The CRC Handbook of Mechanical Engineering. Chapter 2. Engineering Thermodynamics (776125), страница 18
Текст из файла (страница 18)
In a spark-ignition engine a mixture of fuel and air is ignited by a spark© 1999 by CRC Press LLC2-81Engineering ThermodynamicsTABLE 2.15 Sample Calculations for the Rankine and Brayton Cycles of Table 2.14Rankine CycleGiven data:p1 = p4 = 8 kPa (saturated liquid at 1)T3 = 480°C (superheated vapor at 3)p2 = p3 = 8 MPaẆnet = 100 MWIdeal cycle: ηt = ηp = 100%Actual cycle: ηt = 85%, ηp = 70%ParameterabIdeal CycleActual Cyclex4h2 (kJ/ kg)0.794181.9a0.873185.4m˙ ( kg/h )η (%)2.86 × 10539.73.38 × 10533.6Q˙ out (MW)151.9197.6E˙ q,out (MW)b8.210.7h2s ≈ h1 + v1∆pEquation 2.86 with T0 = 298 K, Tj = Tsat (8 kPa) = 315 KBrayton CycleGiven data:p1 = p4 = 1 barp2 = p3 = 10 barT3 = 1400 Kηt = ηc = 100%Air-Standard AnalysisCold Air-Standard Analysisk = 1.4T2 (K)T4 (K)574.1787.7579.2725.1W˙ net m˙ ( kJ/kg)η (%)bwr427.2397.545.70.39648.20.414Parameterplug.
In a compression ignition engine air is compressed to a high-enough pressure and temperature thatcombustion occurs spontaneously when fuel is injected.In a four-stroke internal combustion engine, a piston executes four distinct strokes within a cylinderfor every two revolutions of the crankshaft. Figure 2.17 gives a pressure-displacement diagram as itmight be displayed electronically. With the intake valve open, the piston makes an intake stroke to drawa fresh charge into the cylinder.
Next, with both valves closed, the piston undergoes a compression strokeraising the temperature and pressure of the charge. A combustion process is then initiated, resulting ina high-pressure, high-temperature gas mixture. A power stroke follows the compression stroke, duringwhich the gas mixture expands and work is done on the piston. The piston then executes an exhauststroke in which the burned gases are purged from the cylinder through the open exhaust valve.
Smallerengines operate on two-stroke cycles. In two-stroke engines, the intake, compression, expansion, and© 1999 by CRC Press LLC2-82Section 2FIGURE 2.17 Pressure-displacement diagram for a reciprocating internal combustion engine.exhaust operations are accomplished in one revolution of the crankshaft. Although internal combustionengines undergo mechanical cycles, the cylinder contents do not execute a thermodynamic cycle, sincematter is introduced with one composition and is later discharged at a different composition.A parameter used to describe the performance of reciprocating piston engines is the mean effectivepressure, or mep. The mean effective pressure is the theoretical constant pressure that, if it acted on thepiston during the power stroke, would produce the same net work as actually developed in one cycle.That is,mep =net work for one cycledisplacement volume(2.97)where the displacement volume is the volume swept out by the piston as it moves from the top deadcenter to the bottom dead center.
For two engines of equal displacement volume, the one with a highermean effective pressure would produce the greater net work and, if the engines run at the same speed,greater power.Detailed studies of the performance of reciprocating internal combustion engines may take into accountmany features, including the combustion process occurring within the cylinder and the effects ofirreversibilities associated with friction and with pressure and temperature gradients.
Heat transferbetween the gases in the cylinder and the cylinder walls and the work required to charge the cylinderand exhaust the products of combustion also might be considered. Owing to these complexities, accuratemodeling of reciprocating internal combustion engines normally involves computer simulation.To conduct elementary thermodynamic analyses of internal combustion engines, considerable simplification is required. A procedure that allows engines to be studied qualitatively is to employ an airstandard analysis having the following elements: (1) a fixed amount of air modeled as an ideal gas isthe system; (2) the combustion process is replaced by a heat transfer from an external source and generallyrepresented in terms of elementary thermodynamic processes; (3) there are no exhaust and intakeprocesses as in an actual engine: the cycle is completed by a constant-volume heat rejection process;(4) all processes are internally reversible.The processes employed in air-standard analyses of internal combustion engines are selected torepresent the events taking place within the engine simply and mimic the appearance of observed© 1999 by CRC Press LLC2-83Engineering Thermodynamicspressure-displacement diagrams.
In addition to the constant volume heat rejection noted previously, thecompression stroke and at least a portion of the power stroke are conventionally taken as isentropic. Theheat addition is normally considered to occur at constant volume, at constant pressure, or at constantvolume followed by a constant pressure process, yielding, respectively, the Otto, Diesel, and Dual cyclesshown in Table 2.16.Reducing the closed system energy balance, Equation 2.8, gives the following expressions for heatand work applicable in each case shown in Table 2.16:W12= u1 − u2m(< 0)W34= u3 − u4m(> 0)Q41= u1 − u4m(< 0)Table 2.16 provides additional expressions for work, heat transfer, and thermal efficiency identified witheach case individually. The thermal efficiency, evaluated from Equation 2.9, takes the formη = 1−Q41 mQA mEquations 1 to 6 of Table 2.7 together with data from Table A.8, apply generally to air-standard analyses.In a cold air-standard analysis the specific heat ratio k for air is taken as constant.
Equations 1′ to 6′ ofTable 2.7 apply to cold air-standard analyses, as does Equation 4′ of Table 2.8, with n = k for theisentropic processes of these cycles.Referring to Table 2.16, the ratio v1/v2 is the compression ratio, r. For the Diesel cycle, the ratio v3/v2is the cutoff ratio, rc . Figure 2.18 shows the variation of the thermal efficiency with compression ratiofor an Otto cycle and Diesel cycles having cutoff ratios of 2 and 3. The curves are determined on a coldair-standard basis with k = 1.4 using the following expression:η = 1−k1 rc − 1 r k −1 k (rc − 1) (constant k )(2.98)where the Otto cycle corresponds to rc = 1.As all processes are internally reversible, areas on the p-v and T-s diagrams of Table 2.16 can beinterpreted, respectively, as work and heat transfer.
Invoking Equation 2.10 and referring to the p-vdiagrams, the areas under process 3-4 of the Otto cycle, process 2-3-4 of the Diesel cycle, and processx-3-4 of the Dual cycle represent the work done by the gas during the power stroke, per unit of mass.For each cycle, the area under the isentropic process 1-2 represents the work done on the gas during thecompression stroke, per unit of mass. The enclosed area of each cycle represents the net work done perunit mass. With Equation 2.15 and referring to the T-s diagrams, the areas under process 2-3 of the Ottoand Diesel cycles and under process 2-x-3 of the Dual cycle represent the heat added per unit of mass.For each cycle, the area under the process 4-1 represent the heat rejected per unit of mass. The enclosedarea of each cycle represents the net heat added, which equals the net work done, each per unit of mass.© 1999 by CRC Press LLC2-84TABLE 2.16 Otto, Diesel, and Dual Cycles(a) Otto Cycle(b) Diesel Cycle(c) Dual CycleW23=0mW23= p2 (v 3 − v 2 )mW2 x= 0,mQ23= u3 − u 2mQ23= h3 − h2mWx 3= p3 (v3 − v2 ),m© 1998 by CRC Press LLCu 4 − u1u3 − u 2η = 1−u 4 − u1h3 − h2η = 1−u 4 − u1Qx 3= h3 − hxm(u x − u2 ) + (h3 − hx )Section 2η = 1−Q2 x= u x − u2mEngineering Thermodynamics2-85FIGURE 2.18 Thermal efficiency of the cold air-standard Otto and Diesel cycles, k = 1.4.Carnot, Ericsson, and Stirling CyclesThree thermodynamic cycles that exhibit the Carnot efficiency (Equation 2.12) are the Carnot, Ericsson,and Stirling cycles shown in Figure 2.19.
Each case represents a reversible power cycle in which heatis added from an external source at a constant temperature TH (process 2-3) and rejected to the surroundings at a constant temperature TC (process 4-1). Carnot cycles can be configured both as vapor powercycles and as cycles executed by a gas in a piston-cylinder assembly (see, e.g., Moran and Shapiro,1995). Carnot cycles also can be executed in systems where a capacitor is charged and discharged, aparamagnetic substance is magnetized and demagnetized, and in other ways. Regardless of the type ofdevice and the working substance used, the Carnot cycle always has the same four internally reversibleprocesses in series: two isentropic processes alternated with two isothermal processes.The Ericsson and Stirling cycles also consist of four internally reversible processes in series: heatingfrom state 1 to state 2 (at constant pressure in the Ericsson cycle and at constant volume in the Stirlingcycle), isothermal heating from state 2 to state 3 at temperature TH, cooling from state 3 to state 4 (atconstant pressure in the Ericsson cycle and at constant volume in the Stirling cycle), and isothermalcooling from state 4 to state 1 at temperature TC.
An ideal regenerator allows the heat input required forprocess 1-2 to be obtained from the heat rejected in process 3-4. Accordingly, as in the Carnot cycle allthe heat added externally occurs at TH and all of the heat rejected to the surroundings occurs at TC.The Ericsson and Stirling cycles are principally of theoretical interest as examples of cycles thatexhibit the same thermal efficiency as the Carnot cycle: Equation 2.12.
However, a practical engine ofthe piston-cylinder type that operates on a closed regenerative cycle having features in common withthe Stirling cycle has been under study in recent years. This engine, known as the Stirling engine, offersthe opportunity for high efficiency together with reduced emissions from combustion products becausethe combustion takes place externally and not within the cylinder as in internal combustion engines. Inthe Stirling engine, energy is transferred to the working fluid from products of combustion, which arekept separate.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
