adiabatic. Energy: w p = h - h ; Entropy: = ince incompreible it i eaier to find work (poitive in) a w P = v dp = v (P - P ) = 0.065(600.) =.8 Btu/lbm 778 h = h + w P = 97.97 +.8 = 99.77 Btu/lbm C.V. Boiler: q H = h - h = 0.06-99.77 = 0. Btu/lbm C.V. ubine: w = h - h, = = =.6 = 0.87 + x.79 => x = 0.7, h = 97.97 + 0.7 09.78 = 8.8 Btu/lbm w = 0.06-8.8 = 60. Btu/lbm η CYCLE = (w - w P )/q H = (60. -.8)/0. = 0.5 C.V. Condener: q L = h - h = 8.8-97.97 = 75.9 Btu/lbm Boiler urbine Q B W W P Condener Q
.68E Conider a olar-energy-powered ideal Rankine cycle that ue water a the working fluid. Saturated vapor leave the olar collector at 50 F, and the condener preure i 0.95 lbf/in.. Determine the thermal efficiency of thi cycle. H O ideal Rankine cycle CV: turbine State : able F.7. h = 9. Btu/lbm, =.579 Btu/lbm R = =.579 = 0.96 + x.856 => x = 0.785 h = 68.0 + 0.785 06.98 = 879.5 Btu/lbm w = h - h = 9. - 879.5 =.6 Btu/lbm w P = vdp v (P - P ) = 0.06(.5 0.95) = 0. Btu/lbm 778 w NE = w - w P =.6-0. =. Btu/lbm h = h + w P = 68.0 + 0. = 68. Btu/lbm q H = h - h = 9. - 68. =.7 Btu/lbm η H = w NE /q H =./.7 = 0.78 Solar collector Q RAD urbine W P W Condener Q
.69E A Rankine cycle ue ammonia a the working ubtance and powered by olar energy. It heat the ammonia to 0 F at 800 pia in the boiler/uperheater. he condener i water cooled, and the exit i kept at 70 F. Find (, P, and x if applicable) for all four tate in the cycle. NH ideal Rankine cycle State : able F.8., = 70 F, x = 0, P = 8.85 pia, CV Pump: h = 0. Btu/lbm, v = 0.6 ft /lbm w P = h - h = vdp v (P - P ) = 0.06(800 8.85) 778 =.7 Btu/lbm h = h + w P = 0. +.7 =.8 Btu/lbm = h f [we need the computer oftware to do better (P, = ) ] State : 0 F, 800 pia : uperheated vapor, =.95 Btu/lbm => = 7.8 F CV: turbine = =.95 = 0.59 + x 0.9589 => x = 0.9788 P = P = 8.85 pia, = = 70 F Solar collector Q RAD urbine W P W Condener Q
.70E A upply of geothermal hot water i to be ued a the energy ource in an ideal Rankine cycle, with R-a a the cycle working fluid. Saturated vapor R-a leave the boiler at a temperature of 80 F, and the condener temperature i 00 F. Calculate the thermal efficiency of thi cycle. Solution: CV: Pump (ue R-a able F.0) P = 8.9 pia, P = P = 00. pia h = 8.6 Btu/lbm, = 0.0 Btu/lbm R h = 08.86 Btu/lbm, v = 0.087 ft /lbm w P = h - h = vdp v (P -P ) CV: Boiler CV: urbine = 0.087(00. - 8.9) = 0.67 Btu/lbm 778 h = h + w P = 08.86 + 0.67 = 09.5 Btu/lbm q H = h - h = 8.6-09.5 = 7.8 Btu/lbm = = 0.0 x = (0.0-0.89)/0.7 = 0.9 h = 76.08 Btu/lbm, Energy Eq.: w = h - h = 8.76 Btu/lbm w NE = w - w P = 8.76-0.67 = 7.605 Btu/lbm η H = w NE / q H = 7.605/7.8 = 0.0 W Q H W P, in. Q L
.7E Do Problem.70 with R- a the working fluid. Standard Rankine cycle with propertie from the R- table, h = 9.67 Btu/lbm, v = 0.00 ft /lbm, P = 0.6 pia, P = P = 55.8 pia, h = 0.07 Btu/lbm, = 0.9 Btu/lbm R CV: Pump w P = v (P -P ) = 0.00 (55.8-0.6) = 0.89 Btu/lbm 778 CV: urbine = CV: Boiler h = h + w P = 9.67 + 0.89 = 0.6 Btu/lbm x = (0.9-0.079)/0.0 = 0.9 h = 0.885 Btu/lbm, w = h - h = 8.85 Btu/lbm q H = h - h = 0.07-0.6 = 69.9 Btu/lbm η H = (w w P )/q H = (8.85-0.89)/57. = 0.0 W Q H W P, in. Q L
.7E A maller power plant produce 50 lbm/ team at 00 pia, 00 F, in the boiler. It cool the condener with ocean water coming in at 55 F and returned at 60 F o that the condener exit i at 0 F. Find the net power output and the required ma flow rate of the ocean water. Solution: he tate propertie from able F.7. and F.7. : 0 F, x = 0: h = 78.0 Btu/lbm, v = 0.067 ft /lbm, P at =.8 pia : 00 pia, 00 F: h = 577. Btu/lbm, =.7989 Btu/lbm R C.V. Pump Reverible and adiabatic. Energy: w p = h - h ; Entropy: = ince incompreible it i eaier to find work (poitive in) a w p = v dp = v (P - P ) = 0.067 (00 -.) =.9 Btu/lbm 778 C.V. urbine : w = h - h ; = = =.7989 = 0.7 + x (.80) => x = 0.9 => h = 78.0 + 0.9 (0.8) = 08.95 Btu/lbm w = 577. 08.95 = 558.5 Btu/lbm Ẇ NE = ṁ(w w p ) = 50 (558.5.9) = 7 866 Btu/ C.V. Condener : q L = h - h = 08.95-78.0 = 90.9 Btu/lbm Q. L = ṁq L = 50 90.9 = 7 07 Btu/ = ṁ ocean C p ṁ ocean = Q. L / C p = 7 07 / (.0 5) = 909 lbm/ Boiler urbine Q B W W P Condener Q
.7E he power plant in Problem.67 i modified to have a uperheater ection following the boiler o the team leave the uper heater at 600 lbf/in., 700 F. Find the pecific work and heat tranfer in each of the ideal component and the cycle efficiency. Solution: For thi cycle from able F.7 State : Superheated vapor h = 50.6 Btu/lbm, =.587 Btu/lbm R, State : Saturated liquid h = 97.97 Btu/lbm, v = 0.065 ft /lbm C.V. Pump: Adiabatic and reverible. Ue incompreible fluid o w P = v dp = v (P - P ) = 0.065(600.) =.8 Btu/lbm 778 h = h + w P = 95.8 Btu/lbm C.V. Boiler: q H = h - h = 50.6-97.97 = 5.65 Btu/lbm C.V. ubine: w = h - h, = = =.587 Btu/lbm R = 0.87 + x.79 x = 0.87, h = 97.97 + 0.87 09.78 = 96.75 Btu/lbm w = 50.6-96.75 =.87 Btu/lbm η CYCLE = (w - w P )/q H = (.87 -.8)/5.65 = 0.7 C.V. Condener: q L = h - h = 96.75-97.97 = 88.8 Btu/lbm P v
.7E Conider a imple ideal Rankine cycle uing water at a upercritical preure. Such a cycle ha a potential advantage of minimizing local temperature difference between the fluid in the team generator, uch a the intance in which the high-temperature energy ource i the hot exhaut ga from a gaturbine engine. Calculate the thermal efficiency of the cycle if the tate entering the turbine i 8000 lbf/in., 00 F, and the condener preure i 0.95 lbf/in.. What i the team quality at the turbine exit? Solution: For the efficiency we need the net work and team generator heat tranfer. State : = 0.96 Btu/lbm R, h = 68.0 Btu/lbm State : h = 57.5 Btu/lbm, =.78 Btu/lbm R C.V. Pump. For thi high exit preure we ue able F.7. to get tate. Entropy Eq.: = => h = 9.69 Btu/lbm w p = h - h = 9.69 68.0 =.65 Btu/lbm C.V. urbine. Aume reverible and adiabatic. Entropy Eq.: = =.78 = 0.96 + x.856 Steam generator: x = 0.75 h = 68.0 + x 06.98 = 75.9 Btu/lbm, w = h - h = 796. Btu/lbm Very low for a turbine exhaut q H = h - h = 57.5 9.69 = 55.8 Btu/lbm w NE = w w p = 796..65 = 77.6 Btu/lbm η = w NE /q H = 77.6 / 55.8 = 0.5 P 8000 pia v 0.95 pia
.75E Conider an ideal team reheat cycle in which the team enter the high-preure turbine at 600 lbf/in., 700 F, and then expand to 50 lbf/in.. It i then reheated to 700 F and expand to.5 lbf/in. in the low-preure turbine. Calculate the thermal efficiency of the cycle and the moiture content of the team leaving the low-preure turbine. Solution: Baic Rankine cycle with a reheat ection. For thi cycle from able F.7 State : Superheated vapor h = 50.6 Btu/lbm, =.587 Btu/lbm R, State : Saturated liquid h = 97.97 Btu/lbm, v = 0.065 ft /lbm C.V. Pump: Adiabatic and reverible. Ue incompreible fluid o w P = v dp = v (P - P ) 5 = 0.065(600.) =.8 Btu/lbm 778 h = h + w P = 95.8 Btu/lbm 6 C.V. ubine : w = h - h, = = =.587 Btu/lbm R => h = 08.9 Btu/lbm w = 50.6-08.9 =.69 Btu/lbm C.V. ubine : w = h 5 - h 6, 6 = 5 State 5: h 5 = 76.55 Btu/lbm, 5 =.7568 Btu/lbm R State 6: 6 = 5 =.7568 = 0.87 + x 6.79 => x 6 = 0.909 h 6 = 97.97 + 0.909 09.78 = 06.89 Btu/lbm w = 76.55 06.89 = 9.66 Btu/lbm w,tot = w + w =.69 + 9.66 = 9.5 Btu/lbm C.V. Boiler: q H = h - h = 50.6-97.97 = 5.65 Btu/lbm q H = q H + h 5 - h = 5.65 + 76.55 08.9 = 0. Btu/lbm η CYCLE = (w,tot - w P )/q H = (9.5.8)/0. = 0.5
.76E Conider an ideal team regenerative cycle in which team enter the turbine at 600 lbf/in., 700 F, and exhaut to the condener at.5 lbf/in.. Steam i extracted from the turbine at 50 lbf/in. for an open feedwater heater. he feedwater leave the heater a aturated liquid. he appropriate pump are ued for the water leaving the condener and the feedwater heater. Calculate the thermal efficiency of the cycle and the net work per pound-ma of team. From able F.7. h 5 = 50.6 Btu/lbm, 5 5 =.587 Btu/lbm R h = 97.97 Btu/lbm, v = 0.065 ft /lbm Interpolate to get h = 0.67 Btu/lbm, v = 0.0809 ft /lbm S. GEN. P FW HR URBINE. 6 7 COND. P C.V. Pump: w P = 0.065(50.) 778 = 0. Btu/lbm = h h h = h + w P = 98. Btu/lbm C.V. Pump: 600 pi 5 50 pi 6. pi 7 w P = 0.0809(600-50)/778 =.507 Btu/lbm h = h + w P =.8 Btu/lbm C.V. urbine (high preure ection) nd law: 6 = 5 =.587 Btu/lbm R => h 6 = 08.9 Btu/lbm CV: feedwater heater, call the extraction fraction y = ṁ 6 /ṁ Continuity Eq.: ṁ = ṁ 6 + ṁ, Energy Eq.: ṁ 6 h 6 + ṁ h = ṁ h y 6 h 6 + ( - y 6 )h = h y 6 = (h h )/(h 6 h ) y 6 = (0.67 98.)/(08.9 98.) = 0.09
CV: urbine from 5 to 7 7 = 5 x 7 = (.587-0.87)/.79 = 0.87 h 7 = 97.97 + 0.87 09.78 = 96.75 Btu/lbm w = (h 5 - h 6 ) + ( - y 6 )(h 6 - h 7 ) = (50.6 08.9) + 0.7909(08.9-96.75) = 6.87 Btu/lbm CV: pump w P = ( - y 6 )w P + w P = 0.7909 0. +.507 =.855 Btu/lbm w NE = w - w P = 6.87 -.855 = 6.0 Btu/lbm CV: team generator q H = h 5 - h = 50.6.8 = 08. Btu/lbm η H = w NE /q H = 6/08. = 0.56
.77E A cloed feedwater heater in a regenerative team power cycle heat 0 lbm/ of water from 00 F, 000 lbf/in. to 50 F, 000 lbf/in.. he extraction team from the turbine enter the heater at 600 lbf/in., 550 F and leave a aturated liquid. What i the required ma flow rate of the extraction team? 6 6a C.V. Feedwater Heater From the team table F.7: F.7.: h = 7.6 Btu/lbm F.7.: h =. Btu/lbm F.7.: h 6 = 55.6 Btu/lbm Interpolate for thi tate F.7.: h 6a = 7.56 Btu/lbm Energy Eq.: ṁ h + ṁ 6 h 6 = ṁ h + ṁ 6 h 6a Since all four tate are known we can olve for the extraction flow rate ṁ 6 = ṁ h - h 7.6 -. lbm = 0 =. h 6a - h 6 7.56-55.6
.78E A team power cycle ha a high preure of 600 lbf/in. and a condener exit temperature of 0 F. he turbine efficiency i 85%, and other cycle component are ideal. If the boiler uperheat to 00 F, find the cycle thermal efficiency. State : State : h = 79.5 Btu/lbm, =.897 Btu/lbm R h = 78.0 Btu/lbm, v = 0.067 ft /lbm C.V. Pump: w P = vdp v (P - P ) = h h = 0.067(600.8) /778 =.79 Btu/lbm h = h + w P = 78.0 +.79 = 79.8 Btu/lbm C.V. urb.: w = h - h, = +,GEN Ideal: S = =.897 Btu/lbm R = 0.7 + x S.80 => x S = 0.905, h S = 78.0 + x S 0.8 = 07.9 Btu/lbm => w,s = 79.5-07.9 = 69.58 Btu/lbm Actual: C.V. Boiler: w,ac = η w,s = 0.85 69.58 = 587.8 Btu/lbm q H = h - h = 79.5 79.8 = 659.7 Btu/lbm η = (w,ac - w P )/q H = (587.8 -.79)/659.7 = 0.5 P ac ac v
.79E he team power cycle in Problem.67 ha an ientropic efficiency of the turbine of 85% and that for the pump it i 80%. Find the cycle efficiency and the pecific work and heat tranfer in the component. State numbered a in fig. of text. CV Pump: w P,S = v (P - P ) = 0.065(600.)/778 =.8 Btu/lbm w P,AC =.8/0.8 =.5 Btu/lbm h = h + w P,AC = 97.97 +.5 = 00. Btu/lbm CV urbine: w,s = h - h, = =.6 Btu/lbm R = =.6 = 0.87 + x.79 => x = 0.7, h = 97.97 + 0.7 09.78 = 8.8 Btu/lbm w,s = 0.06-8.8 = 60. Btu/lbm w,ac = h - h AC = 60. 0.85 = 06. h AC = 897.86 Btu/lbm (till two-phae) CV Boiler: q H = h - h = 0.06-00. = 0.9 Btu/lbm q L = h AC - h = 897.86-97.97 = 799.9 Btu/lbm η CYCLE = (w - w P )/q H = (06. -.5)/0.9 = 0.75 Compared to (60.-.8)/0. = 0.5 in the ideal cae. Boiler urbine Q B W P Condener W Q ac tate and ac nearly the ame
.80E Steam leave a power plant team generator at 500 lbf/in., 650 F, and enter the turbine at 90 lbf/in., 65 F. he ientropic turbine efficiency i 88%, and the turbine exhaut preure i.7 lbf/in.. Condenate leave the condener and enter the pump at 0 F,.7 lbf/in.. he ientropic pump efficiency i 80%, and the dicharge preure i 50 lbf/in.. he feedwater enter the team generator at 50 lbf/in., 00 F. Calculate the thermal efficiency of the cycle and the entropy generation of the flow in the line between the team generator exit and the turbine inlet, auming an ambient temperature of 77 F. S. GEN. 6 5 URBINE. η = 0.88 P COND. 500 pia 90 pia 650 F 65 F 5.7 pia 5 6 η S = 0.88, η SP = 0.80 h = 8.0, h =.0 Btu/lbm S = =.575 = 0.68 + x S.7686 => x S = 0.7975 h S = 88. + 0.797 5 05. = 905.8 Btu/lbm w S = h - h S =.0-905.8 = 08. Btu/lbm w = η S w S = 0.88 08. = 59. Btu/lbm h = h - w =.0-59. = 95.8 Btu/lbm w SP = 0.0666(50-.7) =.55 Btu/lbm 778 w p = w SP /η SP =.55/0.80 =.9 Btu/lbm q H = h - h 6 = 8.0-68. = 59.9 Btu/lbm η H = w NE /q H = (59. -.9)/59.9 = 0.8 C.V. Line from to : w = /0, Energy Eq.: q = h - h = - 8 = - Btu/lbm Entropy Eq.: + gen + q/ 0 = => gen = - -q/ 0 =.575 -.586 - (-/56.7) = 0.05 Btu/lbm R
.8E A boiler deliver team at 500 lbf/in., 000 F to a two-tage turbine a hown in Fig..7. After the firt tage, 5% of the team i extracted at 00 lbf/in. for a proce application and returned at 50 lbf/in., 90 F to the feedwater line. he remainder of the team continue through the low-preure turbine tage, which exhaut to the condener at.5 lbf/in.. One pump bring the feedwater to 50 lbf/in. and a econd pump bring it to 500 lbf/in.. Aume the firt and econd tage in the team turbine have ientropic efficiencie of 85% and 80% and that both pump are ideal. If the proce application require 5000 Btu/ of power, how much power can then be cogenerated by the turbine? : h = 90., =.600 Btu/lbmR C.V. urbine : Rev and adiabatic S = able F.7. Sup. vapor h S = 6.6 Btu/lbm w,s = h - h S =.7 Btu/lbm w,ac = 07.5 Btu/lbm h AC = h - w,ac = 8.6 ac: P, h AC AC =.68 Btu/lbm R P B Proc. 5000 Btu/ 7 P 6 5 C.68 0.87 5: 5S = AC x 5S = = 0.8.79 h 5S = 97.97 + x 5S 09.78 = 957.0 Btu/lbm w,s = h AC - h 5S = 6. Btu/lbm w,ac = 60.9 = h AC - h 5AC h 5AC = 0. Btu/lbm 7: Compreed liquid ue at. liq. ame : h 7 = 58.0 Btu/lbm; C.V. proce unit. Aume no work only heat out. q PROC = h AC - h 7 = 5. Btu/lbm ṁ = Q. /q PROC = 5000/5. =. lbm/ = 0.5 ṁ O ṁ O = ṁ = 7.776 lbm/, C.V. otal turbine ṁ 5 = ṁ - ṁ =. lbm/ Ẇ = ṁ h - ṁ h AC - ṁ 5 h 5AC = 760 Btu/
Brayton Cycle.8E A large tationary Brayton cycle ga-turbine power plant deliver a power output of 00 000 hp to an electric generator. he minimum temperature in the cycle i 50 R, and the maximum temperature i 900 R. he minimum preure in the cycle i atm, and the compreor preure ratio i to. Calculate the power output of the turbine, the fraction of the turbine output required to drive the compreor and the thermal efficiency of the cycle? Brayton: ẇ NE = 00 000 hp P = atm, = 50 R P /P =, = 900 R Solve uing contant C P0 : P P = atm Compreion in compreor: = Implemented in Eq.8. = ( P P ) k- k = 50() 0.86 = 8.6 R w C = h - h = C P0 ( - ) = 0. (8.6-50) = 6. Btu/lbm Expanion in turbine: = Implemented in Eq.8. = ( P P ) k- k = 900( )0.86 = 6. R w = h - h = C P0 ( - ) = 0.(900-6.) = 68.8 Btu/lbm w NE = w - w C = 68.8-6. =.7 Btu/lbm ṁ = Ẇ NE /w NE = 00 000 5/.7 = lbm/h Ẇ = ṁw = 65 600 hp, w C /w = 0.96 Energy input i from the combutor q H = C P0 ( - ) = 0.(900-8.6) = 0. Btu/lbm η H = w NE /q H =.7/0. = 0.50
.8E A Brayton cycle produce 000 Btu/ with an inlet tate of 60 F,.7 pia, and a compreion ratio of 6:. he heat added in the combution i 00 Btu/lbm. What are the highet temperature and the ma flow rate of air, auming cold air propertie? Solution: Efficiency i from Eq..8 η = Ẇ net Q. = w net = - r -(k-)/k = - 6-0./. = 0.57 q H p H from the required power we can find the needed heat tranfer Q. H = Ẇ net ṁ = Q. H / q H / η = 000 0.57 = 5 59 Btu/ 5 59 Btu/ = = 6.99 lbm/ 00 Btu/lbm emperature after compreion i = r (k-)/k p = 50 6 0./. = 8 R he highet temperature i after combution = + q H /C p = 8 + 00 0. = 85 R
.8E Do the previou problem with propertie from table F.5 intead of cold air propertie. Solution: With the variable pecific heat we mut go through the procee one by one to get net work and the highet temperature. From F.5: h =.8 btu/lbm, o =.607 Btu/lbm R he compreion i reverible and adiabatic o contant. From Eq.8.8 = o = o + R ln (P P ) =.607 + 5. 778 ln6 =.808 Btu/lbm R back interpolate in F.5 =.5 R, h = 7.58 Btu/lbm Energy equation with compreor work in w C = - w = h - h = 7.58 -.8 = 50. Btu/lbm Energy Eq. combutor: h = h + q H = 7.58 + 00 = 67.6 Btu/lbm State : (P, h): = 600 R, o =.05 Btu/lbm R he expanion i reverible and adiabatic o contant. From Eq.8.8 = o = o + Rln(P /P 5. ) =.05 + ln(/6) =.855 778 = 97 R, h = 6. Btu/lbm Energy equation with turbine work out w = h - h = 67.6-6. = 58. Btu/lbm Now the net work i w net = w - w C = 58. 50. = 08. Btu/lbm he total required power require a ma flow rate a ṁ = Ẇ net 000 Btu/ = = 67. lbm/ w net 08. Btu/lbm
.85E An ideal regenerator i incorporated into the ideal air-tandard Brayton cycle of Problem.8. Calculate the cycle thermal efficiency with thi modification. Solution: P v x y P = atm Compreion ratio P P = Max temperature = 900 R he compreion i reverible and adiabatic o contant. From Eq.8. = ( P P ) k- k = 50() 0.86 = 8.6 R w C = h - h = C P0 ( - ) = 0. (8.6-50) = 6. Btu/lbm Expanion in turbine: = Implemented in Eq.8. = ( P P ) k- k = 900( )0.86 = 6. R w = h - h = C P0 ( - ) = 0.(900-6.) = 68.8 Btu/lbm w NE = w - w C = 68.8-6. =.7 Btu/lbm Ideal regenerator: X = = 6. R q H = h - h X = 0.(900-6.) = 68.8 Btu/lbm = w η H = w NE /q H =.7/68.8 = 0.60
.86E An air-tandard Ericon cycle ha an ideal regenerator a hown in Fig. P.6. Heat i upplied at 800 F and heat i rejected at 68 F. Preure at the beginning of the iothermal compreion proce i 0 lbf/in.. he heat added i 75 Btu/lbm. Find the compreor work, the turbine work, and the cycle efficiency. Identify the tate Heat upplied at high temperature Heat rejected at low temperature = = 800 F = 9.7 R = = 68 F = 57.7 R Beginning of the compreion: P = 0 lbf/in Ideal regenerator: q = - q q H = q w = q H = 75 Btu/lbm η H = η CARNO H. = - L / H = - 57.7/9.7 = 0.775 w net = η H q H = 0.775 75 =. Btu/lbm q L = -w C = 75 -. = 6.88 Btu/lbm P P P v P P
.87E he turbine in a jet engine receive air at 00 R, 0 lbf/in.. It exhaut to a nozzle at 5 lbf/in., which in turn exhaut to the atmophere at.7 lbf/in.. he ientropic efficiency of the turbine i 85% and the nozzle efficiency i 95%. Find the nozzle inlet temperature and the nozzle exit velocity. Aume negligible kinetic energy out of the turbine. Solution: C.V. urbine: h i = 560.588 Btu/lbm, o i =.99765 Btu/lbm R, e = i hen from Eq.8.8 o e = o i + R ln(p e/p i ) =.99765 + 5. 778 ln (5/0) =.876 Btu lbm R able F.5 e = 8 R, h e = 8.7 Btu/lbm, Energy eq.: w, = h i - h e = 560.588-8.7 =. Btu/lbm Eq.9.7: w,ac = w, η = 88.96 = h i - h e,ac h e,ac = 7.6 able F.5 e,ac = 509 R, o e =.897 Btu/lbm R C.V. Nozzle: h i = 7.6 Btu/lbm, o i =.897 Btu/lbm R, e = i hen from Eq.8.8 o e = o i + R ln(p e/p i ) =.897 + 5. 778 able F.5 e, = 99.6 R, h e, = 9. Btu/lbm ln (.7 5 ) =.85 Btu lbm R Energy Eq.: Eq.9.0: (/)V e, = h i - h e, = 7.6-9. = 80. Btu/lbm (/)V e,ac = (/)V e, η NOZ = 76.9 Btu/lbm V e,ac = 507 76.9 = 95 ft/ Recall Btu/lbm = 5 07 ft /
Otto, Dieel, Stirling and Carnot Cycle.88E Air flow into a gaoline engine at lbf/in., 50 R. he air i then compreed with a volumetric compreion ratio of 8:. In the combution proce 560 Btu/lbm of energy i releaed a the fuel burn. Find the temperature and preure after combution. Solution: Solve the problem with contant heat capacity. Compreion to : = From Eq.8. and Eq.8. = (v /v ) k- = 50 8 0. = 0.6 R P = P (v /v ) k = 8. = 57. lbf/in Combution to at contant volume: u = u + q H = + q H /C v = 0.6 + 560/0.7 = 55 R P = P ( / ) = 57. (55 / 0.6) = 96 lbf/in P v v
.89E o approximate an actual park-ignition engine conider an air-tandard Otto cycle that ha a heat addition of 800 Btu/lbm of air, a compreion ratio of 7, and a preure and temperature at the beginning of the compreion proce of lbf/in., 50 F. Auming contant pecific heat, with the value from able F., determine the maximum preure and temperature of the cycle, the thermal efficiency of the cycle and the mean effective preure. Solution: P v v State : v = R /P = 5. 50 =.5 ft /lbm, v = v /7 =.076 ft /lbm he compreion proce, reverible adiabatic o then ientropic. he contant i implemented with Eq.8.5 leading to Eq.8. and 8. P = P (v /v ) k = (7). = 98. lbf/in = (v /v ) k- = 50(7) 0. = 0.7 R he combution proce with contant volume, q H = 800 Btu/lbm = + q H /C V0 = 0.7 + 800/0.7 = 5789 R P = P / = 98. 5789/0.7 = 0 lbf/in Cycle efficiency from the ideal cycle a in Eq..8 η H = - ( / ) = - 50/0.7 = 0.5 o get the mean effective preure we need the net work w NE = η H q H = 0.5 800 =.8 Btu/lbm P m eff = w NE.8 778 = = 88 lbf/in v -v (.5-.076)
.90E A gaoline engine ha a volumetric compreion ratio of 0 and before compreion ha air at 50 R,. pia in the cylinder. he combution peak preure i 900 pia. Aume cold air propertie. What i the highet temperature in the cycle? Find the temperature at the beginning of the exhaut (heat rejection) and the overall cycle efficiency. Solution: Compreion. Ientropic o we ue Eq.8.-8. P = P (v /v ) k =. (0). = 06.5 pia = (v /v ) k- = 50 (0) 0. = 06. R Combution. Contant volume = (P /P ) = 06. 900/06.5 = 86 R Exhaut. Ientropic expanion o from Eq.8. = / (v /v ) k- = / 0 0. = 86 /.59 = 57 R Overall cycle efficiency i from Eq..8, r v = v /v η = r -k v = 0-0. = 0.60 Comment: No actual gaoline engine ha an efficiency that high, maybe 5%.
.9E A for troke gaoline engine ha a compreion ratio of 0: with cylinder of total diplacement 75 in. the inlet tate i 500 R, 0 pia and the engine i running at 00 RPM with the fuel adding 750 Btu/lbm in the combution proce. What i the net work in the cycle and how much power i produced? Solution: Overall cycle efficiency i from Eq..8, r v = v /v η H = r -k v = 0-0. = 0.60 w net = η H q H = 0.60 750 = 5.5 Btu/lbm We alo need pecific volume to evaluate Eq..5 to.7 v = R / P = 5. 500 / (0 ) = 8.5 ft /lbm P meff = w net v v = w net v ( r v ) = Now we can find the power from Eq..7 Ẇ = P meff V dipl RPM 60 = 9 hp Recall hp = 550 lbf-ft/. 5.5 8.5 0.9 778 = 6. pia 75 = 6. 00 60 = 6 00 lbf-ft/
.9E It i found experimentally that the power troke expanion in an internal combution engine can be approximated with a polytropic proce with a value of the polytropic exponent n omewhat larger than the pecific heat ratio k. Repeat Problem.89 but aume the expanion proce i reverible and polytropic (intead of the ientropic expanion in the Otto cycle) with n equal to.50. Firt find tate and. baed on the inlet tate we get v = v = R /P = 5. 50/ =.5 ft /lbm v = v = v /7 =.076 ft /lbm After compreion we have contant lead to Eq.8. and Eq.8. P = P (v /v ) k = (7). = 98. lbf/in = (v /v ) k- = 50(7) 0. = 0.7 R Contant volume combution = + q H /C V0 = 0.7 + 800/0.7 = 5789 R P = P / = 98. 5789/0.7 = 0 lbf/in Proce to : Pv.5 = contant. P = P (v /v ).5 = 0(/7).5 = 55.78 lbf/in = (v /v ) 0.5 = 5789(/7) 0.5 = 88 R For the mean effective preure we need the net work and therefore the induvidual proce work term w = P dv = R( - )/( -.) = -5.(0.7-50)/(0. 778) = -0.96 Btu/lbm w = P dv = R( - )/( -.5) = -5.(88-5789)/(0.5 778) = 9.8 Btu/lbm w NE = 9.8-0.96 = 90.8 Btu/lbm η CYCLE = w NE /q H = 90.8/700 = 0.88 P meff = w NE /(v -v ) = 90.8 778/(.5 -.076) = 69.5 lbf/in Notice a maller w NE, η CYCLE, P meff compared to ideal cycle.
.9E In the Otto cycle all the heat tranfer q H occur at contant volume. It i more realitic to aume that part of q H occur after the piton ha tarted it downward motion in the expanion troke. herefore conider a cycle identical to the Otto cycle, except that the firt two-third of the total q H occur at contant volume and the lat one-third occur at contant preure. Aume the total q H i 700 Btu/lbm, that the tate at the beginning of the compreion proce i lbf/in., 68 F, and that the compreion ratio i 9. Calculate the maximum preure and temperature and the thermal efficiency of thi cycle. Compare the reult with thoe of a conventional Otto cycle having the ame given variable. P 5 v v 5 P =, = 57.67 R r V = v /v = 7 q = Btu 700 = 66.7 lbm q = Btu 700 =. lbm v P = P (v /v ) k = (9). = 8.8 lbf/in = (v /v ) k- = 57.67(9) 0. = 70.7 R = + q /C V0 = 70.7 + 66.7/0.7 = 000 R P = P ( / ) = 8.8 000/70.7 = 887. lbf/in = P = + q /C P0 = 000 +./0. = 97 R v 5 v = = (P v v /P ) ( / ) = 88. 57.67 97 = 7. 5 = (v /v 5 ) k- = 97(/7.) 0. = 5 R q L = C V0 ( 5 - ) = 0.7(5-57.67) = 9.9 Btu/lbm η H = - q L /q H = - 9.9/700 = 0.579 Standard Otto cycle: η H = - (9) -0. = 0.585
.9E A dieel engine ha a bore of in., a troke of. in. and a compreion ratio of 9: running at 000 RPM (revolution per minute). Each cycle take two revolution and ha a mean effective preure of 00 lbf/in.. With a total of 6 cylinder find the engine power in Btu/ and horepower, hp. Solution: Work from mean effective preure. P meff = w net / (v max - v min ) -> w net = P meff (v max - v min ) he diplacement i V = πbore 0.5 S = π 0.5. = 5.05 in Work per cylinder per power troke W = P meff (V max - V min ) = 00 5.05 / ( 778) =.575 Btu/cycle Only every econd revolution ha a power troke o we can find the power Ẇ = W N cyl RPM 0.5 ( cycle min ) (min 60 ) ( Btu cycle ) =.575 6 000 0.5 (/60) = 5.75 Btu/ = 5.75 600/5. hp = 6 hp
.95E At the beginning of compreion in a dieel cycle = 50 R, P = 0 lbf/in. and the tate after combution (heat addition) i 600 R and 000 lbf/in.. Find the compreion ratio, the thermal efficiency and the mean effective preure. Solution: Compreion proce (ientropic) from Eq.8.-8. P = P = 000 lbf/in => v /v = (P /P ) /k = (000/0) 0.7 =. = (P /P ) (k-)/k = 50(000/0) 0.857 = 70.6 R Expanion proce (ientropic) firt get the volume ratio v /v = / = 600/70.6 =.768 v /v = v /v = (v /v )(v /v ) =./.768 = 6.9 he exhaut temperature follow from Eq.8. = (v /v ) k- = 600*6.9-0. = 99 R q L = C V ( - ) = 0.7(99-50) =.7 Btu/lbm q H = h - h = C P ( - ) = 0.(600-70.6) = 7. Btu/lbm η = - q L /q H = -.7 / 7. = 0.58 w net = q net = 7. -.7 = 58. Btu/lbm v max = v = R /P = 5. 50/(0 ) = 6.6675 ft /lbm v min = v max (v /v ) = 6.6675 /. = 0.55 ft /lbm P meff = [58./(6.6675-0.55)] (778/) = 9.8 lbf/in P v P v Remark: hi i a too low compreion ratio for a practical dieel cycle.
.96E Conider an ideal air-tandard dieel cycle where the tate before the compreion proce i lbf/in., 6 F and the compreion ratio i 0. Find the maximum temperature(by iteration) in the cycle to have a thermal efficiency of 60%. Dieel cycle: P =, = 5.67 R, v /v = 0, η H = 0.60 From the inlet tate and the compreion we get = (v /v ) k- = 5.67(0) 0. = 7. R v = 5. 5.67 =.89 ft /lbm, v =.89 0 Contant preure combution relate v and v = v / = 0.695 /7. = 0.00099 he expanion then give interm of = ( v v ) k- = (.89 0.00099 ) 0. = 0.05. Now thee relate to the given efficiency - η H = 0.60 = - k( - ) = - 0.05-5.67.( -7.). = 0.695 ft /lbm 0.05. - 0.56 + 7.5 = 0 rial and error on thi non-linear equation 500 R: LHS = -5.5, 5500 R: LHS = 5.0, 550 R: LHS = -0.5 Linear interpolation, = 555 R
.97E Conider an ideal Stirling-cycle engine in which the preure and temperature at the beginning of the iothermal compreion proce are.7 lbf/in., 80 F, the compreion ratio i 6, and the maximum temperature in the cycle i 000 F. Calculate the maximum preure in the cycle and the thermal efficiency of the cycle with and without regenerator. P v v v v Ideal Stirling cycle = = 80 F P =.7 lbf/in v v = 6 = = 000 F v = P = P v /v =.7 6 = 88. V = V P = P / = 88. 60 50 w = q = R ln (v /v ) = (5./778) 60 ln 6 = 0. Btu/lbm = 0.8 lbf/in q = C V0 ( - ) = 0.7(60-50) = 8. Btu/lbm v w = q = -R ln = - 5. 50 ln 6 = -66. Btu/lbm v 778 w NE = 0. - 66. = 5.9 Btu/lbm η NO REGEN = 5.9 0.+8. = 0.7, η WIH REGEN = 5.9 0. = 0.78
.98E An ideal air-tandard Stirling cycle ue helium a working fluid. he iothermal compreion bring the helium from 5 lbf/in., 70 F to 90 lbf/in.. he expanion take place at 00 R and there i no regenerator. Find the work and heat tranfer in all four procee per lbm helium and the cycle efficiency. Subtance helium F.: R = 86 ft-lbf/lbmr, C v = 0.75 Btu/lbm R v /v = v /v = P /P = 90/5 = 6 -> : - w = - q = P dv = R ln (v /v ) = 86 50 ln(6)/778 = 7.5 Btu/lbm -> : w = 0; q = C P ( - ) = 0.75(00-50) = 8. -> : w = q = R ln(v /v ) = 86 00 ln(6)/778 = 866.8 Btu/lbm -> : w = 0; q = C P ( - ) = -8. Btu/lbm η Cycle = w net / q H = -7.5 + 866.0 8. + 866.8 = 0.58
.99E he air-tandard Carnot cycle wa not hown in the text; how the diagram for thi cycle. In an air-tandard Carnot cycle the low temperature i 500 R and the efficiency i 60%. If the preure before compreion and after heat rejection i.7 lbf/in., find the high temperature and the preure jut before heat addition. Solution: Carnot cycle efficiency from Eq.7.5 η = 0.6 = - H / L H = L /0. = 500/0. = 50 R Jut before heat addition i tate and after heat rejection i tate o P = 00 kpa and the ientropic compreion i from Eq.8. P = P ( H / L ) k- =.7( 50 500 ).5 = 6. lbf/in OR if we do not ue contant pecific heat, but ue able F.5 in Eq.8.8 P = P exp[( o -.65 o )/R] =.7 exp[.857 ] = 89 lbf/in 5. / 778 P v H L
.00E Air in a piton/cylinder goe through a Carnot cycle in which L = 80. F and the total cycle efficiency i η = /. Find H, the pecific work and volume ratio in the adiabatic expanion for contant C p, C v. Carnot cycle: η = - L / H = / H = L = 50 = 60 R Adiabatic expanion to : Pv k = contant w = (P v - P v )/( - k) = [R/(-k)]( - ) = u - u = C v ( - ) = 0.7(60-50) = 8.68 Btu/lbm v /v = ( / ) /(k - ) =.5 = 5.6 P v H L
.0E Do the previou problem.00e uing able F.5. Air in a piton/cylinder goe through a Carnot cycle in which L = 80. F and the total cycle efficiency i η = /. Find H, the pecific work and volume ratio in the adiabatic expanion for contant C p, C v. Carnot cycle: η = - L / H = / H = L = 50 = 60 R w = u - u = 90. - 9.6 = 97.97 Btu/lbm Adiabatic expanion to : = Eq.8.8 o = o + R ln P P able F.5 for tandard entropy P = exp[( o P - o )/R] = exp[.6979-.96 ] = 0.086 5./778 Ideal ga law then give v v = P P = 50 60 0.086 = 8.09 P v H L
Refrigeration Cycle.0E A car air-conditioner (refrigerator) in 70 F ambient ue R-a and I want to have cold air at 0 F produced. What i the minimum high P and the maximum low P it can ue? Since the R-a mut give heat tranfer out to the ambient at 70 F, it mut at leat be that hot at tate. From able F.0.: P = P = P at = 85.95 pia i minimum high P. Since the R-a mut aborb heat tranfer at the cold air 0 F, it mut at leat be that cold at tate. From able F.0.: P = P = P at =.9 pia i maximum low P. Ideal Ref. Cycle cond = 70 F = evap = 0 F Ue able F.0 for R-a
.0E Conider an ideal refrigeration cycle that ha a condener temperature of 0 F and an evaporator temperature of 5 F. Determine the coefficient of performance of thi refrigerator for the working fluid R- and R-. Ideal Ref. Cycle cond = 0 F = evap = 5 F Ue able F.9 for R- Ue computer table for R- R- R- h, Btu/lbm 77.80 0.95 = 0.68 0.705 P, lbf/in 5..0, F 7.9 6.87 h, Btu/lbm 9.07.90 h =h, Btu/lbm.5.6 -w C = h -h. 8.95 q L = h -h.7 6.5 β =q L /(-w C )..0
.0E he environmentally afe refrigerant R-a i one of the replacement for R- in refrigeration ytem. Repeat Problem.0 uing R-a and compare the reult with that for R-. Ideal refrigeration cycle cond = 0 F = evap = 5 F Ue able F.0 for R-a or computer table C.V. Compreor: Adiabatic and reverible o contant State : able F.0. h =67. Btu/lbm, = 0.5 Btu/lbm R State : = and P = 6. pia = P = P at 0 F Interpolate => h = 8.6 Btu/lbm and =.8 F Energy eq.: w C = h - h = 8.6-67. = 7.0 Btu/lbm Expanion valve: h = h =.6 Btu/lbm Evaporator: q L = h - h = 67. -.6 = 5.86 Btu/lbm Overall performance, COP β = q L /w C = 5.86 / 7.0 =.
.05E Conider an ideal heat pump that ha a condener temperature of 0 F and an evaporator temperature of 0 F. Determine the coefficient of performance of thi heat pump for the working fluid R-, R-, and ammonia. Ideal Heat Pump cond = 0 F evap = 0 F Ue able F.8 for NH Ue able F.9 for R- Ue computer table for R- R- R- NH h, Btu/lbm 80. 07.8 69.58 = 0.665 0.8.769 P, lbf/in 7. 7.6 86.5, F. 60. 9. h, Btu/lbm 9.0.7 79.5 h =h, Btu/lbm 6.0 5.7 78.8 -w C = h -h 0.58.89 99.9 q H = h -h 5.995 76.6 50.67 β =q H /(-w C ) 5.98 5.5 5.
.06E he refrigerant R- i ued a the working fluid in a conventional heat pump cycle. Saturated vapor enter the compreor of thi unit at 50 F; it exit temperature from the compreor i meaured and found to be 85 F. If the compreor exit i 00 pia, what i the ientropic efficiency of the compreor and the coefficient of performance of the heat pump? R- heat pump: = 85 F EVAP = 50 F S State : able F.9. h = 08.95 Btu/lbm, = 0.80 Btu/lbm R State : h = 6.55 Btu/lbm Compreor work: w C = h h = 6.55 08.95 = 7.575 Btu/lbm Ientropic compreor: S = = 0.80 Btu/lbm R State : (P, ) S = 60 F, h S = 0.8 Btu/lbm Ideal compreor work: w C = h S - h = 0.8 08.95 =.87 Btu/lbm he efficiency i the ratio of the two work term η S COMP = w C w C =.87 7.575 = 0.675 he condener ha heat tranfer a (h = h f at 00 pia) and a coefficient of performance of q H = h - h = 6.55-8.0 = 78.505 Btu/lbm β = q H /w C =.7
.07E Conider an air tandard refrigeration cycle that ha a heat exchanger included a hown in Fig. P.7. he low preure i.7 pia and the high preure i 00 pia. he temperature into the compreor i 60 F which i and in Fig..8, and = 6 = -60 F. Determine the coefficienct of performance of thi cycle. Solution: 6 q L COMP q H EXP 5 5 6 Standard air refrigeration cycle with = = 60 F = 59.67 R, P =.7 pia, P = 00 pia = 6 = -60 F = 99.67 R We will olve the problem with cold air propertie. Compreor, ientropic = o from Eq.8. k- = (P /P ) k = 59.67 (00/.7) 0.857 = 095.5 R w C = -w = C P0 ( - ) = 0. (095.5-59.67) = 8. Btu/lbm Expanion in expander (turbine) k- 5 = 5 = (P 5 /P ) k = 99.67 (.7/00) 0.857 = 89.58 R w E = C P0 ( - 5 ) = 0. (99.67-89.58) = 50. Btu/lbm Net cycle work w NE = 50. - 8. = -87.78 kj/kg q L = C P0 ( 6-5 ) = w E = 50. Btu/lbm Overall cycle performance, COP β = q L /(-w NE ) = 50. / 87.78 = 0.57
Availability and Combined Cycle.08E Find the flow and fluxe of exergy in the condener of Problem.7E. Ue thoe to determine the nd law efficiency. A maller power plant produce 50 lbm/ team at 00 pia, 00 F, in the boiler. It cool the condener with ocean water coming in at 55 F and returned at 60 F o that the condener exit i at 0 F. Find the net power output and the required ma flow rate of the ocean water. Solution: ake the reference tate at the ocean temperature 55 F = 5.7 R he tate propertie from able F.7. and F.7.. Ref. tate.7 lbf/in, 55 F, h 0 =.06 Btu/lbm, 0 = 0.058 Btu/lbm R 6 cb 5 State : 0 F, x = 0: h = 78.0 Btu/lbm, = 0.7 Btu/lbm R, State : 00 pia, 00 F: h = 577. Btu/lbm, =.7989 Btu/lbm R C.V. urbine : w = h - h ; = = =.7989 = 0.7 + x (.80) => x = 0.9 => h = 78.0 + 0.9 (0.8) = 08.95 Btu/lbm C.V. Condener : q L = h - h = 08.95-78.0 = 90.9 Btu/lbm Q. L = ṁq L = 50 90.9 = 7 07 Btu/ = ṁ ocean C p ṁ ocean = Q. L / C p = 7 07 / (.0 5) = 909 lbm/ he pecific flow exergy for the two tate are from Eq.0. neglecting kinetic and potential energy ψ = h - h 0-0 ( - 0 ), ψ = h - h 0-0 ( - 0 ) he net drop in exergy of the water i Φ. water = ṁ water [h h o ( )] = 50 [ 08.95 78.0 5.7 (.7989 0.7)] = 7 07 50 = 5 Btu/ he net gain in exergy of the ocean water i
Φ. ocean = ṁ ocean [h 6 h 5 o ( 6 5 )] = ṁ ocean [C p ( 6 5 ) o C p ln( 6 5 ) ] = 909 [.0 (60 55) 5.7.0 ln = 7 07 6 88 = 9 Btu/ he econd law efficiency i η II = Φ. ocean / Φ. water = 9 5 = 0.05 59.7 + 60 59.7 + 55 ] In reality all the exergy in the ocean water i detroyed a the 60 F water mixe with the ocean water at 55 F after it flow back out into the ocean and the efficiency doe not have any ignificance. Notice the mall rate of exergy relative to the large rate of energy being tranferred.
.09E (Adv.) Find the availability of the water at all four tate in the Rankine cycle decribed in Problem.7. Aume the high-temperature ource i 900 F and the low-temperature reervoir i at 65 F. Determine the flow of availability in or out of the reervoir per pound-ma of team flowing in the cycle. What i the overall cycle econd law efficiency? Ref. tate.7 lbf/in, 77 F, h 0 = 5.08 Btu/lbm, For thi cycle from able F.7 0 = 0.0877 Btu/lbm R State : Superheated vapor h = 50.6 Btu/lbm, =.587 Btu/lbm R, State : Saturated liquid h = 97.97 Btu/lbm, v = 0.065 ft /lbm C.V. Pump: Adiabatic and reverible. Ue incompreible fluid o w P = v dp = v (P - P ) = 0.065(600.) =.8 Btu/lbm 778 h = h + w P = 95.8 Btu/lbm C.V. Boiler: q H = h - h = 50.6-97.97 = 5.65 Btu/lbm C.V. ubine: w = h - h, = = =.587 Btu/lbm R = 0.87 + x.79 x = 0.87, h = 97.97 + 0.87 09.78 = 96.75 Btu/lbm w = 50.6-96.75 =.87 Btu/lbm η CYCLE = (w - w P )/q H = (.87 -.8)/5.65 = 0.7 C.V. Condener: q L = h - h = 96.75-97.97 = 88.8 Btu/lbm P v From olution to.: = 0.797, = 0.75 =, = =.587 Btu/lbm R h = 9.0, h = 95.8, h = 50.6, h = 9. Btu/lbm
ψ = h - h 0-0 ( - 0 ) ψ = 9.0-5.08-56.67(0.797-0.0877) =.6 Btu/lbm ψ =.9, ψ = 500.86, ψ = 7.9 Btu/lbm ψ H = ( - 0 / H )q H = 0.605 5.79 = 759.65 Btu/lbm ψ L = ( - 0 / 0 )q C = 0/ η II = w NE / ψ H = (9.7 -.8)/759.65 = 0.56 Notice H >, L < =, o cycle i externally irreverible. Both q H and q C over finite.
.0E Find the flow of exergy into and out of the feedwater heater in Problem.76E. State : x = 0, h = 97.97 Btu/lbm, v = 0.065 ft /lbm, = 0.797 State : x = 0, h = 0.67 Btu/lbm, = 0.999 Btu/lbm R State 5: h 5 = 50.5 Btu/lbm, 5 =.587 Btu/lbm R State 6: 6 = 5 =.587 Btu/lbm R => h 6 = 08.9 Btu/lbm C.V Pump P w P = h - h = v (P - P ) = 0.065(50.5) = 0. Btu/lbm 778 => h = h + w P = 97.97 + 0.9 = 98. Btu/lbm = = 0.797 Btu/lbm R C.V. Feedwater heater: Call ṁ 6 / ṁ tot = x (the extraction fraction) Energy Eq.: ( - x) h + x h 6 = h x = h - h h 6 - h = 0.67-98. 08.9-98. = 0.09 FWH 6 x -x Ref. State:.7 pia, 77 F, o = 0.0877 Btu/lbm R, h o = 5.08 Btu/lbm ψ = h - h o - o ( - o ) = 98. - 5.08 56.67(0.797-0.0877) = 6.5 Btu/lbm ψ 6 = 08.9-5.08-56.67(.587-0.0877) = 59. Btu/lbm ψ = 0.67-5.08-56.67(0.999-0.0877) = 68.6 Btu/lbm he rate of exergy flow caled with maximum flow rate i then Φ. /ṁ = ( - x) ψ = 0.7909 6.5 = 5.57 Btu/lbm Φ. 6 /ṁ = xψ = 0.09 59. = 75.09 Btu/lbm 6 Φ. /ṁ = ψ = 68.6 Btu/lbm he mixing i detroying 5.57 + 75.09 68.6 =.6 Btu/lbm of exergy
.E Conider the Brayton cycle in problem.8e. Find all the flow and fluxe of exergy and find the overall cycle econd-law efficiency. Aume the heat tranfer are internally reverible procee and we then neglect any external irreveribility. Solution: Efficiency i from Eq..8 η = Ẇ NE / Q. H = w net = - r -(k-)/k = - 6-0./. = 0.57 q H p from the required power we can find the needed heat tranfer Q. H = Ẇ net / η = 000 / 0.57 = 5 59 Btu/ ṁ = Q. H / q H = 5 59 (Btu/) / 00 Btu/lbm = 6.99 lbm/ emperature after compreion i = r (k-)/k p = 59.67 6 0./. = 8 R he highet temperature i after combution = + q H /C p = 8 + 00 0. = 85 R For the exit flow I need the exhaut temperature = r p (k-)/k = 85 6 0.857 = 7.8 R he high exergy input from combution i Φ. H = ṁ(ψ - ψ ) = ṁ[h h ( )] = 6.99 [00 56.67 0. ln ( 85 )] = 7 895 Btu/ 8 Since the low exergy flow out i lot the econd law efficiency i η II = Ẇ NE /Φ. H = 000 / 7 895 = 0.78 Φ. flow out = ṁ(ψ - ψ o ) = ṁ[h h o ( o )] = 6.99 [ 0.(7.8 56.7) 56.7 0. ln ( 7.8 56.7 Φ. flow in = ṁ(ψ - ψ o ) = ṁ[h h o ( o )] = 6.99 [ 0.(60 77) 56.7 0. ln ( 59.7 ) ] =. Btu/ 56.7 ) ] = 05 Btu/
.E Conider an ideal dual-loop heat-powered refrigeration cycle uing R- a the working fluid, a hown in Fig. P.. Saturated vapor at 0 F leave the boiler and expand in the turbine to the condener preure. Saturated vapor at 0 F leave the evaporator and i compreed to the condener preure. he ratio of the flow through the two loop i uch that the turbine produce jut enough power to drive the compreor. he two exiting tream mix together and enter the condener. Saturated liquid leaving the condener at 0 F i then eparated into two tream in the neceary proportion. Determine the ratio of ma flow rate through the power loop to that through the refrigeration loop. Find alo the performance of the cycle, in term of the ratio Q L /Q H. URB. COMP. 6 7 BOIL. COND. 5 P E V A P.. Q L 5 7 6 P h Computer table for F lbf/in Btu/lbm Btu/lbm R propertie. 0.89 77.7 68.88 P =P =P SA at 0 F - 5. 68.88 P 5 =P 6 =P SA at 0 F 0 5..5 0.067 5 = =0.68 88 0.89.5 h =9.77 5-5. 0.067 5 Pump work: 6 0 5. 89.06 0.5 9 -w P = h 5 -h 7 0 5. 0.5 9 v 5 (P 5 -P ) -w P = 0.09(5. - 5.) 778 = 0.89 h 5 =.5 + 0.89 =.5 Btu/lbm (-x 7 ) = 0.6 79-0.5 9 0.095 = 0.0 0 0.095 = 0.87 h 7 = 87.8-0.87(5.) = 8.97 Btu/lbm
CV: turbine + compreor Continuity Eq.: ṁ = ṁ, ṁ 6 = ṁ 7 Energy Eq.: ṁ h + ṁ 6 h 6 = ṁ h + ṁ 7 h 7 ṁ /ṁ 6 = 89.06-8.97 9.77-77.7 = 7.69.006 = 0.55, ṁ 6 /ṁ =.8 CV: pump: -w P = v (P 5 -P ), h 5 = h - w P CV evaporator: Q. L = ṁ (h -h ), CV boiler: Q. H = ṁ 6 (h 6 -h 5 ) β = Q. ṁ L /Q. H = (h -h ) 77.7-.5 =.8(89.06-.5) = 0.6 ṁ 6 (h 6 -h 5 )
.E Conider an ideal combined reheat and regenerative cycle in which team enter the high-preure turbine at 500 lbf/in., 700 F, and i extracted to an open feedwater heater at 0 lbf/in. with exit a aturated liquid. he remainder of the team i reheated to 700 F at thi preure, 0 lbf/in., and i fed to the lowpreure turbine. he condener preure i lbf/in.. Calculate the thermal efficiency of the cycle and the net work per pound-ma of team. 5: h 5 = 56.66, 5 =.6 7: h 7 = 78.7, 7 =.785 : h = h f =.59, v = 0.0788 C.V. 5 = 6 => h 6 = 09.76 w = h 5 - h 6 = 56.66-09.76 = 6.9 Btu/lbm C.V. Pump -w P = h - h = v (P - P ) = 0.06(0 - ) = 0.5 P 5 -x => h = h - w P = 9.7 + 0.5 = 9.08 Btu/lbm C.V. FWH x HR 6 7 8 -x COND. P 5 700 F 7 x h 6 + ( - x) h = h x = h - h.59-9.08 = h 6 - h 09.76-9.08 = 0.958 C.V. Pump 6 8 pi -w P = h - h = v (P - P ) = 0.0788(500-0)(/778) =.6 Btu/lbm => h = h - w P =.59 +.6 =.85 Btu/lbm q H = h 5 - h + ( - x)(h 7 - h 6 ) = 0.8 + 5. = 78. Btu/lbm C.V. urbine 7 = 8 => x 8 = (.785-0.7)/.76 = 0.9 h 8 = h f + x 8 h fg = 9.7 + 0.9 0. = 05. w = h 7 - h 8 = 78.7-05. =.97 w net = w + ( - x) w + ( - x) w P + w P = 6.9 + 75.8-0.85 -.6 =.5 kj/kg η cycle = w net / q H =.5 / 78. = 0.57
.E In one type of nuclear power plant, heat i tranferred in the nuclear reactor to liquid odium. he liquid odium i then pumped through a heat exchanger where heat i tranferred to boiling water. Saturated vapor team at 700 lbf/in. exit thi heat exchanger and i then uperheated to 00 F in an external ga-fired uperheater. he team enter the turbine, which ha one (open-type) feedwater extraction at 60 lbf/in.. he ientropic turbine efficiency i 87%, and the condener preure i lbf/in.. Determine the heat tranfer in the reactor and in the uperheater to produce a net power output of 000 Btu/. 700 lbf/in 6 6 00 F SUP. H. URBINE. 60 lbf/in Q 5 7 5 8 7 7 REAC. HR. lbf/in P P COND. 8 Ẇ NE = 000 Btu/, η S = 0.87 -w P = 0.066(60 - )/778 = 0.8 Btu/lbm h = h - w P = 69.7 + 0.8 = 69.9 Btu/lbm -w P = 0.0778(700-60)/778 =.06 Btu/lbm h = h - w P = 6. +.06 = 6. Btu/lbm 7S = 6 =.768, P 7 => 7S = 500.8 F, h 7S = 8. h 7 = h 6 - η S (h 6 - h 7S ) = 65.8-0.87(65.8-8.) = 7.9 8S = 6 =.768 = 0.6 + x 8S.85 => x 8S = 0.886 h 8S = 69.7 + 0.886 06 = 987.9 Btu/lbm h 8 = h 6 - η S (h 6 - h 8S ) = 65.8-0.87(65.8-987.9) = 070.8 CV: heater: cont: m + m 7 = m =.0 lbm, t law: m h + m 7 h 7 = m h m 7 = (6.-69.9) / (7.9-69.9) = 0.59 CV: turbine: w = (h 6 - h 7 ) + ( - m 7 )(h 7 - h 8 ) = 65.8-7.9 + 0.87(7.9-070.8) = 55.7 Btu/lbm CV pump: w P = m w P + m w P = -(0.87 0.8 +.06) = -. Btu/lbm w NE = 55.7 -. = 5.5 Btu/lbm => ṁ = 000/5.5 =.97 lbm/ 8
CV: reactor Q. REAC = ṁ(h 5 -h ) =.97(0-6.) = 85.7 Btu/ CV: uperheater Q. SUP = ṁ(h 6 - h ) =.97(65.8-0) = 85 Btu/ 5
.5E Conider an ideal ga-turbine cycle with two tage of compreion and two tage of expanion. he preure ratio acro each compreor tage and each turbine tage i 8 to. he preure at the entrance to the firt compreor i lbf/in., the temperature entering each compreor i 70 F, and the temperature entering each turbine i 000 F. An ideal regenerator i alo incorporated into the cycle. Determine the compreor work, the turbine work, and the thermal efficiency of the cycle. 0 REG I.C. 5 CC 6 9 COMP COMP URB URB P /P = P /P = P 6 /P 7 = P 8 /P 9 = 8.0 P = lbf/in = = 70 F, 6 = 8 = 000 F Aume cont. pecific heat = and = k- = = (P /P ) k = 59.67(8) 0.857 = 959. R otal compreor work 7 8 CC -w C = (-w ) = C P0 ( - ) = 0.(959. - 59.67) = 06. Btu/lbm Alo 6 = 7 and 8 = 9 7 = 9 = 6 otal turbine work P 7 P 6 k- k = 59.67 8 0.857 = 57.9 R w = w 67 = C P0 ( 6-7 ) = 0.(59.67-57.9) = 58.85 Btu/lbm w NE = 58.85-06. =.55 Btu/lbm 5 6 7 0 8 9 Ideal regenerator: 5 = 9, 0 =
q H = (h 6 - h 5 ) + (h 8 - h 7 ) = C P0 ( 6-5 ) = 0.(59.67-57.9) = w = 58.85 Btu/lbm η H = w NE /q H =.55/58.85 = 0.6
.6E Repeat Problem.5, but aume that each compreor tage and each turbine tage ha an ientropic efficiency of 85%. Alo aume that the regenerator ha an efficiency of 70%. S = S = 959. R, -w CS = 06. 7S = 9S = 57.9 R, w S = 58.85 6 -w C = -w SC /η SC =.7 Btu/lbm 5 7S -w = -w =.7/ =.5 Btu/lbm S S = = + (-w /C P0 ) = 59.67 +.5/0. = 05. R w = η w S = 9.5 Btu/lbm 8 9 7 9S 7 = 9 = 6 - (+w 67 /C P0 ) = 59.67-9.5/ 0. = 5 R η REG = h 5 - h 5-5 - 05. = = h 9 - h 9-5 - 05. = 0.7 5 = 76.7 R q H = C P0 ( 6-5 ) + C P0 ( 8-7 ) = 0.(59.67-76.7) + 0.(59.67-5) = 8.7 Btu/lbm w NE = w + w C = 9.5 -.7 = 06.8 Btu/lbm η H = w NE /q H = 06.8/8.7 = 0.7
.7E Conider a mall ammonia aborption refrigeration cycle that i powered by olar energy and i to be ued a an air conditioner. Saturated vapor ammonia leave the generator at 0 F, and aturated vapor leave the evaporator at 50 F. If 000 Btu of heat i required in the generator (olar collector) per pound-ma of ammonia vapor generated, determine the overall performance of thi ytem. NH aborption cycle: at. vapor at 0 F exit the generator. Sat. vapor at 50 F exit the evaporator 0F q H = q GEN = 000 Btu/lbm NH 50 F out of generator. GEN. EXI EVAP EXI q L = h - h = h G 50 F - h F 0 F = 6.8-78.79 = 5.9 Btu/lbm q L /q H = 5.9/000 = 0.85