SOLUION MANUAL CHAPER
CONEN SUBSECION PROB NO. In-ext Concept Quetion a-g Concept problem - Brayton cycle, ga turbine - Regenerator, Intercooler, nonideal cycle 5-9 Ericon cycle 0- Jet engine cycle -5 Air tandard refrigeration cycle 5-59 Otto cycle 60-8 Dieel cycle 8-9 Stirling and Carnot cycle 95-0 Atkinon and Miller cycle 0-09 Combined cycle 0- Availability or Exergy Concept 5- Review -0 Problem re-olved with the Pr, vr function from A.7.:, 7, 7 ee alo 79, 80, 9, 0 and 7
In-ext Concept Quetion
.a he Brayton cycle ha the ame procee a the Rankine cycle, but the - and P-v diagram look very different; why i that? he Brayton cycle have all procee in the uperheated vapor (cloe to ideal ga) region. he Rankine cycle croe in over the two-phae region..b I it alway poible to add a regenerator to the Brayton cycle? What happen when the preure ratio i increaed? No. When the preure ratio i high, the temperature after compreion i higher than the temperature after expanion. he exhaut flow can then not heat the flow into the combutor..c Why would you ue an intercooler between compreor tage? he cooler provide two effect. It reduce the pecific volume and thu reduce the work in the following compreor tage. It alo reduce the temperature into the combutor and thu lower the peak temperature. hi make the control of the combution proce eaier (no autoignition or uncontrollable flame pread), it reduce the formation of NOx that take place at high temperature and lower the cooling requirement for the chamber wall.
.d he jet engine doe not produce haft work; how i power produced? he turbine produce jut enough haft work to drive the compreor and it make a little electric power for the aircraft. he power i produced a thrut of the engine. In order to exhaut the gae at high peed they mut be accelerated o the high preure in the turbine exit provide that force (high P relative to ambient). he high P into the turbine i made by the compreor, which puhe the flow backward, and thu ha a net reulting force forward on the blade tranmitted to the haft and the aircraft. he outer houing alo ha a higher preure inide that give a net component in the forward direction.
.e How i the compreion in the Otto cycle different from the Brayton cycle? he compreion in an Otto cycle i a volume reduction dictated by the piton motion. he phyical handle are the volume V and V. he compreion in a Brayton cycle i the compreor puhing on the flow o it determine the preure. he phyical control i the preure P determined by how much torque you drive the haft with.
.f How many parameter do you need to know to completely decribe the Otto cycle? How about the Dieel cycle? Otto cycle. State ( parameter) and the compreion ratio CR and the energy releae per unit ma in the combution, a total of parameter. With that information you can draw the diagram in Figure.8. Another way of looking at it i four tate (8 propertie) minu the four proce equation ( =, v = v, = and v = v ) give unknown. Dieel cycle. Same a for the Otto cycle namely parameter. he only difference i that one contant v proce i changed to a contant P proce.
.g he exhaut and inlet flow procee are not included in the Otto or Dieel cycle. How do thee neceary procee affect the cycle performance? Due to the preure lo in the intake ytem and the dynamic flow proce we will not have a much ma in the cylinder or a high a P a in a reverible proce. he exhaut flow require a lightly higher preure to puh the flow out through the catalytic converter and the muffler (higher back preure) and the preure lo in the valve o again there i a lo relative to a reverible proce. Both of thee procee ubtract a pumping work from the net work out of the engine and a lower charge ma give le power (not necearily lower efficiency) than otherwie could be obtained.
Concept-Study Guide Problem
. I a Brayton cycle the ame a a Carnot cycle? Name the four procee. No. he Brayton cycle approximate a ga turbine. -: An ientropic compreion (contant ) Compreor -: An iobaric heating (contant P) Combutor -: An ientropic expanion (contant ) urbine -: An iobaric cooling, heat rejection (contant P) Heat exchanger Comment: hi cycle i the ame a the Rankine cycle procee, but it take place in the ideal ga region of tate. he lat proce doe not exit in the real ga turbine which i an open cycle.
. Why i the back work ratio in the Brayton cycle much higher than in the Rankine cycle? Recall the expreion for haft work in a teady flow device w = v dp he pecific volume in the compreor i not o much maller than the pecific volume in the turbine of the Brayton cycle a it i in the pump (liquid) compared to turbine (uperheated vapor) in the Rankine cycle.
. For a given Brayton cycle the cold air approximation gave a formula for the efficiency. If we ue the pecific heat at the average temperature for each change in enthalpy will that give a higher or lower efficiency? he pecific heat are increaing function of temperature. A the expreion for the efficiency i from p.77 η = h h h h = C P - C P - he average from to i lower than the average from to and therefore the ratio of the two pecific heat i lower than one yielding a higher efficiency than the cold air approximation give.
. Doe the efficiency of a jet engine change with altitude ince the denity varie? No, jut like the tandard Brayton cycle the imple model performance depend only on the compreion ratio..5 Why are the two turbine in Figure.7-8 not connected to the ame haft? Such a configuration give a little more flexibility in the control of the cycle under varying load. he two turbine would then not have to run at the ame peed for variou power output level.
.6 Why i an air refrigeration cycle not common for a houehold refrigerator? he capacity of the air cycle per ma flowing through the ytem i very mall compared with the vapor compreion cycle. he cycle alo include the expander which i one more piece of equipment that will add cot and maintenance requirement to the ytem.
.7 Doe the inlet tate (P, ) have any influence on the Otto cycle efficiency? How about the power produced by a real car engine? Very little. he efficiency for the ideal cycle only depend on compreion ratio when we aume cold air propertie. he u are lightly non-linear in o there will be a mall effect. In a real engine there are everal effect. he inlet tate determine the denity and thu the total ma in the chamber. he more ma the more energy i releaed when the fuel burn, the peak P and will alo change which affect the heat tranfer lo to the wall and the formation of NOx (enitive to ). he combution proce may become uncontrollable if i too high (knocking). Some increae in P like that done by a turbo-charger or uper-charger increae the power output and if high, it mut be followed by an intercooler to reduce. If P i too high the loe tart to be more than the gain o there i an optimum level..8 For a given compreion ratio doe an Otto cycle have higher or lower efficiency than a dieel cycle? hi i actually not clear from the formula o we need to refer to the - diagram where we ee that the average at which the heat i added i higher for the Otto cycle than the dieel cycle for the ame compreion ratio. However ince the dieel cycle run with much higher compreion ratio than the Otto cycle mot typical dieel cycle have higher efficiency than typical Otto cycle.
.9 How many parameter do you need to know to completely decribe the Atkinon cycle? How about the Miller cycle? Four parameter for the Atkinon cycle. A total of 8 propertie minu four known proce equation give unknown. he Miller cycle ha one additional proce o that require one more parameter for a total of five. hat i 0 propertie with 5 known proce equation leaving 5 unknown.
.0 Why would one conider a combined cycle ytem for a power plant? For a heat pump or refrigerator? Dual cycle or combined cycle ytem have the advantage of a maller difference between the high and low range for P and. he heat can be added at everal different temperature reducing the difference between the energy ource and the working ubtance. he working ubtance vapor preure at the deired can be reduced from a high value by adding a topping cycle with a different ubtance or have a higher low preure at very low temperature.
. Can the exhaut flow from a ga turbine be ueful? Uually the temperature in the exhaut flow i fairly high compared to ambient condition o we could ue the hot exhaut flow energy. It could be ued directly for heating purpoe in proce application or heating of building. A a topping cycle it can be ued to heat/boil water for ue in a Rankine cycle.. Where may a heat engine driven refrigerator be ueful? Any remote location where electricity i not available. Since a rotating haft i available in a car engine the car A/C unit i driven by a belt engaged with a magnetic clutch o you do not have to have an electric motor.
. Since any heat tranfer i driven by a temperature difference, how doe that affect all the real cycle relative to the ideal cycle? Heat tranfer are given a Q. = CA o to have a reaonable rate the area and the temperature difference mut be large. he working ubtance then mut have a different temperature than the ambient it exchange energy with. hi give a maller temperature difference for a heat engine with a lower efficiency a a reult. he refrigerator or heat pump mut have the working ubtance with a higher temperature difference than the reervoir and thu a lower coefficient of performance (COP). he maller CA i, the larger mut be for a certain magnitude of the heat tranfer rate. hi can be a deign problem, think about the front end air intake grill for a modern car which i very mall compared to a car 0 year ago.
Brayton Cycle, Ga urbine
. In a Brayton cycle the inlet i at 00 K, 00 kpa and the combution add 670 kj/kg. he maximum temperature i 00 K due to material conideration. Find the maximum permiible compreion ratio and for that the cycle efficiency uing cold air propertie. Solution: Combution: h = h + q H ; w = 0 and max = = 00 K = - q H /C P = 00 670/.00 = 5.7 K Reverible adiabatic compreion lead to contant, from Eq.8.8: k P / P = ( / ) k- = (5.7/00).5 = 7.6 Reverible adiabatic expanion lead to contant, from Eq.8.8 k- = (P /P ) k = / = 00 00 5.7 = 675.8 K For net work we get w = C P ( - ) =.00(00 675.8) = 56. kj/kg w C = C P ( - ) =.00(5.7 00) =.6 kj/kg w net = w - w C = 56. -.6 = 9.7 kj/kg η = w net / q H = 9.7 / 670 = 0.7
.5 A Brayton cycle ha compreion ratio of 5: with a high temperature of 600 K and the inlet at 90 K, 00 kpa. Ue cold air propertie and find the pecific heat addition and pecific net work output. Brayton cycle o thi mean: Minimum : = 90 K Maximum : = 600 K Preure ratio: P /P = 5 Compreion in compreor: = Implemented in Eq.8. k- = (P /P ) k = 90(5) 0.86 = 68.65 K Energy input i from the combutor q H = C P0 ( - ) =.00 (600-68.65) = 975. kj/kg Do the overall cycle efficiency and the net work η = Ẇ net Q. H = w net q = - r -(k-)/k = - 5-0./. = 0.587 H p w NE = η q H = 0.587 975. = 55. kj/kg P P = 00 kpa
.6 A large tationary Brayton cycle ga-turbine power plant deliver a power output of 00 MW to an electric generator. he minimum temperature in the cycle i 00 K, and the maximum temperature i 600 K. he minimum preure in the cycle i 00 kpa, and the compreor preure ratio i to. Calculate the power output of the turbine. What fraction of the turbine output i required to drive the compreor? What i the thermal efficiency of the cycle? Solution: Brayton cycle o thi mean: Minimum : = 00 K Maximum : = 600 K Preure ratio: P /P = Solve uing contant C P0 P P = 00 kpa Compreion in compreor: = Implemented in Eq.8. k- = (P /P ) k = 00() 0.86 = 68. K w C = h - h = C P0 ( - ) =.00 (68. - 00) = 9.5 kj/kg Expanion in turbine: = Implemented in Eq.8. k- = (P /P ) k = 600 (/) 0.86 = 75. K w = h h = C P0 ( ) =.00 (600 75.) = 85. kj/kg w NE = 85. - 9.5 = 5.7 kj/kg Do the overall net and cycle efficiency ṁ = Ẇ NE /w NE = 00000/5.7 = 95. kg/ Ẇ = ṁw = 95. 85. = 66. MW w C /w = 9.5/85. = 0.99 Energy input i from the combutor q H = C P0 ( - ) =.00 (600-68.) = 965.7 kj/kg η H = w NE /q H = 5.7/965.7 = 0.50
.7 Conider an ideal air-tandard Brayton cycle in which the air into the compreor i at 00 kpa, 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, value from able A.5. Determine the compreor work, the turbine work, and the thermal efficiency of the cycle. Solution: P v P P = 00 kpa Compreion ratio P P = Max temperature = 00 o C ṁ = 0 kg/ he compreion i reverible and adiabatic o contant. From Eq.8. = P P k- k = 9.() 0.86 = 596.8 K Energy equation with compreor work in w C = - w = C P0 ( - ) =.00(596.8-9.) = 0.8 kj/kg he expanion i reverible and adiabatic o contant. From Eq.8. = k- P P k = 7. 0.86 = 67.7 K Energy equation with turbine work out w = C P0 ( - ) =.00(7. - 67.7) = 70. kj/kg Scale the work with the ma flow rate Ẇ C = ṁw C = 08 kw, Energy added by the combution proce Ẇ = ṁw = 70 kw q H = C P0 ( - ) =.00(7. - 596.8) = 779.5 kj/kg η H = w NE /q H = (70. - 0.8)/779.5 = 0.509
.8 Repeat Problem.7, but aume variable pecific heat for the air, table A.7. Conider an ideal air-tandard Brayton cycle in which the air into the compreor i at 00 kpa, 0 C, and the preure ratio acro the compreor i :. he maximum temperature in the cycle i 00 o C, and the air flow rate i 0 kg/. Aume contant pecific heat for the air, value from able A.5. Determine the compreor work, the turbine work, and the thermal efficiency of the cycle. Solution: From A.7: h = 9.6 kj/kg, o = 6.8597 kj/kg K he compreion i reverible and adiabatic o contant. From Eq.8.8 = o = o + Rln(P /P ) = 6.8597 + 0.87ln = 7.559 = 590 K, h = 597. kj/kg Energy equation with compreor work in w C = - w = h - h = 597. - 9.6 = 0.6 kj/kg he expanion i reverible and adiabatic o contant. From Eq.8.8 From A.7: h = 8. kj/kg, o = 8.5055 kj/kg K = o = o + Rln(P /P ) = 8.5055 + 0.87ln(/) = 7.797 = 7.8 K, h = 75. kj/kg Energy equation with turbine work out w = h - h = 8. - 75. = 7 kj/kg Scale the work with the ma flow rate Ẇ C = ṁw C = 06 kw, Energy added by the combution proce q H = h - h = 8. - 597. = 885.9 kj/kg Ẇ = ṁw = 70 kw w NE = w - w C = 7-0.6 = 8. kj/kg η H = w NE /q H = 8./885.9 = 0.8
.9 A Brayton cycle ha inlet at 90 K, 90 kpa and the combution add 000 kj/kg. How high can the compreion ratio be o the highet temperature i below 700 K? Ue cold air propertie to olve. P v P P = 90 kpa Compreion ratio P / P Max temperature = 700 K Combution add q = h h Let u work back from tate to a q = h h = C P0 ( - ) = - q / C P0 = 700 000/.00 = 70 K he compreion i reverible and adiabatic o contant. From Eq.8. = k- P P P k P = k- = 70 90.5 =.9 k
.0 A Brayton cycle produce net 50 MW with an inlet tate of 7 o C, 00 kpa and the preure ratio i :. he highet cycle temperature i 600 K. Find the thermal efficiency of the cycle and the ma flow rate of air uing cold air propertie. Inlet tate i tate, P /P = and = 600 K. Compreion: Reverible and adiabatic o contant from Eq.8.- = (P /P ) (k-)/k = 90 () 0.857 = 66.8 K η CYCLE = w NE /q H = / = Combution: contant preure 90 66.8 = 0.595 q H = h h = C P0 ( ) =.00(600 66.8) = 987.56 K w NE = η CYCLE q H = 0.595 987.55 = 5.9 kj/kg ṁ = Ẇ/ w NE = 50 000 kw / 5.9 kj/kg = 95.6 kg/ Expanion: ientropic = / (P /P ) (k-)/k = 600/ 0.857 = 75.8 K > Since > ome regeneration can be done. We could find (not needed) η CYCLE = (P /P ) (k-)/k = / = 0.65
. A Brayton cycle produce MW with an inlet tate of 7 o C, 00 kpa, and a compreion ratio of 6:. he heat added in the combution i 960 kj/kg. What are the highet temperature and the ma flow rate of air, auming cold air propertie? Solution: emperature after compreion i = r (k-)/k p = 90 6 0./. = 60.5 K he highet temperature i after combution = + q H /C p = 60.5 + 960.00 = 596.5 K Efficiency i from Eq.. η = Ẇ net Q. = w net q = - r -(k-)/k = - 6-0./. = 0.57 H p H From the required power we can find the needed heat tranfer Q. H = Ẇ 000 net / η = 0.57 = 5 59 kw ṁ = Q. H / q H = 5 59 kw/ 960 kj/kg = 6.66 kg/
. Do Problem. with propertie from table A.7. 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 A.7.: h = 90. kj/kg, o = 6.85 kj/kg K he compreion i reverible and adiabatic o contant. From Eq.8.8 = o = o + Rln(P /P ) = 6.85 + 0.87 ln6 = 7.609 = 6.9 K, h = 6 kj/kg Energy equation with compreor work in w C = - w = h - h = 6-90. = 50.57 kj/kg Energy Eq. combutor: h = h + q H = 6 + 960 = 60 kj/kg State : (P, h): = 7 K, o = 8.588 kj/kg K he expanion i reverible and adiabatic o contant. From Eq.8.8 = o = o + Rln(P /P ) = 8.588 + 0.87ln(/6) = 7.798 = 7.8 K, h = 75. kj/kg Energy equation with turbine work out w = h - h = 60-75. = 89.89 kj/kg Now the net work i w net = w - w C = 89.89 50.57 = 99. kj/kg he total required power require a ma flow rate a ṁ = Ẇ net 000 kw w = net 99. kj/kg = 8.0 kg/
. Solve Problem.5 uing the air table A.7 intead of cold air propertie. Brayton cycle o thi mean: Minimum : = 90 K Maximum : = 600 K Preure ratio: P /P = 5 Compreion in compreor: = Implemented in Eq.8.8 P P = 00 kpa = o = o + Rln(P /P ) = 6.85 + 0.87 ln5 = 7.6 = 60.9 K, h = 69. kj/kg w C = h h = 69. 90. = 8.9 kj/kg Energy input i from the combutor q H = h - h = 757. - 69. = 7.99 kj/kg he expanion i reverible and adiabatic o contant. From Eq.8.8 = o = o + Rln(P /P ) = 8.6905 + 0.87 ln(/5) = 7.9 = 8.0 K, h = 85. kj/kg Energy equation with turbine work out w = h - h = 757. - 85. = 9.9 kj/kg Now the net work i w net = w - w C = 9.9 8.9 = 57 kj/kg [ Comment: Cycle efficiency η = Ẇ net Q. H = w net 57 q = H 7.99 = 0.508 ]
. Solve Problem. with variable pecific heat uing able A.7. A Brayton cycle inlet i at 00 K, 00 kpa and the combution add 670 kj/kg. he maximum temperature i 00 K due to material conideration. What i the maximum allowed compreion ratio? For thi compreion ratio calculate the net work and cycle efficiency uing cold air propertie. Solution: Combution: h = h + q H ; w = 0 and max = = 00 K h = h - q H = 77.8-670 = 607.8 kj/kg From able A.7. 600 K; o = 7.5768 ; = 00 K; o = 6.8696 kj/kg K Reverible adiabatic compreion lead to contant, from Eq.8.8: P / P = exp[ ( o - o )/R ] = exp(.68) =.75 Reverible adiabatic expanion lead to contant, from Eq.8.8 o = o + R ln(p / P ) = 8.596 + 0.87 ln( /.75) = 7.688 kj/kgk From able A.7. by linear interpolation w = h - h = 77.8-65.97 = 6.8 kj/kg w C = h - h = 607.8-00.7 = 07. kj/kg w net = w - w C = 6.8-07. =.5 kj/kg η = w net / q H =.5 / 670 = 0.8 66.6 K, h = 65.97 kj/kg
Regenerator, Intercooler, and Non-ideal Cycle
.5 Would it be better to add an ideal regenerator to the Brayton cycle in problem.0? Inlet tate i tate, P /P = and = 600 K. Compreion: Reverible and adiabatic o contant from Eq.8.- = (P /P ) (k-)/k = 90 () 0.857 = 66.8 K Expanion: ientropic = / (P /P ) (k-)/k = 600/ 0.857 = 75.8 K > Since > ome regeneration can be done. We could find (not needed) η CYCLE = (P /P ) (k-)/k = / = 0.65
.6 A Brayton cycle with an ideal regenerator ha inlet at 90 K, 90 kpa with the highet P, a 70 kpa, 700 K. Find the pecific heat tranfer and the cycle efficiency uing cold air propertie. P v x y P = 90 kpa Compreion ratio P / P Max temperature = 700 K Combution add q H = h h x Let u work from tate to a = P P k- k Energy equation with compreor work in = 90 (70/90) 0.857 = 60.5 K w C = h - h = C P0 ( - ) =.00(60.5-90) =.5 kj/kg = q L he expanion i reverible and adiabatic o contant. From Eq.8. = k- P P k = 700 90 70 0.857 = 86.9 K > Since the exhaut > a regenerator can be ued. Now the heat tranfer added i q H = h h x = C P0 ( - x ) =.00(700 86.9) = 886.6 kj/kg = w η H = w NE /q H = (886.6 -.5)/ 886.6 = 0.66 We could alo have found efficiency a η H = P P k- k = = 60.5 700 = 0.65
.7 An ideal regenerator i incorporated into the ideal air-tandard Brayton cycle of Problem.7. Find the thermal efficiency of the cycle with thi modification. Conider an ideal air-tandard Brayton cycle in which the air into the compreor i at 00 kpa, 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, value from able A.5. Determine the compreor work, the turbine work, and the thermal efficiency of the cycle. Solution: P v x y P = 00 kpa Compreion ratio P P = Max temperature = 00 o C ṁ = 0 kg/ he compreion i reverible and adiabatic o contant. From Eq.8. = P P k- k = 9.() 0.86 = 596.8 K Energy equation with compreor work in w C = h - h = C P0 ( - ) =.00(596.8-9.) = 0.8 kj/kg he expanion i reverible and adiabatic o contant. From Eq.8. = k- P P k = 7. 0.86 = 67.7 K Energy equation with turbine work out Ideal regenerator: w = C P0 ( - ) =.00(7. - 67.7) = 70. kj/kg X = = 67.7 K q H = h - h X =.00(7. - 67.7) = 70. kj/kg = w η H = w NE /q H = (70. - 0.8)/70. = 0.565 We could alo have found efficiency a η H = P P k- k =
.8 Conider an ideal ga-turbine cycle with a preure ratio acro the compreor of to. he compreor inlet i at 00 K and 00 kpa, and the cycle ha a maximum temperature of 600 K. An ideal regenerator i alo incorporated into the cycle. Find the thermal efficiency of the cycle uing cold air (98 K) propertie. Solution: P v x y P = 00 kpa Compreion ratio P P = Max temperature = 600 K he efficiency with an ideal regenerator ( = x ) i from page 85 η H = P P k- k 00 = 600 0.857 = 0.69
.9 A two-tage air compreor ha an intercooler between the two tage a hown in Fig. P.9. he inlet tate i 00 kpa, 90 K, and the final exit preure i.6 MPa. Aume that the contant preure intercooler cool the air to the inlet temperature, =. It can be hown that the optimal preure, P = (P P ) /, for minimum total compreor work. Find the pecific compreor work and the intercooler heat tranfer for the optimal P. Solution: Optimal intercooler preure P = 00 600 = 00 kpa : h = 90. kj/kg, o = 6.85 kj/kg K C.V. C: w C = h - h, = leading to Eq.8.8 o = o + R ln(p /P ) = 6.85 + 0.87 ln = 7. kj/kg K = 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, P /P = P /P o = o + R ln(p /P ) = o, o we have = hu we get w C = w C =.6 kj/kg P v 600 kpa 00 kpa 00 kpa
.0 Aume the compreor in Problem. ha an intercooler that cool the air to 0 K operating at 500 kpa followed by a econd tage of compreion to 600 kpa. Find the pecific heat tranfer in the intercooler and the total combined work required. Solution: C.V. Stage : => Reverible and adiabatic give contant which from Eq.8. give: CV. Intercooler => Contant preure cooling = (P /P ) (k-)/k = 90 (500/00) 0.857 = 59. K w cin = C P ( ) =.00(59. 90) = 87.0 kj/kg q out = h h = C P ( ) =.00 (59. 0) = 9.8 kj/kg C.V. Stage : => Reverible and adiabatic give contant which from Eq.8. give: = (P /P ) (k-)/k = 0 (600/500) 0.857 = 60. K w cin = C P ( - ) =.00(60. 0) = 0.6 kj/kg w tot = w c + w c = 87 + 0.6 = 8 kj/kg he intercooler reduce the work for tage a i lower and o i pecific volume. he reduction in work due to the intercooler i haded in the P-v diagram. P 5 5 v 600 kpa 500 kpa 00 kpa
. he ga-turbine cycle hown in Fig. P. i ued a an automotive engine. In the firt turbine, the ga expand to preure P 5, jut low enough for thi turbine to drive the compreor. he ga i then expanded through the econd turbine connected to the drive wheel. he data for the engine are hown in the figure and aume that all procee are ideal. Determine the intermediate preure P 5, the net pecific work output of the engine, and the ma flow rate through the engine. Find alo the air temperature entering the burner, and the thermal efficiency of the engine. a) Conider the compreor = = P P k- k = 00(6) 0.86 = 500.8 K -w C = -w = C P0 ( - ) =.00(500.8-00) = 0.6 kj/kg Conider then the firt turbine work w = -w C = 0.6 = C P0 ( - 5 ) =.00(600-5 ) 5 = 99. K 5 = P 5 = P P 6 5 k k- = 600 99. 600.5 = 75 kpa b) 6 = 5 6 = 5 k = 99. 00 75 0.86 = 958.8 K P 5 k- he econd turbine give the net work out w = C P0 ( 5-6 ) =.00(99. - 958.8) =. kj/kg ṁ = Ẇ NE /w = 50/. = 0.9 kg/ c) Ideal regenerator = 6 = 958.8 K q H = C P0 ( - ) =.00(600-958.8) = 6.8 kj/kg η H = w NE /q H =./6.8 = 0.687
. Repeat Problem.9 when the intercooler bring the air to = 0 K. he corrected formula for the optimal preure i P = [ P P ( / ) n/(n-) ] / ee Problem 9., where n i the exponent in the aumed polytropic proce. Solution: he polytropic proce ha n = k (ientropic) o n/(n - ) =./0. =.5 P = 00 (0/90).5 = 75. kpa k- C.V. C: = = (P /P ) k = 90 (75./00) 0.857 = 5.67 K w C = h - h = C p ( ) =.00(5.67 90) = 6. kj/kg C.V. Cooler: q OU = h - h =.00(5.67 0) =. kj/kg k- C.V. C: = = (P /P ) k = 0 (600/75.) 0.857 = 5.67 K w C = h - h = C p ( ) =.00(5.67 0) =. kj/kg
. Repeat Problem.6, but include a regenerator with 75% efficiency in the cycle. A large tationary Brayton cycle ga-turbine power plant deliver a power output of 00 MW to an electric generator. he minimum temperature in the cycle i 00 K, and the maximum temperature i 600 K. he minimum preure in the cycle i 00 kpa, and the compreor preure ratio i to. Calculate the power output of the turbine. What fraction of the turbine output i required to drive the compreor? What i the thermal efficiency of the cycle? Solution: Both compreor and turbine are reverible and adiabatic o contant, Eq.8. relate then to P auming contant heat capacity. k- Compreor: = (P /P ) k = 00() 0.86 = 68. K w C = h - h = C P0 ( - ) =.00 (68. - 00) = 9.5 kj/kg k- urbine = = (P /P ) k = 600 (/) 0.86 = 75. K w = h h = C P0 ( ) =.00 (600 75.) = 85. kj/kg w NE = 85. - 9.5 = 5.7 kj/kg ṁ = Ẇ NE /w NE = 00 000/5.7 = 95. kg/ Ẇ = ṁw = 95. 85. = 66. MW w C /w = 9.5/85. = 0.99 x x' P = 00 kpa For the regenerator η REG = 0.75 = h X - h h X' - h = X - X - 68. - = 75. - 68. X = 7.7 K urbine and compreor work not affected by regenerator. Combutor need to add le energy with the regenerator a q H = C P0 ( - X ) =.00(600 7.7) = 879.8 kj/kg η H = w NE /q H = 5.7/879.8 = 0.58
. An air compreor ha inlet of 00 kpa, 90 K and bring it to 500 kpa after which the air i cooled in an intercooler to 0 K by heat tranfer to the ambient 90 K. Aume thi firt compreor tage ha an ientropic efficiency of 85% and it i adiabatic. Uing contant pecific heat and find the compreor exit temperature and the pecific entropy generation in the proce. C.V.: Stage air, Steady flow Ientropic compreor i done firt. Proce: adiabatic: q = 0, reverible: gen = 0 Energy Eq.6.: -w C = h h, Entropy Eq.9.8: = Aume contant C P0 =.00 from A.5 and ientropic lead to Eq.8. k- = (P /P ) k = 90(500/00) 0.86 = 59. K w C = h - h = C Po ( ) =.00(90 59.) = 70 kj/kg Now the actual compreor work become w C ac = w C /η C = -70/0.85 = -00 kj/kg = h h ac = C Po ( - ac ) ac = w C ac /C Po = 90 + 00/.00 = 89. K gen = ac = C Po ln( ac ) Rln( P P ) =.00 ln(89./90) 0.87(500/00) = 0.06 kj/kg-k -W intercooler C Q
.5 A two-tage compreor in a ga turbine bring atmopheric air at 00 kpa, 7 o C to 500 kpa, then cool it in an intercooler to 7 o C at contant P. he econd tage bring the air to 000 kpa. Aume both tage are adiabatic and reverible. Find the combined pecific work to the compreor tage. Compare that to the pecific work for the cae of no intercooler (i.e. one compreor from 00 to 000 kpa). Solution: C.V. Stage : => Reverible and adiabatic give contant which from Eq.8. give: C.V. Stage : => = (P /P ) (k-)/k = 90 (500/00) 0.857 = 59. K w cin = C P ( - ) =.00(59. 90) = 87.0 kj/kg Reverible and adiabatic give contant which from Eq.8. give: = (P /P ) (k-)/k = 00 (000/500) 0.857 = 65.7 K w cin = C P ( - ) =.00(65.7 00) = 65.96 kj/kg w tot = w c + w c = 87 + 65.96 = 5 kj/kg he intercooler reduce the work for tage a i lower and o i pecific volume. C.V. One compreor => 5 Reverible and adiabatic give contant which from Eq.8. give: 5 = (P 5 /P ) (k-)/k = 90 (000/00) 0.857 = 559.88 K w in = C P ( 5 - ) =.00(559.88 90) = 7 kj/kg P 5 v 5 000 kpa 500 kpa 00 kpa he reduction in work due to the intercooler i haded in the P-v diagram.
.6 Repeat Problem.6, but aume that the compreor ha an ientropic efficiency of 85% and the turbine an ientropic efficiency of 88%. Solution: Brayton cycle o thi mean: Minimum : = 00 K Maximum : = 600 K Preure ratio: P /P = Solve uing contant C P0 P P = 00 kpa Ideal compreor: = Implemented in Eq.8. Actual compreor k- = (P /P ) k = 00() 0.86 = 68. K w C = h - h = C P0 ( - ) =.00 (68. - 00) = 9.5 kj/kg w C = w SC /η SC = 9.5/0.85 = 99. kj/kg = C P0 ( - ) = + w c /C P0 = 00 + 99./.00 = 697.8 K Ideal turbine: = Implemented in Eq.8. Actual turbine k- = (P /P ) k = 600 (/) 0.86 = 75. K w = h h = C P0 ( ) =.00 (600 75.) = 85. kj/kg w = η S w S = 0.88 85. = 79. kj/kg = C P0 ( - ) = - w /C P0 = 600-79./.00 = 85.9 K Do the overall net and cycle efficiency w NE = w - w C = 79. - 99. = 9.7 kj/kg ṁ = Ẇ NE /w NE = 00000/9.7 = 86.0 kg/ Ẇ = ṁw = 86.0 79. =. MW w C /w = 99./79. = 0.5 Energy input i from the combutor q H = C P0 ( - ) =.00(600-697.8) = 905.8 kj/kg η H = w NE /q H = 9.7/905.8 = 0.86
.7 A ga turbine with air a the working fluid ha two ideal turbine ection, a hown in Fig. P.7, the firt of which drive the ideal compreor, with the econd producing the power output. he compreor input i at 90 K, 00 kpa, and the exit i at 50 kpa. A fraction of flow, x, bypae the burner and the ret ( x) goe through the burner where 00 kj/kg i added by combution. he two flow then mix before entering the firt turbine and continue through the econd turbine, with exhaut at 00 kpa. If the mixing hould reult in a temperature of 000 K into the firt turbine find the fraction x. Find the required preure and temperature into the econd turbine and it pecific power output. C.V.Comp.: -w C = h - h ; = Reverible and adiabatic give contant which from Eq.8. give: = (P /P ) (k-)/k = 90 (50/00) 0.857 = 5.7 K h = 7.75 kj/kg, -w C = 7.75-90. = 57. kj/kg C.V.Burner: h = h + q H = 7.75 + 00 = 67.75 kj/kg = 50 K C.V.Mixing chamber: ( - x)h + xh = h MIX = 06. kj/kg x = h - h MIX 67.75-06. h - h = 67.75-7.75 = 0.50 Ẇ = Ẇ C,in ẇ = -w C = 57. = h - h h = 06. - 57. = 888.9 kj/kg = 860 K P = P MIX ( / MIX ) k/(k-) = 50 (860/000).5 = 65 kpa = 5 5 = (P 5 /P ) (k-)/k = 860 (00/65) 0.857 = 65 K h 5 = 66. kj/kg w = h - h 5 = 888.9-66. = 7.7 kj/kg
.8 A ga turbine cycle ha two tage of compreion, with an intercooler between the tage. Air enter the firt tage at 00 kpa, 00 K. he preure ratio acro each compreor tage i 5 to, and each tage ha an ientropic efficiency of 8%. Air exit the intercooler at 0 K. Calculate the temperature at the exit of each compreor tage and the total pecific work required. Solution: State : P = 00 kpa, = 00 K State : = 0 K P = 5 P = 500 kpa; P = 5 P = 500 kpa Energy Eq.: w c + h = h => w c = h - h C P ( - ) Ideal C contant, Eq.8.: = (P /P ) (k-)/k = 75. K w c = h - h C P ( - ) = 76.0 kj/kg, Actual Eq.9.8: w c = w c /η = 76/0.8 =.6 kj/kg = + w c /C P = 5.7 K Ideal C contant, Eq.8.: = (P /P ) (k-)/k = 55.6 K w c = h - h C P ( - ) = 9. kj/kg; Actual Eq.9.8: w c = w c /η = 5.9 kj/kg = + w c / C P = 565 K otal work in: w = w c + w c =.6 + 5.9 = 50.5 kj/kg P ac ac v ac ac
.9 Repeat the quetion in Problem. when we aume that friction caue preure drop in the burner and on both ide of the regenerator. In each cae, the preure drop i etimated to be % of the inlet preure to that component of the ytem, o P = 588 kpa, P = 0.98 P and P 6 = 0 kpa. Solution: k- P k P a) From olution.: = = 00(6) 0.86 = 500.8 K w C = -w = C P0 ( - ) =.00(500.8-00) = 0.6 kj/kg P = 0.98 600 = 588 kpa, P = 0.98 588 = 576. kpa k 5 = P 5 = P ( 5S / ) k- = 576.( 99. 600 ).5 = 60. kpa b) P 6 = 00/0.98 = 0 kpa, 6S = 5 6 = 5 k- P 6 P 5 k = 99. 0 9.8 0.86 = 975. K w S = C P0 ( 5-6 ) =.00(99. - 975.) = 5.7 kj/kg ṁ = Ẇ NE /w NE = 50/5.7 = 0.5 kg/ c) = 6 = 975. K q H = C P0 ( - ) =.00 (600-975.) = 67. kj/kg η H = w NE /q H = 5.7/67. = 0.678
Ericon Cycle
.0 Conider an ideal air-tandard Ericon cycle that ha an ideal regenerator a hown in Fig. P.0. he high preure i MPa and the cycle efficiency i 70%. Heat i rejected in the cycle at a temperature of 50 K, and the cycle preure at the beginning of the iothermal compreion proce i 50 kpa. Determine the high temperature, the compreor work, and the turbine work per kilogram of air. P P P v P P P = P = MPa = = 50 K P = 50 kpa q = - q (ideal reg.) q H = q & w = q H r p = P /P = 0 η H = η CARNO H. = - L / H = 0.7 = = H = 67 K P q L = w C = v dp = R ln P = 0.87 50 ln 000 50 = 90.56 kj/kg w = q H = - v dp = -R ln(p /P ) = 65. kj/kg
. An air-tandard Ericon cycle ha an ideal regenerator. Heat i upplied at 000 C and heat i rejected at 80 C. Preure at the beginning of the iothermal compreion proce i 70 kpa. he heat added i 700 kj/kg. Find the compreor work, the turbine work, and the cycle efficiency. Solution: Identify the tate Heat upplied at high temperature Heat rejected at low temperature Beginning of the compreion: Ideal regenerator: = = 000 C = 7.5 K = = 80 C = 5.5 K P = 70 kpa q = - q q H = q = 700 kj/kg w = q H = 700 kj/kg η H = η CARNO = - 5.5 7.5 = 0.76 w NE = η H q H = 0.76 700 = 505.8 kj/kg w C = q L = q H - w NE = 700 505.8 = 9. kj/kg P P P v P P
Jet Engine Cycle
. he Brayton cycle in Problem.6 i changed to be a jet engine. Find the exit velocity uing cold air propertie. A large tationary Brayton cycle ga-turbine power plant deliver a power output of 00 MW to an electric generator. he minimum temperature in the cycle i 00 K, and the maximum temperature i 600 K. he minimum preure in the cycle i 00 kpa, and the compreor preure ratio i to. Calculate the power output of the turbine. What fraction of the turbine output i required to drive the compreor? What i the thermal efficiency of the cycle? Solution: Brayton cycle o thi mean: Minimum : = 00 K Maximum : = 600 K Preure ratio: P /P = Solve uing contant C P0 P 5 P = 00 kpa Compreion in compreor: = Implemented in Eq.8. k- = (P /P ) k = 00() 0.86 = 68. K w C = h - h = C P0 ( - ) =.00 (68. - 00) = 9.5 kj/kg urbine (w = w C ) and nozzle: 5 = = Implemented in Eq.8. k- 5 = (P 5 /P ) k = 600 (/) 0.86 = 75. K h h 5 = C P0 ( 5 ) =.00 (600 75.) = 85. kj/kg /V 5 = h h 5 w C = 85. - 9.5 = 5.7 kj/kg V 5 = 000 5.7 = 0 m/
. Conider an ideal air-tandard cycle for a ga-turbine, jet propulion unit, uch a that hown in Fig..9. he preure and temperature entering the compreor are 90 kpa, 90 K. he preure ratio acro the compreor i to, and the turbine inlet temperature i 500 K. When the air leave the turbine, it enter the nozzle and expand to 90 kpa. Determine the velocity of the air leaving the nozzle. Solution: BURNER COMPR. URBINE P 5 NOZ 5 P = 90 kpa C.V. Compreor: Reverible and adiabatic = From Eq.8.5, 8. = P P k- k = 90 () 0.857 = 66. K w C = h - h C P0 ( ) =.00 (66. 90) = 7.7 kj/kg C.V. urbine: w = h - h = w C and = = - w C /C P0 = 500-7.7/.00 = 7.6 K C.V. Nozzle: 5 = = 5 = k- P 5 P Now the energy equation o from Eq.8. k = 500 90 0.857 60 = 705.7 K (/)V 5 = h - h 5 C P0 ( 5 ) =.00 (7.6 705.7) = 69.77 kj/kg V 5 = 000 69.77 = 969 m/
. Solve the previou problem uing the air table. Conider an ideal air-tandard cycle for a ga-turbine, jet propulion unit, uch a that hown in Fig..9. he preure and temperature entering the compreor are 90 kpa, 90 K. he preure ratio acro the compreor i to, and the turbine inlet temperature i 500 K. When the air leave the turbine, it enter the nozzle and expand to 90 kpa. Determine the preure at the nozzle inlet and the velocity of the air leaving the nozzle. C.V. Compreor: Reverible and adiabatic = From Eq.8.8 o = o + R ln(p /P ) = 6.85 + 0.87 ln = 7.596 kj/kg K From A.7 h = 67. kj/kg, = 609. K w C = h - h = 67. - 90. = 6.8 kj/kg C.V. urbine: w = h - h = w C and = h = h - w C = 65.8-6.8 = 09 kj/kg o = 8.7 kj/kg K, = 7 K P = P exp[( o - o )/R] = 60 exp[ (8.7-8.608)/0.87 ] C.V. Nozzle: 5 = = = 60 exp(-0.885) = 5.8 kpa o from Eq.8.8 o 5 = o + R ln(p 5/P ) = 8.608 + 0.87 ln (/) = 7.8567 kj/kgk => From A.7 5 = 778 K, h 5 = 798. kj/kg Now the energy equation (/)V 5 = h - h 5 = 50.8 V 5 = 000 50.8 = 0 m/ BURNER COMPR. URBINE P 5 NOZ 5 P = 90 kpa
.5 he turbine ection in a jet engine receive ga (aume air) at 00 K, 800 kpa with an ambient atmophere at 80 kpa. he turbine i followed by a nozzle open to the atmophere and all the turbine work drive a compreor. Find the turbine exit preure o the nozzle ha an exit velocity of 800 m/. Hint: take the CV around both turbine and nozzle. Solution: C.V. Reverible and adiabatic turbine and nozzle. hi give contant, from Eq.8. we can relate the and P State : 00 K, 800 kpa State 5: 80 kpa; 5 = Eq.8.: Energy: 5 = (P 5 /P ) (k-)/k = 00(80/800) 0.857 = 6.56 K h + 0 = h 5 + (/)V 5 + w = h + w w = h h 5 (/)V 5 C P ( 5 ) (/)V =.00(00 6.56) (/) 800 /000 = 580.75 0 = 60.75 kj/kg C.V. Nozzle alone to etablih tate (ame a tate 5 and ). h = h 5 + (/)V 5 = h w = 5 + (/)V 5 /C P = 6.56 + 0/.00 = 90.9 K P = P ( / ) k/(k-) = 800 (90.9/00).5 = 0.7 kpa URBINE w NOZZLE 5 5
.6 Given the condition in the previou problem what preure could an ideal compreor generate (not the 800 kpa but higher). C.V. Reverible and adiabatic turbine and nozzle. hi give contant, from Eq.8. we can relate the and P State : 00 K, 800 kpa State 5: 80 kpa; 5 = Eq.8.: Energy: 5 = (P 5 /P ) (k-)/k = 00(80/800) 0.857 = 6.56 K h + 0 = h 5 + (/)V 5 + w = h + w w = h h 5 (/)V 5 C P ( 5 ) (/)V =.00(00 6.56) (/) 800 /000 = 580.75 0 = 60.75 kj/kg C.V. Compreor w c = h - h = w = 60.75 kj/kg = + w c / C P = 70 + 60.75/.00 = 59.7 K Reverible adiabatic compreor, contant give relation in Eq.8. P = P ( / ) k/(k-) = 85 (59.7/70).5 = 899 kpa URBINE w NOZZLE 5 5
.7 Conider a turboprop engine where the turbine power the compreor and a propeller. Aume the ame cycle a in Problem. with a turbine exit temperature of 900 K. Find the pecific work to the propeller and the exit velocity. BURNER COMPR. URBINE P 5 NOZ 5 P = 90 kpa C.V. Compreor: Reverible and adiabatic = From Eq.8.5, 8. = P P k- k = 90 () 0.857 = 66. K w C = h - h C P0 ( ) =.00 (66. 90) = 7.7 kj/kg C.V. urbine: w = h - h = w C + w prop and = w prop = C P0 ( ) w C =.00(500 900) 7.7 = 7.7 kj/kg C.V. Nozzle: 5 = = 5 = k- P 5 P Now the energy equation o from Eq.8. k = 500 90 0.857 60 = 705.7 K (/)V 5 = h - h 5 C P0 ( 5 ) =.00 (900 705.7) = 95.08 kj/kg V 5 = 000 95.08 = 65 m/
.8 Conider an air tandard jet engine cycle operating in a 80 K, 00 kpa environment. he compreor require a haft power input of 000 kw. Air enter the turbine tate at 600 K, MPa, at the rate of 9 kg/, and the ientropic efficiency of the turbine i 85%. Determine the preure and temperature entering the nozzle at tate. If the nozzle efficiency i 95%, determine the temperature and velocity exiting the nozzle at tate 5. Solution: C.V. Shaft: Ẇ = ṁ(h - h ) = Ẇ C CV urbine: h - h = Ẇ C / ṁ = 000/9 =. kj/kg = w a / C p = 600./.00 = 57. K Work back to the ideal turbine condition Eq.9.7: w = w a / η =./0.85 = 5.8 = h - h = C p ( - ) = 600 5.8/.00 = 079. K, P = P ( / ) k/(k-) = 000 (079./600).5 = 50. kpa C.V. Nozzle: Firt the ideal (reverible and adiabatic i.e. contant, Eq.8.) 5 = (P 5 /P ) (k-)/k = 57. (00/50.) (k-)/k = 79 K Energy Eq.: (/)V 5 = h - h 5 = C p ( - 5 ) =.00(57. - 79) = 0. kj/kg Now conider the actual nozzle Eq.9.0: 0.5V 5a = η(0.5v 5 ) = 08.6 kj/kg V 5a = 000 08.6 = 90 m/ 5a = - 0.5V 5a /C p = 57. 08.6 /.00 = 750 K
.9 Solve the previou problem uing the air table. Solution done with the air table A.7: C.V. Shaft: Ẇ = ṁ(h - h ) = Ẇ C CV urbine: h - h = Ẇ C / ṁ = 000/9 =. kj/kg h = 757.. =.9 kj/kg Work back to the ideal turbine condition Eq.9.7: w a = w C =. w = w a / η = 5.8 = h - h h =.5 6 K, o = 8.09 kj/kg K - = 0 = o - o - R ln(p /P ) 0 = 8.09-8.6905-0.87 ln(p /000) => P = 50 kpa State from A.7.: h =.9, = 9.8 K, o = 8.76 kj/kg K Firt conider the reverible adiabatic (ientropic) nozzle o from Eq.8.8 5 - = 0 = o 5 - o - R ln(p 5/P ) o 5 = 8.76 + 0.87 ln(00/50) = 7.8960 kj/kg K able A.7.: 5 = 808. K, h 5 = 8.0 kj/kg 0.5V 5 = h - h 5 =.9-8.0 = 8.9 kj/kg Now conider the actual nozzle Eq.9.0: 0.5V 5a = η(0.5v 5 ) = 57.8 kj/kg V 5a= 957 m/ h 5a = h - 0.5V 5a =.9 57.8 = 855. kj/kg 5a 80 K
.50 A jet aircraft i flying at an altitude of 900 m, where the ambient preure i approximately 55 kpa and the ambient temperature i 8 C. he velocity of the aircraft i 80 m/, the preure ratio acro the compreor i : and the cycle maximum temperature i 50 K. Aume the inlet flow goe through a diffuer to zero relative velocity at tate. Find the temperature and preure at tate. Solution: x 5 P = 55 kpa Ambient X = -8 o C = 55. K, P X = 55 kpa = P 5 alo V X = 80 m/ Aume that the air at thi tate i reveribly decelerated to zero velocity and then enter the compreor at. P /P = & = 50 K C.V. Diffuer ection, tate i the tagnation tate. EnergyEq.: = X + V X 000 Cp = 55. + (80) 000.00 = 9. K X k Eq.8.: P = P X k- = 55 9. 55..5 = 90.5 kpa
.5 he turbine in a jet engine receive air at 50 K,.5 MPa. It exhaut to a nozzle at 50 kpa, which in turn exhaut to the atmophere at 00 kpa. 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. C.V. urbine: Firt the ideal (reverible and adiabatic i.e. contant, Eq.8.) e = i (P e /P i ) = 50 (50/500) (k-)/k = 79. K Energy Eq.: w, = h i - h e = C p ( i - e ) =.00 (50-79.) = 50.8 kj/kg Eq.9.7: w,ac = w, η = 7. kj/kg = h i - h e,ac = C p ( i - e,ac ) e,ac = 50 7. /.00 = 8. K, C.V. Nozzle: Firt the ideal (reverible and adiabatic i.e. contant, Eq.8.) e = i (P e /P i ) (k-)/k = 8. (00/50) (k-)/k = 6. K Energy Eq.: (/)V e, = h i - h e, =.00(8. - 6.) = 90.66 kj/kg Eq.9.0: V e,ac = (/)V e,ac = (/)V e, η NOZ = 8. kj/kg 000 8. = 60 m/
.5 Solve the previou problem uing the air table. Solution uing air table A.7: C.V. urbine: h i = 6.7, o i = 8.90, e = i then from Eq.8.8 o e = o i + R ln(p e/p i ) = 8.90 + 0.87 ln (50/500) = 7.8798 kj/kg K able A.7. e = 796 K, h e = 87.9 kj/kg, Energy Eq.: w, = h i - h e = 6.7-87.9 = 58.8 kj/kg Eq.9.7: w,ac = w, η = kj/kg = h i - h e,ac h e,ac = 895.7 e,ac = 866 K, o e = 7.970 kj/kg K C.V. Nozzle: h i = 895.7 kj/kg, o i = 7.970 kj/kgk, e = i then from Eq.8.8 o e = o i + R ln(p e/p i ) = 7.970 + 0.87 ln (00/50) = 7.700 kj/kgk able A.7. e, = 68 K, h e, = 69. kj/kg Energy Eq.: (/)V e, = h i - h e, = 895.7-69. = 0.6 kj/kg Eq.9.0: V e,ac = (/)V e,ac = (/)V e, η NOZ = 9.7 kj/kg 000 9.7 = 60 m/
.5 An afterburner in a jet engine add fuel after the turbine thu raiing the preure and temperature due to the energy of combution. Aume a tandard condition of 800 K, 50 kpa after the turbine into the nozzle that exhaut at 95 kpa. Aume the afterburner add 50 kj/kg to that tate with a rie in preure for ame pecific volume, and neglect any uptream effect on the turbine. Find the nozzle exit velocity before and after the afterburner i turned on. Solution: Before afterburner i on: After afterburner i on: : 800 K; 50 kpa and : 95 kpa : v = v and : 95 kpa v P = 95 kpa Aume reverible adiabatic nozzle flow, then contant from Eq.8. = (P /P ) (k-)/k = 800 (95/50) 0.857 = 606.8 K Energy Eq.: (/)V = C P ( - ) V = C P ( - ) = 00(800-606.8) = 6.8 m/ Add the q AB at aumed contant volume then energy equation give = + q AB /C v = 800 + 50/0.77 = 7.6 K v = v => P = P ( / ) = 50 7.6/800 = 6. kpa Reverible adiabatic expanion, again from Eq.8. = (P /P ) (k-)/k = 7.6 (95/6.) 0.857 = 97.7 K V = C P ( - ) = 00(7.6-97.7) = 0 m/
Air-tandard refrigeration cycle
.5 An air tandard refrigeration cycle ha air into the compreor at 00 kpa, 70 K with a compreion ratio of :. he temperature after heat rejection i 00 K. Find the COP and the lowet cycle temperature. From the ientropic compreion/expanion procee k- / = (P /P ) k = 0.857 =.687 = / = /.687 = 9. K he COP, Eq..5, i β = = [.687 - ] =.7 -
.55 A tandard air refrigeration cycle ha -0 o C, 00 kpa into the compreor, and the ambient cool the air down to 5 o C at 00 kpa. Find the lowet temperature in the cycle, the low pecific heat tranfer and the pecific compreor work. Solution: State : 5 o C = 08. K, 00 kpa he lowet i at tate which we can relate to tate k- = (P /P ) k = 08. 00 00 0.857 = 0.9 k = 68. o C q = h h = C p ( ) =.00 (-0 + 68.) = 58.5 kj/kg k- = (P /P ) k = 6. 00 00 0.857 = 9. K w c = h h = C p ( ) =.00 (6. 9.) = 8.6 kj/kg
.56 he formula for the COP auming cold air propertie i given for the tandard refrigeration cycle in Eq..5. Develop the imilar formula for the cycle variation with a heat exchanger a hown in Fig... Definition of COP: β = q L w net = q L q H - q L = q H q - L Auming an ideal heat exchanger h = h 6 and h = h o q L = w E From the refrigeration cycle we get the ratio of the heat tranfer a q H q = C p( - ) L C p ( 6-5 ) = - - = ( / ) 5 5 ( / 5 ) he preure ratio are the ame and we have ientropic compreion/expanion P P = P P = k/(k-) 5 = k/(k-) 5 o now we get = q H o 5 q = L = 5 = (k-)/k 5 r P and the COP reduce to β = = 5 (k-)/k r P
.57 Aume a refrigeration cycle a hown in Fig.. with reverible adiabatic compreor and expander. For thi cycle the low preure i 00 kpa and the high preure i. MPa with contant preure heat exchanger, ee Fig.. - diagram. he temperature are = 6 = -50 o C and = = 5 o C. Find the COP for thi refrigeration cycle. Solution: q 6 L COMP q H EXP 5 5 6 Standard air refrigeration cycle with P = 00 kpa, P =. MPa We will olve the problem with cold air propertie. Compreor, ientropic = o from Eq.8. k- = (P /P ) k = 88.(00/00) 0.86 = 6 K w C = -w = C P0 ( - ) =.00(6-88.) = 6 kj/kg Expanion in expander (turbine) k- 5 = 5 = (P 5 /P ) k =.(00/00) 0.86 = 0.9 K w E = C P0 ( - 5 ) =.00(. - 0.9) = 8.7 kj/kg Net cycle work w NE = w E - w C = 8.7-6.0 = -07. kj/kg q L = C P0 ( 6-5 ) = w E = 8.7 kj/kg Overall cycle performance, COP β = q L /w NE = 8.7 / 07. = 0.57
.58 Repeat Problem.57, but aume that helium i the cycle working fluid intead of air. Dicu the ignificance of the reult. A heat exchanger i incorporated into an ideal air-tandard refrigeration cycle, a hown in Fig. P.57. It may be aumed that both the compreion and the expanion are reverible adiabatic procee in thi ideal cae. Determine the coefficient of performance for the cycle. Solution: q 6 L COMP q H EXP Standard air refrigeration cycle with helium and tate a = = 5 o C = 88. K, P = 00 kpa, P =. MPa = 6 = -50 o C =. K Compreor, ientropic = o from Eq.8. k- = (P /P ) k = 88. 00 00 0.0 = 88. K w C = -w = C P0 ( - ) = 5.9(88. - 88.) = 80. kj/kg Expanion in expander (turbine) k- 5 = 5 = (P 5 /P ) k =. 00 00 0.0 = 77.7 K w E = C P0 ( - 5 ) = 5.9(. - 77.7) = 755.5 kj/kg Net cycle work w NE = 755.5-80. = -08.6 kj/kg q L = C P0 ( 6-5 ) = 5.9(. - 77.7) = 755.5 kj/kg Overall cycle performance, COP β = q L /w NE = 755.5/08.6 = 0.69 Notice that the low temperature i lower and work term higher than with air. It i due to the higher heat capacity C P0 and ratio of pecific heat ( k = /). he expene i a lower COP requiring more work input per kj cooling. 5 5 6
.59 Repeat Problem.57, but aume an ientropic efficiency of 75% for both the compreor and the expander. Standard air refrigeration cycle with = = 5 o C = 88. K, P = 00 kpa, P =. MPa = 6 = -50 o C =. K We will olve the problem with cold air propertie. Ideal compreor, ientropic S = o from Eq.8. k- S = (P /P ) k = 88.(00/00) 0.86 = 6 K w SC = -w = C P0 ( S - ) =.00(6-88.) = 6 kj/kg he actual compreor w C = w SC / η SC = 6/0.75 =.6 kj/kg Expanion in ideal expander (turbine) k- 5 = 5S = (P 5 /P ) k =.(00/00) 0.86 = 0.9 K w E = C P0 ( - 5 ) =.00(. - 0.9) = 8.7 kj/kg he actual expander (turbine) w E = η SE w SE = 0.75 8.7 = 89.0 kj/kg = C P0 ( - 5 ) =.00(. - 5 ) 5 =.5 K 5S w NE = 89.0 -.6 = -5.6 kj/kg q L = C P0 ( 6-5 ) =.00(. -.5) = 89.0 kj/kg β = q L /(-w NE ) = 89.0/5.6 = 0.58 5 S 6
Otto Cycle
.60 A troke gaoline engine run at 800 RPM with a total diplacement of.l and a compreion ratio of 0:. he intake i at 90 K, 75 kpa with a mean effective preure of 600 kpa. Find the cycle efficiency and power output. Efficiency from the compreion ratio η = CR -k = 0-0. = 0.60 he power output come from peed and diplacement in Eq.. Ẇ = P meff V dipl RPM 60 800 = 600 0.00 60 =.6 kw
.6 A troke gaoline. L engine running at 000 RPM ha inlet tate of 85 kpa, 80 K and after combution it i 000 K and the highet preure i 5 MPa. Find the compreion ratio, the cycle efficiency and the exhaut temperature. Solution: Combution v = v. Highet and P are after combution. CR = v /v = v /v = ( P / P ) = Efficiency from the compreion ratio Expanion η = CR -k = 8.5-0. = 0.57 80 5000 000 85 = 8.5 = (v /v ) k- = [ / CR ] k- = 000 (/ 8.5) 0. = 860.5 K P v v
.6 Find the power from the engine in Problem.6 Solution: Combution v = v. Highet and P are after combution. CR = v /v = v /v = ( P / P ) = Efficiency from the compreion ratio η = CR -k = 8.5-0. = 0.57 80 5000 000 85 = 8.5 Compreion: = (CR) k- = 80 (8.5) 0. = 650.8 K q H = u - u = C v ( ) = 0.77 (000 650.8) = 967. kj/kg v = R /P = 0.87 80/85 = 0.95 m /kg Diplacement and then mep from net work v - v = v - v /CR = v [ (/CR)] = 0.806 m /kg P meff = wnet/(v v ) = η q H /( v - v ) = 0.57 967./0.806 = 66.9 kpa otal power from Eq.. and troke cycle Ẇ = P meff V dipl RPM 60 000 = 66.9 0.00 60 = 6.5 kw P v v
.6 Air flow into a gaoline engine at 95 kpa, 00 K. he air i then compreed with a volumetric compreion ratio of 8:. In the combution proce 00 kj/kg of energy i releaed a the fuel burn. Find the temperature and preure after combution uing cold air propertie. Solution: Solve the problem with contant heat capacity. Compreion to : = From Eq.8. and Eq.8. = (v /v ) k- = 00 8 0. = 689. K P = P (v /v ) k = 95 8. = 76 kpa Combution to at contant volume: u = u + q H = + q H /C v = 689. + 00/0.77 = 50 K P = P ( / ) = 76 (50 / 689.) = 68 kpa P v v
.6 A. L gaoline engine run at 500 RPM with a compreion ratio of 9:. he tate before compreion i 0 kpa, 80 K and after combution it i at 000 K. Find the highet and P in the cycle, the pecific heat tranfer added, the cycle efficiency and the exhaut temperature. hi i a baic -troke Otto cycle. Compreion: = (CR) k- = 80 (9) 0. = 67. K P = P CR k = 0 9. = 866.96 kpa Combution v = v. Highet and P are after combution. = 000 K, P = P / = 866.96 000 67. = 57. kpa q H = u u = C v ( ) = 0.77 (000 67.) = 950.5 kj/kg Efficiency from the compreion ratio η = CR -k = 9-0. = 0.585 Expanion with volume ratio equal compreion ratio = CR -k = 000 9-0. = 80.5 K
.65 Suppoe we reconider the previou problem and intead of the tandard ideal cycle we aume the expanion i a polytropic proce with n =.5. What are the exhaut temperature and the expanion pecific work? hi i a modified -troke Otto cycle. If a polytropic expanion with n =.5 intead then mod = CR -n = 000 9-0.5 = 667 K hi mean a heat lo out and thu le work (ee diagram) w = R n ( mod ) = 0.87.5 (667 000) = 765. kj/kg P v mod mod v
.66 A gaoline engine ha a volumetric compreion ratio of 8 and before compreion ha air at 80 K, 85 kpa. he combution generate a peak preure of 6500 kpa. Find the peak temperature, the energy added by the combution proce and the exhaut temperature. Solution: Solve the problem with cold air propertie. Compreion. Ientropic o we ue Eq.8.-8. P = P (v /v ) k = 85(8). = 56 kpa = (v /v ) k- = 80(8) 0. = 6. K Combution. Contant volume = (P /P ) = 6. 6500/56 = 677 K q H = u - u C v ( - ) = 0.77 (677 6.) = 58 kj/kg Exhaut. Ientropic expanion o from Eq.8. = /8 0. = 677/.97 = 65 K P v v
.67 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 kpa, 0 C. Auming contant pecific heat, with the value from able A.5, determine the maximum preure and temperature of the cycle, the thermal efficiency of the cycle and the mean effective preure. Solution: P v v Compreion: Reverible and adiabatic o contant from Eq.8.- P = P (v /v ) k = 90(7). = 7 kpa = (v /v ) k- = 8. (7) 0. = 66.6 K Combution: contant volume = + q H /C V0 = 66.6 + 800/0.77 = 7 K P = P / = 7 7 / 66.6 = 6958 kpa Efficiency and net work η H = - / = - 8./66.5 = 0.5 w net = η H q H = 0.5 800 = 97.8 kj/kg Diplacement and P meff v = R /P = (0.87 8.)/90 = 0.909 m /kg v = (/7) v = 0.90 m /kg P meff = w NE 97.8 v -v = 0.909-0.9 = 58 kpa
.68 A. L minivan engine run at 000 RPM with a compreion ratio of 0:. he intake i at 50 kpa, 80 K and after expanion it i at 750 K. Find the highet in the cycle, the pecific heat tranfer added by combution and the mean effective preure. P v v Compreion: Reverible and adiabatic o contant from Eq.8.- = (v /v ) k- = 80 (0) 0. = 70. K Expanion: ientropic, i the highet and wa given = CR k- = 750 0 0. = 88.9 K q H = u u = C V0 ( - ) = 0.77 (88.9 70.) = 86.5 kj/kg Efficiency and net work η H = - / = - 80/70. = 0.60 w net = η H q H = 0.60 86.5 = 509.6 kj/kg Diplacement and P meff v = R /P = (0.87 80)/50 =.607 m /kg v = (/0) v = 0.607 m /kg P meff = w NE 509.6 v -v =.607-0.607 = 5. kpa
.69 A gaoline engine take air in at 90 K, 90 kpa and then compree it. he combution add 000 kj/kg to the air after which the temperature i 050 K. Ue cold air propertie (i.e. contant heat capacitie at 00 K) and find the compreion ratio, the compreion pecific work and the highet preure in the cycle. Solution: Standard Otto Cycle Combution proce: = 050 K; u = u q H = q H / C vo = 050 000 / 0.77 = 655. K Compreion proce P = P ( / ) k/(k-) = 90(655./90).5 = 56 kpa CR = v / v = ( / ) /(k-) = (655. / 90).5 = 7.67 w = u - u = C vo ( - ) = 0.77(655. - 90) = 6 kj / kg Highet preure i after the combution where v = v o we get P = P / = 56 050 / 655. = 88 kpa P v v
.70 Anwer the ame three quetion for the previou problem, but ue variable heat capacitie (ue table A.7). A gaoline engine take air in at 90 K, 90 kpa and then compree it. he combution add 000 kj/kg to the air after which the temperature i 050 K. Ue the cold air propertie (i.e. contant heat capacitie at 00 K) and find the compreion ratio, the compreion pecific work and the highet preure in the cycle. Solution: Standard Otto cycle, olve uing able A.7. Combution proce: = 050 K ; u = 75.7 kj/kg u = u - q H = 75.7-000 = 75.7 kj/kg = 960.5 K ; o = 8.0889 kj/kg K Compreion to : = From Eq.8.8 0 = o - o - R ln(p /P ) = o - o - R ln(τ v / v ) = 8.0889-6.85-0.87 ln(960.5/90) - 0.87 ln(v /v ) Solving => v / v =.78 Comment: hi i much too high for an actual Otto cycle. - w = u - u = 75.7-07. = 58.5 kj/kg Highet preure i after combution where v = v o we get P = P / = P ( / )(v / v ) = 90 (050 / 90).78 = 5 9 kpa P v v
.7 A four troke gaoline engine ha a compreion ratio of 0: with cylinder of total diplacement. L. the inlet tate i 80 K, 70 kpa and the engine i running at 00 RPM with the fuel adding 800 kj/kg 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.., r v = v /v = 0 η H = r -k v = 0-0. = 0.60 w net = η H q H = 0.60 800 = 08.6 kj/kg We alo need pecific volume to evaluate Eq..9 to. v = R / P = 0.87 80 / 70 =.8 m /kg P meff = w net v v = w net v ( r v ) = Now we can find the power from Eq.. Ẇ = P meff V dipl RPM 60 08.6 = 08.8 kpa.8 0.9 00 = 08.8 0.00 60 =. kw
.7 A gaoline engine receive air at 0 C, 00 kpa, having a compreion ratio of 9: by volume. he heat addition by combution give the highet temperature a 500 K. ue cold air propertie to find the highet cycle preure, the pecific energy added by combution, and the mean effective preure. Solution: P v v Compreion: Reverible and adiabatic o contant from Eq.8.- P = P (v /v ) k = 00 (9). = 67. kpa = (v /v ) k- = 8.5 (9) 0. = 68.89 K Combution: contant volume P = P ( / ) = 67. 500 / 68.89 = 796. kpa q H = u u = C vo ( - ) = 0.77 (500 68.89) = 0.6 kj/kg Efficiency, net work, diplacement and P meff η H = - / = - 8.5/68.89 = 0.587 w net = η H q H = 0.587 0.6 = 76.9 kj/kg v = R /P = 0.87 8.5 / 00 = 0.86 m /kg, v = v /9 = 0.0909 m /kg P meff = w net 76.9 v v = 0.86-0.0909 = 055 kpa
.7 A gaoline engine ha a volumetric compreion ratio of 0 and before compreion ha air at 90 K, 85 kpa in the cylinder. he combution peak preure i 6000 kpa. 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 = 85 (0). = 5. kpa = (v /v ) k- = 90 (0) 0. = 78.5 K Combution. Contant volume = (P /P ) = 78.5 6000/5. = 07 K Exhaut. Ientropic expanion o from Eq.8. = / (v /v ) k- = / 0 0. = 07 /.59 = 8.9 K Overall cycle efficiency i from Eq..9, r v = v /v η = r -k v = 0-0. = 0.60 Comment: No actual gaoline engine ha an efficiency that high, maybe 5%.
.7 Repeat Problem.67, but aume variable pecific heat. he ideal ga air table, able A.7, are recommended for thi calculation (or the pecific heat from Fig. 5.0 at high temperature). Solution: able A.7 i ued with interpolation. = 8. K, u = 0. kj/kg, o = 6.8 kj/kg K Compreion to : = From Eq.8.8 0 = o - o - R ln(p /P ) = o - o - R ln(τ v / v ) o - R ln(τ / ) = o + R ln(v /v ) = 6.8 + 0.87 ln 7 = 7.698 hi become trial and error o etimate firt at 600 K and ue A.7.. LHS 600 = 7.576-0.87 ln(600/8.) = 7.609 (too low) LHS 60 = 7.609-0.87 ln(60/8.) = 7.860 (too high) Interpolate to get: = 607. K, u = 0.5 kj/kg => - w = u - u = 8. kj/kg, u = 0.5 + 800 = 0.5 => = 575.8 K, o = 9.859 kj/kgk P = 90 7 575.8 / 8. = 570 kpa Expanion to : = From Eq.8.8 a before o - R ln(τ / ) = o + R ln(v /v ) = 9.859 + 0.87 ln(/7) = 8.77 hi become trial and error o etimate firt at 00 K and ue A.7.. LHS 00 = 8.589-0.87 ln(00/575.8) = 8.709 (too low) LHS 50 = 8.57-0.87 ln(50/575.8) = 8.760 (too high) Interpolation = 6.6 K, u = 6.9 kj/kg Net work, efficiency and mep w = u - u = 0.5-6.9 = 09.6 kj/kg w net = w + w = 09.6-8. = 855. kj/kg η H = w net / q H = 855. / 800 = 0.75 v = R /P = (0.87 8.)/90 = 0.909 m /kg v = (/7) v = 0.90 m /kg P meff = w net v v = 855. / (0.909-0.9) = 05 kpa
.75 An Otto cycle ha the lowet a 90 K and the lowet P a 85 kpa, the highet i 00 K and combution add 00 kj/kg a heat tranfer. Find the compreion ratio and the mean effective preure. Solution: Identify tate: = 90 K, P = 85 kpa, = 00 K, q H = 00 kj/kg Combution: q H = u u = C vo ( ) = 00 kj/kg = q H / C vo = 00 00/0.77 = 76.6 K Compreion: CR = r v = v /v = ( / ) /(k-) = 76.6 90.5 = 9.9 Overall cycle efficiency i from Eq.., r v = v /v = 9.9 η H = r -k v = / = 90/76.6 = 0.60 w net = η H q H = 0.60 00 = 7. kj/kg We alo need pecific volume to evaluate Eq..9 to. v = R / P = 0.87 90 / 85 = 0.979 m /kg v = v /CR = 0.0986 m /kg P meff = w net v v = 7. 0.979 0.0986 kj/kg m = 89 kpa /kg
.76 he cycle in the previou problem i ued in a. L engine running 800 RPM. How much power doe it produce? Identify tate: = 90 K, P = 85 kpa, = 00 K, q H = 00 kj/kg Combution: q H = u u = C vo ( ) = 00 kj/kg = q H / C vo = 00 00/0.77 = 76.6 K Compreion: CR = r v = v /v = ( / ) /(k-) = 76.6 90.5 = 9.9 Overall cycle efficiency i from Eq.., r v = v /v = 9.9 η H = r -k v = / = 90/76.6 = 0.60 w net = η H q H = 0.60 00 = 7. kj/kg We alo need pecific volume to evaluate Eq..9 to. v = R / P = 0.87 90 / 85 = 0.979 m /kg v = v /CR = 0.0986 m /kg P meff = w net v v = 7. 0.979 0.0986 Now we can find the power from Eq.. Ẇ = P meff V dipl RPM 60 kj/kg m = 89 kpa /kg 800 = 89 0.00 60 = 9.5 kw
.77 When methanol produced from coal i conidered a an alternative fuel to gaoline for automotive engine, it i recognized that the engine can be deigned with a higher compreion ratio, ay 0 intead of 7, but that the energy releae with combution for a toichiometric mixture with air i lightly maller, about 700 kj/kg. Repeat Problem.67 uing thee value. Solution: P v v Compreion: Reverible and adiabatic o contant from Eq.8.- P = P (v /v ) k = 90(0). = 60.7 kpa = (v /v ) k- = 8.5(0) 0. = 7. K Combution: contant volume = + q H / C vo = 7. + 700 / 0.77 = 08 K P = P ( / ) = 60.7 08 / 7. = 9797 kpa Efficiency, net work, diplacement and P meff η H = - / = - 8.5/7. = 0.60 w net = η H q H = 0.6 700 = 0. kj/kg v = R /P = 0.87 8.5/90 = 0.909 m /kg, v = v /0 = 0.090 m /kg P meff = w net v v = 0. / (0.909-0.090) = 55 kpa
.78 A gaoline engine ha a volumetric compreion ratio of 9. he tate before compreion i 90 K, 90 kpa, and the peak cycle temperature i 800 K. Find the preure after expanion, the cycle net work and the cycle efficiency uing propertie from able A.7.. Compreion to : = v r = v r (v /v ) v r = 96.7/9 =.89 680 K, P r 0.78, u = 96.9 P = P (P r /P r ) = 90 (0.78 / 0.995) = 880 kpa w = u - u = 07.9-96.9 = -89.75 kj/kg Combution to : q H = u - u = 86. - 96.9 = 989.9 kj/kg P = P ( / ) = 880 (800 / 680) = 976 kpa Expanion to : = v r = v r 9 =. 9 = 0.78 = 889 K, P r = 57.77, u = 665.8 kj/kg P = P (P r /P r ) = 976 (57.77 / 05) = 7.5 kpa w = u - u = 86. - 665.8 = 80.5 kj/kg w NE = w + w = 80.5-89.75 = 50.8 kj/kg η = w NE /q H = 50.8/989.9 = 0.56 P v v
.79 Air flow into a gaoline engine at 95 kpa, 00 K. he air i then compreed with a volumetric compreion ratio of 8:. In the combution proce 00 kj/kg of energy i releaed a the fuel burn. Find the temperature and preure after combution uing cold air propertie. Solution: Solve the problem with variable heat capacity, ue A.7. and A.7.. P v v Compreion to : = From A.7. v r = v r 8 = 79.9 8 =.6, = 67 K, u = 9.5 kj/kg, P r = 0 P = P Compreion to : P r 0 P = 95 r.6 = 705 kpa u = u + q H = 9.5 + 00 = 79.5 kj/kg = 8 K P = P ( / ) = 705 8 67 = 566 kpa
.80 Solve Problem.70 uing the P r and v r function from able A7. A gaoline engine take air in at 90 K, 90 kpa and then compree it. he combution add 000 kj/kg to the air after which the temperature i 050 K. Ue the cold air propertie (i.e. contant heat capacitie at 00 K) and find the compreion ratio, the compreion pecific work and the highet preure in the cycle. Solution: Standard Otto cycle, olve uing able A.7. and able A.7. Combution proce: = 050 K ; u = 75.7 kj/kg u = u - q H = 75.7-000 = 75.7 kj/kg = 960.5 K ; v r = 8.66 Compreion to : = From the v r function v /v = v r /v r = 95.6 8.66 =.78 Comment: hi i much too high for an actual Otto cycle. - w = u - u = 75.7-07. = 58.5 kj/kg Highet preure i after combution P = P / = P ( / )(v / v ) = 90 (050 / 90).78 = 5 9 kpa P v v
.8 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.67 but aume that the expanion proce i reverible and polytropic (intead of the ientropic expanion in the Otto cycle) with n equal to.50. See olution to.67 except for proce to. = 7 K, P = 6.958 MPa v = R /P = v = 0.9 m /kg, v = v = 0.909 m /kg Proce: Pv.5 = contant. P = P (v /v ).5 = 6958 (/7).5 = 75.7 kpa = (v /v ) 0.5 = 7(/7) 0.5 = 8.9 K w = Pdv = R -. ( - ) = 0.87-0. (606.6-8.5) = -9. kj/kg w = Pdv = R( - )/( -.5) = -0.87(8.9-7)/0.5 = 6.5 kj/kg w NE = 6.5-9. = 877. kj/kg η CYCLE = w NE /q H = 877./800 = 0.87 P meff = w net v v = 877./(0.909-0.9) = kpa Note a maller w NE, η CYCLE, P meff compared to an ideal cycle.
.8 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 that the total q H i 00 kj/kg, that the tate at the beginning of the compreion proce i 90 kpa, 0 C, 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 v 5 5 v v P = 90 kpa, = 0 o C r V = v /v = 7 a) q = (/) 00 = 00 kj/kg; q = 00/ = 700 kj/kg b) P = P (v /v ) k = 90(9). = 95 kpa = (v /v ) k- = 9.5(9) 0. = 706 K = + q /C V0 = 706 + 00/0.77 = 660 K P = P / = 95(660/706) = 750.8 kpa = P = + q /C P0 = 660 + 700/.00 = 57 K v 5 v P v = v = P = 750.8 90 9.5 57 = 7. 5 = (v /v 5 ) k- = 57(/7.) 0. = 50 K q L = C V0 ( 5 - ) = 0.77(50-9.5) = 886. kj/kg η H = - q L /q H = - 886./00 = 0.578 Std. Otto Cycle: η H = - (9) -0. = 0.585, mall difference
Dieel Cycle
.8 A dieel engine ha an inlet at 95 kpa, 00 K and a compreion ratio of 0:. he combution releae 00 kj/kg. Find the temperature after combution uing cold air propertie. Solution: Compreion proce (ientropic) from Eq.8.- = (v / v ) k- = 00 0 0. = 99. K Combution at contant P which i the maximum preure w = P (v - v ) q = u - u + w = h - h = C po ( - ) = + q / C po = 99. + 00 /.00 = 89 K P v P v
.8 A dieel engine ha a tate before compreion of 95 kpa, 90 K, and a peak preure of 6000 kpa, a maximum temperature of 00 K. Find the volumetric compreion ratio and the thermal efficiency. Solution: Standard Dieel cycle and we will ue cold air propertie. Compreion proce (ientropic) from Eq.8.-8.: (P /P ) = (v /v ) k = CR. CR = v /v = (P /P ) /k = (6000/95) /. = 9. = (P /P ) k-/k = 90 (6000/95) 0.857 = 97.9 K Combution and expanion volume v = v / = v /( CR) ; v = v Expanion proce, ientropic from Eq.8. Efficiency from Eq.. = (v /v ) k- = [ / (CR )] k- = 00 [ 00/(9. 97.9) ] 0. = 06.6 K η = k - - = 06.6 90. 00 97.9 = 0.69 P v P v
.85 Find the cycle efficiency and mean effective preure for the cycle in Problem.8 Solution: Compreion proce (ientropic) from Eq.8.- = (v / v ) k- = 00 0 0. = 99. K Combution at contant P which i the maximum preure w = P (v - v ) q = u - u + w = h - h = C po ( - ) = + q / C po = 99. + 00 /.00 = 89 K Combution and expanion volume v = v / = v /( CR) ; v = v Expanion proce, ientropic from Eq.8. Efficiency from Eq.. = (v /v ) k- = [ / (CR )] k- = 89 [ 89/(0 99.) ] 0. = 96 K η = k - - = 96 00. 89 99. = 0.6 w net = η q H = 0.6 00 = 8. kj/kg v = R /P = 0.906 m /kg P meff = = w net w net v max v = min v v /CR 8. 0.906 ( /0) kj/kg m = 957 kpa /kg
.86 A dieel engine ha a compreion ratio of 0: with an inlet of 95 kpa, 90 K, tate, with volume 0.5 L. he maximum cycle temperature i 800 K. Find the maximum preure, the net pecific work and the thermal efficiency. Solution: Compreion proce (ientropic) from Eq.8.- = (v / v ) k- = 90 0 0. = 96 K P = 95 (0). = 697.5 kpa ; v = v /0 = R /(0 P ) = 0.0805 m /kg - w = u - u C vo ( - ) = 0.77 (96-90) = 8. kj/kg Combution at contant P which i the maximum preure P = P = 698 kpa ; v = v / = 0.0805 800/96 = 0.0805 m /kg w = P (v - v ) = 698 (0.085-0.0805) =.5 kj/kg q = u - u + w = h - h = C po ( - ) =.00(800-96) = 8. kj/kg Expanion proce (ientropic) from Eq.8. = ( v / v ) 0. = 800 (0.0805 / 0.876) 0. = 698 K w = u - u C vo ( - ) = 0.77 (800-698) = 790. kj/kg Cycle net work and efficiency w net = w + w + w =.5 + 790. - 8. = 550.5 kj/kg η = w net / q H = 550.5/ 8. = 0.65 P v P v
.87 A dieel engine ha a bore of 0. m, a troke of 0. m 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 kpa. With a total of 6 cylinder find the engine power in kw and horepower, hp. Solution: Work from mean effective preure, Eq..9. P meff = he diplacement i w net v max v min => w net = P meff (v max - v min ) V = πbore 0.5 S = π 0. 0.5 0. = 0.00086 m Work per cylinder per power troke, Eq..0 W = P meff (V max - V min ) = 00 0.00086 kpa m =.096 kj/cycle Only every econd revolution ha a power troke o we can find the power, ee alo Eq.. Ẇ = W N cyl RPM 0.5 (cycle / min) (min / 60 ) (kj / cycle) =.096 6 000 0.5 (/60) = kw = 6 hp he converion factor from kw to hp i from able A. under power.
.88 A uper charger i ued for a two troke 0 L dieel engine o intake i 00 kpa, 0 K and the cycle ha compreion ratio of 8: and mean effective preure of 80 kpa. he engine i 0 L running at 00 RPM find the power output. Solution: he power i from Eq.. Ẇ = m wnet N cyl RPM /60 = m P meff (v v ) N cyl RPM/60 = P meff V dipl RPM/60 = 80 0.00 00 / 60 = 7.7 kw
.89 At the beginning of compreion in a dieel cycle = 00 K, P = 00 kpa and after combution (heat addition) i complete = 500 K and P = 7.0 MPa. Find the compreion ratio, the thermal efficiency and the mean effective preure. Solution: Standard Dieel cycle. See P-v and - diagram for tate number. Compreion proce (ientropic) from Eq.8.-8. P = P = 7000 kpa => v / v = (P /P ) / k = (7000 / 00) 0.7 =.67 = (P / P ) (k-) / k = 00(7000 / 00) 0.857 = 88. K Expanion proce (ientropic) firt get the volume ratio v / v = / = 500 / 88. =.8 v / v = v / v = (v / v )( v / v ) =.67 /.8 = 7 he exhaut temperature follow from Eq.8. = (v / v ) k- = 500 ( / 7) 0. = 688.7 K q L = C vo ( - ) = 0.77(688.7-00) = 78.5 kj/kg q H = h - h C po ( - ) =.00(500-88.) = 67 kj/kg Overall performance η = - q L / q H = - 78.5 / 67 = 0.587 w net = q net = q H - q L = 67-78.5 = 95.5 kj/kg v max = v = R / P = 0.87 00 / 00 = 0.05 m /kg v min = v max / (v / v ) = 0.05 /.67 = 0.0 m /kg P meff = w net v max v min = 95.5 / (0.05-0.0) = 997 kpa P v P v Remark: hi i a too low compreion ratio for a practical dieel cycle.
.90 Do problem.8, but ue the propertie from A.7 and not the cold air propertie. A dieel engine ha a tate before compreion of 95 kpa, 90 K, and a peak preure of 6000 kpa, a maximum temperature of 00 K. Find the volumetric compreion ratio and the thermal efficiency. Solution: Compreion: = => from Eq.8.8 o = o + R ln(p / P) = 6.85 + 0.87 ln(6000/95) = 8.05 kj/kg K A.7. => = 907.6 K; h = 9.6 kj/kg; h = 755.8 kj/kg; o = 9.9586 kj/kg K q H = h - h = 755.8 9.6 = 8. kj/kg CR = v /v = ( / )(P /P ) = (90/907.6) (6000/ 95) = 0.8 Expanion proce o = o + R ln(p / P ) = o + R ln( / ) + R ln(v /v ) v /v = v /v = (v /v ) ( / ) = ( / ) (/CR) = (00/907.6) (/0.8) = 0.0 o - R ln( / ) = o + R ln(v /v ) = 9.958 + 0.87 ln 0.0 = 8.65 rial and error on ince it appear both in o and the ln function = 00 K LHS = 8.05 0.87 ln (00/00) = 8.66 = 50 K LHS = 8.90 0.87 ln (50/00) = 8.58 Now Linear interpolation = 95 K, u = 08.6 kj/kg q L = u - u = 08.6 07.9 = 8.08 kj/kg η = (q L / q H ) = (8.08/8.) = 0.55 P v P v
.9 Solve Problem.8 uing the P r and v r function from able A7. A dieel engine ha a tate before compreion of 95 kpa, 90 K, and a peak preure of 6000 kpa, a maximum temperature of 00 K. Find the volumetric compreion ratio and the thermal efficiency. Solution: Compreion: = => From definition of the P r function P r = P r (P /P ) = 0.9899 (6000/95) = 6.5 A.7. => = 907 K; h = 9.0 kj/kg; h = 755.8 kj/kg; v r = 0.8 q H = h - h = 755.8 9.0 = 8.8 kj/kg CR = v /v = ( / )(P /P ) = (90/907) (6000/ 95) = 0.9 Expanion proce v r = v r (v / v ) = v r (v / v ) = v r (v / v ) ( / ) = v r CR ( / ) = 0.8 0.9 (907/00) =.0675 Linear interpolation = 9.8 K, u = 08. kj/kg q L = u - u = 08. 07. = 80.9 kj/kg η = (q L / q H ) = (80.9/8.8) = 0.55 P v P v
.9 he world larget dieel engine ha diplacement of 5 m running at 00 RPM in a two troke cycle producing 00 000 hp. Aume an inlet tate of 00 kpa, 00 K and a compreion ratio of 0:. What i the mean effective preure? We have parameter for the cycle:, P and CR we need one more, o thi come from the total rate of work. Ẇ = m wnet N cyl RPM /60 = m P meff (v v ) N cyl RPM/60 = P meff V dipl RPM/60 P meff = Ẇ 60 / V dipl RPM = 00 000 0.76 kw 60 5 m 00 With thi information we could now get all the cycle tate etc. = 895 kpa A imilar engine i in thi container hip hown at dock. he engine wa build up inide the hip a it i too large to put in finihed. he crane are on the port dock.