Lecture 40: Air standard cycle, internal combustion engines, Otto cycle

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ME 200 Thermodynamics I Spring 206 Lecture 40: Air standard cycle, internal combustion engines, Otto cycle Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 200240, P. R. China Email : liyo@sjtu.edu.cn Phone: 86-2-34206056; Fax: 86-2-34206056.

Air Standard Cycles Air standard cycles are idealized cycles based on the following approximations: A fixed amount of air modeled as an ideal gas (working fluid). The combustion process is replaced by a heat transfer from an external source. There are no exhaust and intake processes as in an actual engine. The cycle is completed by a constant-volume heat transfer process taking place while the piston is at the bottom dead center position. All processes are internally reversible. Cold air-standard analysis The specific heats are assumed constant at T a..2

Continue Air Standard Cycles Cycles under consideration:» Carnot cycle maximum cycle efficiency» Otto cycle Spark-ignition engine (SI engine)» Diesel cycle Compression-ignition engine (CI engine)» Dual cycle modern CI engine» Brayton cycle gas turbines» Other cycles that we want to analyze: Stirling cycle Ericsson cycle.3

Ideal Gas Model Review.4

Ideal Gas Model Review.5

Polytropic Processes on p v and T s Diagrams pv n c n 0 p c n p / v c n k pv k c s c n RT pv c T c n v c.6

Continue Air Standard Cycles Air Standard Carnot Cycle: T T H p 2 T L p 2 3 q L q H 4 s =s 2 s 3 =s 4 Question: p 3 w net s p4 q T s H H 23 q T s L L 4 w q q net H L net th,carnot qh th,carnot w q T s th,carnot q T s T T H H L L L 4 H H 23 How to have isothermal heat transfer with air as the working fluid? L H q q q.7

Continue Air Standard Carnot Cycle Answer: p Must have work! For an ideal gas, is T = constant, then p v = constant! p 2 p 3 p 2 s = const. s = const. 3 T H = const. p 4 4 v 2 v v 3 v 4 T L = const. v.8

Continue Air Standard Carnot Cycle Physical Devices: 2 3 4 q L w net q H 2 isentropic compression 2 3 isothermal expansion q H 3 isentropic expansion isothermal compression q L 4 4.9

ME 200 Thermodynamics I Spring 205 Otto Cycle.0

Otto Cycle--Internal Combustion Engine The stroke ::: the distance the piston moves in one direction Bore ::: cylinder diameter Clearance volume ::: minimum volume Top dead center ::: a position when the piston moves to the minimum cylinder volume Bottom dead center ::: the position when the piston moves to the maximum cylinder volume. Displacement volume ::: The volume swept out by the piston as it moves from the top dead center to the bottom dead center position. Compression ratio r ::: the volume at bottom dead center divided by the volume at top dead center r V V max min V V BDC TDC.

Otto Cycle--Internal Combustion Engine Ignition.2

Continue Internal Combustion Engines Spark ignition (SI):» combustion initiated by spark» air and fuel can be added together» Light and lower in cost, used in automobiles Compression ignition (CI):» combustion initiated by auto ignition» requires fuel injection to control ignition» large power, heavy trucks, locomotives, ships.3

An automotive engine with the combustion chamber exposed.4

Spark ignition (SI):.5

Continue Internal Combustion Engines Mean effective pressure, MEP MEP Wnet net work for one cycle V V displacement volume max min Notes:» MEP would produce the same net work with constant pressure as for actual cycle (includes both expansion and compression)» want high MEP (high power density).6

Otto Cycle The ideal cycle for spark ignited engines: 2 2 3 TDC w in 3 q H 4 w out 4 q L BDC.7

Ideal Otto cycle Continue Otto Cycle» -2 Isentropic compression» 2-3 Constant-volume heat addition» 3-4 Isentropic expansion» 4- Constant-volume heat rejection.8

Continue Otto Cycle Otto cycle energy balances:» Assuming: W = 0 during heat transfer processes, KE = PE = 0 Air is ideal gas, constant specific heats w net = w out w in = q in q out q in = u 3 u 2 = c v (T 3 T 2 ) q out = u 4 u = c v (T 4 T ).9

Continue Otto Cycle Otto cycle efficiency: w q q q th q q q net in out out in in in k For isentropic process : pv constant, with k For process 2 : p v p v RT2 k v p2 v2 T2 v k v RT 2 p T v2 v k k 2 2 k k v v2 T2 v v k k v2 v T v2 v2 k T4 T c (T T ) T v 4 T 3 c v(t3 T 2) T2 c c p v T 2.20

Continue Otto Cycle Continue Otto cycle efficiency: Compression ratio r V V max min V V BDC TDC k k T 2 V V BDC For m = constant: T V2 VTDC For process 3 4, using the same analysis: k k T 3 V 4 V BDC r T4 V3 VTDC k r k Then, T T T T or T T T T 2 3 3 4 4 2 r th k =.2

Continue Otto Cycle r th k Answer: Increase in r results in:» an increase in T for heat addition and» a decrease in T for heat rejection Typical r for gasoline engines Question: Why increase in th with increase in r? Note: Need to consider material limitations as function of T and p.22

Example : Air-Standard Otto Cycle Known: r = 8, state : p =00 kpa, T =290 K, q in = 800 kj/kg Find: p, T and v at all state points, w net and η th, MEP Assumptions: ) Air-Standard assumptions. 2) ΔKE=ΔPE =0. 3)Variable c p Analysis: cold-air-standard assumptions.23

w q q q th q q q net in out ou in in in.24