Internal Combustion Engines
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1 Internal Combustion Engines Reading Problems , 8-45, 8-52 Definitions 1. spark ignition: a mixture of fuel and air is ignited by a spark plug applications requiring power to about 225 kw (300 HP) relatively light and low in cost 2. compression ignition engine: air is compressed to a high enough pressure and temperature that combustion occurs when the fuel is injected applications where fuel economy and relatively large amounts of power are required The Gasoline Engine 1
2 conversion of chemical energy to mechanical energy can obtain very high temperatures due to the short duration of the power stroke Air Standard Cycle ASSUMPTIONS: air is an ideal gas with constant c p and c v no intake or exhaust processes the cycle is completed by heat transfer to the surroundings the internal combustion process is replaced by a heat transfer process from a TER all internal processes are reversible heat addition occurs instantaneously while the piston is at TDC Definitions Mean Effective Pressure (MEP): The theoretical constant pressure that, if it acted on the piston during the power stroke would produce the same net work as actually developed in one complete cycle. MEP = net work for one cycle displacement volume = W net V BDC V TDC The mean effective pressure is an index that relates the work output of the engine to it size (displacement volume). 2
3 Otto Cycle the theoretical model for the gasoline engine consists of four internally reversible processes heat is transferred to the working fluid at constant volume The Otto cycle consists of four internally reversible processes in series 1 2 isentropic compression or air as the piston moves from BDC to TDC 2 3 constant volume heat addition to the fuel/air mixture from an external source while the piston is at TDC (represents the ignition process and the subsequent burning of fuel) 3 4 isentropic expansion (power stroke) 4 1 constant volume heat rejection at BDC 3
4 The Otto cycle efficiency is given as η =1 T 1 T 2 =1 ( V2 V 1 ) k 1 =1 ( ) 1 k V1 If we let r = V 1 = V 4 V 3 = compression ratio Then η Otto =1 r 1 k Why not go to higher compression ratios? there is an increased tendency for the fuel to detonate as the compression ratio increases the pressure wave gives rise to engine knock can be reduced by adding tetraethyl lead to the fuel not good for the environment 4
5 Diesel Cycle an ideal cycle for the compression ignition engine (diesel engine) all steps in the cycle are reversible heat is transferred to the working fluid at constant pressure heat transfer must be just sufficient to maintain a constant pressure If we let r = V 1 = compression ratio = V 4 r v = V 3 = cutoff ratio injection period then the diesel cycle efficiency is given as η Diesel =1 1 r k 1 ( )( 1 r k k v 1 r v 1 ) 5
6 Where we note η Diesel =1 1 ( )( 1 r k v 1 ) r k 1 k r v 1 }{{} =1 in the Otto Cycle Comparison of the Otto and the Diesel Cycle η Otto >η Diesel for the same compression ratio but a diesel engine can tolerate a higher ratio since only air is compressed in a diesel cycle and spark knock is not an issue direct comparisons are difficult Dual Cycle (Limited Pressure Cycle) this is a better representation of the combustion process in both the gasoline and the diesel engines in a compression ignition engine, combustion occurs at TDC while the piston moves down to maintain a constant pressure 6
7 Dual Cycle Efficiency Given r = V 1 = compression ratio r v = V 4 V 3 = cutoff ratio r p = P 3 P 2 = pressure ratio η Dual =1 r p r k v 1 [(r p 1) + kr p (r v 1)] r k 1 Note: if r p =1we get the diesel efficiency. 7
8 Stirling Cycle 8
9 reversible regenerator used as an energy storage device possible to recover all heat given up by the working fluid in the constant volume cooling process all the heat received by the cycle is at T H and all heat rejected at T L η Stirling =1 T L /T H (Carnot efficiency) With perfect regeneration Q H = T H (s 2 s 1 ) Q L = T L (s 3 s 4 ) 9
10 η = W net Q H = Q H Q L Q H =1 Q L Q H =1 T L(s 3 s 4 ) T H (s 2 s 1 ) (1) From Gibb s equation Tds = du + Pdv = c v dt + Pdv if T = constant Tds = Pdv ds = Pdv Integrating gives T = Rdv v s 3 s 4 = R ln ( ) v3 v 4 = R ln ( ) v2 v 1 = s 2 s 1 Therefore s 3 s 4 = s 2 s 1, and Eq. 1 gives η =1 T L T H Carnot efficiency 10
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