OUTCOME 2 INTERNAL COMBUSTION ENGINE PERFORMANCE. TUTORIAL No. 4 HEAT ENGINE CYCLES

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1 UNI 6: ENGINEERING HERMODYNAMICS Unit code: D/60/0 CF leel: 5 Cedit alue: 5 OUCOME INERNAL COMBUSION ENGINE PERFORMANCE UORIAL No. HEA ENGINE CYCLES Be able to ealuate the efomance of tenal combustion enges Second law of themodynamics: statement of law; schematic eesentation of a heat enge to show heat and wok flow Heat enge cycles: Canot cycle; Otto cycle; Diesel cycle; dual combustion cycle; Joule cycle; oety diagams; Canot efficiency; aistandad efficiency Pefomance chaacteistics: enge tials; dicated and bake mean effectie essue; dicated and bake owe; dicated and bake themal efficiency; mechanical efficiency; elatie efficiency; secific fuel consumtion; heat balance Imoements: tubochagg; tubochagg and tecoolg; coolg system and exhaust gas heat ecoey systems Defe AIR SANDARD CYCLES. Identify the ideal cycle fo a gien tye of enge. Exla and sole oblems fo the OO cycle Exla and sole oblems fo the DIESEL cycle Exla and sole oblems fo the Dual Combustion cycle Exla and sole oblems fo the JOULE cycle D.J.Dunn

2 . HEOREICAL CYCLES FOR ENGINES Intenal combustion enges fall to two gous, those that use a sakg lug to ignite the fuel sak ignition enges and those that use the natual temeatue of the comessed ai to ignite the fuel comession ignition enges. Anothe way to gou enges is to those that use nonflow ocesses and those that use flow ocesses. Fo examle, nonflow ocesses ae used iston enges. Flow ocesses ae used gas tube enges. heoetical cycles ae made u of ideal themodynamic ocesses to esemble those that occu a eal enge as closely as ossible. Many of these cycles ae based on ai as the wokg fluid and ae called AIR SANDARD CYCLES. Befoe lookg at ai standad cycles, we should biefly eise the Canot Cycle fom tutoial.. HE CARNO CYCLE he most efficient way of tansfeg heat to o out of a fluid is at constant temeatue. All the heat tansfe the Canot cycle is at constant temeatue so it follows that the Canot cycle is the most efficient cycle ossible. he heat tansfe to the cycle occus at a hot temeatue hot and the heat tansfe out of the cycle occus at a colde temeatue cold. he themodynamic efficiency was sown to be gien as follows. cold th hot None of the followg cycles can hae an efficiency geate than this when oeatg between the same temeatues limits. D.J.Dunn

3 SPARK IGNIION ENGINE. HE OO CYCLE he ideal cycle is named afte Count N.A.Otto. It eesents the ideal cycle fo a sak ignition enge. In an ideal sak ignition enge, thee ae fou ocesses as follows. Fig. COMPRESSION SROKE Ai and fuel ae mixed and comessed so aidly that thee is no time fo heat to be lost. Figue A In othe wods the comession is adiabatic. Wok must be done to comess the gas. IGNIION Just befoe the ot of maximum comession, the ai is hot and a sak ignites the mixtue causg an exlosion Figue B. his oduces a aid ise the essue and temeatue. he ocess is idealised as a constant olume ocess the Otto cycle. EXPANSION OR WORKING SROKE he exlosion is followed by an adiabatic exansion ushg the iston and gig out wok. Figue C EXHAUS At the end of the wokg stoke, thee is still some essue the cylde. his is eleased suddenly by the oeng of an exhaust ale. Figue D his is idealised by a constant olume do essue the Otto cycle. In stoke enges a second cycle is efomed to ush out the oducts of combustion and daw fesh ai and fuel. It is only the owe cycle that we ae concened with. D.J.Dunn

4 he fou ideal ocesses that make u the Otto cycle ae as follows. to he ai is comessed eesibly and adiabatically. Wok is ut and no heat tansfe occus. Fig. to he ai is heated at constant olume. No wok is done. = mc Fig. to he ai exands eesibly and adiabatically with no heat tansfe back to its oigal olume. Wok outut is obtaed. Fig. to he ai is cooled at constant olume back to its oigal essue and temeatue. No wok is done out = mc Fig.5 D.J.Dunn

5 D.J.Dunn 5 he total heat tansfe to the system dug one cycle is nett = out he total wok outut e cycle is Wnett Fom the st. Law of themodynamics nett = Wnett EFFICIENCY mc mc W out nett Fo the ocess to we may use the ule Fo the ocess to we may similaly wite whee is the olume comession atio It follows that and and that then Sce this theoetical cycle is caied out on ai fo which =. then the efficiency of an Otto Cycle is gien by 0. his shows that the themal efficiency deends only on the comession atio. If the comession atio is ceased, the efficiency is imoed. his tun ceases the temeatue atios between the two isentoic ocesses and exlas why the efficiency is imoed.

6 WORKED EXAMPLE No. An Otto cycle is conducted as follows. Ai at 00 kpa and 0oC is comessed eesibly and adiabatically. he ai is then heated at constant olume to 500oC. he ai then exands eesibly and adiabatically back to the oigal olume and is cooled at constant olume back to the oigal essue and temeatue. he olume comession atio is 8. Calculate the followg. i. he themal efficiency. ii. he heat ut e kg of ai. iii. he net wok outut e kg of ai. i. he maximum cycle essue. c = 78 kj/kg =.. R = 87 J/kg K SOLUION Remembe to use absolute temeatues thoughout. Sole fo a mass of kg. =0 +7=9K =500+7=77K =8 mc W nett o 56.5% 67. K x J / kg kj / kg x kj / kg Fom the gas law we hae x x 77 9 x x 77 x 8.8 MPa 9 If you hae followed the ciles used hee you should be able to sole any cycle. D.J.Dunn 6

7 SELF ASSESSMEN EXERCISE No. ake C = 0.78 kj/kg K, R = 87 J/kg K and =. thoughout.. An Otto cycle has a olume comession atio of 9/. he heat ut is 500kJ/kg. At the stat of comession the essue and temeatue ae 00 kpa and 0oC esectiely. Calculate the followg. i. he themal efficiency. 58.5% ii. he maximum cycle temeatue. 50 K. iii. he maximum essue..7 MPa. i. he net wok outut e kg of ai. 9 kj/kg.. Calculate the olume comession atio equied of an Otto cycle which will oduce an efficiency of 60%. 9.88/ he essue and temeatue befoe comession ae 05 kpa and 5oC esectiely. he net wok outut is 500 kj/kg. Calculate the followg. i. he heat ut. 8 kj/kg. ii. he maximum temeatue. 906 K iii. he maximum essue. 6.6 MPa.. An Otto cycle uses a olume comession atio of 9.5/. he essue and temeatue befoe comession ae 00 kpa and 0oC esectiely. he mass of ai used is.5 gams/cycle. he heat ut is 600 kj/kg. he cycle is efomed 000 times e mute. Deteme the followg. i. he themal efficiency. 59.%. ii. he net wok outut.. kj/cycle iii. he net owe outut. 05 kw.. An Otto cycle with a olume comession atio of 9 is equied to oduce a net wok outut of 50 kj/cycle. Calculate the mass of ai to be used if the maximum and mimum temeatues the cycle ae 00oC and 0oC esectiely..5 kg. 5. he wokg of a etol enge can be aoximated to an Otto cycle with a comession atio of 8 usg ai at ba and 88 K with heat addition of MJ/kg. Calculate the heat ejected and the wok done e kg of ai. 87 kj/kg and 9 kj/kg. Now let's moe on to study enges with comession ignition. D.J.Dunn 7

8 COMPRESSION IGNIION ENGINES he ention of comession ignition enges, commonly known as diesel enges, was cedited to Rudolf Diesel, although many othe eole woked on simila enges. he basic cile is that when high comession atios ae used, the ai becomes hot enough to make the fuel detonate without a sak. Diesel's fist enge used coal dust blasted to the combustion chambe with comessed ai. his deeloed to blastg oil with comessed ai. In moden enges the fuel oil is jected diectly to the cylde as fe dolets. hee ae two ideal cycles fo these enges, the Diesel Cycle and the Dual Combustion Cycle.. DUAL COMBUSION CYCLE his is the ai standad cycle fo a moden fast unng diesel enge. Fist the ai is comessed isentoically makg it hot. Fuel jection stats befoe the ot of maximum comession. Afte a shot delay which fuel accumulates the cylde, the fuel wams u to the ai temeatue and detonates causg a sudden ise essue. his is ideally a constant olume heatg ocess. Futhe jection kees the fuel bung as the olume ceases and oduces a constant essue heatg ocess. Afte cut off, the hot ai exands isentoically and then at the end of the stoke, the exhaust ale oens oducg a sudden do essue. his is ideally a constant olume coolg ocess. he ideal cycle is shown figue 6. Fig. 6 he ocesses ae as follows. eesible adiabatic isentoic comession. constant olume heatg. constant essue heatg. 5 eesible adiabatic isentoic exansion. 5 constant olume coolg. D.J.Dunn 8

9 he analysis of the cycle is as follows. he heat is sulied two stages hence = mc + mc he heat ejected is out = mc 5 he themal efficiency may be found as follows. out mc mc 5 mc 5 he fomula can be futhe deeloed to show that k k k is the OLUME COMPRESSION RAIO. = / is the CU OFF RAIO. = / k is the atio /. Most students will fd this adequate to sole oblems conceng the dual combustion cycle. Geneally, the method of solution oles fdg all the temeatues by alication of the gas laws. D.J.Dunn 9

10 D.J.Dunn 0 hose equig a detailed analysis of the cycle should study the followg deiation fomula. temeatues the efficiency Substitutefo all Isentoicexansion to 5 Constant essueheatg to note note heatg to Constant olume Isentoiccomession to of the temeatues tems Obta all out k k k k k k k k k k k k k k Note that if =, the cycle becomes an Otto cycle and the efficiency fomulae becomes the same as fo an Otto cycle.

11 WORKED EXAMPLE No. In a dual combustion cycle, the comession stats fom ba and 0 o C. he comession atio is 8/ and the cut off atio is.5. he maximum cycle essue is 60 K. he total heat ut is kj e cycle. Calculate the followg. i. he themal efficiency of the cycle. ii. he net wok outut e cycle. Check that the efficiency does not contaene the Canot cile. SOLUION Known data. = 0 +7 = 9 K he hottest temeatue is = 60K. =.5 = 8 =. η γ k β.7 kβ γ k γkβ.7. x.7 x.5 η 0.68 o 68% 9 x 8 γ 0. 9K 60 8 K.5.7 x.5. x 8 0. W nett = x = 0.68 x = 0.68 kj e cycle. he Canot efficiency should be highe. cold hot he figue of 0.68 is lowe so the Canot cile has not been contaened. D.J.Dunn

12 WORKED EXAMPLE No. A dual combustion cycle has a comession atio of 8/. he maximum essue the cycle is 9 MPa and the maximum temeatue is 000oC. he essue and temeatue befoe comession is 5 kpa and 5oC esectiely. Calculate the followg. i. the cut off atio. ii. the cycle efficiency. iii. the net wok outut e kg of ai. Assume =. C =.005 kj/kgk C = 0.78 kj/kg K. SOLUION Known data. = 98 K = 7 K = = 9 MPa = 5 kpa / = / = 8 = = 98 x 8 = 97 K Cut off atio β β but 7 x x 0 x 98 x 5 x 0 γ 0. 5 x K but 6 9 x 0 x 98 5 x 0 5 x 8 soβ = mc + mc m= kg = =.5 kj/kg out = mc 5 out = = 9 kj/g 96K η W nett out out o 65% kj/kg D.J.Dunn

13 . HE DIESEL CYCLE he Diesel Cycle ecedes the dual combustion cycle. he Diesel cycle is a easonable aoximation of what haens slow unng enges such as lage mae diesels. he itial accumulation of fuel and sha detonation does not occu and the heat ut is idealised as a constant essue ocess only. Aga conside this cycle as beg caied out side a cylde fitted with a iston. he and s cycles diagams ae shown figue 7 Fig. 7 eesible adiabatic isentoic comession. constant essue heatg. eesible adiabatic isentoic exansion. constant olume coolg. out mc mc he cycle is the same as the dual combustion cycle without the constant olume heatg ocess. In this case sce k= the efficiency is gien by the followg fomula. D.J.Dunn

14 WORKED EXAMPLE No. An enge usg the Diesel Cycle has a comession atio of 0/ and a cut off atio of. At the stat of the comession stoke the ai is at ba and 5 o C. Calculate the followg. i. he ai standad efficiency of the cycle. ii. he maximum temeatue the cycle. iii. he heat ut. i. he net wok outut. SOLUION Initial data. = =0 =. c = 78 J/kg K fo ai =88 K = ba. he maximum temeatue is and the maximum essue is and. γ β η γ β γ η. η x. x 0 W η W nett.69 x. x. γ mc η 88 x 0 β o 6.7% 95.5 K 95. x 909 K kj nett x kj D.J.Dunn

15 SELF ASSESSMEN EXERCISE No. Use c = 0.78 kj/kg K, c =.005 kj/kg K and =. thoughout.. Daw a and s diagam fo a Diesel Cycle. he efomance of a comession ignition enge is to be comaed to the Diesel cycle. he comession atio is 6. he essue and temeatue at the begng of comession ae ba and 5oC esectiely. he maximum temeatue the cycle is 00 K. Calculate the followg. i. he cut off atio..7 ii. he ai standad efficiency. 66%. A Dual Combustion Cycle uses a comession atio of /. he cut off atio is /. he temeatue and essue befoe comession is 80 K and ba esectiely. he maximum temeatue 000 K. Calculate the followg. i. he net wok outut e cycle. 680 kj/kg. ii. he themal efficiency %.. A Dual Combustion Cycle uses a comession atio of 0/. he cut off atio is.6/. he temeatue and essue befoe comession is 0oC and ba esectiely. he maximum cycle essue is 00 ba. Calculate the followg. i. he maximum cycle temeatue. K. ii. he net wok outut e cycle. 86 kj/kg. iii. he themal efficiency %. D.J.Dunn 5

16 GAS URBINES A gas tube enge nomally buns fuel the ai that it uses as the wokg fluid. Fom this ot of iew it is an tenal combustion enge that uses steady flow ocesses. Figue 8 shows a basic design. Fig.8 he ai is dawn fom atmoshee and comessed. his makes it hotte. he comessed ai is blown to a combustion chambe and fuel is buned it makg it een hotte. his makes the olume cease. he hot ai exands out of the chambe though a tube focg it to eole and oduce owe. he ai becomes colde as it exands and eentually exhausts to atmoshee. he temeatue do oe the tube is lage than the temeatue cease oe the comesso. he tube oduces moe owe than is needed to die the comesso. Net owe outut is the esult. In the basic system, the tube is couled diectly to the comesso and the owe outut is taken fom the same shaft. he ideal ai standad cycle is the Joule Cycle. D.J.Dunn 6

17 . HE JOULE CYCLE he Joule Cycle is also known as the constant essue cycle because the heatg and coolg ocesses ae conducted at constant essue. he cycle is that used by a gas tube enge but could conceiably be used a closed system. We may daw the layout block diagam fom as shown figue 9 Figue 9 hee ae ideal ocesses the cycle. Reesible adiabatic isentoic comession equig owe ut. P = H/s = mc Constant essue heatg equig heat ut. = H/s = mc Reesible adiabatic isentoic exansion oducg owe outut. P out = H/s = mc Constant essue coolg back to the oigal state equig heat emoal. out = H/s = mc he essue olume, essue enthaly and temeatueentoy diagams ae shown figue 0 Fig. 0 D.J.Dunn 7

18 D.J.Dunn 8 he efficiency is found by alyg the fist law of themodynamics. mc mc P P P P out nett th out out nett nett It assumed that the mass and the secific heats ae the same fo the heate and coole. It is easy to show that the temeatue atio fo the tube and comesso ae the same. is the essue comession atio fo the tube and comesso.. sce 0.86 th th his shows that the efficiency deends only on the essue atio which tun affects the hottest temeatue the cycle.

19 WORKED EXAMPLE No. 5 A gas tube uses the Joule cycle. he essue atio is 6/. he let temeatue to the comesso is 0oC. he flow ate of ai is 0. kg/s. he temeatue at let to the tube is 950oC. Calculate the followg. i. he cycle efficiency. ii. he heat tansfe to the heate. iii. he net owe outut. =. C =.005 kj/kg K SOLUION P th th nett mc P nett 6 8 x 6 0. x kw K o 0% 0. x.005 x kw D.J.Dunn 9

20 SELF ASSESSMEN EXERCISE No. =. and C =.005 kj/kg K thoughout.. A gas tube uses the Joule cycle. he let essue and temeatue to the comesso ae esectiely ba and 0oC. Afte constant essue heatg, the essue and temeatue ae 7 ba and 700oC esectiely. he flow ate of ai is 0. kg/s. Calculate the followg. i. he cycle efficiency. ii. he heat tansfe to the heate. iii. he net owe outut. Answes.7 %, 06.7 kw and 88.6 kw. A gas tube exands daws kg/s of ai fom atmoshee at ba and 0 o C. he combustion chambe essue and temeatue ae 0 ba and 90oC esectiely. Calculate the followg. i. he Joule efficiency. ii. he exhaust temeatue. iii. he net owe outut. Answes 8. %, 67.5 K and 9 kw. A gas tube daws 7 kg/s of ai fom atmoshee at ba and 5 o C. he combustion chambe essue and temeatue ae 9 ba and 850oC esectiely. Calculate the followg. i. he Joule efficiency. ii. he exhaust temeatue. iii. he net owe outut. Answes 6.7 %, 599 K and.96 MW D.J.Dunn 0