.5 Conider an ideal team regeneratie cycle in which team enter the turbine at.0 Ma, 00 C, and exhaut to the condener at 0 ka. Steam i extracted from the turbine at 0.8 Ma for an open feedwater heater. he feedwater leae the heater a aturated liquid. he appropriate pump are ued for the water leaing the condener and the feedwater heater. Calculate the thermal efficiency of the cycle and the net work per kilogram of team. hi i a tandard Rankine cycle with an open FWH a hown in Fig..0 C.V ump w = h - h = ( - ) = 0.000(800-0) = 0.798 kj/kg => h = h + w = 9.8 + 0.798 = 9.6 kj/kg C.V. FWH Call ṁ 6 / ṁ tot = x (the extraction fraction) C.V ump CV Boiler: CV urbine ( - x) h + x h 6 = h x = h - h h 6 - h = 7. - 9.6 89.6-9.6 = 0.958 w = h - h = ( - ) = 0.005(000-800) =.5 kj/kg h = h + w = 7. +.5 = 7.55 kj/kg q H = h 5 - h = 0.8-7.55 = 507. kj/kg nd Law 7 = 6 = 5 = 6.9 kj/kg K 6, 6 => h 6 = 89.6 kj/kg (uperheated apor) 6.9-0.69 7 = 6 = 5 = 6.9 => x 7 = = 0.86 7.50 => h 7 = 9.8 + x 7 9.8 = 9.55 kj/kg urbine ha full flow in H ection and fraction -x in L ection Ẇ / ṁ 5 = h 5 - h 6 + ( - x) (h 6 - h 7 ) w = 0.8 89.6 + ( - 0.988) ( 89.6 9.55) = 899. ha the full flow and ha the fraction -x of the flow w net = w - ( - x) w - w = 899. - ( - 0.988)0.798.5 = 896. kj/kg cycle = w net / q H = 896. / 507. = 0.57
.6 A 0 kg/ teady upply of aturated-apor team at 500 ka i required for drying a wood pulp lurry in a paper mill. It i decided to upply thi team by cogeneration, that i, the team upply will be the exhaut from a team turbine. Water at 0 C, 00 ka, i pumped to a preure of 5 Ma and then fed to a team generator with an exit at 00 C. What i the additional heat tranfer rate to the team generator beyond what would hae been required to produce only the deired team upply? What i the difference in net power? Deired exit State : = 500 ka, at. ap. => x =.0, = 5.9 C h = h g = 78.7 kj/kg, = g = 6.8 kj/kg-k Inlet State: 0 C, 00 ka h = h f = 8.9 kj/kg, = f = 0.0000 m /kg Without Cogeneration; he water i pumped up to 500 ka and then heated in the team generator to the deired exit. C.V. ump: w w/o = ( - ) = 0. kj/kg h = h + w w/o = 8. kj/kg C.V. Steam Generator: q w/o = h - h = 66. kj/kg With Cogeneration; he water i pumped to 5 Ma, heated in the team generator to 00 C and then flow through the turbine with deired exit tate. C.V. ump: w w = d = ( - ) =.9 kj/kg h = h + w w = 88.85 kj/kg C.V. Steam Generator: Exit 00 C, 5 Ma => h = 95.6 kj/kg q w = h - h = 95.6-88.85 = 06.8 kj/kg C.V.: urbine, Inlet and exit tate gien w t = h - h = 95.6-78.7 = 6.9 kj/kg Comparion Additional Heat ranfer: q w - q w/o = 06.8-66. =. kj/kg Q. extra = ṁ(q w - q w/o ) = kw Difference in Net ower: w diff = (w t - w w ) + w w/o, w diff = 6.9 -.9 + 0. =. kj/kg Ẇ diff = ṁw diff = kw By adding the extra heat tranfer at the higher preure and a turbine all the extra heat tranfer can come out a work (it appear a a 00% efficiency)
Brayton Cycle, Ga urbine.68 Conider an ideal air-tandard Brayton cycle in which the air into the compreor i at 00 ka, 0 C, and the preure ratio acro the compreor i :. he maximum temperature in the cycle i 00 C, and the air flow rate i 0 kg/. Aume contant pecific heat for the air, alue from able A.5. Determine the compreor work, the turbine work, and the thermal efficiency of the cycle. = 00 ka Compreion ratio = Max temperature = 00 o C ṁ = 0 kg/ he compreion i reerible and adiabatic o contant. From Eq.8. = k- k = 9.() 0.86 = 596.8 K Energy equation with compreor work in w C = - w = C 0 ( - ) =.00(596.8-9.) = 0.8 kj/kg he expanion i reerible and adiabatic o contant. From Eq.8. k- = k = 7. 0.86 = 67.7 K Energy equation with turbine work out w = C 0 ( - ) =.00(7. - 67.7) = 70. kj/kg Scale the work with the ma flow rate Ẇ C = ṁw C = 08 kw, Ẇ = ṁw = 70 kw Energy added by the combution proce q H = C 0 ( - ) =.00(7. - 596.8) = 779.5 kj/kg H = w NE /q H = (70. - 0.8)/779.5 = 0.509
.77 A two-tage air compreor ha an intercooler between the two tage a hown in Fig..77. he inlet tate i 00 ka, 90 K, and the final exit preure i.6 Ma. Aume that the contant preure intercooler cool the air to the inlet temperature,. It can be hown that the optimal preure, = ( ) /, for minimum total compreor work. Find the pecific compreor work and the intercooler heat tranfer for the optimal. Optimal intercooler preure = : h = 90., o = 6.85 00 600 = 00 ka C.V. C: w C = h - h, = leading to Eq.8.8 o = o R ln / ) = 6.85 + 0.87 ln = 7. = 0. K, h =.05 kj/kg w C =.05-90. =.6 kj/kg C.V. Cooler: = h = h q OU = h - h = h - h = w C =.6 kj/kg C.V. C: =, = and ince o = o, / = / o = o R ln / ) = o, o we hae = hu we get w C = w C =.6 kj/kg 600 ka 00 ka 00 ka
Ericon Cycle.85 Conider an ideal air-tandard Ericon cycle that ha an ideal regenerator a hown in Fig..85. he high preure i Ma and the cycle efficiency i 70%. Heat i rejected in the cycle at a temperature of 00 K, and the cycle preure at the beginning of the iothermal compreion proce i 00 ka. Determine the high temperature, the compreor work, and the turbine work per kilogram of air. = = Ma = = 00 K = 00 ka q = - q (ideal reg.) q H = q & w = q H r p = / = 0 H = CARNO H. = - L / H = 0.7 = = H = 000 K q L = -w C = d = R ln = 0.87 00 ln 000 00 = 98.5 w = q H = - d = -R ln( / ) = 660.8 kj/kg
.88 he turbine ection in a jet engine receie ga (aume air) at 00 K, 800 ka with an ambient atmophere at 80 ka. he turbine i followed by a nozzle open to the atmophere and all the turbine work drie a compreor receiing air at 85 ka, 70 K with the ame flow rate. Find the turbine exit preure o the nozzle ha an exit elocity of 800 m/. o what preure can the compreor bring the incomming air? C.V. Reerible and adiabatic turbine and nozzle. hi gie contant, from Eq.8. we can relate the and State : 00 K, 800 ka State : 80 ka; = Eq.8.: Energy: = ( / ) (k-)/k = 00(80/800) 0.857 = 6.56 K h + 0 = h + (/)V + w = h + w w = h - h - (/)V C ( - ) - (/)V =.00(00 6.56) (/) 800 /000 = 580.75 0 = 60.75 kj/kg C.V. Nozzle alone to etablih tate. h = h + (/)V = + (/)V /C = 6.56 + 0/.00 = 90.9 K = + ( / ) k/(k-) = 800 (90.9/00).5 = 0.7 ka C.V. Compreor w c = h e - h i = w = 60.75 kj/kg e = i + w c / C = 70 + 60.75/.00 = 59.7 K Reerible adiabatic compreor, contant gie relation in Eq.8. e = i ( e / i ) k/(k-) = 85 (59.7/70).5 = 899 ka URBINE NOZZLE
.95 o approximate an actual park-ignition engine conider an air-tandard Otto cycle that ha a heat addition of 800 kj/kg of air, a compreion ratio of 7, and a preure and temperature at the beginning of the compreion proce of 90 ka, 0 C. Auming contant pecific heat, with the alue from able A.5, determine the maximum preure and temperature of the cycle, the thermal efficiency of the cycle and the mean effectie preure. Compreion: Reerible and adiabatic o contant from Eq.8.- = ( / ) k = 90(7). = 7 ka = ( / ) k- = 8. 7) 0. = 66.6 K Combution: contant olume = + q H /C V0 = 66.6 + 800/0.77 = 7 K = / = 7 7 / 66.6 = 6958 ka Efficiency and net work H = - / = - 8./66.5 = 0.5 w net = H q H = 0.5 800 = 97.8 kj/kg Diplacement and meff = R / = (0.87 8.)/90 = 0.909 m /kg = (/7) = 0.90 m /kg meff = w NE 97.8 = = 58 ka - 0.909-0.9
.07 A dieel engine ha a bore of 0. m, a troke of 0. m and a compreion ratio of 9: running at 000 RM (reolution per minute). Each cycle take two reolution and ha a mean effectie preure of 00 ka. With a total of 6 cylinder find the engine power in kw and horepower, hp. Work from mean effectie preure, Eq..5. w net meff = => w max net = meff ( max - min ) min he diplacement i V = Bore 0.5 S = 0. 0.5 0. = 0.00086 m Work per cylinder per power troke, Eq..6 W = meff (V max - V min ) = 00 0.00086 ka m =.096 kj/cycle Only eery econd reolution ha a power troke o we can find the power, ee alo Eq..7 Ẇ = W N cyl RM 0.5 (cycle / min) (min / 60 ) (kj / cycle) =.096 6 000 0.5 (/60) = kw = 6 hp he conerion factor from kw to hp i from able A. under power.
Stirling-cycle engine. Conider an ideal Stirling-cycle engine in which the tate at the beginning of the iothermal compreion proce i 00 ka, 5 C, the compreion ratio i 6, and the maximum temperature in the cycle i 00 C. Calculate the maximum cycle preure and the thermal efficiency of the cycle with and without regenerator. Iothermal compreion (heat goe out) = = ( / ) = 00 6 = 600 ka Ideal Stirling cycle = = 5 o C = 00 ka CR = / = 6 = = 00 o C w = q = -R ln( / ) = -0.87 98. ln(6) = -5. kj/kg Contant olume heat addition V = V = / = 600 7./98. = 76 ka q = u u = C o ( - ) = 0.77 (00-5) = 770.8 kj/kg Iothermal expanion (heat come in) w = q = R ln( / ) = 0.87 7. ln6 = 706. kj/kg w net = 706. - 5. = 55.8 kj/kg Efficiency without regenerator, (q and q are coming in from ource) w net 55.8 NO REGEN = q + q 770.8 + 706. = 0.7, Efficiency with regenerator, (Now only q i coming in from ource) WIH REGEN w net q = 55.8 706. = 0.78
. A refrigerator in a laboratory ue R- a the working ubtance. he high preure i 00 ka, the low preure i 0 ka, and the compreor i reerible. It hould remoe 500 W from a pecimen currently at 0 C (not equal to in the cycle) that i inide the refrigerated pace. Find the cycle CO and the electrical power required. State : 0 ka, x =, able B..: h = 9.9 kj/kg, = 0.9685 kj/kg K State : 00 ka, x = 0, able B..: h = 8.57 kj/kg C.V. Compreor Energy Eq.: w C = h - h Entropy Eq.: = + gen = State :. Ma, = = 0.9685 kj/kg, 60 o C, h = 85. kj/kg w C = h - h = 85. 9.9 = 5.9 kj/kg Energy Eq. eaporator: CO Refrigerator: q L = h h = h h = 9.9 8.57 = 58.5 kj/kg = q L w C = 58.5 5.9 =.5 ower: Ẇ IN = Q. L / = 500 W/.5 =.9 W